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Instructor Edition Third Edition
Practicing Biology A Student Workbook Campbell • Reece
Biology, Eighth Edition Jean Heitz and Cynthia Giffen University of Wisconsin, Madison
San Francisco • Boston • New York Capetown • Hong Kong • London • Madrid • Mexico City Montreal • Munich • Paris • Singapore • Sydney • Tokyo • Toronto
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Editor-in-Chief: Beth Wilbur Senior Editorial Manager: Ginnie Simione Jutson Senior Supplements Project Editor: Susan Berge Project Editor: Strawberry Field Publishing, Melanie Field Executive Marketing Manager: Lauren Harp Managing Editor: Michael Early Production Supervisor: Jane Brundage Production Services and Composition: Erin Melloy, S4 Carlisle Publishing Services
Student Edition ISBN-13: 978-0-321-52293-1 Student Edition ISBN-10: 0-321-52293-1
Copyright © 2008 Pearson Education, Inc., publishing as Benjamin Cummings, 1301 Sansome St., San Francisco, CA 94111. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E. Lake Ave., Glenview, IL 60025. For information regarding permissions, call 847/486/2635. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps.
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Contents Introduction to the Instructor’s Guide for Practicing Biology: A Student Workbook
vii
Notes to Instructors
1
Activity 2.1
3
A Quick Review of Elements and Compounds
Activity 3.1 A Quick Review of the Properties of Water Notes to Instructors
8 11
Activity 4.1/5.1
13
How can you identify organic macromolecules?
Activity 4.2/5.2
What predictions can you make about the behavior of organic macromolecules if you know their structure? Notes to Instructors
19 23
Activity 6.1 What makes a cell a living organism? Notes to Instructors
24 27
Activity 7.1
What controls the movement of materials into and out of the cell?
28
Activity 7.2
How is the structure of a cell membrane related to its function? Notes to Instructors
31
Activity 8.1
37
What factors affect chemical reactions in cells?
Activity 8.2
How can changes in experimental conditions affect enzyme-mediated reactions? Notes to Instructors
40 43
Activity 9.1
A Quick Review of Energy Transformations
46
Activity 9.2
Modeling Cellular Respiration: How can cells convert the energy in glucose to ATP?
48
Modeling Photosynthesis: How can cells use the sun’s energy to convert carbon dioxide and water into glucose?
54
Activity 10.1
Activity 10.2 How do C3, C4, and CAM photosynthesis compare? Notes to Instructors
58 62
Activity 11.1 How are chemical signals translated into cellular responses? Notes to Instructors
63 66
Activity 12.1
What is mitosis?
67
Activity 13.1
What is meiosis?
71
Contents
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Activity 13.2 How do mitosis and meiosis differ? Notes to Instructors
75 75
Activity 14.1
A Genetics Vocabulary Review
81
Activity 14.2
Modeling Meiosis: How can diploid organisms produce haploid gametes?
81
A Quick Guide to Solving Genetics Problems
85
Activity 14.3
Activity 14.4 How can you determine all the possible types of gametes? Notes to Instructors
90 92
Activity 15.1
Solving Problems When the Genetics Are Known
93
Activity 15.2
Solving Problems When the Genetics Are Unknown
95
Activity 15.3
How can the mode of inheritance be determined experimentally? Notes to Instructors
99 104
Activity 16.1
106
Is the hereditary material DNA or protein?
Activity 16.2 How does DNA replicate? Notes to Instructors
111 114
Activity 17.1
Modeling Transcription and Translation: What processes produce RNA from DNA and protein from mRNA? Notes to Instructors
115 124
Activity 18.1
How is gene expression controlled in bacteria?
126
Activity 18.2
Modeling the lac and trp Operon Systems: How can gene expression be controlled in prokaryotes?
128
How is gene activity controlled in eukaryotes?
130
Activity 18.3
Activity 18.4 What controls the cell cycle? Notes to Instructors
131 133
Activity 19.1 How do viruses, viroids, and prions affect host cells? Notes to Instructors
133 136
Activity 20.1
How and why are genes cloned into recombinant DNA vectors?
137
Activity 20.2 How can PCR be used to amplify specific genes? Notes to Instructors
139 144
Activity 21.1 How can we discover the sequence of an organism’s DNA? Notes to Instructors
144 148
Activity 22.1
151
How did Darwin view evolution via natural selection?
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How do Darwin’s and Lamarck’s ideas about evolution differ?
155
How would you evaluate these explanations of Darwin’s ideas? Notes to Instructors
157 159
Activity 23.1
162
Activity 22.3
A Quick Review of Hardy-Weinberg Population Genetics
Activity 23.2 What effects can selection have on populations? Notes to Instructors
168 173
Activity 24.1
173
What factors affect speciation?
Activity 24.2 How does hybridization affect speciation? Notes to Instructors
176 180
Activity 25.1
What do we know about the origin of life on Earth?
180
Activity 25.2 How can we determine the age of fossils and rocks? Notes to Instructors
183 186
Activity 26.1
How are phylogenies constructed?
187
Activity 26.2
What is parsimony analysis?
191
Activity 26.3
Put yourself in the professor’s shoes: What questions would you ask? Notes to Instructors
193 195
Activity 27.1
195
How diverse are the Archaea?
Activity 27.2 How has small size affected prokaryotic diversity? Notes to Instructors
198 203
Activity 28.1
How has endosymbiosis contributed to the diversity of organisms on Earth today? Notes to Instructors
203 208
Activity 29.1/30.1 What major events occurred in the evolution of the plant kingdom?
208
Activity 29.2/30.2 What can a study of extant species tell us about the evolution of form and function in the plant kingdom?
210
Activity 29.3/30.3 How are the events in plant evolution related? Notes to Instructors
216 220
Activity 31.1 How diverse are the fungi in form and function? Notes to Instructors
220 224
Activity 32.1/33.1 What can we learn about the evolution of the animal kingdom by examining modern invertebrates?
224
Activity 32.2/33.2 What factors affect the evolution of organisms as they become larger?
229
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Notes to Instructors
234
Activity 34.1
What can we learn about the evolution of the chordates by examining modern chordates? Notes to Instructors
234 240
Activity 35.1
How does plant structure differ among monocots, herbaceous dicots, and woody dicots? Notes to Instructors
241 246
Activity 36.1 How are water and food transported in plants? Notes to Instructors
247 253
Activity 37.1
What do you need to consider in order to grow plants in space (or anywhere else for that matter)? Notes to Instructors
253 255
Activity 38.1
How can plant reproduction be modified using biotechnology? Notes to Instructors
255 258
Activity 39.1 How do gravity and light affect plant growth responses? Notes to Instructors
259 262
Activity 40.1
How does an organism’s structure help it maintain homeostasis? Notes to Instructors
263 267
Activity 41.1 How are form and function related in the digestive system? Notes to Instructors
268 277
Activity 42.1
How is mammalian heart structure related to function?
279
Activity 42.2
How do we breathe, and why do we breathe?
282
Activity 42.3
How are heart and lung structure and function related to metabolic rate? Notes to Instructors
286 290
Activity 43.1
How does the immune system keep the body free of pathogens? Notes to Instructors
290 294
Activity 44.1
What is nitrogenous waste, and how is it removed from the body? Notes to Instructors
298 301
Activity 45.1 How do hormones regulate cell functions? Notes to Instructors
301 305
Activity 46.1
How does the production of male and female gametes differ in human males and females?
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Notes to Instructors
310
Activity 47.1
What common events occur in the early development of animals? Notes to Instructors
310 315
Activity 48.1
How do ion concentrations affect neuron function?
316
Activity 48.2
How do neurons function to transmit information?
319
Activity 48.3
What would happen if you modified a particular aspect of neuron function? Notes to Instructors
324 327
Activity 49.1 How is our nervous system organized? Notes to Instructors
327 331
Activity 50.1
331
How does sarcomere structure affect muscle function?
Activity 50.2
What would happen if you modified particular aspects of muscle function? Notes to Instructors
335 337
Activity 51.1 What determines behavior? Notes to Instructors
337 341
Activity 52.1 What factors determine climate? Notes to Instructors
341 348
Activity 53.1
What methods can you use to determine population density and distribution?
What models can you use to calculate how quickly a population can grow? Notes to Instructors
348
Activity 53.2
Activity 54.1 Activity 54.2
354 360
What do you need to consider when analyzing communities of organisms?
360
What affects can a disturbance have on a community?
365
Activity 54.3
How can distance from the mainland and island size affect species richness? Notes to Instructors What limits do available solar radiation and nutrients place on carrying capacities? Notes to Instructors
368 371
Activity 55.1
Activity 56.1
371 377
What factors can affect the survival of a species or community?
377
Appendix A
An Introduction to Data Analysis and Graphing for a PC
381
Appendix B
An Introduction to Data Analysis and Graphing for a Mac
387
Contents
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Introduction to the Instructor’s Guide for Practicing Biology: A Student Workbook What does Practicing Biology: A Student Workbook contain? The activities in this workbook focus on key ideas, principles, and concepts that are basic to understanding biology. The overall organization follows that of Biology, 8th edition. Key principles or processes developed in activities are often revisited and integrated in subsequent activities. Although the individual activities may vary in the thought processes required and in their specific biological content, the overall goals of this workbook are to: • Allow students to discover what they know and, more important, what they don’t know. • Help students to discover and modify any misconceptions in their understanding of biology. • Provide students with opportunities to synthesize and apply what they have learned to novel situations.
What kinds of activities are included? The activities in Practicing Biology take a number of different forms:
Leading questions. In these activities, students are asked a series of leading questions that are designed to build their basic understanding of principles. Leading questions are generally followed by additional questions that give students the opportunity to apply what they have learned to new situations.
Concept mapping/Diagramming and Drawing. These activities, which include drawing exercises, concept maps, and flow diagrams, are designed to help students organize information and ideas and develop an understanding of how various pieces of information are interrelated.
Modeling. Modeling activities provide students with instructions for building models of dynamic biological processes that occur at the molecular, cellular, and physiological levels. Modeling can help students both develop and test their understanding of processes that are generally invisible to the naked eye.
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Process of Science. Process of science activities encourage students to practice the use of scientific thought processes. They are designed to give students a better understanding of how the knowledge they gain in class can be applied to • propose experiments, • predict possible outcomes of experiments, and • interpret experimental data.
Reviewing. This group of activities provides an opportunity for students to review and integrate key ideas and principles in biology. The reviews are generally followed by activities that require students to apply their knowledge.
Teaching. In the teaching activities, students examine ideas, principles, and concepts from the instructor’s point of view. Most instructors will agree that their understanding of a process, idea, or concept increases when they work to help someone else learn it. These activities give students opportunities to develop deeper understanding by “putting themselves in the instructor’s shoes.”
Data Analysis and Graphing. These activities are designed to give students practice with interpreting graphs, analyzing data, and/or developing graphs from data sets. These skills are integral to both understanding and communicating information in biology. About the Authors Jean Heitz is a Faculty Associate in Zoology at the University of Wisconsin (Madison) and has worked with Introductory Biology 151, and 152 since 1978. Her key roles have been in development of active learning activities for discussion sections and open-ended investigations for laboratory sections. She has also taught Botany/Zoology 969, a graduate course in “Teaching College Biology,” for more than 14 years and has presented workshops at a number of national meetings. Cynthia Giffen joined the University of Wisconsin in 2005, following 5 years of research and teaching at Potomac State College and the University of Maryland. She earned a BS in Environmental Science from Allegheny College and an MS in Marine, Estuarine, and Environmental Sciences at the University of Maryland–Appalachian Lab. Previously, she served as Adjunct Faculty in Introductory Biology at Potomac State College and participated in aquatic and terrestrial ecology research at the University of Maryland. Introduction to the Instructor’s Guide for Practicing Biology
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Together their current research interests include the effects of undergraduate research opportunities on students, the effects of new technologies on student learning, and the development of new activities in introductory biology to engage students. What additional information is available in the Instructor’s Guide? 1. A short discussion of some of the learning techniques used in the activities. 2. Activity-specific “Notes to Instructors.” The notes address these questions: • What is the focus of these activities? • What are the particular activities designed to do? • What misconceptions or difficulties can these activities reveal? The “Notes to Instructors” also include sample answers to the questions posed in each activity. We provide the answers in both Word and HTML formats to allow you to copy and electronically post answers or to print out a hard copy for posting. (Note: To avoid the idea that there is only one correct concept map or model for a given exercise, the Instructor’s Guide does not include examples of models or concept maps.) A Short Discussion of Some of the Learning Techniques Used in the Activities The methods used in the activities can be categorized loosely under these headings: 1. 2. 3. 4. 5.
Leading, or Socratic, questioning Process of science (including problem solving) Modeling Concept mapping, diagramming, and drawing Teaching (or putting yourself in the instructor’s shoes)
1. Leading, or Socratic, questioning As instructors, we ask questions to accomplish several goals: • • • •
Set up a learning climate based on questioning and curiosity Initiate interest or gain attention Focus student learning Discover what students do and don’t know
Questions can also be used for these purposes: • Organize or put information in context • Demonstrate the logic that leads to conceptual understanding • Lead students to discover their own answers to questions How can leading, or Socratic, questioning be used in the classroom? As noted, leading, or Socratic, questioning is used in the classroom to determine what students do and don’t know—that is, the depth of their understanding. More important, it can be used to make our students self-directed learners, to help them develop the skills that will allow them to find answers for themselves, and to encourage them to build and reinforce their conceptual understanding. x
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An example of leading, or Socratic, questioning Assume you are beginning a section on human physiology. You want to discover what your students already know about why we eat and why we breathe. At the same time, you want to help them build a conceptual understanding of why we eat and breathe. Student: Instructor: Student: Instructor: Student: Instructor: Student: Instructor:
You say the important thing for me is to determine what I know and what I don’t know. So how am I supposed to know what I don’t know? Let’s take an example. Do you understand the big picture—that is, do you understand conceptually why we eat and why we breathe? To get energy and building blocks and oxygen. That’s true, but can you be more specific about how eating provides energy? The energy comes from carbohydrates or sugars and fats that we eat. Okay, but what is it in these compounds that supplies energy? The energy of the molecules. So, what is the energy of the molecules and how is it used by the cells? Can you explain this?
At this point, the student may discover that she can’t explain that it is the energy in the C—H bonds and may need to be prompted. As a result, she discovers some of what she doesn’t know. Instructor:
Can we use this C—H bond energy directly to do work in our cells?
Again at this point, the student may not know how to answer. Instructor:
What kinds of bond energy do you know cells use?
Most students will know that the answer is ATP. Others will still not make the connection. Student: Instructor: Student: Instructor: Student: Instructor:
ATP. So how does the energy of sugar molecules get to ATP? Cells break down sugars to make ATP (in cellular respiration). What do you need for cellular respiration to produce ATP? Sugar and oxygen. Why do we eat and why do we breathe?
The students’ answers to each question indicate which part(s) of the overall concept they know or understand. If they don’t already have this conceptual understanding, the series of questions should help them make the logical connections between the overall gross organ-level functions of digestion and respiration and what ultimately goes on at the cellular level. We find that many undergraduates do not understand this connection. Types of student questions The types of questions students ask fall into different categories. • Questions students don’t know the answer to and can’t know unless we tell them. For example: “We ran out of (a supply). Where can I find more of it?” These require simple, direct answers.
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• Questions students should be able to answer for themselves. Students may have to think back to what they know and build on this. These questions may also require students to integrate information they have acquired from different sources. • Questions students have generated from their curiosity. They don’t know the answer, and they can’t figure out the answer without further investigation. Questions that require students to think and integrate available information You’re doing neurophysiology experiments using the live nerves of a cockroach. Student: Instructor: Student: Instructor: Student: Instructor:
Should I rinse off the nerve in distilled water to clean it up before recording the action potential from it? What is the normal environment of the nerve? I don’t know. What’s the normal environment for human nerves? They’re surrounded by extracellular fluid. What characteristics does the extracellular fluid have?
If the student doesn’t know, ask: Is it the same as distilled water? Student: Instructor: Student: Instructor: Student: Instructor: Student: Instructor: Student: Instructor:
No. So, would you expect the cockroach extracellular fluid to be equivalent to distilled water? No. But that doesn’t tell me what to use. What did we use to keep the frog nerve alive for experiments? Frog “Ringer’s” solution (a specific salt solution similar in composition to frog blood or extracellular fluid). Okay, I get it. I need to look for cockroach Ringer’s solution. What would happen if you used the distilled water instead of the cockroach Ringer’s on the nerve tissue? It probably wouldn’t function normally. Why? Because of osmotic differences and ionic differences. Good! I think you really understand.
At this point, you’re probably asking yourself, is it worth it? Why didn’t I just give the answer? The value comes in the longer term. In other words, you’re using leading questions to teach students the process of questioning they should be using to answer their own questions. For this method to be most effective, it is best to let students know what you are doing and why you are doing it—that is, the reasons you have for using leading questions. You will find that you don’t have to demonstrate this process very often for your students to learn it. Over time you are asked far fewer questions of the type the students can answer for themselves. Instead, the questions you are asked tend to be more thoughtful and often arise from students’ curiosity.
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Questions that come from curiosity and from what students don’t know and can’t necessarily know without further investigation In the previous neurobiology example, the students may know from their reading that the nerves in the insects tend to be covered by a sheath. As a result, they may wonder whether or not the sheath protects the insects from osmotic changes. This is something that students wouldn’t necessarily know and that could be explored experimentally. These are the kinds of questions we hope to encourage using the leading, or Socratic, questioning method. Keep in mind that not all questions are good questions. Student: Instructor: Student: Instructor: Student: Instructor:
Should I rinse off the nerve in distilled water to clean it up before recording the action potential from it? What do you think? or Do you think you should rinse off the nerve with distilled water? If I knew, I wouldn’t have asked the question. Did you read your lab manual? Yes, but I can’t find the answer there, and if you talked about it, I don’t remember. You’re right. I didn’t tell you that specifically, but you should be able to figure it out for yourself. Just think, why wouldn’t you want to use distilled water?
This exchange is going nowhere fast because the instructor is not using leading questions. None of the instructor’s questions leads the student to consider the possible effects of using distilled water. None of them reminds the student of what he already knows about diffusion gradients and the need to maintain homeostasis in the cell’s environment for appropriate function. The instructor is simply restating the student’s questions in a slightly different format. As a result, these questions provide the student no further insight; they are dead-end questions. How can we encourage student questions and help students learn how to ask good questions? The best way to encourage questions is to make it clear that you want students to ask questions. It also helps to model the types of questions you want them to ask. The following exercise encourages questioning and helps students learn how to develop good questions. During the first few class sessions and periodically thereafter, we give each student an index card or half a sheet of paper and ask each to write a question that logically follows from the day’s discussion in class. We reserve about 5 minutes at the end of the period for this activity. At the end of class, we collect the cards and read through them to determine the types of questions and general themes. We then pick the best three or four and retype them in large type (36 point or larger) for display on an overhead or PowerPoint presentation.
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At the beginning of the next class we explain that all the questions were valuable and could lead to a variety of interesting discussions. We show the ones we picked out as examples and indicate why these were chosen. For example: “These questions stand out because they exemplify the kinds of questions that • would lead us to learn more about this area, • would help us integrate what we’ve learned with other areas of biology, or • are interesting because they take what we learned and probe into new areas that we may not understand at present.” How much time do students need to answer a question? When using leading questions, you need to consider how much time to give students to respond. Try this test for yourself. Go into someone else’s classroom. Observe how long the instructor waits after asking a question before she requires a response or answers the question herself. On average, most instructors give students less than 3 seconds to respond to questions. If we want thoughtful answers (or any answer at all), we need to give students adequate time to think and respond. At a minimum, we should allow 10 seconds for a response. If asking questions of the whole class, we find that the following steps lead to much better responses: • Give the students a minimum of 10 to 20 seconds to write down their individual responses. • Then give them 2 or 3 minutes to share their thoughts in small groups before they are required to respond. (During this time, wander around the class and “eavesdrop” on conversations. Note the locations of groups with good, unique, or interesting ideas so you can call on some of them later.) • Call on a number of different individuals. Following a procedure like this is especially important when asking questions that require integration or synthesis or that are more open-ended. The amount of time you should give for individual thought and for group discussion will vary depending on the complexity of the question. How do you know how much time is needed? You watch your students and listen to them. When it appears that the majority of individuals are done writing down their ideas, give the class another 10 seconds or so to finish their individual thinking. Then put students in small groups to share their ideas. Monitor the small groups and use the same basic rule for determining how long the small-group discussions should be. 2. Process of Science In this workbook, problem solving is included as one of the processes of science. Simple problem-solving activities often involve mathematical calculations. For example, xiv
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calculations can be used to determine the change in population size per unit time or methods of inheritance in genetics. In addition, we include in this category designing and interpreting experiments and determining what students need to know to solve more open-ended problems—for example, problems in medicine or ecology. In all problem-solving activities, after the sample problems, we provide one or two conceptual questions that are designed to address why it is important to know how to do these problems. Example problem (question 10 in Activity 53.2) A rabbit population has the life table shown here. Age class
Number of survivors
Number of deaths
Mortality rate
Number of offspring per reproducing pair
0–1
100
10
0.10
0
1–2
90
0.33
1.5
2–3
60
30
3–4
30
24
0.80
2.5
6
1.0
0
4–5
2.0
a. Fill in the missing data in the table. b. Owing to a good food supply and a small predator population, the rabbit population is growing by leaps and bounds. The rabbits call a meeting to discuss population control measures. Two strategies are proposed: • Delay all rabbit marriages until age class 2–3 (rabbits never breed until after marriage). • Sterilize all rabbits in age class 3–4. Which of the proposed strategies will be more effective in slowing population growth? Explain your reasoning and show your calculations. Here is an alternative way of introducing this type of problem: Proponents of birth control often disagree on the best method. Some claim that limiting all couples to two children is best. Others argue for a later age of childbearing in addition to a limit of two children. What effect would setting a later age of childbearing have on population growth if each couple goes on to have two children anyway? Note that in both cases all calculations are done in the context of a larger question.
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3. Modeling Modeling helps students understand dynamic processes—for example, mitosis and meiosis, the laws of segregation and independent assortment, the transmission of an action potential along a neuron, and protein synthesis. To understand a dynamic process, students must develop a dynamic (claymation, if you will) model that they can manipulate or move through the various steps of the process. To do this, they need to understand each step of the process. Any lack in understanding or misconception becomes evident and easily can be filled in or modified. To set up a modeling exercise, identify • which process the students need to model, and • which elements of the process must be included in the model. Include a few questions designed to help students realize why it is important to understand the process. For example: People who suffer from bulimia often have reduced electrolyte levels (for example, Na⫹ and K⫹) in their blood and intercellular fluids. Use your model of the action potential to explain the effects reduced K+ levels could have on nerve function. It is important to keep modeling as simple as possible. We have discovered that when we give students a prepackaged kit (for example, a chemical structures kit), they spend the majority of the class time trying to figure out how to use the kit. Students also get the impression that there is only one correct way to use the kit. As a result, they think that any question the instructor asks about their work with the kit is a “test question.” In contrast, when we give them a piece of chalk (to draw membranes and other features) and some containers of playdough, students make the parts as they need them and jump into the process quickly. Each model is different, so it is natural for the instructor to ask questions about them. For example, as the instructor, how can you know why some chromosomes in a meiosis model are red and others are blue unless you ask? It is much easier to talk with students about their models and to discover whether they have any questions or misconceptions. 4. Concept mapping, diagramming, and drawing Concept mapping or diagramming can be used: • to bring a structure of hierarchy and relatedness to what may seem to the student to be a set of disjointed topics, facts, and ideas; and • to serve as a guide for the student to integrate and understand both broad and specific concepts associated with the topic being explored. xvi
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The mechanics of developing a concept map or diagram are simple. The students brainstorm, or the instructor provides, the list of terms to be included in the map or diagram. Students write each term on a separate sticky note or piece of paper. Working in small groups, students organize the terms into a map or diagram that indicates how the terms are associated or related. Students draw lines between related terms and write action phrases on the lines to indicate how the terms are related. Developing this structure helps the students to discover for themselves the multiple associations that exist among the various levels and components of a system or topic. The ability of students to organize and develop a conceptual framework from a list of terms associated with a topic (for example, a list of the various parts of the digestive, circulatory, and respiratory systems) is a direct measure of the students’ understanding of the topic. Often the schemes or maps the students develop resemble flow diagrams. Using the same set of terms, different groups of students may come up with similar concept maps, but the maps are seldom identical. For example, maps developed by students with deeper understanding are generally less linear because they can see many more interrelationships and connections. After the maps are finished, student groups are asked to explain them to each other. This gives the students an opportunity to discuss how and why their maps differ. These discussions help students to understand that that there are different ways of representing the same relationships. The discussions can also uncover misconceptions, which may appear as inappropriate or unsupportable associations in the map’s structure. In other activities, students are asked to develop their own diagrammatic representation or drawing of specific structures and processes. To keep students from simply copying drawings out of the text, we require that the drawings be in the form of stick figures or cartoon characters. We ask that domino-effect processes (1 triggers 2, which leads to 3, and so on) be drawn as Rube Goldberg cartoon-type systems. For example, we ask students to draw a flow diagram (or Rube Goldberg cartoon-type handout) to explain how the various components of the ear interact to produce the sensation of sound in the brain. In another exercise, we ask them to diagram the interactions among cells in the B and T cell immune responses. To let students know what we mean by a Rube Goldberg–type diagram, we provide the following example and remind them of the board game “Mouse Trap.”
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5. Teaching Many of us recognize that one of the best ways to learn something is to teach it. As a result, some of the activities ask students to develop methods to teach each other. These activities are rated among the best learning experiences by the students themselves. However, the students rate the same activities as being among the most difficult and time-consuming. In other activities, we ask students to develop exam questions (including answer choices) to give them a better understanding of how multiple-choice questions are structured. They also discover how difficult it is to write good questions.
Forward to Instructors Since 1978, I have been involved in teaching introductory biology at the University of Wisconsin. During that period of time it has been my good fortune to work directly with more than 25 different faculty members, hundreds of teaching assistants, and more than 7,000 undergraduate students. I have learned a great deal about teaching from all of them. Between 1989 and 1991, Dr. Marion Meyer and I received several small grants from the UW Center for Biology Education (funded by HHMI) to develop programs and courses. 1. A cross-campus Biology TA Training Workshop 2. A graduate-level course for teaching assistants titled “Teaching College Biology” 3. A model discussion/tutorial program (designed for students with little or no experience in the sciences or for students who have had previous difficulty in science courses) 4. A set of introductory investigative lab modules for use in introductory biology
1
Frank Heppner is the author of a number of books and articles, including The Green Book of Grading (Ornis Press, Kingston, RI, 1984), and Professor Farnsworth’s Explanations in Biology (McGraw-Hill, 1990). 2 Sheila Tobias is the author of many articles and books, including “Too Often, College-Level Science Is Dull as Well as Difficult,” Chronicle of Higher Education, (March 27, 1991), and They’re Not Dumb, They’re Different: Stalking the Second Tier (Research Corp, Tucson, AZ, 1990).
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The grants also allowed us to invite Dr. Frank Heppner1 and Dr. Sheila Tobias2 to present all-campus lectures related to improving undergraduate biology education. We developed the Teaching College Biology course based on recognized need, the results of graduate student surveys, readings in learning/teaching theory, and discussions with knowledgeable faculty, staff, and deans from the Colleges of Letters and Sciences, Agricultural and Life Sciences, the School of Education, and the Academic Advancement Program. During the planning and development of this course, it became clear that a great deal of information had been published that identified needs and problem areas in education. On the other hand, at that time, very little had been published that offered solutions to those problems. Similarly, various methods had been proposed to engage students in active learning, for example, concept mapping and small-group learning. However, very few examples existed of how to apply these methods to college biology. As a result, we developed our own activities and exercises. We tested these in our graduate course. Later, we served as advisors and our graduate students served as teaching assistants in development of the model discussion/tutorial program designed to help students learn biology, not just memorize facts. What we learned from these efforts became the basis for our continuing development of both TA training methods and activities designed to help students learn major principles, concepts, and processes in introductory biology. When Cynthia Giffen joined UW in 2005, she and I immediately formed a great working partnership. She brings a fresh perspective to the course and many ideas for new activities and additional teaching techniques. While at UW, she has made strides to increase active learning in lecture and to bring new technologies, such as podcasting, into introductory biology. Her experience with ecology and biostatistics has allowed us to develop additional data and graphical analysis questions and activities that help students understand the experimental nature of science. Many of the ideas and examples used in this workbook are offshoots of these efforts. All of the activities were designed to more actively involve students in constructing their own understanding of basic principles and concepts in biology.
Notes and Resources Since 1990 the educational literature on teaching for student learning and long-term retention has grown exponentially. I cannot claim to be an expert in the educational literature. However, for those of you who would like more information on the methods used in this workbook, or the educational/psychological theories behind the methods, I present (in no particular order) the following annotated list of books and articles. Introduction to the Instructor’s Guide for Practicing Biology
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The following articles can all be found in: Costa, Arthur L., Developing Minds: A Resource Book for Teaching Thinking, 3rd edition (Association for Supervision and Curriculum Development, Alexandria, VA, 2001). Paul, Richard, “Dialogical and Dialectical Thinking,” pp. 427–436. This article explains the differences between didactic (teaching by telling and learning by memorization) and dialogical and dialectical thinking or multilogical or critical thinking. It also describes four interrelated skills teachers need: 1. 2. 3. 4.
How to identify and distinguish multilogical from monlogical problems and issues How to teach Socratically How to use dialogical and dialectical thought to master content How to assess dialogical and dialectical thought
Jackson, Thomas, “The Art and Craft of ‘Gently Socractic Inquiry,’” pp. 459–465. This article is written for the K–6 teacher. However, it includes good ideas on creating a community of inquiry including cognitive tools students need to develop inquiry skills based on intellectual rigor. Ennis, Robert, “Goals for a Critical Thinking Curriculum and Its Assessment,” pp. 44–46. This is a brief and very succinct discussion of what the author calls “a useable, comprehensive and defensible set of critical thinking goals . . . to provide a useful basis on which to build curriculum and assessment procedures.” Hyerle, David, “Visual Tools for Mapping Minds,” pp. 401–407. This article provides an overview of various types of visual tools that can be useful for “representing and understanding patterns and interdependencies and systems,” all of which can facilitate long-term retention of learning. Handlesman, J., Houser B., and Kriegel, H., Biology Brought to Life: A Guide to Teaching Students to Think Like Scientists (Times Mirror Higher Education Group, Dubuque, IA, 1997). This Instructor’s guide to the corresponding lab manual includes considerable practical information on how to apply group learning in a classroom setting. Some of the key literature on cooperative and group learning is also reviewed. Johnson D. W., Johnson R. T., and E. Johnson Holubec, Circles of Learning: Cooperation in the Classroom (Interaction Book Company, Edina, MN, 1993).
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Introduces the reader to a wide range of small-group and cooperative learning mechanisms as well as information on using the methods. Novak, J. D., and Gowin, D. B., Learning How to Learn (Cambridge Press, 1984). Tishman S., Perkins, D. N., and Jay, E., The Thinking Classroom: Learning and Teaching in a Culture of Thinking (Allyn and Bacon, Boston, 1995). As stated by the authors: “This book explores six dimensions of good thinking and how to take a cultural approach to teaching them. These six dimensions are: 1. 2. 3. 4. 5. 6.
a language of thinking thinking dispositions mental management [metacognition] the strategic spirit higher order knowledge transfer”
It explains core ideas and theories and provides examples of how these might be put into practice in the classroom. Uno, G. E. Handbook on Teaching Undergraduate Science Courses (Saunders College Publishing, 1999). This book provides considerable practical information on teaching science. A complete chapter (Chapter 5) is devoted to how students learn. Two other chapters deal with inquiry instruction and critical thinking skills. Billson, J. M., “The College Classroom as a Small Group: Some Implications for Teaching and Learning,” Teaching Sociology, 14 (July, 1986), pp. 143–151. In this article, the author develops 15 principles of group process and development in the classroom and provides suggestions for their implemention. Blackwell, P. J., “Student Learning: Education’s Field of Dreams,” Phi Delta Kappan, 84(5), p. 362. As stated in the abstract, “Blackwell urges schools of education to shift their emphasis to the knowledge base about student learning and she provides seven benchmarks for programs that will produce high quality teachers who understand how students learn.” Greenwald, N. “Learning from Problems,” The Science Teacher, (April 2000), pp. 28–32.
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In this article, the author examines the use of problem-based learning when teaching for understanding. Halpern, D. F., and Hakel, M. D., “Applying the Science of Learning to the University and Beyond,” Change, 35(4)(2003), p. 36. This article includes a short summary of the types of teaching required for long-term retention and transfer of knowledge among college students. This includes brief notes on the use of concept maps and other alternative presentation formats for learning for retention. Hufford, T. L., “Increasing Academic Performance in an Introductory Biology Course,” Bioscience, 41(2), pp. 107–108. The author reports on changes made to improve student learning in introductory biology. Among these were reducing class size, increasing and promoting cooperative learning both in and outside of class, and providing help with study and test-taking skills. Leonard, W. H., “How Do College Students Best Learn Science?” Journal of College Science Teaching (May, 2000), pp. 385–388. This article provides a brief description of constructivist learning and support for its use in teaching college science. In addition, it points out the need to use a diversity of teaching methods to better fit the needs of today’s diverse undergraduate student body.
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Notes to Instructors Chapter 2 The Chemical Context of Life Chapter 3 Water and the Fitness of the Environment What is the focus of these activities? Living organisms function in the real world, so they are subject to all the laws of chemistry and physics. In addition, biological organisms and systems are variable. No two organisms are exactly alike, and no two systems are identical in form or function. As a result, our analysis of such systems tends to deal with statistical averages or probabilities. This means that it is difficult to understand biological systems without having a good basic understanding of chemistry, physics, and math (including probability and statistics). The vast majority of introductory biology students have studied inorganic chemistry in their high school and first-year college chemistry courses. Many students compartmentalize their knowledge, however. In some cases, the compartmentalization is so extreme that the students feel uncomfortable dealing with chemical formulas and ideas outside of chemistry classes. Therefore, it is generally useful to review some of the basic ideas in chemistry and, at the same time, demonstrate how they can be applied to understanding biological systems.
What are the particular activities designed to do? Activity 2.1 A Quick Review of Elements and Compounds The questions in this activity are designed to help students review and understand: • atomic/molecular number, mass number, and atomic/molecular weight and how they can be used to determine the reactivity of elements; • various types of chemical bonds and how they affect the structure and energetics of molecules; and • the difference between a mole and a molar equivalent and how a knowledge of these can be used in biological applications. Activity 3.1 A Quick Review of the Properties of Water The questions in this activity are designed to help students review and understand the properties of water and how they support life. Students are asked to review these key properties: • • • • • •
H2O molecules are cohesive; they form hydrogen bonds with each other. H2O molecules are adhesive; they form hydrogen bonds with polar surfaces. Water is a liquid at normal physiological (or body) temperatures. Water has a high specific heat. Water has a high heat of vaporization. Water’s greatest density occurs at 4°C.
Notes to Instructors
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In addition, students review pH and how it is related to both the ionization constant of pure water and the concentration of H+ ions in a solution.
What misconceptions or difficulties can these activities reveal? Activity 2.1 Question 1: Many students don’t understand that nutrients for plants are inorganic and most nutrients for animals (heterotrophs) are organic. Questions 2 and 3: Most students know how to balance a chemical equation. Fewer understand the relationship between molecules of a substance and moles of that substance. Similarly, most students can recite what a mole is; however, the majority have not thought about how that knowledge can be applied. Therefore, much of this first activity is devoted to making it clear that a balanced equation indicates not only the number of molecules required but also the number of moles required. It also explains why moles can be substituted for molecules in such equations. Question 4: Some students have difficulty understanding that a solution’s concentration or molarity does not change if you aliquot or subdivide the solution into smaller volumes. To test this, ask your students: “There is 10% sugar in this solution. If I pour half of it into one beaker and the other half into another beaker, what percent sugar will I have in each beaker?” More than half of the students will automatically answer 5%. Questions 5 and 6: These questions are designed to help students understand how a knowledge of balanced equations and molar equivalents can be useful in biology. Questions 7 and 8: The answers go into a little more detail than does Biology, 8th edition. Students obviously shouldn’t be asked to know the specific electronegativity of each of the elements. However, using concrete numbers may help students understand how their electronegativity is related to the type of bonds formed between elements. Activity 3.1 Most students have no difficulty stating the properties of water and the definition of pH. On the other hand, not all of them have a good understanding of how these properties are related to biological and other phenomena. Therefore, some questions ask students to relate pH values to actual concentrations of H+ ions in solution and to relate the properties of water to common experiences they have had in class or in life.
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Answers Activity 2.1 A Quick Review of Elements and Compounds 1. Table 2.1 (page 32) lists the chemical elements that occur naturally in the human body. Similar percentages of these elements are found in most living organisms. a. In what abiotic (nonlife) chemical forms are these elements often found in nature?
b. In what chemical form(s) do animals need to obtain these elements?
c. In what chemical form(s) do plants need to obtain these elements?
With the exception of oxygen and water, animals obtain the majority of these elements in the form of organic compounds.
These elements are most commonly found as CO2, N2, and O2 in the atmosphere and as H20, PO4, and S compounds on Earth.
Plants can obtain C as CO2, N as ammonia or nitrite or nitrates, phosphorus as phosphates, sulfur as sulfides or sulfates, and so on. In other words, plants obtain these elements as inorganic compounds.
2. A chemical element cannot be broken down to other forms by chemical reactions. Each element has a specific number of protons, neutrons, and electrons. a. What is the name of the following element, and how many protons, neutrons, and electrons does it have?
11
Na 23
Name Sodium
Number of protons 11
Number of electrons Number of neutrons 11 12
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b. What information do you need to calculate or determine the following? The atomic number of an element
The mass number of an element
The weight in daltons of one atom of an element
The atomic number is equal to the number of protons (or electrons).
The mass number is equal to the number of protons plus the number of neutrons.
You can estimate the weight in daltons as 1 dalton per proton or neutron.Therefore, the weight in daltons of an element is approximately equal to the number of protons plus the number of neutrons.
c. What are the atomic number, mass number, and weight in daltons of the element shown in part a? Atomic number
Mass number
Weight in daltons
11
23
23
3. One mole of an element or compound contains 6.02 1023 atoms or molecules of the element or compound. One mole of an element or compound has a mass equal to its mass number (or molecular weight) in grams. For example, 1 mole of hydrogen gas (H2) contains 6.02 1023 molecules and weighs 2 g. a. What is the weight of 1 mole of pure sodium (Na)? 23 g
b. How many molecules of Na are in 1 mole of Na? 6.02 1023
c. How would you determine how many grams are in a mole of any chemical element or compound? A mole of any chemical element or compound is equal to the mass number in grams of that mole or compound. For example, the mass number of Na is 23; therefore, a mole of Na has a mass of 23 g. The mass number of water is 18; therefore, a mole of water has a mass of 18 g. 4. One atom of Na can combine with one atom of Cl (chlorine) to produce one molecule of NaCl (table salt). a. If Cl has 17 electrons, 17 b. What is the mass number c. How many grams of protons, and 18 neutrons, of NaCl? NaCl equal a mole of what is its mass number? NaCl? 35 58 g 23 + 35 = 58 4
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d. If you wanted to combine equal numbers of Na and Cl atoms in a flask, how much Cl would you have to add if you added 23 g of Na? (Include an explanation of the reasoning behind your answer.) 23 g of Na is equal to 1 mole of Na. A mole contains 6.02 1023 molecules of the substance. To add an equal number of molecules of Cl, you need to add 1 mole of Cl, or 35 g. e. To make a one-molar (1 M) solution of NaCl, you need to add 1 mol of NaCl to distilled water to make a final volume of 1 L (1,000 ml). A 1 M solution is said to have a molarity of 1. If you added 2 moles of NaCl to 1 L of distilled water, you would make a 2 M solution and its molarity would equal 2. You make up a 1 M solution of NaCl. How many molecules of NaCl are in the 1 M NaCl solution?
How many molecules of NaCl are there per ml of the solution?
If you used 1 L of water to make the 1 M solution, you would have 6.02 1023 molecules in the liter.
To calculate the number of molecules per ml, divide 6.02 1023 by 1,000 = 6.02 1020 molecules/ml.
f. Next, you divide this 1 M solution of NaCl into four separate flasks, putting 250 mL into each flask. How many grams of NaCl are in each flask?
How many molecules of NaCl are in each flask?
58/4 = 14.5 g
6.02 1023 divided by 4 = 1.51 1023
How many molecules of NaCl are there per ml of distilled water?
What is the molarity of NaCl in each of the four flasks?
6.02 1020
1M
5. The summary formula for photosynthesis is 6 CO2 + 6 H2O → C6H12O6 + 6 O2 a. How many molecules of carbon dioxide and water would a plant have to use to produce three molecules of glucose (C6H12O6)? For each molecule of glucose produced, 6 molecules of carbon dioxide and 6 molecules of water are consumed. Therefore, the plant would need to use 18 molecules of each.
b. How many moles of carbon dioxide and water would a plant have to use to produce 2 mole of glucose? Because a mole of anything contains the same number of molecules, the plant would need to use 6 times as many moles of carbon dioxide and water, or 12 moles of each.
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c. Refer to the summary formula for photosynthesis. If you know the number of molecules or moles of any of the reactants used (or products produced), how would you calculate the number of molecules or moles of all of the other reactants needed and products produced? If the formula is balanced and if it is a true representation of the overall reactions that occur, then the numbers in front of each reactant and product indicate the molecular or molar equivalents required for the reactions. Note: To represent the actual reactants required and products produced, the overall formula for photosynthesis is more correctly stated as: 6 CO2 + 12 H2O → C6H12O6 + 6 O2 + 6 H2O In most texts, however, this is reduced to 6 CO2 + 6 H2O → C6H12O6 + 6 O2 6. A biologist places a plant in a closed chamber. A sensor in the chamber maintains the carbon dioxide level at the normal atmospheric concentration of 0.03%. Another sensor allows the biologist to measure the amount of oxygen produced by the plant over time. If the plant produces 0.001 mole of oxygen in an hour, how much carbon dioxide had to be added to the chamber during that hour to maintain the atmospheric concentration of 0.03%? For every mole of oxygen produced, 1 mole of carbon dioxide had to be consumed. Therefore, 0.001 mole of carbon dioxide had to be added to maintain a constant level of CO2 in the chamber. 7. O2 and NH3 are both small covalent molecules found in cells. NH3 is extremely soluble in the aqueous environment of the cell, while O2 is relatively insoluble. What is the basis for this difference in solubility between the two molecules? In reaching your answer, draw the structures of the molecules as valence shell diagrams (as in Figure 2.12, page 38). Given these diagrams, consider the types of interactions each molecule could have with water.
H
N
H
O
O
H
Oxygen (O2)
Ammonia (NH3)
Ammonia is a polar molecule much like water. The N in it is relatively negative, and the H’s are relatively positive. Polar substances tend to be more soluble in water. O2, on the other hand, is not polar. 6
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8. Refer to pages 38–41 of Biology, 8th edition, which describe these types of chemical bonds: nonpolar and polar covalent bonds, ionic bonds, hydrogen bonds, and van der Waals interactions. The molecule diagrammed here can also be represented by the formula CH3COOH.
H
O
H C C H
OH
Explain how you could determine which of the bonds between elements in this molecule are polar or nonpolar covalent bonds, ionic bonds, hydrogen bonds, and van der Waals interactions. The best way to determine the bond types is to determine each atom’s electronegativity, or its attraction for electrons. As a general rule, the more filled the outer electron shell of an atom is, the higher is its electronegativity. In addition, the fewer electron shells, the greater the electron negativity. As a result, an atom’s attraction for electrons increases as you go from left to right in the periodic table. Electronegativity values tend to decrease as you go from top to bottom of the periodic table. To determine whether bonds are ionic, polar covalent, or nonpolar covalent, you need to determine the difference in electronegativity between the atoms that make up a molecule. If the difference in electronegativity is small, the bond is likely to be nonpolar covalent. If the difference is very large, the bond is likely to be ionic. Intermediate differences produce polar covalent bonds. The following table lists specific electronegativity values for selected elements. H = 2.1 Li = 1.0
Be = 1.5
B = 2.0
C = 2.5
N = 3.0
O = 3.5
F = 4.0
Na = 0.9
Mg = 1.2
Al = 1.5
Si = 1.8
P = 2.1
S = 2.5
Cl = 3.0
Using specific electronegativity values, you can determine the type of bond: If the difference in electronegativity between two atoms in a compound is less than 0.5, the bond is nonpolar covalent. If the difference is between 0.5 and 1.6, the bond is polar covalent. If the difference is greater than 1.6, the bond is ionic.
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Activity 3.1 A Quick Review of the Properties of Water 1. Compounds that have the capacity to form hydrogen bonds with water are said to be hydrophilic (water loving). Those without this capacity are hydrophobic (water fearing). Is the molecule on the left hydrophilic or hydrophobic? Explain your answer.
H
O
H C C H
This molecule is acetic acid. The COOH group is polar, which makes this molecule hydrophilic.
OH
2. In addition to being polar, water molecules can dissociate into hydronium ions (H3O+, often described simply as H+) and hydroxide ions (OH). The concentration of each of these ions in pure water is 107. Another way to say this is that the concentration of hydronium ions, or H+ ions, is one out of every 10 million molecules. Similarly, the concentration of OH ions is one in 10 million molecules. a. The H+ ion concentration of a solution can be represented as its pH value. The pH of a solution is defined as the negative log10 of the hydrogen ion concentration.What is the pH of pure water? The hydrogen ion concentration of pure water is 107. The log10 of 107 is 7. The negative log10 of 107 is therefore +7. b. Refer to the diagram of the molecule of acetic acid in question 1. The COOH group can ionize to release a H+ ion into solution. If you add acetic acid to water and raise the concentration of H+ ions to 104, what is the pH of this solution? The pH of a solution with a H+ ion concentration of 104 is 4. 3. Life as we know it could not exist without water. All the chemical reactions of life occur in aqueous solution. Water molecules are polar and are capable of forming hydrogen bonds with other polar or charged molecules. As a result, water has the following properties: A. H2O molecules are cohesive; they form hydrogen bonds with each other. B. H2O molecules are adhesive; they form hydrogen bonds with polar surfaces. C. Water is a liquid at normal physiological (or body) temperatures. D. Water has a high specific heat. E. Water has a high heat of vaporization. F. Water’s greatest density occurs at 4°C. 8
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Explain how these properties of water are related to the phenomena described in parts a–h below. More than one property may be used to explain a given phenomenon. a. During the winter, air temperatures in the northern United States can remain below 0°C for months; however, the fish and other animals living in the lakes survive. Water’s greatest density occurs at 4°C. In a lake, the 4°C water sinks below the water that is colder (or warmer). As a result, 0°C water is less dense than 4°C water. In addition, as it freezes, water takes on a crystalline structure and becomes ice. Ice has a density of about 0.92 g/cm3, pure water at 0°C has a density of about 0.99 g/cm3, and pure water at 4°C has a density of 1.0 g/cm3. The ice on top of a lake acts like insulation and, as a result, most deep lakes do not freeze to the bottom. b. Many substances—for example, salt (NaCl) and sucrose—dissolve quickly in water. Water is very polar. The attraction of the polar water molecules for the Na+ and Cl ions of NaCl is strong enough to allow them to dissociate and interact with water molecules (dissolve). c. When you pour water into a 25-mL graduated cylinder, a meniscus forms at the top of the water column. Water is attracted to the polar molecules that make up the glass (or plastic) cylinder. At the same time, they are attracted to each other. As a result, some of the water molecules associate with the polar molecules of the cylinder and are apparently “pulled up” the inside edge of the cylinder. d. Sweating and the evaporation of sweat from the body surface help reduce a human’s body temperature. Water has a high specific heat. The specific heat of water is 1 cal/g/°C. In other words, it takes 1 calorie of heat to change the temperature of 1 g of water 1°C. In addition, water has a high latent heat of vaporization (540 cal/g at 100°C). This can be thought of as the additional heat required to break apart polar water molecules so that they can move from the liquid to the gaseous state. As a result, evaporation (change of water from liquid to gaseous state) carries with it large amounts of heat. e. A bottle contains a liquid mixture of equal parts water and mineral oil. You shake the bottle vigorously and then set it on the table. Although the law of entropy favors maximum randomness, this mixture separates into layers of oil over water. The oil molecules are nonpolar and hydrophobic. The water molecules are polar and cohesive. As a result, the water molecules tend to interact strongly with each other and exclude the oil molecules. The oil layers on top of the water because it is less dense than water. Activity 3.1
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f. Water drops that fall on a surface tend to form rounded drops or beads. Water molecules are cohesive and form hydrogen bonds with each other. As a result, a drop of water tends to bead up or become rounded. g. Water drops that fall on your car tend to bead or round up more after you polish (or wax) the car than before you polished it. The wax (or polish) is hydrophobic and therefore less polar than the surface was likely to be before you polished it. Because the adhesion between the surface and the water molecules is lower, the cohesion of the water molecules for each other appears even more dramatic. h. If you touch the edge of a paper towel to a drop of colored water, the water will move up into (or be absorbed by) the towel. The polar water molecules adhere to the cellulose in the paper towel and cohere to each other. As a result, they are drawn up into the towel. The same mechanism accounts for the movement of water molecules up capillary tubes.
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Notes to Instructors Chapter 4 Carbon and the Molecular Diversity of Life Chapter 5 The Structure and Function of Macromolecules What is the focus of these activities? The activities associated with Chapters 2 and 3 provided students with a review of some of the basics of inorganic chemistry. The activities associated with Chapters 4 and 5 deal with organic chemistry and biochemistry. Although most introductory biology students have had some exposure to inorganic chemistry, fewer have had courses in organic chemistry or biochemistry. Therefore, these activities are designed with these goals in mind: • Help students identify the major differences among the four main types of biological macromolecules: carbohydrates, lipids, proteins, and nucleic acids. • Recognize what functional groups are and how they can modify the general properties and functions of organic compounds. • Use their understanding from the goals above to predict how various structural modifications can affect the behavior of macromolecules.
What are the particular activities designed to do? Activity 4.1/5.1 How can you identify organic macromolecules? This activity is designed to help students easily recognize carbohydrates, lipids, proteins, and nucleic acids when viewed as chemical structures. Students develop some simple rules to make it easier for them to recognize differences among the general chemical structures of carbohydrates, lipids, proteins, and nucleic acids. This type of activity is especially important for the many introductory biology students who have not yet had organic chemistry or biochemistry. Activity 4.2/5.2 What predictions can you make about the behavior of organic macromolecules if you know their structure? In this activity, students examine the general properties of organic macromolecules. In particular, they examine the properties of functional groups and how these can modify the behavior of macromolecules. This includes how modifying functional groups can affect the water solubility and the chemical reactivity of molecules. At the end of this activity, students are asked to predict how various macromolecules could react if placed in specific environments or if modified in specific ways. Answering these questions requires students to integrate and use understanding they have gained from Chapters 2–5.
Notes to Instructors
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What misconceptions or difficulties can these activities reveal? Activity 4.1/5.1 Question 1, Part A: This question asks for the C:H:O ratio of the various macromolecules. For carbohydrates, it is approximately 1:2:1. For many lipids, it is approximately 1:2:very few. And for proteins and nucleic acids, there is no reliable ratio of C:H:O. Many students become upset that the answer for proteins and nucleic acids is “no reliable ratio of C:H:O.” Remind them that they would not have known the C:H:O ratio was not a good predictor for these molecules unless they investigated it. Answers like this just mean we need to look for other ways of identifying these molecules. Some of these other ways include looking for specific functional groups, for example amino and carboxyl groups in amino acids and ribose versus deoxyribose and specific nucleotides to identify RNA versus DNA. Part B: After developing their own rules for identifying macromolecules, most students won’t have difficulty identifying these structures as carbohydrate, protein, lipid, or nucleic acid. It is best to present this activity in that light; that is, let students know that the purpose of the activity is to help them prove it is possible to categorize complex macromolecules using only a few simple rules. Activity 4.2/5.2 Question 1: The table asks students to look at different possible characteristics of R groups. Basic, acidic, or neutral
R group
Polar or nonpolar
Hydrophilic or hydrophobic
This question helps students understand, for example, that something that is polar is also hydrophilic; that is, these designations are not mutually exclusive. In addition, by learning the characteristics of key functional groups, students will have a better understanding of how modifications in macromolecular structure can lead to modifications in function. Question 4: The first three experiments ask students to examine how phospholipids (and glucose) will distribute themselves in different kinds of aqueous environments. Many students have memorized that phospholipids can form bilayers or micelles in aqueous environments, and they have the misconception that phospholipids will organize into micelles or bilayers under any circumstance.
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Answers Activity 4.1/5.1 How can you identify organic macromolecules? Refer to the figure (Some Simple Chemistry) on the next page when doing this activity. Part A. Answer the questions. Then use your answers to develop simple rules for identifying carbohydrates, lipids, proteins, and nucleic acids. 1. What is the approximate C:H:O ratio in each of the following types of macromolecules? Carbohydrates 1:2:1
Lipids 1:2:very few
Proteins There is no reliable C:H:O ratio for proteins.
Nucleic acids There is no reliable C:H:O ratio for nucleic acids.
2. Which of the compounds listed in question 1 can often be composed of C, H, and O alone? Carbohydrates and lipids can often be composed of C, H, and O alone. 3. Which of the compounds can be identified by looking at the C:H:O ratios alone? Only carbohydrates and some lipids can be identified using C:H:O ratios alone. 4. What other elements are commonly associated with each of these four types of macromolecules? Always contain P
Carbohydrates No
Generally contain Yes no P* Always contain N No Generally contain Yes no N Frequently No contain S Generally contain Yes no S
Lipids No (except for phospholipids) Yes (except for phospholipids) No
Proteins No
Nucleic acids Yes
Yes
No
Yes
Yes
Yes
No
No
No
Yes
No
Yes
No
Yes
*
Note: It is possible to find some exceptions in each of these categories where “Yes” is the answer to “Generally contain no ___.” For example, in reaction sequences many compounds undergo phosphorylation. However, if the natural state of the compound does not contain P (for example) the answer to “Generally contain no P” would be yes. Activity 4.1/5.1
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Some Simple Chemistry Basic components
Compound Carbohydrates:
Reaction
CH2OH
Sugars, starches, glycogen, cellulose
O H
C
C OH HO C
C
CH2OH O H
O
C
C
C
CH2OH
O
H
O
C
H2 O
6C hexose
C
H C OH HO
H
(CH2 )n CH3
O
H C
O
H C
O
H C
O
O C H C OH HO C H
C
(CH2 )n CH3
C
(CH2 )n CH3
C
(CH2 )n CH3
O
(CH2 )n CH3
O H C OH HO
+ 1 H2 O
Disaccharide
dehydration reaction
O
H
Fats, oils, waxes, cholesterol
H C
C
H
HO
Lipids:
CH2OH
CH2OH
O H H C H C H HO C HO C C OH
Product
O
(CH2 )n CH3
H
3 H2 O
Glycerol + 3 fatty acids
+ 3 H2 O
Triglyceride or fat dehydration reaction
Proteins:
H
Enzymes, structural proteins
R
amino group
R
carboxyl group
5
HOCH2
+
H H 3C
H
HOCH2 PO4
+
+
C2
peptide bond
OH
H
+ 1 H2 O
Base (Base = A, U, G, or C)
P
P
O
OH
H H
B
B
S
S
P
B P
S
(etc.)
P
RNA
C
3C
R
O
N C C
Dipeptide
1
H
4C
H
OH
H2 O
OH OH Ribose 5
H2N C C
OH
C
H
OH H
OH R
H
O
N C C
dehydration reaction
O
4C
PO4
R
O H
H2N C C
Amino acid
Nucleic acids: DNA, RNA
H
H2N C COOH
1
H
+
Base (Base = A, T, G, or C)
C2
P
S
P
S
P
S
B
B
B
B
B
B
S
P
OH H Deoxyribose
S
P
S
P
(etc.)
hydrogen bonds (etc.)
P
DNA
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5. Functional groups can modify the properties of organic molecules. In the following table, indicate whether each functional group is polar or nonpolar and hydrophobic or hydrophilic. Which of these functional groups are found in proteins and lipids? Found in all Found in Found in many proteins many lipids proteins No In some R In fatty acids groups as terminal reactive group
Functional group —OH
Polar or nonpolar Polar
Hydrophobic or hydrophilic Hydrophilic
—CH2
Nonpolar
Hydrophobic
No
—COOH
Polar
Hydrophilic
Yes
No
—NH2
Polar
Hydrophilic
Yes
No
—SH
Polar
Hydrophilic
No
No Found in cysteine
—PO4
Polar
Hydrophilic
No
Only if they In have been phospholipids phosphorylated
Yes in side groups
Yes
No
6. You want to use a radioactive tracer that will label only the protein in an RNA virus. Assume the virus is composed of only a protein coat and an RNA core. Which of the following would you use? Be sure to explain your answer. a. Radioactive P
b. Radioactive N
c. Radioactive S
d. Radioactive C
To distinguish between protein and RNA in a virus, you could use radioactively labeled S compounds. If you grew viruses on cells with radioactively labeled S compounds, the sulfhydryl groups in the virus’s protein would become labeled but the RNA would not become labeled. 7. Closely related macromolecules often have many characteristics in common. For example, they share many of the same chemical elements and functional groups. Therefore, to separate or distinguish closely related macromolecules, you need to determine how they differ and then target or label that difference. a. What makes RNA different from DNA? RNA contains ribose sugar, whereas DNA contains deoxyribose sugar. In addition, RNA contains uracil and not thymine. DNA contains thymine but not uracil. b. If you wanted to use a radioactive or fluorescent tag to label only the RNA in a cell and not the DNA, what compound(s) could you label that is/are specific for RNA? You could label either ribose or uracil. Activity 4.1/5.1
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c. If you wanted to label only the DNA, what compound(s) could you label? You could label either deoxyribose or thymine. 8. Based on your answers to questions 1–7, what simple rule(s) can you use to identify the following macromolecules? Carbohydrates
Look for a 1:2:1 C:H:O ratio. Many carbohydrates will contain no P, N, or S. Look for a 1:2 ratio of C:H and only very small amounts of O. Most will contain no S. Phospholipids can contain P and N (as part of the choline group; see Figure 5.13 in Biology, 8th edition).
Lipids
Proteins
Look for amino and carboxyl groups. Some contain S. All proteins can be identified by the presence of peptide bonds. (See Figure 5.18 for the structure of a peptide bond.)
Nucleic acids
Look for nucleotides made up of a five-carbon sugar, a phosphate group, and a nitrogenous base. DNA contains phosphate, deoxyribose sugar, and adenine, guanine, cytosine, and thymine. RNA contains phosphate, ribose sugar, and adenine, guanine, cytosine, and uracil.
DNA vs RNA
Part B. Carbohydrate, lipid, protein, or nucleic acid? Name that structure! Based on the rules you developed in Part A, identify the compounds below (and on the following page) as carbohydrates, lipids, amino acids, polypeptides, or nucleic acids. In addition, indicate whether each is likely to be polar or nonpolar, hydrophilic or hydrophobic. H 1) C17H35COOH +
1) lipid (fat or triglyceride)
H
H O C H
C17H35COO C H
H O C H
C17H35COO C H
H O C H
C17H35COO C H
Hydrophobic and nonpolar 2) amino acid The amino and carboxyl group would make this somewhat polar and hydrophilic.
H 2)
HC
C
H+ N
N C H
CH2C COOH NH2
3) a tripeptide made up of 3 amino acids
H
The R groups are hydrophobic with the possible exception of the OH group. The amino and carboxyl groups are hydrophilic and polar.
0H H
O 3)
0 CHCH3
C C N C C HO
H H
H
0 CH2 H N C C
N
H
H
H
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Base
O
HH
2:03 AM
CH2OH O H HOCH2 O H H H O 5) OH H H HO CH2OH HO
HH
O H P HO O O CH2
OH
H
OH H
Base
O
HH
HH
O H HO P O O CH2
Base
O
HH
HH
O H HO P O O CH2
Base
O
HH
O H
HH
OH H
O O
6) H3C
CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C OH
O
N N
7) HO P O CH2 O OH HH HH
OH OH NH2 8) NH2 C
NH
O CH2 CH2 CH2 CH N
C H OH
H
CH2OH O 9)
CH2OH O
OH
CH2OH O
O
O
O
CH2OH O
OH
O OH
4) single strand of 4 bases in DNA
OH
H
H
H
H
10) H C
C
C
C C
OH OH OH OH
OH
This compound is hydrophilic and polar. 6) fatty acid Hydrophobic and nonpolar with the exception of the carboxyl group.
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Because of the phosphate groups DNA tends to be negatively charged and therefore somewhat polar.
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7) ribonucleotide Again the phosphate group is polar. 8) amino acid The amine and carboxyl groups are polar. 9) polysaccharide Polysaccharides made up of many monosaccharide subunits (like this part of a glycogen molecule) are
used for storage. While the hydroxyl groups on this molecule may be somewhat polar, the molecule as whole is relatively insoluble and therefore hydrophobic. 10) 5 carbon sugar This molecule is both polar and hydrophilic.
4.1/5.1 Test Your Understanding A student, Mary, is given four samples and told they are lysine (an amino acid), lactose (a disaccharide), insulin (a protein hormone), and RNA. The samples are in test tubes marked 1, 2, 3, and 4, but Mary doesn’t know which compound is in which tube. She is instructed to identify the contents of each tube. a. In her first test, she tries to hydrolyze a portion of the contents of each tube. Hydrolysis occurs in all tubes except tube 3. b. In Mary’s next test, she finds that tubes 1, 2, and 3 are positive for nitrogen but only tube 2 gives a positive result for the presence of sulfur. c. The last test Mary performs shows that the compound in tube 1 contains a high percentage of phosphate. Based on these data, fill in the following table and explain your answers. Tube number 1
Contents
Explanation
RNA
Like DNA, RNA contains a sugar-phosphate backbone.
2
Insulin
3
Lysine
Sulfur is a component of some amino acid side chains. It is not found in lysine, lactose, or RNA. All of the compounds except lysine are composed of macromolecular monomers joined by dehydration reactions.
4
Lactose
Since the contents of all the other tubes were determined in tests a to c, this tube must contain the lactose, a disaccharide.
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Activity 4.2/5.2 What predictions can you make about the behavior of organic macromolecules if you know their structure? 1. Twenty amino acids are commonly utilized in the synthesis of proteins. These amino acids differ in the chemical properties of their side chains (also called R groups). What properties does each of the following R groups have? (Note: A side chain may display more than one of these properties.) R group
Basic, acidic, or neutral?
Polar or nonpolar? Hydrophilic or hydrophobic?
Neutral
Nonpolar
Hydrophilic
Acidic
Polar
Hydrophilic
Basic
Polar
Hydrophilic
Neutral
Polar
Hydrophilic
a.
CH2 CH CH3
CH3
b.
CH2 –O
C
O
c.
CH2 CH2 CH2 CH2 NH3+ d.
CH2 OH
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2. Polypeptides and proteins are made up of linear sequences of amino acids. In its functional form, each protein has a specific three-dimensional structure or shape. Interactions among the individual amino acids and their side chains play a major role in determining this shape. a. How are amino acids linked together to form polypeptides or proteins? What is this type of bond called? Amino acids are covalently linked together via peptide bonds to form polypeptides or proteins. (See Figure 5.18 in Biology, 8th edition.) b. Define the four structures of a protein. Primary: The linear sequence of amino acids in a polypeptide or protein
c. What kinds of bonds hold each of these structures together? Covalent peptide bonds formed by dehydration reactions hold the individual amino acids together in the polypeptide chain.
Secondary: helix or pleated sheet conformations occurring at regular intervals along the polypeptide
The secondary structure results from H bonding relationships set up between the H attached to the N in one amino acid and the O attached to the C of another amino acid. (See Figure 5.21.)
Tertiary: The folded or functional conformation of a protein
Hydrogen and covalent bonds between side chains (R groups) of various amino acids contribute, as do hydrophobic interactions and van der Waals interactions.
Quaternary: The folded or functional conformation of a protein made up of more than one polypeptide chain
Hydrogen and covalent bonds between side chains (R groups) of various amino acids contribute, as do hydrophobic interactions and van der Waals interactions.
3. Lipids as a group are defined as being hydrophobic, or insoluble in water. As a result, this group includes a fairly wide range of compounds—for example, fats, oils, waxes, and steroids like cholesterol. a. How are fatty acids and glycerol linked together to form fats (triglycerides)? Dehydration reactions between the OH of the carboxyl group on the fatty acid and the OH group on the glycerol molecule bond the fatty acids to the glycerol molecules. b. What functions do fats serve in living organisms? In general, fats are energy storage molecules.
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c. How do phospholipids differ from triglycerides? Phospholipids have one of the OH groups of the glycerol interacting with a phosphate-containing side group—for example, phosphatidylcholine as in Figure 5.13. d. What characteristics do phospholipids have that triglycerides do not have? Phospholipids are amphipathic because the phosphate-containing side group is hydrophilic and the remainder of the molecule is hydrophobic. Triglycerides are hydrophobic.
4.2/5.2 Test Your Understanding Use your understanding of the chemical characteristics of the four major types of macromolecules in living organisms to predict the outcome of the following experiments. Be sure to explain your reasoning. Experiment 1: You stir 10 g of glucose and 10 mL of phospholipids in a 500-mL beaker that contains 200 mL of distilled water. Draw a diagram to show where and how the glucose and phospholipids would be distributed after you let the mixture settle for about 30 minutes. The 10 g of glucose will dissolve in the water and be relatively evenly distributed in the water. The phospholipids will float on the surface of the water. The phospholipids at the water interface will have their hydrophilic phosphate heads in the water and their hydrophobic tails sticking out of the water. Any phospholipids trapped under the water may form micelles, with their fatty acid tails on the interior and their phospholipid heads pointing outward. It is possible that some of the phospholipids will form bilayers, which organize themselves into spheres containing small amounts of water. Experiment 2: You repeat Experiment 1, but this time you stir 10 g of glucose and 10 mL of phospholipids in a different 500-mL beaker that contains 200 mL of distilled water and 100 mL of oil. Draw a diagram to show where and how the glucose, phospholipids, and oil would be distributed after you let the solution settle for about 30 minutes. As in Experiment 1, the 10 g of glucose will dissolve in the water and be relatively evenly distributed in the water. Depending on how vigorously the system is mixed, the phospholipids may still float on the surface of the water. The phospholipids at the water interface will have their hydrophilic phosphate heads in the water and their hydrophobic tails associated with the oil layer above the water. Any phospholipids trapped under the water may form micelles, with their fatty acid tails on the interior and their phospholipid heads pointing outward. It is possible that some of the phospholipids will form bilayers,
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which organize themselves into spheres containing small amounts of water. Any phospholipids trapped in the oil layer may form inverted micelles, with their hydrophilic heads interior and their hydrophobic tails on the surface. Experiment 3: To completely fill a sealed 500-mL glass container that contains 490 mL of distilled water, you inject 10 mL of phospholipids into it. (A small gasket allows the air to leave as you inject the phospholipids.) You shake this mixture vigorously and then let it settle for an hour or more. Draw a diagram to show how the phospholipids would be distributed in the container. Under these circumstances, only micelles (and some phospholipid bilayer spheres) are likely to form. Experiment 4: A globular protein that is ordinarily found in aqueous solution has these amino acids in its primary structure: glutamic acid, lysine, leucine, and tryptophan. Predict where you would find each amino acid: in the interior portion of the protein (away from water) or on the outside of the protein (facing water). (Refer to Figure 5.17, page 79.) Glutamic acid and lysine are electrically charged and will therefore be on the outside of the protein. Leucine and tryptophan are nonpolar and will be inside the protein. Experiment 5: Drawn below is part of the tertiary structure of a protein showing the positions of two amino acids (aspartic acid and lysine). Replacing lysine with another amino acid in the protein may change the shape and function of the protein. Replacing lysine with which type(s) of amino acid(s) would lead to the least amount of change in the tertiary structure of this protein? (Refer to Figure 5.17, page 79.)
Ionic bond
CH2 CH2 CH2 CH2
O
NH3+ –O C CH2 Aspartic acid
Lysine
Aspartic acid has a negatively charged R group. Lysine has a positively charged side group. To cause the least amount of change in the tertiary structure of this protein, you have to replace lysine with an amino acid that contains a positively charged side group, such as arginine or histidine. 22
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Notes to Instructors Chapter 6 A Tour of the Cell What is the focus of this activity? Students sometimes think that introductory biology texts and courses spend too much time dealing with cellular structure and function. It helps to remind them that many species of organisms (probably most species) exist only as single cells. Even complex multicellular organisms such as oak trees and humans start their lives as single-celled zygotes. In other words, all the characteristics of life can be found in single cells. Considering the complexity of some multicellular organisms, how is this possible? How do prokaryotes such as heterotrophic bacteria eat? Why are all multicellular organisms eukaryotic? How can liver and nerve cells perform such different functions? How can different organelles in a cell contain different levels or concentrations of specific ions and other chemicals? Questions like these, and many others, cannot be answered without an understanding of basic cell structure and function.
What is the particular activity designed to do? Activity 6.1 What makes a cell a living organism? The specific questions in this activity are designed to help students review and understand: • the minimum structures or components a cell must contain to be alive, • the function(s) of each part of the cell and how the function(s) is/are related to its structure, and • the relative sizes of cellular structures or components.
What misconceptions or difficulties can this activity reveal? Activity 6.1 Question 2: In general, students understand that animal cells contain mitochondria. Yet, many think that plant cells differ from animal cells because they contain chloroplasts instead of mitochondria. It helps to remind students that plant cells contain both mitochondria and chloroplasts. It is even more effective, however, to ask them how plants make ATP at night, or how root cells, which cannot photosynthesize, make ATP. Given these questions, most students will understand that plants must use ATP even in the dark. Some students don’t understand why this is a problem, however, because they think that ATP can be stored in the cells for use during periods when the plant is not
Notes to Instructors
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photosynthesizing. As a result, you may need to let students know that ATP is not an energy storage molecule. Instead, plants produce sugars (which are converted to macromolecules such as starches) for long-term energy storage. Question 3: Most students understand that a micrometer is one-millionth of a meter and a nanometer is one-billionth of a meter. However, they still may not have a good feel for the relative sizes of molecules and organelles in the cell. This question is designed to give students a better understanding of these relative sizes and help them understand how so many different chemical reactions can occur simultaneously inside a cell.
Answers Activity 6.1 What makes a cell a living organism? 1. Single-celled organisms and individual cells within multicellular organisms can vary greatly in appearance as well as in the functions they perform. Nonetheless, each of these cells is alive and therefore must have some common characteristics. a. At a minimum, what structures or b. What is the function of each structure components must a cell contain to be or component listed in part a? alive? 1. Plasma or cell membrane
1. A selectively permeable cell membrane allows the cell to control what enters and exits the cell. 2. DNA contains the genetic information for the production of all the macromolecules required by the cell—for example, enzymes, carbohydrates, structural proteins, etc. 3. Ribosomes are required for the translation of proteins from mRNA.
2. DNA
3. Ribosomes
c. If you consider the types of single-celled organisms that exist today, which, if any, have a structure similar to your description in part a? Many members of the prokaryotes have a structure similar to that described in part a.
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2. What would you need to add to or change about the cell you described in question 1 to make it: a. A eukaryotic animal cell?
b. A eukaryotic plant cell?
1. Double-membrane-bound nucleus containing chromosomes complexed with histone proteins. 2. Double-membrane-bound mitochondria for use in aerobic cellular respiration. 3. Cytoskeleton. 4. Endoplasmic reticulum and Golgi apparatus. 5. Centrosome for microtubule and spindle production. In animal cells, the centrosome contains a pair of centrioles. 6. Peroxisomes for a variety of functions, including generating hydrogen peroxide from oxygen and degrading it to water. 7. Lysosomes for intracellular digestion of macromolecules
1. Double-membrane-bound nucleus containing chromosomes complexed with histone proteins. 2. Double-membrane-bound mitochondria for use in aerobic cellular respiration. 3. Cytoskeleton. 4. Endoplasmic reticulum and Golgi apparatus. 5. Microtubule organizing center (MTOC) for microtubule and spindle production. In general, plants do not have centrioles. Only a few cell types (e.g. in ferns and gymnosperms) develop centrioles. 6. Peroxisomes for a variety of functions, including generating hydrogen peroxide from oxygen and degrading it to water. 7. Double-membrane-bound chloroplasts for photosynthesis. 8. Central vacuole for storage and for breakdown of waste products. 9. Cell wall. 10. Plasmodesmata, which are connections of cytoplasm from one cell to the next.
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3. To get an idea of the different sizes of various cellular components, do the following calculations: Assume that the cell, its nucleus, and a globular protein—for example, an enzyme—are spherical. In addition, assume the diameter of the protein is 5 nm, the diameter of the cell is 100 m (micrometers), and the diameter of the nucleus is 40 m. If you draw the globular protein as a sphere with a diameter of 2 cm (approximately the diameter of a U.S. penny), what size would each of the following measurements of the cell be if drawn to the same scale (5 nm real length 2 cm)? a. The radius of a microtubule (Refer If 5 nm = 2 cm, then 1 nm = 0.4 cm. Radius to Table 6.1, page 113, in Biology, of microtubule = 25 nm 25 0.4 cm 8th ed.) 10 cm. b. The diameter of the nucleus
40 m 40,000 nm 0.4 cm/nm 160 m (A football field is 100 yards long, or about 91.4 m.)
c. The diameter of the cell
100 m 100,000 nm 0.4 cm/nm 400 m
d. The volume (V 4/3 r3) of the 4/3 (1 cm)3 4.2 cm3 protein 1 nanometer cubed (1 nm3) 1.0 1021 centimeters cubed (cm3). e. The volume of the nucleus
4/3 (80 m)3 2.1 106 m3
f. The volume of the cell
4/3 (200 m)3 3.4 107 m3
g. The volume of the Empire State Building is 1.05 106 m3. How many of your scaled nuclei could fit into the Empire State Building? How many of your scaled cells could fit? The volume of the scaled nucleus is almost 2 times the volume of the Empire State Building and the cell volume is about 20 times greater. h. Do the results of these calculations help you to understand how so much can be going on inside a cell at once? Explain. The calculations give a clearer idea of dimension relationships inside cells. For example, if a single protein molecule is only about 2 cm in diameter, 20,000 protein molecules of this size could be lined up along the diameter of the cell (400 m).
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Notes to Instructors Chapter 7 Membrane Structure and Function What is the focus of these activities? To be alive and to continue to survive in changing environments, organisms must be able to keep their integrity separate from the environment and other organisms. In Activity 6.1, students reviewed the cell as the basic unit of life. In the Chapter 7 activities, students review the functions of membranes that allow them to control the movement of substances into and out of the cells.
What are the particular activities designed to do? Activity 7.1 What controls the movement of materials into and out of the cell? Activity 7.2 How is the structure of a cell membrane related to its function? These activities are designed to help students begin to understand: • the physical and chemical factors that affect the transport of substances into and out of cells (as well as into and out of organelles), and • how the size of solute molecules affects osmotic potential.
What misconceptions or difficulties can these activities reveal? Activity 7.1 Question 1: Many students visualize the phospholipid bilayer as a static structure rather than as a fluid structure. As a result, they have a difficult time understanding how anything could move across a membrane made of phospholipids alone. The difficulty understanding the fluid nature of membranes isn’t that surprising. After all, how does something that is fluid act as a boundary layer? If it is so fluid, why do red blood cells in hypotonic solution swell so much before they burst or lyse? Why don’t the membranes just separate into individual phospholipid units under this pressure? Many of us have similar difficulty understanding how various chemical elements can be solid or dense when each atom is composed of electrons orbiting around a central core of protons and neutrons. Although it doesn’t answer these questions, it helps to remind students of the size differences that exist among cells, phospholipids, and the substances that can move through them. For example, a molecule of oxygen, carbon dioxide, or water is at least 50 times smaller than a phospholipid molecule. In addition, the distances that individual phospholipid molecules move per unit time are very small. For example, Biology, 8th Notes to Instructors
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edition, notes that phospholipids can move about 2 m/second. In comparison, the diffusion rate for substances in water is about 40 m/second. This means that the movement of phospholipids in the membrane is about 20 times slower than the movement of water molecules in diffusion. In other words, phospholipid membranes are fluid but nowhere near as fluid as water. Question 2: Many students don’t understand how molecular size affects osmotic potential. Once they understand this, it will be easier for them to understand why systems don’t store ATP or sugar molecules, but instead store energy in the form of starches and fats. Activity 7.2 Modeling the different types of transport as they occur in a single membrane allows students to recognize that all three types of transport can be going on simultaneously in different parts of the membrane.
Answers Activity 7.1 What controls the movement of materials into and out of the cell? 1. To be alive, most cells must maintain a relatively constant internal environment. To do this, they must be able to control the movement of materials into and out of the cell. What characteristics of the cell membrane determine what gets into the cell and what doesn’t? That is, what determines the permeability of a cell or organelle membrane? To answer these questions, first consider the answers to the following questions:
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a. If a cell membrane were b. What different roles or
c. Why are some cell composed of only a functions do membrane types more permeable phospholipid bilayer, proteins serve? to a substance (for what properties would it example, sodium ions) have? than others? Phosopholipids contain a Membrane proteins allow The difference in hydrophilic phosphate charged and larger permeability is the result head region and a molecules to cross the of differences in the types hydrophobic fatty acid tail membrane. They facilitate of membrane proteins the region. When these form the passage of these cells contain. a bilayer, the phosphate molecules. As a result, heads interact with the diffusion of molecules water and the fatty acid through such membrane tails interact with each other. If you think of this proteins is called as a static, continuous facilitated diffusion. structure, then it is easy to Other membrane proteins see that nothing should get use the energy in ATP to through. Charged move molecules across molecules should be membranes against a repelled or attracted by the concentration phospholipid head ends, gradient.These types of and hydrophilic membrane proteins are substances should be often called pumps and unable to cross the fatty function in active acid region. However, this transport. is not a static structure, so very small uncharged molecules and water can move across a phospholipid bilayer.
Using your understanding of the answers in a-c, now answer these questions: What characteristics of the cell membrane determine what gets into the cell and what doesn’t? That is, what determines the permeability of a cell or organelle membrane? Only very small uncharged or nonpolar molecules are capable of diffusing across the phospholipid bilayer. Specific membrane transport proteins are required for the movement of all other types of molecules into or out of the cell or organelle. Cells and organelles vary in the types of membrane transport proteins they contain. This variability is a function of which genes are active in each cell type.
Activity 7.1
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2. You design an experiment to test the effect(s) various compounds have on the osmotic potential of a model cell. You know that substances dissolved in aqueous or gaseous solutions tend to diffuse from regions of higher concentration to regions of lower concentration. You fill each of three (20 mL) dialysis bags half full with one of these substances: • 5% by weight of glucose in distilled water • 5% by weight of egg albumin (protein) in distilled water • 5% by weight of glass bead (one glass bead) in distilled water The dialysis bag is permeable to water but impermeable to glucose, albumin, and glass bead. a. If the final weight of each bag is 10 g, how many grams of glucose, albumin, and glass bead were added to each bag? Five percent of the weight in each bag is glucose, albumin, or glass bead. Therefore, each of these weighs 0.5 g. b. The molecular weight of the protein is about 45 kilodaltons, and the molecular weight of glucose is about 180 daltons. How can you estimate the number of molecules of glucose in the 5% solution compared to the number of albumin molecules in its 5% solution? Each molecule of protein is 45,000/180 (or 250) times heavier than each molecule of glucose. Therefore, if the weights of glucose and albumin are the same in each of the bags, there are about 250 times more molecules of glucose in 0.5 g than molecules of protein in 0.5 g. c. You put the dialysis bags into three separate flasks of distilled water. After 2 hours, you remove the bags and record these weights: Dialysis bag
Weight
Glucose Albumin Glass bead
13.2 g 10.1 g 10.0 g
How do you explain these results? (Hint: Consider the surface area-to-volume ratio of each of the three substances and review pages 50 and 51 of Biology, 8th edition.)
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Five grams of glucose equals 5g/180g/M = approximately 0.027 M of glucose. Five grams of albumin equals 5g/45,000g/M = approximately 0.00011M of albumin. In other words, 5 g of glucose contains many more molecules of solute than does 5 g of albumin. As a result, the 5 g of smaller glucose molecules have a much larger total surface area than the 5 g of albumin molecules. When molecules dissolve in water, water molecules form a ring of hydration around each of the solute molecules. The water in the ring of hydration in effect reduces the concentration of free water molecules in the solution (relative to the distilled water outside the bag). Free water then moves down its concentration gradient and enters the bag. Because the 5 grams of albumin have a much smaller surface area, much less water is required to form the rings of hydration around the albumin molecules. Therefore, there is a smaller difference in the concentrations of free water molecules inside versus outside the bag. When the concentration difference is smaller, less water enters the bag. In contrast, the glass bead is insoluble in water. It can be thought of as one large “molecule.” Therefore, its surface-area-to-volume ratio is much smaller than for either the albumin or glucose. The amount of free water that is required to form the ring of hydration around the glass bead is very small relative to either albumin or glucose. d. What results would you predict if you set up a similar experiment but used 5% glucose and 5% sucrose? Glucose has a molecular weight of 180 daltons. Sucrose has a molecular weight of 342 daltons. Both would gain water, but the glucose solution would gain more water because, on average, a 5% glucose solution contains twice as many molecules as a 5% sucrose solution. In addition, each glucose molecule is smaller than each sucrose molecule, so its surface-area-to-volume ratio is higher.
Activity 7.2 How is the structure of a cell membrane related to its function? Membranes compartmentalize the different functions of living cells. The cell membrane is a barrier between the cell or organism and its environment. Similarly, within the cell, membranes of organelles separate the different reactions of metabolism from each other. Use the supplies provided in class or devise your own at home to develop a model of a cell membrane. Developing models of systems can help you understand not only their overall structure but also their function(s).
Activity 7.2
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Building the Model • Include in the membrane the phospholipid bilayer (phosphate heads and fatty acid tails) as well as the integral proteins. • Design integral proteins that serve the functions of facilitated diffusion and active transport. • Indicate how the various types of integral proteins might differ in structure and operation. Use the understanding you gain from your model to answer the following questions. 1. Substances can move across the membrane via simple diffusion, facilitated diffusion, or active transport.
Simple diffusion
Facilitated diffusion Active transport
a. Where does it b. Does it require occur in transport membrane? protein? Across the No phospholipid bilayer Through membrane Yes proteins
c. Does it require input of energy?
Through membrane Yes proteins
Yes
No
No
d. What functions might each of the three types of diffusion serve in an independent cell such as a Paramecium or an amoeba? Many possible examples can be used to answer this. For example, water and oxygen could enter these cells via simple diffusion. Fatty acids, produced as a result of digestion in a food vacuole, may also be able to move across the vacuole membrane by simple diffusion. Other macromolecules are likely to be moved across the vacuole membrane via active transport or facilitated transport. Ions may move passively across membranes inside the cell or may be actively transported if concentration differences exist across membranes. e. What functions might each of the three types of diffusion serve in a multicellular organism—for example, a human or a tree? Again, many possible examples can be used here. For example, in humans and trees, various ion concentration gradients are set up via active transport. Water, oxygen, and carbon dioxide move across membranes via simple diffusion. Movement of some macromolecules, such as glucose, can occur via facilitated transport. 32
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2. What would you need to observe or measure to determine whether a substance was moved across a membrane via each type of diffusion? Simple diffusion
Facilitated diffusion
Active transport
No energy required—that is, this can occur in absence of ATP. Can be observed to occur across artificial phopholipid bilayer.
No energy required. Does not occur across a phospholipid bilayer alone. Occurs across a membrane containing appropriate membrane proteins.
Energy required. Occurs only across a membrane containing the appropriate membrane proteins.
3. The ratios of saturated to unsaturated phospholipids in an organism’s membranes can change in response to changes in environmental conditions. a. How do the properties of a membrane that contains a low percentage of unsaturated phospholipids compare with those of a membrane that contains a high percentage of unsaturated phospholipids? Unsaturated phopholipids remain fluid at lower temperatures. Therefore, a membrane with a low percentage of unsaturated phospholipids is less fluid than one with a high percentage of unsaturated phospholipids. b. Considering what you know about the properties of saturated and unsaturated fatty acids, would you expect an amoeba that lives in a pond in a cold northern climate to have a higher or lower percentage of saturated fatty acids in its membranes during the summer as compared to the winter? Explain your answer. For the membranes to remain fluid in cold temperatures, they must contain a higher concentration of unsaturated phospholipids in the winter than in the summer. In the summer, increasing the concentration of saturated phospholipids would prevent the membrane from becoming too fluid. In fact, scientists have measured the percentage of saturated and unsaturated fatty acids in the membranes of the same organisms over the year and have noted that the composition changes to increase the percentage of unsaturated fatty acids in colder weather.
Activity 7.2
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4. A fish is removed from a contaminated lake. You determine that a particular toxin is present in its cells at concentration X = 1500 g/L. You place the fish in a tank full of clean water (X = 0 g/L), and measure the toxin concentration in the fish cells over the next 10 days. a. On the graphs below, predict how the toxin concentrations in the fish and in the water will change over time if: i. the toxin is water soluble ii. the toxin is fat soluble
Toxin g/L
i) Toxin level in fish
1,600
1,600
1,000
1,000
200
200 1
Toxin g/L
i) Toxin level in water
2
3
4
5
6
7
8
1
9 10
2
3
4
5
6
Time in days
Time in days
ii) Toxin level in fish
ii) Toxin level in water
1,600
1,600
1,000
1,000
200
200 1
2
3
4
5
6
7
8
9 10
Time in days
1 2
3
4
5
6
7
8
9 10
7
8
9 10
Time in days
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b. After making your hypothesis, you test it by measuring the toxin levels in the fish at various times during its 10 days in the tank. You observe that the level of toxin in the fish drops from 1,500 g/L to 750 g/L and then stabilizes at 750 g/L. You test the water in the tank and find that after it stabilizes, toxin is present in the water at 750 g/L also. • Which of your predictions fit these data? The hypothesis that the toxin is water soluble fits the data. • Which of the following processes is most likely eliminating the toxin from the fish? i. Passive transport ii. First active, then passive transport iii. First passive, then active transport iv. Active transport Because the levels inside and outside the fish become equal and stabilize there, the data support a passive transport mechanism alone. c. Given the situation in part b, what should you do, in the short term, to continue to reduce the toxin level in the fish below 750 g/L? The easiest method would be to continue to change the water on a daily basis. Each time the water is changed, the concentration of X outside the fish is lower than inside. As a result, X will tend to move from the region of higher concentration (in the fish cells) to lower concentration in the surrounding water. 5. A particular amino acid is transported from the extracellular medium against its concentration gradient. The integral membrane protein that transports the amino acid also binds and transports Na+. Using your model of the cell membrane, develop a transport mechanism that will permit the amino acid uptake to be coupled to the Na+ transport so that the amino acid’s entry is linked only indirectly to ATP hydrolysis. This situation is similar to the one diagrammed in Figure 7.19, on page 79.
Activity 7.2
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Notes to Instructors Chapter 8 An Introduction to Metabolism What is the focus of these activities? Many students study biology as if it were a foreign language. For them, each new topic or idea is distilled into separate definitions. Each is then written on a flash card to be memorized. This approach to studying biology has obvious limitations; the biggest is that it does not require students to make connections and integrate their understanding. Activity 8.1 requires students to integrate information from Chapter 8 on general metabolism and enzyme function with what they learned about protein structure in Chapter 5. Activity 8.2 has them apply their understanding to propose how differences in experimental conditions could affect reaction rates and to analyze the results of some simple experiments in enzyme function.
What are the particular activities designed to do? Activity 8.1 What factors affect chemical reaction in cells? This activity asks the students to construct a concept map that integrates their understanding of protein structure, enzyme function, and general energy transformations in metabolic reactions. This requires that they have a general understanding of enzyme structure and function. Activity 8.2 How can changes in experimental conditions affect enzyme-mediated reactions? This activity asks students to apply their general understanding from Activity 8.1 to specific experimental situations. They must understand how enzyme function (and therefore metabolism) can be affected by changes in substrate and/or product concentration and by changes in the enzyme itself.
What misconceptions or difficulties can these activities reveal? Activity 8.1 If done in small groups in class, the concept mapping activity can be used to reveal both the students’ levels of understanding and any misconceptions they may have. Doing this activity helps many students understand how the tertiary (or quaternary) structure of an enzyme can be modified by modifying a side group on an amino acid. In addition, understanding the impact of changes in the physical or chemical conditions surrounding the enzyme can help students understand why organisms expend so much energy maintaining homeostasis.
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Question 1: Because they tend to memorize terms and definitions separately, many students don’t recognize that a single reaction can simultaneously be spontaneous, catabolic, and exergonic and have a negative G. Question 2: For years the dogma held that all enzymes were proteins. This question can be used to point out that we now know that some enzymes are made of RNA. Question 3: This question points out the types of factors that can affect enzyme function. Students need to make the logical leap that for cells to maintain their metabolism at optimal levels, they must be able to regulate their internal environment. Many students do not make this connection without additional help or suggestions. Activity 8.2 This set of experimental situations gives students practice in making the connection between “knowing something” (for example, knowing the definitions of competitive and noncompetitive inhibitors) and understanding how to use that information to interpret results or to predict results if their understanding is correct. Many students are not comfortable doing this, especially in situations that have more than one possible answer depending on the assumptions the student makes. Therefore, these questions give them practice with developing both assumptions and predictions based on those assumptions.
Answers Activity 8.1 What factors affect chemical reactions in cells? Construct a concept map of general metabolism using the terms in the list below. Keep in mind that there are many ways to construct a concept map. • Begin by writing each term on a separate sticky note or piece of paper. • Then organize the terms into a map that indicates how the terms are associated or related. • Draw lines between terms and add action phrases to the lines to indicate how the terms are related. • If you are doing this activity in small groups in class, explain your map to another group when you finish it. Here is an example:
Sun
provides energy for
Photosynthesis
which occurs in
Plants
Activity 8.1
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Terms peptide bonds proteins helix primary structure secondary structure tertiary structure pleated sheet R groups hydrogen bonds substrate or reactant (ligand)
activation energy G / free energy endergonic exergonic enzymes catalysts competitive inhibitor noncompetitive inhibitor active site product allosteric regulation
activator four-step enzyme-mediated reaction sequence or metabolic pathway (A → B → C → D) intermediate compound end product feedback inhibition
Use the understanding you gained from doing the concept map to answer the questions. 1. Reduced organic compounds tend to contain stored energy in C—H bonds. As a general rule, the greater the number of C—H bonds, the greater the amount of potential energy stored in the molecule. Answer each question in the chart as it relates to the two reactions shown at the top. Be sure to explain the reasoning behind your answers. Reaction 1: Reaction 2: CH4 2 O2 → H2O CO2 6 CO2 6 H2O → C6H12O6 6 O2 (methane)
a. Is the reaction exergonic or endergonic?
Exergonic This reaction releases energy.
Endergonic This reaction requires energy.
b. Is the reaction spontaneous?
Yes The end products are at lower energy levels than the reactants. Catabolic This reaction “breaks down” methane into water and carbon dioxide. Negative Free energy is released.
No The end products are at higher energy levels than the reactants.
c. Is the reaction anabolic or catabolic? d. Is G (the change in free energy) positive or negative?
Anabolic This is a synthetic reaction that “builds up” sugar from water and carbon dioxide. Positive The reaction requires the input of free energy.
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2. All metabolic reactions in living organisms are enzyme mediated. Each enzyme is specific for one (or only a very few similar types of) reaction. Given this, there are approximately as many different kinds of enzymes as there are reactions. a. What characteristics do all enzymes share? All enzymes contain a reactive site. All have a specific conformation that fits them to their substrate(s). All serve to reduce the activation energy of reactions. b. What characteristics can differ among enzymes? Before ribozymes were discovered, we thought that all enzymes were composed of protein. Now we recognize that some RNAs are catalytic also. Some protein enzymes include cofactors, which can be ions or other organic compounds. If a cofactor is organic, it is called a coenzyme. Many vitamins are coenzymes. 3. How can enzyme function be mediated or modified? To answer, complete a and b below. a. What factors can modify enzyme function?
b. What effect(s) can each of these factors have on enzyme function?
Among other factors, the temperature Each enzyme has a range of and pH at which an enzyme is active can temperatures in which it functions. vary. (Refer to Figure 8.18.) The same is true for pH. At temperatures and pHs outside this range, protein enzymes can become denatured. This changes their 3-D configuration and therefore their ability to function. In addition, various enzymes may be inhibited or activated by different factors.
The action of some enzymes can be inhibited by molecules that compete for the active site (competitive inhibition). Other inhibitors can bind at another site and alter the configuration of the active site (noncompetitive inhibition). (See Figure 8.19.) In some cases, activators are required to stabilize the active 3-D form of the enzyme. (See Figure 8.20.)
Activity 8.1
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c. What role(s) can modification of enzyme function play in the cell? To maintain optimal levels of enzyme activity and therefore metabolism, cells must maintain a relatively constant internal environment (homeostasis). On the other hand, cells can control or modify enzyme function to inhibit or enhance the function of a specific enzyme (as noted in part b). As a result, cells can maintain tight control over the levels of various cellular components. For example, an end product of a reaction series can serve as an allosteric inhibitor of the first enzyme in the pathway. This type of control has the effect of quickly and efficiently turning off the entire pathway when the end product occurs in excess (or of turning it on when the end product concentration decreases).
Activity 8.2 How can changes in experimental conditions affect enzyme-mediated reactions? 1. You set up a series of experiments to monitor the rates of a reaction. The reaction is an enzyme-mediated reaction in which A → B + C. For each experiment in this series, you continuously add the reactant A and monitor its concentration so that the amount of A remains constant over time. For each group of experiments, explain how the differences in experimental conditions could affect the reaction. a. You compare two side-by-side experiments. In experiment 1, you use X amount of the enzyme. In experiment 2, you use 2X amount of the same enzyme. If you double the amount of enzyme present, you double the rate of accumulation of the product. The rate at which any specific enzyme operates to make A → B + C does not change. However, the rate at which C accumulates appears to double. For example, if one enzyme can complete the reaction A → B + C in 1 msec, then at the end of 1,000 msec (1 second) we would expect 1,000 A to be converted to 1,000 B + 1,000 C. If two enzymes were working simultaneously, then within 1 second, 2,000 A could be converted to 2,000 B + 2,000 C. (Note: This answer assumes that the amount of A is much greater than the amount of B + C.) b. You compare two side-by-side experiments. In both you use equal amounts of the enzyme. In experiment 3, you allow the products to accumulate over time. In experiment 4, you remove the products from the system as they are produced. Recall that most metabolic reactions are reversible, and for these reactions, the same enzyme that catalyzes the forward reaction can generally catalyze the reverse reaction. As a result, in experiment 4, the apparent rate of the reaction will be constant over time. In other words, to use the example above, each enzyme will break down a unit of A into B and C every millisecond. In experiment 3, however, as the reaction reaches equilibrium levels of A versus B + C, the apparent rate of the reaction will decrease. Each enzyme will still be 40
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operating at the same rate (1 reaction per millisecond); however, some will be catalyzing the forward reaction and some the reverse reaction. At equilibrium, the numbers of enzyme molecules catalyzing the forward versus reverse reactions per unit time should be equal. c. In the next two experiments, you use equal amounts of the enzyme. You run experiment 5 at 20°C and experiment 6 at 25°C. As noted in Figure 8.18, each type of enzyme is functional within a set range of temperatures. The temperature at which the enzyme has its greatest rate of activity or reaction rate is called its optimal temperature. On either side of the optimal temperature (positive or negative), the rate of the enzyme-catalyzed reaction decreases. As a result, exactly how the enzyme in this system will react depends on both its optimal temperature and the range of temperatures over which it is active. For example, if we assume the enzyme’s range of temperature is 0° to 50°C and its optimal temperature is about 35°C, then the rate at which the enzyme works at 25°C will be faster than its rate at 20°C. On the other hand, if the range of temperatures for this enzyme is 40° to 90°C, then at both 20° and 25°C, we will see no activity. d. In two final experiments, you use equal amounts of the enzyme. You run experiment 7 at pH 6 and experiment 8 at pH 8. Similar to part c, each enzyme has an optimal pH and a range of pH in which it is active. As above, exactly how the enzyme will react at pH 6 versus pH 8 will depend on its optimal pH and range of pH. 2. Enzyme function can be inhibited or regulated by the presence of chemicals that mimic either the reactants or the products. a. How do competitive and noncompetitive inhibition of an enzyme differ? Refer to Figure 8.19. A competitive inhibitor can bind at the active site of an enzyme and prevent the substrate from binding. A noncompetitive inhibitor binds at a different site on the enzyme and changes the shape of the active site, which then prevents the substrate from binding. b. What are allosteric enzymes? What function(s) can they serve in reaction sequences? Refer to Figure 8.20. Allosteric enzymes are usually made up of more than one polypeptide. The activity of these enzymes is often controlled by enhancers and inhibitors, which bind at different sites on the enzyme. Binding of an enhancer changes the enzyme’s shape and makes its active site(s) available for catalysis. Binding of an inhibitor also causes a shape change, but one that makes the active site(s) unavailable.
Activity 8.2
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3. An enzyme catalyzes the reaction X → Y + Z. In a series of experiments, it was found that substance A inhibits the enzyme. • When the concentration of X is high and A is low, the reaction proceeds rapidly and Y and Z are formed. • As the concentration of A increases, the reaction slows regardless of whether X is present in high or low concentration. • If the concentration of A is high (relative to X), the reaction stops. • If the concentration of A again decreases, the reaction will ultimately resume. What type of enzyme regulation is described here? Explain or justify your answer. The enzyme in the reaction described, X → Y + Z, is most likely an enzyme that has a site for a noncompetitive inhibitor molecule. If this were a case of competitive inhibition, then the higher the concentration of X (relative to A), the faster the reaction rate should be. The inhibition described is also reversible, as evidenced by the ability of the reaction to resume if the concentration of A decreases. 4. In an enzymatic pathway, A, B and C are intermediates required to make D and 1, 2, and 3 are enzymes that catalyze the designated reactions: 1 2 3 A→B→C→D → E This is analogous to what happens in a factory. In a leather goods factory, for example, the leather (A) is cut (1) into the parts needed for shoes (B). The shoe parts are sewn (2) together (making C), and C is packaged (3) for shipping as D. Now shoe sales are dropping and backpack sales (E) are increasing. As a result, the manager of the factory decides to switch production from shoes to leather backpacks (E). a. Where should the shoe-making process be shut down: step 1, 2, or 3? Explain. The shoe-making process should be shut down at step 1. It makes no sense to produce any of the parts to make shoes if you don’t need them. b. In a cell, if an excess of a chemical product D arises, where should this synthetic pathway be shut down in the cell? If we use the same logic as in part a, then the synthetic pathway A → B → C → D should also be shut down at the first step. c. What type(s) of enzyme regulation is/are most likely to occur in the cell in this type of feedback system? Allosteric enzymes are frequently involved in this type of feedback regulation of synthetic pathways. 42
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Notes to Instructors Chapter 9 Cellular Respiration: Harvesting Chemical Energy Chapter 10 Photosynthesis What is the focus of these activities? In studying both cellular respiration and photosynthesis, many students tend to focus on the details and miss the big picture. They can recite specific reactions that occur in glycolysis and the Krebs cycle, for example, but they don’t understand the overall purpose of these parts of the process. These activities are designed to help students understand the overall purpose of each process and how these processes are interrelated evolutionarily.
What are the particular activities designed to do? Activity 9.1 A Quick Review of Energy Transformations This activity adds consideration of the terms oxidation and reduction to the energy relationships students learned in Chapter 8. Activity 9.2 Modeling Cellular Respiration Activity 9.2 is designed to help students understand: • the overall functions of glycolysis, the Krebs cycle, and oxidative phosphorylation; • how fermentation allows organisms to survive periods of low or no oxygen; and • how the potential energy in a hydrogen ion concentration gradient can be used to generate ATP in oxidative phosphorylation. Activity 10.1 Modeling Photosynthesis Activity 10.2 How do C3, C4, and CAM photosynthesis compare? Activities 10.1 and 10.2 are designed to help students understand: • the roles photosystems I and II and the Calvin cycle play in photosynthesis, and • how and why C4 and CAM photosynthesis differ from C3 photosynthesis.
What misconceptions or difficulties can these activities reveal? Activity 9.1 This activity reviews the information presented in Chapter 8 and helps students integrate into that an understanding of oxidation and reduction reactions that occur in living organisms. Notes to Instructors
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Activity 9.2 By doing this modeling exercise, students will not only learn the definitions of all the terms and structures involved in cellular respiration but also get an understanding of how they function or interact in the cell. Many students don’t understand what they do and, more important, don’t know about the processes of cellular respiration. When students build this kind of visual claymation model, what they don’t know becomes apparent and they have reason to “fill in the blanks.” Visualizing the processes using a model they build for themselves generally leads to a better and more complete understanding. Question 1: To help give students the “big picture,” this question looks at the summary formula for cellular respiration and asks where each reactant is used and where each product is produced in the process. Many students have difficulty answering most parts of this question. Questions 2 and 3: The same observation applies to these questions. Many students concentrate on the details of reactions and don’t appear to understand the overall purpose of each part of the process. These questions are designed to help them put the pieces together. Questions 4, 5, and 6: These questions examine what happens in aerobic cellular respiration when oxygen and, therefore, NAD+ become limiting. We generally teach the various processes of cellular respiration in order from glycolysis to the Krebs cycle to electron transport and oxidative phosphorylation. Many students get the mistaken impression that they must operate in this sort of relay fashion in the cell as well. Only a few introductory students understand that all of these processes are occurring simultaneously in the cell. Fewer yet have a good idea that various molecules and resources in the cell (for example, NAD+) are finite and can be limiting. And it is the rare student who can answer the question: Why does the Krebs cycle stop in the absence of oxygen if oxygen is not required in the Krebs cycle? We hope these questions will help students understand this and the relationships among these processes. Question 7: This question gives students practice with the concept of energy efficiency and methods of calculating it, both of which are difficult for some students. Question 8: Students know that ATP is the “energy currency” of the cell. However, few understand why organisms store energy as starch, fats, or oils instead of as ATP. Students must integrate an understanding of osmotic relationships from Chapter 3 with an understanding of the structure and function of ATP in order to answer this question.
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Question 9: When chemiosmosis is discussed, students are asked to understand that a hydrogen ion concentration gradient has potential energy that can be used to do work—in this case, to drive the synthesis of ATP. The operation of a battery is often used as an example; however, few students seem to understand how batteries work. We provide the example of mixing concentrated acid with water, which can be done in class (under strict supervision) to demonstrate how establishing (and releasing) a hydrogen ion concentration gradient can generate energy (in this case, heat). Questions 10 and 11: These questions ask students to apply what they have learned about molar equivalents (Chapter 2) to solve simple problems in cellular respiration. Question 12: This question asks why living organisms don’t spontaneously combust. Students should be able to understand this if they understand the process of cellular respiration. Activity 10.1 Students have the same difficulties understanding photosynthesis that they have understanding cellular respiration. This activity asks students to model photosynthesis in the chloroplast. Its design and purpose are parallel to those for the cellular respiration modeling activity (Activity 9.2). By building their own model of photosynthesis, students will not only learn the definitions of all the terms and structures involved but also get an understanding of how they function or interact in the cell. More important, students will discover what they don’t know or don’t understand and then remedy the problem. Questions 1, 2, and 3: As in the Activity 9.2, these questions are designed to help students develop the big picture and get a good understanding of the overall purpose of each process. Question 4: Whereas Activity 9.2 looked at what happens when NAD+ becomes limiting in cellular respiration, this question looks at what happens when NADP+ becomes limiting in photosynthesis. Question 5: The question of why plants need to make glucose and store starch as an energy source is addressed. Many students don’t understand why plants can’t just use the ATP directly for processes other than photosynthesis. In fact, they can, as long as they are making excess ATP. When there is little or no sunlight, however, ATP production via photophosphorylation is reduced or halted entirely. In addition, most students don’t understand that the three-carbon compounds produced in the Krebs cycle are not all used to make glucose. All the organic compounds produced by plants use these (or modifications of these) as precursors.
Notes to Instructors
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Activity 10.2 Question 1: This question is set up to allow students to more easily compare, and therefore learn, the similarities and differences among C3, C4, and CAM photosynthesis. Question 2: Many students have difficulty understanding how rubisco can serve as both a carboxylase and an oxidase. This becomes easier to understand when students recognize that rubisco probably arose as a mutation in organisms when the early Earth’s atmosphere was anaerobic. Question 3: This question asks students to develop an evolutionary scheme for glycolysis, the Krebs cycle, oxidative phosphorylation or electron transport, and photosynthesis. Many students have the misconception that the order in which the processes were presented either in lecture or in the text is the order in which they evolved. This will be obvious in their answer to this question. Fewer students will automatically use what they know about each process—for example, what each process requires—to develop a logical scheme of evolution.
Answers Activity 9.1 A Quick Review of Energy Transformations Review Chapter 8 and pages 162–164 of Chapter 9 in Biology, 8th edition. Then complete the discussion by supplying or choosing the appropriate terms. To maintain life, organisms must be able to convert energy from one form to another. For example, in the process of photosynthesis, algae, plants, and photosynthetic prokaryotes use the energy from sunlight to convert carbon dioxide and water to glucose and oxygen (a waste product). The summary reaction for photosynthesis can be written as 6 CO2 + 6 H2O → C6H12O6 + 6 O2 This type of reaction is an oxidation-reduction (or redox) reaction. This reaction is also [anabolic/catabolic] and [endergonic/exergonic]. In redox reactions, electrons (and associated H+ ions) are transferred from one compound or element to another. If one compound or element “loses” electrons and becomes oxidized, another must “gain” electrons and become reduced. For example, in photosynthesis, water becomes [oxidized/reduced] (to O2) and the electrons (and associated H+ ions) it loses in the process [oxidize/reduce] CO2 to glucose. 46
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[Anabolic/Catabolic] reactions “build” more complex molecules from simpler ones. To do this they require energy input. Reactions that require the input of energy are termed [endergonic/exergonic] reactions. The reactions involved in aerobic respiration are also redox reactions: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O In this set of reactions, however, more complex molecules are “broken down” into simpler ones. Glucose is broken down or becomes [oxidized/reduced] (to CO2), and the oxygen becomes [oxidized/reduced] (to water). [Anabolic/Catabolic] reactions break down more complex molecules into simpler ones and in the process release energy. Reactions that release energy that can be used to do work are [endergonic/exergonic]. Therefore, aerobic respiration is a(n) [anabolic/catabolic] process and is [endergonic/exergonic]. [Endergonic/Exergonic] reactions are also said to be spontaneous reactions. Does this mean that if we don’t keep glucose in tightly sealed containers it will spontaneously interact with atmospheric oxygen and turn into carbon dioxide and water? The answer is obviously no. Spontaneous reactions rarely occur “spontaneously” because all chemical reactions, even those that release energy, require some addition of energy—the energy of activation— before they can occur. One way of supplying this energy is to add heat. An example is heating a marshmallow over a flame or campfire. When enough heat is added to reach (or overcome) the activation energy, the sugar in the marshmallow reacts by oxidizing. (Burning is a form of oxidation.) The marshmallow will continue to burn even if you remove it from the campfire. As the marshmallow burns, carbon dioxide and water are formed as products of the reaction, and the energy that was stored in the bonds of the sugar is released as heat. If our cells used heat to overcome activation energies in metabolism, they would probably burn up like the marshmallow did. Instead, living systems use protein catalysts or enzymes to lower the energy of activation without adding heat. In addition, the metabolic breakdown of sugars is carried out in a controlled series of reactions. At each step or reaction in the sequence, a small amount of the total energy is released. Some of this energy is still lost as heat. The rest is converted to other forms that can be used in the cell to drive or fuel coupled endergonic reactions or to make ATP.
Activity 9.1
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Activity 9.2 Modeling cellular respiration: How can cells convert the energy in glucose to ATP? Using your textbook, lecture notes, and the materials available in class (or those you devise at home), model both fermentation (an anaerobic process) and cellular respiration (an aerobic process) as they occur in a plant or animal cell. Each model should include a dynamic (working or active) representation of the events that occur in glycolysis.
Building the Model • Use chalk on a tabletop or a marker on a large sheet of paper to draw the cell membrane and the mitochondrial membranes. • Use playdough or cutout pieces of paper to represent the molecules, ions, and membrane transporters or pumps. • Use the pieces you assembled to model the processes of fermentation and aerobic respiration. Develop a dynamic (claymation-type) model that allows you to manipulate or move glucose and its breakdown products through the various steps of both fermentation and aerobic respiration. • When you feel you have developed a good working model, demonstrate and explain it to another student. Be sure your model of fermentation includes and explains the actions and roles of the following: glycolysis cytoplasm electrons protons glucose NAD+ NADH
ADP Pi ATP pyruvate ethyl alcohol (or lactic acid) substrate-level phosphorylation
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Be sure your model of cellular respiration includes and explains the actions and roles of the following: glucose oxygen carbon dioxide pyruvate acetyl CoA NAD+ NADH FAD FADH2 ADP Pi ATP water
electron transport chain mitochondria inner mitochondrial membrane outer mitochondrial membrane H+ electrons (e) chemiosmosis ATP synthase (proton pumps) cristae proton gradients oxidative phosphorylation substrate-level phosphorylation oxidative phosphorylation
Use your models to answer the questions. 1. The summary formula for cellular respiration is C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + Energy a. At what stage(s) in the b. At what stage(s) in the overall process is each overall process is each of the of the products produced? reactants used? C6H12O6 Glycolysis
6 O2
→
6 CO2
Oxidative Pyruvate Acetyl phosphorylation CoA and Krebs cycle
6 H2O
Energy
Oxidative ATP/glucose phosphorylation Glycolysis (2), Krebs (2 GTP), oxidative phosphorylation (up to 34)
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2. In cellular respiration, the oxidation of glucose is carried out in a controlled series of reactions. At each step or reaction in the sequence, a small amount of the total energy is released. Some of this energy is lost as heat. The rest is converted to other forms that can be used by the cell to drive or fuel coupled endergonic reactions or to make ATP. a. What is/are the overall function(s) of glycolysis?
b. What is/are the overall c. What is/are the overall function(s) of the Krebs function(s) of oxidative cycle? phosphorylation?
Oxidation of glucose to 2 Oxidation of pyruvate. Generates 2 ATP pyruvate/acetyl CoA to and 2 NADH per glucose carbon dioxide. Generates 2 GTP, 6 NADH, and 2 FADH2 per glucose.
3. Are the compounds listed here used or produced in: Glycolysis? Glucose
Oxidation of NADH and FADH2to H2O (and NAD or FAD). Generates H+ ion concentration gradient and therefore ATP.
The Krebs cycle?
Oxidative phosphorylation?
Used
O2
Used
CO2
Produced
H2O
Produced (GTP)
ATP
Produced
ADP Pi
Produced and used
NADH
Produced
NAD
Used
Produced Produced
Used
Used Used
Used
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4. The cell’s supply of ADP, Pi, and NAD+ is finite (limited). What happens to cellular respiration when all of the cell’s NAD+ has been converted to NADH? If NAD is unavailable, the cell is unable to conduct any processes that involve the conversion of NAD+ to NADH. Because both glycolysis and the Krebs cycle produce NADH, both of these processes shut down when there is no available NAD+. 5. If the Krebs cycle does not require oxygen, why does cellular respiration stop after glycolysis when no oxygen is present? When no oxygen is present, oxidative phosphorylation cannot occur. As a result, the NADH produced in glycolysis and the Krebs cycle cannot be oxidized to NAD+. When no NAD+ is available, pyruvate cannot be converted to the acetyl CoA that is required for the Krebs cycle. 6. Many organisms can withstand periods of oxygen debt (anaerobic conditions). Yeast undergoing oxygen debt converts pyruvic acid to ethanol and carbon dioxide. Animals undergoing oxygen debt convert pyruvic acid to lactic acid. Pyruvic acid is fairly nontoxic in even high concentrations. Both ethanol and lactic acid are toxic in even moderate concentrations. Explain why this conversion occurs in organisms. As noted in question 4, when no NAD+ is available, even glycolysis stops. No ATP will be produced and the cell (or organism) will die. The conversion of pyruvic acid (pyruvate) to lactic acid (or ethanol) requires the input of NADH and generates NAD+. This process, called fermentation, allows the cell to continue getting at least 2 ATP per glucose. 7. How efficient is fermentation? How efficient is cellular respiration? Remember that efficiency is the amount of useful energy (as ATP) gained during the process divided by the total amount of energy available in glucose. Use 686 kcal as the total energy available in 1 mole of glucose and 8 kcal as the energy available in 1 mol of ATP. Efficiency of fermentation
Efficiency of aerobic respiration
8 kcal/mole of ATP ⫻ 2 ATP ⫽ 16 kcal 16 kcal/2 moles of ATP ⫽ 2.3% 686 kcal/mole of glucose
8 kcal/mole of ATP ⫻ 38 ATP (maximum) ⫽304 kcal 304 kcal/38 moles of ATP ⫽ 44.3% 686 kcal/mole of glucose
Activity 9.2
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8. a. Why can’t cells store large quantities of ATP? (Hint: Consider both the chemical stability of the molecule and the cell’s osmotic potential.) ATP is highly reactive at normal body temperatures and therefore difficult for cells to store for any period of time. (In the lab, ATP is usually stored at very low temperatures, for example, at ⫺20°C.) In addition, ATP is a relatively small molecule. As a result, if cells could store high concentrations of ATP, their osmotic potential would change. This is also why cells don’t store glucose. The cells would become hypertonic to the fluid around them and could pick up enough water to burst. b. Given that cells can’t store ATP for long periods of time, how do they store energy? Instead of storing ATP, cells tend to store energy as fats, oils, or starches c. What are the advantages of storing energy in these alternative forms? These are very large molecules and, as a result, do not have as great an effect on osmotic potential. They are also much more stable chemically than ATP. 9. To make a 5 M solution of hydrochloric acid, we add 400 mL of 12.5 M hydrochloric acid to 600 mL of distilled water. Before we add the acid, however, we place the flask containing the distilled water into the sink because this solution can heat up so rapidly that the flask breaks. How is this reaction similar to what happens in chemiosmosis? How is it different? a. Similarities
b. Differences
In both processes, as we add the acid to the water, we are generating a difference in concentration between the two, or a H+ ion gradient. As the H+ ions flow down this gradient (that is, mix with the water), they release energy in the form of heat.
Both processes set up a H+ ion concentration gradient. However, in chemiosmosis the energy release is controlled as the H+ ions pass through the ATP synthase molecules and ATP is generated. Some energy is lost as heat, but much of it is captured in the chemical bonds of ATP.
9.2 Test Your Understanding 1. If it takes 1,000 g of glucose to grow 10 g of an anaerobic bacterium, how many grams of glucose would it take to grow 10 g of that same bacterium if it was respiring aerobically? Estimate your answer. For example, if it takes X amount of glucose to grow 10 g of anaerobic bacteria, what factor would you have to multiply or divide X by to grow 10 g of the same bacterium aerobically? Explain how you arrived at your answer. 52
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Aerobic respiration can produce a maximum of 38 ATP per glucose molecule. Anaerobic respiration can produce 2 ATP per glucose molecule. As a result, aerobic respiration is about 19 times more efficient. Therefore, you would need 19 times less glucose if respiring aerobically: 1,000 g of glucose divided by 19 equals approximately 50 g of glucose required if respiration is aerobic. 2. Mitochondria isolated from liver cells can be used to study the rate of electron transport in response to a variety of chemicals. The rate of electron transport is measured as the rate of disappearance of O2 from the solution using an oxygensensitive electrode. How can we justify using the disappearance of oxygen from the solution as a measure of electron transport? Use the balanced equation for aerobic respiration: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + Energy If the final energy produced is 38 ATP, then for every 6 oxygen molecules consumed (or 6 moles of oxygen consumed), we expect 38 molecules of ATP (or moles of ATP) to be produced. 3. Humans oxidize glucose in the presence of oxygen. For each mole of glucose oxidized, about 686 kcal of energy is released. This is true whether the mole of glucose is oxidized in human cells or burned in the air. A calorie is the amount of energy required to raise the temperature of 1 g of water by 1°C; 686 kcal ⫽ 686,000 calories. The average human requires about 2,000 kcal of energy per day, which is equivalent to about 3 mol of glucose per day. Given this, why don’t humans spontaneously combust? As noted in question 9, during cellular respiration, the energy from the oxidation of glucose is not released all at once (as it is in burning). Instead, each of the reactions in glycolysis, the Krebs cycle, and electron transport releases a small amount of the energy stored in the molecules. Much of this energy is captured as NADH, FADH2, ATP, or GTP. Some is lost as heat; however, the heat loss also occurs at each step and not all at once. 4. A gene has recently been identified that encodes for a protein that increases longevity in mice. To function in increasing longevity, this gene requires a high ratio of NAD+/NADH. Researchers have used this as evidence in support of a “caloric restriction” hypothesis for longevity—that a decrease in total calorie intake increases longevity. How does the requirement for a high NAD+/NADH ratio support the caloric restriction hypothesis? A decrease in calorie intake will decrease the rate of glycolysis and the Krebs cycle. Therefore, over a 24-hour period, there will be less NADH produced by glycolysis and the Krebs cycle, and the NAD+/NADH ratio will increase.
Activity 9.1
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5. An active college-age athlete can burn more than 3,000 kcal/day in exercise). a. If conversion of one mole of ATP to ADP + Pi releases about 7.3 kcal, roughly speaking, how many moles of ATP need to be produced per day in order for this energy need to be met? 3000 kcal/day divided by 7.3 kcal/mole of ATP ⫽ 411 moles of ATP b. If the molecular weight of ATP is 573, how much would the required ATP weigh in kilograms? 411 moles of ATP times 573 grams per mole ⫽ 235,503 grams or 235 kilogram (about 518 pounds) c. Explain these results ATP is broken down to ADP ⫽ Pi, which is continuously recycled to ATP during cell respiration.
Activity 10.1 Modeling photosynthesis: How can cells use the sun’s energy to convert carbon dioxide and water into glucose? Activity 10.1 is designed to help you understand: 1. The roles photosystems I and II and the Calvin cycle play in photosynthesis, and 2. How and why C4 and CAM photosynthesis differ from C3 photosynthesis. Using your textbook, lecture notes, and the materials available in class (or those you devise at home), model photosynthesis as it occurs in a plant cell. Your model should be a dynamic (working or active) representation of the events that occur in the various phases of C3 photosynthesis.
Building the Model • Use chalk on a tabletop or a marker on a large sheet of paper to draw the cell membrane and the chloroplast membranes. • Use playdough or cutout pieces of paper to represent the molecules, ions, and membrane transporters or pumps. • Use the pieces you assembled to model the processes involved in C3 photosynthesis. Develop a dynamic (claymation-type) model that allows you to manipulate or move carbon dioxide and water and its breakdown products through the various steps of the process. • When you feel you have developed a good working model, demonstrate and explain it to another student or to your instructor.
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Your model of C3 photosynthesis should include what occurs in photosystems I and II and in the Calvin cycle. For photosystems I and II, be sure your model includes and explains the roles of the following: ATP water and oxygen H+ e⫺
NADP+ NADPH ADP Pi
chemiosmosis ATP synthase e⫺ carriers in thylakoid membranes
Also indicate where in the plant cell each item is required or produced. For the Calvin cycle, be sure your model includes and explains the roles of the following: NADPH ATP
glucose C3 or 3C sugars carbon dioxide
Also indicate where in the plant cell each item is required or produced. After you’ve modeled C3 photosynthesis, indicate how the system would be altered for C4 and CAM photosynthesis. • Indicate where in the cells of the leaf PEP carboxylase exists and how it reacts to capture CO2. Be sure to indicate the fate of the captured CO2. • Do the same for PEP carboxylase in CAM plants. Use your model and the information in Chapter 10 of Biology, 8th edition, to answer the questions. 1. The various reactions in photosynthesis are spatially segregated from each other within the chloroplast. Draw a simplified diagram of a chloroplast and include these parts: outer membrane, grana, thylakoid, lumen, stroma/matrix. Refer to Figure 10.3, page 187, in Biology, 8th edition. a. Where in the chloroplast do the light reactions occur?
In the thylakoid membranes
b. Where in the chloroplast is the chemiosmotic gradient developed?
Across the thylakoid membrane; H+ ions are pumped into the thylakoid space
c. Where in the chloroplast does the Calvin cycle occur?
In the stroma or liquid portion of the chloroplast
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2. In photosynthesis, the reduction of carbon dioxide to form glucose is carried out in a controlled series of reactions. In general, each step or reaction in the sequence requires the input of energy. The sun is the ultimate source of this energy. a. What is/are the overall function(s) of photosystem I?
b. What is/are the overall function(s) of photosystem II?
c. What is/are the overall function(s) of the Calvin cycle?
In noncyclic photosphosphorylation, photosystem I produces NADPH. In cyclic photophosphorylation, photosystem I produces ATP.
Photosystem II generates ATP. To fill the electron hole in photosystem II, water is split into 2 H+, 2e⫺, and 1/2 O2. (The electron from photosystem II fills the electron hole in photosystem I.)
The Calvin cycle uses the ATP and NADPH generated in the light reactions to reduce CO2 to three-carbon compounds in a cyclic series of reactions that regenerates the original five-carbon sugar required to accept the CO2. The three-carbon compounds can be used to make glucose or other organic compounds required by the cells.
3. Are the compounds listed here used or produced in: Photosystem I?
Photosystem II?
Glucose
The Calvin cycle? Produced
O2
Produced from the breakdown of H2O
CO2
Used
H2O
Used to produce 2 H+, 2e⫺, and 1/2 O2
ATP
Produced (in Produced cyclic photophosphoryl ation)
Used
ADP + Pi
Used
Produced
NADPH
Produced
Used
NADP+
Used
Produced
Used
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4. Which light reaction system (cyclic or noncyclic) would a chloroplast use in each situation? a. Plenty of light is available, but the cell contains little NADP+.
b. There is plenty of light, and the cell contains a high concentration of NADP+.
If there is little NADP+, there must be much NADPH. This could occur if the Calvin cycle is not using up the NADPH. For example, if CO2 levels are low, little NADPH will be used to make glucose. Under these circumstances, the system would switch to cyclic photophosphorylation and gain ATP, which can be used both in photosynthesis and in other types of metabolism.
In this case, it appears that NADPH is being used rapidly (therefore the high levels of NADP+). As a result, the system would switch to noncyclic photophosphorylation, which produces both ATP and NADPH.
5. All living organisms require a constant supply of ATP to maintain life. If no light is available, how can a plant make ATP? Keep in mind that it is not always light and that not all cells of a plant are directly exposed to light. For example, cells on the interior of a plant stem and those in the roots have little, if any, exposure to light. Plants, like other eukaryotic organisms on Earth, also contain mitochondria. Plant cells undergo glycolysis in the cytoplasm and transfer acetyl CoA to mitochondria, where it enters the Krebs cycle. The NADH and FADH2 produced during the Krebs cycle then undergo oxidative phosphorylation to produce ATP.
10.1 Test Your Understanding Chloroplast thylakoids can be isolated and purified for biochemical experiments. Shown below is an experiment in which pH was measured in a suspension of isolated thylakoids before and after light illumination (first arrow). At the time indicated by the second arrow, a chemical compound was added to the thylakoids. Examine these data and address the following questions.
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Light
Chemical
8
pH
6
time (minutes)
a. Based on your understanding of the function of the chloroplasts, why does turning on the light cause the pH in the solution outside the thylakoids to increase? Electron transfer (Photosystems II and I) in the thylakoid membrane resulted in pumping of H+ from stroma (outside) to thylakoid (inside). As a consequence, the H+ concentration outside the thylakoids became lower and the pH increased. b. Given the response, the chemical added was probably an inhibitor of: i. oxidative phosphorylation ii. ATP synthase iii. NADPH breakdown iv. Electron transport chain between photosystems II and I v. Rubisco The answer is iv. Disrupting or inhibiting the electron transport chain between photosystems II and I would prevent transport of H+ ions into the thylakoid space. As a result, the concentration of H+ ions would be reduced and the pH would increase.
Activity 10.2 How do C3, C4, and CAM photosynthesis compare? 1. Carbon dioxide enters plant leaves through the stomata, while oxygen (the photosynthetic waste product) and water from the leaves exit through the stomata. Plants must constantly balance both water loss and energy gain (as photosynthesis). This has led to the evolution of various modifications of C3 photosynthesis.
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C3
C4
Draw simplified diagrams of the cross sections of a See Figure 10.3. leaf from a C3, a C4 and a CAM plant. a. How are the leaves similar?
See Figure 10.19.
CAM CAM leaf anatomy is similar to C3 leaf anatomy.
All have stomata, epidermal cells that lack chloroplasts, mesophyll cells with chloroplasts, and veins that conduct water and the products of photosynthesis.
b. How are the C4 plants have large bundle sheath cells not found in the leaves different? others. In C4 plants, the Calvin cycle occurs only in the bundle sheath cells. c. How and when does carbon dioxide get into each leaf?
During daylight hours, when stomata are open
During cooler parts At night, when it is of the day, when cool and stomata stomata are open are open
d. Which enzyme(s) (1) capture carbon dioxide and (2) carry it to the Calvin cycle?
The CO2 is picked up by the enzyme, rubisco, which catalyzes the first step in the Calvin cycle.
PEP carboxylase in the mesophyll cells converts CO2 to a four-carbon organic acid, which is transported to the bundle sheath cells, where it is converted to CO2 and PEP, and rubisco catalyzes the first step in the Calvin cycle.
PEP carboxylase in the mesophyll cells converts CO2 to a four-carbon organic acid, which is transported to the cells’ central vacuoles and can later be converted back to CO2 and PEP. The CO2 can then be picked up by rubisco and used in the Calvin cycle in mesophyll cells.
e. What makes C4 photosynthesis more efficient than C3 photosynthesis in tropical climates? PEP carboxylase is much more efficient than rubisco at picking up CO2. As a result, C4 plants can capture large quantities of CO2 and store it as a four-carbon organic compound in a relatively short period of time. This means that during the hottest parts of the day, the stomata can close to reduce water loss. Even with the stomata closed, however, the Calvin cycle can continue by using the stored CO2. This system also maintains a relatively high ratio of CO2 to O2 in the cells that rely on rubisco, the bundle sheath cells. This greatly reduces the amount of photorespiration in these plants. Activity 10.2
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f. How is CAM photosynthesis advantageous in desert climates? Stomata can be open at night when there is less evaporative loss of water and closed during the day. At night, PEP carboxylase allows desert plants to store CO2 as a four-carbon organic acid. However, the amount that can be stored in the central vacuole of its photosynthetic cells is finite. This stored CO2 can then be used during the day to support the Calvin cycle. 2. Photosynthesis evolved very early in Earth’s history. Central to the evolution of photosynthesis was the evolution of the enzyme rubisco (an abbreviation for ribulose bisphosphate carboxylase oxidase). To the best of our knowledge, all photosynthetic plants use rubisco. Rubisco’s function is to supply carbon dioxide to the Calvin cycle; however, it does this only if the ratio of carbon dioxide to oxygen is relatively high. (For comparison, a relatively high ratio of carbon dioxide to oxygen is 0.03% carbon dioxide to 20% oxygen.) When the carbon-dioxide-to-oxygen ratio becomes low, the role of rubisco switches and it catalyzes photorespiration, the breakdown of glucose to carbon dioxide and water. a. Why could we call photorespiration a “mistake” in the functioning of the cell? Photorespiration could be called a “mistake” because under high O2/CO2 conditions, rubisco breaks down glucose into carbon dioxide and water but no useful energy is gained. b. Rubisco is thought to have evolved when Earth had a reducing atmosphere. How does this help explain the photorespiration “mistake?” When the first photosynthetic organisms arose, the early Earth’s atmosphere contained little, if any, oxygen. Rubisco would have functioned very well under these conditions. It was only later, when the concentration of oxygen in the atmosphere increased considerably, that rubisco’s ability to oxidize glucose became evident.
10.2 Test Your Understanding The metabolic pathways of organisms living today evolved over a long period of time— undoubtedly in a stepwise fashion because of their complexity. Put the following processes in the order in which they might have evolved, and give a short explanation for your arrangement. ___ 4 Krebs cycle ___ 3 Electron transport ___ 1 Glycolysis ___ 2 Photosynthesis
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First, glycolysis is found in all eukaryotes and many prokaryotes. It takes place in the cytoplasm and can occur in the absence of oxygen. Second, photosynthesis produces oxygen as a by-product. Neither the Krebs cycle nor electron transport can occur in the absence of oxygen. Third, electron transport is required to convert NADH to NAD+. Because glycolysis produces 2 ATP (net) and 2 NADH, the addition of electron transport represents an advantage. Organisms can then gain 8 ATP (net) from glycolysis plus electron transport. Fourth, the Krebs cycle cannot occur without a mechanism to convert NADH to NAD+. Electron transport must have evolved before the Krebs cycle.
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Notes to Instructors Chapter 11 Cell Communication What is the focus of this activity? Most students understand that external signals interact with receptors in cells and that the interaction leads to a response by the cell. However, fewer have a good understanding of these processes: • how a protein signal that cannot cross the cell membrane can cause a response, • how very low concentrations of signal molecules can produce high levels of response, and • exactly what a cell does to respond to a signal.
What is the particular activity designed to do? Activity 11.1 How are chemical signals translated into cellular responses? In this activity, students model and compare the functions of a G-protein receptor system and a tyrosine-kinase receptor system. In addition, they are asked to use their knowledge of enzyme function from Chapter 8 to understand how a signal-transduction pathway can amplify the response to a single signal molecule.
What misconceptions or difficulties can this activity reveal? Activity 11.1 Modeling the G-protein receptor system and the tyrosine-kinase receptor system does not reveal misconceptions; rather, it tends to fill in missing information. Most students at the introductory level have little understanding of these systems. Questions 1 and 2: These questions ask students to look back at their two models and consider how they are similar and how they differ. Although engaging in this type of comparative process seems standard to those of us who have been working in the sciences for years, it is not something that introductory students do automatically. Posing these types of questions helps students learn not only to ask themselves the questions but also to organize and clarify their own understanding of the individual processes they model. Question 3: Because these pathways are called signal-transduction pathways, many students seem to get the idea (or misconception) that once each carrier or enzyme in a given pathway “transduces” or moves the signal on to the next carrier or enzyme, its job is done. This question focuses students’ attention on Figure 11.15, page 219, to help them understand the process of signal amplification—in other words, to understand that once a single enzyme in the pathway is activated, it can catalyze more than one reaction and the product of that reaction can catalyze more than one, and so on. 62
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Answers Activity 11.1 How are chemical signals translated into cellular responses? Chapter 11 in Biology, 8th edition, describes at least four kinds of signal receptors. Three of these—G-protein-linked receptors, tyrosine-kinase receptors, and ion-channel receptors—are plasma membrane proteins. Protein receptors found in the cytoplasm, or nucleus, of the cell are the fourth type. Some signals (for example, a protein hormone) interact with signal receptors in the cell membrane to initiate the process of signal transduction. This often involves changes in a series of different relay molecules in a signal-transduction pathway. Ultimately, the transduced signal initiates an intracellular response. Other types of signals (for example, steroid hormones) can diffuse through the cell membrane and interact with intracellular receptors. For example, testosterone interacts with its receptor in the cell’s cytoplasm, enters the nucleus, and causes the transcription of specific genes. To help you understand how signal transduction occurs in cells, develop dynamic (claymation-type) models of both a G-protein receptor system and a tyrosine-kinase receptor system. Use playdough or cutout pieces of paper to represent all the structural components and molecules listed here under each system. G-Protein Receptor System Tyrosine-Kinase Receptor System signal protein G-protein-linked receptor plasma membrane inactive and active G protein GTP and GDP inactive and active enzyme signal-transduction pathway
signal protein tyrosine-kinase receptor plasma membrane inactive and active relay proteins ATP and ADP signal-transduction pathway
Use your models to show how signal reception by each of the systems can lead to the release of Ca+ from the endoplasmic reticulum. Demonstrate and explain your models to another student group or to your instructor. Then use your models to answer the questions on the next page.
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1. How are these two systems similar? Consider both structural similarities and similarities in how the systems function. In both systems, the receptor proteins are bound in the cell’s membrane. Binding of signal molecules to the receptors activates them. Activated receptor(s) interact with inactive relay protein(s) and activate them. The role of the activated relay protein(s) is to activate other protein(s) to produce the cellular response. 2. How are the two systems different? Consider both structural differences and differences in how the systems function. The G-protein-linked receptor protein is a single unit that becomes functional when activated by its signal molecule. Two tyrosine-kinase receptor proteins must be activated by signal molecules and aggregate to become activated. The activated G-protein-linked receptor protein activates the G protein, which is also membrane bound, by converting an associated GDP to GTP. The activated G protein then moves along the membrane and activates a specific membrane-bound enzyme, which produces the cellular response. The activated tyrosine-kinase receptor aggregate can activate up to ten different specific relay proteins inside the cell and therefore produce multiple responses. The activated relay proteins are not membrane bound. Each type of activated relay molecule can activate a different transduction pathway and produce a different cellular response. 3. Both systems can generate elaborate multistep signal-transduction pathways. These pathways can greatly amplify the cell’s response to a signal; the more steps in the pathway, the greater the amplification of the signal. Explain how this amplification can occur. (Review Figure 11.15, page 219, in Biology, 8th edition.) In a signal-transduction pathway, each activated enzyme or second messenger has the potential to catalyze more than one reaction. Each of its reaction products similarly has the potential to trigger more than one reaction. As a result, the effects produced by a single signal molecule can be greatly amplified.
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11.1 Test Your Understanding Humans have the ability to detect and recognize many different aromatic chemicals by smell. Many of these chemicals are present in concentrations less than 1 ppm (part per million) in the air. For example, the majority of humans can detect and recognize chlorine at a concentration of about 0.3 ppm. a. What characteristics of olfactory (smell) receptors would you look for or propose to explain this ability? Proposing that olfactory receptors are G-coupled protein receptors would be reasonable here. In fact, this is borne out by the literature. The G-coupled receptor multi-step cascade allows amplification of and therefore detection of stimuli available in extremely low concentration, in this case the chemical, chlorine. b. Dogs are known to have a much better sense of smell than humans. Given this, what differences may exist in their olfactory system (as compared to humans)? Here students could propose either greater expression of receptors in the olfactory tissue of dogs or a greater surface area of olfactory tissue.
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Notes to Instructors Chapter 12 The Cell Cycle Chapter 13 Meiosis and Sexual Life Cycles What is the focus of these activities? Most students can recite what happens in each phase of mitosis and meiosis. However, many have difficulty translating those descriptions into visual pictures of a cell with a particular number of chromosomes.
What are the particular activities designed to do? Activity 12.1 What is mitosis? Activity 13.1 What is meiosis? Activity 13.2 How do mitosis and meiosis differ? These activities are designed to give students practice in translating their knowledge of what goes on in the various phases of mitosis and meiosis into visual representations. Activity 13.2 asks students to compare events in each of the various phases of mitosis and meiosis and determine similarities and differences.
What misconceptions or difficulties can these activities reveal? Most students don’t have difficulty reciting what events occur in each stage of mitosis or meiosis. If you ask them to draw what is occurring in each of these stages and give a specific chromosome complement (as in question 3 in both Activities 12.1 and 13.1), however, many have a difficult time. Two common reasons for this are: • The students do not understand how many chromosomes the cell contains. For example, if a question indicates that a eukaryotic cell has a full complement of eight chromosomes, many students may not understand this means that the cell has eight total chromosomes, or four pairs of chromosomes. • The students have memorized the list of events that occur in each stage, but they have not translated this into a real understanding of the events. In either case, asking students to draw cells in different stages of cell division will give them a better understanding of the overall process.
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Activity 12.1 Question 2: Some students don’t understand that mitosis and meiosis occur only in eukaryotes. This question is meant to point out that mitosis does not occur in prokaryotes, for example, bacteria. Question 7: Students are often confused about how to count the number of chromosomes in a cell. A general rule or convention is that chromosomes should be counted by the number of centromere regions present. Using this convention, we count two chromatids attached to a common centromere region as one chromosome. When sister chromatids separate to opposite poles, each daughter chromosome has its own centromere region and is now counted separately. To help avoid confusion, this question asks students to indicate both the number of centromeres visible and the number of chromatids attached to centromeres. Activity 13.2 Most students learn mitosis and meiosis by memorizing the stages of each in order. Few realize that the stages were named because of similarities early microscopists saw. This activity is designed to demonstrate that many of the events in a given phase—for example, metaphase of mitosis, meiosis I, and meiosis II—are the same. Once students understand that, they can focus on the general processes that occur in each phase. To distinguish between similar phases in mitosis and meiosis, students need only to remember what makes one different from the other.
Answers Activity 12.1 What is mitosis? What is mitosis? 1. What is the overall purpose of mitosis? The purpose of mitosis is to produce daughter cells that are identical to the parent cell. To do this, the cells must first duplicate all of their chromosomes. Then the chromosomes must be equally divided among the daughter cells such that each has the same complement (number and kinds) of chromosomes as the parent cell. 2. In what types of organism(s) and cells does mitosis occur? Mitosis occurs in all eukaryotic organisms.
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3. What type of cell division occurs in bacteria? Bacteria undergo a type of cell division called fission. Fission involves duplication of the DNA, or genophore, and subdivision of the cell into two daughters, each of which contains a copy of the DNA from the parent. What are the stages of mitosis? 4. The fruit fly, Drosophila melanogaster, has a total of eight chromosomes (four pairs) in each of its somatic cells. Somatic cells are all cells of the body except those that will divide to form the gametes (ova or sperm). Review the events that occur in the various stages of mitosis. Keep in mind that the stages of cell division were first recognized from an examination of fixed slides of tissues undergoing division. On fixed slides, cells are captured or frozen at particular points in the division cycle. Using these static slides, early microscopists identified specific arrangements or patterns of chromosomes that occurred at various stages of the cycle and gave these stages names (interphase, prophase, and so on). Later work using time-lapse photography made it clear that mitosis is a continuous process. Once division begins, the chromosomes move fluidly from one phase to the next. Assume you are a microscopist viewing fruit fly cells that are undergoing mitosis. Within each of the circles (which represent cell membranes) on the following page, draw what you would expect to see if you were looking at a cell in the stage of mitosis indicated. If no circle is present, draw what you would expect to see at the given stage.
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Prophase
Prometaphase
Metaphase
Anaphase
Telophase
Cytokinesis
Daughter cells in interphase
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What are the products of mitosis? 5. How many cells are produced at the end of a single mitotic division? Two cells are produced at the end of a single mitotic division. 6. How many different kinds of cells are produced at the end of a single mitotic division? Only one kind of cell is produced. Two daughter cells are produced, but they are identical to each other and to the parent cell that gave rise to them. 7. Six centromeres are observed in a prophase cell from another species of insect. a. How many pairs of chromosomes does this organism contain? Three pairs b. For each stage of mitosis, indicate the number of centromeres you would expect to find and the number of copies of chromosomes attached to each centromere. Stage of mitosis:
Number of centromeres visible per cell
Prophase
6
Anaphase
12
Number of chromosome copies attached to each centromere 2 1
12.1 Test Your Understanding Haplopappus is an annual flowering plant that grows in deserts. It is of interest because its 2n number is only four. a. This means that cells in the vegetative parts of the plant that are not undergoing mitosis have how many DNA molecules in their nuclei? There would be 4 DNA molecules or 4 total chromosomes in cells not undergoing division. b. During metaphase of mitosis, how many DNA molecules would be in the nucleus? During metaphase, there would be 8 DNA molecules in the nucleus. DNA would have duplicated in the S phase of interphase prior to division.
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Activity 13.1 What is meiosis? What is meiosis? 1. What is the overall purpose of meiosis? The purpose of meiosis is to reduce the diploid chromosome number by half to the haploid number. Note that this is a very specific half of the chromosomes. The haploid cell contains one member of every pair of chromosomes found in the diploid parent cell. 2. In what types of organism(s) does meiosis occur? Meiosis occurs in eukaryotes in cells that will produce gametes (ova or sperm). What are the stages of meiosis?
3. The fruit fly, Drosophila melanogaster, has a total of eight chromosomes (four pairs) in each of its somatic cells. Somatic cells are all cells of the body except those that will divide to form the gametes (ova or sperm). Review the events that occur in the various stages of meiosis. Keep in mind that the stages of cell division were first recognized from an examination of fixed slides of tissues undergoing division. On fixed slides, cells are captured or frozen at particular points in the division cycle. Using these static slides, early microscopists identified specific arrangements or patterns of chromosomes that occurred at various stages of the cycle and gave these stages names (interphase, prophase I, and so on). Later work using time-lapse photography made it clear that meiosis is a continuous process. Once division begins, the chromosomes move fluidly from one phase to the next. Assume you are a microscopist viewing fruit fly cells that are undergoing meiosis. Within each of the circles (which represent cell membranes) on the next pages, draw what you would expect to see if you were looking at a cell in the stage of meiosis indicated. If no circle is present, draw what you would expect to see at the given stage.
Activity 13.1
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Prophase I
Prometaphase I
Metaphase I
Anaphase I
Cytokinesis
Telophase I
Note: Whether or not the chromosomes uncoil or decondense at this stage will vary by species.
Daughter cells
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Follow one daughter cell through meiosis II.
Prophase II
Prometaphase II
Metaphase II
Anaphase II
Cytokinesis
Telophase II
Daughter cells
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What are the products of meiosis? 4. Consider a single cell going through meiosis. a. How many cells are produced at the end of meiosis? A single cell going through meiosis produces four daughter cells by the end of meiosis. b. How many chromosomes, and which chromosomes, does each of the daughter cells contain? Each daughter cell has half the number of chromosomes in the parental cell. Each daughter cell contains one member of each pair of chromosomes found in the parent cell. 5. Six centromeres are observed in a prophase I cell from another species of insect. a. How many pairs of chromosomes does this organism contain? Three pairs b. For each stage of meiosis indicate the number of centromeres you would expect to find and the number of copies of chromosomes attached to each centromere. Number of chromosome copies attached to each centromere
Number of centromeres visible per cell
Stage of meiosis: Anaphase I
6
2
Prophase II
3
2
13.1 Test Your Understanding Nondisjunction of sex chromosomes during human gamete formation may lead to individuals with sex chromosome trisomy. An individual with the sex chromosome trisomy of XXY may have resulted from nondisjunction occurring in (Circle T if true, F if false): T/F 1. meiosis I in the father’s sperm production True—Meiosis I in sperm production would result in some gametes with both an X and a Y and some with neither an X nor aY. If an XY sperm fertilized an egg carrying an X chromosome, an XXY individual would be produced.
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meiosis II in the father’s sperm production False—Nondisjunction during meiosis II of sperm production would produce sperm with either 2 Xs, 2 Ys, or no X or Y. Any of these sperm fertilizing an egg with an X chromosome would not produce an XXY individual. meiosis I in the mother’s egg production True—Nondisjunction in meiosis I of egg production could produce eggs with 2 Xs or no X. If fertilized by a sperm carrying a Y chromosome an XXY individual would be produced. meiosis II in the mother’s egg production True—Nondisjuction in meiosis II of egg production could produce eggs with 2 Xs or no X. If fertilized by a sperm carrying a Y chromosome, an XXY individual would be produced.
Activity 13.2 How do mitosis and meiosis differ? Review the processes of mitosis and meiosis in Chapters 12 and 13 of Biology, 8th edition, then fill in the chart. Keep in mind that the stages of cell division were first recognized from an examination of fixed slides of tissues undergoing division. On fixed slides, cells are captured or frozen at particular points in the division cycle. Using these static slides, early microscopists identified specific arrangements or patterns of chromosomes that occurred at various stages of the cycle and gave these stages names (interphase, prophase, and so on). Later work using time-lapse photography made it clear that mitosis and meiosis are continuous processes. Once division begins, the chromosomes move fluidly from one phase to the next. 1. What events occur during each phase of mitosis and meiosis?
Interphase Mitosis For example: G1—cell growth S—DNA duplication G2—cell growth
Prophase Chromosomes coil and condense. Nuclear membrane breaks down. Spindle forms.
Metaphase For example: Duplicated chromosomes, each with two sister chromatids, line up independently on the metaphase plate.
Anaphase Sister chromatids move to opposite poles of the spindle.
Telophase and cytokinesis The events of telophase are the opposite of those in prophase. Cytokinesis is division of the two daughter nuclei into separate cells.
(continues on next page) Activity 13.2
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Meiosis I G1—cell growth S—DNA duplicatio n G2—cell growth
Chromosomes coil and condense. Homologous chromosomes synapse. Nuclear membrane breaks down. Spindle forms.
Synapsed pairs of chromosomes, each with two sister chromatids, line up on the metaphase plate
Members of each homologous pair separate to opposite poles of the spindle.
Chromosom es do not generally uncoil. Nuclear membrane reforms and spindle breaks down. Cytokinesis is division of the two daughter nuclei into separate cells.
Meiosis II There may be a short G phase to prepare the cell for the next division phase. DNA does NOT duplicate.
Chromosomes coil and condense. Nuclear membrane breaks down. Spindle forms.
Duplicated chromosomes each with two sister chromatids, line up independently on the metaphase plate.
Sister chromatids move to opposite poles of the spindle.
The events of telophase are the opposite of these in prophase. Cytokinesis is division of the two daughter nuclei into separate cells.
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2. Fill in the chart to summarize the major similarities and differences in the two types of cell division (mitosis vs. meiosis). For similarities, include the event(s) that always happen(s) at that stage, no matter which of the cell division cycles you’re describing. Interphase a. What This phase similarities is identical do you for mitosis see? and meiosis I.
Prophase Chromosomes always coil and condense. Spindle always forms. Nuclear membrane always breaks down.
Metaphase Something always lines up on the equator or metaphase plate of the spindle.
Anaphase Something always moves to opposite poles of the spindle.
Telophase Nuclear membrane reforms Spindle breaks down.
b. What differences do you see?
In prophase I, homologous chromosomes synapse.
In mitosis and metaphase II, individual chromosomes line up. In metaphase I, synapsed pairs line up.
In mitosis and anaphase II, sister chromatids separate and move to opposite poles. In anaphase I, members of each homologous pair separate to opposite poles.
Chromosomes usually don’t uncoil during telophase I of meiosis.
No DNA duplication in interphase of meiosis II. G phase may be shortened.
(continues on next page)
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c. If the amount of DNA in a somatic cell equals C during G1 of interphase, then how much DNA is present in the cell during each phase of mitosis and meiosis? Amount of DNA in:
Interphase
Mitosis
Prophase
Metaphase
Anaphase
Telophase
C in G1 2 C 2 C in G2
2C
2C
2C (cytokinesis reduces the amount to C)
Meiosis I
C in G1 2 C 2 C in G2
2C
2C
2C (cytokinesis reduces the amount to C)
Meiosis II
C in G1 C in C G2
C
C
C (cytokinesis reduces the amount to 1/2 C)
3. How do the similarities in prophase of mitosis and meiosis compare with the similarities in telophase of mitosis and meiosis? As noted, telophase can be thought of as the opposite of prophase. In other words, what is done in prophase is undone in telophase. 4. At what stage(s) does/do most of the differences among mitosis, meiosis I, and meiosis II occur? Why do these differences exist? The primary differences between mitosis and meiosis occur as a result of synapsis in prophase I. It is synapsis that allows the members of homologous pairs to separate to opposite poles and that reduces the chromosome number at the end of meiosis I to half that of the original cell.
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Notes to Instructors Chapter 14 Mendel and the Gene Idea What is the focus of these activities? Meiosis is the basis for Mendel’s laws of segregation and independent assortment. If students have a good understanding of meiosis and Mendel’s laws, they should be able to demonstrate or model them. They should be able to determine the types of gametes an individual can produce and the probability of each type.
What are the particular activities designed to do? Activity 14.1 A Genetics Vocabulary Review Activity 14.2 Modeling meiosis: How can diploid organisms produce haploid gametes? Activity 14.3 A Quick Guide to Solving Genetics Problems Activity 14.4 How can you determine all the possible types of gametes? The activities for Chapter 14 are designed to help students integrate their understanding of meiosis and Mendelian genetics using modern terminology. Activity 14.1 provides a quick review of some modern terminology. Activity 14.2 requires students to integrate their understanding of meiosis (Chapter 13) and of basic Mendelian principles (Chapter 14) to develop a dynamic model of meiosis. Activity 14.3 provides a quick review of some of the basic rules for solving genetics problems. Activity 14.4 provides students with a mechanism for determining the type(s) of gametes that can be produced when the genotype of an organism is known.
What misconceptions or difficulties can these activities reveal? Activity 14.2 Modeling Meiosis Almost all students will quickly discover that this activity is not as easy as it first seems. They will also discover that they have difficulty translating the information provided in the activity into a total number of chromosomes, placement of genes and alleles on chromosomes, and so on. If you give students time, however, they will work these problems out for themselves. A misconception that crops up in a large percentage of the students’ models concerns what an X-shaped chromosome represents. Because students (and researchers) generally see only duplicated chromosomes under the microscope (the X-shaped ones), many students have the misconception that this X shape represents a single unduplicated chromosome.
Notes to Instructors
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Some students will join a maternal and a paternal chromosome to form the X-shaped (duplicated) chromosome prior to cell division. When they do this, half of the X is usually made from one color of playdough and the other half is a different color. Ask them what the two colors represent. Most will say it’s a duplicated chromosome, and one color represents the maternal chromosome (of the pair) and the other represents the paternal chromosome. Then ask why they are connected to each other and point out that they were separate in the gametes that produced this cell. If left to think about this for a few minutes, most of the student groups will correct themselves. Most students think the Y chromosome is not shaped like an X when duplicated. In general, students are unaware they have any of these problems until they are asked to develop a dynamic model of the process. The questions in this modeling activity ask students how many different kinds of gametes a cell from this individual (genotype CcBb) produces at the end of meiosis. Students are to assume no crossing over occurs. The correct answer is two. However, most students have learned that an individual heterozygous for two genes can produce four different kinds of gametes. When asked why they have only two different kinds, most can’t answer immediately. Again, give students a few minutes and then return and ask the question again. Most will self-correct. If not, ask them to demonstrate Mendel’s law of independent assortment using their cell. Activities 14.3 and 14.4 Many introductory students have not developed good strategies for solving genetics problems involving more than one gene. These activities are designed to help students understand that if the genes involved are not linked, solving genetics problems one gene at a time is generally the easiest and most accurate method.
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Answers Activity 14.1 A Genetics Vocabulary Review Mendel did not know anything about chromosomes, genes, or DNA. Because modern genetics uses vocabulary that assumes students today understand these ideas, it’s helpful to review some key terms. Match each commonly used genetics term with its appropriate definition or example. Terms
Definitions and Examples
e heterozygous b homozygous g monohybrid cross c autosomal h genotype d phenotype
a. b. c. d. e. f.
Blue-eyed blonde mates with brown-eyed brunette BB or bb not on sex chromosomes blue or brown eyes Bb locus on a chromosome that codes for a given polypeptide* g. Blonde mates with brunette h. BB, Bb, or bb i. Males have only one for each gene on the X chromosome
f gene d allele a dihybrid cross
* Note: Though it is true that a gene can code for a polypeptide, it is important to remember that not all genes code for polypeptides. Some code for mRNAs that produce polypeptides, but others code for other forms of RNA—for example, rRNA and tRNA.
Activity 14.2 Modeling meiosis: How can diploid organisms
produce haploid gametes? Integrate your understanding of meiosis (Chapter 13) and of basic Mendelian principles (Chapter 14) to develop a dynamic model of meiosis. When you’ve completed the model, use it to explain what aspects of meiosis account for Mendel’s laws of segregation and independent assortment.
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Building the Model Working in groups of three or four, construct a dynamic (claymation-type) model of meiosis for the organism described on the next page. You may use the materials provided in class or devise your own. What genetic and chromosomal traits does your organism have? 1. Your individual is male/female (choose one). Females are XX and males are XY. For simplicity, assume that the individual is diploid with 2n ⫽ 6, including the sex chromosomes. On one pair of autosomes (the nonsex chromosomes), the individual is heterozygous for hair color (B ⫽ brown and dominant, b ⫽ blonde and recessive). On another pair of autosomes, the organism is heterozygous for hair structure (C ⫽ curly and dominant, c ⫽ straight and recessive). Assume further that the individual’s mother was homozygous dominant for both traits and the father was homozygous recessive for both. a. Is your individual’s hair curly or straight? Brown or blonde? The individual described is heterozygous for both traits. Therefore, s/he has curly, brown hair. b. What did the individual’s mother’s hair look like? What did the father’s hair look like? The individual’s mother was homozygous dominant and therefore had curly, brown hair. The father was homozygous recessive and had straight, blonde hair. c. What chromosomes and alleles were in the egg and the sperm that gave rise to your individual? The egg contained an X chromosome, a number 1 chromosome with a brown hair gene, and a number 2 chromosome with a curly gene. The sperm contained an X if your individual is female or a Y if your individual is male. It also contained a number 1 chromosome with the blonde hair gene and a number 2 chromosome with the straight hair gene. What does the nucleus contain? To answer this question, develop a model of a cell from your individual. • Use chalk on a tabletop or a marker on a large sheet of paper to draw a cell’s membrane and its nuclear membrane. The nucleus should be at least 9 inches in diameter. • Use playdough or cutout pieces of paper to represent your individual’s chromosomes. Indicate the placement of genes on the chromosomes. Put all the chromosomes from your individual into the nucleus. 82
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• Make a key for your model that indicates how alleles are designated and which of the chromosomes are maternal versus paternal contributions. Then develop a model of the meiotically active cell. • Make an identical copy of the original cell. This will be the “active” cell—that is, the one that undergoes meiosis. • Using the “active” cell only, develop a dynamic model of meiosis. To do this, actively move the chromosomes of this one cell through a complete round of meiosis in a sex cell. (Sex cells are the cells of the body that give rise to gametes: ova or sperm.) • Use your model to demonstrate meiosis to another student group or to your instructor. Then use your model to answer the questions on the next page. When developing and explaining your model, be sure to include definitions or descriptions of all these terms and structures: diploid 2n/n chromosome chromatid chromatin centromere (kinetochore) autosome sex chromosome sex cell
autosome crossing over synapsis recessive allele dominant allele genotype maternal paternal spindle
spindle fibers nuclear membrane nucleolus phenotype heterozygous homozygous law of segregation law of independent assortment
What are the products of meiosis? 2. From a single sex cell going through meiosis, how many daughter cells are produced? Four cells are produced by the end of meiosis. 3. For your model organism or individual (defined in question 1), how many different kinds of gametes can be produced from a single cell undergoing meiosis? (Assume no crossing over occurs.) A single cell from this individual that undergoes meiosis (with no crossing over) produces two different kinds of gametes. 4. Your individual is heterozygous for two genes on separate pairs of homologous chromosomes. His/her genotype is CcBb. Given this information alone, how many
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different kinds of gametes could this individual produce? (Again. assume no crossing over occurs.) A heterozygous individual like this could, on average, produce four different kinds of gametes in equal proportions. The kinds of gametes are CB, Cb, cB, and cb. 5. Compare your answer to question 4 with your answer to question 3. How do the numbers of different kinds of gametes in your answers compare? Explain any difference. Any single cell going through meiosis (no crossing over) produces only two types of gametes (maximum). However, that individual has the potential to produce four different kinds of gametes. The particular combination of C and B alleles in the gametes is a result of how the chromosomes line up at metaphase I. They could line up at metaphase I in either of these ways: B b bB B C Cc c
A B Bb b C Cc c
If they line up as in box A, the gamete types produced are BC and bc. If they line up as in box B, the gamete types produced are bC and Bc. Because each way of lining up is equally probable, half the time (statistically) they will line up as in A and half the time as in B. As a result, on average, all four types of gametes are expected to occur in equal proportions.
14.2 Test Your Understanding What aspect(s) of meiosis account(s) for: 1. Mendel’s law of segregation? Mendel’s law of segregation states that although each organism contains two traits (today known as alleles) for a given character (today known as a gene), only one allele is found in each gamete. Synapsis of homologous chromosomes in prophase I and their separation to opposite poles in anaphase I separate or segregate alleles of a given gene into different gametes. 2. Mendel’s law of independent assortment? Mendel’s law of independent assortment states that the pairs of traits that control each character act independently of each other in gamete formation. (Today we
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know that the assumption is that these traits or alleles are on separate pairs of homologous chromosomes.) In other words, how one set of traits on one pair of homologous chromosomes segregates in gametes does not affect how another set of traits on a different pair of homologous chromosomes segregates. The alignment of homologous chromsome pairs relative to each other during metaphase I of meiosis is random. As noted in question 5, how any given pair of chromosomes lines up on the metaphase plate during metaphase I is independent of how any other pair lines up.
Activity 14.3 A Quick Guide to Solving Genetics Problems Over the years, rules have been developed for setting up genetics problems and denoting genes and their alleles in these problems. This activity provides a quick review of some of these rules. After you have read through all of this material, complete Activities 14.4, 15.1, and 15.2.
Basic Assumptions to Make When Solving Genetics Problems 1. Are the genes linked? If the problem does not (a) indicate that the genes are linked or (b) ask whether the genes are (or could be) linked, then you should assume that the genes are not linked. 2. Are the genes sex-linked? Similarly, if the problem does not (a) indicate that the genes are sex-linked (that is, on the X chromosome) or (b) ask whether the genes are (or could be) on the X chromosome (or Y chromosome), then you should assume that the genes are on autosomes and are not sex-linked. 3. Is there a lethal allele? If a gene is lethal, then you should assume that the offspring that get the lethal allele (if dominant) or alleles (if homozygous recessive) do not appear; that is, they are not born, do not hatch, and so on. Therefore, they are not counted among the offspring. An obvious exception is lethal genes that have their effect late in life. If this is the case, however, it should be noted in the question. 4. Are the alleles dominant, recessive, or neither? Unless the problem states otherwise, assume that capital letters (BB, for example) designate dominant alleles and lowercase letters (bb, for example) indicate recessive alleles. When
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there is codominance or incomplete dominance, the alleles are usually designated by the same capital letter and each one is given a superscript (for example, CRCW in Figure 14.10, page 272, of Biology, 8th edition). 5. How are genotypes written? Assume a gene for fur color in hamsters is located on the number 1 pair of homologous autosomes. Brown fur (B) is dominant over white fur (b). The genotype for fur color can be designated in different ways: a. The alleles can be shown associated with the number 1 chromosome. In this notation, an individual heterozygous for this gene is designated as |B|b.
b. Most commonly, this notation is simplified to Bb.
In problems that involve sex-linked genes, the chromosomes are always indicated—for example, XAXa and XaY. 6. What information do you need to gather before trying to solve a genetics problem? Before trying to solve any problem, answer these questions: a. What information is provided? For example: • What type of cross is it? Is it a monohybrid or dihybrid cross? • Are the genes sex-linked or autosomal? • Linked or unlinked? b. What does the information provided tell you about the gene(s) in question? For example: • What phenotypes can result? • How many alleles does the gene have? • Are the alleles of the gene dominant? Recessive? Codominant? c. Does the question supply any information about the individuals’ genotypes? If so, what information is provided? • Grandparent information? • Parental (P) information? • Gamete possibilities? • Offspring possibilities? 86
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Solving Genetics Problems 1. What is a Punnett square? Punnett squares are frequently used in solving genetics problems. A Punnett square is a device that allows you to determine all the possible paired combinations of two sets of characteristics. For example, if you wanted to determine all the possible combinations of red, blue, and green shirts with red, blue, and green pants, you could set up this Punnett square:
Pants
Shirts
Red pants Blue pants Green pants
Red shirt
Blue shirt
Green shirt
Red shirt and red pants Red shirt and blue pants Red shirt and green pants
Blue shirt and red pants Blue shirt and blue pants Blue shirt and green pants
Green shirt and red pants Green shirt and blue pants Green shirt and green pants
Similarly, if you wanted to determine the probability of a male (XY) and a female (XX) having a son or a daughter, you would first determine the possible gametes each could produce and then set up a Punnett square to look at all the possible combinations of male and female gametes. Here, meiosis dictates that the female’s gametes get one of her X chromosomes or the other. In the male, the gametes get either the X chromosome or the Y. As a result, the Punnett square would look like this: Female’s gamete possibilities Male’s gamete possibilities
X
X
X
XX
XX
Y
XY
XY
2. If you know the parents’ genotypes, how can you determine what types of offspring they will produce? a. Autosomal genes: For an autosomal gene that has the alleles A and a, there are three possible genotypes: AA, Aa, and aa.
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All possible combinations of matings and offspring for two individuals carrying the autosomal gene with alleles A and a are shown in the figure below. If you know how to solve these six crosses you can solve any problem involving one or more autosomal genes.
AA X AA Ova Sperm
A
A
A
AA
AA
A
AA
AA
AA X aa * a
a
A
Aa
Aa
A
Aa
Aa
AA X Aa * A
a
A
AA
Aa
A
AA
Aa
aa X aa a
a
a
aa
aa
a
aa
aa
aa X Aa *
Aa X Aa A
a
A
AA
Aa
a
Aa
aa
A
a
a
Aa
aa
a
Aa
aa
* Note: If you take sex into account there are actually nine possible combinations of matings: Female genotypes Male genotypes
AA
Aa
aa
AA
AA x AA
AA x Aa
AA x aa
Aa
Aa x AA
Aa x Aa
Aa x aa
aa
AA x aa
aa x Aa
aa x aa
Because the results of reciprocal autosomal matings—e.g., AA male with aa female and aa male with AA female are the same—only one of each reciprocal type is included in the six combinations above. b. Sex-linked genes: For sex-linked genes that have two alleles, e.g., w+ and w, females have three possible genotypes: Xw+Xw+, Xw+Xw, and XwXw. Males have only two possible genotypes: Xw+Y and XwY. All the possible combinations of matings and offspring for a sex-linked trait are listed in the next figure. If you know how to solve these six single-gene crosses, then you can solve any genetics problem involving sex-linked genes.
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All possible combinations of matings for two individuals with a sex-linked gene are shown in the figure below. Fill in the Punnet squares to determine all possible combinations of offspring.
AA X AA Ova Sperm
A
A
A
AA
AA
A
AA
AA
AA X aa * a
a
A
Aa
Aa
A
Aa
Aa
AA X Aa * A
a
A
AA
Aa
A
AA
Aa
aa X aa a
a
a
aa
aa
a
aa
aa
aa X Aa *
Aa X Aa A
a
A
AA
Aa
a
Aa
aa
A
a
a
Aa
aa
a
Aa
aa
c. Multiple genes: Remember, if genes are on separate chromosomes, then they assort independently in meiosis. Therefore, to solve a genetics problem involving multiple genes, where each gene is on a separate pair of homologous chromosomes: • Solve for each gene separately. • Determine probabilities for combination (multiple-gene) genotypes by multiplying the probabilities of the individual genotypes. Example: What is the probability that two individuals of the genotype AaBb and aaBb will have any aabb offspring? To answer this, solve for each gene separately. A cross of Aa ⫻ aa could produce the following offspring: A
a
a
Aa
aa
a
Aa
aa
1/2
Aa and 1/2 aa offspring
Activity 14.3
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A cross of Bb ⫻ Bb could produce the following offspring: B
b
B
BB
Bb
b
Bb
bb
1
⁄4 BB, 1⁄2 Bb, and 1⁄4 bb offspring
The probability of having any aabb offspring is then the probability of having any aa offspring times the probability of having any bb offspring. The probability is 1⁄2 ⫻ 1⁄4 ⫽ 1⁄8.
Activity 14.4 How can you determine all the possible types of gametes? To solve genetics problems in which genotypes are given, you must first know what types of gametes each organism can produce. 1. How many different kinds of gametes can individuals with each of the following genotypes produce? a. AA
1 kind of gamete ⫽ A
b. aa
1 kind of gamete ⫽ a
c. Aa
2 kinds of gametes ⫽ either A or a in equal proportions
d. AaBB
2 kinds of gametes ⫽ either AB or aB in equal proportions
e. AaBb
4 kinds of gametes ⫽ AB, Ab, aB, and ab in equal proportions
f. AaBbCC
4 kinds of gametes ⫽ ABC, AbC, aBC, and abC in equal proportions
g. AaBbCc
8 kinds of gametes ⫽ ABC, ABc, AbC, Abc, aBC, abC, aBc, and abc in equal proportions
h. AaBbCcDdEeFf 32 different kinds of gametes in equal proportions 2. Based on your answer in question 1, propose a general rule for determining the number of different gametes organisms like those described in question 1 can produce. Number of different kinds of gametes ⫽ 2n, where n ⫽ number of heterozygous alleles (genes). Here the assumption is that the different genes are on separate pairs of homologous chromosomes. 90
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3. Two individuals have the genotypes AaBbCcDd. a. How many different types of gametes can each produce? 2n ⫽ 24 ⫽ 16 different kinds of gametes Alleles: B b
A AB Ab
a aB ab
Alleles: D d
C CD Cd
c cD cd
b. What are these gametes? One way of figuring this out is to take two genes at a time. CD cD Cd cd
AB ABCD ABcD ABCd ABcd
aB aBCD aBcD aBCd aBcd
Ab AbCD AbcD AbCd Abcd
ab abCD abcD abCd abcd
Then here are all the possible combinations: c. You set up a Punnett square using all the possible gametes for both individuals. How many “offspring squares” are in this Punnett square? If you used 16 gametes across the top of the Punnett square and 16 down the side, you have 256 offspring possibilities in this table. d. If you completed this Punnett square, how easy would it be to find all the “offspring squares” that contain the genotype AaBBccDd? It wouldn’t be very easy. On the other hand, it would be very easy to make a mistake in filling in or counting the offspring squares in the table. e. Given that the genes are all on separate pairs of homologous chromosomes, what other method(s) could you use to determine the probability of these individuals having any offspring with the genotype AaBbccDd? You could handle each gene pair as a separate cross. For example, the cross AaBbCcDd ⫻ AaBbCcDd becomes Aa ⫻ Aa
Bb ⫻ Bb
Cc ⫻ Cc
and
Dd ⫻ Dd
Each of these crosses has similar results; that is, each produces 1⁄4 homozygous dominant offspring, 1⁄2 heterozygous offspring, and 1⁄4 homozygous recessive offspring. Therefore, the probability that any of their offspring will be AaBbccDd is 1⁄2 ⫻ 1⁄2 ⫻ 1⁄2 ⫻ 1⁄2 ⫽ 1⁄32. Activity 14.4
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Notes to Instructors Chapter 15 The Chromosomal Basis of Inheritance What is the focus of these activities? Many students have difficulty solving genetics problems. This is especially true for problems that include both autosomal and sex-linked genes.
What are the particular activities designed to do? Activity 15.1 Solving problems when the genetics are known This activity is designed to give students practice in solving autosomal genetics problems, sex-linked genetics problems, and problems that involve both autosomal and sex-linked genes. Activity 15.2 Solving problems when the genetics are unknown The types of questions presented in Activity 15.1 provided students with practice solving problems when the genetics of the parents are known. Activity 15.2 asks students to discover the genetics of individuals by setting up and analyzing the results of controlled crosses.
What misconceptions or difficulties can these activities reveal? Activity 15.1 Question 2b: The answer is zero. Given that, many students automatically think this question was designed to trick them. You may present the following scenario to point out the value of zero as an answer: Assume various members of your family have been born with a genetic disorder that is lethal by age 25. You want to know the probability that you have this gene and can pass it on to your offspring. How would you feel if the genetic counselor told you that you had zero probability of having the trait? (Note: A zero probability is usually given only when a genetic counselor has a direct test for the presence of the gene. If, on the other hand, pedigree analysis indicates that neither of an individual’s parents carry the gene, then the counselor is likely to indicate that the individual has a one in a million chance of carrying the gene. In this example, one in a million is the rate of spontaneous mutation of the normal allele to the mutant allele.) Questions 3 and 4: Many students have difficulty solving sex-linked genetics problems because they try to solve them using the alleles alone. In other words, they do not indicate X and Y chromosomes in their Punnett squares. Many others who do solve the problem correctly with the Punnett square have difficulty determining what the question is asking. For example, is it asking what proportion of all offspring have a certain genotype? Or is it asking what proportion of the males alone, for example, have a certain genotype? 92
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Activities 15.2 and 15.3 Both of these activities are designed to give students practice in some of the actual types of problems/situations that might be encountered by geneticists. For Activity 15.3, only one way of solving each problem is presented. A number of other approaches could be used, however.
Answers Activity 15.1 Solving Problems When the Genetics Are Known Refer to Activity 14.3 and to Chapters 14 and 15 in Biology, 8th edition, to complete this activity. 1. An organism that has the genotype AaBbCc is crossed with an organism that has the genotype AABbCc. Assume all genes are on separate sets of chromosomes (that is, they are not linked). a. What is the probability that any of the offspring will have the genotype AABBCC? (Hint: To get the answer, consider the six possible types of autosomal crosses. Determine the individual probabilities of getting AA offspring from the monohybrid cross. Then do the same to determine the probabilities of getting BB offspring and CC offspring. Multiply these probabilities together.) The probability of any offspring being AA ⫽ 1⁄2, BB ⫽ 1⁄4, and CC ⫽ 1⁄4. Therefore, AABBCC ⫽ 1⁄2 ⫻ 1⁄4 ⫻ 1⁄4 ⫽ 1⁄34. b. What is the probability that any of the offspring will have the genotype AaBbcc? The probability of any offspring being Aa ⫽ 1⁄2, Bb ⫽ 1⁄ 2, and cc ⫽ 1⁄4. Therefore, AaBbcc ⫽ 1⁄2 ⫻ 1⁄2 ⫻ 1⁄4 ⫽ 1⁄16. 2. Consider the cross AaBbCcddEe ⫻ AABBccDDEe. a. What is the probability that any offspring will have the genotype AaBBCcDdEE? 1 ⁄2 ⫻ 1⁄2 ⫻ 1⁄2 ⫻ 1 ⫻ 1⁄4 ⫽ 1⁄32 b. What is the probability that any offspring will have the genotype AABBCCDDee? 1 ⁄2 ⫻ 1⁄2 ⫻ 0 ⫻ 0 ⫻ 1⁄4 ⫽ 0
Activity 15.1
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3. In fruit flies (Drosophila melanogaster), the most common eye color is red. A mutation (or allele) of the gene for eye color produces white eyes. The gene is located on the X chromosome. a. What is the probability that a heterozygous red-eyed female fruit fly mated with a white-eyed male will produce any white-eyed offspring? Xw+Xw ⫻ XwY ⫽ heterozygous red female crossed with a white male Xw+
Xw
Xw
Xw+ Xw
Xw Xw
Y
Xw+Y
Xw Y
Half of the offspring will be white-eyed. Half of the females will be white-eyed, and half of the males will be white-eyed. b. What is the probability that the mating in part a will produce any white-eyed females? The probability that the cross will produce any white-eyed females is 1⁄2. (Note: The question is asking the probability that any of the offspring will be white-eyed and female. It is not asking how many of the females will have white eyes.) c. What is the probability that this mating will produce any white-eyed males? Similarly, the probability of producing any white-eyed males is 1⁄2. 4. A heterozygous brown-eyed human female who is a carrier of color blindness marries a blue-eyed male who is not color-blind. Color blindness is a sex-linked trait. Assume that eye color is an autosomal trait and that brown is dominant over blue. What is the probability that any of the offspring produced have the following traits? Bb X+Xcb (female) ⫻ bb X+Y (male) Use the two separate crosses: Bb ⫻ bb and X+Xcb ⫻ X+Y a. Brown eyes 1⁄2 b. Blue eyes 1⁄2 c. Color blindness 1⁄4 d. Color-blind males 1⁄4 e. Brown-eyed, color-blind males 1⁄2 ⫻ 1⁄4 ⫽ 1⁄18
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f. Blue-eyed, color-blind females 1⁄2 ⫻ 0 ⫽ 0 g. What is the probability that any of the males will be color-blind? 1 ⁄2 (Note: This question asks only about the males, not about all of the offspring. If we look at all of the offspring we find 1/4 will be color-blind males.) h. Why do males show sex-linked traits more often than females? Males have only one X chromosome. The X chromosome carries many more genes than does the Y chromosome. For example, in humans, the X carries a few thousand genes and the Y carries only a few dozen genes. Females have two alleles for every gene on the X chromosome. Females have the recessive phenotype only when both Xs carry the recessive allele. In contrast, for most genes on the X chromosome, males need to have only the recessive allele to show or display the recessive phenotype.
Activity 15.2 Solving Problems When the Genetics Are Unknown An understanding of Mendelian genetics allows us to determine the theoretical probabilities associated with normal transmission of autosomal and sex-linked alleles during reproduction. This understanding provides us with strategies for solving genetics problems. In real-life situations, geneticists use these strategies to determine the genetics behind specific phenotypic traits in organisms. They do this by conducting controlled crosses of experimental organisms (e.g. Drosophila) or by analyzing family pedigrees (as for humans). Controlled Crosses Two problems are presented below. In each, you are given: a. “Wild population”—the phenotypic characteristics of a wild population of fruit flies that were trapped randomly on a remote island.
b. “Cross 1, 2, etc.”—the phenotypic characteristics of offspring from a controlled cross. The phenotypes of the parents are indicated after each cross—e.g., “Cross 1: Male Ambler ⫻ Female Wild Type.”
Activity 15.2
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For each of the problems, analyze the results in each cross and answer the questions that follow. 1. Problem One Wild population Male Female Total
Wild type
Ambler
Total
33 31 64
17 19 36
50 50 100
Ambler
Total
24 31 55
53 50 113
Cross 1: Male Ambler ⴛ Female Wild Type Wild type Offspring Vial 1 29 Male 29 Female 58 Total
a. What does cross 1 tell you about dominance versus recessiveness of the alleles? Because you get equal numbers of both phenotypes, it is impossible to determine if one is dominant over the other. b. What does cross 1 tell you about placement of the alleles on autosomes vs. sex chromosomes? Because the numbers of males and females in each phenotype is approximately the same, it is again impossible to determine if the allele is on an autosome or a sex chromosome. Cross 2: Female Ambler ⴛ Male Wild Type Wild type
Ambler
Total
0
32
32
Female
32
0
32
Total
32
32
64
Offspring Vial 2 Male
a. What does cross 2 tell you about dominance versus recessiveness of the alleles? Because all males are ambler and all females are wild type, this indicates that ambler is recessive. b. What does cross 2 tell you about placement of the alleles on autosomes vs. sex chromosomes? (In your answer show the chromosomal genotypes for the parents in this cross.) It indicates that the alleles are on the X chromosome. The parents in this cross must have been XamXam female and X+Y. 96
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2. Problem Two Mt ⫽ Monocle; Bt ⫽ Bifocal; Tr ⫽Trifocal; Sp ⫽ Spinner; Sh ⫽ Shing Wild Population
Mt, Sp
Mt, Sh
Bt, Sp
Bt, Sh
Tr, Sp
Tr, Sh
Total
Male
10
6
6
0
22
3
47
Female
19
1
9
1
20
4
54
Total
29
7
15
1
42
7
101
Cross 1: Bifocal, Spinner Female ⴛ Monocle, Shiny Male Mt ⫽ Monocle; Bt ⫽ Bifocal; Tr ⫽Trifocal; Sp ⫽ Spinner; Sh ⫽ Shing Offspring Mt, Sp Mt, Sh Bt, Sp Bt, Sh Tr, Sp Tr, Sh Vial 1 Male Female Total
0 0 0
0 0 0
0 0 0
0 0 0
31 34 65
34 38 72
Total 65 72 137
a. What does cross 1 tell you about dominance versus recessiveness of the alleles? These results indicate that Mt and Bt are codominant and the heterozygote is Tr. However, these results don’t indicate whether Sp or Sh is dominant since both appear in relatively equal numbers. b. What does cross 1 tell you about placement of the alleles on autosomes vs. sex chromosomes? Since all males and females are hybrid, the Mt and Bt alleles must be autosomal. It is unclear whether the Sp and Sh alleles are autosomal vs sex linked because numbers of males and females with each trait are similar. You could get the same results if the genes were either sex linked or autosomal. Cross 2: Monocle, Spinner Female ⴛ Trifocal, Spinner Male Mt ⫽ Monocle; Bt ⫽ Bifocal; Tr ⫽Trifocal; Sp ⫽ Spinner; Sh ⫽ Shing Offspring Vial 2
Mt, Sp
Mt, Sh
Bt, Sp
Bt, Sh
Tr, Sp
Tr, Sh
Total
8
8
0
0
8
8
32
Female
23
0
0
0
15
0
38
Total
31
8
0
0
23
8
70
Male
Activity 15.2
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a. What does cross 2 tell you about dominance versus recessiveness of the alleles? This confirms that the alleles for Mt and Bt are codominant MtMt ⫻ MtBt ⫽>__ MtMt and __ MtBt offspring (Tr). Because we get some Sh from an Sp ⫻ Sp cross, Sh must be recessive. b. What does cross 2 tell you about placement of the alleles on autosomes vs. sex chromosomes? This indicates that the Sp and Sh alleles are on the X chromosome. The female must be XspXsh and the male must be XspY. The offspring are therefore: Xsp Xsp Xsp (Sp females) Xsp Y (Sp males)
Xsp Y
Xsh Xsp Xsh (Sp females) Xsh Y (Sh males)
We should see no Sh females and the ratios of the others as in this table: 2 Sp females: 1 Sp males: 1 Sh males.
Analysis of Pedigrees Analyze the pedigree and answer the questions that follow. The diagram below shows a pedigree of three generations in a family. Black circles/squares indicate persons with a genetic disorder. A square indicates a male and a circle indicates a female. The two males in generation 1 are siblings. A
B
Generation 1 Generation 2 Generation 3
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3. Looking only at the generation 2 offspring (of the two generation 1 brothers), what can you say about the gene(s) controlling the genetic disorder? Is the disorder caused by a gene that is dominant or recessive, autosomal or sex-linked? The gene is most likely dominant. If it is dominant, the gene may be either autosomal or sex-linked based on these data alone. There is a chance that the gene is recessive. However, for this to be true, the two brothers would both have to mate with a heterozygous female in order to produce the offspring in generation 2. This is much less likely but still a possibilty. 4. What additional information do you gain from examining the generation 3 offspring? The mating between two affected individuals (lineage A – Generation 2) produces one unaffected male offspring. If the disorder were caused by an autosomal recessive gene, all of the offspring in this cross would be homozygous recessive and have the disorder. Because one male does not have it, the disorder must be caused by a dominant allele. Given the information in generations 2 and 3, it is likely that the allele is also sex-linked (since all daughters of affected males have the disorder and males only have the disorder if their mother had it). However, given that this is a relatively small population, there is still a possibility that the disorder is autosomal, dominant.
Activity 15.3 How can the mode of inheritance be determined experimentally? Outline the experimental crosses you would need to make to solve each problem. 1. Three new traits have been discovered in a population of Drosophila: • Tapping (a behavioral mutant in which the fly taps one foot constantly) • Single stripe (a pigmentation change that leads to a long stripe down the fly’s back) • Angular (causes angular bends in bristles that are normally straight) The positions of the three genes on the chromosomes are unknown. Given two pure breeding (homozygous) lines and using an initial cross of normal, normal, normal females with tapping, single stripe, angular males, describe the appropriate genetic experiments needed to establish whether any of these traits are caused by genes that are: a. Autosomal or sex-linked First, mate normal, normal, normal homozygous females with tapping, single stripe, angular males. The phenotypes of the F1 individuals will indicate which Activity 15.3
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alleles are dominant. Next, mate the F1 males with females that are homozygous recessive for all three traits, and mate the F1 females with males that are homozygous recessive for all three traits. Examine the ratio of phenotypes for each trait in the offspring as a whole. Compare the ratio for the offspring as a whole with the ratio for each sex. For example, if we assume normal is dominant in all cases, then the crosses would look like this: AABBCC ⫻ aabbcc All F1 ⫻ AaBbCc F1 crosses with homozygous recessive mates: AaBbCc ⫻ aabbcc Male a a OR a
X Xa
A Aa Aa
a aa aa
XA XA Xa XA Xa
Y XaY XaY
Female a a OR a
X Y
A Aa Aa
a aa aa
XA XA Xa XAY
Xa Xa Xa XaY
Note that the ratios of offspring phenotypes are the same for an autosomal cross of a heterozygous male with a homozygous recessive female and for a heterozygous female with a homozygous recessive male. In both cases, half of the offspring are Aa and half are aa. If the gene is sex-linked, however, the results differ. When a male F1 showing the dominant phenotype is mated with a recessive phenotype female, all the females show the dominant phenotype and all the males show the recessive phenotype. When a female F1 showing the dominant phenotype is mated with a recessive phenotype male, half of the males and half of the females show the recessive phenotype. b. Linked on the same chromosome or unlinked If the genes are not linked, we expect the probability of offspring with a given set of phenotypes—for example, normal, one stripe, angular—to be equal to the product of the individual probabilities for each occurring as separate crosses. For example, if the genes are autosomal, then the F1 mating is AaBbCc ⫻ aabbcc. If the genes are not linked, we expect to see 1⁄2 normal, 1⁄2 one stripe, and 1⁄2 angular among the offspring. The probability of all these characteristics showing in the same offspring is 1⁄2 ⫻ 1⁄2 ⫻ 1⁄2 ⫽ 1⁄18. If A and B are linked, we get different results. The F1 cross becomes AB/ab ⫻ ab/ab → 1⁄2 AB/ab and 1⁄2 ab/ab (if no crossing over occurs) and Cc ⫻ cc → 1⁄2 Cc and 1⁄2 cc. The following table lists the combinations of the offspring:
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1
⁄2 AB/ab 1 ⁄4 AaBbCc 1 ⁄4 AaBbcc
1
⁄2 Cc ⁄2 cc
1
1
⁄2 ab/ab ⁄4 aabbCc 1 ⁄4 aabbc 1
If no crossing over occurs, normal, one stripe, and angular offspring do not appear. 2. A genetics student chose a special project involving a three-gene cross to check the relative positions and map distances separating three genes in Drosophila that she thought were all on the third chromosome. To do this, she mated Drosophila females that were homozygous for the recessive genes cu (curled), sr (striped), and e (ebony) with males that were homozygous for the wild type, cu⫹ (straight), sr⫹ (not striped), and e⫹ (gray). She then mated (testcrossed) the F1 females with homozygous recessive curled, striped, ebony males. Here are the phenotypic results of the testcross: straight, gray, not striped curled, ebony, striped straight, ebony, striped curled, gray, not striped straight, ebony, not striped curled, gray, striped straight, gray, striped curled, ebony, not striped Total
820 810 100 97 80 90 1 2 2,000
a. How are the three genes arranged on the chromosomes? The three genes appear to be linked on the same chromosome.
Activity 15.3
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b. What evidence allows you to answer the question in part a? If one of the genes was not linked, we would expect to see results similar to those calculated in part b of question 1, where we looked at the results we would get if A and B were linked but C wasn’t. Instead, we see many more of the parental phenotypes (straight, gray, not striped and curled, ebony, striped) than any other type. The other types are therefore most likely the result of crossovers. For example, the F1 chromosomes might look like this: ______ e+______ sr+______ + + + –cu ______ e ______ sr ______ –cu______ e______ sr______ –cu______ e______ sr______ –cu
+
A crossover between the cu+ and the e on one homologous chromosome and the cu and the e on the other would result in some offspring that have these phenotypes: Curled, gray, not striped and straight, ebony, striped + + + –cu______ e ______ sr ______ –cu ______ e______ sr______ –cu______ e______ sr______ –cu______ e______ sr______ c. If any of the genes are linked, how far apart are they on the chromosome? How can you determine this? First, look at the double crossovers to determine how the genes are arranged on the chromosome. The offspring phenotypes that occur in the smallest numbers are most likely to be the result of double crossovers. They are straight, gray, striped 1/2000 curled, ebony, not striped 2/2000 For these phenotypes to be the result of double crossovers, the order of genes on the chromosome has to be + + + –cu ______ sr ______ e –cu______ sr______ e Given this order, the following phenotypes occurred because of crossovers between cu+ and sr+: straight, ebony, striped 100 curled, gray, not striped 97 Subtotal 197
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These phenotypes occurred because of crossovers between sr+ and e+: straight, ebony, not striped curled, gray, striped Subtotal
80 90 170
If we add the double crossovers to each subtotal (because each represents an additional crossover at each of these sites), then the percent crossover between cu+ and sr+ ⫽ 197 + 3 ⫽ 200/2000 (⫻ 100) ⫽ 10%. The percent crossover between sr+ and e+ ⫽ 170 + 3 ⫽ 173/2000 (⫻ 100) ⫽ 8.7%. Because 1% crossover is said to be equivalent to one centimorgan in distance, the cu+ and sr+ genes are 10 centimorgans apart and the Jsr+ and e+ genes are 8.7 centimorgans apart.
Activity 15.3
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Notes to Instructors Chapter 16 The Molecular Basis of Inheritance What is the focus of these activities? Almost all introductory biology students know that DNA is the hereditary material in living cells. Many of them have a difficult time visualizing its overall structure, however, and how that structure and the characteristics of associated enzymes determine its mode of replication.
What are the particular activities designed to do? Activity 16.1 Is the hereditary material DNA or protein? This activity is designed to help students organize and review the experiments and thought processes that lead first to an understanding that DNA is the hereditary material and later to the structure of the DNA double helix. Activity 16.2 How does DNA replicate? This activity is designed to give students a better understanding of both the overall process of DNA replication and the experimental evidence used to support the semiconservative model of replication.
What misconceptions or difficulties can these activities reveal? Activity 16.1 This activity asks students to review and integrate the evidence from a series of experiments that together demonstrated that DNA (not protein) is the hereditary material. The vast majority of introductory biology students already know that DNA is the hereditary material (and have known this “all their lives”). As a result, they often “can’t see the point in rehashing old experiments.” It is often necessary to be very explicit about what you want them to learn from this review. You may need to state that you want students to understand the logic behind the experiments in addition to the evidence that the experiments provide. In addition, many students don’t understand the experiments of Meselson and Stahl. We offer a couple of reasons: 1. Many don’t understand why growing bacteria in a medium that contains only 15N means that the DNA the bacteria produce will ultimately contain 15N in its nucleotides. It is useful to let them know that, like plants, many bacteria are capable of manufacturing macromolecules (like nucleotides) from inorganic precursors.
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2. Students may not understand why DNA that contains 15N will layer out in a centrifuge tube lower than DNA that contains 14N. They generally don’t understand that the centrifuge tube contains a density gradient established prior to the addition of DNA and that centrifugation separates compounds based on their density. (Note: A density gradient can be set up by centrifuging a concentrated solution of CsCl in an ultracentrifuge for up to 24 hours.) Activity 16.2 Students tend to encounter a number of difficulties, misconceptions, and missing conceptions as they model DNA replication. Here are several possible problems: 1. Many students don’t understand what 5⬘ and 3⬘ mean relative to the structure of DNA. To explain this, it is useful to draw a deoxyribose molecule and show how the carbons are numbered clockwise in the ring (from the oxygen). The phosphate group is attached to the number 5 carbon, the base is attached to the number 1 carbon, and the number 3 carbon is the only one left with an open hydroxyl group. As a result, one nucleotide is attached to the next by adding to this hydroxyl group on the number 3 carbon. The chain of nucleotides appears to grow in the direction of the number 3 carbon—that is, from the 5⬘ end to the 3⬘ end. 2. Many students want to know why DNA polymerase can add nucleotides only to an open hydroxyl group at the 3⬘ end of a nucleotide. Remind them that enzymes operate in a lock-and-key type of operation. The active site has a specific configuration and can operate only in this configuration, which in this case dictates that DNA polymerase can add only onto the 3⬘ end of a nucleotide. 3. The majority of students have difficulty understanding both why and how Okasaki fragments are produced on the lagging strand. When they understand the limitations of DNA polymerase, the why is clear. The how becomes more apparent as they work through the model. 4. Students may not understand how RNA primers on the lagging strand are removed by a second type of DNA polymerase. Again, the model helps explain this. Knowing that this removal is done by a second polymerase also helps students understand why these fragments need to be ligated together by the action of ligase. 5. Most students are able to understand which strand of replicating DNA is the leading strand and which is the lagging strand if they begin replication at one end of a DNA strand. It is more difficult for them to visualize this, however, if replication begins in the middle of the strand. It may be useful to have students look at what happens on one side of the replication bubble first and then look at the other side. Finally, they can combine the two images to get the overall picture.
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Answers Activity 16.1 Is the hereditary material DNA or protein? Accumulating and Analyzing the Evidence Build a concept map to review the evidence used to determine that DNA was the genetic material, the structure of DNA, and its mode of replicaton. Keep in mind that there are many ways to construct a concept map. • First, develop a separate concept map for each set of terms (A to D on the next page). Begin by writing each term on a separate sticky note or sheet of paper. • Then organize each set of terms into a map that indicates how the terms are associated or related. • Draw lines between the terms and add action phrases to the lines to indicate how the terms are related. Here is an example: DNA Nucleus
contains
Chromosomes
made of
Proteins
• After you have completed each of the individual concept maps, merge or interrelate the maps to show the overall logic used to conclude that DNA (not protein) is the hereditary material. • When you have completed the overall concept map, answer the questions.
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Terms Map A Griffith mice S strain of Streptococcus R strain of Streptococcus live heat-killed transformation Avery, McCarty, and MacLeod DNA protein
Map C Watson and Crick X-ray crystallography Chargaff’s rule purine structure pyrimidine structure H bonds phosphate sugar backbone Map D Meselson and Stahl conservative dispersive semiconservative nucleic acid bases 14 N 15 N bacteria density equilibrium centrifugation replication
Map B Hershey and Chase bacteria bacteriophage (phage) (only a protein and DNA) 35 S 32 P Waring blender high-velocity centrifugation
1. In the early to mid-1900s, there was considerable debate about whether protein or DNA was the hereditary material. a. For what reasons did many researchers assume that protein was the genetic material? Biologists understood that chromosomes segregated to opposite poles in mitosis and that the chromosome number was halved in meiosis. They also knew that chromosomes were made of both protein and DNA. Chemistry had revealed that proteins were made up of about 20 different amino acids. In contrast, DNA was composed of only four different nucleotides: adenine, thymine, guanine, and cytosine. Proteins were also known to have a more complex structure. As a result, the general feeling was that protein was more likely than DNA to be the genetic material.
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b. What key sets of experiments led to the c. What evidence did each experiment understanding that, in fact, DNA and provide? not protein was the hereditary material? Griffith experimented with a lethal S strain These experiments indicated that some of bacteria and a nonlethal R strain of the factor from the heat-killed S strain was same bacterial species. S strain injected able to transform live R into S bacteria. into mice killed them. R strain didn’t, and neither did heat-killed S strain. If heatkilled S and live R strains were mixed and injected simultaneously, however, the mice died. When autopsied, live S bacteria were found. Avery, McCarty, and MacLeod repeated Griffith’s experiments and confirmed his results. They then separated and purified the various components of the heat-killed S strain: carbohydrates, proteins, and DNA. They added each of the purified compounds separately to R bacteria. Only the DNA fraction caused R bacteria to transform to S bacterial. Hershey and Chase knew that T2 phage were made of only protein and DNA. They also knew that T2 phage could somehow cause bacteria to produce more T2 phage. Hershey and Chase grew two different cultures of phage, one on cells that contained amino acids labeled with radioactive sulfur, and the other on cells that contained nucleotides labeled with radioactive phosphorus. They used these viruses to infect unlabeled bacteria. They added 35S phage to bacteria, waited a few minutes, and then blended the mixture in a Waring blender. When they centrifuged the mixture, bacterial cells pelleted to the bottom of the tube. Any viruses or virus parts that were not attached to the cells remained in the supernate. In a second set of experiments, they did the same with the 32 P-labeled viruses and a new culture of the bacteria.
These experiments indicated that it was the DNA that caused the transformation and not the protein. Critics argued that the DNA fraction was not pure, however, that it contained some protein. They thought it was the associated protein and not the DNA that caused the transformation.
These experiments showed that only the 32 P- labeled viruses were associated with the bacterial cells that pelleted out in the centrifuge. If the researchers cultured these cells further, they could show that new viruses were produced. As a result, they demonstrated that it was the DNA in the viruses that was transferred to the bacteria and caused the production of new viruses. (There was arguably some protein contamination in these experiments, too, but by this time, enough evidence had accumulated that most were willing to accept this experiment as conclusively settling the argument—that is, showing that DNA is the hereditary material.)
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2. Watson and Crick were the first to correctly describe the structure of DNA. What evidence did they use to do this? How did they use this evidence to put together or propose the structure of DNA? Watson and Crick collected the following evidence: • Chargaff’s rule indicated that the amount of adenine in DNA was equal to the amount of thymine and that the amount of guanine was equal to the amount of cytosine. • The chemical structure of the four nucleotides was known. It was clear from their structure that adenine and thymine were each capable of forming two hydrogen bonds with other compounds and that guanine and cytosine were each capable of forming three hydrogen bonds. • Rosalind Franklin’s X-ray crystallography data allowed Watson and Crick to determine the width of the DNA molecule, the fact that it was helical, and the distance between turns on the helix. • Chemical analysis of DNA indicated that it also contained phosphate and deoxyribose sugar. Watson and Crick used this evidence to build scale models of the molecules that make up DNA—that is, of adenine, guanine, cytosine, and thymine, deoxyribose sugar, and phosphate groups. Using these models and what they knew to be the distance across the DNA molecule and the distance between turns of the helix, they pieced together a model that not only “fit” the evidence but also suggested a method of replication. (See pages 308–310 of Biology, 8th edition, for further details.) 3. How did the results of Meselson and Stahl’s experiments show that DNA replicates semiconservatively? To answer this, answer the following questions. a. Diagram the results that would be expected for each type of replication proposed. Meselson and Stahl grew bacteria for many generations in a medium containing heavy nitrogen (N15). The bacteria used the heavy nitrogen to make the nitrogenous bases of their DNA. The scientists isolated the DNA from some of these bacteria and centrifuged it on a density gradient. The DNA banded out in a single heavy-density layer. DNA from a different culture of bacteria grown on N14 medium only banded out in a single lighter-density layer. Meselson and Stahl then took some of the N15 bacteria, placed them in N14 medium, and allowed them to remain there for one DNA replication cycle. Next, they took some of these bacteria and allowed them to go through one more DNA replication cycle in N14 medium. See Figure 16.10 on page 312 in Biology, 8th edition, for the proposed outcomes of the experiment if replication was conservative versus dispersive versus semiconservative.
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b. What evidence allowed Meselson and Stahl to eliminate the conservative model? If the DNA replicated conservatively, after one replication cycle the two DNA molecules produced should have been the original DNA molecule and an entirely new DNA molecule. If the original DNA was labeled with N15 and the new molecule was labeled with N14, these should have banded out during centrifugation at two different levels (two different density bands), one light and one heavy. Instead the DNA was all found in a single intermediate-density band. This eliminated the conservative model as a possibility. c. What evidence allowed them to eliminate the dispersive model. If the dispersive model were correct, after the first division cycle they would expect to find one band of DNA at an intermediate density. At the end of the second replication cycle (in N14 medium), they would expect to see one band again at a density level a bit closer to that of the N14 DNA alone. Instead, after the second replication in N14 medium, they found the DNA in two bands—one that was at the N14 level and another that was intermediate between the N14 and N15 levels. This eliminated the dispersive model and supported the semiconservative model.
16.1 Test Your Understanding An E. coli cell that contains a single circular chromosome made of double-stranded DNA is allowed to replicate for many generations in 15N medium until all of the E. coli cells’ DNA is labeled with 15N. One E. coli cell is removed from the 15N medium and is placed into medium in which all of the available nucleotides are 14N labeled. The E. coli cell is allowed to replicate until eight E. coli are formed. 1. Given this situation, which of the following is true? a. Some 15N-labeled DNA will be found in all eight cells. b. Some 15N-labeled DNA will be found in only four of the cells. c. Some 15N-labeled DNA will be found in only two of the cells. d. Some 15N-labeled DNA will be found in only one of the cells. c is the correct answer. See the drawing on the next page.
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2. Draw the sequence of events that occurred to explain your answer.
15
N 15N
15
N 14N
14
N 15N
15
N 14N
14
N 14N
Note: Although it is drawn in linear format here, the DNA in E. coli would be circular.
Activity 16.2 How does DNA replicate? Working in groups of three or four, construct a dynamic (working or active) model of DNA replication. You may use the materials provided in class or devise your own.
Building the Model • Develop a model of a short segment of double-stranded DNA. • Include a key for your model that indicates what each component represents in the DNA molecule—for example, adenine, phosphate group, deoxyribose. • Create a dynamic (claymation-type) model of replication. Actively move the required bases, enzymes, and other components needed to model replication of your DNA segment. Your model should describe the roles and relationships of all the following enzymes and structures in replication: parental DNA nucleotide excision repair daughter DNA mutation antiparallel strands single-stranded DNAbinding proteins leading strand telomeres
lagging strand telomerase 5⬘ end 3⬘ end 3⬘ → 5⬘ versus 5⬘ → 3⬘ nitrogenous bases A, T, G, C replication fork replication bubble
Okazaki fragment DNA polymerase helicase DNA ligase primase RNA primers origin of replication
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Use your model to answer the questions. 1. Explain how Meselson and Stahl’s experiments support the idea that DNA replication is semiconservative. Three alternative models for replication of DNA were possible. See Figure 16.10 on page 312 for diagrams of the conservative, semiconservative, and dispersive models. Meselson and Stahl grew bacteria in a culture medium that contained nucleotides labeled with heavy nitrogen (one extra neutron added), or 15N. After many generations, the DNA in the bacteria was completely labeled with 15N nucleotides. They grew other bacteria in only 14N-labeled nucleotides. If they disrupted (broke open or lysed) the bacteria, they could extract the DNA. They could then layer the DNA on top of a CsCl gradient in a centrifuge tube. When they centrifuged this tube, the DNA settled out or layered at the density (in the CsCl solution) that was equal to its own density. When they followed this procedure with the 15N-labeled DNA, it settled out into a layer at a higher density than when they used the 14N-labeled DNA. Meselson and Stahl then removed some of the bacteria from this culture medium and placed them in a medium that contained only 14N-labeled nucleotides. In this culture medium, the bacteria were known to replicate every 20 minutes. They let the bacteria replicate two times in this medium. They extracted their DNA, layered the DNA on CsCl gradients in centrifuge tubes, centrifuged the DNA, and looked to see where it layered or settled out in the density gradient. Only if the semiconservative model was true would they find two layers of DNA, one at the 14N density level and one at a level intermediate between 14N and 15N. (Refer to Figure 16.11, page 312.) (Note: To develop a CsCl gradient in a centrifuge tube, a concentrated solution of CsCl is centrifuged for up to 24 hours in an ultracentrifuge. Today, different-colored beads of known density can be added to the solution. When a compound of unknown density is layered on top of this density gradient and then centrifuged, the location of the band relative to the beads allows researchers to easily identify the density of the compound.) 2. A new form of DNA is discovered that appears to be able to replicate itself both in the 3⬘ → 5⬘ direction and in the 5⬘ → 3⬘ direction. If this is true, how would this newly discovered DNA replication differ from DNA replication as we know it? No Okasaki fragments would be found in this new form of DNA replication. 3. Amazingly, an alien species of cellular organism is found alive in the remains of a meteorite that landed in the Mojave Desert. As a scientist, you are trying to determine whether this alien life-form uses DNA, protein, or some other type of compound as its hereditary material. a. What kinds of experiments would you propose to determine what the hereditary material is?
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There are a number of different ways of doing this. Here is one possibility: You could try to grow some of the organism’s cells in culture and then observe the cells. Do they have a nucleus and chromosomes? Do the chromosomes behave like chromosomes in eukaryotes on Earth? Can the chromosomes be observed to separate during cell divison? If not, what does separate to the daughter cells? Based on your findings, do some chemical analysis of the possible hereditary material to determine whether it is protein, DNA, or some other compound. b. Assuming that the hereditary material turns out to be similar to our DNA, describe the simplest experiments you could run to try to determine if it is double-stranded like our DNA, triple-stranded, or something else. Here is one possible way of doing this: You could do X-ray crystallography of the DNA from the organism to check the distance across the DNA molecule and the distance between turns of the helix (assuming it is a helix). You would also need to do an analysis of the relative amounts of adenine, thymine, guanine, and cytosine in the cell as well as the relative amounts of phosphate and deoxyribose. Then use the results of your analysis to build a model similar to that of Watson and Crick. 4. Some researchers estimate that the mutation rate for any given gene (or its DNA) in certain strains of bacteria is about 108. This means that one error or mutation in a given gene is introduced for every 100-million cell divisions. a. What can cause mistakes in replication? An incorrect or mismatched nucleotide can be added during replication. Alteratively, X-rays, UV light, or chemicals can modify the DNA so that errors are perpetuated during subsequent replication events. b. How are such mistakes normally corrected? DNA polymerase inserts new nucleotides during replication. It also checks or proofreads each new nucleotide against the template for mismatches. If a mismatch is found, DNA polymerase removes it and replaces it with a correct nucleotide. Errors introduced by X-rays and other factors are generally repaired by DNA repair enzymes in the cell. (For an example, see Figure 16.18, page 318.)
Activity 16.2
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Notes to Instructors Chapter 17 From Gene to Protein What is the focus of this activity? It is very difficult for many introductory biology students to sort out and visualize active processes at the molecular level. Although viewing animations of such processes can help some students, most need to build active models for themselves to discover what it is they understand and, more important, what they don’t. After modeling DNA replication in Activity 17.1, most students won’t have difficulty modeling transcription; however, modeling and understanding the process of translation are more challenging for many students.
What is this particular activity designed to do? Activity 17.1 Modeling transcription and translation: What processes produce RNA from DNA and protein from mRNA? This activity allows students to build a visual model of transcription in the nucleus and translation of the transcript in the cytoplasm. Building an active model gives students a better understanding of these dynamic processes.
What misconceptions or difficulties can this activity reveal? Activity 17.1 Students tend to encounter a number of difficulties, misconceptions, and missing conceptions as they model both transcription and translation. Here are several possible problems: 1. Most students are able to give the definition of a gene (for example, a gene is the region of DNA that produces an RNA molecule). However, many have difficulty defining a gene in terms of the bases in a template strand of DNA. 2. Many students want to know “how we know which strand is the template strand.” We often answer this by saying, “the one that makes the mRNA transcript.” However, students may not realize that this means that the only way we know which strand is the template strand is by comparing the base sequence of the transcript with that of each DNA strand.
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3. In translation, many students have difficulty understanding what occurs at each of the various sites (A, P, and E) on the ribosome. Modeling what happens as each codon moves first into the A site, then the P site, and so on helps them sort this out. 4. Many students don’t understand “where the amino acids in the cell come from” or “where they are in the cell.” It’s often useful to remind them that the food they eat is digested or broken down into its monomers, including amino acids. The amino acids enter the bloodstream and are picked up by cells. Actively metabolizing cells will contain amino acids in their cytoplasm. 5. Students often ask how the various aminoacyl tRNA synthetases are able to find the correct amino acids and tRNA molecules. Again, it is useful to remind them that aminoacyl tRNA synthetases are enzymes. Ask them what properties enzymes have. This will help them understand that each is specific for a particular type (and therefore 3-D shape) of amino acid and tRNA. However, many students will still need to be reminded that interactions between these enzymes and their amino acid and tRNA molecules occur as a result of random interactions within the cell. 6. The modeling activity helps students to better understand why a stop codon actually stops polypeptide formation and allows the mRNA and ribosome to dissociate.
Answers Activity 17.1 Modeling transcription and translation: What processes produce RNA from DNA and protein from mRNA? Create a model of the processes of transcription and translation. Your model should be a dynamic (working or active) representation of the events that occur first in transcription in the nucleus and then in translation in the cytoplasm. For the purposes of this activity, assume there are no introns in the mRNA transcript. When developing and explaining your model, be sure to include definitions or descriptions of the following terms and structures:
Activity 17.1
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gene DNA nucleotides: A, T, G, and C versus A, U, G, and C RNA modification(s) after transcription mRNA RNA polymerase poly(A) tail 5⬘ cap translation protein synthesis ribosome (large versus small subunit) A, P, and E sites tRNA rRNA
start codon (methionine) aminoacyl-tRNA synthetase amino acids (see Figure 17.4, page 329, in Biology, 8th edition) peptidyl transferase polypeptide energy codons stop codons anticodons initiation elongation termination polypeptide
Building the Model • Use chalk on a tabletop or a marker on a large sheet of paper to draw a cell’s plasma membrane and nuclear membrane. The nucleus should have a diameter of about 12 inches. • Draw a DNA molecule in the nucleus that contains the following DNA sequence: Template strand 3⬘ TAC TTT AAA GCG ATT 5⬘ Nontemplate strand 5⬘ ATG AAA TTT CGC TAA 3⬘ • Use playdough or cutout pieces of paper to represent the various enzymes, ribosome subunits, amino acids, and other components. • Use the pieces you assembled to build a dynamic (claymation-type) model of the processes of transcription and translation. • When you feel you have developed a good working model, use it to explain the processes of transcription and translation to another student or to your instructor. Use your model of transcription and translation to answer the questions. 1. How would you need to modify your model to include intron removal? Your explanation should contain definitions or descriptions of the following terms and structures: pre-mRNA exons RNA splicing spliceosome introns 116
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To answer this question, review Figures 17.10 and 17.11 and the accompanying text on pages 334–336 of Biology, 8th edition. 2. If 20% of the DNA in a guinea pig cell is adenine, what percentage is cytosine? Explain your answer. If 20% is adenine, then 20% is thymine. The remaining 60% is composed of cytosine and guanine in equal percentages so that each makes up 30% of the DNA. 3. A number of different types of RNA exist in prokaryotic and eukaryotic cells. List the three main types of RNA involved in transcription and translation. Answer the questions to complete the chart. a. Types of RNA
b. Where are they produced? In the nucleus, from specific genes (often called structural genes) on the DNA
mRNA
tRNA
Other genes in the nuclear DNA code for tRNA molecules
rRNA
Still other genes in the nuclear DNA code for rRNA molecules.
c. Where and how do they function in cells? mRNA functions in the cytoplasm, where it is translated into protein. The mRNA carries the information in codons that determine the order of amino acids in a protein. tRNA molecules function in the cytoplasm in translation Each tRNA molecule can combine with a specific amino acid. Complementary base pairing of tRNA molecule with a codon in the A site of the ribosome brings the correct amino acid into position in the growing polypeptide chain. rRNA molecules combine with protein to form the ribosomes, which serve as the base for interactions between mRNA codons and tRNA anticodons in traslation in the cytoplasm. (See Figure 17.18, page 324)
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4. Given your understanding of transcription and translation, fill in the blanks below and indicate the 5⬘ and 3⬘ ends of each nucleotide sequence. Again, assume no RNA processing occurs. Nontemplate strand of DNA:
5⬘ A T G T A T G C C A A T G C A 3⬘
Template strand of DNA:
–⬘ T – – – – – – – – – – – – – – –⬘
mRNA:
–⬘ A – – – – U – – – – – – – – – –⬘
Anticodons on complementary tRNA:
–⬘ – – – / – – – / – – – / – – – / – – – / –⬘
Template strand of DNA:
3⬘ T A C A T A C G G T T A C G T 5⬘
mRNA: tRNA:
5⬘ A U G U A U G C C A A U G C A 3⬘ 3⬘ U A C / A U A / C G G / U U A / C G U / 5⬘
5. Scientists struggled to understand how four bases could code for 20 different amino acids. If one base coded for one amino acid, the cell could produce only four different kinds of amino acids (41). If two bases coded for each amino acid, there would be four possible choices (of nucleotides) for the first base and four possible choices for the second base. This would produce 42 or 16 possible amino acids. a. What is the maximum number of three-letter codons that can be produced using only four different nucleotide bases in DNA? 43, or 64 b. How many different codons could be produced if the codons were four bases long? 44, or 256 Mathematical logic indicates that at least three bases must code for each amino acid. This led scientists to ask: • How can we determine whether this is true? • Which combinations of bases code for each of the amino acids? To answer these questions, scientists manufactured different artificial mRNA strands. When placed in appropriate conditions, the strands could be used to produce polypeptides. 118
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Assume a scientist makes three artificial mRNA strands: (x) 5⬘ AAAAAAAAAAAAAAAAAAAAAAAAAA 3⬘ (y) 5⬘ AAACCCAAACCCAAACCCAAACCCAAA 3⬘ (z) 5⬘ AUAUAUAUAUAUAUAUAUAUAUAUAU 3⬘ When he analyzes the polypeptides produced, he finds that: x produces a polypeptide composed entirely of lysine. y produces a polypeptide that is 50% phenylalanine and 50% proline. z produces a polypeptide that is 50% isoleucine and 50% tyrosine. c. Do these results support the three-bases-per-codon or the four-bases-per-codon hypothesis? Explain. Only if there were three bases per codon would both y and z produce only two different kinds of amino acids in equal proportions. In fact, each strand would produce the two in alternating order; for example, the z strand would produce a polypeptide chain of isoleucine followed by tyrosine followed by isoleucine, then tyrosine, and so on. d. This type of experiment was used to discover the mRNA nucleotide codons for each of the 20 amino acids. If you were doing these experiments, what sequences would you try next? Explain your logic. There are many possible ways to answer this question. One possibility follows: Continue as above and make the remaining three types of mRNA made up of only one type of nucleotide—that is, poly G, poly U, or poly C. Then make all possible combinations of the nucleotides taken two at a time—for example, GCGC, CGCG, AGAG, and so on. Next, make other combinations of nucleotides taken three at a time—for example, AAAGGGAAAGGG and so on. Continue with combinations of nucleotides taken four at a time—for example, AAAAUUUUAAAAUUUUAAAAUUUU and so on. In this last example, if the codon for each amino acid is three bases long, these combinations of nucleotides should give you a maximum of three different types of amino acids in equal proportions or percentages. 6. Now that the complete genetic code has been determined, you can use the strand of DNA shown here and the codon chart in Figure 17.4 on page 329 in Biology, 8th edition, to answer the next questions. Original template strand of DNA: 3⬘ TAC GCA AGC AAT ACC GAC GAA 5⬘ a. If this DNA strand produces an mRNA, what does the sequence of the mRNA read from 5⬘ to 3⬘? mRNA ⫽ 5⬘ AUG CGU UCG UUA UGG CUG CUU 3⬘ Activity 17.1
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b. For what sequence of amino acids does this mRNA code? (Assume it does not contain introns.) Sequence of amino acids: methionine-arginine-serine-leucine-tryptophanleucine-leucine c. The chart lists five point mutations that may occur in the original strand of DNA. What happens to the amino acid sequence or protein produced as a result of each mutation? (Note: Position 1 refers to the first base at the 3⬘ end of the transcribed strand. The last base in the DNA strand, at the 5⬘ end, is at position 21.) Original template strand: 3⬘ TAC GCA AGC AAT ACC GAC GAA 5⬘
i.
Mutation
Effect on amino acid sequence
Substitution of T for G at position 8.
This changes the codon in mRNA to a stop codon; translation stops at this point. A shorter (truncated) polypeptide is produced and this shortened polypeptide is likely to be nonfunctional.
ii. Addition of T between positions 8 and 9.
Serine is still incorporated as the third amino acid, but the amino acids that follow all differ from the sequence in part b above. This is a frameshift mutation.
iii. Deletion of C at position 15.
The first four amino acids in the chain are not affected. The fifth amino acid becomes cysteine and the subsequent amino acids are also changed from part b.
iv. Substitution of T for C at position 18.
The original mRNA codon, CUG, and the one resulting from the substitution, CUA, both code for leucine, so no change occurs in the polypeptide sequence.
v.
Leucine is still inserted as the sixth amino acid in the polypeptide. However, since we’re given only a part of the sequence, it is uncertain what the next amino acid in the chain will be.
Deletion of C at position 18.
vi. Which of the mutations produces the greatest change in the amino acid sequence of the polypeptide coded for by this 21-base-pair gene? The addition of T between positions 8 and 9 still leaves the third amino acid intact; however, all amino acids after that are different. In the substitution of T for G at position 8, a stop codon is inserted and only the first two amino acids are unaltered. As a result, this mutation produces the greatest change. 120
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7. Sickle-cell disease is caused by a single base substitution in the gene for the beta subunit of hemoglobin. This base substitution changes one of the amino acids in the hemoglobin molecule from glutamic acid to valine. Look up the structures of glutamic acid (glu) and valine (val) on page 79 of Biology, 8th edition. What kinds of changes in protein structure might result from this substitution? Explain. Glutamic acid is polar, and valine is nonpolar. Being polar, the glutamic acid molecule would have been able to interact with water and other polar molecules, but the valine molecule cannot. As a result, unlike glutamic acid, valine is more likely to have an interior position in the hemoglobin molecule. 8. Why do dentists and physicians cover patients with lead aprons when they take mouth or other X-rays? As noted in this and other chapters, X-rays, UV light, and many chemicals can damage DNA. Such damage can result in point mutations such as base substitutions, deletions, and insertions. These mutations can cause cancer. If they occur in the cells that will produce the gametes, the mutations can be passed on to offspring. As a result, lead aprons are used to shield the rest of your body from any stray radiation.
17.1 Test Your Understanding During DNA replication, which of the following would you expect to be true? Explain your answers. T/F 1. More ligase would be associated with the lagging strand than with the leading strand. True—Okazaki fragments on the lagging strand would need to be connected by ligase. T/F 2. More primase would be used for the lagging strand than for the leading strand. True—Each new Okazaki fragment would be started by addition of a short RNA sequence by primase. T/F 3. More helicase would be associated with the lagging strand than with the leading strand. False—Both strands would need to be unwound equally. T/F 4. DNA ligase links the 3⬘ end of one Okazaki fragment to the 5⬘ end of the another Okazaki fragment in the lagging strand. True—The end of the first Okazaki fragment would be 3⬘; the next fragment would begin at the 5⬘. T/F 5. In the lagging strand, the enzyme DNA polymerase III which produces the next Okazaki fragment also removes the short segment of primer RNA on the previous Okazaki fragment. False—A different DNA polymerase removes the primer. Activity 17.1
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6. You obtain a sample of double-stranded DNA and transcribe mRNA from this DNA. You then analyze the base composition of each of the two DNA strands and the one mRNA strand, and get the following results. The numbers indicate percentage of each base in the strand: A
G
C
T
U
strand 1
40.1
28.9
9.9
0.0
21.1
strand 2
21.5
9.5
29.9
39.1
0.0
strand 3
40.0
29.0
9.7
21.3
0.0
a. Which of these strands must be the mRNA? Explain. Strand 1 must be the mRNA because it contains U (uracil), which is found only in RNA. b. Which one is the template strand for the mRNA? Explain. Strand 2 is the complement to strand 1. Therefore, it must be the template strand for the mRNA. 7. In a new experiment, you obtain a different sample of double-stranded DNA and transcribe mRNA from this DNA. You then analyze the base composition of each of the two DNA strands and the one mRNA strand, and get the following results. The numbers indicate percentage of each base in the strand: A
G
C
T
U
strand 1
29.1
39.9
31.0
0.0
0.0
strand 2
0.0
30.0
39.8
30.2
0.0
strand 3
29.4
39.4
31.2
0.0
0.0
a. Which of these strands could be the mRNA? Explain. Either strand 1 or 3 could be the mRNA. b. Which one must be the template strand for the mRNA? Explain. Strand 2 must be DNA and it must also be the template strand for the mRNA because it is complementary to both strands 1 and 3.
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8. Cystic fibrosis transmembrane conductance regulator (CFTR) proteins function in cell membranes to allow chloride ions across cell membranes. Individuals with cystic fibrosis (CF) have abnormal CFTR proteins that do not allow Cl⫺ to move across cell membranes. Chloride channels are essential to maintain osmotic balance inside cells. Without properly functioning Cl⫺ channels, water builds up inside the cell. One result is a thickening of mucous in lungs and air passages. You are doing research on a different disease, and you hypothesize that it may also be due to a defect in an ion channel in the cell membrane. a. Diagram or model production of a normal membrane ion channel. b. Based on you understanding of cell membrane structure and function, propose at least three different alterations that could result in a nonfunctional or missing ion channel. c. What questions would you need to answer to determine which of these may be correct? This question is designed to allow students to integrate their understanding of general cell structure and function with that of genetic mutation and control. a. Briefly, the diagram should include: i. DNA/gene(s) for the channel protein ii. RNA production and processing, including the target sequence on the gene to get it to the rER iii. Properly functioning Golgi that • finalizes protein structure and • functions to move the protein to the membrane, e.g., via targeted vesicles b. Alterations that could result in a nonfunctional or missing channel could include: i. Error/mutation in the DNA that results in faulty or nonexistent/terminated protein ii. Error in target sequence iii. Error in function of Golgi and/or vesicle transport to membrane c. Questions to be answered could include: i. Is the DNA sequence of the gene altered? ii. Is there an error in the splicing of the RNA or its targeting mechanism? iii. Is there an error in the translation of the RNA? iv. Are the Golgi functioning correctly • in final folding, etc., of the protein channel, and • in delivery of the protein to the membrane?
Activity 17.1
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Notes to Instructors Chapter 18 The Genetics of Viruses and Bacteria What is the focus of these activities? Many students try to memorize these systems instead of understanding the logic associated with their operation. These activities can be used to point out that it is much easier to understand how these systems work if students look at the logic behind their operation. Having students build models/diagrams of the systems and their operation is very useful in this regard.
What are the particular activities designed to do? Activity 18.1 How is gene expression controlled in bacteria? Activity 18.2 Modeling the lac and trp operon systems: How can gene expression be controlled in prokaryotes? Activity 18.1 provides students with a mechanism to sort out similarities and differences in the control of gene expression for inducible versus repressible operons. Activity 18.2 asks students to develop a visual model of how gene expression can be controlled in prokaryotes. Both activities are designed to help students develop a better understanding of how bacteria control gene activity. In particular, they should help clarify both what operons are and how operons act to control gene expression. Activity 18.3 How is gene activity controlled in eukaryotes? Activity 18.3 asks students to integrate information from both Chapter 18 and earlier chapters to develop an understanding of the various levels (or mechanisms for) controlling gene expression in eukaryotes. For example, by this time, students understand the DNA → RNA → protein sequence. However, they may not have considered how many different types of control can exist in this general pathway or which of the available types of control allow for fast versus slow responses by the organism. Activity 18.4 What controls the cell cycle? Activity 18.4 asks students to review the major controls or checkpoints that regulate cell division. Cancer, uncontrolled cell division, can result if these controls fail.
What misconceptions or difficulties can these activities reveal? Activity 18.1 Activity 18.2 Students tend to encounter a number of difficulties, misconceptions, and missing conceptions in their understanding of what operons are and how they function to control gene expression. We present three possible problems. 124
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1. Many students try to memorize these systems. Modeling is used to help students develop and understand the logic associated with the systems’ operation. As students work on the chart comparing the two types of operons and as they model the system, it is helpful to ask questions that focus on the logic. For example, if specific enzymes function to break down compound X, is it energetically efficient (or logical from an energy standpoint) for the cell to produce the enzymes when compound X is not present? Is it energetically efficient for the cell to produce glucose by hydrolyzing lactose if a supply of glucose is already present in the cell? Bacteria are capable of producing the amino acids they need from inorganic precursors; for example, E. coli produce tryptophan as the end product of a series of five reactions. This requires considerable energy expenditure. What would happen if the bacteria came upon a source of ready-made tryptophan in their environment? What type of control would allow the cell to simultaneously shut down production of all the enzymes required for tryptophan synthesis? 2. Students often have difficulty understanding why there are two levels of control for the lac operon. Again, it helps to direct students to look at these from an energetics viewpoint. As noted, if lactose is not present, there is no need for the cell to produce the enzymes for its digestion. However, is it energetically efficient for the cell to produce the enzymes for lactose digestion when glucose is readily available in the cell? One level of control determines whether or not the genes for lactose digestion can be transcribed. Another level of control (the cAMP-CRP complex) determines the binding efficiency of the RNA polymerase to the promoter site. 3. Many students don’t understand why cAMP levels vary relative to the level of glucose in the cell. To understand this, they need to realize that when glucose levels are high, the rate of metabolism and cell respiration in the cell is also likely to be high. As a result, the cell’s supply of AMP is converted to ADP and the ADP is converted to ATP. Under these conditions, the ratio of AMP to ATP in the cell is low and the amount of cAMP is correspondingly low. When glucose supplies run low, the cell uses the energy stored in its ATP by converting it to ADP. To gain additional energy, the ADP is converted to AMP. The ratio of AMP to ATP increases, and the level of cAMP in the cell also increases. Activity 18.3 Before you assign this activity, it is a good idea to remind students that they already know the DNA → RNA → protein pathway. Given this understanding and their knowledge of how DNA, RNA, and protein function in the cell, you can ask them to work in small groups. Their assignment is to use simple logic (don’t look in the textbook) to answer these questions: • Under what conditions would a cell find it necessary to control expression of its genes? • Which of these types of control would occur over the long term and which would have to occur quickly (in the short term)? Notes to Instructors
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Collect ideas from the students and write them on the board. Then ask the following question: • Given the answers to the previous questions, what types of control might exist in cells to deal with the longer-term versus shorter-term responses? Doing this exercise before assigning Activity 19.1 serves two purposes: • It teaches students the types of questions they should be asking themselves as they study. • It helps students understand how to put what they learn into a logical context that makes learning (and remembering what they learn) easier. Activity 18.4 Many students are unaware of how complex the cell cycle controls are. It is often counterintuitive to many that some of the controls actually function to disable or kill cells with damaged control systems.
Answers Activity 18.1 How is gene expression controlled in bacteria? Fill in the chart to organize what we know about the lac and trp operons. Operon: Is the metabolic pathway anabolic or catabolic? What regulatory genes are associated with the operon and what functions does each serve?
lac
trp
Catabolic Hydrolyzes or breaks down lactose(a disaccharide) into glucose and galactose (two sixcarbon surgars)
Anabolic Synthesizes tryptophan from precursors
Genes: lacI promotor, CRP binding site operator
Genes: trpR promoter, operator
Functions: lacI produces an active repressor protein that binds to the operator in the absence of lactose. The promotor is the site where RNA polymerase binds to the DNA. Binding of RNA polymerase is enhanced when cAMP interacts with
Functions: trpR produces an inactive repressor protein that binds to the operator only when complexed with excess tryptophan. RNA polymerase binds at the promotor site and transcribes the genes for making tryptophan as long as tryptophan is not is excess.
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lac CRP and the complex bind at the CRP site on the promotor region. As a result, the enzymes for lactose digestion are produced only when needed and only when glucose is not present.
trp
What structural genes are included in each operon and what does each produce?
Genes: lacZ lacy lacA
Products: LacZ codes for the -galactosidase enzyme, which digest lactose into glucose and galactose. LacY codes for a permease to allow lactose to enter the cell. LacA codes for transacetylase. Its function is unclear.
Genes: tryA tryB tryC tryD tryE
Products: These five genes code for the five enzymes required to convert a precursor molecule to tryptophan (one of the amino acids required for protein synthesis)
Is the operon inducible or repressible?
The lac operon is inducible.
The trp operon is repressible.
Is the repressor protein produced in active or inactive from?
The repressor protein is produced in its active form. (The active form binds to the operator and stops transcription of the structural genes)
The repressor protein is produced in its inactive form.
The repressor protein becomes active when it interacts with:
The repressor is active until it complexes with lactose (or allolactose). Then it becomes inactive.
The repressor is inactive until it complexes with excess tryptophan in the cell. Tryptophan changes the configuration of the repressor, and it is capable to binding to the operator and stopping transcription of the enzymes that synthesize or make tryptophan.
Activity 18.1
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Activity 18.2 Modeling the lac and trp operon systems: How can gene expression be controlled in prokaryotes? Using the information in Activity 18.1 and in Chapter 18 of Biology, 8th edition, construct a model or diagram of the normal operation of both the lac and trp operon systems. In your models or diagrams, be sure to include these considerations: regulatory and structural genes inducible versus repressible control anabolic versus catabolic enzyme activity negative versus positive controls Use your model to answer the questions. 1. Under what circumstances would the lac operon be “on” versus “off”? The trp operon? The lac operon would be off when there is no lactose in the cell. The lac operon would be on when lactose is present and there is little or no glucose in the cell. However, the lac operon would be off (or operating at very low levels) even when lactose is present if sufficient glucose is simultaneously present. The trp operon would be off when excess tryptophan is readily available to the cell. It would be on at all other times. 2. How are the lac and trp operons similar (in structure, function, or both)? Both have regulatory genes that produce repressor proteins that can interact with the operator and shut down transcription of the structural genes. 3. What are the key differences between the lac and trp operons? The lac operon is inducible; the presence of lactose induces production of the enzymes needed for lactose digestion. The trp operon is repressible; it is ordinarily on, producing tryptophan, which is needed for protein production by the cells. It is turned off or repressed only when an excess of tryptophan is available to the cell. The lac operon is controlled by both a regulatory protein, which interacts with the operator and blocks RNA polymerase action, and a CRP site. RNA polymerase does not attach effectively to the operator unless CRP (complexed with cAMP) is attached at the CRP site. Once attached, it enhances the interaction of RNA polymerase with the promoter region. cAMP levels in cells tend to be low when glucose is present. As a result, even if lactose is present at relatively high levels, this second control keeps production of the enzymes for digesting lactose at very low levels if glucose is also present in the cells. 128
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4. What advantages are gained by having genes organized into operons? As they are needed, both systems are set up to simultaneously turn on (or off) all of the genes required in a metabolic pathway. This is much more efficient than having each gene under independent control. 5. Strain X of E. coli contains a mutated lac regulatory gene on its bacterial genome. As a result, the gene produces a nonfunctional lac repressor protein. You add a plasmid (an extra circular piece of double-stranded DNA) to these cells. The plasmid contains a normal regulatory gene and a normal lac operon. Build a model or diagram of what one of these modified E. coli cells would look like. Then answer the questions and use your model or diagram to explain your answers. a. Before the addition of the plasmid, would the E. coli strain X cells be able to produce the enzymes for lactose digestion? Explain. Yes. The lacI gene ordinarily produces an active repressor protein that inhibits production of the genes for lactose digestion. In this case, this gene is mutated so that it cannot produce the repressor protein. b. After the addition of the plasmid, would the plasmid’s lac operon produce the enzymes for lactose digestion constitutively (all the time) or only when lactose was the available sugar source? Explain. The plasmid contains a normal regulatory gene and a normal lac operon. As a result, the plasmid’s lac operon should produce the genes for lactose digestion only when lactose was the available sugar source. c. After the addition of the plasmid, would the bacterial genome’s lac operon produce the enzymes for lactose digestion constitutively or only when lactose was the available energy source? Explain. The regulatory gene on the plasmid could produce enough repressor protein molecules to affect both the plasmid operon and the bacterial chromosome’s operon. So, after addition of the plasmid, the bacterial genome’s lac operon would produce enzymes for lactose digestion only when lactose was the available energy source. d. If equal amounts of lactose and glucose were present in the cell, would the lac operon in the bacterial DNA be off or on? Would the lac operon on the introduced plasmid be off or on? Explain. In combination with cAMP, CAP (catabolic activator protein) is an activator of transcription for the lac operon. When glucose is present, cAMP levels in the cell tend to be low. As a result, cAMP does not interact with CAP and CAP is unable to bind to the DNA. Therefore, even if lactose is present, little lac mRNA will be synthesized from either the E. coli or plasmid DNA. Activity 18.2
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Activity 18.3 How is gene activity controlled in eukaryotes? Human genes cannot all be active at the same time. If they were, all the cells in our bodies would look the same and have the same function(s). For specialization to occur, some genes or gene products must be active while others are turned off or inactive. 1. In eukaryotes, gene expression or gene product expression can be controlled at several different levels. Indicate what types of control might occur at each level of gene or gene product expression. Note: The table presents a representative sample of the various types of control mechanisms. It does not include all possible mechanisms of control. Level
Types of control
a. The gene or DNA itself
Whole chromosomes or parts of chromosomes can be inactivated by DNA packing as heterochromatin. Transcription from DNA can also be controlled by enhancing or inhibiting the action of various proteins that interact with the DNA and/or RNA polymerase to make it more or less likely to produce an mRNA transcript.
b. The mRNA product of the gene
At the mRNA level, controls can affect whether the pre-mRNA is processed into mRNA, how long the mRNA is active in translation, and whether the mRNA is capable of attaching to the ribosome and being translated.
c. The protein product of the mRNA
The protein product may be made in an inactive form that becomes active only in specific environments. For example, digestive enzymes are usually produced in inactive form and activated only under specific conditions of pH or in the presence of specific activator molecules. If produced in active form, the protein may be inactivated by the presence of a molecule that changes its allosteric form. The protein product may have a very limited life span and, as a result, have a limited time of functioning in the cell.
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18.3 Test Your Understanding Single-celled organisms such as Amoeba and Paramecia often live in environments that change quickly. Which of the following types of control allow organisms like Amoeba to respond most quickly to frequent short-term environmental changes? Explain your reasoning. a. b. c. d.
Control of mRNA transcription from DNA Control of enzyme concentration by controlling the rate of mRNA translation Control of activity of existing enzymes Control of the amount of DNA present in the cell
The answers to this question may vary depending on how frequently the changes occur and how one defines “short-term.” For example, overall, controlling the activity of existing enzymes allows the organism to respond very quickly to short-term changes in environmental conditions. However, maintaining a number of different enzymes in the cell is energetically efficient only if the short-term changes occur frequently over time. If the changes occur infrequently and do not result in the death of the organism but rather reduced function, then it is energetically more efficient to instead control the rate of mRNA transcription from DNA.
Activity 18.4 What controls the cell cycle? 1. Checkpoints in the normal cell cycle prevent cells from going through division if problems occur—for example, if the DNA is damaged. a. What forms do the checkpoints take? That is, how do they control whether or not cell division occurs? A number of different cyclins are produced during interphase. Three of the major checkpoints in mitosis—the G1, G2, and M phase checkpoints—appear to be controlled by Cdks (cyclin-dependent kinases). The ratio of Cdks to their cyclins determines their activity. The activities of the different activated Cdks control the cell cycle. The cell can move past a given stage of the cycle only when the appropriate checkpoint’s Cdk has been activated.
b. In the space below, develop a handout or diagram to explain how these checkpoints work under normal conditions. Your diagram should include a description of each checkpoint, where it acts in the cell cycle, and what each does to control cell division. Refer to Figure 12.17, page 240, of Biology, 8th edition, which shows the molecular mechanisms associated with activity of the cyclin-Cdk complex called Activity 18.4
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MPF. Here is an easy way to repesent the G1 and G2 control points and their actions: Following Figure 12.17, draw the inner circle of arrows, with a diameter of about 3 inches, on a piece of paper. On another piece of paper, draw a circle about 4 inches in diameter, and around this draw another circle about 6 inches in diameter. Draw and label the lines that show the activity of cyclins versus Cdk (Figure 12.17) between these two lines. Cut out the center of this double circle. Place the circle over the cell cycle arrow diagram you drew first to create the same diagram you see in Figure 12.17. Next, substitute the words “G1 checkpoint” for MPF on the drawing. Now rotate the top sheet of paper so that the G1 checkpoint lines up with its position in G1 on the arrow diagram. Do the same for the G2 checkpoint. c. Cancer results from uncontrolled cell division. Explain how mutations in one or more of the checkpoints might lead to cancer. The basic function of each checkpoint is to determine whether or not the cell is functioning normally and should enter into or continue through division. If these checkpoints break down, they could allow any cell to continue through the cell cycle—that is, to divide. Cancer is uncontrolled cell division, so cancer can result from a breakdown of the operation of these checkpoints.
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Notes to Instructors Chapter 19 Viruses What is the focus of these activities? Although almost all students recognize that viruses exist and are pathogenic organisms, few have a good understanding of how viruses cause disease and why they are so difficult to treat.
What are the particular activities designed to do? Activity 19.1 How do viruses, viroids, and prions affect host cells? Activity 19.1 provides students with a mechanism to sort out what viruses are and how they affect host cells. Given an understanding of various pathogen characteristics and replication requirements, students are asked to consider what aspects of their structures or life cycles could be interrupted and therefore serve as effective treatments against infection.
What misconceptions or difficulties can these activities reveal? Activity 19.1 Like most people, many students do not understand that antibiotics are designed to treat bacterial infections. Antibiotics do this by attacking bacteria-specific structures or functions—e.g., 70s ribosomes, bacterial cell wall structures, or bacteria-specific enzymes. As a result, antibiotics (antibacterials) are not effective against viruses.
Answers Activity 19.1 How do viruses, viroids, and prions affect host cells? 1. By definition, viruses are obligate intracellular parasites. What does this mean? An obligate intracellular parasite is one that can neither survive long term nor reproduce outside its host. 2. In general, how are viruses classified? Viruses are often classified based on the type of nucleic acid they contain—e.g., ss or ds RNA or ss or ds DNA. See Chapter 19 of Biology, 8th edition.
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3. What is reverse transcriptase? Reverse transcriptase is the enzyme that allows retroviruses to make DNA from ⫺RNA strands and then + and ⫺RNA from the DNA. a. Where was reverse transcriptase first found? It was discovered in retroviruses. b. Of what use is reverse transcriptase to viruses? Of what use is it to scientists? It is useful in genetic recombination to make cDNA from mRNA and avoid the problem of eucaryotic introns. Reverse transcriptase allows viruses that invade eukayotic cells to make DNA from their RNA genome and use that DNA to make more viruses or to incorporate into the host’s DNA. 4. Why must viruses invade other cells to reproduce? a. Describe the general process. Include a discussion of lytic and lysogenic viral cycles. Some viruses are only lytic. They enter the cell and take over the cell’s machinery to make more viruses. Then they lyse the cell to allow the new viruses to escape and attack other cells. Some viruses also have a lysogenic stage and alternate between lysogenic and lytic stages throughout their life cycle. In the lysogenic stage, the virus incorporates the viral DNA into the host DNA and remains quiescent there, replicating with the host’s DNA. It can later come out of the host’s DNA and enter a lytic cycle. Environmental conditions that stress the host cell may trigger this. b. Which types of viruses are more likely to have a lysogenic phase? A number of double-stranded DNA viruses and retroviruses that can produce double stranded DNA can become lysogenic (incorporate into the host DNA). RNA viruses that do not produce DNA as templates for their reproduction are unable to incorporate into the host DNA. 5. More than 100 different viruses can cause the “common cold” in humans. Many of these are rhinoviruses. Other viruses—influenza viruses—cause the flu. While there are many different antibiotics for treating bacterial infections, there are relatively few drugs available to treat viral infections. Explain. Antibiotics used to counteract bacterial infections target bacteria-specific structures or functions such as 70s ribosomes, bacterial cell wall structures, or bacteria-specific enzymes. It is more difficult to find something that a virus requires for its function and replication that is not also required by the host for its normal metabolic functions. When a virus-specific coat protein or enzyme, such as reverse transcriptase, is found, high rates of mutation often make it difficult to develop a drug or vaccine against it. (For example, reverse transcriptase has an error rate of about 1/8000 bases where 134
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other DNA polymerase and RNA polymerase error rates are about 1/million or so. In addition, reverse transcriptase has no proofreading capacity.) 6. a. How do viruses, viroids and prions differ in terms of both composition and function? Composition: Viruses ⫽ nucleic acid core with protein coat or capsid. Viroid ⫽ bare RNA. Prion ⫽ altered 3-D structure of normally active protein. Function: Viruses take over cell machinery and ultimately kill the cells they invade. Viroids can do the same in plants. Prions accumulate in brain tissue and destroy it. The complete mechanisms of prion replication and action are still unknown. b. Compared to viral infections, do these differences make it easier or harder to treat viroid and prion infections? Explain your reasoning. Because viroids are basically bare RNA and prions are altered proteins, there are no easy ways of attacking them without simultaneously attacking the native RNAs and proteins required for normal cell function. As a result, these pathogens are even more difficult than viruses to treat.
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Notes to Instructors Chapter 20 Biotechnology What is the focus of this activity? Almost all students have heard of DNA technology and know that DNA typing (or profiling or fingerprinting) can be used in forensic analysis. Yet, fewer have a good understanding of the methods that are used, why these methods are used (that is, what they allow us to do), and what types of results can be expected.
What is the particular activity designed to do? Activity 20.1 How and why are genes cloned into recombinant DNA vectors? This activity is designed to help students understand: • how genes from other species can be cloned by incorporating them into bacterial plasmids, and • how such recombinant bacteria can be used to produce a wide range of products, including human insulin. Activity 20.2 How can PCR be used to amplify specific genes? This activity is designed to help students understand: • how using known primers and PCR allows us to amplify only targeted genes, and • how DNA fingerprinting using RFLPs or STRs can be applied in forensics.
What misconceptions or difficulties can this activity reveal? Activity 20.1 Students tend to encounter a number of difficulties, misconceptions, and missing conceptions that become evident as they try to build a model to demonstrate how genes can be cloned by incorporation into bacterial plasmids. Here are three possible problems: 1. The news media can help educate the public, but in some cases, it can also cause confusion. Some students have been “convinced” by what they read that gene cloning and recombinant technology are “bad” things with no redeeming value. To help counter this idea, this activity has students investigate the production of human insulin (Humulin) by bacteria. 2. If it hasn’t surfaced before this time, you may notice that some students have the idea that the DNA in bacteria is single-stranded. This misconception often comes from their misunderstanding or misinterpreting statements like: “The bacterial chromosome is composed of a single circular DNA molecule.”
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3. Many students don’t understand how restriction enzymes “cut” DNA. It is helpful to mention that the “cuts” are in the phosphate sugar backbone of the DNA molecule. These cuts are similar to the gaps left between Okazaki fragments. When the phosphate sugar backbone is cut, the hydrogen bonds between complementary bases are the only forces holding the structure together. Changes in the physiological conditions of the medium, including heating the solution, can cause these hydrogen bonds between complementary bases to break and produce the “sticky ends” required to interact with the complementary “sticky ends” of the gene of interest. Activity 20.2 Though the overall process of PCR is not conceptually complex, it is difficult to visualize without modeling or diagramming it. This exercise asks student to diagram what happens in several rounds or cycles of PCR. Having students do this will help them understand the whole process and how known primers can be used to specifically target genes for amplification.
Answers Activity 20.1 How and why are genes cloned into recombinant DNA vectors? Develop a model to explain how a human gene can be cloned into a bacterial plasmid. Your model should be a dynamic (working or active) representation of the events that need to occur in order to • clone the insulin gene into a bacterial plasmid, and • transform the plasmid into E. coli. When you develop and explain your model, be sure to include definitions or descriptions of the following terms and components: restriction enzyme(s) ligase plasmid DNA human mRNA (the mRNA for insulin) reverse transcriptase
transformation E. coli marker genes (an antibiotic resistance gene) cloning vector
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Building the Model • Use chalk on a tabletop or a marker on a large sheet of paper to draw at least two test tubes and an E. coli cell’s plasma membrane. The E. coli should have a diameter of at least 12 inches. The test tubes should have a width of at least 6 inches. Use the test tubes for producing the insulin gene and for cloning the gene into the plasmid. Then transform your recombinant plasmid into the E. coli cell. • Use playdough or cutout pieces of paper to represent the enzymes, RNA molecules, and other components. • Use the pieces you assembled to develop a dynamic (claymation-type) model to demonstrate how a gene can be cloned into a plasmid and how the plasmid can then be transformed into a bacterial cell. • When you feel you have developed a good working model, demonstrate it to another student or to your instructor. Use your model to answer the questions. 1. Prior to recombinant gene technology, the insulin required to treat diabetes was obtained from the pancreases of slaughtered farm animals. Because the insulin was from other species, some humans developed immune responses or allergic reactions to it. As recombinant gene technology advanced, researchers explored the possibility of incorporating the human insulin gene into a plasmid that could be transformed into E. coli. If this technology was successful, the E. coli would produce human insulin that could be harvested from the bacterial culture medium. Researchers first needed to isolate the gene for insulin. To do this, they isolated mRNA (rather than DNA) from the beta cells of human pancreas tissue. Using reverse transcriptase, they made double-stranded DNA molecules that were complementary to the mRNA molecules they extracted from the pancreas cells. a. Based on what you know about eukaryotic chromosomes and genes, why did researchers choose to isolate mRNA rather than DNA? Eukaryotic genes contain introns, which must be excised from any pre-mRNA transcript produced. Bacterial cells do not contain the machinery (spliceosomes) required for doing this. b. What further adjustments might researchers need to make in the DNA molecules produced by reverse transcriptase before the molecules could be incorporated into bacterial plasmids? To incorporate the DNA molecules into the plasmid DNA, researchers would need to add the “sticky ends” for a specific restriction site. They would need to cut the plasmid DNA with the same restriction enzyme. When they combined the
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plasmid DNA with the insulin DNA, some of the plasmid DNA could then combine with the insulin DNA and incorporate it. For the bacteria to produce the mRNA for the insulin gene, researchers would also need to include a promoter site upstream of the insulin gene. c. Not all the DNA molecules produced by reverse transcription from pancreatic mRNA contained the gene for insulin. Some contained other genes. What mechanisms can be used to locate those bacterial colonies that picked up plasmids containing any of the genes produced by reverse transcription from pancreatic mRNAs? Grow the bacteria on a medium containing an antibiotic and X-gal. The plasmid selected has an antibiotic resistance gene on it and a lacZ gene. The restriction site used lies in the lacZ gene. As a result, only cells that have picked up plasmids will grow on medium containing the antibiotic. Cells with intact lacZ genes—that is, cells that picked up plasmids without an additional gene inserted into the lacZ site—will produce -galactosidase. The -galactosidase hydrolyzes X-gal and produces a blue compound. Colonies of these cells appear blue in color. Therefore, only the white bacterial colonies on the plate will contain plasmids carrying inserted genes. d. What mechanisms can be used to locate bacterial colonies that picked up only plasmids containing the insulin gene? A replica plate or blot of these white colonies can be made (see Figure 20.7, page 402). The filter used to make the replica separates the DNA into single strands and can be probed with labeled strands of DNA complementary to the desired gene. The label may be radioactive or fluorescent so that the colonies that carry the insulin gene can be easily identified on the blot. These colonies are then removed from the original plate and grown in culture medium. The bacteria from these colonies can be used to produce human insulin. Note: the bacteria produce the insulin, but do not use it themselves. (Note: Today almost all the insulin used for the treatment of human diabetes is produced using recombinant technology.)
Activity 20.2 How can PCR be used to amplify specific genes? 1. Assume you are using PCR to make multiple copies of a gene (shaded in grey below). DNA containing gene of interest: 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ 5⬘ ATATTTCTGAATGTTTAAACAGGGGTAAAACG3⬘ Activity 20.2
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On separate sheets of paper, describe the overall process and diagram the results you would obtain for 1, 2 and 3 rounds of PCR replication using the primers, ATGTT and CCATT. (Note: For simplicity we are showing DNA primers that are only 5 bases in length. In actual use, the DNA primers used are at least 17 bases long. This length is used to help reduce the risk that the primer anneals with [base pairs with] anything other than the specific segment of DNA to be amplified.) The explanation and diagram should look something like the following: a. Heat is applied (heat to 95°C) to separate the strands of the DNA molecule(s). 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ (Temperature ⫽ 95°C) 5⬘ ATATTTCTGAATGTTTAAACAGGGGTAAAACG3⬘ b. The temperature is reduced to between 45° and 60°C. 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ (Temperature ⫽ 50°C) 5⬘ ATATTTCTGAATGTTTAAACAGGGGTAAAACG3⬘ c. This allows the DNA primers to anneal. 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ 5⬘ ATGTT CCATT 5⬘ 5⬘ ATATTTCTGAATGTTTAAACAGGGGTAAAACG3⬘ d. and e. The Taq polymerase adds DNTPs to the open 3’ ends of the DNA primers. 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ 5⬘ ATGTTTAAACAG 3⬘ 3⬘ TTTGTCCCCATT 5⬘ 5⬘ ATATTTCTGAATGTTTAAACAGGGGTAAAACG3⬘ TOTAL Copies: 2n ⫽ 21 ⫽ 2 (n ⫽ number of cycles of PCR completed) f. In the next cycle, heating, cooling, primers annealing, and Taq polymerase elongating the DNA strands occurs again, but now on both of the daughter products of the first cycle: 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ ATGTT 140
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CCATT 5⬘ ATGTTTAAACAGGGGTAAAACG3⬘ 3⬘TATAAAGACTTACAAATTTGTCCCCATT 5⬘ ATGTT CCATT 5⬘ ATATTTCTGAATGTTTAAACAGGGGTAAAACG3⬘ The products of the second cycle are: 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ 5⬘ ATGTTTAAACAGGGGTAAAACG3⬘ 3⬘ TACAAATTTGTCCCCATT5⬘ 5⬘ ATGTTTAAACAGGGGTAAAACG3⬘ 3⬘TATAAAGACTTACAAATTTGTCCCCATT 5⬘ 5⬘ATGTTTAAACAGGGGTAA3⬘ 3⬘ TATAAAGACTTACAAATTTGTCCCCATT5⬘ 5⬘ ATATTTC TGAATGTTTAAACAGGGGTAAAACG3⬘ TOTAL Copies ⫽ 2n ⫽ 22 ⫽ 4 g. In the third cycle, heat separates the double strands of DNA. The system is then cooled to allow DNA primers to anneal, and the Taq polymerase produces the following 8 products: 3⬘ TATAAAGACTTACAAATTTGTCCCCATTTTGC5⬘ 5⬘ATGTT CCATT5⬘ 5⬘ ATGTTTAAACAGGGGTAAAACG3⬘ 3⬘ TACAAATTTGTCCCCATT5⬘ 5⬘ATGTT CCATT5⬘ 5⬘ ATGTTTAAACAGGGGTAAAACG3⬘ 3⬘TATAAAGACTTACAAATTTGTCCCCATT 5⬘ 5⬘ATGTT CCATT5⬘ 5⬘ATGTTTAAACAGGGGTAA3⬘ 3⬘ TATAAAGACTTACAAATTTGTCCCCATT5⬘ 5⬘ATGTT CCATT5⬘ 5⬘ ATATTTC TGAATGTTTAAACAGGGGTAAAACG3⬘ TOTAL Copies ⫽ 2n ⫽ 23 ⫽ 8 Activity 20.2
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2. PCR (polymerase chain reaction) is often used in forensics to amplify small amounts of DNA found at crime scenes. The amplified DNA is then tested for differences in RFLP (restriction fragment length polymorphisms) or STR (single tandem repeat) lengths. a. Explain what RFLPs and STRs are. RFLPs (restriction fragment length polymorphisms) are defined as the different restriction fragment patterns produced when DNA is cut using specified restriction enzymes. An individual’s RFLPs are inherited in a Mendelian fashion. STRs (single tandem repeats) are very short lengths of DNA (a few base pairs) that are repeated many times within single gene loci. These are also genetically inherited. In forensics, the variations in STRs in satellite DNA are generally used to identify individuals. b. How do STRs compare for unrelated individuals versus for closely related individuals (for example, parent and child or brother and sister)? STRs are genetically inherited. Therefore, since half of a child’s DNA comes from each parent, each of the STRs found in the child should be found in at least one of the parents. Similarly, on average, brothers and sisters share at least a quarter of their genes. Therefore, you would expect at least one-fourth of the STRs to compare between siblings. The same relationships should hold true for RFLPs. c. How reliable are these types of DNA fingerprinting for identifying individuals? What factors affect their reliability? Only a small portion of the DNA is selected for DNA fingerprinting. The specific RFLP or STR sites on the DNA that are commonly used tend to be highly variable among unrelated individuals. Probability estimates indicate that for any two unrelated individuals, there is a 1 in 100,000 to 1 in a billion chance that their DNA will match based on chance alone. Factors that can affect the reliability of DNA fingerprinting are how closely related the individuals are, the number of markers used in the tests, and the frequency of the specific markers in the population. For example, a specific ethnic group may share a marker more frequently than individuals outside of that ethnic group. This problem may be overcome by using more markers.
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20.2 Test Your Understanding 1. Which of the following sequences (on one strand of a double stranded DNA molecule) is likely to be a cleavage site for a restriction enzyme? Explain your answer. a. b. c. d.
CGTACC ATGTCG GATATG TGCGCA
The correct answer is d. The complementary strand for d would read: ACGCGT, i.e., the reverse of the sequence in d. 2. After undergoing electrophoresis, the gel in the figure below shows the RFLP analysis of DNA samples obtained from a crime scene. Bloodstains on a suspect’s shirt (B) were analyzed and compared with blood from the victim (V) and from the suspect (S). Are the bloodstains on the shirt from the victim or from the suspect? Explain.
V
S
B
If you examine the lane that contains the blood stains, you can find bands that do not match a band in either the victim’s or the suspect’s lane. As a result, the blood that contains this band could not have come from the suspect or the victim.
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Notes to Instructors Chapter 21 Genomes and Their Evolution What is the focus of this activity? While the Sanger method for sequencing DNA and the modifications that follow are conceptually fairly simple, most students don’t understand them. As noted previously, in order for most students to understand unfamiliar processes, they need to build models for themselves to discover what it is they understand and, more important, what they don’t.
What is this particular activity designed to do? Activity 21.1 How can we discover the sequence of an organism’s DNA? This activity allows students to interpret the results of a classic Sanger method for sequencing a DNA molecule only 20 bp long. It then asks them to translate these results into the results they would expect using more modern fluorescently tagged ddNTP methods.
What misconceptions or difficulties can this activity reveal? Because much of the methodology and information in this area is relatively new, students tend to lack conceptions in this area rather than have misconceptions. Walking them through how sequencing is done and how it is interpreted should overcome this.
Answers Activity 21.1 How can we discover the sequence of an organism’s DNA? Bacterial genomes have between 1 million and 6 million base pairs (Mb). Most plants and animals have about 100 Mb; humans have approximately 2,900 Mb. Individual chromosomes may therefore contain millions of base pairs. It is difficult to work with DNA sequences this large, so for study purposes the DNA is broken into smaller pieces (approximately 500 to 1,000 bp each). These pieces are sequenced and then the sequenced pieces are examined and aligned based on overlapping sequence homology at their ends.
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By comparing the DNA sequences among organisms, scientists can determine • what parts of the genomes are most similar among organisms and are therefore likely to have evolved earliest, • what key differences exist in the genomes that may account for variations among related species, and • what differences within species exist that may account for development of specific types of disease. The following activity has been designed to help you understand how genomes are sequenced and how the sequence information may be used. 1. In 1980, Frederick Sanger was awarded the Nobel Prize for inventing the dideoxy method (or Sanger method) of DNA sequencing. A double-stranded DNA segment approximately 700 bp in length is heated (or treated chemically) to separate the two strands. The single-stranded DNA that results is placed into a test tube that contains a 9-to-1 ratio of normal deoxynucleotides to dideoxynucleotides. A dideoxynucleotide has no ⫺OH group at either 2⬘ or 3⬘ carbon. As a result, whenever any dideoxynucleotide (abbreviated ddNTP) is added to the growing DNA strand, synthesis stops at that point. If the ratio of normal to dideoxynucleotides is high enough, where the dideoxynucleotide (rather than the normal deoxynucleotide) will be included in the sequence is random. You set up each of four test tubes as noted below: Tube number
Deoxynucleotides
Dideoxynucleotide
1
dATP, dTTP, dGTP, dCTP
ddATP
2
dATP, dTTP, dGTP, dCTP
ddTTP
3
dATP, dTTP, dGTP, dCTP
ddGTP
4
dATP, dTTP, dGTP, dCTP
ddCTP
All tubes contain the same single-stranded DNA molecules and the same primers. All other components required for DNA replication, such as enzymes, are present in each tube. You allow the replication to continue for the same length of time in each tube. At the end of the time period, you extract the DNA from each tube and run it on an agarose gel. You dye the gel with ethidium bromide and observe the following banding patterns on the gel. (Note: For this demonstration, we are using a DNA strand that is only 20 bases in length.)
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ddTTP tube
ddGTP tube
ddCTP tube
— — — — — — — — — — — — — — — — — — — — a. Which band in the gel contains the shortest DNA strand? What is the identity of its terminal ddNTP? The last band in the ddCTP column contains the shortest DNA strand. (Remember, the shorter the DNA the faster it moves in the gel.) The terminal ddNTP is C. b. Which band contains the next shortest DNA strand? What is the identity of its terminal ddNTP? The last band in the ddTTP column is the next shortest. The terminal ddNTP here is T. c. Continue reading the terminal ddNTP of each band from shortest to longest to determine the linear sequence of nucleotides in the DNA strand complement. What is the sequence? The sequence (reading from shortest to longest) would be: CTGCTTAATCGTATGCGATT. 2. The Sanger method has been modified so that each ddNTP used is now flagged with an identifying fluorescent tag. Assume that you run the same experiment that you did earlier. However, this time you combine all of the different nucleotides (both dNTPs and ddNTPs) in the same test tube. You run the products of the reaction on an agarose gel. Indicate the bands you would see on the gel below using the appropriate colors: ddTTP fluoresces red; ddGTP, yellow; ddCTP, blue and ddATP, green. 146
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Band sequence for combined experiment T red T red A green G yellow C blue G yellow T red A green T red G yellow C blue T red A green A green T red T red C blue G yellow T red C blue
3. To help determine evolutionary relationships among different groups of organisms, scientists compare gene sequences of highly conserved genes. What are “highly conserved genes”? Give examples and indicate what is “highly conserved” and why. Highly conserved genes are genes that have experienced little change throughout evolution. Highly conserved genes are essential for the survival of organisms. For example, the genes coding for enzymes involved in glycolysis and the genes for ribosomal RNA are both highly conserved. Most changes in the sequence of these genes would result in a nonfunctional gene product. As a result, the organism would be unable to survive. Therefore, over time, only mutations that did not significantly alter the function or end product of these genes were retained or conserved. Many of these types of alterations may have occurred in the “wobble” site of codons. The greater the number of these differences in sequence, the more distantly related the organisms are; the fewer the differences, the more closely related. 4. What types of DNA do scientists use to determine individual identities of organisms within the same species? Why do they use this type of DNA? Repetitive D—for example, simple sequence DNA or short tandem repeat DNA—is used for identifying individuals of the same species. This DNA is noncoding and as a result can be highly variable from individual to individual.
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Notes to Instructors Chapter 22 Descent with Modification: A Darwinian View of Life What is the focus of these activities? Biology, 8th edition, defines natural selection as “differential success in the reproduction of different phenotypes resulting from the interaction of organisms with their environment.” The text goes on to state that “evolution occurs when natural selection causes changes in relative frequencies of alleles in the gene pool.” Taken at face value, these two sentences are unambiguous, direct statements of fact. Then why do so many students have difficulty understanding what evolution is and how it can operate to change species over time? Often their previous experiences in life and education have left them with preexisting alternate conceptions (or misconceptions) about how biology and life work. How can we change students’ misconceptions? It would be easy if all we had to do was tell them what the current evidence indicates. However, the vast majority of students don’t learn new concepts and ideas simply by hearing about them in lecture. To learn new concepts and modify their existing ideas, students have to work with the evidence, test it, and prove to themselves that their new understanding works and has value. For this to happen, we need to address these questions: • • • •
Where do student misconceptions and problems about evolution originate? How can we determine what misconceptions our students have? How can we help students alter their misconceptions? How can we evaluate our students’ levels of understanding? What diagnostics can we use?
What are the particular activities designed to do? Activity 22.1 How did Darwin view evolution via natural selection? Activity 22.2 How do Darwin’s and Lamarck’s ideas about evolution differ? Activity 22.3 How would you evaluate these explanations of Darwin’s ideas? These activities are designed to help students understand the logic Darwin used to develop his theory of evolution via natural selection. They focus on how Darwin’s theory for the mechanism behind evolution differs from other theories—in particular, Lamarck’s.
What misconceptions or difficulties can these activities reveal? Activity 22.1 Activity 22.2 Activity 22.3 It should come as no surprise that some students’ misconceptions come from our own teaching and lectures, textbooks, and popular videos and movies dealing with evolution. 148
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For example, have you ever caught yourself saying, “In order to survive, this organism needed to evolve (some structure or capability)”? One slip of the tongue like this can undo volumes of evidence. As a result, we need to be very careful and very precise in how we express evolutionary ideas. If we find ourselves saying, “this organism needed to evolve . . . ,” we must immediately correct ourselves. We also have to be aware that certain key words used in explaining evolution may have very different meanings to our students. For example, the words adaptation, fitness, primitive, and advanced have very different meanings in common use compared to their meanings in the study of evolution. When I use the terms primitive and advanced, I make sure to let my students know that primitive is what came first and advanced is what came later; that is, in evolution, these are chronological terms, and no value is implied. For example, the chairs you’re sitting in are advanced; the Louis the 14th chairs in museums are primitive. Which are better? In some environments, the more primitive have the advantage (are better), and in other situations, the more advanced have the advantage. We also need to stop using the same examples of evolution over and over again. By the time students have reached college (if they’ve had any education in evolution), they’ve studied evolution at least three different times in both grade school and college. You can bet that each of those times, the key examples used were the peppered moth (Biston betularia) and Darwin’s finches. Students start to think that these are the only examples, and this does not make a very convincing argument for evolution as a unifying principle of biology. We need to use many more and varied examples of evolution. To uncover students’ misconceptions and alter them, we need to find out what each individual student’s understanding is. We also have to recognize that simply presenting students with the evidence one more time will not necessarily alter their existing understanding. For students to change their ideas, we need to provide them with the basic concepts and with novel examples, exercises, and problems that challenge their existing conceptions. If you assign these activities, keep in mind that the best way to find out what students are really thinking is to have them answer some of the questions in class (lecture or discussion). Have students discuss them in small groups. Wander among the groups during their discussions to answer any questions they may have. Without appearing to do so, listen in on their discussions. You will learn more about possible misconceptions this way than if you collect activities as written assignments. What students write down is not necessarily what they are really thinking. Instead, they often write down what they think the instructor wants to hear. The activities for Chapter 22 reveal some common misconceptions: • Many students have a Lamarkian view that organisms are modified by their environment. • Some think that evolution and natural selection lead directly to speciation. Notes to Instructors
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• Some think that nature purposefully selects for traits that are beneficial for survival. • Students may confuse physiological adaptations with evolutionary adaptations or be unable to distinguish between them. • Many students think that evolution leads to perfection in organisms. In other words, they think that because organisms today all appear well suited to their environment, this must have been planned. Otherwise, they can’t understand why we don’t see organisms that aren’t well suited. Given these misconceptions, when we teach the basics of evolution and natural selection, we need to make sure that we include the following ideas: 1. Mutation is random. Almost all mutations that could have occurred in the past probably did occur. This begs the question of why we don’t see all these mutations in the fossil record or among organisms. Most were so deleterious that they died during development. 2. Variability exists in populations, and only the variability that is heritable can be acted upon by natural selection. This variability includes some genes that are currently advantageous, some that are disadvantageous, and some that are neither; that is, some genes are neutral. Because students often think that a gene can be only one of these, I provide the following examples: • Currently eye color in humans is a neutral mutation. It is neither advantageous nor disadvantageous to survival and reproduction. • In chickens, the frizzle gene affects the feathers. Chickens with this trait have feathers that don’t interlock and, as a result, the chickens can’t control loft and insulation well. The feathers on these chickens stick up all over at odd angles from the body. In temperate zones, frizzle chickens can’t maintain heat well in winter and tend to die. In other words, in temperate zones, this mutation is disadvantageous. Environmental conditions can and did change over the course of life’s history on Earth. This means that some genes that were advantageous could have become disadvantageous, and vice versa. It also means that some genes that were previously neutral could have become either disadvantageous or advantageous. Here are two examples: • Consider the eye color trait. Assume that the hole in the ozone layer becomes even larger, and more ultraviolet light penetrates to Earth. People with blue eyes have no pigment in their iris, so UV light can penetrate much farther into their eyes and cause blindness. Blue-eyed people would find themselves at a disadvantage for survival compared to brown-eyed people. 150
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• To use a real example, consider the frizzle chicken. Normal chickens sent to the tropics to be used in agriculture don’t survive well in the heat. Frizzle chickens do very well, however. Ask your students, “So, is the frizzle gene advantageous or disadvantageous?” 3. It is not that advantageous genes are selected for, so much as that disadvantageous genes are selected against. For example, if you have a mutation that prevents adequate function of cellular respiration, you die. Keep in mind however, that what is a disadvantageous gene in one environment might be advantageous in another. 4. Note that these activities can be used to probe student understanding. They require students to recognize that they may need to modify their ideas and hypotheses based on differences in the evidence that is available at any one time.
Answers Activity 22.1 How did Darwin view evolution via natural selection? Darwin is remembered not because he was the first to propose that evolution occurs. Many others had presented this idea before. Instead, he is remembered for defining the mechanism behind evolution—that is, the theory of natural selection. To do this, Darwin integrated, or put together, information from a wide range of sources. Some of this information was provided by others; some he observed on his own. Working alone or in groups of three or four, construct a concept map of Darwin’s view of evolution via natural selection. Be sure to include definitions or descriptions of all the terms in the list below. Keep in mind that there are many ways to construct a concept map. • Begin by writing each term on a separate sticky note or piece of paper. • Then organize the terms into a map that indicates how the terms are associated or related. • Draw lines between terms and add action phrases to the lines that indicate how the terms are related. • When you finish your map, explain it to another group of students. Here is an example:
Embryology
provides evidence for
Evolution
Activity 22.1
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Terms fact biogeography gradualism uniformitarianism theory Galápagos Islands evolution homology natural selection analogy
Darwin vertebrate limb structure species population individual variability paleontology Malthus population size environment or resources
fit individuals fossil record embryology taxonomy selective (domestic) breeding limited resources struggle for existence reproduction extinction
Use the understanding you gained from creating the concept map to answer the questions. 1. In the 1860s, what types of evidence were available to indicate that evolution had occurred on Earth? Evidence was available from studies of taxonomy (classification of organisms based on similarities in morphology, for example), paleontology (fossil evidence that demonstrated how organisms had changed over geologic time), and biogeography (the study of the distribution of organisms in the present and the past). 2. How did knowledge of mechanisms of artificial selection (used in developing various strains of domesticated animals and plants) help Darwin understand how evolution could occur? Darwin knew that artificial selection could lead to dramatic changes in the phenotypes of individuals in a species population. As a result, he thought that natural selection could similarly lead to changes in populations. These changes would tend to occur more gradually, however. Similar to artificial selection, natural selection would eliminate or reduce the numbers of some variants in the population because these variants either didn’t survive or did not reproduce as well as others. 3. Based on his studies, Darwin made a number of observations; they are listed in the chart. Complete the chart by answering how Darwin made the observations.
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How did Darwin make this observation? That is, what did he read or observe that gave him this understanding?
a. All species populations have the reproduction potential to increase exponentially over time.
Darwin read Malthus’s essay on the potential for human populations to grow at a rate far beyond the capacity of their food supply and other resources. He applied this concept to other populations and concluded that all natural populations have this potential.
b. The number of individuals in natural populations tends to remain stable over time.
Although this may not be true for the human population, it is true for most natural populations of organisms. Darwin observed that in nature, the number of organisms per species in a given area tends to remain relatively constant over time.
c. Environmental resources are limited.
This was made obvious in Mathus’s essay. It was also obvious to Darwin as he observed natural populations. A given area has only so much food, only so many nesting sites, and so on.
d. Individuals in a population vary in their characteristics.
Darwin was an amateur naturalist even before his voyage on the Beagle. He was also a pigeon breeder. Both of these experiences led him to understand that there is considerable variation in populations of organisms.
e. Much of this variation is heritable.
As a pigeon breeder, Darwin could demonstrate this to himself. He also had a good understanding of the mechanism of artificial selection as it applied to animal husbandry and agriculture.
4. Based on these observations, Darwin made a number of inferences. Which of the observation(s) in question 3 allowed Darwin to make each inference? Inference
Observations that led to the inference
a. Production of more individuals than the environment can support leads to a struggle for existence such that only a fraction of the offspring survive each generation.
Darwin combined his understanding of the first three observations just noted to make this inference.
b. Survival for existence is not random. Those individuals whose inherited traits best fit them to the environment are likely to leave more offspring than less fit individuals.
Using the first inference and the observations that variation exists and some of it is heritable, Darwin made the logical assumption or inference that the less fit variants would not survive as well. As a result, they would be removed or reduced in the population.
c. The unequal ability of individuals to survive and reproduce leads to a gradual change in the population, with favorable characteristics accumulating over the generations.
This is a logical extension of the previous two inferences. Removal of unfit individuals (or their reduction relative to others) would obviously lead to gradual change in the population over time.
Activity 22.1
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5. Based on these observations and inferences, how did Darwin define fitness? Biology, 8th edition, defines fitness as “the relative contribution an individual makes to the gene pool of the next generation.” Organisms less able to survive and reproduce are less fit. In contrast, the remaining organisms are considered more fit. 6. How did Darwin define evolution? Darwin preferred to talk about descent with modification, or the idea that all organisms are related through descent from some unknown ancestor. Evolution was, as a result, defined as a gradual change in species over time. 7. What is the unit of natural selection—that is, what is selected? What is the unit of evolution—that is, what evolves? The individual organism is the unit that is selected because it is the individual organism that either dies or survives, reproduces or does not reproduce. The unit of evolution is the species population. 8. In a population of mice, some individuals have brown fur and some have black fur. At present, both phenotypes are equally fit. What could happen to change the relative fitness of the two phenotypes in the population? For example, what could cause individuals with brown fur to show reduced fitness relative to individuals with black fur? There are different possible ways of answering this question. Here is one: If both populations are equally fit at present, we can assume that neither is more or less threatened by predation as a result of their color. In other words, both colors are equally visible (or invisible) to predators in the current environmental circumstances. Perhaps the ground and ground cover are a mix of both black and brown patches. If something occurs to change the background color of the environment, the colors of the mice may become more or less apparent. For example, assume a small colony of the mice is transported to another area where the ground and ground cover are a mix of black and green. In this environment, the brown mice would be more visible to predators than the black mice. As a result, the brown mice would become less fit. 9. Assume you discover a new world on another planet that is full of organisms. a. What characteristics would you look for to determine that these organisms arose as a result of evolutionary processes? Again, this question has many possible answers. One approach is to look for the same types of evidence that Darwin used. For example, if you look at taxonomy, can you develop groupings of organisms based on similarities in morphology? Is there a fossil record on the planet? Does the fossil record show a gradual change in species over time? Do any of the fossil organisms look similar to existing organisms? Is there any evidence that species in close proximity to each other appear more closely related than species at great distances from each other? (You could also go beyond what Darwin knew and use more modern techniques of 154
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DNA analysis, assuming they have DNA or some similar hereditary material, and look for similarities among species.) b. What characteristics would you look for to determine that these organisms did not arise as the result of evolutionary processes? You would look for evidence that indicated there was no genetic relationship among different species. If the organisms did not arise as a result of evolutionary processes, there would be no reason for them to share any similarities in morphology, development, physiology, or molecular biology. 10. Why is it incorrect to say: Vertebrates evolved eyes in order to see? Natural selection occurs in the present. Organisms that survive must have traits that allow them to survive under existing conditions. As a result, each mutation required to produce the eye must have made those individuals (relative to others without the mutations) more fit. Relative to the less fit individuals, they survived better and produced more offspring.
Activity 22.2 How do Darwin’s and Lamarck’s ideas about evolution differ? Early in the 1800s Lamarck proposed a theory of evolution. He suggested that traits acquired during an organism’s life—for example, larger muscles—could be passed on to its offspring. The idea of inheritance of acquired characteristics was popular for many years. No such mechanism is implied in Darwin’s theory of evolution via natural selection, however. After Darwin published his work, scientists conducted many experiments to disprove the inheritance of acquired traits. By the middle of the 20th century, enough data had accumulated to make even its most adamant supporters give up the idea of inheritance of acquired characteristics. Given your understanding of both Lamarck’s and Darwin’s ideas about evolution, determine whether the statements on the next page are more Lamarckian or more Darwinian. If the statement is Lamarckian, change it to make it Darwinian. Here are two example statements and answers.
Examples A. The widespread use of DDT in the mid-1900s put pressure on insect populations to evolve resistance to DDT. As a result, large populations of insects today are resistant to DDT. Answer: This is a Lamarckian statement. DDT worked only against insects that had no DDT-resistance genes. The genes for DDT resistance had to be present for insects to survive DDT use in the first place. Activity 22.2
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Suggested change: Wide-scale use of DDT in the mid-1900s selected against insects that had no resistance to DDT. Only the insects that were resistant to DDT survived. These insects mated and passed their resistance genes on to their offspring. As a result, large populations of insects today are resistant to DDT. B. According to one theory, the dinosaurs became extinct because they couldn’t evolve fast enough to deal with climatic changes that affected their food and water supplies. Answer: This is a quasi-Lamarckian statement. Organisms do not purposefully evolve. (Genetic recombination experiments are perhaps an exception.) Once you are conceived, your genes are not going to change; that is, you are not going to evolve. The genetic composition of a species population can change over time as certain genotypes are selected against. Genes determine phenotypes. The environmental conditions may favor the phenotype produced by one genotype more than that produced by another. Suggested change: According to one theory, the dinosaurs became extinct because their physiological and behavioral characteristics were too specialized to allow them to survive the rapid changes in climate that occurred. The climatic changes caused changes in the dinosaurs’ food and water supplies. Because none of the dinosaurs survived, the genes and associated phenotypes that would have led to their survival must not have been present in the populations.
Statements 1. Many of the bacterial strains that infect humans today are resistant to a wide range of antibiotics. These resistant strains were not so numerous or common prior to the use of antibiotics. These strains must have appeared or evolved in response to the use of the antibiotics. Answer: This is a quasi-Lamarckian statement. Although the strains evolved in response to the use of antibiotics (the antibiotics killed off the strains that did not have genes for resistance), the strains did not appear in response to the antibiotics. If no resistance genes were present when antibiotics were applied, all would have died off. Suggested change: Many of the bacterial strains that infect humans today are resistant to a wide range of antibiotics. These resistant strains were not as numerous or common prior to the use of antibiotics. Antibiotic use must have selected against those bacterial strains that did not have resistance genes, leaving only those with resistance to survive. 2. Life arose in the aquatic environment and later invaded land. Once animals came onto land, they had to evolve effective methods of support against gravity and locomotion in order to survive. 156
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Answer: This is a Lamarckian statement. If the animals were not already able to support themselves and move in gravity, they would not have survived on land. Suggested change: Life arose in the aquatic environment and later invaded land. The animals that came onto land had to have previously evolved effective methods of support against gravity and locomotion in order to survive. 3. A given phenotypic trait—for example, height, speed, tooth structure—(and therefore the genes that determine it) may have positive survival or selective value, negative survival or selective value, or neutral (neither positive nor negative) survival or selective value. Which of these it has depends on the environmental conditions the organism encounters. Answer: This statement is Darwinian. Each of the variants we see in phenotype has a specific fitness and, as a result, a selective value under the existing environmental conditions. 4. The children of bodybuilders tend to be much more athletic, on average, than other children because the characteristics and abilities gained by their parents have been passed on to the children. Answer: This is a Lamarckian statement. The parents cannot pass on traits they acquired during their lifetimes. They can pass on only the genes that they have. Suggested change: The children of bodybuilders tend to be more athletic, on average, than other children. Bodybuilders may tend to have specific genes for these traits, or they may train their children to become athletic more than other people do.
Activity 22.3 How would you evaluate these explanations of Darwin’s ideas? Unfortunately, even today some people get or give the impression that acquired characteristics can be inherited. As a result, we need to be very careful about how we state our understanding of evolution and evolutionary theory. To test understanding of Darwin’s ideas, this question was included on an exam. 4-point question: In two or three sentences describe Darwin’s theory of descent with modification and the mechanism, natural selection, that he proposed to explain how this comes about. Four student answers to the question are given. Based on what you know about Darwinian evolution and natural selection, evaluate and grade how well each answer represents Darwin’s ideas. For any answer that does not receive full credit (4 points) be sure to indicate why points were lost. Activity 22.3
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Student 1. Darwin saw that populations increased faster than the ability of the land to support them could increase, so that individuals must struggle for limited resources. He proposed that individuals with some inborn advantage over others would have a better chance of surviving and reproducing offspring and so be naturally selected. As time passes, these advantageous characteristics accumulate and change the species into a new species. (Note: The following are the grades the author of the workbook would have given for these answers.) Grade: 4 This student demonstrates a good understanding of how fitness, resource limitation, population growth, and natural selection can cause changes in a species population over time. In addition, s/he is able to put the ideas into his/her own words. (Note: Darwin’s Origin of Species didn’t deal directly with the production of new species. However, Darwin did indicate that all species could be arranged on a tree of life. This can be explained only if new species arise over time.) Student 2. Darwin’s theory of evolution explains how new species arise from already existing ones. In his mechanism of natural selection, organisms with favorable traits tend to survive and reproduce more successfully, while those that lack the traits do not. Beneficial traits are passed on to future generations in this manner, and a new species will be created in the end! Grade: 3 As noted above, Darwin did not explain how new species arise from already existing ones. The rest of the answer indicates an understanding of the definition of natural selection. However, not enough information is provided to know whether the student understands the factors associated with natural selection—for example, overproduction of offspring, limited resources, heritability of traits. Student 3. Descent with modification using natural selection was Darwin’s attempt at explaining evolution. An organism is modified by its surroundings, activities, and lifestyle. These modifications, by natural selection, make the organism better suited to its life. Grade: 0 In the first sentence, the student restates the information available in the question without providing any more information or clarification. The second sentence indicates that the student has a Lamarckian view of how organisms are changed over time. S/he mentions natural selection in the last sentence but does not define it or indicate how it will make the species “better suited to its life.” Student 4. Darwin’s theory states that organisms can become modified by environmental conditions or use or disuse features and that the modifications can be passed down to succeeding generations. He proposes that nature selects for a characteristic trait that is beneficial to the survival of the organisms and that organisms would pass on this trait. Grade: 0 The first sentence of this answer indicates that this student has a Lamarckian view of evolution. In contrast, the second sentence provides a fairly good description of natural selection. No credit is given, however, because this student obviously cannot distinguish between the Darwinian and Lamarckian views of evolution. 158
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Notes to Instructors Chapter 23 The Evolution of Populations What is the focus of these activities? Both genetics and ecology (more specifically, natural selection) play roles at many levels in the continued evolution of life. Chapter 23 presents these ideas: 1. Genetic mutation in nature occurs randomly. Mutations in genes can produce changes in • the form, function, or both of individual organisms; • an organism’s environmental requirements; and • the ways different types or species of organisms interact with one another. 2. Change in environmental conditions can affect an organism’s ability to survive and reproduce; that is, it can affect natural selection. • Organisms with mutations that decrease their ability to survive and/or reproduce will produce fewer offspring. As a result, the relative frequencies of these genes decrease in the next generation of the population. • Organisms with mutations that either have no effect on or improve their ability to survive and/or reproduce will produce more offspring than organisms without those mutations. As a result, the relative frequencies of these mutated genes increase in the next generation of the population. Students need to understand, however, that the environment does not directly select for and increase the number of advantageous phenotypes or organisms (and their genes). It is more appropriate to think of alterations in gene frequency occurring because of the loss of specific phenotypes (and the genes that cause them). The activities can be used to generate discussion about all these ideas.
What are the particular activities designed to do? Activity 23.1 A Quick Review of Hardy-Weinberg Population Genetics This activity is designed to help students understand the logic behind the Hardy-Weinberg theorem. It also gives them practice using the theorem to determine whether a population is evolving.
Notes to Instructors
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Activity 23.2 What effects can selection have on populations? This activity is designed to help students understand how natural selection can lead to the evolution of a population. They will learn why and how certain assumptions can be made in studying the evolution of populations.
What misconceptions or difficulties can these activities reveal? Activity 23.1 Most of the difficulties students have solving Hardy-Weinberg problems result from a lack of understanding of basic probability. Question 4, Part A: Some students are concerned about which allele is given the frequency p and which is given the frequency q. It helps to tell them that p is usually the frequency of the dominant allele and q is the frequency of the recessive allele. Be sure to note, however, that there is no hard-and-fast rule. For example, if two alleles are codominant, one is still assigned the frequency p and the other is given the frequency q. For question 4e, many students know what the binomial expansion is. They may even recognize that (p2 + 2pq + q2 = 1) is the binomial expansion of (p + q = 1). However, many will not understand what this mathematical reality has to do with random mating. For this reason, it helps to set up a Punnett square with the frequencies for female gametes on one side and the frequencies for male gametes on the other. Combining the gametes gives all possible genotypes of offspring for the subsequent generation. The probability of a specific type of female gamete pairing with a specific type of male gamete is then the product of the individual probabilities. When the probabilities of the possible genotypes are added together, we get p2 + 2pq + q2 = 1. Using the Punnett square raises another difficulty: how to determine what p and q are for the males alone and the females alone in a population. Pose the following problem in class: “Assume the frequency of the b allele in the population is 10% and the frequency of the B allele is 90%. If females make up half of the population, what is the frequency of the b allele in the female population alone?” A large percentage of the class will answer 5%. Push the issue further: “OK, here’s another example. Let’s assume that I have a cheesecake that contains 50% fat. If I cut it in half, what percent fat does each half contain? And if I cut it into quarters, what percent fat does each piece contain?” Eventually students will begin to realize that the percent of fat (or the frequency of alleles) doesn’t change when the population is divided in two. In other words, if there is 50% fat in the whole cheesecake, then there is 50% fat in each piece (no matter how small). And if 10% of the alleles in the population are b alleles, then the female half of the population should contain 10% b and 90% B alleles. Question 6, Part B: Many students don’t understand why one can randomly assign only two alleles to each diploid organism in a population. For example, assume we want to know the allele frequency in an isolated subpopulation of birds. The population contains only six individuals: one of genotype TT, two of genotype Tt, and three of genotype tt. 160
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To determine the allele frequencies of T and t, we randomly assign two alleles to each individual. We then calculate the frequencies of the alleles: Number of individuals who are:
Number of T alleles
Number of t alleles
1 = TT
2
0
2 = Tt
2
2
3 = tt
0
6
4/12 = 1/3 = 33.3%
8/12 = 2/3 = 66.7%
Allele frequency:
It is helpful to explain that the assumption here is that each male and each female in the population, on average, produces the same number of gametes. Given that assumption, it doesn’t matter how many each produces. All of the gametes from TT individuals will contain Ts. Half of the gametes from Tt individuals will contain T and the other half will contain t. All of the gametes from tt individuals will contain t. Therefore, for ease of calculation, we assign the smallest possible number of gametes or alleles to each individual, and that number is two. Activity 23.2 To answer any of these questions, students have to make some assumptions. In particular, they have to ask themselves whether or not the characteristic under selection could be heritable. If the characteristic is heritable and selection removes some types more often than others, then the population is likely to change over time. Scenarios I, IIa and b, and IIIa and b are designed to require students to rethink the assumptions they made in light of additional information. Changing assumptions in response to the addition of new information seems logical to instructors, but the same is not true for many students. Some students balk at making any assumptions and make statements like: “How am I supposed to know whether this characteristic is heritable?” Other students get upset if the additional information seems to contradict an earlier assumption they made: “You can just change your assumptions whenever you want?” As a result, it is useful to have the students do this activity in small groups in class. Then collect their ideas and discuss them in large group. Note: For scenario I, some students may answer that no selection is going on in this situation. When asked to explain, they may say, “There is no evidence of selection against any of the mice.” Given the information provided, this is true. However, these students need a nudge to recognize that all living organisms, plants included, can be affected by selection. After working through several of the scenarios, some students may get the idea that answering these types of questions can be done by formula: First assume that the characteristic being affected is heritable, and then ask what would happen to the population over time if many organisms with a particular characteristic were removed. This strategy works well only if the Notes to Instructors
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characteristics could be heritable. In some cases (for example, scenario IV), however, it is not reasonable to assume the characteristic is heritable. That it is not heritable usually becomes obvious when you take the example to extremes. For example, if you take scenario IV to the extremes, then over time, the population would contain fewer and fewer very young and very old individuals and more and more middle-aged individuals. If you explain this to the class, some will recognize that there’s something wrong with this idea. Others may have to be nudged further. Ask: “Which individuals in the population are most likely to breed? Given this, how would the number of newborns in the population be likely to change over time?”
Answers Activity 23.1 A Quick Review of Hardy-Weinberg Population Genetics Part A. Review Chapter 23 of Biology, 8th edition. Then complete the discussion by filling in the missing information. If evolution can be defined as a change in gene (or more appropriately, allele) frequencies, is it conversely true that a population not undergoing evolution should maintain a stable gene frequency from generation to generation? This was the question that Hardy and Weinberg answered independently. 1. Definitions. Complete these definitions or ideas that are central to understanding the Hardy-Weinberg theorem. a. Population: An interbreeding group of individuals of the same species. b. Gene pool: All the alleles contained in the gametes of all the individuals in the population. c. Genetic drift: Evolution (defined as a change in allele frequencies) that occurs in small populations as a result of chance events. 2. The Hardy-Weinberg theorem. The Hardy-Weinberg theorem states that in a population that is not (is/is not) evolving, the allele frequencies and genotype frequencies remain constant from one generation to another. 3. Assumptions. The assumptions required for the theorem to be true are listed on page 472 of Biology, 8th edition, and are presented here in shortened form. a. The population is very large. b. There is no net migration of individuals into or out of the population. c. There is no net mutation; that is, the forward and backward mutation rates for alleles are the same. For example, A goes to a as often as a goes to A. 162
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d. Mating is at random for the trait/gene(s) in question. e. There is no selection. Offspring from all possible matings for the trait/gene are equally likely to survive. 4. The Hardy-Weinberg proof. Consider a gene that has only two alleles, R (dominant) and r (recessive). The sum total of all R plus all r alleles equals all the alleles at this gene locus or 100% of all the alleles for that gene. Let p = the percentage or probability of all the R alleles in the population Let q = the percentage or probability of all the r alleles in the population If all R + all r alleles = 100% of all the alleles, then p + q = 1 (or p = 1 ⫺ q or q = 1 ⫺ p) (Note: Frequencies are stated as percentages [e.g., 50%] and their associated probabilities are stated as decimal fractions [e.g., 0.5].) Assume that 50% of the alleles for fur color in a population of mice are B (black) and 50% are b (brown). The fur color gene is autosomal. a. b. c. d. e.
What percentage of the gametes in the females (alone) carry the B allele? 50% What percentage of the gametes in the females (alone) carry the b allele? 50% What percentage of the gametes in the males carry the B allele? 50% What percentage of the gametes in the males carry the b allele? 50% Given the preceding case and all the Hardy-Weinberg assumptions, calculate the probabilities of the three possible genotypes (RR, Rr, and rr) occurring in all possible combinations of eggs and sperm for the population.
Male gametes and probabilities
Female gametes and probabilities R ( p)
r (q)
R ( p)
RR (p 2 ) _____
Rr (p q ) _____
r (q)
Rr (p q ) _____
rr (q 2 ) _____
Because the offspring types represent all possible genotypes for this gene, it follows that p2 + 2pq + q2 = 1 or 100% of all genotypes for this gene
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Part B. Use your understanding of the Hardy-Weinberg proof and theorem to answer the questions. 1. According to the Hardy-Weinberg theorem, p + q = 1 and p2 + 2pq + q2 = 1. What does each of these formulas mean, and how are the formulas derived? p + q = 1: If you add all the dominant alleles for a gene to all the recessive alleles for the gene, you get all of the alleles for that gene, or 100% of the alleles for the gene. (Note: This assumes the gene has only two alleles.) p2 + 2pq + q2 = 1: If you combine all the individuals that are homozygous dominant for a gene with all the heterozygotes and homozygous recessive individuals for that gene, you have counted or combined all the individuals in the population that carry that gene. 2. Assume a population is in Hardy-Weinberg equilibrium for a given genetic autosomal trait. What proportion of individuals in the population are heterozygous for the gene if the frequency of the recessive allele is 1%? Assume that D is the dominant allele and d is the recessive allele. Because all the alleles are either d or D, if the frequency of the d alleles is 1% or 1/100 (= q), then the frequency of the D alleles must be 99% or 99/100 (= p). The frequency of heterozygous individuals in the population is 2pq or 2(99/100)(1/100) = 198/10,000. 3. About one child in 2,500 is born with phenylketonuria (an inability to metabolize the amino acid phenylalanine). This is known to be a recessive autosomal trait. a. If the population is in Hardy-Weinberg equilibrium for this trait, what is the frequency of the phenylketonuria allele? Assume P is the normal allele and p is the phenylketonuria allele. The frequency of homozygous pp individuals in the population is then equal to q2, which is 1/2,500. The frequency of the p allele is the square root of 1/2,500 = 1/50 or 2%. b. What proportion of the population are carriers of the phenylketonuria allele (that is. what proportion are heterozygous)? Heterozygotes should occur in the frequency 2pq: 2pq = 2(1/50)(49/50) = 98/2,500. 4. In purebred Holstein cattle, about 1 calf in 100 is spotted red rather than black. The trait is autosomal and red is a recessive to black. a. What is the frequency of the red alleles in the population? If 1 calf out of 100 is spotted red, then the frequency of the recessive red genotype is 1/100 = q2. Therefore, q (the frequency of the red allele) = 1/10 or 10%. b. What is the frequency of black homozygous cattle in the population? p2 = (1 ⫺ q)2 = (9/10)2 = 81/100 164
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c. What is the frequency of black heterozygous cattle in the population? 2pq = 2(1/10) (9/10) = 18/100 5. Assume that the probability of a sex-linked gene for color blindness is 0.09 = q and the probability of the normal allele is 0.91 = p. This means that the probability of X chromosomes carrying the color blindness allele is 0.09 and the probability of X chromosomes carrying the normal allele is 0.91. a. What is the probability of having a color-blind male in the population? Remember, males have only one X chromosome. Color blindness is a sex-linked gene and is found on the X chromosome and not on the Y. As a result, a male has a 0.09 probability of having the X with the color blindness allele and a 0.91 probability of having an X with the normal allele. For sex-linked genes, males display the allele frequency. b. What is the probability of a color-blind female? Unlike males, females have two copies of the X chromosome. As a result, they display the genotype frequency for genes on the X. A color-blind female is homozygous recessive for the color blindness allele. The frequency of the homozygous recessive genotype is q2, or (0.09)2 = 0.0081. 6. The ear tuft allele in chickens is autosomal and produces feathered skin projections near the ear on each side of the head. This gene is dominant and is lethal in the homozygous state. In other words, homozygous dominant embryos do not hatch from the egg. Assume that in a population of 6,000 chickens, 2,000 have no ear tufts and 4,000 have ear tufts. What are the frequencies of the normal versus ear tuft alleles in this population? This gene is lethal in the homozygous condition. Therefore, homozygotes that are produced do not hatch and do not appear in the population. The population contains 4,000 heterozygous tufted chickens and 2,000 homozygous normal chickens. Because we don’t know how many eggs did not hatch (or how many of these contained homozygous tufted chickens), we need to calculate the allele frequencies by assigning alleles to the existing population. Number of T (tufted) alleles
Number of t (normal) alleles
4,000 tufted chickens
4,000 T
4,000 t
2,000 normal chickens
______
4,000 t
Total alleles
4,000 T
8,000 t
Allele frequencies
4,000/12,000
8,000/12,000
0.33
0.66
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If these are in Hardy-Weinberg equilibrium, we would expect the following offspring in the next generation: Frequency of alleles in eggs Frequency of alleles in sperm
T (p = 1/3)
t (q = 2/3)
T (p = 1/3)
TT (1/9)
Tt (2/9)
t (q = 2/3)
Tt (2/9)
Tt (4/9)
In the next generation, when you remove the homozygous lethals, the frequency of Tt and tt genotypes would be equal. This indicates that the assumption is incorrect. In other words, the population is not in Hardy-Weinberg equilibrium. 7. How can one determine whether or not a population is in Hardy-Weinberg equilibrium? What factors need to be considered? To determine whether a population is in Hardy-Weinberg equilibrium, you need to be able to calculate the numbers of individuals in the population that are homozygous versus heterozygous for the alleles. If you know the frequencies of each genotype, you can calculate the allele frequencies (as in question 6). Given the allele frequencies, you can calculate the genotype frequencies that would be expected if the population were in Hardy-Weinberg equilibrium. Then compare these values to the known values for the population. In reality, this is difficult to do because if alleles show dominance, it is hard to distinguish the homozygous dominants from the heterozygotes. As a result, we tend to look at the frequency of the homozygous recessive phenotype in a population. If this remains relatively constant from one generation to the next, we use it as evidence to assume that the population is in Hardy-Weinberg equilibrium. 8. Is it possible for a population’s genotype frequencies to change from one generation to the next but for its gene (allele) frequencies to remain constant? Explain by providing an example. There are a number of ways that this is possible. Here is one example of how it could occur: Assume inbreeding has occurred in two populations of mice. In one population, the mice have become homozygous AA at a particular gene locus. In the other population, the mice have become homozygous aa at that gene locus. Equal numbers of aa and AA mice happen to migrate to a new habitat. The frequency of the A allele is 0.5 and the frequency of the a allele is 0.5 in this new population. All possible combinations of matings of these mice are listed in the following table.
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AA males
aa males
AA females
AA ⫻ AA
all AA offspring
AA ⫻ aa
all Aa offspring
aa females
AA ⫻ aa
all Aa offspring
aa ⫻ aa
all aa offspring
Each of these matings should occur with equal probability. Therefore, each of the offspring types should occur with equal probability. As a result, 1/4 of the offspring will be AA, 1/2 will be Aa, and 1/4 will be aa. In this case, within one generation the population has gone to Hardy-Weinberg equilibrium for genotypes; however, the allele frequency has remained the same.
23.1 Test Your Understanding In each of the following scenarios, choose which assumption of the Hardy-Weinberg Law is being violated. 1. In a particular region of the coast, limpets (a type of mollusc) live on near shore habitats that are uniformly made up of brown sandstone rock. The principle predators of these limpets are shorebirds. The limpets occur in two morphs, one with a light-colored shell and one with a dark-colored shell. The shorebirds hunt by sight and are able to see the light ones on the dark sandstone easier than the dark ones. No selection—The shorebirds are selectively taking the most visible (light) limpets. 2. In Chen caerulescens (a species of goose), the white body form, the snow goose and the blue body form, the blue goose, occasionally coexist. In these areas of contact, white-by-white and blue-by-blue matings are much more common than white-by-blue matings. Random mating—Selective mating is occurring when white by white and blue by blue mating are more common than white by blue. 3. Prior to the Mongolian invasions which occurred between the 6th and 16th centuries, the frequency of blood type B across Europe was close to zero. The frequency of blood type B among the Mongols was relatively high. Today, it is possible to see fairly high frequencies of blood type B in the Eastern European countries and a gradual decrease in the frequency of blood type B as one moves from the Eastern European countries to the Western European countries, such as France and England. No migration—Mongolian invasions were more common in Eastern European countries than Western European countries. Offspring of Eastern Europeans and Mongols with blood type B led to the variability in distribution of the blood type in Europe.
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Activity 23.2 What effects can selection have on populations? 1. What effects can natural selection have on populations? For example, what types of selection can occur in a population, and how does each affect a population? Selection can be directional, stabilizing, or disruptive. If selection removes individuals at one of the extremes of a phenotypic characteristic, then the mean for the characteristic can appear to move in a more positive (or more negative) direction—directional selection. Alternatively, selection can act against both extremes at the same time, and the characteristic will appear to stabilize around the mean—stabilizing selection. If selection acts against the mean or average phenotype, it can lead to the production of a polymorphic population—disruptive selection. 2. Examine the scenarios on the following pages. For each scenario: a. Decide whether or not natural selection is operating. In doing this, indicate whether there is variability in the population(s). If no, what does this imply about evolution? If yes, what is the nature of the variation? For example, what characteristics must the variation have for selection to operate on it? b. Is there any indication that members of the population(s) differ in fitness? If no, what does this imply about the operation of natural selection? If yes, describe the difference in fitness. c. Given your answers to parts a and b, what trend should characterize the future behavior or composition of the population(s)? Be sure to indicate any assumptions you make in answering the questions. The key to analyzing these scenarios is to ask first what is being selected against and then whether or not this phenotype or trait could have a genetic basis; that is, could the trait be inherited genetically? If the answer is yes, you are likely to see selection. Then you need to determine what type of selection. Scenario I. A particular species of mouse feeds on the seeds of a single species of cherry tree. When the mice eat a seed, they digest it completely. The mice choose seeds of intermediate and large sizes, leaving the very small seeds of the cherry tree uneaten.
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a. Selection is occurring. The variation is in cherry seed size. For selection to be operating, the assumption is that seed size is a heritable trait. b. Individual seeds do differ in fitness. The mice are selecting against the very large and intermediate-sized seeds. Only the very small seeds are not eaten. Therefore, many more small seeds will be left (relative to other sizes) to produce the next generation of cherry trees. c. If this selection occurs long term, future generations of cherry trees will produce more small-sized seeds. Scenario IIa. Small island A contains three separate populations of a single species of cherry tree. The seed size varies between trees. That is, some trees produce seeds that are all in the small size ranges, others produce seeds all in the middle size ranges, and others produce seeds in the large size ranges. A small population of mice is introduced to the island. The mice eat cherries and are the only predators on the cherry trees. When the mice eat a cherry, they completely digest it and the pit or seed inside it. The mice choose medium and large seeds and leave the smallest seeds uneaten. a. Selection is occurring. The variation is in cherry seed size. For selection to be operating, the assumption is that seed size is a heritable trait. b. Individual seeds do differ in fitness. The mice are selecting against the very large and medium seeds. Only the small seeds are not eaten. Therefore, many more small seeds will be left (relative to other sizes) to produce the next generation of cherry trees. c. If this selection occurs long term, future generations of cherry trees will produce more small seeds. Furthermore, since the three populations are each made up of trees with different sized seeds, the population with small seeds will become much larger over time compared to the other two populations. Scenario IIb. Would your answer for Scenario IIa change given the following information? Explain. As you continue to study the populations of trees, you note that the viability of the seeds varies with size such that the viability of the small seeds is less than that of the middle-sized seeds, which is less than that of the largest seeds. a. Again, selection is occurring. The variation is in cherry seed size. For selection to be operating, the assumption is that seed size is a heritable trait. b. Individual seeds differ in fitness. The mice are selecting against the very large and middle-sized seeds. Only the small seeds are not eaten. Therefore, many more small-sized seeds will be left (relative to the other sizes). In addition, there is a difference in fitness related to seed size. The larger the seed, the more likely it is to survive and produce a new tree. c. In this scenario, you need to consider both the effects of selection by mice and the difference in relative fitness of the seeds associated with size. If the mice are not eating all of the large and middle-sized seeds, some will remain to produce Activity 23.2
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individuals in the next generation. In the future populations, the proportion of trees producing each seed type depends on both the numbers of large, middlesized, and small seeds that are uneaten and the relative viability of each of these. Scenario IIIa. Small island B contains three separate populations of a single species of cherry tree. Unlike the species on island A, in this species seed size varies within trees. That is, each tree produces seeds that range in size from large to small. A small population of mice is introduced to the island. These are the only predators on the cherry trees. When the mice eat a cherry, they digest it and the pit or seed completely. The mice choose medium and large seeds and leave the smallest seeds uneaten. a. Again, selection could be occurring if you assume that seed size is a heritable trait. If each flower on a tree is independently pollinated to produce its seeds, then the individual seeds on a tree can have different combinations of genes. However, if you noticed that the most shaded parts of the tree produced cherries with small seeds and the sunniest parts of the tree produced the cherries with the largest seeds, you might reconsider your assumption. In that case, seed size may be the result of available nutrients or growth conditions and not the result of genetic differences. b. If seed size varies with nutrition (as opposed to genetics), the selection against large and small seeds is not necessarily selection against any particular genotype. Then there is no genetic difference in fitness. c. If there are no selection and no difference in fitness among the surviving seeds, then the population should remain relatively stable in genetic composition over time. Scenario IIIb. Would your answer for Scenario IIIa change given the following information? Explain. As you continue to study these populations, you note that the viability of the seeds does not vary with size. Over time, however, you find that the trees that grow from the smallest seeds produce fewer large seeds. a. This information provides additional evidence that seed size is heritable and not the result of growth conditions. As a result, selection is occurring. b. Small seeds or individuals would survive more frequently (compared to medium and large seeds, which would be preferentially eaten and therefore removed from the population). c. Over time, you would expect to see more offspring from small-sized seeds and, as a result, more production of small-sized seeds. Scenario IV. After a severe spring ice storm, about half of the finches (small birds) in a population are found dead. Examination of the dead birds indicates that they vary in age from young to old. About 60% of the dead are new fledglings (just left the nest); about 20% are over 3 years of age (old for this species). 170
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a. I’d argue that this is not selection. The vast majority of the dead birds are the very young (just fledged and without parental support) and the very old. Individuals in these age categories are less able to survive based on their level of experience and general health. These characteristics are not necessarily based on genetic differences in these individuals compared to the rest of the population that survived. b. If this were selection, it would be against young and old ages. If we took this to extremes, we would have to argue that over time each subsequent generation would have more middle-aged individuals (compared to the numbers of young and old individuals). Because middle-aged individuals must have at one time been young, this doesn’t make logical sense. c. Without any further information to indicate that only specific types of young and old birds died, we cannot say that genetic selection occurred. Therefore, we would expect to see no significant changes in the genetics of this population over time.
23.2 Test Your Understanding The figure below shows the distribution of a phenotypic characteristic in a population of organisms. Also shown are three portions (A, B, and C) of that distribution.
A
B
C
Low or less High or more Characteristic For the situations described in questions 1 to 3 below, which portion(s) of the population will be selected against and least likely to have their genes represented in the next generation? Explain your answers. Use the following set of answers A. Portion A only B. Portion B only C. Portion C only D. Portions A and C simultaneously E. Portions B and C simultaneously
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1. Male sticklebacks with bright red coloring are favored by female stickleback as mates. However, the bright red color makes the males more likely to be seen by predators. D. In this example, the bright red color is both selected against (by predators) and for (by females during mating). As a result, there will be a relative equilibrium set up between these two forces, which will tend to stabilize the color at some intermediate level. 2. On Island Z in the Galápagos the plant population contains only two species. One of the two plant species produces very large seeds and the other produces small seeds. A species of seed-eating finch has lived on the island for many years. This established species has large beaks and prefers large seeds. A small population of a different species of seed eating finch has migrated to the island. The beak size in the new species varies over a continuum from relatively small to large. How will evolution of the new finch species be affected? C. The established species with large beaks are likely to be better competitors for the large seeds. As a result, the members of the new species, that can only eat large seeds are likely to be at a disadvantage relative to those members of the new species that can eat smaller seeds. As a result, selection in the new species will be against larger-beaked birds. 3. In a species of plant-eating land snail, the shell color is variable and ranges from very light to intermediate to very dark in color. The snails are preyed upon by birds that use sight to find their prey. A small population of these snails is moved to an island where the food plants in their preferred habitat are either very light in color or very dark. B. In this scenario the landscape is either light or dark. As a result, light snails would be less visible on light landscape and dark snails would be less visible on dark landscapes. However, intermediate shades of snails would be visible on all landscapes and therefore selected against (seen and eaten) more often than either light or dark snails.
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Notes to Instructors Chapter 24 The Origin of Species What is the focus of this activity? As noted earlier, certain key words used in discussions of evolution may have very different meanings to students. For example, adaptation, fitness, primitive, and advanced have very different meanings in common use compared to their meanings in the study of evolution. Many other words used in discussing evolution and speciation may be new to students—for example, anagenesis, cladogeneis, allopatric, and sympatric. In this activity, students learn the meanings of these words in context as they examine how speciation occurs.
What is the particular activity designed to do? Activity 24.1 What factors affect speciation? This activity is designed to help students understand how evolutionary changes in a population can lead to speciation. As they examine how speciation can occur, students also learn key terms used to describe processes associated with speciation. Activity 24.2 How does hybridization affect speciation? This activity is designed to help students understand how environmental differences and interactions between species in zones of hybridization function either to reinforce species differences or to fuse species.
Answers Activity 24.1 What factors affect speciation? 1. The Galápagos Archipelago consists of a dozen islands all within 64 km of their nearest neighbor. From 1 to 11 of the 13 species of Darwin’s finches live on each island. Many evolutionary biologists believe that if there had been only one island, there would be only one species of finch. This view is supported by the fact that Cocos Island is isolated (by several hundred kilometers of open ocean) from the other islands in the archipelago and only one species of finch is found there. a. How does the existence of an archipelago promote speciation? Explain and provide an example.
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The distance between islands makes it very difficult for organisms on one island to migrate to another island. For example, it is likely that only a few birds made it to one of the Galápagos Islands from the mainland (Ecuador). Birds without phenotypes that allowed them to survive on the island were eliminated. Phenotypes that survived increased in number. Because conditions on the island differed from those on the mainland, this population diverged from the population on the mainland. As genetic changes accumulated in this island population, the population may have become reproductively isolated from the mainland species and could then be considered a new species. As the population on the first island became more numerous, a few birds from the island may have successfully migrated to the next island in the archipelago. (Many may have tried, but only a few succeeded.) The few that migrated would establish a new colony of birds. If the environmental conditions on the new island differed from those on the old island, the phenotypes that survived could also differ. Over time, the colony on the second island could diverge enough genetically to become reproductively isolated from the first island population. At this point, it would be considered another new species. This type of progressive migration to the various islands could result in the formation of a new species on each of the islands. Additional migration events between islands could account for the presence of more than one species per island. For more than one species to exist on an island, however, each species must have adopted a separate niche. b. Is the mode of speciation that occurred on these islands more likely to have been allopatric or sympatric? Explain. This mode of speciation is more likely allopatric. For genetic changes to accumulate differentially in two populations, gene flow must be limited between the populations. For these species, allopatric isolation of species populations is the most likely mechanism for limiting gene flow. c. Is the type of speciation seen on the Galápagos Archipelago more likely to be the result of anagenesis or cladogenesis? Explain. It is more likely the result of cladogenesis. Anagenesis is the gradual change of one species into another over time. The species may change over time; however, at any point in time, only one species is present. Cladogenesis is the divergence of a species into two or more species. 2. Hybrids formed by mating two different species are often incapable of reproducing successfully with each other or with the members of their parent populations. Explain why this is the case. (Hint: Consider what you know about chromosome numbers and meiosis.) In diploid animals, meiosis produces gametes that each contain half of the chromosomes of the parent cell. Each gamete receives a very specific half of the 174
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chromosomes, however. Each gamete receives one member of each homologous pair. As a result, when the zygote forms, it contains a complete set of chromosomes (two members of each pair). The partitioning of one member of each homologous pair to each gamete occurs as a result of synapsis and separation of the members of each pair to opposite poles at anaphase I of meiosis. Hybrid organisms have one member of each homologous pair of chromosomes from the male parent and one member of each homologous pair from the female parent. The haploid number of chromosomes may differ for the two species. In addition, during meiosis, synapsis of chromosomes from different species may not occur. As a result, the division of chromosomes to opposite poles at anaphase I of meiosis occurs randomly. There is no control over which chromosomes will appear in the gametes. Combinations of gametes from hybrids most often result in zygotes that are aneuploid for a number of the chromosomes and therefore nonviable. Combinations of gametes from hybrids with gametes from either of the parental types also result in the production of nonviable aneuploids. 3. Because most hybrids can’t reproduce, their genes (and the genes of their parents) are removed from the population. Only the genes of individuals who breed with members of their own species remain in the population. This implies that there is a strong selective advantage for genes that enable individual organisms to recognize members of their own species. Today a wide range of reproductive isolating mechanisms has been identified. Each of the following scenarios describes a reproductive isolating mechanism. Indicate whether each is a prezygotic or postzygotic isolating mechanism. Explain your answers. (Note: A prezygotic isolating mechanism by definition prevents the formation of the zygote. A postzygotic isolating mechanism affects the viability or reproduction of zygotes that are produced.) a. Crickets use species-specific chirp patterns to identify a mate of their own species. This is a prezygotic mechanism because it prevents matings between different species. b. Two species of butterfly mate where their ranges overlap and produce fertile offspring, but the hybrids are less viable than the parental forms. Hybrids are produced; however, the hybrids are less viable than either parent species. As a result, this is a postzygotic isolating mechanism. c. Two species of a plant cannot interbreed because their flowers differ in size and shape and require pollination by different species of bee. This is a prezygotic mechanism because it prevents pollen from one plant from fertilizing the other species of plant. As a result, it blocks zygote formation. Activity 24.1
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d. Two species of firefly occupy the same prairie and have similar flash patterns, but one is active for a half-hour around sunset while the other doesn’t become active until an hour after sunset. This is another prezygotic isolating mechanism. The behavior of the two species prevents them from interacting and, as a result, prevents the production of hybrid zygotes. 4. Many of our most successful grain crops arose as hybrids; most are also allopolyploids. These crops can successfully reproduce. Explain. An allopolyploid can be produced by hybridization between two species to form a zygote. If the chromosomes of the zygote duplicate for the first mitotic division but that division fails, the zygote becomes an allopolyploid. In other words, the zygote contains two copies of each of the chromosomes donated by each of the parents. These copies can synapse in meiosis I and separate to opposite poles at anaphase I. As a result, meiosis produces gametes, each with one member of each pair of chromosomes. If this type of allopolyploidy occurred in a sexually reproducing animal, the story would stop there. The likelihood that a sexually reproducing allopolyploid animal would find another similar allopolyploid of the opposite sex to mate with is about zero. On the other hand, many plants produce both pollen (containing male gametes) and eggs. If such a plant becomes allopolyploid and is capable of self-fertilization, it automatically becomes a new species.
Activity 24.2 How does hybridization affect speciation? As noted in the text, two species of toads, Bombina bombina and Bombina variegata, share a hybrid zone that is 4,000 km in length and only 10 km wide. The frequency of alleles specific to B. bombina decreases from close to 100% on one edge of the hybrid zone to 0% on the opposite edge. Similarly, the frequency of B. variegata–specific alleles decreases across the hybrid zone (beginning at the opposite edge) from close to 100% to 0%. 1. On the following graph, map the general percentage of each type of species-specific genes across the hybrid zone. Use an X to indicate the frequency of B. bombina–specific genes and an O to indicate the frequency of B. variegata–specific genes.
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O O
X
% of genes in the population
X O X
O
X 0%
O B.b edge of range
B.v edge of range
2. How would you predict this gene distribution would change over time if: a. reinforcement occurs? 100%
X X X O X O X O X O O O
% of genes in the population
0% B.b edge of range
B.v edge of range
Note: In the graph above, each X or O represents the percentage of B.b or B.v alleles in a sampled individual. If reinforcement occurs and is extreme, each toad would contain either entirely B.b-specific genes or entirely B.v-specific genes— that is, there would be no hybridization or mixing of genes.
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b. fusion occurs? 100% % of genes in the population
X X X O X O X O X O O O
0% B.b edge of range
B.v edge of range
Note: In the graph above, each X or O represents the percentage of B.b or B.v alleles in a sampled individual. If fusion or weakening of reproductive barriers occurs, so much gene flow may occur that the gene pools of the two species will become alike. In the extreme, a new hybrid species would replace the separate species. c. Hypothetically, which (a or b) is more likely to occur if all environmental conditions across the two species ranges are similar? Explain. Similar environmental conditions across the range imply no selective advantage for one set of genes over the other. As a result, fusion would be more likely. This assumes hybrids are viable and not at a selective disadvantage for survival compared to the parent types. d. Hypothetically, which (a or b) is more likely to occur if the environmental conditions vary gradually across the species ranges such that one end of the range was much colder than the other for example? Explain. Very different environmental conditions at the ends of the hybrid range imply that the two species are differently adapted. As a result, reinforcement would be favored over fusion. This assumes that even if hybrids are viable, they are at a selective disadvantage for survival at the edges of the hybrid range compared to the parent types. e. As noted in the text, what changes in allele frequencies have been recorded over this hybrid range in the last 20 years? What does this indicate? In reality, there has been little change observed in the hybrid ratio for these two species over the past 20 years. The authors propose that this may be the result of extensive gene flow from outside the zone.
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24.2 Test Your Understanding Read the following statement. Then, based on your knowledge of cell biology, genetics and evolution, decide to agree or disagree with the statement. Whichever you decide, write a short paragraph that provides solid evidence defending your choice. “When you think about sexual reproduction, it makes no sense. After all, evolution selects for organisms that are best fit. In a population of sexually reproducing organisms, a mutant that reproduced asexually would increase its representation more quickly than the wild type, that is, it would have a higher fitness. So we should expect asexual reproduction to be much more widespread among eukaryotic species than it is.” While there are many ways of addressing this, a reasonable student answer may be similar to the following: Because all members of an asexually reproducing species can produce offspring, one could argue that asexual reproduction results in a more rapid population expansion than sexual reproduction (where only the females produce the offspring). However, this does not necessarily mean that it is more beneficial for the population to only reproduce asexually. The recombination of genes that occurs in meiosis and fertilization in sexual reproduction generates much of the genetic variation on which natural selection acts. Asexual reproduction is limited in the sense that many fewer new genetic material/ combinations are introduced into the offspring. Offspring are for the most part clones of their parents. It is genetic variation that provides new combinations of genes that may allow for future adaptation and evolution in changing environments.
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Notes to Instructors Chapter 25 The History of Life on Earth What is the focus of these activities? All the evidence we have been able to gather indicates that life on Earth began about 3.5 billion years ago. For the first 1.5 billion years, life-forms on Earth were similar to today’s prokaryotes.
What are the particular activities designed to do? Activity 25.1 What do we know about the origin of life on Earth? Students can use this activity to review the types of evidence that exist for the origin of life on Earth as well as current ideas about the characteristics of early life-forms. Activity 25.2 How can we determine the age of fossils and rocks? This activity is designed to help students understand some of the methods used to age the Earth.
What misconceptions or difficulties can these activities reveal? Activity 25.1 Students often have difficulty “putting all the pieces together.” This type of activity will help them integrate the various ideas and theories about the evolution of life on the early Earth. Activity 25.2 While the math involved here is fairly simple, many students still have difficulty with it. Actually doing a few problems will help them understand not only how to do the calculations, but the theoretical reasoning behind them.
Answers Activity 25.1 What do we know about the origin of life on Earth? Construct a concept map of conditions on the early Earth and the origin of life-forms. Be sure to include definitions or descriptions of all the terms in the list below. Keep in mind that there are many ways to construct a concept map. • Begin by writing each term on a separate sticky note or piece of paper. • Then organize the terms into a map that indicates how the terms are associated or related. 180
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• Draw lines between terms and add action phrases to the lines that indicate how the terms are related. • If you are doing this activity in small groups in class, explain your map to another group of students when you are done. Here is an example: and therefore
no oxygen
no ozone layer resulting in
which contained
a reducing atmosphere
The early Earth had
high levels of UV radiation
(and)
Terms no oxygen reducing atmosphere high-oxygen atmosphere sunlight electrical discharge (for example, lightning) amino acids ozone layer sugars nucleic acids DNA inorganic compounds organic compounds carbon dioxide
protobiont micelle phospholipid bilayer ammonia phospholipids water Stanley Miller methane molecular clocks heat heterotrophs autotrophs high levels of UV radiation
low levels of UV radiation prokaryotes RNA world Eukarya energy source carbon source mode of nutrition anaerobic bacteria cyanobacteria (blue-green algae) Gram stain antibiotics penicillin
Use the understanding you gained from creating the concept map to answer the questions. 1. Modern theory suggests that the early (pre-life) atmosphere on Earth was a reducing one. Why (for what reasons) is it believed that oxygen was not present when life formed on Earth? One key piece of evidence is the lack of oxidized iron in rocks that are more than 2.7 billion years old. Therefore, oxygenic photosynthesis is thought to have evolved about 2.7 billion years ago. As oxygen accumulated in the oceans, iron oxides were formed and precipitated out. Eventually enough oxygen was formed not only to Activity 25.1
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saturate the ocean water but also to outgas or move from the ocean into the atmosphere. As oxygen levels rose in the atmosphere, iron compounds on land also oxidized. (Note: Rust is a form of oxidized iron.) 2. What proposed energy sources existed on this early (pre-life) Earth? Energy sources on the early Earth included geothermal energy, volcanic energy, and lightning. 3. In the 1950s, Stanley Miller performed a set of experiments to determine whether life could have evolved given the conditions stated in the answers to questions 1 and 2. a. How was the experiment designed? Miller created a sealed system. The only compounds in the system were water and a number of the different atmospheric gases thought to have been present on the early Earth—for example, methane, ammonia, and hydrogen. Miller used heat to vaporize the water. The steam passed through the chamber that contained the atmospheric gases. The vapor cooled, and the water was collected and recirculated to be heated again to produce steam. b. What were the necessary controls? The system had to be meticulously cleaned and sterilized to be sure it contained no organic molecules. This included evacuating all the air out of the system and replacing it with the ammonia, methane, and hydrogen. Another control was an identical system that was provided with no energy source. This would allow Miller to determine whether the macromolecules produced were contaminants or direct results of the experiment. In subsequent experiments, Miller also varied the concentrations and presence or absence of various compounds to determine which were required to produce specific types of macromolecules. c. What was produced in the experiment? In the first experiment, organic molecules, including some amino acids, were found in the system after only about 10 days of operation. In subsequent experiments, Miller was able to produce additional amino acids, carbohydrates, nucleotide bases, and other macromolecules by adjusting the gas composition of the atmosphere in the system. d. What did the results imply about the possible origin of life on Earth? Miller’s experiments showed that organic compounds could be produced in an anaerobic, reducing atmosphere. The results implied that life could have evolved under these conditions as well. 182
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e. There is general agreement that life must have evolved in the oceans originally and only much later invaded land. What factors of the physical environment on the early and evolving earth support these ideas? Changes in which of these were essential to allow life to survive on land? The early Earth’s atmosphere contained little or no oxygen. No ozone layer existed in the upper atmosphere. Though many of the early organisms on Earth did not require oxygen, with no ozone layer, the Earth was bombarded by high levels of UV light, which can easily mutate DNA and therefore kill organisms. Some scientists estimate that life on land was not possible until the atmosphere reached a minimum of about 10% oxygen and the ozone layer was formed. f. Most of us can’t imagine a world without oxygen. However, as you learned earlier, chemically oxygen is a powerful oxidizing compound. What effect(s) would the increase in oxygen levels of the atmosphere have on the organisms that existed at that time? Most anaerobic organisms (e.g., prokaryotes) alive at the time would be unlikely to have mechanisms to counteract the oxidizing effects of oxygen. As a result, most of them would have died off. Only those that lived in anaerobic environments that were not exposed to oxygen and those that had random mutations that equipped them with antioxidant capabilities (e.g., peroxidases) would have survived.
Activity 25.2 How can we determine the age of fossils and rocks? To determine the age of fossils and rocks, scientists determine the amounts of radioactive compounds and their stable daughter products present in the sample. Radioactive elements are known to decay into stable daughter compounds at specific rates. A number of radioactive compounds, their stable daughter compounds and their half lifes are shown in the table below. Radioactive Compound
Stable daughter compound
Half life
Carbon 14
Nitrogen 14
5370 years
Potassium 40
Argon 40
1.25 billion years
Rubidium 87
Strontium 87
48.8 billion years
Thorium 232
Lead 208
14 billion years
Uranium 235
Lead 207
704 million years
Uranium 238
Lead 206
4.47 billion years
Activity 25.2
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For dating rocks, the potassium-argon method is often used because: • Argon is a gas. When rock is molten, any existing argon gas will escape. As a result, newly formed rocks contain no argon. • As the 40potassium naturally occurring in the rock decays to 40argon, the 40argon is trapped in pocket in the rock. The ratio of 40potassium to 40argon can be measured to determine the rock’s age. For dating organic material, carbon-nitrogen dating is often used because: • The ratio of radioactive to nonradioactive carbon dioxide in the atmosphere is fairly constant over time. As a result, the levels of these elements in organic tissue remain relatively constant as long as the organism is alive. • Once the organism is dead, no new inputs of carbon occur and the existing radioactive-carbon-to-nonradioactive-carbon ratio will decrease over time as the radioactive carbon decays. 1. In one half life, half of the original radioactive compound will decay into its stable daughter compound. a. If a newly formed rock contains 100 units of 40potassium (40K), how many units of 40pottassium would it contain after 1.25 billion years? It would contain 50 units. b. How many units of 40argon (40Ar) would the rock contain when newly formed vs. after 1.25 billion years? The rock would contain 0 units of 40argon when newly formed and 50 units after 1.25 billion years. c. After the 1.25 billion years, what would be the ratio of 40K to 40Ar in the rock? Half of the 40K would be gone. The ratio would be 50 units of 40K to 50 units of 40 Ar, or 50:50 or 1:1. d. After 2.5 billion years, what would be the ratio of 40K to 40Ar? After the next 1.25 billion years, 25 of the 50 remaining 40K molecules would decay to 40Ar. In other words, only 1/4; of the original 40K would be present and the ratio would now be 25 units of 40K to 75 units of 40Ar or 1:3. 2. You are fossil hunting and find a trilobite fossil in an old riverbed. You have it radiometrically dated and are told it is 275 million years old. a. In this amount of time, how many half lifes of 40K have elapsed? 275 million divided by 1.25 billion = 2.75 ⫻ 108 / 1.25 ⫻ 109 = about 2 ⫻ 10⫺1 or 0.2 half-lives. 184
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b. Given your answer in a, what would be the ratio of 40K to 40Ar found in the fossil remains? In 0.2 half-lives, 0.2 of a half-life has elapsed. Therefore, 0.2 of half of the 40K would have decayed to 40Ar, or 0.1 or 1/10 would have decayed to 40Ar. As a result, the ratio of 40K to 40Ar would be 9 to 1. 3. You want to date some fabric that you have discovered at an archeologic dig. a. What method of dating would be best for this? Describe the general procedure for dating the cloth. Since human civilizations arose fairly recently in geologic time, it would be best to use carbon dating for the cloth sample. The lab would need to test the relative proportions of carbon 14 to carbon 12 in the sample. Carbon 14: carbon 12 ratios in organic material will decrease over time proportional to the half-life of carbon 14. b. Assume the cloth is 2,000 years old. How would the level of radioisotope used change in this period of time? Two thousand years is 2000/5730 or about 0.35 of the half-life for carbon 14. As a result, 0.35 ⫻ 0.5, or about 0.18 or 18%, of the normal level of carbon 14 would have been removed, leaving the carbon 14 level at about 82% of normal.
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Notes to Instructors Chapter 26 Phylogeny and the Tree of Life What is the focus of these activities? One of the most valuable lessons we can teach students is that their minds should always be open to new ideas and evidence. Scientific study is based on the accumulation of evidence and information in order to understand how organisms work, evolve, and function. As a result, scientific understanding is not static. New discoveries continually lead to new ideas and new ways of looking at existing evidence. This is true for all areas of science. In recent years, it has been especially true for the field of systematics.
What are the particular activities designed to do? Activity 26.1 How are phylogenies constructed? This activity is designed to help students get a better understanding of how taxonomy and phylogeny differ. They will learn how phylogenies (as cladograms) are developed and what we can do when new discoveries change our ideas about phylogenies. Activity 26.2 What is parsimony analysis? In this activity, students apply parsimony analysis to three bird species using morphological characters and compare their analysis to the analysis in the text that is based on a sequence of DNA. This is designed to help them address the question of which phylogeny is more correct. The answer is that each is correct for the data used. Activity 26.3 Put yourself in the professor’s shoes: What questions would you ask? This activity asks students to put themselves in the instructor’s shoes to see how exam questions are developed to test understanding. Students will learn to sort out the major ideas and concepts in this section and then use these concepts and ideas to develop exam questions that test understanding.
What misconceptions or difficulties can these activities reveal? Activity 26.1 Activity 23.2 pointed out that many students are unfamiliar with, and therefore uncomfortable with, making assumptions about the possible heritability of phenotypic characteristics. In Activity 26.1, many students will be similarly uncomfortable with the idea that “what we know” in science can change in light of new evidence or information. These same students will be uncomfortable trying to answer question 4b: “The phylogenies developed using DNA sequence analysis may differ from those constructed using morphology and physiology. How do scientists know which way is more correct?” 186
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This is a good place to reintroduce the idea that not only are organisms evolving but also that our understanding of them and all of life evolves as we gather new evidence. In other words, what we know or teach as fact today may be modified in the future. Activity 26.2 This activity reinforces the idea that our present understandings may be modified in the future. To help students understand that this is what science is all about, present this example: Today it is a fact that there is no cure for AIDS. In the future, I don’t think any of us would be upset if new evidence proves this fact false.
Answers Activity 26.1 How are phylogenies constructed? Construct a modified concept map to relate the ideas of phylogeny and systematics listed below to the phylogenetic tree on the next page. • Begin by writing each term on a separate sticky note or piece of paper. • Then indicate how the terms are associated or related to each other and to the phylogenetic tree on the next page. • Be sure to include definitions or descriptions of all the terms as you use them to explain these relationships. Terms clade cladistics phylogenetic tree homology analogy
monophyletic polyphyletic paraphyletic convergent evolution shared primitive character
shared derived character outgroup ingroup taxonomy phylogeny
Activity 26.1
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Legs
lig al
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Amniotic egg
I
J
K
Feathers
s) or
d a s r i ke an s lia les au sna ne i s i i d do s, ud s) co od s s) pi ard st rtle ro oc ve rd Le (liz C (cr A (bi Te (tu
Jaws; Two sets of paired appendages; Mineralized (ossified) skeletons and teeth
Vertebral column
C
Lungs or lung derivitives
E
F
G
H
Milk
ia ) al ls m ma m m a a M (m
nd ,a
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Osteichthyes (bony fishes)
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Use the understanding you gained from creating the concept map to answer the questions. 1. Compare the taxonomy of a group with its phylogeny (in general terms). Taxonomy
Phylogeny
a. Definition or purpose of a:
A taxonomy is a system for naming and classifying organisms.
A phylogeny categorizes organisms based on their evolutionary relatedness.
b. Types of characters used to develop a:
Most taxonomies are based on visible characteristics of organisms—for example similarities in morphology.
Phylogenetic relationships are determined based on historical information—for example, the fossil record and, more recently, similarities in DNA sequence.
c. What similarities could there be between the taxonomy of a given group and its phylogeny?
Phylogenetically related organisms often share similarities in morphology. As a result, many taxonomies reflect phylogenetic relationships as well.
d. What are the key differences between the taxonomy of a given group and its phylogeny?
By definition, taxonomies are developed to categorize and name organisms. There is no specific intent to determine phylogenetic relationships in the development of a taxonomy.
2. On the phylogenetic tree shown earlier, are the groups that contain humans, whales, crocodiles, and birds monophyletic, polyphyletic, or paraphyletic? Explain. The group containing humans, whales, crocodiles, and birds would be considered polyphyletic because no common ancestor is included. An example of a monophyletic group: The group circled with the solid line includes a common ancestor, and all the other members in the group are descended from this common ancestor. Another example of a polyphyletic group: The group circled with the simple dashed line includes two sets of organisms, each descended from a separate common ancestor. An example of a paraphyletic group: The group circled with the heavy dashed and dotted line is a paraphyletic. If you did not know the other species in the complete cladogram existed, this group would appear to be a monophyletic group. However, Activity 26.1
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because the other species do exist and are related to the common ancestor of this group, the group is paraphyletic. 3. Considering only the individual representative organisms in the phylogenetic tree (e.g., bird, whale, frog), which can be used as good examples of analogy or convergent evolution? As good examples of homology? Explain your reasoning. The fish and whale are a good example of convergent evolution. Fish evolved in aquatic habitats. The whale evolved from a mammalian terrestrial ancestor that took up an aquatic habitat. The ability of large animals to survive in an aquatic habitat is directly related to their abilities to move through that habitat efficiently to both gather food and avoid predators. As a result, over time, the body forms not suited to efficient movement in water were eliminated. Both the fish and whale have similar body forms. The body is torpedo-shaped, and the fins and tail are used as broad paddles for locomotion and orientation. The crocodile, bird, human, and whale are all homologous with regard to their pentadactyl limb structure. Although their specific structure is modified from species to species, each species has one upper limb bone, two lower limb bones, and five digits or phalanges (fingers or toes). 4. In recent years, DNA sequence analysis has been used in developing phylogenetic relationships among organisms. a. What type of DNA has been used most commonly in this analysis? Why was this type chosen over others? The genes for the small subunit ribosomal RNA molecule tend to change relatively slowly. As a result, they are used to determine relationships among groups of organisms that are not closely related. When scientists examine closely related groups, they tend to use mitochondrial DNA instead. Mitochondrial DNA mutates at a faster rate and is therefore more likely to provide evidence of more recent divergences or evolutionary changes that occur among closely related species. b. The phylogenies developed using DNA sequence analysis may differ from those constructed using morphology and physiology. How do scientists know which method is more correct? If the mechanisms used to develop competing phylogenies are both valid, there is no easy way to know which is correct. As a result, until more data become available, scientists may fuse the different results and produce a reconciled phylogeny. 5. Based on DNA sequence analysis three major domains of life have been proposed What are the three major domains of life? What sets of characteristics place organisms into one domain versus another? 190
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Major domains of life
Key characteristics For a more complete list, see Table 27.2.
Bacteria
Bacteria display the prokaryote cell structure. Bacteria have a circular chromosome. Bacterial cell walls contain peptidoglycan. All bacterial species use the same type of RNA polymerase. Bacteria use formyl methionine as their initiator amino acid to start protein synthesis.
Archaea
Archaea display the prokaryote cell structure. Archaea have a circular chromosome. Archaeal cell walls do not contain peptidoglycan. Several types of RNA polymerase can be found among the different archaeal species. Archaea use methionine as the initiator amino acid in translation.
Eukarya
Eukaryotes have the eukaryotic cell structure, which includes a double-membrane-bound nucleus and membrane-bound organelles. Eukaryotes have linear chromosomes. Eukaryotic cell walls (for example, around plant cells) do not contain peptidoglycan. Methionine is the initiator amino acid in translation.
Activity 26.2 What is parsimony analysis? To determine relatedness among species, parsimony analysis is often used. In Figure 26.15 of Biology, 8th edition, parsimony analysis is applied to matched DNA sequences from three species of birds. The same type of analysis can be applied to morphological and developmental characteristics of species. Review Figure 26.15 and the parsimony method used. Then conduct your own parsimony analysis using the morphological characteristics of these birds in the table below. 1. Begin by recording in the table below the morphological characteristics of the three bird species in Figure 26.15. 2. On the three possible phylogenies for these species (below the table) indicate the number of changes that must have occurred for each of the proposed phylogenies to be correct.
Activity 26.2
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Traits
A
B
C
D
Species
Eye ring (light/dark)
Red bar on breast (yes/no)
Red wing bar (yes/no)
I II III Ancestral
D L L D
N N Y N
N Y N Y
E
Light bar on Beak shape tail tip (curved/point) (yes/no) Y N N N
P C P C
I
I
III
II
III
II
III
II
I
3. Based on your analysis, which of the phylogenies is most parsimonious? How does this result compare to the result given in Figure 26.15? Using the characteristics in the table, the second proposed phylogeny is most parsimonious. This required six changes in morphology where the first required eight and the third required seven. In Figure 26.15, the first proposed phylogeny was most parsimonious. 4. Which of the proposed phylogenies (the one you developed or the one in Figure 26.15) is more correct? Explain your answer. Each is equally correct for the data used. Keep in mind that any analysis is only as good as the data used. In both cases, only a few characteristics/bases were used. In reality, many more characteristics/bases are used to make inferences about phylogenetic relationships. While some phylogenies based on analyses of DNA sequence data agree with those based on morphological and behavioral characteristics, in others there is disagreement. Because such analyses are based on comparison of a limited number of genes, scientists are cautious in declaring one more correct than the other. Rather, they will state the differences and what these may imply about our previous ideas. Scientists will seldom make definitive statements about phylogenies based on DNA sequence analysis until a significant amount of research on additional gene sequences is shown to support the findings.
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Activity 26.3 Put yourself in the professor’s shoes: What questions would you ask? One of the best ways to study for an exam is to put yourself in the professor’s shoes. For example, ask yourself: What questions would I ask about the material if I were Professor __________. Asking and answering such questions are good practice for taking the actual exam. They also help you to better understand and organize the major ideas and concepts you have studied. Write three exam questions designed to test how well a student understands the major concepts in Chapters 22–26. Indicate the correct answer to each question and also tell the reason why each alternative answer is incorrect. Your questions should be of the following types: I. Problem solving or application of a concept or principle to a problem For example: The little-known hypothetical organism Skyscra parius is a long-necked animal that feeds on the leaves of Australian trees that grow to heights of 30 feet. Being a hooved animal, S. parius cannot climb trees, so it feeds much like modern-day giraffes do. Fossil evidence indicates that the ancestors of S. parius had fairly short necks. Read the arguments presented below. For each, indicate whether or not the factors described could have affected neck length or tree height over the course of evolution of S. parius. [A = factors described could have affected neck length; B = factors described probably did not affect neck length] A B A
1. S. parius ancestors likely demonstrated significant variation in neck length, with some having shorter necks and others having longer necks. 2. When first born, juveniles of S. parius were much shorter than adults, so they were not able to compete successfully with adults that had longer necks. 3. Female S. parius preferentially mate with longer-necked males.
II. Translation: the ability to recognize concepts restated in a different form or to restate concepts in a different form For example: In what ways are the structure and function of the angiosperm seed and the amniotic egg (in this example, the chicken’s egg) similar? In what ways are they different? T/F 1. Both the angiosperm seed and the chicken’s egg contain stored food for the early development of the embryo. T/F 2. When released from the plant or laid by the chicken, both the angiosperm seed and the chicken’s egg contain a partially developed diploid embryo/offspring generation. T/F 3. The seed coat of the seed and the shell of the egg help prevent desiccation (water loss). T/F 4. Seeds may remain dormant and viable for hundreds of years; the same is true of chicken eggs. Activity 26.3
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Write your exam questions in the spaces provided. Exam Question 1:
Answer:
Exam Question 2:
Answer:
Exam Question 3:
Answer:
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Notes to Instructors Chapter 27 Bacteria and Archaea What is the focus of these activities? Activity 27.1 How diverse are the Archaea? Relatively recent studies indicate that prokaryotes are incredibly diverse, so diverse that they have been split into two domains: the Archaea and the Eubacteria. This activity helps students see that it is likely that early life-forms were more similar to the Archaea. In other words, life on the early Earth was very different from what we see today. Activity 27.2 How has small size affected prokaryotic diversity? This activity is designed to help students understand how small size has limited morphological diversity but promoted biochemical (metabolic) diversity among the prokaryotes.
What misconceptions or difficulties can these activities reveal? Activity 27.2 We often state that the prokaryotes are more diverse metabolically and the eukaryotes are more diverse morphologically. Then we go on to indicate that a wide variety of metabolic types exist in the prokaryotes. As instructors, we understand that any single species of prokaryote has a specific type of metabolism. When we talk about the variety of metabolic types, we mean that different species of prokaryotes can display very different types of metabolism. Some students don’t have this same understanding, however. They have the impression that a single species can display many different types of metabolism.
Answers Activity 27.1 How diverse are the Archaea? 1. The Archaea are divided into two major groups: the Euryarchaeota and the Crenarchaeota.
Notes to Instructors
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a. What characteristics are used to place organisms into each of these groups? Archaean group
Characteristics
Euryarchaeota
This group includes the methanogens, many extreme halophiles, and some extreme thermophiles. The methanogens are strictly anaerobic. The extreme halophiles are aerobic and require high environmental salt concentrations (>20%). Many nonextremophile species of Euryarcheota have also been discovered, for example, in soil samples and lake sediments. Most of these species were discovered by “genetic prospecting”—use of Euryarcheota DNA primer sequences and PCR on soil samples. As a result, we know that these species exist, but at present do not know much about their biochemistry or ecology.
Crenarchaeota
Most of the thermophilic species of Archaea fall into this category. Optimal temperatures for these species are between 60 and 80C. As with the Euryarcheota, many nonextremophile species of Crenarcheota have also been discovered. As with the newly discovered Euryarcheota, at present, we have information about the DNA sequences of these species, but little information about their biochemistry or ecology.
b. Two additional groups of the Archaea have been proposed. What new groups are these?
What characteristics do members of these groups have?
Korarcheota
Based on comparison of DNA sequences, these species found in hot springs in Yellowstone Park appear to be the oldest lineage of Archaea known. The DNA and proteins of these species have modifications that allow them to survive and function at temperatures above 100°C.
Nanoarcheota
Four species have been discovered in hydrothermal vents and hot springs. These are distinguished by their size, only 0.4 m in diameter, and the fact that their DNA contains only about 500,000 bp.
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2. There is great diversity in the ways different species of microbe • obtain energy for metabolic functions, and • obtain carbon for building the macromolecules of life. Fill in the chart to indicate how these characteristics are used to describe the nutritional type or nutritional classification of organisms. (Refer also to Table 27.1 on page 564 in Biology, 8th edition.) Plants
Animals
Bacteria and/or Archaea
Energy source
Light
Organic compounds
Light
Inorganic chemicals
Light
Organic compounds
Carbon source
Inorganic CO2
Organic compounds
CO2
CO2
Organic compounds
Organic compounds
Mode of Nutrition
Autotroph
Heterotroph Photoautotroph
Chemoautotroph
Photoheterotroph
Chemoheterotroph
27.1 Test Your Understanding 1. Bacteria first appear in the fossil record about 3.5 billion years ago. Humans first appear in the fossil record only a few million years ago. Given this, which group would you say is more highly evolved? Answers to this question depend a great deal on how one defines evolved. Many kinds of comparisons can be made. We provide examples below. a. What kinds of arguments or evidence would you use to support the idea that bacteria are more highly evolved? If we consider organisms that are closely fit (in their metabolism, for example) to their environment as more highly evolved, then many of the species of bacteria are more highly evolved than humans. b. How would you support the idea that humans are more highly evolved? If we use the term evolved to indicate major changes in morphology that allow organisms to survive in a number of different habitats, then humans are more highly evolved.
Activity 27.1
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2. Given what you know about the origin of life on Earth, you want to look for life on other planets. a. What characteristics of the planet’s environment would you look for? Explain your reasoning. Although we know that life can exist in the absence of oxygen, to the best of our knowledge it cannot exist in the absence of water. Therefore, one of the first characteristics of the environment to look for would be the presence of free water. b. What kind(s) of life would you look for? Explain your reasoning. To the best of our knowledge, free oxygen is not present on other planets. Therefore, it would be best to look for life-forms that have characteristics similar to those of anaerobic archaeal species. c. What tests or probes would you use to find the kind of life you proposed in question b? Explain your reasoning. To answer this question, you need to make the assumption that life on other planets may have evolved similarly to life on Earth. This assumption is supported by Miller’s experiments, which indicated that key organic compounds present on Earth can be produced under reducing atmospheric conditions. Given that assumption, you could look for some of the compounds that we know are common to life. For example, you could analyze soil or other samples for the presence of polymers of carbon—carbohydrates, proteins, lipids, or nucleic acids. On the assumption that the life-forms have DNA or RNA as their genetic material, you could use radioactive DNA or RNA probes. Making these similar to highly conserved genes in Archaea and other organisms would give you a better chance for finding possible homology.
Activity 27.2 How has small size affected prokaryotic diversity? It is often said that bacteria tend to be more diverse biochemically and eukaryotes tend to be more diverse morphologically. Answer questions 1–7. Then write a summary argument (question 8) to support the statement above. In other words, write an argument describing how small size has limited morphological diversity but promoted biochemical (metabolic) diversity among the prokaryotes. 1. All living organisms must maintain a relatively constant internal environment. Maintaining this environment means that a certain concentration of each substance must be maintained per unit volume of the cell. The ability of the cell to maintain a specific concentration of a substance is affected by
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• the ability of the substance to diffuse through the membrane, • the overall size and shape of the cell, and • the maximum amount of the substance a given area of membrane can transport per unit time. a. How do surface-area-to-volume (SA/V) ratios change as the size and shape of cells and organisms change? To answer this, calculate the SA and V of a cube 1 mm on a side. Then do the same for cubes that are 2 mm and 4 mm on a side and compare their SA/V ratios. Cubes:
1-mm square
2-mm square
4-mm square
Linear dimension
1 mm
2 mm
4 mm
Surface area (SA)
6(1 mm2) = 6 mm2
6(2 mm2) = 24 mm2
6(4 mm2) = 96 mm2
Volume (V)
(1 mm)3
(2 mm)3 = 8 mm3
(4 mm)3 = 64 mm3
SA/V ratio
6 mm2/1 mm3
24 mm2/8 mm3 = 3 mm2/1 mm3
96 mm2/64 mm3 = 1.5 mm2/1 mm3
b. In general, how does surface area change as linear dimensions increase twofold? Surface area increases as a function of the square of the increase in linear dimension. For example, the surface area of a 1-mm square is 6 mm2. The surface area of a 2-mm square is 24 mm2. Two millimeters in linear dimension is two times 1 mm. The increase in surface area should then be 22 (the number of times bigger that 2 mm is than 1 mm2). In fact, 24 mm2 is four times greater than 6 mm2. Similarly, the 4-mm square is two times the linear dimensions of the 2-mm square and its surface area (96 mm2) is four times the surface area (24 mm2) of the 2-mm square. On the other hand, the 4-mm square is four times the linear dimensions of the 1-mm square. Therefore, its surface area should be 42 or 16 times that of the 1-mm square, and it is (96/6 = 16). c. In general, how does volume change as linear dimensions increase twofold? Volume changes as the cube of the difference in linear dimension. If linear dimensions double, the volume increases by 23 or 8 times. The volume of the 1-mm cube is 1 mm3, the volume of the 2-mm cube is 8 mm3, and the volume of the 4-mm cube is 64 mm3. Each doubling in linear dimensions produces an eightfold (23) increase in volume. If you quadruple the linear dimensions from 1 mm to 4 mm, the volume increases by 43 or 64 times.
Activity 27.2
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d. In general, how do SA/V ratios change as linear dimensions increase twofold? If you compare the SA/V ratios among the cubes, you can see that as linear dimension increases twofold (two times), the SA/V ratio is halved. (Note: This assumes that the “organisms” under consideration retain the same general shape—in this case, square.) 2. Assume a bacterium is 10 m in linear dimension. Fill in the chart. a. If modeled as a cube, what would its SA, V, and SA/V ratio be? b. If modeled as a sphere, what would its SA, V, and SA/V ratio be? c. What are the SA and V values and the SA/V ratios for a cube-shaped eukaryotic cell that is 100 m in linear dimension? a. 10-m bacterium as b. 10-m bacterium a cube as a sphere
c. 100-m eukaryote, cube-shaped
SA
6(10 m)2 = 600 m2
6(100 m)2 = 6 104 m2
V
(10 m)3 = 1,000 m3 4/3 r3 = 523.3 m3
SA/V ratio
6 m2/10 m3
4r2 = 3.14(5 m)2 = 314 m2
(100 m)3 = 1,000,000 m3
314 m2/523.3 m3 = 6 m2 /100 m3 6.2 m2 /10 m3
3. Assume that every cell requires a minimum of 1 unit of oxygen per m3 per second to stay alive. Fill in the chart. a. How much oxygen must cross each m2 of surface area per second in the 10-m bacterium versus the 100-m eukaryote to keep each alive? b. What effects might this difference have on metabolic rates in these organisms?
10-m bacterium
100-m eukaryote
a. Oxygen/m2 of SA/second
10 units/6 m2/second
100 units/6 m2/second
b. Possible effect(s) on metabolic rate
The metabolic rate can be relatively high.
If there is any difficulty supplying oxygen at this rate, the metabolic rate would be lower (compared to that of the 10-m bacterium).
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4. Given what you know about cell membranes, is there likely to be a maximum upper limit on the number of molecules of a substance that can cross a given area of membrane per unit time? If so, what factors would be involved in determining the maximum upper limit? Because oxygen can diffuse through the phospholipid bilayer, its movement into the cell would be limited only by its rate of diffusion and the total surface area available. If we assume a finite maximum rate of diffusion, then aerobic cells would be limited in their maximum size by this maximum rate. If a substance requires a transport protein to facilitate its movement across the membrane, then its maximum rate of movement into (or out of) cells would also be limited by the total number of available transport proteins in the membrane. As you answer questions 5, 6, and 7, fill in the chart on the next page. 5. a. On average, how large is a prokaryotic genome? b. On average, how many times larger is a eukaryotic genome? c. Are these genomes haploid or diploid? a. Prokaryotic genomes can contain a thousand to several thousand genes. b. In contrast, a eukaryotic genome can contain tens of thousands of genes. For example, the H. influenzae bacterium contains 1,700 genes; A. thaliana (plant) contains 26,000 genes; and the human contains between 30,000 and 40,000 genes. c. Many eukaryotic genomes are diploid. Prokaryotic genomes are haploid. 6. Under ideal conditions, many bacterial cells can reproduce or duplicate themselves within an hour. Some species (for example, E. coli) can reproduce every 20 minutes under ideal conditions. If you inoculate 1 liter of culture medium with one bacterium per milliliter (of medium) at time t = 0, how many bacteria will be present in 1 milliliter of the culture medium after 10 hours? (To do this calculation, assume that in these culture conditions the bacteria duplicate—or double in number—every hour.) In the first hour, the one bacterium will become two. The number will double every hour over the next 9 hours, or it will double during each of the next 29 intervals. As a result, at the end of 10 hours, there will be 29 bacteria in the flask (29 is approximately equal to 103). If the bacteria double every 20 minutes (instead of every hour), there would be 229 or about 108 bacteria/mL after the 10 hours. (Note: This calculation assumes no limitations are placed on the ability of the bacteria to grow and replicate as the number of bacteria increases over time.)
Activity 27.2
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7. Assume that the mutation rate of bacterial cells in culture is 106 to 108. This means that in every 1 million to 100 million cells produced from a single original cell, you would expect to find at least one mutation. How many mutations would you be likely to find in 1 liter of the 10-hour culture of the cells you grew in question 6? Prokaryote
Eukaryote
Genome size
5a. 1,700 genes
5b. 30,000 genes
5c. Haploid versus diploid
Haploid
Diploid
6. Maximum reproduction rate
Every 20 minutes
One hour to many hours
7. Mutation rate per hour
If 106 bacteria contain one mutation, then 1,000 mL of culture containing 103 bacteria per mL would contain about 101 or 1 mutation every 10 hours. If the culture contained 108 bacteria/mL, it would contain 103 or 1,000 mutations in the 10 hours.
The mutation rate per unit of time for eukaryotic cells would appear to be lower because the reproduction rate of these cells is much slower.
8. Using all of the information in this activity, write an argument entitled: “How small size in prokaryotes played a role in limiting their morphological diversity and promoting their biochemical (metabolic) diversity.” There are many possible ways to make this argument. All arguments should mention the fact that the genome of prokaryotes is haploid. As a result, any mutations have immediate effects on the phenotype of the organism. High surface-area-to-volume ratios mean metabolic rates can also be very high. High metabolic rates support rapid reproduction (cell growth and fission), so the apparent rate at which mutations show up in the population is much higher in prokaryotes. (Note: The metabolic rate depends a great deal on whether the prokaryote is aerobic or anaerobic and other characteristics.) Arguments should also consider how the absolute mass or volume of an organism can limit its morphology. A single cell 1 m in diameter can assume only a limited number of different shapes. Multicellular organisms, in contrast, are much less limited in their morphology. Of course, they have many more cells and a considerably greater volume per cell to work with. 202
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Notes to Instructors Chapter 28 Protists What is the focus of this activity? This activity provides students with an overview of the early evolution of eukaryotic, single-celled organisms. The first eukaryotes appear in the fossil record around 2.1 billion years ago. Current evidence indicates that eukaryotic cells arose as a result of endosymbiosis (sometimes multiple endosymbioses) among ancestral prokaryotes. The early eukaryotic cells that resulted underwent great diversification to become the five supergroups of Eukarya.
What is the particular activity designed to do? Activity 28.1 How has endosymbiosis contributed to the diversity of organisms on Earth today? This activity is designed to help students begin to understand the vast diversity of the protists as well as the some of the evidence and thought processes that were involved in the development of the endosymbiotic theory.
Answers Activity 28.1 How has endosymbiosis contributed to the diversity of organisms on Earth today? 1. The protists been called a “catch-all group.” What does this mean? Explain. This is a paraphyletic group, including all eukaryotes except those assigned to the fungi, animal, or plant kingdoms. Some systematists have indicated that this group may include 20 or more different evolutionary lineages or kingdoms. The protists range in size from microscopic to macroscopic. Some of the large brown algae can be 60 m long. About the only thing the group—formerly known as Protista—has in common is their type of reproduction; that is, all species have a single cell as a reproductive cell. (Even in the large kelp and laminaria, single cells on the margins of the fronds undergo meiosis to produce eggs or sperm, which are released into the water.) 2. There are more than 100,000 recognized species of protists; in Biology, 8th edition, they are subdivided into five supergroups and 11 major grades. Even though the protists are extremely diverse, we can recognize some common themes in their evolution, including the evolution of complex cell structure, novel genetic Notes to Instructors
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recombination strategies, and complex life cycles. In addition, this group as a whole has great ecological importance. Provide at least two examples of protists from Chapter 28 in Biology, 8th edition, that illustrate each theme. Note: Any number of different examples can be used to answer the following questions; a few are given here. Theme
Examples
a. Complex cell structure
Euglena Contains light detector, eyespot, contractile vacuole, pyrenoids, and chloroplast. (See Figure 28.7 of Biology, 8th edition, for a diagram and description of their functions.) Paramecium Contains contractile vacuole, oral groove and cell mouth, micro- and macronuclei, and cilia. (See Figure 28.11.)
b. Novel genetic recombination strategies
Paramecium Has micro- and macronuclei. The micronuclei function in sexual recombination and reproduction. The macronuclei function in growth, maintenance, and asexual reproduction. (See Figure 28.11.) Laminaria Displays alternation of generations between gametophyte (haploid) and sporophyte (diploid) multicellular forms. (See Figure 28.16.)
c. Complex life cycles
Plasmodium This malaria-causing organism has a two-host life history. It reproduces asexually in its human host and sexually in its mosquito host. (See Figure 28.10 for additional details.) Cellular slime molds (such as Dictyostelium) Have a life cycle that includes single-celled feeding amoebas (n). In the sexual phase, haploid amoebas can fuse to form a zygote, which immediately undergoes meiosis. Amoebas can also aggregate to form colonies. The aggregate can develop into a stalked asexual fruiting body that gives rise to spores, which can survive unfavorable environments. (See Figure 28.25.)
d. Ecological importance
All the species of photosynthetic protists (in association with the cyanobacteria) provide the basis for the food webs of freshwater and marine environments. The absorptive or fungus-like protists aid in the decomposition and recycling of organic matter (carbon compounds).
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3. In the late 1960s (and since), Lynn Margulis provided considerable evidence for the endosymbiotic theory of the origin of various organelles in eukaryotic cells. a. What is the endosymbiotic theory?
The endosymbiotic theory proposes that eukaryotes originated when ancestral prokaryotes (most likely archaeal species) engulfed the cells of formerly free-living prokaryotes and enclosed them in vacuoles. These endosymbionts became major eukaryotic organelles. b. Which two eukaryotic organelles were proposed to have arisen as endosymbionts?
c. What evidence did Margulis present to support each organelle as an endosymbiont?
Mitochondria
Mitochondria are bounded by two membranes. In addition, they contain their own circular DNA (with no associated histones) and 70s ribosomes. Some of the inner membrane proteins are more similar to those of prokaryotes than eukaryotes.
Chloroplasts
Chloroplasts are also bounded by two membranes. They also contain their own circular DNA (with no associated histones) and 70s ribosomes. Some of their inner membrane proteins are more similar to those of prokaryotes than eukaryotes.
4. The mitochondria of the chromalveolates and the chlorarachinophytes (in the clade Rhizaria) are both thought to have arisen as a result of secondary endosymbiosis. a. What is secondary endosymbiosis? Secondary endosymbiosis is said to occur when a heterotrophic eukaryote engulfs a photosynthetic eukaryote, which then becomes an endosymbiont. b. What evidence supports the idea of secondary endosymbiosis? Evidence includes the following: • The engulfed green alga endosymbiotic in chlorarachniophytes is surrounded by four membranes. It carries out photosynthesis in its own plastids and contains a vestigial nucleus or nucleomorph. • The structure and DNA of Chromalveolata chloroplasts have many similarities to those of red algae. These chloroplasts also have more than two membranes surrounding them.
Activity 28.1
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5. Structurally, Giardia lamblia lacks complete mitochondria and was once thought to be an example of what the earliest living eukaryotes may have looked like. a. If Giardia is similar in structure to the earliest living eukaryotes, what does this imply about the order of evolution of the various eukaryotic organelles (that is, nucleus, cytoskeleton, mitochondria, chloroplasts)? Giardia has several flagella, a primitive cytoskeleton, and a nucleus. However, it contains no plastids and no mitochondria. Giardia’s structure would indicate that the nucleus and cytoskeleton in eukaryotes evolved prior to the endosymbiosis of mitochondria or chloroplasts. b. More recently, we have discovered that the Diplomonads (e.g., Giardia) and the Parabasalids (the other group of Excavates) have modified mitochondria. i. What types of mitochondria do they have? Diplomonads contain modified mitochondria called mitosomes, which lack electron transport chains. As a result, they are anaerobic and rely primarily on glycolysis. Parabasilids contain modified mitochondria called hydrogenosomes, which allow for anaerobic respiration but release hydrogen as a waste product. ii. What does this evidence indicate about the evolution of the various eukaryotic organelles? Evidence that both groups have modified mitochondria means that it is still uncertain which came first—the nucleus, the cytoskeleton, or the mitchondrion.
28.1 Test Your Understanding The metabolic pathways of organisms living today evolved over a long period of time— undoubtedly in a stepwise fashion because of their complexity. Considering everything you have learned to date about the evolution of life on Earth, put the following in the order in which they might have evolved, and provide an explanation for your arrangement. 4 prokaryotes capable of performing the Krebs cycle 5.5 eukaryotes capable of performing the Krebs cycle 2.5 prokaryotes capable of performing electron transport 5.5 eukaryotes capable of performing electron transport 1 prokaryotes capable of performing glycolysis 2.5 prokaryotes capable of performing photosynthesis 7 eukaryotes capable of performing photosynthesis
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This question is similar to the question at the end of the Chapter 10 activities. However, it is more detailed in that it requires integrating the endosymbiotic theory with the basic logic behind which biochemical pathway must have evolved first. 1—First, glycolysis in prokaryotes. This process is found in all eukaryotes and many prokaryotes. It takes place in the cytoplasm and can occur in the absence of oxygen. Because eukaryotes arose as endosymbionts with prokaryotes, the prokaryotes capable of photosynthesis and oxidative respiration must have evolved after the prokaryotes capable of these. 2.5—Second and third, prokaryotic photosynthesis requires electron transport, which cannot occur in the absence of oxygen. Both are ranked 2.5 (rather than 2 and 3) because photosynthesis requires electron transport. Similarly, it is difficult to envision an independent role for electron transport. Electron transport is required to convert NADH to NAD+. Because glycolysis produces 2 ATP (net) and 2 NADH, the addition of electron transport, which uses oxygen as a final electron acceptor, represents an advantage. Organisms can then gain 8 ATP (net) from glycolysis plus electron transport. 4—The Krebs cycle in prokaryotes cannot occur without a mechanism to convert NADH to NAD+. Electron transport must have evolved before the Krebs cycle. 5.5—Eukaryotes capable of glycolysis, the Krebs cycle, and electron transport (as oxidative phosphorylation) are thought to have evolved as a result of endosymbiosis of an amitochondrial organism and an aerobic bacterium. 7—Some of the eukaryotes with mitochondria are thought to have undergone another endosymbiosis with a cyanobacterium to produce chloroplasts in photosynthetic eukaryotic cells.
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Notes to Instructors Chapter 29 Plant Diversity I: How Plants Colonized Land Chapter 30 Plant Diversity II: The Evolution of Seed Plants What is the focus of these activities? By about 500 million years ago, increasing amounts of oxygen led to the formation of an ozone layer in Earth’s upper atmosphere. The ozone layer prevented much of the sun’s ultraviolet light from reaching Earth, and the land became habitable. To take advantage of this new habitat, however, organisms had to be able to survive in an environment in which water was not always readily available. This meant that these organisms had to be able to support themselves against gravity and deal with greater variations in environmental conditions—for example, temperature, chemistry, and light availability. Life on land also required new methods of gamete protection and transport.
What are the particular activities designed to do? Activity 29.1/30.1 What major events occurred in the evolution of the plant kingdom? Activity 29.2/30.2 What can a study of extant species tell us about the evolution of form and function in the plant kingdom? Activity 29.3/30.3 How are the events in plant evolution related? These activities are designed to help students understand the major events that took place in the evolution of the plant kingdom as well as the logic that was used in the development of our understanding of evolution.
Answers Activity 29.1/30.1 What major events occurred in the evolution of the plant kingdom? Construct a concept map that describes the early evolution of plant life on Earth. Be sure to include relationships among all the organisms and factors in the list on the next page. Keep in mind that there are many ways to construct a concept map. • Begin by writing each term on a separate it note or piece of paper. • Then organize the terms into a map that indicates how the terms are associated or related. • Draw lines between terms and add action phrases to the lines that indicate how the terms are related. 208
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Here is an example:
mitochondria
chloroplasts that became
that became
cyanobacteria
aerobic bacteria which were chimera of prokaryotic ancestors including
ancestral eukaryotes evolved from
ancestral chlor ophytes gave rise to
ancestral charophyceans If you are doing this activity in small groups in class, explain your map to another group of students when you finish it. Terms bryophytes pteridophytes lycophytes endosymbiont anaerobic bacteria cyanobacteria (blue-green algae) chloroplast mitochondria vascular tissue waxy cuticle charophyceans chlorophytes
alternation of generations megaspore microspore nonvascular plants seedless seeds angiosperm gymnosperm flowers xylem phloem microphyll megaphyll
spore gametophyte sporophyte egg gametangia root stem flagellated sperm archegonium antheridium pollen grain ovule
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Use the understanding you gained from constructing the concept map to answer the questions. 1. Describe the major problems ancestral land plants had to overcome before they could make the transition from water to land. Water was less available on land, and most of the available water was not on the surface. In addition, desiccation (water loss) was a problem in the atmosphere. On land, there were greater variations than in the ocean—for example, in environmental temperature and chemistry. On land, new types of transport mechanisms were required—for example, for food, water, gametes, and so on. Water was no longer available to support the plants’ “bodies” against gravity. In water, most of the ultraviolet light is filtered out in the first few centimeters of the water column. On land, the higher UV levels could easily bleach chlorophyll and damage or mutate nucleic acids. 2. Describe the major solutions to the problems in question 1 that can be found in today’s land plants. In other words, what mutations occurred that allowed organisms to make the transition to land? To reduce evaporative loss, a waxy cuticle covers the surface of cells in contact with the environment. Flavonoids, pigments that selectively filter UV light, are found in all land plants. Sporopollenin evolved to form resistant walls on land spores to protect them from desiccation. All land plants have protective gametangia (for example, archegonia or antheridia in lower plants) that protect haploid gametes against both desiccation and UV light. This feature must have evolved prior to or concomitant with the move onto land. Air spaces and guard cells with associated stomata provide additional protection against desiccation while also allowing for gas exchange in photosynthesis. The evolution of lignin added structural support to cells and supported them against gravity. Subsequent evolution of xylem and phloem allowed for greater efficiency in food and water transport. The seed, with its waxy coating and low water content, enabled the next generation of embryos to withstand environmental fluctuations in water availability and temperature until conditions were suitable for germination.
Activity 29.2/30.2 What can a study of extant species tell us about the evolution of form and function in the plant kingdom? Fill in the chart on the next pages to compare the major features of key groups of land plants with one another and with the charophyceans. 210
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Yes Yes No No
Yes
Some have air spaces but no guard cells No No
Yes
Yes
Vascular tissue
Stems containing vascular tissue
Roots or rhizomes No No
Stomata
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True leaves (contain vascular tissue)
Sh
Antheridia
K /N
Archegonia
l
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Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Microsporophylls, microspores, and empryo sacs
Microsporophylls, microspores, and pollen
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Angiosperms
Yes (pollen releases No flagellated sperm in some species)
Macrosporophylls, macrospores, and archegonia
Microsporophylls, microspores, and pollen
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Gymnosperms
Yes
No
No
No
No
No
No
No
No
No
Yes
Yes
2:07 AM
Yes
Yes
Yes
Cuticle
Yes
Yes
Yes
Pterophytes
Charophyceans (green algae)
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Flagellate sperm
Yes
Yes
Jacketed gametangia
Yes
Yes
Chlorophylls a and b
Yes
Lycophytes
Yes
Bryophytes
Peroxisomes
Feature
Plant Group
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Sporophyte and gametophyte both independent No
Yes
Sporophyte dependent on gametophyte for energy
Gametophyte dependent on sporophyte
No
No
Fruit
Sporophyte dominant
No
Flower
No
Yes
No (except in very early stages of development)
Yes
No
No
No
No
No
Yes
No (except in very early stages of development)
Yes
No
Yes
No
No
Yes
No
No
No
No No
Yes
Yes
Gymnosperms
No
No
Pterophytes
Yes
No
No
Yes
No
Yes
Yes
Yes
Yes
Angiosperms
No
No
Yes
No
Yes
No
No
No
No
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Yes
No
Seed
No
Lycophytes
Charophyceans (green algae)
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Gametophyte dominant
No
Bryophytes
Pollen
Feature
Plant Group
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Use the information in the chart you have completed to answer the questions. 1. Some of the major plant groups are listed in the following table from most primitive to most advanced. For each group, indicate what major characteristics make it more advanced than the preceding group. For example, how are ferns more advanced than mosses?
Plant group
Charophyceans Bryophytes
Lycophytes
Pterophytes
Gymnosperms
Angiosperms
Advance(s)
Peroxisomes
Jacketed
Dominance
Rhizomes
Pollen grain
Flower
over
help reduce
gametangia
of
increase the
removes the
attracts
preceding
photores-
(archegonia
sporophyte
ability to
need for water
animal
group
piration, and
and
generation
pick up
for transport of
pollinators
chlorophyll
antheridia).
helps reduce
water from
male gametes.
and increases
b is less
cuticle, and
the effect of
the soil.
Seed allows the
the
susceptible to
air spaces
UV mutation
True leaves
next-genaration
probability
UV bleaching.
help protect
of DNA.
(with
sporophyte
of
from
Vascular
vascular
offspring to
outcrossing
desiccation.
tissue,
tissue)
begin
and
xylem, and
increase
development
fertilization.
phloem
photosynthe
on a
Fruit
increase
tic surface
sporophyte
increases
support
areas and
parent plant.
dispersal
against
therefore
Seed prevents
probability.
gravity and
food
desiccation and
Offspring
provide
production.
allows embryo
(seeds) are
greater
to survive until
transported
efficiency of
conditions are
to new
water and
appropriate for
location
food
germination.
(away from
transport.
Seed also
the parent
includes food
plant) by
source for
animals.
early growth.
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2. How do the bryophytes differ from the seedless vascular plants? How are they similar? Refer to the previous table. Differences between bryophytes and the seedless vascular plants are: a. the dominant phase of the life cycle. In bryophytes the gametophyte stage is dominant; in seedless vascular plants the sporophyte is dominant. b. the absence/presence of vascular tissue. Bryophytes contain no vascular tissue. As a result, they do not have true stems, roots, or leaves. c. the absence of guard cells in bryophytes. Both bryophytes and seedless vascular plants have air spaces in their structure for gas exchange. However, the bryophytes do not have guard cells associated with their air spaces. Guard cells allow plants to control the size of the stomata. In most other major features (listed in the table), these two groups are similar. 3. The life cycle of all land plants includes an alternation of generations between a multicellular gametophyte phase and a multicellular sporophyte phase. Diagram the life cycle of a seed plant. Refer to Figures 30.6 and 30.10 in Biology, 8th edition. See also Figure 13.6 on page 252. a. What cellular division process always precedes formation of the gametophyte generation? Meiosis b. What cellular process always precedes formation of the sporophyte generation? Fertilization forms the zygote (the first cell in the sporophyte generation), and mitosis of the zygote forms the multicellular sporophyte. c. If the sexual generation gives rise to the gametes, what part of an angiosperm is sexual? The gametophyte generation, which is haploid, gives rise to the gametes and is therefore the sexual generation. d. If the sexual generation gives rise to the gametes, what part of a bryophyte moss is sexual? The gametophyte, or green, leaflike moss structures, give rise to the gametes in archegonia and antheridia. 214
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4. Until the evolution of the seed plants, land plants were dependent on the availability of water for reproduction. Explain why this was true. Explain how seed plants overcame the need for water in reproduction. Seed plants produce pollen rather than flagellated sperm. Pollen do not require water to move from one plant to another. Pollen can be carried by the wind or by animals to other plants. 5. Pollen, seeds, flowers, and fruits are considered among the most advanced characteristics in the plant kingdom. What evolutionary advantage(s) does each of these offer (relative to what existed before)? Pollen: As noted in question 4, pollen no longer requires free water for transport. Seeds: The seed contains a partially developed 2n embryo (the product of fertilization of the egg by a sperm from the pollen tube), a food store for its early development, and a protective coating (derived from the integuments surrounding the ovule) that allows it to survive periods of desiccation. You can think of the seed as an embryo, with a packed lunch and a raincoat, that is sent out into the world to make it on its own. A seed has two main advantages over a spore: The embryo is in a more advanced state of development, and the larger food store both allows for longer periods of dormancy and supports early growth phases of the seedling following germination. Upon germination, the food store in the seed allows it to continue developing roots, stem, and leaves until the plant breaks through the soil surface and is capable of photosynthesis. Flowers: The flower provides protection and nourishment for both the gametophyte (the embryo sac as it is developing) and the next-generation sporophyte (the developing seed after fertilization occurs). In addition, the flower often serves to attract specific pollinators that increase the possibility of successful pollination. Fruits: The fruit serves no function for the developing embryo or the seed. In fact, the fruit tends to contain hormones that prevent germination of the seeds (that is, maintain seed dormancy). Only after the fruit has rotted away (or is eaten away) can the seeds germinate. The fruit is generally thought to be a mechanism used by angiosperms to coopt or bribe animals to disperse their seeds (contained in the fruit). In some cases, this system has evolved to the point that certain species’ seeds will not germinate unless they have passed through the digestive tract of a particular animal species.
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Activity 29.3/30.3 How are the events in plant evolution related? 1. Working in groups of three or four, assign each student in the group one of the following events in plant evolution to research. Then give each student 5 minutes to report the results of his or her review to the other members of the group. Note: The following are minimal answers to these questions. Students may come up with other valid ideas in addition to these.
Events in Plant Evolution I. Evolution of vascular tissue Mutations in some land plants gave rise to vascular tissue. What advantage(s) did these plants have compared with land plants that did not contain any vascular tissue? The evolution of vascular tissue provided plants with support against the force of gravity. As a result, they could grow taller. When space is limited, taller plants have greater amounts of light energy available to them (compared to shorter plants, which may be shaded). As a result, taller plants would have faster rates of metabolism and growth. The addition of vascular tissue also allowed for quicker, more efficient transport of water and food throughout the plants. This also increased the potential for faster metabolism and growth. II. Evolution of roots and leaves Mutations in some land plants gave rise to roots, leaves, or both. What advantage(s) did these plants have compared with plants that did not contain roots or leaves? Roots increased a plant’s surface area for water absorption from the soil. Leaves with vascular tissue provided increased surface area for photosynthesis. As a result, some cells could specialize in functions other than photosynthesis (for example, roots for water absorption). III. The trend toward reduction of the gametophyte generation Mutations in some land plants gave rise to life cycles in which the gametophyte generation was reduced. What advantage(s) did these plants have compared with plants that did not have a reduced gametophyte generation? Reduction and enclosure of the gametophyte generation within the sporophyte helped protect the reproductive phase from mutations of the DNA, which could be caused by increased amounts of ultraviolet and cosmic radiation present on land (as compared to the ocean environment). A mutation in a haploid genome shows immediately. A mutation in a diploid genome can be masked by the presence of a nonmutated gene on a homologous chromosome.
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IV. Evolution of the seed Mutations in some land plants gave rise to the seed. What advantages did these plants have compared with plants that did not have seeds? The seed contains a partially developed 2n embryo (the product of fertilization of the egg by a sperm from the pollen tube), a food store for its early development, and a protective coating (derived from the integuments surrounding the ovule) that allows it to survive periods of desiccation. You can think of the seed as an embryo, with a packed lunch and a raincoat, that is sent out into the world to make it on its own. A seed has two main advantages over a spore: The embryo is in a more advanced state of development, and the larger food store both allows for longer periods of dormancy and supports early growth phases of the seedling following germination. Upon germination, the food store in the seed allows it to continue developing roots, stem, and leaves until the plant breaks through the soil surface and is capable of photosynthesis. 2. Work together as a group to determine how these events in plant evolution (I to IV) might be related. For example, which would have to come first (in evolution), which next, and so on? Another way to look at this question is to consider which of these events paved the way (or made it possible) for the other events to occur. Be sure to state evidence for your proposed evolutionary scheme. To do this, it is useful to ask yourself questions like these: Would it have been possible for the seed to evolve without vascular tissue first having evolved? Leaves? And so on. If yes, how could this have occurred? If no, why not? Because roots and leaves contain vascular tissue, vascular tissue must have evolved before roots and leaves. When we look at the evolution of dominance of the sporophyte generation versus evolution of the seed, it is most likely that the sporophyte generation became dominant before the seed evolved. This order is supported by fossil evidence, but even if it weren’t, this sequence seems most logical for the protection of both the gametophyte generation and the subsequent or offspring sporophyte generation. In combination, it is again logical (and supported by the fossil record) that the vascular system and leaves and roots evolved prior to the seed because the production of seeds is costly in terms of energy. 3. Incorporate the following true observations into your analysis of how the events in evolution (I to IV) could be related. a. The fossil record (spore evidence and so on) indicates that the first plants on Earth were similar to modern-day bryophytes. This indicates that the first plants were likely to have a dominant gametophyte generation and no vascular tissue. b. The very large marine brown algae (for example, giant kelp) can grow to heights of 30 feet or more. These algae have both leaflike and stemlike structures and Activity 29.3/30.3
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are held to the bottom of the sea by a holdfast. When examined microscopically, the algae are found to contain transport vessels that are similar to phloem in function. These giant kelp do not contain any xylemlike vessels, however, nor do they have roots. This provides evidence for the coevolution of transport systems. In the ocean, a food transport system evolved, and this allowed organisms such as the giant kelp to evolve. With efficient food transport systems, photosynthetic parts of the plant nearer the surface of the water were able to supply food to nonphotosynthetic parts, which were not exposed to sufficient light energy. (Note: No water transport system evolved. Fitness would not be increased by the addition of a water transport system because all parts of the plant were equally exposed to water.) c. The first land plants with xylem and phloem had no leaves or roots. This indicates that there was a selective advantage to having xylem and phloem alone. In other words, the addition of xylem and phloem allowed plants to grow taller. Xylem enabled water to be transported quickly to regions at a distance from the soil. In addition, some parts of the plants could be shaded and still be “fed” by transporting sugars from photosynthetic parts. Taller plants were less likely to be shaded and as a result were able to gather more energy per unit time. Thus, they had better survival and reproductive rates than shorter plants. d. Some seedless vascular sporophyte plants do not release megaspores. Instead, the megaspore divides on the sporophyte (in the old sporangium) to produce the female gametophyte. This female gametophyte produces eggs in archegonia, which are fertilized by sperm produced in antheridia of this or other plants. Neither the female gametophyte (once formed) nor the developing embryo receives nutrition from the old sporophyte plant. This information allows us to recognize how the stage could have been set for evolution of a seed (containing the offspring sporophyte) that was dependent on the parent sporophyte plant. 4. Write an analysis of how events a through d could be related. Each change in evolution builds on what already exists. Observations a, c, and d provide evidence for this. Observation b, on the other hand, indicates that similar environmental conditions often select for similar “evolutionary solutions.” In this case, the evidence indicates that effective means of transport are required for photosynthetic organisms to grow taller. In the aquatic environment, gravity and water availability were not problems. As a result, mutations which produced food transport mechanisms led to greater fitness in some organisms. On land, plants had to deal with all three problems: support against gravity as well as food and water transport. Therefore, on land, mutations which led to better transport of both food and water led to greater fitness and also allowed plants to grow taller. 218
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29.3/30.3 Test Your Understanding Consider a puddle of water in which mosses, ferns, and an angiosperm are growing. In a drop of water taken from the puddle, you observe flagellated sperm. Which of the following statements concerning the sperm could be true and which are definitely false? Consider each statement separately and explain your answers. 1. (T/F) The sperm could have been produced by an antheridium on the moss sporophyte. False—Moss sporophytes produce spores. Antheridia produce sperm on moss gametophytes. 2. (T/F) The sperm could later stop swimming and develop into the male gametophyte of the fern. False—The male gametophyte is formed following germination of a spore. The male gametophyte produces the male gamete or sperm. 3. (T/F) The sperm may be swimming to the archegonium of the fern. This could be true—Sperm from the antheridia of fern gametophytes fertilize the egg in the archegonium to produce the diploid sporophyte. 4. (T/F) The sperm could have been released from a pollen tube of an angiosperm. False—No known angiosperms produce flagellated sperm. (Note: This could be true for some of the more primitive seed plants, e.g. cycads.)
Activity 29.3/30.3
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Notes to Instructors Chapter 31 Fungi What is the focus of this activity? This activity provides students with an overview of the diversity that exists in the kingdom Fungi.
What is this particular activity designed to do? Activity 31.1 How diverse are the fungi in form and function? This activity is designed to help students understand the basic characteristics of the organisms in the kingdom Fungi as well as the diversity of life-forms and functions that exist in the kingdom Fungi.
Answers Activity 31.1 How diverse are the fungi in form and function? 1. a. What is the basic body plan of most fungi? The body of most fungi is composed of filamentous hyphae that can aggregate to form mats of mycelia (singular, mycelium) below ground or fruiting bodies that often appear above ground. b. Which fungi do not share this basic body plan? The yeasts are unicellular. 2. Fungi may be said to have both plant-like and animal-like characteristics. What plant-like characteristics do fungi have? What animal-like characteristics? Like plants, the cells of fungi are surrounded by cell walls. Also, like many lower plants, they produce haploid spores as a means of reproduction and dispersal. However, the cell walls of fungi are composed of chitin (a substance found in some animal phyla) rather than cellulose. In addition, fungi are heterotrophic (like animals) rather than phototrophic.
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a. Into what five major phyla (or divisions) is the kingdom Fungi divided?
b. On what basis are these divisions made?
Phylum Chytridiomycota
The chytridomycota are primarily aquatic, and, unlike other fungi, their zoospores have flagella.
Phylum Zygomycota
The zygomycota produce resistant structures (zoosporangia) during reproduction.
Phylum Glomeromycota
The glomeromycota are a group of about 160 species, all of which form arbuscular mycorrhizae. The group is thought to be monophyletic and is based on phylogenetic analysis of DNA.
Phylum Ascomycota
The ascomycota produce sexual spores in saclike asci (singular, ascus).
Phylum Basidiomycota
The basiodiomycota produce sexual spores in structures called basidia (singular, basidium).
4. Is it more correct to describe a mushroom as haploid or diploid? Explain. The mushroom, or fruiting body, is for the most part a dikaryon; that is, the cells of the fruiting body contain two haploid nuclei. The dikaryon and fruiting body are the result of the fusion of haploid mycelia of opposite mating types. 5. If you did not know that fungi were primarily terrestrial organisms, what structures or features of the organisms would suggest that they were terrestrial? With the exception of the chytrids, the fungi produce spores. The spores are resistant to desiccation and can be easily distributed by air currents. 6. a. In what ways are fungi important in an ecosystem? Fungi are key decomposers in an ecosystem. As such, they are key recyclers of carbon compounds in the ecosystem. b. In what ways are fungi important to humans? On the negative side, many fungi are pathogens on humans, other animals, or crops. On the positive side, fungi such as yeast are important in the cheesemaking, baking, and brewing industries. They are also the source of a wide range of antibiotics.
Activity 31.1
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7. Lichens are symbiotic associations of a fungus, usually an ascomycete, and an alga, usually a green algal species or a cyanobacterial species. Lichens can often survive in harsh natural environments. a. To what environmental conditions are lichens well adapted? Lichens can be found in many environments, but the best-known lichen species are recognized for their ability to survive in extreme environments. For example, some species are capable of tolerating severe cold and can survive in the tundra. Others are capable of withstanding desiccation and can survive in very dry climates. b. What makes them so well adapted to these conditions? Neither the algae nor the fungus by itself appears able to withstand the severe conditions they can survive when combined as a lichen. In their symbiotic relationship as a lichen species, the alga produces food via photosynthesis and provides some of this to the fungus. The fungus, on the other hand, provides the alga with water and minerals as well as protection from intense sunlight.
31.1 Test Your Understanding Patients with AIDS (caused by human immunodeficiency virus) often acquire opportunistic infections. Imagine a patient with AIDS and two opportunistic infections— a respiratory disease caused by a Mycobacterium and another disease caused by the fungus, Candida. As the patient’s physician, you need to prescribe drugs to counteract the infections. Given the normal anatomical and functional characteristics of each disease organism, what characteristics should a drug combination have to treat each of these three infections while doing the least harm to the patient? There are many possible ways to answer this question. One reasonable student answer is provided here. The treatment will need to address each of the three infections separately to attack the unique characteristics of each of the disease organisms. AIDS—will need continual treatment using several different kinds of drugs as the virus will eventually adapt to most treatments. One kind of treatment could target or inhibit proteins on the virus that allow it to link to receptors on the cell and enter or leave the cell. Alternatively, the drug may inhibit reverse transcriptase and prevent the virus from making DNA from its RNA and therefore from replicating. In reality, treatment of AIDS is very difficult because its DNA can integrate into human chromosomes and lie dormant, hidden for long periods of time. In addition, the virus that causes AIDS undergoes very 222
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rapid evolution via mutation. As existing drugs destroy the virus mutants that are susceptible, those that not susceptible can continue to infect. Mycobacterium—The best option for treating a bacterial infection is an antibiotic that targets a specific bacterial trait not found in humans. For example, a drug such as penicillin compromises the peptidoglycan wall of many bacteria. However, mycobacterium is not affected by penicillin. Therefore, it would be better to choose a drug such as erythromycin, which interferes with bacterial protein synthesis by affecting the 70s ribosomes of bacteria (which differ from the 80s ribosomes of eukaryotes). Candida—Like humans, yeasts are eukaryotes. Given the many similarities in eukaryotic cell structure and function, the fungal infection is more difficult to treat than the bacterial infection. However, it may be possible to target the fungal cell walls (since humans have no cell walls to damage). It may also be possible to destroy the fungus’ exoenzymes that are needed to break down nutrients. Each of these treatments is designed to affect a specific characteristic of the disease organism that is not found in humans and therefore should have minimal effect on the human.
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Notes to Instructors Chapter 32 An Introduction to Animal Evolution Chapter 33 Invertebrates What is the focus of this activity? In studying biodiversity, many students try to memorize each of the phyla and its characteristics. What students memorize, however, they tend to forget. This activity provides students with a different way of learning phylum characteristics. It asks them to recognize and learn where evolutionary advances occur among the phyla.
What is this activity designed to do? Activity 32.1/33.1 What can we learn about the evolution of the animal kingdom by examining modern invertebrates? This activity is designed to help students understand how studying the existing traits of organisms can shed light on their evolution. They will also discover how an understanding of the evolutionary changes that occur between phyla can make it easier to learn the characteristics of the phyla. Activity 32.2/33.2 What factors affect the evolution of organisms as they become larger? Activity 27.2 addressed the difference in oxygen needs per m2 of membrane surface as cells increased from about 10 m in diameter (prokaryote) to 100 m in diameter (eukaryote). This activity looks at how these same ideas apply as multicelled organisms become larger. It is designed to help students see how changes in the sizes of multicelled organisms affect their surface-area-to-volume ratios. As a result, students should recognize that simple alterations in morphology allowed organisms to maintain relatively high surface-area-to-volume ratios and thereby maximize their exchange rates with the environment.
Answers Activity 32.1/33.1 What can we learn about the evolution of the animal kingdom by examining modern invertebrates? Fill in the chart on the next two pages to organize the major characteristics of key invertebrate phyla.
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Radial symmetry, no cephalization, sessile lifestye
No real symmetry, no cephalization, sessile lifestyle
Symmetry? Cephalization?
Tissue level
Quasi -tissue level
Tissue versus organ level development
Bilateral symmetry with concomitant appearance of cephalization and locomotion
Yes
Tissue level, with the beginnings of organs in the reproductive system
3; ectoderm, mesoderm, and endoderm The mesoderm allows for evolution of the circulatory system and true muscles.
→
→
Earthworms, leeches, fan worms (segmented worms)
Annelida
l
→
→
Longitudinal Longitudinally arranged muscles and circular muscles only
Organ Level
→
Ascaris, horsehair worms (round worms
Nematoda
→
Increased complexity of muscle system
→
→
Clams, oysters, snails
Mollusca
→
→
→
→
Insects, crabs, lobsters, crayfish
→
→
→
→
Starfish, sea cucumbers, sea urchins, sand dollars
Arthropoda Echinodermata
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No; have epitheliomuscular cells
2; ectoderm and endoderm
Doesn’t apply
Number of tissue layers in embryo
Planaria, flukes, tapeworms (flatworms)
Platyhelminthes
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True muscle cells? No
Hydra, corals, jellyfish
Cnidaria
Sponges
Porifera
Examples of organisms
Characteristics
Key invertebrate phyla
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None, cell-to-cell communication
Other
Type?
None, B cell communication
Nervous system?
None
Platyhelminthes
Nerve net
Diffusion and some facilitation of circulation via fluid movement in gastrovascular cavity
Ladder-like nervous system ganglion in head end
Diffusion and some facilitation of circulation via fluid movement in gastrovascular cavity.
Gastrovascular Gastrovascular cavity, some cavity more highly extracellular branched digestion and some intracellular
None
Cnidaria
Lateral nerve cords
Diffusion and some facilitation of circulation via fluid movement in hydrostatic skeleton
One-way digestive tract complete from mouth to anus, allows for more energy input per unit time and evolutionarily allows for specialization along tract
Pseudocoelom, gut not lined with mesoderm embyonically
Nematoda
Increased complexity
Closed circulatory system, heart(s), arteries, capillaries, and veins
→
→
Annelida
→
Some species have an open circulatory system, and some have a closed circulatory system (e.g., squid and octopus)
→
→
Mollusca
→
All species have an open circulatory system
→
→
→
Species have a closed circulatory system
→
Entercocoelom formed by outpocketing off gut
Arthropoda Echinodermata
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Type?
None, primarily diffusion
Intra-cellular digestion
None
Porifera
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Ciculatory system?
Type?
Digestive tract?
Type?
Coelom?
Characteristics
Key invertebrate phyla
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Using the information in the chart and in Chapters 32 and 33 of Biology, 8th edition, answer the questions. 1. What set of characteristics is shared by all of the invertebrate animal phyla in the chart? All are multicellular heterotrophs. 2. What unique combination of characteristics defines each of the invertebrate phyla as separate from the other phyla? If you read down each column, you can determine the characteristics of each phylum. No single characteristic defines each phylum. Rather, it is the specific combination of characteristics that is unique to and defines each phylum. 3. If you compare the characteristics of one phylum of the invertebrates with the next, what key differences separate the groups from each other? To answer this question, you need to compare the characteristics in a given column/phylum with those in the next. Those characteristics that change from one column to the next represent the key differences that separate the groups. For example, annelids differ from nematodes in that annelids have both longitudinal and circular muscles and a schizocoelom, whereas nematodes have only longitudinal muscles and a pseudocoelom. In all other respects (at least those listed in this chart), the two phyla are similar. 4. a. Looking across the rows, what major trends appear to occur in the evolution of various organs or organ systems in these animal groups? This chart is set up to allow you to visualize how the various systems changed in the evolution of the various phyla. The arrangement also allows you to visualize major trends in this evolution. For example, if you look at the number of body layers in the embryo, you can see a change from no apparent differentiation of body layers in the development of sponges, to the appearance of two distinct body layers in the embryos of Cnidaria, and ultimately to the appearance of three body layers in the embryos of Platyhelminthes (and all subsequent animal phyla). If you look at the evolution of the phyla this way, you don’t need to memorize all the characteristics of each phylum. Rather, you need to remember only where major changes were introduced. For example, all Platyhelminthes have three body layers and so does every phylum following it. The mesoderm layer allowed for evolution of the circulatory system and true muscles. Platyhelminthes have true muscle cells and all phyla following it do as well; however, the types of muscles may vary. b. What developmental evidence is used to link Annelids, Arthropods, and Molluscs evolutionarily? Developmentally, Annelids, Arthropods, and Molluscs are all protostomes. They all display spiral determinate cleavage. In the development of the digestive tract, Activity 32/33.1
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the blastopore becomes the mouth. The coelom forms as a split in the mesoderm developmentally. c. What evidence is used to separate these phyla from the Echinoderms and Chordates? Echinoderms and Chordates are deuterostomes. This means that, developmentally, they undergo radial indeterminate cleavage. In the development of the digestive tract, the mouth forms from a secondary opening opposite the blastopore and the coelom forms from mesodermal outpocketings off the primitive gut or archenteron. d. Does this analysis provide evidence for or against the statement: “Evolution adds onto or modifies what already exists”? Explain. As is apparent in the chart, modifications are built upon what already exists. For example, we see an evolution from no clear separation of body layers in the sponge embryo to two body layers in Cnidaria and then three in Platyhelminthes. The addition of the third layer, the mesoderm, allows for the evolution of muscle cells and then the elaboration of the muscle system through succeeding phyla. 5. The chart organizes the major groups of animals based on grade, or shared body plan features. What changes would you need to make in this organization to reflect the possible phylogenetic relationships uncovered using molecular evidence? To answer this: a. On a separate sheet of paper, redraw the chart to reflect the new phylogenetic relationships based on molecular evidence. The new chart should show that recent DNA analyses indicate that, among the bilateria, the nematodes and arthropods should be grouped together as the Ecdysozoa; the echinoderms and chordates should be listed among the Deuterostomia; and the remaining phyla, among the Lophotrochozoa. See Figure 32.11. This differs considerably from the previous organization based on morphological and developmental evidence alone (Figure 32.10). b. What specific molecular characteristics/data are being used to determine evolutionary relationships among animal phyla? As noted in the text, the new molecular phylogenies “are based on mitochondrial genes, ribosomal genes, Hox genes, and dozens of protein-encoding genes. Collectively, these studies indicate that there are three major clades of bilaterally symmetric animals: Deuterostomia, Lophotrochozoa, and Ecdysozoa (see Figure 32.11).”
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6. How would your answers to questions 2, 3, and 4 differ (if at all) when the chart is redrawn and filled in to reflect changes in relationships based on molecular evidence? The answers to questions 2 and 3 would not change. Some of the evolutionary trends noted in question 4 would be altered. In particular, the rearrangement would alter trends in the evolution of muscle organization, the coelom, the circulatory system, and the nervous system. 7. In biological terms, a group of organisms is said to be successful if it is represented by a large number of species or if the mass of all the organisms in the group is large. (In both cases, “large” is determined relative to other groups or organisms.) Given this definition of success, which of the major groups of animals would you argue is the most successful? Be sure to provide evidence for your argument. Answers may vary depending on the criteria chosen. For example, did you choose success based on the total mass of organisms, the total number of organisms, or the total number of different species of organisms within a group? On one hand, because nematodes are found as ectoparasites and endoparasites on almost every living organism, they could be considered the most successful. On the other hand, if you look at the total number of known species, you might argue that arthropods are the most successful. Any reasonable argument is acceptable as long as it is backed up with evidence that indicates why the group chosen should be considered most successful.
Activity 32.2/33.2 What factors affect the evolution of organisms as they become larger? In the evolution of life on Earth organisms have evolved from single celled to multicelled; small to large; simpler to more complex. Keep in mind, that this apparent increase in complexity occurs because evolution adds on to or modifies what already exists. As you discovered in Activity 27.2, surface-area-to-volume ratios and the need for organisms to gain or lose substances across their surface areas have put constraints on the evolution of cell structure and function. These same constraints affect the ability of multicellular organisms to survive. Given their small size, most of the evolution in unicellular organisms has involved changes in cell chemistry and/or internal cellular structure. In contrast the evolution of larger and larger multicellular organisms is evidenced primarily as changes in both external and internal morphology. 1. A quick review and extension of Activity 27.2: How has small size affected prokaryotic diversity?
Activity 32.2/33.2
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a. How do surface-area-to-volume (SA/V) ratios change as the size and shape of cells and organisms change? To answer this, calculate the SA and V of a cube 1 mm on a side. Then do the same for cubes that are 2 mm and 4 mm on a side and compare their SA/V ratios. Cubes:
1 mm square
Linear dimension 1 mm
2 mm square
4 mm square
Ratios
2 mm
4 mm
1:2:4
Surface area (SA) 6(1 mm2) = 6 mm2 6(2 m2) = 24 mm2 6(4 mm2) = 96 mm2 1:4:16 Volume (V)
(1 mm)3
(2 mm)3 = 8 mm3 (4 mm)3 = 64 mm3
1:8:64
SA/V ratio
6 mm2/1 mm3
24 mm2/8 mm3 = 3 mm2/1 mm3
6:3:1.5
96 mm2/64 mm3 = 1.5 mm2/1 mm3
b. What do the ratios mean? As linear dimensions increase by 2 times, SA increases by 4 times, but volume increases by 8 times. As a result, where 6 units of surface area were supporting one unit of volume in the 1-mm cube, only 3 units of surface area are available to support each unit of volume in a cube that is twice the size in linear dimensions. Put another way, as the linear dimensions increase by a factor of 2, the surface areas increase as the square of the increase in linear dimensions (4), and the volume increases as the cube of the increase in linear dimensions (8). c. Is this graph of the change (in surface area compared to volume) linear or exponential? To answer this question, complete the following table and then graph the surface areas and volumes from this table and the previous table. Cube linear dimension:
8 mm
16 mm
32 mm
64 mm
Suface area
666 = 384
6 16 16 = 1536
6 32 32 = 6144
6 64 64 = 24,576
Volume
83 = 512
163 = 4096
323 = 32,768
643 = 262,144
Students can graph these data by hand or they can use a software graphing program (e.g., Excel). Because it is the most widely used commercial software for this purpose, we have included a primer on the use of Excel for data analysis and graphing at the back of this workbook. There are many other similar software packages available. Please feel free to use the primer as a template for making your own student primer for another software program.
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Data: SA (mm2) 6 24 96 384 1536 6144 24576
V (mm3) 1 8 64 512 4096 32768 262144
Graph output: The graph indicates that the data are not linear. Surface area and volume increase as the square and cube of the linear dimensions increase in size. This produces a special type of exponential relationship, the “power” relationship.
300000
y = 0.068x1.5 R2 = 1
Volume
250000 200000 150000 100000 50000 0 0
30000
10000 20000 Surface area
d. What effect does this relationship between surface area and volume have on the ability of larger and larger multicellular organisms to support the metabolic needs of all parts of their bodies, for examples, to supply oxygen needed for cell respiration? It is obvious that as organisms become larger, the SA/V ratio decreases dramatically with each doubling of size such that organisms only 64 mm in diameter (a little more than 2.5 inches) have each mm3 of their volume supported by only 0.09 mm2 of surface area. Compared to the 6 mm2 of surface area per 1-mm3 volume in the 1-mm cube, this larger organism has 66 times less surface area supporting each mm3 of its volume. Maintaining adequate flow of oxygen (etc.) to the cells obviously becomes more difficult as organisms become larger. Activity 32.2/33.2
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2. In the previous examples, as the “organism” sizes changed, their shapes remained constant (as cubes). How would changing the shape of the organism affect SA/V ratios? Use modeling clay or playdough to make a cube that is 2.5 cm (1 inch) on a side. The cube will obviously have a constant mass or volume. Using your model, devise three ways to change the SA/V ratio of this “organism.” If possible, find one way to reduce the SA/V ratio and two ways to increase the SA/V ratio. For each of your solutions to increase or decrease the SA/V ratio, calculate the actual change in SA. Enter the data you collect/calculate in the chart. Shape change
Effect on SA/V ratio
Model 1 Model 2 Model 3
The answers will depend on the methods each student uses to change the shape. For example, the cubes can be flattened somewhat or a lot. The more the cube is flattened, the larger the surface area becomes relative to the volume. Alternatively, the cube may be cut partway through one or more times to produce additional surface area. As the number of cuts increases, the relative surface-area-to-volume ratio increases. 3. Review the chapters on fungal, animal and plant diversity in your textbook. For each of the models that you developed to increase SA/V ratios, find an example of a fungus, plant and/or animal, or a particular organ system, that uses the same method to increase SA/V ratios and describe it below. This type of increase in SA/V ratios can be found in the following organism(s)/organ systems:
Proposed model–increase SA/V by:
There will be many possible answers; a few are given on the following page.
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For students who flattened the cube to increase the SA/V ratio, examples among organisms include cnidaria, flatworms, and leaves, all of which have maximized SA/V ratios by keeping their “bodies” thin. For students who have cut into the clay to make many small projections or flaps, examples include the gill structures of fish and other organisms or the fanlike structures of tube worms. If students have poked holes into the clay, this is similar to the structure of lacy sea fans, the trachea of insects, and the alveoli in lungs.
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Notes to Instructors Chapter 34 Vertebrates What is the focus of this activity? As noted in Activity 32.1/33.1, when studying biodiversity, many students try to memorize each phylum and class and their characteristics. Again, this activity provides students with a different way of learning these characteristics. It asks them to recognize and learn where evolutionary advances occur among the various subphyla and classes of chordates.
What is this activity designed to do? Activity 34.1 What can we learn about the evolution of the chordates by examining modern chordates? This activity is designed to help students understand the relationships that exist among the basic characteristics of the major subphyla and classes of chordates. Students will also learn how studying the traits of modern chordates can shed light on their evolution.
Answers Activity 34.1 What can we learn about the evolution of the chordates by examining modern chordates? Fill in the chart on the next two pages to organize the major characteristics of key groups of chordates.
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None
None
Jaw?
None
None; notochord present in adult form
Cartilaginous skeleton
None
Jaws present
Cartilaginous vertebral column
Cartilaginous skeleton
Sharks, rays, skates
→
Bony vertebral column
Bony skeleton
Bony fish—for example bass, trout, tuna
Chondrichthyes Osteichthyes
→
→
→
→
→
→
→
→
→
→
→
Birds
Lizards, snakes, turtles
Frogs, Mice, salamanders rabbits, humans
→
Aves
Mammalia Reptilia
Amphibia
Subphylum Vertabrata
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None; notochord present in adult form
None
Type of skeleton
Lamprey and hagfish
Agnatha Amphioxus
Subphylum Cephalochordata
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Vertebral column? None; notochord present only in larval form
Sea squirts
Examples of organisms
Characteristics
Subphylum Urochordata
Characteristics of phylum chordata
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Respiratory structure?
Diffusion across body surface and pharyngeal gill basket
Two pairs of lateral fins (pectoral and pelvic)
Gills Respiration primarily via diffusion across the body surface
Gills
Tetrapods; four limbs, each with one upper and two lower limb bones and five phalanges (pentadactyl limb structure). Limbs are positioned out to the sides of the body.
Amphibia
Gills covered by operculum
Tetrapods; four limbs, each with one upper and two lower limb bones and five phalanges. Limbs are positioned more laterally than in mammals but less than in amphibians.
Reptilia
Lungs
Keratinized Scales (waterproofed) containing skin keratin
Tetrapods; four limbs, each with one upper and two lower limb bones and five phalanges. Limbs are positioned directly beneath the body.
Mammalia
Gills in aquatic Lungs forms or gills in early development with lungs in terrestrial forms
Bony scales Skin
Two pairs of lateral fins (pectoral and pelvic)
Chondrichth Osteichthyes yes
Permeable Toothlike epidermis scales or skin
None or a single pair (some fossil species)
Agnatha
Lungs
Feathers made of keratin
Tetrapods; four limbs, each with one upper and two lower limb bones and five phalanges. Hind limbs are positioned directly beneath the body.
Aves
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Body covering?
None
Cephalochordata
Subphylum Vertabrata
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Paired limbs? Structure? Limb position?
None
Characteristics Urochordata
Subphylum Cephalochordata
Subphylum Urochordata
Characteristics of phylum chordata
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Sexual; internal fertilization amniotic egg in monotremes Hair, mammary glands, diaphragm
Sexual; external fertilization
Many species undergo metamorphosis from aquatic larval form to terrestrial adult.
Sexual; external fertilization
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Other
Type of reproduction
Sexual; internal fertilization amniotic egg
Sexual; internal fertilization amniotic egg
three-chambered fourchambered heart, except heart crocodiles and alligators, which have fourchambered heart
Closed circulatory system;
Aves
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Sexual; Sexual; external external fertilization fertilization
fourchambered heart
Closed circulatory system;
Reptilia
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Heart structure?
threechambered heart
Closed circulatory system;
two-chambered twoheart chambered heart
Closed circulatory system;
twochambered heart
Closed circulatory system;
Circulatory system?
Mammalia
Closed circulatory system;
Amphibia
Closed circulatory system;
Chondrichthyes Osteichthyes
Subphylum Vertabrata Agnatha
chordata
Cephalo-
Subphylum
Characteristics
Urochordata
Subphylum
Characteristics of phylum chordata
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Using the information in the table, answer the questions. 1. a. According to Biology, 8th edition, what three characteristics are common to all chordates? All chordates have a notochord, pharyngeal gill slits, and a dorsal hollow nerve chord at some stage in their life cycle. b. Why don’t we find all three of these chordate characteristics in our own bodies? The notochord and pharyngeal gill slits are present in the developing embryo. However, as development continues, they are replaced by other structures. The notochord is replaced by the vertebral column. The cells associated with the pharyngeal gill slits continue in development to form parts of the jaw and ear structures. 2. The chart organizes the chordates based on grade or shared body plan features. If you compare the characteristics of one group with the next, what key differences separate the groups from each other? To answer this question, you need to compare the characteristics in a given column/phylum with those in the next. Those characteristics that change from one column to the next represent the key differences that separate the groups. For example, Chondrichthyes is separated from Agnatha by the presence of paired lateral fins and toothlike scales. 3. What unique combination of characteristics defines each group as separate from the others? If you read down each column, you can determine the characteristics of each phylum. No single characteristic defines each phylum. Rather, it is the specific combination of characteristics that is unique to and defines each phylum. 4. a. Looking across each of the rows in the chart, what major trends do you see in the evolution of the different organs and organ systems? This chart is set up to allow you to visualize how the various systems changed in the evolution of the subphyla and classes of chordates. This arrangement also enables you to visualize major trends in this evolution. For example, evolution of limbs began in Chondrichthyes as paired fins. Amphibians evolved pentadactyl limb structure and this became highly modified in the evolution of reptiles, birds, and mammals. b. Does this analysis provide evidence for or against the statement: “Evolution adds onto or modifies what already exists?” Explain. As is apparent in this chart, modifications are built upon what already exists. For example, we see an evolution from a two-chambered heart to three- and four-chambered hearts as the classes of chordates evolved. 238
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5. a. What major changes in structure and function are seen in terrestrial groups as compared with aquatic groups? One of the major changes is in reproduction. Evolution of the amniotic egg freed terrestrial organisms from the water environment for reproduction. Another obvious change is from the use of gills to the use of lungs for respiration on land. Lungs provide more protection from desiccation. b. Can these changes be related to differences in natural selection on land versus in an aquatic environment? Explain. Early forms of gills probably evolved in an aquatic system and allowed fish to supplement respiration by gulping air. However, the continued evolution of lung structure to the forms we see in terrestrial organisms today is unlikely to have occurred in an aquatic system. Similarly, where desiccation is not a problem, it is unlikely that a structure like the amniotic egg would have evolved.
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Notes to Instructors What is the overall focus of activities associated with Chapters 35 to 51? Activities from earlier chapters and from the chapters on the diversity of organisms allowed students to develop an understanding of these topics: • The events that occurred in the evolution of multicellular organisms in general and of plants and animals in particular • The effects that surface-area-to-volume (SA/V) ratios, rates of osmosis and diffusion, and metabolic requirements per unit volume per unit time can have on the evolution of the form and function of organisms • The effects that different sets of selective forces (aquatic versus terrestrial) can have on the evolution of the form and function of organisms The activities in Chapters 35–51, which cover the units “Plant Form and Function” and “Animal Form and Function” in Biology, 8th edition, are designed with these objectives: • These activities build on the understanding of how the structure of each system is related to (or in a sense, determines) its function. • In addition, they are designed to help students develop a good understanding of both the basic function(s) of each system and the interrelationships among systems.
Chapter 35 Plant Structure, Growth, and Development What is the focus of this activity? In this activity, students are asked to compare the basic form of monocots and dicots (both herbaceous and woody) to learn key similarities and differences in their structure and function.
What is this particular activity designed to do? Activity 35.1 How does plant structure differ among monocots, herbaceous dicots, and woody dicots? The questions in this activity are designed to help students review and understand the basic form and function of the angiosperm plant body as well as how that basic form is modified in monocots, herbaceous dicots, and woody dicots.
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Answers Activity 35.1 How does plant structure differ among monocots, herbaceous dicots, and woody dicots? The chart on the next page shows a drawing of a generic plant. The arrows indicate various plant parts or organs. In the columns to the right of the plant, draw cross sections of the plant at the points indicated by the arrows. To help you visualize the differences in structure among different types of plants, in Column I draw the cross sections assuming the plant is a monocot. In Column II draw the cross sections assuming the plant is a herbaceous dicot. Finally, in Column III draw the cross sections as if the plant is a woody dicot. Be sure to label your drawings.
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pith
Root
Stem
phloem xylem
phloem xylem
phloem
xylem
phloem
xylem
endodermis
dorsal
dorsal
endodermis
phloem xylem
phloem xylem
Same as Column I
phloem
xylem
nd
xylem
xylem endodermis
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2
2nd pholem
2
carbium
2nd pholem
Same as column 1
Same as column 2
Same as column 1
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Petiole
Leaf
Same as Column I
Apical meristem
Column 3 Cross sections if plant is a woody dicot
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Column 2 Cross sections if plant is a herbaceous dicot
Column 1 Cross sections if plant is a monocot
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Use the information in your drawings to answer the questions on the next pages.
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1. In Column I, connect the cross sections of the monocot by drawing a line (color 1) from the water transport tissue in one section to the water transport tissue in the next, and so on. Draw another line (color 2) to connect the food transport tissue from one section to the next. Do the same for the herbaceous and the woody dicot cross sections. How do the positions of these tissues change from one cross section to the next? As students make these connections, they should see: • In leaves organized horizontally, the xylem lies dorsal to or on top of the phloem within each vascular bundle. • In all vascular bundles, the xylem is more centrally located than the phloem. • In dicots, the xylem and phloem are organized around the periphery of the stem, but are located in the center of roots. • In monocots, the xylem and phloem bundles are scattered throughout the cross section of the stem, but are centrally located in the root. 2. Use the information in the cross sections to fill in the chart below. Note: A distinguishing feature is one that is found in only the given type of organism. For example, a distinguishing feature of mammals is the presence of hair: Hair is a characteristic of all mammals and is not found in any other animals. Distinguishing feature(s) of
Monocot
Apical meristem shoot vs. root
Leaf
Stem
Root
Vascular traces (veins) in the leaf are arranged in parallel. In a cross section of the leaf you would expect to see the vascular bundles evenly spaced and of approximately equal size across the section.
Vascular bundles are scattered throughout the cross section.
Vascular bundles form a ringlike pattern around a central pith. (Vascular bundles are surrounded by a pericycle, which is surrounded by an endodermal layer.)
The apical meristem of the root is covered by a protective root cap. In addition to the apical meristem, which gives rise to the leaves, many monocots have intercalary meristems, which lie near the base of the stem. For example, grass blades or leaves continue to grow after mowing or grazing. This growth is from their intercalary meristems.
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Vascular traces (veins) in the leaf are arranged in a netlike pattern.
Vascular bundles are arranged in a ring between a thin outer cortex layer and a central region of pith.
There is no central pith. Xylem forms a crosslike pattern in the center of the root. Phloem sits in the arms of the cross. As in monocots, this central region is surrounded by a pericycle, which is surrounded by an endodermal layer.
As in monocots, the apical meristems of the roots are generally protected by a root cap. There is generally no protective tissue layer over the apical meristems of the aerial portions of the plant.
Vascular cambium is continuous and produces secondary xylem to the inside and In a cross secondary section of the phloem to the leaf you outside. would not Variations in expect to see water the vascular availability bundles (wet vs. dry evenly spaced. seasons) can In addition, result in the bundles growth rings. would likely be of varying size.
A ring of vascular cambium forms between the central xylem and phloem and lays down secondary xylem to the inside and secondary phloem to the outside.
As in monocots, the apical meristems of the roots are generally protected by a root cap. There is generally no protective tissue layer over the apical meristems of the aerial portions of the plant.
In a cross section of the leaf you would not expect to see the vascular bundles evenly spaced. In addition, the bundles would likely be of varying size.
Woody dicot
Vascular traces (veins) in the leaf are arranged in a netlike pattern.
New root growth is similar in cross section to that of herbaceous dicots.
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3. If cross sections were not available, what other characteristics of the plants as a whole could you use to determine whether each was a monocot, a herbaceous dicot, or a woody dicot? Monocot leaves have parallel veins (vascular bundles); dicot leaves have netlike veins. Monocot floral parts are generally in multiples of three; dicot floral parts are in multiples of four or five. Monocots tend to have fibrous root systems; dicots usually have taproot systems. Older woody dicots could obviously be distinguished from herbaceous dicots by the presence of woody lateral growth, including bark. Very young woody dicots would be difficult to distinguish from herbaceous dicots. 4. A cartoon shows a man going to sleep in a hammock suspended between two relatively short trees. The second frame of the cartoon shows the man waking 20 years later and finding his hammock 15 feet higher off the ground. Critique this drawing in terms of what you know about the growth pattern of trees. If the tree is a dicot, then its growth in height occurs by the addition of cells at the apical meristems. As a result, the position of the hammock should not change over time.
35.1 Test Your Understanding How is the general morphology of the various organs of a plant correlated with the function(s) of those organs? For example, why are leaves generally thin and flat? What structural advantage is provided by stems having their vascular tissue arranged in a ring near the periphery of the stem? What characteristics of roots and root growth dictate that the vascular tissue be more centrally located? Thin flat leaves have very high surface-area-to-volume ratios. As a result, they maximize surface area, enabling them to expose more photosynthetic cells to sunlight. Their large surface area contains many stomata, which allow for gas exchange associated with photosynthesis. Stems often have their strongest tissues (xylem and phloem) organized around the periphery (rather than centralized). On the other hand, roots tend to have their strongest tissues (xylem and phloem) centralized. This makes stems (which support the aerial parts of the plant) much more rigid and roots (which must be able to grow around obstacles in the ground) much more able to bend. You can demonstrate this difference using a piece of paper. First fold the paper lengthwise two times. Take another piece of paper and roll it into a tube of the same diameter as the folded paper. Have a volunteer hold each by one end and add coins to the opposite end of each. The folded paper will bend under much less weight than the tube.
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Notes to Instructors Chapter 36 Transport in Vascular Plants What is the focus of this activity? In this activity, students will examine the basic structure and function of transport systems in higher plants.
What is this particular activity designed to do? Activity 36.1 How are water and food transported in plants? This activity is designed to help students review and understand • how water and specific dissolved minerals can enter and be transported in xylem, • how sugars are transported in phloem, and • how the functions of these two transport systems are interrelated.
What misconceptions or difficulties can this activity reveal? Activity 36.1 To understand transport in plants, students must have a good understanding of the properties of water and of osmotic relationships (Chapter 3) and the properties of cell membranes (Chapter 7). It is helpful to remind students of these topics and refer them to the appropriate text chapters and activities. To help students understand how water can be pulled up a tree by transpirational loss, you can use a juice box analogy. When you put negative partial pressure on the straw in a juice box (by sucking on it), you pull juice up the straw. If the negative partial pressure is great enough, you will even collapse the sides of the juice box. This analogy also helps students understand how a tree can decrease in diameter during periods of high transpiration (see “Test Your Understanding,” question 9). Many students don’t understand how water in the xylem can move laterally into the phloem. They may assume that xylem vessels and tracheids are laterally impermeable. Figure 35.10, Water-Conducting Cells of the Xylem, in Biology, 8th edition, demonstrates that both tracheids and xylem vessels contain pits in their lateral walls. These pits allow for the lateral movement of water. Explaining lateral movement sometimes further confuses students, who wonder: If this is true, how can water move up the xylem? Aren’t the lateral pits like holes in the side of a straw, which make it much more difficult to draw water up the straw? You can demonstrate the situation in class. Give each student two straws, and have different 246
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students cut a different-sized hole (from pin holes to up to 1/8 inch across) in the side of one of their straws. Now have them test how much suction they need to exert to pull water up both straws. Students will find that it is still possible (though harder) to pull water up the straw with the hole. Next, to simulate the environment of the xylem inside a plant, have students wrap the cut straw with wet paper toweling (analogous to the water-filled cell walls of a plant). They will discover that the amount of energy they need to exert to pull water up in this straw is similar to the exertion for the uncut straw. In addition, this demonstration should help students understand how a difference in water potential between phloem and xylem can lead to movement of some of the water laterally in addition to vertically. Some students have difficulty understanding how water could initially be drawn up into a tree that is many meters tall. They may assume that the tree started out 50 meters tall with empty xylem. It helps to remind them that even the tallest trees started out as small seedlings. As the trees grew, the xylem and the water columns inside the xylem grew along with them.
Answers Activity 36.1 How are water and food transported in plants? Using the cross sections of the plants you developed in Activity 35.1 you identified the location of water-conducting xylem in roots, stems, and leaves. In this activity, you will use playdough, cutout pieces of paper, chalk, or some other material to create a model of how that transport occurs. Be sure to include all the terms and concepts in the list. Use your model to demonstrate how both water and K+ ions are transported from the soil into the xylem of the plant and to the leaves of the plant. Terms endodermis oxidative phosphorylation Casparian strip oxygen plasmodesmata respiration stele
channels root hairs ADP symplastic transport ATP apoplastic transport epidermis
symplast lateral transport apoplast absorption xylem versus
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Use your model to answer the questions. 1. How do water’s properties of adhesion and cohesion help maintain the flow of water in the xylem of a plant? Water molecules cohere to one another and adhere to the sides of the xylem vessels or tracheids. As negative pressure is applied to pull the water up in the xylem, each water molecule therefore pulls on the next in the column. As a result, the entire water column rises. 2. a. If water flows from a region of more positive (higher) water potential to a region of more negative (lower) water potential, how does the water potential in the root compare to that in the soil outside the root? The water potential in the soil must be higher than it is in the root. b. How does the water potential in the air compare to that in the leaf of a plant undergoing transpiration? If transpiration is occurring, the air must have a lower water potential than the leaf. 3. A student uses an AUB tube for a series of experiments. Sides A and B of the tube are separated by a membrane that is permeable to water but not to sugar or starch. What results would you expect under the experimental conditions given below? Explain your answers in terms of osmotic potential, water potential, and the equation =P+S (Hint: Solute pressure is always negative and a 0.1 M solution of any substance has S = 0.23. Therefore, a 0.2 M solution would have S = 0.23 2 = 0.46.) Experiment a: A solution of 10 g of sucrose in 1,000 g of water (the molecular weight of sucrose is 342) is added to side A. An equal volume of pure water is added to side B. What will happen to the concentrations of water and sugar in the two sides over time? Explain. A solution with 10 g of sucrose per 1,000 g of water is equivalent to 10 g of sucrose divided by 342 g per mole, or 0.03 mole of sucrose/1,000 g of water. If 0.1 mole of a solution has S = 0.23, then 0.03 mole has S = 0.23 0.3 = 0.069: = P + S = 0 + (0.069) = 0.069 Because the of pure water (on the other side of the U tube) is zero, water will move from this region of higher water potential to the side with the sucrose, which has a lower water potential. 248
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Experiment b: A solution of 10 g of soluble starch in 1,000 g of water is added to side A. Assume the molecular weight of soluble starch is about 63,000. An equal volume of pure water is added to side B. What will happen to the concentrations of water and starch in the two sides over time. Explain how this compares with the results in Experiment a. A solution with 10 g of starch per 1,000 g of water is equivalent to 10 g of starch divided by 63,000 g per mole, or 0.00016 mole of starch/1,000 g of water. If 0.1 mole of a solution has S = 0.23, then 0.0016 mole has S = 0.23 0.0016 = 0.0004. Because the of pure water (on the other side of the U tube) is zero, water will move from this region of higher water potential to the side with the starch, which has a lower water potential. This movement will be much slower, however. In fact, it will be 0.069/0.0004 = 172.5 times slower. 4. Fertilizer generally contains nitrogen and phosphorus compounds required by plants. The nitrogen is often in the form of nitrates, and the phosphorus is in the form of phosphates. Based on what you know about chemistry and water potential, why would overfertilizing lead to the death of plants? Adding nitrates and phosphates to the soil will decrease the water potential in the soil. If the water potential in the soil equals the water potential in the root, water will not flow into the root. Water will actually leave the root if the water potential in the soil is lower than it is in the root. 5.
a. One of the most common ways of killing a plant is overwatering. Why does overwatering kill a plant? Roots, like all other living tissues of the plant, require oxygen for cellular respiration. If you overwater a plant, you fill all the air spaces in the soil with water. Root tissue cannot survive without oxygen. b. If overwatering kills plants, why can you sprout roots from cuttings of stems in water? Oxygen from the atmosphere can diffuse into the water in a jar or vase. As a result, the water will contain enough oxygen to allow cellular respiration to continue in the roots. (If the jar is clear and in the sun, green portions of the root underwater will also photosynthesize and add additional oxygen to the water.) However, air spaces in soil are narrow and penetrate deep into the soil. When these are filled with water, the surface area available for the diffusion of oxygen into the water is greatly reduced.
6. Xylem cells are dead when functional. Why must phloem cells be alive when functional? Phloem cells must be capable of loading sugars against a concentration gradient. To do this, they must have a semipermeable membrane and must expend ATP. As a result, they must be alive. Activity 36.1
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What forces bring about:
How are these forces generated in the plant?
a. xylem conduction? The major force is the negative partial pressure generated in the xylem.
Transpirational loss of water at the leaf surface (and elsewhere) on the plant generates this negative partial pressure.
b. phloem conduction? phloem conduction? The major force is a positive hydrostatic pressure in these cells.
This force is generated by the difference in water potential in phloem versus xylem. Sucrose is actively transported into phloem (to levels as high as 25%). This potential difference pulls water out of the xylem and into the phloem, generating the hydrostatic pressure that moves substances in phloem.
8. Refer to your diagram of the cross-sectional structure of a typical angiosperm leaf from Activity 35.1. Explain how this structure (that is, the type and placement of cells, and so on) is correlated with the activities of the leaf as they relate to photosynthesis, water conservation, and food and water transport. When the stomata are open in the leaf, carbon dioxide can enter and oxygen and water vapor can exit. It is this water loss through the stomata that generates the transpirational pull on the xylem. Closing the stomata reduces water loss; however, it also reduces the amount of carbon dioxide available to the plant (C3 plant). Xylem vessels and phloem sieve cells lie in close contact. This facilitates the movement of water from xylem to phloem to produce the hydrostatic pressure required for the movement of substances in phloem.
36.1 Test Your Understanding
1. Scientists have measured the circumference of trees at 2 A.M. and at 2 P.M. If they collect measurements when the ground has adequate moisture and the days are sunny and dry, they find that the circumference (and therefore the diameter) of the tree trunk is smaller at 2 P.M. than at 2 A.M. From your knowledge of the mechanisms of water transport, suggest the reasons for this decrease in circumference. On bright sunny days, the transpirational pull generated by evaporation from the leaves can be so great that it partially collapses (pulls in) the walls of the xylem vessels. This is analogous to what happens when you drink out of a juice box. You can put so much suction on the straw that you collapse (pull inward) the sides of the box. In a tree, this pull can be measured as an actual decrease in circumference (and, as a result, in diameter). 250
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2. Outline an experiment that would allow you to determine a. how fast a substance is carried in the xylem, b. in what direction the substance flows in the xylem, and c. what percentages of solutes are in the xylem at various distances away from the leaves or roots. There are a number of different ways to design an experiment. Xylem is dead when functional. As a result, you can insert taps and/or measuring devices into it. For an example of an experiment, place taps and/or measuring devices into the xylem along one side of a branch. Place these at 10-cm intervals in a straight line from near the leaf end to the opposite end of the branch. In one now-classic experiment, Bruno Huber inserted temperature-recording devices at intervals along a tree branch. He heated the water in the xylem at various locations along the branch and then recorded how (if at all) the temperature of the water changed at the other recording devices. This allowed him to determine that the water in the xylem moved vertically. By measuring the time it took for the heated water to move from one recorder to the next, he could also determine the speed of transport in xylem. If you wanted to determine the amount of solutes present in xylem at various distances away from roots or leaves, you could put taps into the xylem at 10-cm (or other) intervals. Open the taps periodically and remove some of the fluid in the xylem. Analyze the fluid from the different taps for contents (for example, for the percentages of various minerals). 3. When researchers have tried to tap into phloem cells during experiments, they find that the disrupted phloem immediately stops functioning. However, aphids can pierce through plant tissues with their mouthparts and locate individual phloem cells. Once inside a phloem cell, the aphids are essentially force-fed phloem sap. If the aphid body is removed from the mouthparts, phloem sap will continue to flow and can be collected (as honeydew).
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You place two aphids on a plant you are using for radioactive tracer experiments. One is located 5 cm above leaf A and the other is located 5 cm below leaf A. Once the aphids begin feeding, you remove the aphid bodies, allowing you to sample phloem sap from the mouthparts remaining in the plant. You then cover leaf A with a plastic bag and inject 14CO2 (radioactive carbon dioxide) into the plastic bag continuously for 10 minutes, during which time leaf A continues to conduct photosynthesis. You start collecting honeydew from the two aphids’ mouthparts from the time the 14 CO2 is first injected into the bag and every 30 seconds thereafter for a period of one hour. If you analyze the phloem samples for 14C, where would you expect to find it? Explain your reasoning. The 14CO2 would be used by the cells in the leaf to make sugars in photosynthesis. These sugars would be loaded into the phloem. Because flow of sugars in phloem is from source (higher concentration) to sink (lower concentration), you would expect to find some 14C in the phloem sampled from the aphid mouthparts below the leaf. This would occur because sugars produced by the leaves are needed to feed the nonphotosynthetic parts of the plant (e.g., roots). However, you would also see some 14C in the samples from the aphid mouthparts above the leaf. The growing bud at the apical meristem is also a sink requiring sugars from photosynthetic parts of the plant. This region would require considerable sugars if it were forming a flower bud and ultimately a fruit. 4. On a separate sheet of paper, outline an experiment that would allow you to determine: a. how fast a substance is carried in the phloem, b. in what direction the substance flows in the phloem, and c. what percentages of solutes are in the phloem at various distances away from the leaves or roots. Different methods could be used. One is to sample phloem sap at different distances from the leaves and determine the concentration of sugar present. If you added a tracer, you could determine the rate and direction of flow in phloem. For example, you could cover the leaves of a plant with a plastic bag and inject radioactive carbon dioxide (14CO2) into the bag. The carbon will be incorporated into sugar during photosynthesis. Over time, you could sample the sap from each of the aphid mouthparts and determine • when and where the radioactivity first appears, • how quickly the radioactivity moves from the first aphid mouthpart to the other aphid mouthparts (rate of movement equals the time it takes the tracer to move from one point to the next divided by distance between the two points, or aphid mouthparts), and • the direction the sap moves (determined by where the tracer appears over time). 252
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Notes to Instructors Chapter 37 Plant Nutrition What is the focus of this activity? This activity focuses student attention on the fact that “plant foods” are usually inorganic; that is, unlike animals, most plants are capable of making all of their organic compounds from inorganic precursors.
What is this particular activity designed to do? Activity 37.1 What do you need to consider in order to grow plants in space (or anywhere else for that matter)? This activity is designed to help students understand the basic requirements for plant growth and how knowledge of these requirements can help them solve novel problems— for example, how to grow plants in space.
Answers Activity 37.1 What do you need to consider in order to grow plants in space (or anywhere else for that matter)? Long-range, human-operated space travel and space stations may someday become a reality. Before this can occur, however, we will have to develop sustainable methods of agriculture suitable for use in space. One of the key methods being investigated is hydroponics, growing plants in water supplemented with nutrients. You are assigned to a team working on the design of plant growth systems for use in a space station. 1. What types of plants would you choose to grow? Explain the reasoning behind your choices. There are many possible answers to this question. All answers should include how much room the proposed crop plant requires for growth, and how much of the crop plant produced can be consumed or otherwise used by humans (or other organisms on the space station). For example, it makes sense to grow plants that produce a high ratio of edible to inedible matter per plant. A wheat plant, for instance, produces a fairly low ratio of edible to inedible matter per plant. By contrast, a potato plant produces a much higher ratio of edible to inedible matter.
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2. When you set up the growth system: Note: Minimal answers to the following questions are provided. a. What nutrients would you b. What atmosphere would c. Which of the need to add to the water? you need to maintain? requirements in parts a List what you would List what components you and b could be recycled? need, and why each would need to maintain in What could not be would be necessary. the atmosphere and why recycled? Explain. each would be necessary. Refer to Table 37.1 of Biology, 8th edition, for the complete list of both macronutrients and micronutrients required for plant growth. This table also indicates the major functions of each nutrient.
The atmosphere in the space station would have to be similar to the atmosphere on Earth. This would be a requirement for both human life and successful plant growth. Plants require carbon dioxide for photosynthesis and oxygen for cellular respiration. If the oxygen-to-carbon-dioxide ratio became too high, then C3 plants would undergo photorespiration instead of photosynthesis. This would reduce plant growth and crop yield. You would obviously have to balance the number of plants versus animals (and other carbon-dioxide producers such as microbes) on the station to control the oxygen-to-carbon-dioxide ratios in the atmosphere over time. (If you chose to grow legumes, symbionts associated with plant roots— for example, Rhizobium in legume roots—could fix atmospheric nitrogen, which would reduce the need for additional N fertilizer.)
Oxygen and carbon dioxide could be recycled between the plants and animals on the station. Water in the atmosphere (and in the urine of animals) could be recycled. Many of the minerals in the plants could be recycled by composting inedible parts of the plants and animal feces. These could be converted to inorganic nutrients for plants by the actions of a variety of microbes. Without the microbes, large quantities of organic waste would build up and, as a result, most compounds other than oxygen, carbon dioxide, and water could not be recycled.
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Notes to Instructors Chapter 38 Angiosperm Reproduction and Biotechnology What is the focus of this activity? All plants display an alternation of generations between a multicellular haploid or gametophyte generation and a multicellular diploid or sporophyte generation. Evolutionarily there has been a general trend in the plant kingdom toward reduction of the haploid generation. This reduction is maximum in the angiosperms, where the female gametophyte is reduced to the embryo sac (seven cells, eight nuclei). The male gametophyte is reduced to even fewer cells: the germinated pollen grain, which contains a generative nucleus, and two sperm nuclei.
What is this particular activity designed to do? Activity 38.1 How can plant reproduction be modified using biotechnology? This activity is designed to help students understand the basic structure and function of male and female gametophytes in angiosperms. They will also learn how biotechnology differs from artificial selection and cloning of plants.
Answers Activity 38.1 How can plant reproduction be modified using biotechnology? Answer questions 1–7. Then in question 8 you will develop a concept map that links the information from all seven answers. 1. Draw a general diagram of the life cycle of a seed plant. Indicate which steps are haploid and which are diploid. Refer to Figure 13.6, part B, on page 252 of Biology, 8th edition. 2. Draw and label all parts of a complete flower. Indicate the functions of the major parts. A complete flower is diagrammed in Figure 38.2 of Biology, 8th edition.
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A complete flower includes sepals, petals, stamens, and carpel(s). • Sepals enclose and protect the developing flower. • Petals attract the attention of pollinators. • Anthers contain the male microsporangium. • Carpel(s) contain the female megasporangium. 3. Draw and label a mature ovule. Include the micropyle, integuments, nucellus, synergids, polar nuclei, egg, and anti-podals. Indicate the functions of each of these structures. See Figure 38.3 of Biology, 8th edition. 4. Define microsporogenesis and megasporogenesis. In what portion(s) of the flower does each of these processes occur? What is the end product of each process? Microsporogenesis occurs in the sporangia of the anther in flowers. Microsporocytes (2n) undergo meiosis to produce haploid microspores; each then develops into a pollen grain. Megasporogenesis occurs in the sporanium of the ovule in flowers. The megasporocyte (2n) undergoes meiosis to form a haploid megaspore (plus three other haploid cells that do not survive). The megaspore undergoes mitosis to produce the embryo sac, which contains the egg, synergids, antipodal cells, and polar nuclei. 5. What is pollination? How does it differ from fertilization? Pollination occurs when pollen lands on the stigma of a genetically compatible plant. Fertilization occurs when a sperm nucleus from the pollen tube joins with the egg nucleus in the embryo sac. In double fertilization, the other sperm nucleus (from the pollen tube) joins with the polar nuclei to form the endosperm nucleus. 6. What stages of the life cycle are eliminated or bypassed when plants are cloned naturally? When plants are cloned on the farm or in the laboratory? The gametophyte generation is bypassed in both situations. Natural cloning occurs when, for example, plants send out runners or new sprouts arise from underground roots or rhizomes. Similarly, cuttings or small pieces of plants can be grown into complete plants on the farm or in the lab. In all of these cases, the genetics of the new plants are identical to those of the parent. (Note: Some variations can arise due to spontaneous mutations.) Biology, 8th edition, includes other examples as well.
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7. What does the science of plant biotechnology do that artificial selection and/or cloning practices don’t do? Biotechnology can add genes from other organisms to plants. For example, a gene from Bacillus thuringiensis has been added to some crop plants. It produces the Bt toxin, which acts as an insecticide. 8. Construct a concept map that links all of the information in questions 1–7. Do this on a separate piece of paper.
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Notes to Instructors Chapter 39 Plant Responses to Internal and External Signals What is the focus of this activity? In Activity 37.1, students looked at some of the nutrient conditions required to develop sustainable agriculture in space. In this activity, students are asked to refine their ideas to include the effects of gravitropism, phototropism, and photoperiodism.
What is this particular activity designed to do? Activity 39.1 How do gravity and light affect plant growth responses? Students apply what they have learned about gravitropism, phototropism, and photoperiodism to propose methods for improving systems for growing plants in space.
What misconceptions or difficulties can this activity reveal? Activity 39.1 Many students have difficulty understanding how higher concentrations of auxin can affect stems or shoots differently than they affect roots. It helps to remind students that although stems and roots are continuous structures, they are distinctly different plant organs. As such, they can respond differently to a given chemical signal. This situation is analogous to the mammalian digestive tract. The esophagus, stomach, and small and large intestines are all continuous; however, each has a different structure and function. In addition, each can respond differently to a single hormone.
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Answers Activity 39.1 How do gravity and light affect plant growth responses? Review Chapter 39 of Biology, 8th edition, and your answers to Activity 37.1. Then answer the questions. 1. One of the problems associated with growing plants in space is lack of gravity. a. How does gravity affect the normal growth of a plant’s roots, stems, and other parts? Explain the mechanisms involved. Under the influence of gravity, auxin accumulates on the lower side of the root and stem. Higher auxin concentrations stimulate cell elongation in the stem but inhibit cell elongation in the root. In the shoot region, a concentration of auxin between 108 and 104 stimulates cell elongation. According to the acid growth hypothesis, auxin concentrations in this range stimulate proton pumps, which lower the pH in the cell wall. This activates enzymes that break cross-links between cellulose molecules and allow the cell to elongate.
b. How would a lack of gravity affect normal plant growth? Normal seed germination and seedling growth begin underground in the absence of light. Under these conditions, the seedling relies on the gravitropic responses of the shoot and root to orient these above ground and underground, respectively. In the absence of gravity, auxin concentrations in the root and shoot would not vary, and growth of the shoot away from gravity and growth of the root toward gravity would not occur.
c. Propose mechanisms to overcome the problems associated with a lack of gravity. Plant orientation is also affected by light. (See question 2.) The response to light can help counteract the lack of gravity. In addition, if plants are grown hydroponically, a dense meshwork mat can be used to prevent roots from growing above water.
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d. What design modifications would you need to make to support plants with different photoperiods—for example, long-day versus short-day plants? If the different plant species you were growing had different photoperiods, you would need to grow them in separate chambers. You could then set up different timer controls in each chamber to regulate the number of hours of light (versus dark) per day (24 hours).
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c. What kind of physical environment would you need to maintain appropriate phototropic responses among plants? In space or in the lab, you would need to set up banks of overhead lights. Light bulbs that are rich in red and blue wavelengths should be used. A hydroponic growth system should be set up to include a meshwork mat to support seeds or seedlings in early growth and to support the stems and maintain the roots below the mat (in the hydroponic solution) in later growth.
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b. What light characteristics would you use to maximize plant growth per unit time? In photosynthesis, the chlorophyll molecules in photosystems I and II respond to or absorb light in the red and blue wavelengths of the visible spectrum. As a result, if nothing else is limiting, the more red and blue light available, the faster the photosynthetic rate per unit time.
a. How do phototropism and photoperiodism differ? Phototropism is the growth of a plant toward or away from light. In general, shoots grow toward light (are positively phototropic) and roots grow away from light (are negatively phototropic). Again, this response appears to be the result of assymetric distribution of auxin. (Refer to Figures 39.5 and 39.6.) Photoperiodism is a physiological response to changes in day length. For example, the physiological responses that cause flowering in some species of plants are triggered by changes in day length.
2. Another problem with growing plants in space relates to a plant’s light requirements and phototropic responses versus the photoperiods required for the plant to flower and produce fruit.
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39.1 Test Your Understanding 1. Compared to the wild type, a mutant plant that overproduces cytokinin will probably have ______ lateral branches. Fill in the blank with the best answer and explain your answer. a. more b. fewer c. the same number of The answer is a. Adding cytokinins changes (decreases) the auxin-to-cytokinin ratio and stimulates lateral branching. 2. Combining the mutation in question 1 above with one that causes the overproduction of auxin would ______ the effect you chose above. Fill in the blank with the best answer and explain your answer. a. increase b. decrease c. not change The answer is b. Adding auxin would increase the auxin-to-cytokinin ratio and counteract the action of the added cytokinin. 3 and 4. A species of long-day plant will flower when the days are over 14 hours long. Which of the following light cycles would cause this plant species to flower? Answer True if the cycle causes flowering and False if it does not. Explain your answers. T/F
T/F
3. Continuous white light day and night, interrupted every 4 hours by flashes of red light. True—A long-day plant blooms when the Pfr/Pr ratio is high. Continuous white light would keep the Pfr/Pr ratio high. Flashes of red light would do the same. 4. A daily cycle of 9 hours of light followed by three periods of 4 hours and 50 minutes of uninterrupted dark, and 10 minutes each of red light and white light. True—The light period and each of the flashes of red light converts all of the Pr to Pfr. This maintains the high Pfr/Pr ratio required for flowering. Knowing this, greenhouses can save considerable amounts of money by not having to keep the lights on as long to stimulate flowering in long-day plants.
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Notes to Instructors Chapter 40 Basic Principles of Animal Form and Function What is the focus of this activity? All organisms, whether single-celled or complex and multicelled, must be able to maintain homeostasis. For each cell in the organism to remain alive, it must be continuously supplied with food and oxygen; wastes must be removed, pH must be maintained within a specific range, and so on. As a result, maintaining homeostasis is dependent on the surface-area-to-volume ratio of individual cells, the rates of diffusion or transport of substances across the surface area, and the general availability of these substances to the cells. These requirements effectively direct and limit animal structural and/or organ system design.
What is this particular activity designed to do? Exercise 40.1 How does an organism’s structure help it maintain homeostasis? The questions in this activity are designed to help students review and understand how surface-area-to-volume ratios can affect cell structure and function.
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Answers Activity 40.1 How does an organism’s structure help it maintain homeostasis? 1. To remain alive, an organism must be able to maintain homeostasis of its internal environment relative to the external environment. What behaviors, structure(s) or system(s) are of primary importance in maintaining homeostasis in the following situations in amoeba versus mammal? Situation a. Change in environmental: pH
temperature
Amoeba
Mammal
Behavioral: Amoebas can move to avoid areas that have unfavorable pH or temperature. Structural and physiological: The only barrier between amoebas and their environment is the cell membrane. Changes in the percentage of saturated versus nonsaturated fatty acids in the cell membrane can occur in response to changes in environmental temperature. pH responses depend on the specific pH encountered and the permeability and structural characteristics of the membrane. Some amoebas can form and surround themselves with a protective shield when they encounter adverse conditions. (Note: In all cases, the ability to respond depends on how quickly conditions change.)
Behavioral: Mammals can avoid areas that have unfavorable pH or temperature. Structural and physiological: The skin of mammals is waterproof and contains some chemical buffering compounds. It acts as a barrier between the environment and the organism. Sweat glands in many mammals allow for evaporative cooling during periods of high temperature. In many species, fur provides insulation. Subcutaneous fat also provides insulation.
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Situation b. Reception of stimuli, for example: light
touch
c. Response to stimuli
Amoeba
Mammal
Amoebas do not have specific subcellular organelles designed to respond to either light or touch. Rather, the cell as a whole is capable of receiving these stimuli.
As multicellular organisms, mammals have specialized parts of the body for various functions. For example, they have eyes that perceive differences in light levels and in colors or wavelengths of visible light. They also have specialized touch receptors that respond to differences in pressure.
Amoebas are negatively phototropic. When touched or disturbed, they often pull in all pseudopodia and round up. Given this, they are obviously capable of responding to light and touch stimuli. The response is a whole-cell response.
Both the eye and touch receptors respond to stimuli by sending nerve impulses (action potentials) to the central nervous system (CNS). In the CNS, the signals are interpreted and a response is generated by nerve signals sent to many parts of the body. For example, a strong touch can result in the mammal pulling away from the stimulus, looking to see what caused the stimulus, crying out in response, and so on.
2. Cells must be bathed continuously in an aqueous medium to take in oxygen and nutrients and get rid of waste products via diffusion. Diffusion is efficient over only short distances. In fact, diffusion is efficient only for a distance of about three cell diameters maximum (approx. 200 to 300 m). Note the following times required to diffuse specific distances: Diffusion Distance (m) 1 10 100 1,000 (1 mm) 10,000 (1 cm)
Time Required for Diffusion 0.5 msec 50 msec 5000 msec (5 sec) 500,000 msec (8.3 min) 50,000,000 msec (14 hr)
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a. Graph the data in the table and calculate the mathematical relationship between the increase in distance and the corresponding increase in time required to diffuse that distance. You can do this by hand or you can use a software program with graphing capability to do this. Include your graph and your calculations below. In general, for each 10-fold increase in distance, a 100-fold increase in time for diffusion is required. In other words, as the distance increases by X times, the time to diffuse that distance increases by X 2. See the graph below for the specific formula. micrometers 1 10 100 1000 10000
milliseconds 0.5 50 5000 500000 50000000
Diffusion Distance vs. Time y = 0.5x2 R2 = 1
60000000 Time (msec)
50000000
Diffusion Distance vs. Time Power (Diffusion Distance vs. Time)
40000000 30000000 20000000 10000000 0 0
5000 1000 1500 0 0
Diffusion distance (um)
b. How much time would be required for oxygen to diffuse 5 m? 200 m? The difference between 1 and 5 m is 5. We square this value (5 5 25) to determine the difference in the rate of diffusion (25 33 0.5 msec 12.5 msec). Similarly, the difference between 100 m and 200 m is 2; 2 squared 4; 4 5 seconds (to diffuse 100 m) 20 seconds (to diffuse 200 m).
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3. What happens to the surface-area-to-volume (SA/V) ratio of a three-dimensional object (such as a cell) as its linear dimension increases? For example, how does the SA/V ratio of a sphere or cube change as the linear dimensions increase? (Formulas for a sphere: surface area = 4r2; volume = 4/3 r3.) (Also see Activity 7.1.) As the linear dimensions of an object increase by two times, the surface area increases by the square (of the increase in linear dimensions) and the volume increases by the cube (of the increase in linear dimensions). (Note: This assumes the object retains the same shape—for example, that it remains spherical as it increases in size.)
40.1 Test Your Understanding Propose the effect(s) the physical properties of diffusion are likely to have on the structure and function of epithelia and epithelial cells and digestive and circulatory systems as animals become larger and larger. In your answer, consider how SA/V ratios change as organisms become larger and the effect(s) this change is likely to impose on the structure and function of organisms. See the description of various types of epithelial cells in Chapter 40 of Biology, 8th edition. Also, refer to Figure 40.5. In general, barrier-type cells are columnar and have fairly low SA/V ratios. Squamous cells tend to be flatter and allow for more efficient diffusion to occur. In the digestive and circulatory systems, villi, microvilli, capillary beds, and other structures all allow for massive expansion of the surface area across which diffusion of substances can occur. Remember, the larger the volume of an organism, the larger its surface area must be to support that volume—that is, to support metabolism or life. However, this becomes a catch-22 of sorts because animals need to move, not only to get food, mates, and so on, but also to avoid predation. However as they become larger, they need proportionately larger surface areas for exchange with the environment. Those with mutations that led to both more efficient locomotion and exchange—for example, those with mutations that allowed internalization of surface area, as alveloli in lungs, and so on—survived. Animals without these mutations died out.
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Notes to Instructors Chapter 41 Animal Nutrition What is the focus of this activity? Many of the organ systems in mammals are designed to handle the bulk flow of substances into and out of the body. The digestive system is designed to handle the bulk processing and digestion of food. The food we eat is composed of macromolecules—for example, proteins, carbohydrates, and fats that were produced by other organisms. These macromolecules are species-specific. In contrast, the monomers that make up the macromoleules are used almost universally among organisms. The function of digestion is to break food down into its monomers, which can then be taken into the body and used to build our own species-specific macromolecules. As a result, all the processing and digestion of other species’ macromolecules goes on outside of our cells, in the space inside our digestive tract. This space, the lumen of the tract, can therefore be considered a part of the external environment. We have internalized it, or brought it inside of our bodies, to increase the efficiency of food processing and digestion.
What is this particular activity designed to do? Activity 41.1 How are form and function related in the digestive system? This activity is designed to help students understand: • the basic structure and function of the digestive system in animals, and • the roles the various parts of the system play and how these parts are modified to best play these roles.
What misconceptions or difficulties can these activities reveal? Activity 41.1 Question 2: If you haven’t encountered it already, you will discover that many students do not understand “how bacteria eat.” They understand that bacteria are decomposers and some are parasites, but in general they don’t know how “eating” occurs. This question is designed to force the point. This question and subsequent questions demonstrate the basic commonalities that exist in methods of digestion across a wide range of organisms, including bacteria, amoebas, and humans. Question 3: The figure reduces the digestive tract to the basics to allow students to see the direct connections between its various parts. For many students this understanding becomes blurred when they view realistic diagrams of the actual organ system. The more linear view presented here helps them see how the system can be understood as a continuous disassembly line. Notes to Instructors
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Question 5: Many students don’t understand that the food that enters the digestive tract is technically not inside the body. It is contained in a space that exists within the body. To be in the body requires that a substance cross a cell membrane. Understanding this difference also helps students understand later (Chapter 43) why foreign macromolecules anywhere else in the body produce an immune response. Question 7: Most students understand that enzymes operate using a lock-and-key-type system of recognizing and interacting with their substrates. Many don’t realize, however, that this means that enzymes work on surfaces. As a result, the larger the surface area, the greater the number of enzymes that can be digesting the food at the same time. Question 9: Students may not understand the roles that the hepatic portal vein and the liver play in maintaining homeostasis of the blood (and therefore the body in general). In that case, it will be difficult for them understand how insulin and glucagon can act to control blood sugar levels (see Chapter 45).
Answers Activity 41.1 How are form and function related in the digestive system? 1. What is the overall function of digestion (as a whole)? Digestion breaks down complex carbohydrates, nucleic acids, proteins, and lipids into their component parts (for example, monosaccharides, amino acids, nucleotides, glycerol, and fatty acids). The component parts are not recognized as “foreign” by the body and can be used in the body’s metabolism (no matter what the original source). 2.
a. How do bacteria eat? b. How do amoebas eat? When they encounter a food source, Amoebas trap small food items in bacteria secrete digestive enzymes food vacuoles and then secrete into the environment around the digestive enzymes into the vacuoles food source. The digested food— (often by combining a vacuole monosaccharides, amino acids, and containing hydrolytic enzymes with a other components—is picked up by food vacuole). the bacteria by transport across their cell membranes.
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How are their eating styles similar to what we see in humans? How are they different? c. Similarities with bacteria
d. Differences from bacteria
In both humans and bacteria, digestive enzymes are secreted to the outside of the cells. Digestion is extracellular.
In humans a part of the outside environment is held within the body (the digestive tract). As a result, the products of digestion are contained and can be picked up by the circulatory system (or lacteal system). In bacteria, it is possible for the products of digestion to diffuse away from the bacteria and be lost.
e. Similarities with f. Differences amoeba from amoeba As with bacteria, humans and amoebas secrete digestive enzymes to digest food. Both humans and amoebas enclose a portion of the external environment inside their "bodies." Food is trapped in this space and enzymes are added.
In amoebas, the food vacuoles are intracellular, so digestion is intracellular. In humans, digestion occurs outside of cells. Digested food is picked up by the circulatory system or the lacteal system (lipids only) for delivery to the rest of the body.
3. Label the parts of the mammalian digestive tract in the diagram, and state the major function(s) of each part. small intestive large intestine
mouth and teeth esophagus
stomach
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How does the structure of each part reflect its function?
a. Mouth/teeth b. Esophagus
c. Stomach
d. Small intestine
e. Large intestine
The mouth and teeth function to capture food, to break it down into smaller pieces, and thus to increase the surface-areato-volume ratio of the food. This increases the efficiency of digestion because digestive enzymes act only on surfaces. Salivary amylase begins the digestion of starch to smaller polysaccharides.
The stomach is muscular and has heavy walls. It allows us to consume large quantities of food at a given time. The muscular contractions of the stomach, in combination with the action of hydrochloric acid (secreted by cells in the wall of the stomach), break down large food particles into smaller ones and increase the SA/V ratio of the food particles. Pepsin (a proteindigesting enzyme) is also added and begins the digestion of protein into polypeptides. Another enzyme, rennin, denatures and coagulates milk protein so that it remains in the tract longer and can be digested by other enzymes.
The majority of enzymatic digestion occurs in the small intestine. See Figure 41.13 on page 887 for a complete listing of the various enzymes, their substrates, and products. The small intestine is long relative to the animal's body length. The walls of the small intestine are highly folded. The folds increase surface area. The surface of the folds contains many villi (fine projections). The surface of each of the villi contains many microvilli. All of these structures serve to increase the surface area for efficient absorption of digested nutrients.
The primary function of the large intestine is to reclaim or reabsorb the water remaining in the undigested material that leaves the small intestine. About 90% of this water is returned to the body. The large intestine also excretes excess salts and calcium into the feces. In addition, it houses a population of bacteria that produce some important vitamins (e.g., vitamin K) required by humans. As a result, the large intestine contains much less internal surface area and is considerably shorter than the small intestine. It is also more muscular, especially the lower part, which is under some voluntary control.
The esophagus is muscular and capable of peristalsis (rhythmic, wavelike contraction). Peristalsis moves food from the mouth to the stomach. (Peristalsis works even if you are eating or drinking while standing on your head.)
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4. Make a larger copy of the diagram in question 3. Use this larger diagram to model or trace what happens to a food particle from the time it enters the mouth until its indigestible remains are egested, or eliminated. Include all of the following terms in your discussion. Also, note on the diagram the function of each and where it is found in the digestive system. carbohydrate fat nucleic acid protein cardiac orifice epiglottis pyloric sphincter pharynx pepsinogen
lipase pepsin salivary amylase bile dipeptidase saliva amylase pancreas nuclease
gastric acid (HCl) bolus microvilli lacteals capillaries feces intestinal bacteria hepatic portal vein fiber or roughage
Using the understanding of the structure and function of the digestive system you gained from the model in question 4, answer the questions. 5. The mammalian digestive tract has been called an extension of the outside world that you enclose in your body. a. What does this statement mean? Consider what would happen if you swallowed a marble. Is the marble ever “inside your body”? If you swallowed a marble, it could pass through your entire digestive tract and never once cross a cell membrane. Technically, it was never “inside your body.” The lumen or space within the digestive tract can be considered an extension of the external environment. b. At what point in the digestive process is food officially inside the body? Food is not officially “inside the body” until it has crossed a cell membrane. Only after a food subunit (for example, an amino acid) crosses the membrane of a villus (in the small intestine) is the food officially inside the body. c. How is mammalian digestion more efficient than the type of digestion seen in bacteria? In bacteria, some of the secreted enzymes may diffuse away from the food source. Similarly, some of the products of digestion may diffuse away and not be picked up by the bacteria. In the mammalian digestive tract, the enzymes and food are “trapped” together. This makes digestion more efficient. In addition, uptake of the products of digestion is more efficient. Activity 41.1
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6. Digestion (in humans and many other animals) is both physical and chemical. Among the chief chemical agents of digestion are the digestive enzymes. What do enzymes do to food? Digestive enzymes break food down into its component macromolecules—for example, monosaccharides, amino acids, nucleotides or glycerol, and fatty acids. Each enzyme is capable of a specific type of hydrolytic chemical reaction. For example, certain amylases break carbohydrates down into disaccharides. Then disaccharases break specific disaccharides down into monosaccharides. 7. Have you ever heard the old adage: “Be sure to chew your food 20 times before swallowing?” a. What, if any, effect would this chewing have on how well the digestive system functions? Keep in mind that enzymes work only on the surfaces of food particles. Explain. Chewing your food breaks it down into smaller pieces. This increases the surface-area-to-volume ratio of the total mass. Keep in mind that food moves through the digestive tract at a fairly constant rate. This means that it is available to each part of the digestive system for only a specific period of time. Because enzymes work only on surfaces, the larger the surface area of food available, the more enzymes that can be functioning simultaneously and the faster the apparent rate of digestion. In addition, the faster the rate, the more likely that all of the available food will be digested in a given period of time. b. How does the function of the teeth complement the function of one of the digestive chemicals in the stomach? Be sure to name the specific chemical in your answer. Both the teeth and the hydrochloric acid in the stomach break down larger pieces of food into smaller pieces. Again, this increases the available surface area for enzyme action. 8. Although enormous quantities of various enzymes are added to the contents of the duodenum of the small intestine, no traces of enzymatic activity are left in the intestinal contents when they pass into the large intestine (colon). Why? What happens to the enzymes? Enzymes are made of protein. When no food remains for digestion, the enzymes will digest whatever protein is available. In other words, when no food remains, they will digest each other. 9. Most of the blood that leaves the digestive tract of a human is collected into a series of veins that merge to form the hepatic portal vein. The hepatic portal vein carries blood to the liver, where the hepatic portal vein divides again into a system of venules and then liver capillaries. The liver capillaries drain into the hepatic vein, 272
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which carries blood to the vena cava. The vena cava carries blood from the body to the right atrium of the heart. Some of the products of digestion enter a different system of transport, the lacteal system. The lacteal system bypasses the liver and carries its contents directly to the right atrium of the heart. a. Which products of digestion are carried in the blood to the liver? The monosaccharides, amino acids, and nucleic acids are carried in the blood to the liver. b. Which products of digestion are carried via the lacteal system? The fatty acids and glycerol are carried in the lacteal system. c. During the first hour after a heavy meal, how does the concentration of glucose in the blood going from the small intestine to the liver compare to the concentration entering the right atrium of the heart? If we assume the meal contained carbohydrates, then the blood going from the small intestine to the liver contains a higher concentration of glucose than the blood going from the liver to the right side of the heart. The liver regulates the amount of glucose that enters the blood to help maintain it at a homeostatic level of about 0.1%. d. Similarly, how does the concentration of amino acids compare? The liver also regulates the concentration of amino acids in the circulating blood. Excess amino acids are deaminated (the amine group is removed), and the rest of the molecule is converted to a form that can be stored—for example, as fat. e. How does the concentration of fat leaving the small intestine compare to the concentration in the right atrium? The products of lipid digestion (as well as some small fats) enter the lacteal system directly from the small intestine. The lacteal system empties directly into the right side of the heart. As a result, the concentration of fats (and fatty acids) leaving the intestine is about the same as the concentration entering the heart. 10. How does an herbivore such as a cow extract the glucose from the cellulose in its diet? What characteristics of the structure and function of the digestive tract of a ruminant suit it for this function? Symbiotic prokaryotes and protists are housed in the cow’s rumen. The cow chews its food, usually hay or grass. The chewed food or cud is swallowed and passes into the rumen, where the symbiotic prokaryotes and protists digest the cellulose. This cud may be regurgitated several times and rechewed to increase the available surface area of the food. Finally, the cud with its associated microorganisms is swallowed into the omasum (a sort of prestomach area), where excess water is removed. Activity 41.1
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It continues into the abomasum, where digestion of the bacteria and digestible parts of the grass begins. A colony of prokaryotes and protists remains in the rumen. These organisms secrete fatty acids, which can be absorbed and used by the cow. In other words, the cellulose is digested by the microorganisms and used to manufacture more microorganisms. The cow itself digests some of the microorganisms and this provides the cow with additional protein. In the rumen, the microorganisms also produce excess fatty acids. These excess fatty acids are secreted into the rumen, where they are picked up by the cow’s circulatory system and provide additional nutrition.
41.1 Test Your Understanding 1. A peanut butter and jelly sandwich contains a variety of carbohydrates, fats, and proteins. Complete the graph below to indicate the relative percentages of carbohydrate, fat, and protein that remain in this ingested food as it progresses from your mouth through your digestive tract. Explain your reasoning. Key: Protein = xxxxx; Carbohydrate = ---------; Fat = oooooooo
100%
Percent remaining
0% Mouth- esophagus-stomach-small intestine-large intestine-rectum
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A small amount of the carbohydrates has been digested by the salivary amylase in the mouth. The amount digested depends on the length of time the food remained in the mouth and esophagus. Salivary amylase is denatured in the acid environment of the stomach. Therefore, no further carbohydrate digestion will occur in the stomach. Some of the proteins are digested to polypeptides by pepsin in the stomach. (Pepsin is denatured in the basic environment of the small intestine.) However, the vast majority of digestion (of carbohydrates, polypeptides, proteins, and fats) occurs and is completed in the small intestine. 2. A good rough generalization is that the more meat in the diet of a species of animal, the shorter its intestine. In comparison, herbivores have long intestines (length always being relative to total body length). How can this be explained? In general, vegetation or plant materials are more difficult to digest than meat (animal muscle protein). Although the majority of digestive enzymes are added to the duodenum, they continue to function as the food passes along the digestive tract. In a normally functioning digestive tract, food passes along at a set pace. Therefore, the longer the various parts of the digestive tract, the longer the time available for digestion to occur. Similarly, a longer chamber containing symbiotic bacteria (for example, the cecum) allows for a longer time available for the digestion of cellulose. As a result, the relative length of the digestive system of herbivores is greater than that of carnivores. 3. What would happen to the normal function of the digestive tract if part of one of the following organs was removed or greatly reduced in size (for example, as the result of surgery following an accident)? How would a person’s eating habits need to change to accommodate the reduction in size? a. Stomach If part of the stomach were removed, you would not be able to eat very much at any given time. The volume consumed per unit time would have to be reduced. To maintain the same amount of energy input per day, you would need to eat many small meals. b. Small intestine If part of the small intestine were removed, your ability to digest food would be reduced. It is likely that you would need to eat much more, but since you would be getting less in the way of digested food, your diet would need to be supplemented with amino acids, fatty acids, and monosaccharides. c. Large intestine If part of the large intestine were removed, you would be less able to remove water from the digestive system so more water would be lost from the body. As a result, you would need to consume more water.
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4. A friend of yours has come up with a new idea for people who don’t have time to eat. He has developed a high-powered Blenderizer that breaks food up into very small particles. He has tested his product on amoeba and Paramecia. Both of these single-celled organisms can pick up these small particles and incorporate them into food vacuoles. Your friend is hoping to market the Blenderizer to busy people who could blenderize their food and use an IV bag and tube to run the food directly into their blood system. Because people could hide the system under their clothes, they could essentially eat any time without it interfering with meetings and other activities. Your friend comes to you for advice and possible financial support to get his idea off the ground. What would you say to him? Anyone who used this system would probably die. The proteins and other speciesspecific macromolecules of the food would be recognized by the immune system as foreign. A massive immune response could block the circulatory system and lead to stroke or heart attacks. To avoid these problem, digestion breaks down the polypeptides, polysaccharides, and other macromolecules in food into their constituent monomers, which are used across species.
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Notes to Instructors Chapter 42 Circulation and Gas Exchange What is the focus of these activities? All the cells of the body need to be continuously supplied with oxygen and food to support cellular respiration (Chapter 9). Wastes need to be removed. Oxygen and carbon dioxide are transported by diffusion into and out of cells. Food and wastes are transported via facilitated diffusion or active transport (Chapter 7). In unicellular and very small organisms, all the cells may be in direct contact with the environment. In complex multicellular organisms, many of the cells are not in direct contact with the environment. In addition, the cells of multicellular organisms may be specialized for different functions (and couldn’t capture or process food even if they were in contact with the environment). These cells would die without a system to supply them with their needs and remove metabolic wastes. The circulatory system is the delivery system. As noted in Chapter 41, the digestive system supplies the required nutrients to the circulatory system. The respiratory system supplies it with oxygen and removes the carbon dioxide. In other words, understanding these systems requires that students understand and integrate what they have learned previously with what they learn in this chapter.
What are the particular activities designed to do? Activity 42.1 How is mammalian heart structure related to function? This activity is designed to give students an understanding of the normal function of the circulatory system as well as how normal function can be modified by specific diseases or disorders. Activity 42.2 How do we breathe, and why do we breathe? This activity is designed to give students an understanding of how and why we breathe and some of the mechanisms that exist to control breathing rates. Activity 42.3 How are heart and lung structure and function related to metabolic rate? This activity asks students to integrate their understanding of the structure and function of both the circulatory and respiratory systems with their understanding of cellular metabolism and metabolic rate.
What misconceptions or difficulties can these activities reveal? Activity 42.1 Question 1: Many students have difficulty understanding the overall flow pattern from the left side of the heart to the body and back to the right side of the heart, to the lungs, and Notes to Instructors
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then back to the left side of the body. They may be confused by realistic diagrams and pictures of the heart that show all the major vessels “coming out the top of the heart.” You can simplify the system for students by drawing a schematic heart that has a vessel coming out of the side of each chamber. For example:
Right
Left
As you do this, let students know that your drawing is anatomically incorrect; however, have them use the drawing to explain circulation. Then ask them to tell you what is incorrect about your drawing. Question 5: If the students have worked out and understand how to track blood flow in the body, they will have little difficulty with this question compared to those who memorized the pattern of flow. Activity 42.2 Question 1: Most students understand how we breathe. Many fewer make the connection between why we breathe and the requirements of cellular respiration. If you ask students why we breathe, most will automatically reply, “Because we need oxygen to live.” If you then ask them why we need oxygen to live, many can’t answer, but those who do often say, “Because our cells need oxygen.” That begs the question: Why do our cells need oxygen? Again, many students find this difficult. It may take several more questions to get them to connect the need for oxygen with the functioning of the electron transport chain in oxidative phosphorylation. Questions 2 and 3: Students may have difficulty reading oxygen dissociation graphs because they try to read them from left to right (the way most graphs are read). It is useful to point out that it is easier to understand these graphs if you read them from right to left. For example, at about 100 mm Hg partial pressure, hemoglobin is about 100% saturated with oxygen (all four sites on each hemoglobin molecule are bound to an oxygen molecule). So, 100 mm Hg is the approximate partial pressure of oxygen in the lungs. The blood moves from the lungs to the heart and out to the tissues. The tissues have used up much of the oxygen available to them. As a result, the partial pressure of oxygen in the tissues may be about 60 to 70 mm Hg. When the oxygen-saturated blood comes in contact with these tissues, it unloads or gives up (reading from right to left) about 25% of its oxygen to the tissues.
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Answers Activity 42.1 How is mammalian heart structure related to function? 1. Diagram and describe the path a red blood cell takes from a capillary in your big toe to your heart and back to your big toe. At each point in the pathway, indicate whether the red blood cell is most likely picking up or losing oxygen. Indicate also the relative blood pressure that part of the path is likely to have. Include all of the following terms in your diagram. aorta venules veins arteries arterioles capillaries right atrium left atrium
right ventricle left ventricle coronary arteries heart internal organs (for example, lungs, digestive tract) skeletal muscle
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2. The degree of musculature differs in these chambers of the heart: atria, right ventricle, left ventricle. a. Draw the heart and its chambers, including differences in musculature.
right alrium
right ventricle
b. Explain why the c. Include below the differences in musculature functions of the SA node might exist by explaining (pacemaker), the AV the normal functions of node, and the AV and each chamber. semilunar valves. The walls of the atria, both right and left, are fairly thin, indicating little musculature. Blood flows through the atria into the ventricles. When the atria contract, they push the blood they contain only a few centimeters into the ventricles. This overfills the ventricles and helps initiate the ventricular contraction. The right ventricle pumps its contents to the lungs. The left ventricle pumps its contents out the aorta and to the rest of the body. The difference in musculature of the two ventricles is correlated with the amount of effort required to move the blood through the circulatory pathway associated with each ventricle.
left alrium
left ventricle
The signal from the SA node (also called the pacemaker) triggers the contraction of the atria. This signal is delayed at the AV node, which allows the atria to empty before the ventricles are signaled to contract. The atrioventricular (AV) valves lie between the atria and the ventricles. These are one-way valves that prevent blood from flowing back into the atria when the ventricles contract. The semilunar valves lie between the left ventricle and the aorta and between the right ventricle and the pulmonary artery. These are also one-way valves that prevent blood from flowing backward into the ventricles.
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42.1 Test Your Understanding 1. While in utero, the wall between the two atria of the human fetal heart is not complete. An opening, the foramen ovale, allows blood from the two atria to mix. Normally, at birth, this hole seals over and the two atria are separated from each other. What would be the consequences to the infant if this hole did not seal over at birth? If the foramen ovale did not seal at birth, oxygenated blood from the lungs (left atrium) would mix with deoxygenated blood from the body (right atrium). This condition is called “blue baby syndrome.” The blood going to the body carries less oxygen than normal. As a result, the baby turns blue during times of higher metabolic activity, which require greater amounts of oxygen input per unit time. 2. One of the most common congenital defects of the cardiovascular system is called “transposition of the great arteries.” In infants who have this defect, the pulmonary artery exits from the heart where the aorta should and the aorta exits where the pulmonary artery should. All other circulatory connections are normal. a. Diagram and describe the circulation of blood in an infant who has this genetic defect. When the positions of the aorta and pulmonary artery are reversed, two separate circulatory systems are set up. One pumps the blood from the left side of the heart via the pulmonary artery to the lungs and back via the pulmonary vein to the left side of the heart. The other pumps the blood from the right side of the heart via the aorta to the body and back to the right side of the heart via the anterior and posterior vena cava. b. What type of treatment would such an infant need? Unless something is done immediately, the infant will die. An operation to reverse the arteries takes a relatively long time. In the short term, the physician opens the foramen ovale to allow at least some oxygen to be pumped to the body.
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Activity 42.2 How do we breathe, and why do we breathe? This activity is designed to give you an understanding of how and why we breathe and some of the mechanisms that control the rates of breathing. How do we breathe? Explain the mechanics of breathing. To do this, diagram (or model) and explain the general movement of oxygen and carbon dioxide into and out of the lungs, lung alveoli, blood, and tissues. Be sure to include all of the following terms in your diagram. diaphragm diffusion CO2 removal O2 demand lungs alveoli expired air inspired air oxygen
carbon dioxide brain PCO2 sensors medulla oblongata lung tissue cells skeletal muscle cells carbohydrate + O2 → CO2 + H2O + energy CO2 + H2O ↔ H2CO3 ↔ HCO3⫺ + H+
Use your diagram or model to answer the questions. 1. Explain briefly why we breathe. Or, more specifically, where is what we breathe in used in the body and where is what we breathe out produced in the body? We breathe in to supply our cells with oxygen for use in cellular respiration— specifically, oxygen is the ultimate electron acceptor in the electron transport chain (oxidative phosphorylation). We breathe out to rid our body of carbon dioxide, which is a waste product of cellular respiration. The sugars that undergo cellular respiration are broken down into carbon dioxide and water.
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2. Oxygen is transported in the blood in reversible combination with hemoglobin. a. How does the level of carbon dioxide in the blood contribute to blood pH? (Write the chemical reaction.) CO2 + H2O → H2CO3 (carbonic acid) → H+ + HCO3⫺ In other words, the higher the carbon dioxide concentration in the blood, the higher the H+ ion concentration and the lower the pH. (Note: This is also the reason carbonated beverages are more acidic than noncarbonated ones.) b. How does the level of CO2 in the blood affect the affinity of hemoglobin for oxygen? That is, how does the concentration of CO2 in the blood change its ability to carry oxygen (the effect of the Bohr shift)? What is the Bohr shift? The conformation or shape of the hemoglobin molecule changes in response to pH. At lower physiological pHs, this shape change causes hemoglobin to give up more of its oxygen to the tissues (at a given partial pressure) than it would at a higher physiological pH. This pH-induced shift in ability to hold onto (or give up) oxygen is termed the Bohr shift. c. Is the Bohr shift more likely to occur after exercise or after hyperventilation (rapid breathing with no exercise involved)? Explain. Exercise increases the amount of ATP required per unit time (compared to no exercise). ATP production via cellular respiration produces carbon dioxide as a waste product. The addition of more carbon dioxide lowers the pH in the tissue. The resulting conformational change in the hemoglobin delivers more oxygen to tissues undergoing exercise (or rapid cellular respiration) than to other tissues. Hyperventilation would have the opposite effect and would raise the pH of the blood.
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O2 saturation of hemoglobin (%)
O2 saturation of hemoglobin (%)
3. Compare these oxygen dissociation curves for hemoglobin.
100 80 60 40 20 0
O2 saturation of hemoglobin (%)
0
20
40
60
100 80 60 40 20 0 0
80 100
20
40
60
80 100
PO2 (mm Hg)
PO2 (mm Hg)
Normal adult
Reduced pH conditions Note: The solid line indicates the curve for reduced pH conditions.
100
Fetus
80 Mother
60 40 20 0 0
20
40
60
80 100
PO2 (mm Hg) Fetal vs. Adult
a. How do the curves differ? i. Normal versus reduced pH? Adult hemoglobin becomes completely saturated with oxygen at 80 mm Hg (in the lungs). The graph indicates that at normal pH when blood saturated with oxygen encounters tissues at 60 mm Hg, the Hb releases about 15% of its oxygen. However, because of the conformational change in Hb that occurs at low pH, under low-pH conditions, the Hb instead releases about 20% of its oxygen at 60 mm Hg.
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ii. Fetus versus adult? Adult hemoglobin becomes completely saturated with oxygen (carries as much as it can) at a partial pressure of 80 mm Hg. This is approximately the partial pressure of oxygen in the human lung. In contrast, fetal hemoglobin becomes saturated at about 60 mm Hg. This allows the fetus to pull more oxygen from the mother’s blood than it could if the fetus had the same type of hemoglobin as the mother. b. What would happen if fetal and adult hemoglobin had the same affinities for oxygen? Oxygenated blood from the mother’s left ventricle passes through other tissues before it reaches the placenta. As a result, the partial pressure of oxygen in the mother’s blood is less than 80 mm Hg. If the hemoglobin of the mother and fetus had the same affinities for oxygen, the fetal blood would never be saturated with oxygen. In other words, oxygen transport to the fetus would be reduced. As a result, the fetal metabolism would be reduced, which could affect all of development. 4. If they don’t get what they want, some small children threaten to “hold their breath until they die.” Children who are very strong-willed may turn blue, but they will not die because they will pass out and the autonomic breathing response will take over. What types of control are involved in holding your breath? Why (for what physiological reason) will you pass out if you hold your breath too long? When you hold your breath the carbon dioxide level in your blood increases and pH decreases. Changes in blood pH (carbon dioxide levels) are sensed in the medulla of the brain. If the carbon dioxide level gets too high, the medulla triggers a nervous response that causes you to pass out. The medulla’s involuntary control of the rate of breathing is then restored. It is interesting to note that the same thing can happen when you hyperventilate. When you hyperventilate you reduce the carbon dioxide level in the blood (and therefore increase the blood pH). This is generally recognized by the medulla, and involuntary breathing is reduced or stops. If you continue to force hyperventilation, again you can reduce the carbon dioxide level so much (or conversely increase the oxygen level so much) that it causes peripheral arteries to constrict. This reduces blood flow to the brain (and other parts of the body) and causes you to pass out. Breathing stops temporarily until the carbon dioxide levels in the blood build up and involuntary control by the medulla takes over.
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Activity 42.3 How are heart and lung structure and function related to metabolic rate? 1. Describe the simplest type of closed circulatory system. Include the function of each part of the system and how the structure of each part is related to its function. The simplest closed circulatory system would include a heart, a muscular organ that pumps blood to arteries. The arteries branch into arterioles and capillaries (where the exchange occurs). From the capillaries the blood continues on in venules and then veins back to the heart. Arteries are more muscular than veins to withstand the higher blood pressure resulting from the force of the heart’s contraction. Capillaries are one-cell-wall thick to allow for diffusion of substances into and out of the capillaries. 2. If you want to increase the rate of metabolism in an organism, you need to increase: a) the rate of food and oxygen delivery to the cells (or body) and b) the rate of removal of waste products from the cells (or body). In the evolution of the vertebrates, what modifications of the circulatory system and/or circulatory function have occurred to increase metabolic rate? • Fish have two-chambered hearts (one atrium and one ventricle). Deoxygenated blood is pumped from the heart to the gills and oxygenated blood then travels from the gills to the rest of the body. Blood pressure is fairly high as the blood leaves the heart. However, as the arteries branch to form arterioles and as the arterioles branch into capillaries, the cross-sectional area becomes so great that the blood pressure is reduced to about zero. As a result, the blood flow from the gills to the rest of the body can be relatively slow even when aided by muscular contractions of the body wall. As noted, low blood flow means reduced delivery of food and oxygen to cells and therefore lower metabolic rates. • Amphibians and many reptiles (exceptions include the crocodilians and birds) have a three-chambered heart (two atria and one ventricle). Deoxygenated blood from the body is pumped via the right atrium to the lungs. In passing through the lung capillaries, the blood loses pressure. However, the oxygenated blood is returned to the left atrium where it is pumped out to the body. Because there is only one ventricle, there is the potential for some mixing of oxygenated and deoxygenated blood. This increase in blood pressure increases the rate of flow of blood to the cells. However, the mixing of oxygenated and unoxygenated blood in the heart reduces the rate at which oxygen is delivered to the cells and as a result can reduce metabolic rate. • Birds and mammals have essentially two hearts (two atria and two ventricles). The right atrium receives blood from the body and the right ventricle pumps it to the lungs. The oxygenated blood from the lungs is transported to the left atrium and is then pumped out to the body by the left ventricle. In this system, blood pressure and therefore flow rate are high throughout the arterial system and there is no mixing of oxygenated and deoxyenated blood. As a result, higher levels of 286
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oxygen can be delivered per unit time to the cells (or body) and a higher level of metabolism can be maintained. 3. Both reptiles and mammals obtain their energy via aerobic respiration: C6H12O6 + 6O2 → 6CO2 + 6H2O + 36ATP Both also need to be able to maintain internal homeostasis in order to survive. a. Given this, would you expect the lungs of the lizard (an ectotherm) to have more or fewer alveoli per unit area than the lungs of the rat (an endotherm)? Assume the overall SA/V ratios of the rat and lizard are about the same. Explain the physiological reasoning behind your answer. The lizard will have fewer alveoli per unit area of lung tissue than the rat. This is reasonable because, on average, the ectothermic lizard requires less energy per unit time than the rat. Similar to humans, the rat’s normal body temperature is generally higher than the environmental temperature and its enzymes and so on are functional at this higher temperature. Maintaining this higher body temperature requires higher levels of cellular respiration and, as a result, higher levels of oxygen and food intake. b. On average, would you expect the heart rate of the lizard to be higher or lower than the heart rate of the rat? Explain the physiological reasoning behind your answer. The argument here is similar to that above. To maintain a high metabolic rate (e.g., high body temperature) requires rapid delivery of food and oxygen to cells and removal of wastes from cells. As a result, the heart rate of the rat is expected to be higher than that of the lizard. 4. Based on your knowledge of SA/V ratios, metabolic rates and cardiopulmonary physiology, would you expect the following to be higher or lower in a human infant as compared to an adult? Explain the physiological reasoning behind your answer(s). a. body temperature Body temperature is lost across surface area. Since an infant has a much higher SA/V ratio than an adult, it will tend to lose heat much more rapidly than an adult. (The same is true for a small warm-blooded animal compared to a different species of larger animal.) b and c. heart rate and oxygen consumption rate/unit mass of body To maintain a constant body temperature, the infant will require a higher metabolic rate. As a result, both heart rate and oxygen consumption rate/unit mass of body will be higher in the infant than in the adult. This is why most infants weighing less than 20 pounds require about 900 dietary calories per day. If adults consumed calories at this rate, a 200-pound human would require 9,000 calories per day. Activity 42.3
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42.3 Test Your Understanding 1 to 3. At rest, an average human’s mean arterial blood pressure (MABP) is 120/80. As a person increases activity however a variety of modifications must be made to support the increased activity. For example, assume you are late for an exam and run four blocks to get there on time. Which of the following would occur during your four-block run? Explain your answers. T/F
1. As you run, your muscles will require more oxygen per unit time than when you walk. True—Active muscle requires more ATP per unit time and, as a result, more oxygen per unit time for cellular respiration.
T/F 2. The percentage of carbon dioxide in your blood will increase and your blood pH will increase. False—It is true that there will be higher levels of carbon dioxide in the blood. However, higher levels of carbon dioxide will produce higher levels of carbonic acid, which will reduce the blood pH. T/F
3. Your MABP will increase because both the heart rate and stroke volume will increase. True—To increase the rate of delivery of both food and oxygen to the muscles, both heart rate and stroke volume will increase.
Refer to the graph below of dissociation curves for two mammalian hemoglobins to answer questions 4 to 7. B A
% Saturation of Hb with O2
t2
t1
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At oxygen tension t0 both hemoglobins have released most of their oxygen. True—The percent saturation of both hemoglobins is near zero on the graph. At oxygen tension t1 hemoglobin B is carrying less oxygen than hemoglobin A. False—At t1, Hb B has a higher percent oxygen saturation of Hb than Hb A does.
T/F 6. If comparison of curves A and B illustrates the Bohr effect in a single species, then curve B represents the dissociation at the higher concentration of carbon dioxide (lower pH) False—At higher carbon dioxide levels or lower pH, you would expect the Hb to release more of its oxygen to the active tissues. Therefore, curve A is a better representation of what happens under lower pH. 7. In addition to carbon dioxide, contracting muscles cells produce a large amount of heat energy. It has been demonstrated that an increase in temperature, independent of carbon dioxide levels, encourages the blood to give up its oxygen to exercising tissues. Given this information would you expect: a or b below to be true? Explain your answer. a. Curve A (of the graph) would be the lower temperature curve and B would be the higher temperature curve. b. Curve B (of the graph) would be the lower temperature curve and A would be the higher temperature curve. Choice b would be correct—again, curve A releases more of its oxygen at the same partial pressures of oxygen compared to curve B.
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Notes to Instructors Chapter 43 The Immune System What is the focus of this activity? Many aspects of the immune system function in a series of domino effect–like responses. In other words, one action leads to another, which leads to another, and so on. This is often unclear and confusing to students. Modeling or diagramming the responses helps students to understand the individual triggers and their responses.
What is this particular activity designed to do? Activity 43.1 How does the immune system keep the body free of pathogens? This activity is designed to give students an understanding of how the immune system recognizes foreign cells or substances and acts to remove them from the body.
What misconceptions or difficulties can this activity reveal? Activity 43.1 Question 1: All students understand that bacterial and other infections are “bad”; however, fewer have consciously thought about why (for what reasons) they are “bad.” This question requires them to consider the reasons.
Answers Activity 43.1 How does the immune system keep the body free of pathogens? Draw a Rube Goldberg cartoon-type diagram or develop a dynamic (claymation-type) model to demonstrate how the components of the immune system interact to rid the body of a pathogen—for example, a bacterial cell or a viral particle. Be sure to explain the function of each “actor” in the system. Your diagram or model should include all the terms below.
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Here is an example of a Rube Goldberg–type drawing: lf
Terms bacterium or virus particle helper T cell receptor helper T cell cytotoxic T cell active cytotoxic T cell macrophage B cell memory helper T cell
memory B cell memory T cell plasma cell interleukins (or cytokines) CD4 protein MHC molecules antibody antigen
epitope thymus bone marrow hypothalamus fever clonal expansion self versus nonself
After you have completed your model or diagram, use what you have learned to answer the questions on the next page. 1. What are pathogens? Why do we need to prevent them from colonizing our bodies? If pathogens do manage to colonize, what effects can they have? Pathogen is a generic term for any disease-causing organism. Pathogens include disease-causing bacteria, viruses, fungi, and protists. The cells of our bodies contain all the components required for life. These components support the lives of our own cells and can easily support the lives of pathogens, which can compete with our own cells and potentially destroy them. Pathogens can have the following effects: • They overgrow and put pressure on specific membranes or cavities. When this occurs in the ear, for example, eardrums may rupture. When it occurs in the sinuses, the pressure builds up to cause blockage and pain.
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• They produce toxins that can destroy our cells. For example, some strains of bacteria produce exotoxins that can destroy cellular structure. • They parasitize cells and destroy their normal structure and function. Many fungal infections—for example, athlete’s foot—are parasitic. 2. What general defense mechanisms does the body use to help prevent colonization by pathogens? For example, what general defense mechanisms are involved in local inflammatory responses? The skin and mucous membranes act as physical barriers to the entry of pathogens. In addition, glandular secretions include lysozymes and keep the skin at an acidic pH (between 3 and 5), which inhibits the growth of many microbes. A localized injury (for example, a cut or sliver) releases prostaglandins and histamines, which cause increased blood flow and swelling in the area. The clotting mechanism helps prevent invasion by additional pathogens. Phagocytic white blood cells move into the region quickly and consume the pathogens to prevent their colonization. The pus that forms at the site of an infection is an accumulation of these phagocytic cells. (Refer also to pages 935–936 in Biology, 8th edition, for a more detailed description.) 3. In specific immunity, how do B cell responses differ from T cell responses? B cell responses Individual B cells respond to specific types of foreign antigens by secreting antibodies that interact with the antigen and cause cells to agglutinate or clump together. This agglutination tags or identifies the antigens for removal by phagocytic macrophages. Other antibodies can interfere with the function of antigens by binding with them and blocking their actions.
T cell responses When body cells become infected with a pathogen, a piece of foreign protein from the pathogen interacts with an MHC molecule. (Cell infection can occur by phagocytosis or by endocytosis.) The MHC-antigen complex migrates to the surface of the cell's membrane. Individual T cells bearing complementary antigen receptors on their membrane surfaces interact or bind with the antigen displayed on the MHC. Cytotoxic T cells bind to foreign proteins displayed on class I MHC molecules. Helper T cells bind to antigens displayed on class II MHC molecules. Cytotoxic T cells act directly by killing the infected cell. Helper T cells respond by dividing and producing a clone of activated helper T cells and memory helper T cells. The activated helper T cells produce cytokines (for example, interleukin-2), which help the appropriate B cells to differentiate into antibody-producing cells. They also help cytotoxic T cells to become active in killing the infected cells.
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4. If about 105 genes are available in the human genome to produce proteins, how can we produce more than 10 ⫻ 106 different kinds of Ab receptors (proteins) on B cells? Although Biology, 8th edition, doesn’t explain this in any detail, the information that the immune system can produce more types of B cells than there are genes in the genome begs this question. Each of the variable regions of the antibody (on both light and heavy chains) is made up of three parts. Each is coded by a different region on the DNA. The DNA segments for the three separate parts of each variable region undergo random recombination in the production of B cells. As a result, many fewer genes are required to code for the 10 ⫻ 106 different kinds of Ab receptors (proteins) on B cells. 5. How does HIV affect the immune system? HIV (human immunodeficiency virus) targets cells that display both the CD4 receptor and a chemokine receptor. These two receptors are found on helper T cells. Infection by HIV ultimately leads to the death of helper T cells and as a result destroys immune responses triggered by helper T cells. Infected individuals cannot fight off infections and may die.
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Notes to Instructors Chapter 44 Osmoregulation and Excretion What is the focus of this activity? The digestive system and the respiratory system function in the bulk flow of nutrients and gases into the body. The excretory system functions to maintain the body’s water balance and to remove metabolic wastes from the body—in particular, nitrogenous wastes. All three of these systems involve the organism internalizing a part of the external environment to gain control over these functions.
What is this particular activity designed to do? Activity 44.1 What is nitrogenous waste, and how is it removed from the body? This activity is designed to help students understand the structure and function of the mammalian kidney.
What misconceptions or difficulties can this activity reveal? Activity 44.1 Most students can work through how the filtrate is formed and how various substances are later removed from the filtrate. Many don’t understand what happens to these substances after they are removed from the filtrate, however. It helps to remind students that the efferent arterioles subdivide to form a capillary network around the loops of Henle. These drain into the renal vein. The substances removed from the filtrate either diffuse back into the capillaries or are actively transported into them. As students study how the concentration gradient of NaCl is set up in the medulla, some wonder why this region doesn’t become a solid block of NaCl. After all, the diagrams all show NaCl being actively pumped out of the filtrate in the medullary regions. Again, it helps to remind students that the loops of Henle are surrounded by a network of capillaries. Substances in the interstitial fluid of the kidney can diffuse into the capillary network. Blood flowing from the afferent arteriole and into the capillary network picks up salt and loses water as it moves into the medulla. As it flows from the capillary network to the renal vein, the blood reverses direction. As it encounters progressively lower extracellular salt concentrations, the blood in the capillaries loses salt and picks up water. A type of countercurrent exchange system is set up. In association with the active transport of NaCl out of the tubules, this sets up a gradient of salt concentration within the medulla.
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Answers Activity 44.1 What is nitrogenous waste, and how is it removed from the body? In the space below, draw a longitudinal section of a mammalian kidney. Next to this, draw a blowup of a nephon (including Bowman’s capsule and the loop of Henle) and its associated collecting duct. Be sure to include the afferent arteriole, glomerulus, and efferent arteriole that are associated with this nephron. You may do your drawing in chalk on a tabletop or blackboard if they are available. Refer to Figure 44.14 on page 963 of Biology, 8th edition. Note: The diameter of the afferent arteriole can be changed in response to blood pressure, (etc.) to maintain a relatively constant hydrostatic pressure in the glomeruli. This and the presence of aquaporins in the glomerular capillary results in movement of fluid from the glomerular capillaries into the Bowman’s capsule. Use your drawing and your understanding of the operation of the kidney to answer the questions. 1. Define excretion, and indicate how it differs from elimination. Excretion is the removal of nitrogenous waste from the body. Elimination is the removal of indigestible substances from the digestive tract. These substances are included in the feces. 2. The removal of nitrogenous wastes (excess nitrogen) is a special problem in most animals. a. Where does the nitrogenous waste come from? The primary source of nitrogenous waste is deamination of amino acids. This occurs when excess protein is consumed or when dead or damaged cells and other substances are removed from the body. b. What is it about the chemistry of nitrogen that makes it difficult for most animals to deal with? Amine groups are easily converted to ammonia (NH3), which is very soluble in water (NH3 H2O → NH4 OH). It is also toxic, however. As a result, large quantities of water must be readily available to organisms that excrete ammonia. For most terrestrial animals, water availability is limited. In these animals, the ammonia produced by deamination is converted to a less toxic form, either urea or uric acid. Activity 44.1
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3. Work through parts a and b, and then use the information you gather there to answer the question in part c. a. Describe the composition of the newly filtered solution that enters Bowman’s capsule. Then compare it to the composition of the blood entering and leaving the glomerulus. The solution that enters Bowman’s capsule contains water and anything soluble in water that is found in the blood. Among other things, this solution includes salts, water-soluble medicines and drugs, sugar, vitamins, and nitrogenous wastes. The concentrations of these substances in the fluid entering Bowman’s capsule are similar to their concentrations in the blood that enters the afferent arteriole. b. Starting with the solution that escapes into Bowman’s capsule from the glomerulus, describe the changes that occur in its composition as it moves through each of these regions: i. Proximal convoluted tubule In this region, NaCl and nutrients (for example, sugars) are actively transported out of the filtrate and into the cortex of the kidney. Bicarbonate ions, potassium ions, and water passively diffuse into the cortex. They are picked up by capillaries and returned to the blood. In this same region, excess H ions are actively transported into the tubule. Ammonia produced by the surrounding cells diffuses passively into the tubule in response to increased H concentrations and neutralizes the pH of the fluid in the tubule. ii. Loop of Henle In the thick segment of the ascending portion of the loop, NaCl is actively transported out of the fluid and into the medulla of the kidney. In the thin (lower) portion of the ascending loop, NaCl diffuses out of the tubule. In the descending portion of the loop, water passively diffuses out of the tubule and into the medulla of the kidney. iii. Distal convoluted tubule In this region, NaCl and bicarbonate ions are actively transported out of the tubule. Excess potassium ions and H ions are actively transported into the tubule. Water leaves by passive diffusion into the cortex of the kidney. iv. Collecting duct The cortical and outer medullary regions of the collecting duct are permeable to water but not to NaCl or urea. Excess NaCl is actively transported out of the duct into the medulla. In the lower medullary region, the duct becomes permeable to urea. If the concentration of urea in the duct is high, some passively diffuses out into the medulla. As the collecting duct passes through the high concentration of salt (and urea) in the medulla, the filtrate solution loses water passively and the filtrate becomes very concentrated. 296
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v. Urinary bladder The collecting ducts empty into the renal pelvis, which leads to the ureter. The fluid in the ureter can be stored in the bladder. Little change occurs in the fluid after it reaches the ureter. During periods of extended dehydration, some of the water may be removed from the bladder by passive diffusion into surrounding tissues. c. Now explain how the general function of the kidney enables it to “remove” (a better expression would be “let out”) from the body a wide variety of unfamiliar substances (drugs, inorganic molecules, or ions of many kinds) that the body has never encountered before. After answering this, explain why “let out” from the body is a better expression than “remove.” The membranes of the capillaries of the glomerulus contain numerous aquaporins, which makes them much more permeable to water. In association with the hydrostatic pressure in the glomerulus, this causes water and water-soluble substances in the blood to move into Bowman’s capsule. It is there that the filtrate has been “let out” of the body. This is a nonselective process. Anything that is soluble in water is filtered out at this point. (Remember that to be in the body, a substance must cross a cell membrane.) As this filtrate moves through the nephron and collecting duct, the cells surrounding it actively transport needed components from the filtrate back into the blood. What is not reclaimed simply exits the body. 4. It is useful to consider the excretory system (along with the digestive and gasexchange systems) as primarily involved in bulk exchange with the external environment. The excretory system could also be interpreted as a specialized part of the external surface of the organism, which in its own way encloses and modifies part of the environment. Describe how this is true for the human kidney. For example: a. Where in the kidney does the organism end and the environment begin? The organism ends at the glomerulus, where the fluid crosses the membrane of the capillaries. From that point on, the route taken by urine from Bowman’s capsule to the urethra (the external opening of the excretory system) is unobstructed. This pathway can also be thought of as an extension of the external environment into the organism. b. Are changes in the glomerular filtrate changes in the organism, changes in the environment, or both? The filtrate in the tubule is technically in the external environment. Changes in the fluid occur by the addition of substances from the cells of the organism that surround the proximal and distal convoluted tubule. In addition, changes occur by the removal of substances from the filtrate in these same regions and in the region of the loop of Henle and the collecting duct. Activity 44.1
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c. What do your answers in parts a and b indicate about the possible evolutionary origins of the kidney? As noted on pages 961–963 of Biology, 8th edition, excretory systems in animals appear to be variations on a tubular theme. Such tubes could have formed as invaginations of the body surface. In all these cases, the cells that make up the tubules and the cells surrounding the tubules have some control over what is placed into the tubule and what is returned to the body. Any substance that is not returned to the body by active or passive diffusion across a membrane remains in the tube and is ultimately lost to the external environment. 5. a. What is the difference between hydrostatic pressure and osmotic pressure? Hydrostatic pressure is the pressure exerted by a fluid (for example, water) on the walls of the containing vessel. Water tends to move from a region of high hydrostatic pressure to a region of lower hydrostatic pressure. In contrast, osmotic pressure is a measure of the force required to counteract the movement of water from a region of low solute concentration to one of higher solute concentration. Water tends to move from a region of low osmotic potential (or low solute concentration) to a region of higher osmotic potential (or higher solute concentration). b. Where in the human excretory system is hydrostatic pressure responsible for moving water across a membrane or layer of cells? Hydrostatic pressure is responsible for moving fluid from the glomerulus into Bowman’s capsule. c. Where in the excretory system is osmotic pressure responsible for such movement? Osmotic pressure is responsible for moving water from regions of lower solute concentration to regions of higher solute concentrations. For example, it is responsible for moving water from the collecting duct into the surrounding tissues.
44.1 Test Your Understanding 1. a. You examine the kidney structure and function of a two species of mice, one from the desert and another from a meadow or grassland. What differences would you expect to find? Use drawings of the two systems and their functions to explain your reasoning. To survive in the desert, organisms need to conserve much more water than do organisms that live in grasslands. Water is removed from the collecting duct as it passes through the concentration gradient set up in the medulla. The longer the loop of Henle and the longer the collecting duct, the greater the potential to
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remove water from the urine. As a result, you would expect the kidney of a mammal that evolved to live in the desert to have longer loops of Henle and longer collecting ducts than the kidney of a mammal that evolved in a more temperate region. b. Would you expect the excretory systems of organisms that live in the sea to resemble more closely those of animals that live in deserts or animals that live in fresh water? Explain. The fluids of fish living in the sea contain a lower concentration of salts than is found in seawater. As a result, they tend to lose water to their surrounding environment. Natural selection would tend to remove those organisms incapable of conserving water. As a result, those organisms with a kidney structure more capable of conserving water survived. As in question 1, in both marine and desert animals more water is conserved by the kidney when the loops of Henle and the collecting ducts are long. 2. Many medicines are taken orally—that is, swallowed and absorbed from the digestive system. Aspirin and many antibiotics are examples. Even though these drugs may be very different chemically, the instructions for taking them often say “Repeat dose every 4 to 6 hours.” What is the reasoning behind this dose rate? In other words, why do you need to take the medicine every 4 to 6 hours? Medicines and anything else that is soluble in the blood are filtered out into Bowman’s capsule. Unlike vitamins and sugars, medicines are not actively transported back into the blood. As a result, to maintain a given concentration of the medicine in the blood, you need to repeat the dose periodically throughout the day. Indicate whether the following are true or false. Explain your reasoning. Questions 3 to 5. In a normal, healthy human, urine: T/F 3. includes nitrogenous waste products formed following deamination of amino acids. True—Deamination of amino acids is the primary source of nitrogenous wastes (e.g., urea). T/F 4.
functions to rid mammals of excess glucose. False—The glucose that enters the nephron system is actively or passively transported into the interstitial fluid and from there moved into the peritubular capillaries.
T/F
production involves a countercurrent principle of exchange. True—However, unlike other countercurrent systems (e.g., in gills and duck feet), this involves energy expenditure. NaCl is actively transported out of the upper part of the ascending loop of Henle. This maintains a high salt concentration in the inner medulla and allows formation of a concentrated urine.
5.
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Questions 6 to 8. Injection of an individual with antidiuretic hormone (ADH) would cause: T/F 6. increased urine volume. False—ADH would make the distal tubules and collecting ducts more permeable to water. This allows water in the tubules/ducts to move back into the interstitial fluid. It would result in a more concentrated, lower volume of urine. T/F 7.
decreased filtration at the glomerulus. False—ADH has no effect on the pressure filtration that forces fluid from the glomerulus into Bowman’s capsule.
T/F
an increase in permeability of the collecting ducts. True—As noted, the collecting ducts and the distal tubules are the targets of ADH, which functions to increase their permeability to water.
8.
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Notes to Instructors Chapter 45 Hormones and Endocrine System What is the focus of this activity? This activity requires students to review their understanding of the general mechanisms of cellular communication (introduced in Chapter 11 of Biology, 8th edition) and use that understanding to explain how peptide- versus cholesterol-derived hormones act in the body. How hormones can be affected by the digestive system is also addressed as we consider methods of administration.
What is this particular activity designed to do? Activity 45.1 How do hormones regulate cell functions? This activity is designed to help students understand how hormones act within cells to produce a response.
Answers Activity 45.1 How do hormones regulate cell functions? This activity is designed to help you understand how hormones act within cells to produce a response.
Building the Model Working in groups of three or four, construct a dynamic (claymation-type) model of hormone action in cells. You may use the materials provided in class or devise your own. Step 1. Use chalk on a tabletop or blackboard to draw a eukaryotic cell. Your cell should be at least 18 inches in diameter. Be sure your drawing includes the cell membrane, the nuclear membrane, and the DNA inside the nucleus. Assume that your eukaryotic cell responds to two different hormones: • Hormone 1 is a protein-derived hormone. The cell responds to hormone 1 by increasing production of substance X. • Hormone 2 is a cholesterol-derived hormone. The cell responds to hormone 2 by decreasing production of substance Y.
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Step 2. Answer the following questions. 1. What structures or components do you need to add to your model to allow hormone 1 to react and increase production of substance X? Because hormone 1 is protein derived, you need to add a signal-transduction pathway. This includes a plasma-membrane-bound signal-receptor, its associated signal-transduction pathway, and the target molecule that causes the response. The target molecule could be an enzyme, which is substance X. The effect on the target molecule would then be activation of substance X. Alternatively, the target molecule could act to initiate transcription of the gene for substance X. (Refer to Figure 45.5 for an overview, and review Chapter 11.) 2. What structures or components do you need to add to your model to allow hormone 2 to decrease production of substance Y? Hormone 2 is a cholesterol-derived hormone, so it can freely cross the plasma and nuclear membranes. Within the nucleus, you need to add a signal receptor molecule. When this molecule interacts with hormone 2, it shuts down the transcription of substance Y. Now use playdough or cutout pieces of paper to make your hormones, cell membrane proteins, and any other proteins you need. Indicate their placement on the membrane or cell. Include a key for your model that indicates how different hormones and proteins are designated. Step 3. Using claymation, demonstrate how each of the two hormones is likely to produce its response. Use the understanding you gained from your model to answer the questions. 3. In medical applications, the type of hormone dictates the mode of administration— for example, oral versus injection, and so on. a. How would you need to administer hormone 1 to an organism deficient in this hormone? Because hormone 1 is protein derived, it is likely to be digested if taken orally. As a result, protein hormones—for example, insulin—must be administered by injection. b. How would you need to administer hormone 2? Cholesterol-derived hormones are generally not digested. If taken orally, they pass from the digestive tract into the lacteal system and move from there into the circulatory system. An example is birth control pills, which contain combinations of progestin and estradiol. 302
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4. Hormones often act in an antagonistic fashion. That is, one hormone will initiate a certain response while another inhibits that response. Illustrate this process using insulin and glucagon as examples of antagonistic hormones. Insulin and glucagon act to regulate blood sugar concentrations within a fairly narrow range. Increases in blood sugar concentration—for example, after a meal— stimulate the beta cells of the pancreas to secrete insulin. Insulin triggers the cells of the liver and body to take up more glucose and store it as glycogen. Both the blood sugar level and the release of insulin drop. Low blood sugar levels cause the alpha cells of the pancreas to respond by releasing glucagon into the blood. Glucagon triggers the liver to break down glycogen and release glucose into the blood. (See also Figure 45.12.)
45.1 Test Your Understanding 1. Assume you are trying to characterize one of the hormones that is involved in the neuroendocrine regulation of milk production. You make extracts of blood and mammary tissues of normal, lactating (milk-producing) animals and assay these extracts by injecting them into animals (each of whose hypothalamus has been surgically removed) and look for restoration of milk production. Indicate whether each of the following findings is consistent with the hormone being a steroid, a peptide, or either type. Explain your reasoning. a. It is found in blood. Either type is found in the blood. b. It is found in the cytoplasm of mammary tissue cells. Only steroid-type hormones are able to cross the cell membrane. c. It is found associated with receptors. Both types associate with receptors to produce a response. The difference lies in the location of the receptors. d. It is found associated with protein complexes that contain G protein. Only peptide or protein hormones associate with G protein complexes. e. It is found in nuclear extracts of cells. Only steroid hormones are able to enter the cell and the nucleus.
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2 to 4. Glucagon and insulin are hormones that act homeostatically to maintain glucose levels in the body. Are the following statements True or False concerning insulin and glucagon. Explain your reasoning. T/F 2. Their production and release is stimulated by trophic hormones from the anterior pituitary. False—The pancreas itself monitors levels of glucose in the blood and interstitial fluid. The beta cells are stimulated to release insulin when blood glucose levels are elevated (above about 90 mg/100 ml), and the alpha cells release glucagon when blood glucose levels fall below normal levels. T/F
3.
They are examples of antagonistic hormones. True—As noted above, they operate antagonistically to maintain blood glucose levels at about 90 mg/100 ml.
T/F
4.
They act by stimulating storage or release of glucose from cells. True—Insulin causes almost all body cells (except brain cells) to take up glucose. It also acts to slow glycogen breakdown in the liver. On the other hand, glucagon has the opposite effect on these cells and tissues.
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Notes to Instructors Chapter 46 Animal Reproduction What is the focus of this activity? This activity highlights both the similarities and differences in the reproductive systems of human males and females.
What is this particular activity designed to do? Activity 46.1 How does the production of male and female gametes differ in human males and females? This activity is designed to help students understand how gamete production is controlled in mammals and in humans in particular.
What misconceptions or difficulties can this activity reveal? Activity 46.1 Many students do not understand the mechanisms that control gamete production in human males versus females. In addition, although most students know about birth control, many do not understand how the various methods—in particular, birth control pills—work to prevent pregnancy.
Answers Activity 46.1 How does the production of male and female gametes differ in human males and females? This activity is designed to help you understand how gamete production is controlled in mammals and particularly in humans. In human males and females, the production of gametes and the hormones estrogen, progesterone, and testosterone is ultimately controlled by actions of the hypothalamus.
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Using all the terms below, diagram the control of gamete and sex hormone production first in a human male and then in a human female. Be sure to explain the role(s) of each term in your diagram. hypothalamus anterior pituitary LH FSH ovary or testes follicle or seminiferous tubule corpus luteum or Leydig cells
progesterone estrogen or testosterone secondary sex characteristics primary sex characteristics negative feedback egg and polar bodies or sperm
Human male: Please refer to the following figures in Biology, 8th edition: Figure 46.11 on page 1,005, Figure 46.12 on page 1,008, and Figure 46.13 on page 1,010. Human female: Please refer to the following figures in Biology, 8th edition: Figure 46.10 on page 1,004, Figure 46.12 on page 1,009, and Figure 46.14 on page 1,011. Use your diagrams to answer the questions. 1. In both males and females, the hypothalamus produces GnRH, which stimulates the pitutitary to release LH and FSH. Fill in the chart. Hormone
a. In males causes:
b. In females causes:
LH
LH stimulates the Leydig cells of the testes to produce testosterone and other androgens (steroid hormones) that promote spermatogenesis. (See Figure 46.13 on page 1,010.)
LH stimulates ovulation. Following ovulation, LH causes the old follicle to become the corpus luteum, which produces higher levels of progesterone.
FSH
FSH stimulates the Sertoli cells in seminiferous tubules. These cells nourish the developing sperm cells.
FSH stimulates the development of several follicles in the ovary. Each follicle contains a developing egg cell. Usually only one of the follicles survives. Estrogen production by the follicles also increases under the influence of FSH.
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2. In both males and females, the testes or ovaries produce additional hormones. Hormones
Function of these hormones produced in the gonads
a. Males produce:
b. Females produce:
Testosterone and other androgens
Estrogens and progesterone
Prior to puberty, these hormones are responsible for development of the primary sex characteristics—the testes and associated structures and glands. Following puberty, increased production of testosterone and the other androgens is responsible for development of secondary sex characteristics—for example, a deep voice, facial and pubic hair, and muscle growth. The hormones also function to stimulate sperm production.
Prior to puberty, these hormones are responsible for development of the primary sex characteristics—the ovaries and associated structures. Following puberty, increased production of estrogens and progesterone is responsible for development of secondary sex characteristics. The hormones are also involved in preparing the uterus to receive the developing embryo.
3. Most birth control methods are designed to keep the egg and sperm from uniting to form a zygote. Many birth control pills or patches used by human females contain a combination of estrogen and progesterone. How do they keep sperm from uniting with egg? Explain the mechanism. The set of 28 pills usually includes 21 pills that contain a combination of estradiol (a form of estrogen) and progesterone. The last 7 pills often contain iron supplements but no hormones. During the first 21 days, the elevated levels of progesterone inhibit the production of GnrRH, which in turn inhibits the production of FSH. No follicles are stimulated to mature, so no developing egg is available for fertilization. The estrogen in the pills causes the uterine lining to respond by increasing glandularization and vascularization. When the hormone level drops (in the last 7 days), the lining of the uterus is sloughed off (as in a normal menstrual cycle). 4. Efforts to make a male contraceptive pill (analogous to the pills used by females) have not been very successful. Given what you know about the similarities and differences in male and female gamete production, propose why this might be the case. In human males, the levels of testosterone (and other androgens) directly affect secondary sexual characteristics, sex drive, and spermatogenesis. Unlike in females, Activity 46.1
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existing feedback loops between testosterone (and other androgens), GnRH, FSH, and LH tend to keep all of these hormones at relatively constant levels. Although increasing the level of testosterone may reduce FSH levels, it may not stop spermatogenesis. In addition, varying the levels of testosterone could produce undesirable behavioral and physiological effects. 5. Fertilization generally occurs in the upper third of the oviduct, and development of the fetus occurs in the uterus. In some relatively rare cases, however, developing embryos have attached to the outside of the uterus and developed there for the full nine months of pregnancy. a. Given the anatomy of the female reproductive system, can you explain how this could happen? In the human female, the fallopian tubes or oviducts are not directly attached to the ovary. Eggs released from the ovary are drawn into the oviduct by the action of cilia on the inner epithelial lining of the oviduct. In some instances, eggs can escape and migrate into the body cavity. If the egg becomes fertilized and then escapes, it could implant on the outside of the uterus. Once implanted the developing fetus is capable of forming extraembryonic membranes and interacting with the uterine wall to form the placenta. While not common, development to full term may occur. b. What modifications of normal birthing procedures (if any) would have to be made in such cases? Because the baby is not inside the uterus, it cannot be delivered normally (through the vaginal opening). Instead, the baby would have to be removed surgically through an opening in the mother’s abdominal wall (cesarean section).
46.1 Test Your Understanding 1 to 3. In an experiment, an adult rat’s testes, including the vascular connections, were transplanted to the wall of the abdomen. Connections of the testes to the reproductive tract were cut/severed. Following recovery, which of the following would be true for this rat? Explain your answers. T/F 1. The rat would have lowered sexual activity due to loss of testosterone. False—With the circulatory/vascular connections intact, the testes would receive appropriate hormonal signals (LH from the anterior pituitary) for testosterone production. 308
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T/F 2. The rat would have normal sexual activity but be unable to produce any ejaculate. False—Sexual activity would be normal and an ejaculate would be produced. However, this would contain only the fluids produced by the accessory glands (i.e., the seminal vesicles, prostate gland and bulbourethral gland). T/F 3. The rat would have normal sexual activity but have no sperm in the ejaculate. True—As noted above, an ejaculate would be formed, but no sperm would be present because the connection to the testes was severed. 4. A girl begins to develop breasts and pubic hair at the age of four. Given these symptoms, her physician orders a CT scan (imaging procedure) to look for an endocrine tumor. Which organ would he most likely not investigate as the cause? Explain your reasoning. a. hypothalamus b. pituitary c. ovary d. uterus Hyperactivity of the hypothalamus, pituitary, or ovary, in the form of a tumor, could produce early onset of sexual maturation or puberty by overproducing hormones involved in puberty. However, because the uterus produces no hormones, a tumor in this region would not result in early puberty. 5 to 8. Assume that women can be vaccinated against the following hormones. Each vaccine is designed to completely neutralize the target hormone’s activity. Which vaccine(s) would prevent pregnancy? Explain your answers. A ⫽ Would prevent pregnancy; B ⫽ would not prevent pregnancy 5. A vaccine against LH A—LH (luteinizing hormone) is required for gamete production. 6. A vaccine against CG A—CG (chorionic gonadotropin) is required to maintain the corpus luteum until the placenta has matured sufficiently to take over its function. 7. A vaccine against estrogen A—Estrogen is required to prepare the uterine wall to support the embryo. 8. A vaccine against prolactin B—Prolactin, while it is required for mammary gland growth and milk production, is not required for successful pregnancy and birth. Activity 46.1
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Notes to Instructors Chapter 47 Animal Development What is the focus of this activity? Chapter 21 provided a review of how genes act to control development. Chapter 47 reviews some of the major morphological changes that occur in the development of many animals. Similarities in early development are often used as evidence for evolutionary relationships among organisms.
What is this particular activity designed to do? Activity 47.1 What common events occur in the early development of animals? This activity provides a brief overview of major events that occur in the early development of vertebrates and many other animals.
Answers Activity 47.1 What common events occur in the early development of animals? The early stages in the development of all vertebrates (and many other animals) include zygote formation, cleavage, blastula formation, gastrula formation, and organogenesis (for example, neurulation). Among the major aspects of development are cell adhesion/ recognition, cell growth (in number and/or size), cell induction, and cell determination. 1. What key events occur at each stage of development? Developmental stage
What occurs during this stage?
What is the influence or effect on the subsequent development of the embryo?
a. Cleavage
In vertebrates, cell division is radial and indeterminate. The first mitotic cell divisions that follow fertilization occur rapidly and are not accompanied by cell growth. The
In many animals (mammals are an exception), the zygote has a definite polarity. As a result, the cleavage divisions create cells that differ in their cytoplasmic contents (e.g., cytoplasmic determinants: mRNAs and proteins present in the egg cell). (Continued on next page)
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chromosomes duplicate and are partitioned to daughter cells, but the cytoplasm is simply partitioned among the daughter cells. As a result, these divisions are called cleavage.
These differences can affect the subsequent development of the daughter cells.
b. Blastula formation
As cleavage continues, the blastula forms. The blastula is a hollow ball of cells surrounding a fluid-filled cavity, or blastocoel.
Cells on the outside of the blastula are in direct contact with the external environment. Those on the inside are in contact with the fluid-filled blastocoel. Those that lie between are only in contact with other cells. As a result, the cells experience differences in chemical and physical factors that can affect how each develops.
c. Gastrula formation
Blastula cells (near what will become the blastopore) begin to divide rapidly. These cells undergo changes in their shape, motility, and adhesion properties. Overall, there is a general migration of cells into the blastocoel to form the archenteron (primitive gut). Ultimately, gastrulation results in the formation of the germ layers: ectoderm, endoderm, and mesoderm.
These cellular movements and rearrangements produce new gradients in chemical signals and new types of cell-cell interactions between ectoderm and mesoderm and between mesoderm and endoderm. Extracellular glycoproteins, such as fibronectin, have been shown to play a role in directing the cell movements of gastrulation. By the late gastrula stage, the fates of many of the cells have been determined. In general, ectodermal cells develop into the nervous system and the epidermis. The endodermal cells become the digestive tract and associated organs (liver and pancreas).Mesodermal cells give rise to the muscles, the circulatory system, the kidneys, and the dermis. In deuterostomes (echinoderms and chordates), the anus forms near the old blastopore and the mouth forms from a new opening (opposite the site of the blastopore). (Continued on next page)
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Developmental stage
What occurs during this stage?
What is the influence or effect on the subsequent development of the embryo?
d. Organogenesis (for example, neural tube formation)
Organogenesis is basically the formation of the organs from the germ layers set up in the gastrula. The notochord and neural tube (nervous system) are among the first organ systems to develop in vertebrates.
The development of one structure is often triggered by or dependent on signals (chemical or physical) produced by other structures. For example, in early development, the cells of the dorsal lip of the blastopore appear to initiate the development of the neural tube and a number of other organs.
2. Many animal species share these similarities in early development. Yet, the stage at which the individual cells of the embryo lose their totipotency can vary considerably among these species. a. What does it mean to say that a cell is totipotent? A cell that is totipotent is undifferentiated and has the potential to differentiate into any of a number of different types of cell. b. What factors can affect the point at which a cell loses its totipotency—that is, when its fate becomes determined? Among the factors that affect when a cell loses it totipotency are: • the distribution of cytoplasmic determinants in the cells, • the presence of “organizing centers or regions” that appear to trigger the development of surrounding cells, • an unequal distribution of chemical signals, and • specific cell-cell interactions encountered by the cell. Each of these factors appears to have some effect on which genes are activated in the cells during development and, as a result, how the cells differentiate.
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47.1 Test Your Understanding 1 and 2.
Choose the graph that best fits each situation. A
B
C
D
E
time
time
time
time
time
F
time
1. Which graph best describes the change in size of individual cells of a vertebrate embryo from the time of zygote formation to the end of cleavage? Explain your answer. D—During cleavage cells divide rapidly, but there is little or no intervening cell growth between mitoses. As a result, the cells become smaller and smaller with each division. 2. Which graph best describes the change in the number of cells in the embryo from fertilization to gastrulation? Explain your answer. B—The overall number of cells in the embryo increases continuously from fertilization to gastrulation. 3 to 6. While the mechanics of gastrulation may differ among various organisms, the overall objectives and problems are the same. Which of the following are accomplished by the end of gastrulation? T/F 3. formation of the three germ layers—ectoderm, mesoderm, and endoderm True T/F 4. establishment of the embryonic axis (anterior to posterior) True T/F 5. determination of the fates of the individual cells of the gastrula True T/F 6. formation of the neural tube False—This occurs as part of organogenesis
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7. In a now classic experiment, Spemann and Mangold took cells from the dorsal lip of the blastopore from one frog embryo (the donor) and transplanted them to an area opposite the dorsal lip in another frog embryo (the host). As a result, the host embryo developed two heads, one at the site of the “new” or transplanted dorsal lip and the other at the site of the host’s original dorsal lip. A thorough examination indicated that the second head was formed from host cells. Which of the following developmental cues or mechanisms was most likely the trigger (or cause) for the generation of the second head. Explain your answer. a. cytoplasmic determinants that are unequally distributed in the host embryo b. gradients set up in the egg that gave rise to the host embryo c. hox genes (or homeobox genes) present in the embryo receiving the transplant d. cell-cell signalling between the transplanted dorsal lip and the host embryo e. all of the above d—If the donor cells had not been transplanted to the host, the host would have developed normally. As a result, the altered development must have been triggered by some chemical or cell-cell signal from the transplanted or donor tissue.
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Notes to Instructors Chapter 48 Nervous Systems What is the focus of these activities? In vertebrates, the endocrine and nervous systems maintain primary control over homeostasis. In the endocrine system, messages (hormone signals) are transmitted through the circulatory system to all cells of the body. Only the cells that have appropriate receptors are able to respond, however. In the nervous system, messages (action potentials) are sent via neurons that have direct connections to the responding cells. As a result, response rates can be much faster. In fact, the speed of transmission along myelinated neurons can exceed 150 m/sec (540 km/hr, or about 335 miles/hr). These activities focus on how information is transmitted along neurons.
What are the particular activities designed to do? Activity 48.1 How do ion concentrations affect nerve function? This activity is designed to give students an understanding of both the experimental system and the thought processes used by early investigators studying how neurons function. Students also get practice using the Nernst equation in this activity. Activity 48.2 How do neurons function to transmit information? This activity is designed to help students understand how an action potential can be generated and transmitted along the neuron. They will also learn how an action potential generated in a presynaptic neuron can initiate an action potential in a postsynaptic neuron. Activity 48.3 What would happen if you modified a particular aspect of neuron function? The questions in this activity are designed to help students test their understanding of nervous system function by asking them to propose what would happen if some part of it were damaged. What would the system still be able to do? What would it be unable to do?
What misconceptions or difficulties can these activities reveal? Activity 48.1 Most students have no idea why some axons in squid should be much larger than others. In addition, they often have the mistaken idea that the squid “giant axon” is from a giant squid, as opposed to its being an axon of very large diameter. Having them work through some of the basic information available to early researchers in the field should help them overcome these misconceptions.
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Activity 48.2 Activity 48.3 Many students have difficulty understanding that an action potential is recorded at a single point along an axon and results from highly localized changes in ion concentrations. These changes require the movement of only a very few ions through gated channels in the membrane. Students may not understand that the Na ions that enter at a given point on the axon can diffuse within the axon. As a result, the ions can bring the next point on the axon to threshold and generate an action potential at that point. You may need to clarify for students that this type of chain reaction continuously regenerates the action potential (or conducts the action potential) along the axon. Problems in understanding often come up as students are doing the modeling exercise suggested in Activity 48.2. Many of the questions that follow the modeling exercise can be used to explain these ideas.
Answers Activity 48.1 How do ion concentrations affect neuron function? Much of our understanding of neuron function was based on studies of the squid giant axon. Squid move through the water by contracting the muscles of the mantle. This compresses water inside the mantle that is forced out the siphon. Squid can change the direction of movement by directing the flow from the siphon either forward or backward (relative to the head or anterior end). 1. In which direction would the squid move if water flow from the siphon was directed toward the head end of the organism? The squid would move backward (or toward its tail end). Conversely, if the siphon were directed backward so that it pointed toward the tail end of the organism, the squid would move in the forward (toward the head end) direction. 2. When squid are startled or in danger, they can simultaneously contract all muscles of the mantle to jet water forcefully out of the siphon and escape rapidly. Assume the mantle of the squid is 30 cm in length (about 12 inches). The brain sends a signal to major nerve ganglia in the mantle, which relay the signals to axons innervating the mantle muscles. For all muscles of the mantle to contract simultaneously, all nerve signals sent along these axons must reach all parts of the mantle at the same time.
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ganglion
siphon
mantle
a. On the diagram of the squid above, draw and number three neurons. Assume all are simultaneously stimulated by a single motor neuron from the brain. Neuron 1 innervates the mantle muscles nearest the brain. Neuron 2 innervates the muscles in the midregion of the mantle. Neuron 3 innervates the muscles at the tail end of the mantle. This diagram would show only the location of the axons relative to the ganglion. b. In invertebrates, like the squid, how must the nervous system be structured to allow both the muscles nearest the brain and those farthest from it to contract simultaneously? Modify your drawing to indicate any differences in the size or structure of the three neurons that would be required. Explain your reasoning. Invertebrates do not contain myelinated nerves. As a result, speed of nerve transmission is modulated by changes in the size of the different axons/neurons innervating the muscles, in this case, those innervating the muscles of the mantle. In general, the drawing should indicate that neuron 1 is shortest and narrowest in diameter. Neuron 2 is of intermediate length and intermediate diameter. Neuron 3 would be longest and have the greatest diameter (approximately 1 mm in the squid). 3. Researchers discovered that they could remove these “giant axons” of the squid. With some skill, the axons could be maintained outside the body if held in ion solutions of the same concentrations as the squid’s extracellular fluids. If stimulated with a microelectrode, these isolated axons would generate action potentials. By recording the potential difference between an electrode in the axon versus one in the fluid bathing the axon, the scientists could also record the potential difference at rest and any change in potential difference that occurred as a nerve impulse, or action potential, was generated.
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The ion concentrations the researchers recorded for the squid giant axon are listed in the table below. Ion concentrations (mM) for intra- and extracellular fluids of the squid axon Intracellular (mM) Extracellular (mM) Potassium (K)
400
20
Sodium (Na )
50
440
Chloride (Cl)
40 – 150
560
Calcium (Ca)
1 104
10
Some of these ion concentrations for the human are listed in the table below. Ion concentrations (mM) for intra and extracellular fluids of the human axon Intracellular (mM) Extracellular (mM) Potassium (K)
140
5
Sodium (Na)
15
150
Using the Nernst equation, your text book calculates that in humans the equilibrium potential for potassium is 90 mV and that for sodium is 62 mV. What are the equilibrium potentials for potassium and sodium for the squid The Nernst Equation Eion
z ion valance (e.g., Na 1, Cl 1)
62mV [ion]out log z [ion]in
EK 62/1 log(20/400) 80 mV ENa 62/1 log(440/50) 58 mV Values in this range are typical for most neurons. The difference in actual concentrations of the ions (human system vs. squid) can be accounted for because the squid is a marine organism and the sodium and potassium concentrations in the ocean are approximately 469 mM and 10 mM, respectively. 4. In a typical neuron, which ion has the greatest influence on the membrane potential at rest? In other words, flux of which ion contributes the most to the resting membrane potential? Explain why this occurs. At rest almost all sodium voltage-gated channels are closed. However, while most of the potassium voltage-gated channels are closed, some potassium channels are open. As a result, potassium can move down its chemical concentration gradient out of the axon. However, the resting potential is seldom equal to the equilibrium potential for
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potassium because the outward flow of potassium is counterbalanced by the electric potential difference set up that is the result not only of the potassium gradients but also of the anion and other cation gradients. In addition, some very small amount of sodium does leak across (out to in) down its concentration gradient. 5. In calculating the resting and action potentials of the axon, we generally don’t worry about the concentration differences of calcium or chloride ions. Explain. The concentration differences are set up by active transport of the ions across the neuron membranes. The same is true for all such concentration differences. In general, we are not concerned with the concentration differences of calcium and chloride ions because the neuron membrane is essentially impermeable to these under normal conditions. 6. What effects would the following changes in extracellular K concentration be likely to have on the resting membrane potential of neurons in a human? a. Change extracellular K from 5 mM to 2 mM EK = 62/1 log(2/140) = 113 mV This would hyperpolarize the cells, making it more difficult to generate an action potential. b. Change extracellular K from 5 mM to 10 mM EK = 62/1 log(20/140) = 52 mV The normal potassium equilibrium potential is 90 mV. Changing the external potassium concentration to 20 mM would most likely change the equilibrium potential enough to bring it above threshold. As a result, all neurons in the body would fire and death would be the result. (Note: Death by lethal injection often involves sedatives and muscle relaxers in combination with intravenous delivery of a high concentration of potassium.)
Activity 48.2 How do neurons function to transmit information? Working in groups of three or four, construct a dynamic (claymation-type) model of the transmission of an action potential along a neuron and then across a synapse to generate an action potential in a postsynaptic neuron. When developing and explaining your model, be sure to include definitions or descriptions of the following terms and structures.
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Neurons or Parts of Neurons dendrite axon cell body synaptic vesicles presynaptic neuron postsynaptic neuron
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Ions
Gates
K Na negative organic ions Ca
voltage-gated ion channels Na gates or channels K gates or channels Ca gates or channels
Building the Model • Using chalk on a tabletop or a marker on a large sheet of paper, draw the membranes of two neurons and the synaptic region between them. Each neuron should each be at least 4 inches across and a foot or more in length. • Identify the axon, cell body, and dendrite(s) on your drawing. • Make the ions and gates from playdough or cutout pieces of paper. Indicate the placement of gates in the membranes and ions inside the membrane versus outside the membranes. • Include a key for your model that indicates how ions and gates are differentiated from each other. You may use color coding for the ions and gates. • Start the model by “initiating” an action potential in the axon hillock. • Indicate how this action potential is propagated along the axon and how it can lead to production of an action potential in the postsynaptic neuron. • When you have completed your model, explain it to another student or to your instructor. Use your understanding of how action potentials are generated and propagated to answer the questions. 1. All cells maintain an ionic (and therefore electrical) potential difference across their membranes. In most cells, this potential difference is between 50 and 100 mV. That is, the inside of the cells is more negative than the outside by 50 to 100 mV. Although all cells in the body maintain this potential difference across their membranes, only certain cells (for example, neurons) are capable of generating action potentials. a. How is this potential difference across the cell membrane generated? Sodium-potassium pumps in cell membranes actively pump Na ions out of cells and K ions into cells. In general, for every three Na ions pumped out of a cell, two K ions are pumped into the cell. As a result, the inside of the cell contains fewer positively charged ions than the outside; that is, the inside of the cell is 320
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more negative than the outside of the cell. This difference in ion distribution can be measured as a voltage or potential difference across the cell membrane. For the squid axon, the potential difference is 70 millivolts (mV). b. What characteristics of membranes allow cells to concentrate or exclude ions? Cell membranes are selectively permeable. Ions such as Na and K cannot diffuse across the phospholipid bilayer. Ions must either be actively transported or diffuse through specific ion channels in the membranes. The ratio of Na to K ion channels can also affect the relative concentrations of the ions in cells. c. What is it about neurons (nerve cells) that make their properties different from those of other cells? In other words, what enables nerve cells to produce action potentials? All cells generate a potential difference across their membranes. However, only nerve cells have voltage-gated ion channels that open and close in response to changes in the potential difference across the membrane. Postsynaptic neurons also contain chemically gated ion channels that open and close in response to neurotransmitters released from presynaptic neuron terminals. d. How is an action potential started and propagated? Refer to Figure 48.10 of Biology, 8th edition, and keep in mind that all of the following responses are recorded at a single site on the axon. • Local depolarizations in the neuron membrane open some of the Na gates and Na ions diffuse into the axon. When the potential difference at that site on the membrane reaches the threshold potential (55 mV for the squid axon), the Na activation gates at the site all open and Na ions diffuse in (down the concentration gradient). During this time, the K gates remain closed. • When the potential difference across the membrane at this site on the axon reaches 30 mV, the Na inactivation gates close and the K channels open. • K ions move out of this region of the axon down the concentration gradient, and this region or site is repolarized. • When the potential difference reaches 70 mV (resting potential), the K gates are stimulated to close. The K gates are slow in closing, however, which produces the characteristic undershoot. During this time, both the Na activation and inactivation gates are closed. This is the refractory period. When the Na inactivation gates are closed, the neuron is insensitive to depolarization. This limits the number of action potentials that can be generated per unit time. The refractory period also helps explain why action potentials normally travel in only one direction along the axon.
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e. Is any direct or indirect energy input required to generate an action potential? If so, when and where is the energy used? Other than whatever energy is required to generate a depolarizing stimulus, no direct energy input is required to generate an action potential. Energy is required indirectly, however, to fuel the action of the sodium-potassium pumps, which generate the potential difference across the membrane. f. What happens in time and space (along the axon) once an action potential begins? In part d, we looked at what happened at a single site on the axon as it was depolarized to generate an action potential. At the height of the action potential (30 mV), the inside of the axon is more positive than the outside, at least at this one point. The Na ions that enter the axon at this site diffuse in all directions from their point of entry. As they diffuse, they change the potential difference across the membrane. The next site on the axon reaches threshold potential and its Na activation gates open. An action potential is generated. The Na ions that enter at this site diffuse in all directions and depolarize the next part of the axon, generating an action potential. This repeats along the length of the axon. (Note: Although the Na ions diffuse in all directions along the inside of the axon, the action potential is generated in only one direction. The refractory period prevents the generation of an action potential in the opposite direction.) g. What factors ultimately limit the ability of the nervous system to respond (that is, to continue to generate impulses)? Ultimately, the operation of the sodium-potassium pump limits the ability of the system to respond. The sodium-potassium pump establishes the potential difference (70 mV, for example) across the neuron membrane. In experimental situations, isolated neurons are used to generate and record action potentials. When the sodium-potassium pumps in the neuron membrane are poisoned (or are nonfunctional), the axon is still able to generate action potentials (in some cases up to several hundred action potentials). This occurs because the number of Na and K ions that move across the membrane during an action potential is very small (on the order of eight or ten ions per gate per action potential). On the other hand, the Na and K ion concentrations inside and outside the neuron are measured in millimoles. (See Figure 48.10.) If a molar solution of K ions contains 6.02 1023 K ions per liter, then a millimolar solution contains 6.02 1020 K ions per liter. As a result, many action potentials can be generated before the potential difference across the membrane is reduced enough to prevent their generation. 2. If an axon is stimulated in the middle of its length, nervous signals (action potentials) will move out from the point of stimulus in both directions. Normally, however, nerve signals move in only one direction along neurons. Explain. 322
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Normally, the refractory period prevents the action potential from moving in more than one direction. When an axon is artificially stimulated at any point along its length, however, no refractory period exists, so action potentials are generated in both directions. 3. Whether or not an action potential is generated in a postsynaptic neuron depends on a number of factors. What are they? What ultimately determines whether or not an action potential is generated in the postsynaptic neuron? Whether or not an action potential is generated in a postsynaptic neuron depends on • the type of neurotransmitter—that is, whether it is excitatory or inhibitory, • the number and/or frequency of excitatory versus inhibitory inputs to the postsynaptic neuron per unit time, and • how close the synapses are to the axon hillock. If the combination of all these inputs depolarizes the postsynaptic neuron to threshold level, an action potential is generated. 4. Diffusion is efficient over only very short distances. In fact, as you can see in this table, diffusion is efficient only for distances of about 1 to 100 m. Diffusion Distance (m) 1 10 100 1,000 (1 mm)
Time Required for Diffusion 0.5 msec 50 msec 5 sec 8.3 min
a. How wide is a synapse? The average synapse is between 10 and 20 nm wide. b. If a synapse were two times as wide, what effect would it have on the transmission of nerve signals from one neuron to the next? How would this change affect the response time of an organism? If the synapse were two times as wide, the time for diffusion would increase by the square, or four times. It would take longer for the organism to respond. (Note: Although the response time would be somewhat longer, it would not necessarily be four times longer. Response time is also affected by the time required for a given neurotransmitter vesicle to release its contents into the synaptic cleft as well as by the need for the neuron to integrate or sum all of the inputs it receives.)
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5. If you examine neuron transmission within an organism, you discover that every action potential generated is stereotyped; for example, every action potential reaches the same maximum height and the same minimum height. In addition, the generation of action potentials is an all-or-none phenomenon. That is, once the potential difference across the membrane reaches threshold, an action potential will be generated. Given this, how does the nervous system signal differences in intensity of signal? Intensity of signal is generally determined by the number of action potentials fired per unit time along a given neuron, the total number of neurons firing per unit time, or both.
Activity 48.3 What would happen if you modified a particular aspect of neuron function? In the following questions, test your understanding of the various parts of the nervous system by asking yourself what would happen if a certain part was damaged. What would the system still be able to do? What would it be unable to do? 1. Some nerve gases and insect poisons work by destroying acetylcholine esterase. Acetylcholine esterase is normally present in acetylcholine synapses and acts to degrade acetylcholine. What is likely to happen to nervous transmission in insects exposed to this type of insect poison? In the central nervous system (of vertebrates), acetylcholine can cause inhibition or excitation of the postsynaptic neuron. Which it does depends on the type of postsynaptic receptor present. In neuromuscular junctions (voluntary muscle), acetylcholine is excitatory. In heart muscle, the effect of acetylcholine is to reduce the rate and strength of heart muscle contractions. If acetylcholine esterase is destroyed, any acetylcholine released into a synapse will remain there. As a result, the postsynaptic neuron will be continuously stimulated by acetylcholine to either depolarize (if its receptor is excitatory) or hyperpolarize (if its receptor is inhibitory). For example, after an action potential is initiated in a muscle fiber, the muscle will contract continuously. The heart rate will slow. Inability to control the contraction and relaxation of muscles (such as the diaphragm) will quickly result in death. 2. The pufferfish (fugu) contains the poison tetrodotoxin. Some shellfish produce a paralytic poison called saxotoxin. Both of these poisons block the Na channels in neurons. What specific effects could these toxins have on neuron function? If the Na channels were blocked, axons would be unable to generate or propagate action potentials.
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3. A type of spider (the funnel-web spider) produces a toxin that blocks the Ca channels. a. Can a neuron exposed to this toxin fire an action potential? Explain. Neurons exposed to this toxin should still be able to generate action potentials. b. Can a neuron transmit a signal across the synapse using neurotransmitters? Explain. In chemical synapses, an influx of Ca2 is required to cause the synaptic vesicles to fuse with the membrane of the presynaptic neuron. If these Ca2 channels are blocked, no neurotransmitters are released into the synapse. Therefore, the signal cannot be transmitted across the synapse. 4. You isolate a section of a squid giant axon and arrange an experiment so that you can change the solution bathing the axon. You insert an electrode into the axon and place another electrode outside the cell so that you can measure the potential across the cell membrane. With the axon bathed in normal extracellular fluid, you observe a resting potential of 70mV and action potentials, when stimulated, that reach 55 mV. mM concentration of each ion Normal concentrations Ion Na
K
Experimental concentration in (a)
Inside neuron
Outside neuron
Inside neuron
Outside neuron
50
440
50
440
400
20
400
40
a. You change the solution bathing the neuron by increasing the K concentration to 40 mM. What effect will this have on the neuron? For example, will it depolarize the membrane and make it easier to start an action potential? Will it hyperpolarize the membrane and make it more resistant to starting an action potential? Or will it have no effect? Explain your answer. To answer this question, keep these facts in mind: • The axon’s permeability to K ions is much greater than its permeability to Cl ions, which is much greater than to Na ions. • At rest, the concentrations of negative ions inside and outside the axon are approximately equal. • The equilibrium potential for each ion is dependent both on its concentration gradient and on the electrical gradient. K ions cannot be added without simultaneously adding a counterbalancing anion—for example, Cl. As a result, the addition of KCl adds both K and Cl ions to the outside of the membrane. This addition disrupts both the K and Cl Activity 48.3
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equilibrium potentials. The axon is much more permeable to K ions (than to Cl or Na ions). As a result, K ions move more quickly into the axon (moving down the electrical gradient) to reestablish its equilibrium potential. Therefore, the result will be to depolarize the axon and make it more likely to generate an action potential. b. What would happen if, instead of adding more K to the outside, you added more Na to the fluid bathing the neuron? Explain. As in part a, Na ions cannot be added without adding a counterbalancing anion—for example, Cl. The addition disrupts the Na and Cl equilibrium potentials. As before, the axon is more permeable to Cl (than to Na), so Cl ions would move into and hyperpolarize the axon. In both parts a and b, the sodium-potassium pump will ultimately restore the Na and K gradients.
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Notes to Instructors Chapter 49 Nervous Systems What is the focus of these activities? Maintaining homeostasis is central to maintaining life. Control of homeostasis is the function of both the nervous system and the endocrine system. In Chapter 45 (Hormones and the Endocrine System) students learned that much of endocrine control is based on feedback loops as well as antagonistic action of hormones. In this activity, students examine how the nervous system is set up to control not only the gross movements of organisms (somatic nervous system) but also the finer homeostatic responses of smooth muscle, cardiac muscle, and glands that support homeostasis (autonomic nervous system).
What is this activity designed to do? This activity allows students to sort out and organize the functions of the peripheral nervous system. In doing this it also helps them to understand how interactions between the sympathetic and parasympathetic divisions of the autonomic system maintain homeostasis.
Activity 49.1 How is our nervous system organized? Neurons are single cells composed of processes called dendrites, which carry information to the neuron’s cell body, and an axon, which carries information away from the cell body. In simple terms, nerves are bundles of dendrites, or axons or mixed axons and dendrites. Ganglia are collections of nerve cell bodies and brains are collections of ganglia. Processing and integration of the information occurs as a result of interaction within and between ganglia. 1. Responding to stimuli requires at a minimum: a. b. c. d.
reception of a stimulus which triggers an action potential in a sensory neuron which synapses with a motor neuron which causes a response in a muscle or gland
To coordinate activity an interneuron is often found between the sensory and motor neurons. The interneuron can function to send information to the central nervous system and as a result lead to a response by the whole organism.
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In the space below, draw this minimum system. Be sure to name each neuron and label each of its parts. Sensory neuron cell body
dendrite
synapse axon CNS spinal cord
Stimulus
axon
dendrite
cell body
muscle motor neuron
2. The blink response is one type of simple reflex response that occurs, for example, in response to water splashing up out of the sink or to a snowball aimed at your face. When these events happen, you blink and only afterwards do you “realize” why you blinked. a. What would you have to add to your diagram above to allow you to “realize” why you blinked? You would need to add an interneuron or another synapse to the sensory axon that led to an interneuron that would carry the information to the brain to allow you to “realize” what happened. b. Why does it take you longer to “realize” why you blinked than it takes for the blink reaction to occur? The simple reflex (sensory to motor neuron) cuts out the “middle man,” the interneuron, which allows reception and interpretation functions of the brain to occur. This means that the protective response (the blink) can occur very quickly. Most such reflexes are designed to protect or ensure the livelihood of the organism. 3. What composes the central nervous system (CNS) and how does the CNS differ in general form and function from the peripheral nervous system (PNS)? The CNS is the integration and processing center. As such, it contains large collections of cell bodies and interneurons. The PNS contains the sensory neurons, which carry information from the sensors to the CNS and motor neurons, which carry information from the CNS to the effectors.
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4. The peripheral nervous system is subdivided into the somatic, autonomic and enteric nervous systems. a. What is the general function of each? PNS subdivision:
General function:
Somatic nervous system
Carries signals to and from skeletal muscles, primarily in response to external stimuli.
Autonomic nervous system
Functions to regulate the internal environment by controlling function of smooth and cardiac muscle and glands.
Enteric nervous system
Controls secretion and function of the digestive tract (including peristalsis), the pancreas, and the gall bladder. While it can function on its own to do this, it is normally regulated by interaction between the sympathetic and parasympathetic divisions of the autonomic system.
b. The autonomic system is again further subdivided into the sympathetic and parasympathetic systems. How do the general functions of each of these divisions interact to maintain homeostasis? Autonomic nervous system divisions:
Functions: See Figure 49.8 also.
Sympathetic
Responses by the sympathetic nervous system are often called “fight or flight” responses. These tend to up-regulate or increase conversion of glycogen to glucose for energy generation; to cause secretion of epinephrine; and to increase heart rate, blood pressure, and flow of blood to skeletal muscles and brain.
Parasympathetic
The parasympathetic system acts antagonistically to the sympathetic system for the most part. When stimuli to the sympathetic system (e.g., stress, danger) are removed, the parasympathetic system promotes “calming” activities (e.g., reduced heart rate and blood pressure and increased digestion and glycogen production). One exception to this is in control of reproductive organs, where the sympathetic and parasympathetic systems complement each other (or work in concert).
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49.1 Test Your Understanding Fibromyalgia affects about 2% of the U.S. population. Its primary symptoms are chronic pain, difficulty sleeping and fatigue that is not relieved by sleep. Other reported symptoms include: jaw clenching at night, dryness of mouth and eyes, an increased need to urinate, irritable bowels, muscle numbness and tingling, and headaches. 1. Several researchers have proposed that fibromyalgia results from failed regulation of part of the autonomic nervous system. If this is true, failed regulation of which division of the autonomic nervous system could account for these symptoms? Explain your reasoning. The sympathetic system is responsible for arousal and energy generation while the paraympathetic system acts antagonistically to the sympathetic system. As a result, students may think that overactivity of the parasympathetic system could produce the symptoms of fibromyalgia. Symptoms not accounted for by action of the parasympathetic system would be dry mouth, muscle problems, and headaches. 2. To test their ideas, the researchers recorded heart rate of both controls and individuals with fibromyalgia for 24-hour periods of normal activity. They observed that patients with fibromyalgia showed greatly elevated heart rates both while active and asleep. Does the addition of this information change or support the hypothesis you developed in 1 above? Given this information, the primary symptoms appear to be the result of hyperactivity of the sympathetic system. Hyperactivity of the sympathetic system could elevate the heart rate and blood pressure and lead to difficulty sleeping and chronic fatigue. Prolonged exposure to high levels of adrenaline associated with hyperactivity of the sympathetic system also leads to a reduced stress response. The researchers propose that this may be the cause of the irritable bowel and bladder as well as limb numbness and dryness of eyes and mouth.
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Notes to Instructors Chapter 50 Sensory and Motor Mechanisms What is the focus of these activities? Skeletal muscles are made up of bundles of muscle fibers (muscle cells). Each muscle fiber contains numerous myofibrils arranged parallel to one another. As a result, the degree of contraction of a given skeletal muscle is determined by which of its muscle fibers contract at any given time. Still, the maximum contraction that can be achieved by a skeletal muscle is almost always about 70% of its uncontracted or relaxed length. This is determined by the structure of the sarcomere. To understand muscle function, students need to understand both the overall structure and function of muscles and their subcellular structure.
What are the particular activities designed to do? Activity 50.1 How does sarcomere structure affect muscle function? The activity is designed to help students review and understand the basic structure and function of the sarcomere and how these constrain muscle function. Activity 50.2 What would happen if you modified particular aspects of muscle function? These questions are designed to have students test their understanding of muscle function by asking themselves what would happen if some part of the muscular system were damaged. What would the system still be able to do? What would it be unable to do?
Answers Activity 50.1 How does sarcomere structure affect muscle function? Working in groups of three or four, use playdough or cutout pieces of paper to construct a dynamic model of the sarcomere and demonstrate the sliding-filament model for muscle contraction.
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When developing and explaining your model, be sure to include definitions or descriptions of the following terms and structures. thick filament thin filament actin myosin Z line H zone cross-bridges
sarcomere contracted length extended length troponin complex tropomyosin Ca
Alternatively, assign at least three people to model different parts of the sarcomere and demonstrate its action in muscle contraction. Assign a different role to each person, and assign specific actions to each role. When you have completed your model, explain it to another student or to your instructor. Use your understanding of the model to answer the questions. 1. How are sarcomeres arranged within muscle fibers (and therefore within muscles)? Sarcomeres are arranged linearly in myofibrils. Myofibrils are arranged linearly in muscle fibers. Muscle fibers are organized into muscles. Within a given muscle, the muscle fibers are all oriented in the same direction. (See Figure 50.25 of Biology, 8th edition.) 2. Describe the sliding-filament model of muscle action. For example, what interactions power the movement of filaments in association with each other? Include troponin complex, Ca binding sites, tropomyosin, myosin binding site, and actin in your discussion. Refer to Figures 50.26, 50.27, and 50.28. In the sarcomere, thick myosin filaments lie between thin actin filaments. The actin filaments are attached to the Z lines of the sarcomere. The thick filaments contain myosin heads, which alternate between low- and high-energy states. In its low-energy state, a myosin head is bound to an ATP molecule. The ATP is hydrolyzed to generate the myosin head’s high-energy state. The actin filaments are composed of two strands of actin intertwined to form a helix. Long strands of tropomyosin are wound around the actin helix. At rest, the tropomyosin strands block the myosin binding sites on the actin. Action potentials in the sarcoplasmic reticulum and T tubules cause the sarcoplasmic reticulum to release Ca2 into the cytoplasm of the myofibrils. When Ca2 binds to the troponin complexes (which are situated at intervals along the tropomyosin strands), they change configuration. The configuration change alters the position of the tropomyosin strands and exposes the myosin binding sites. The myosin heads hydrolyze ATP and pull the actin filaments toward the center of the sarcomere— that is, toward the M line. The muscle contracts. 332
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3. How do your answers to questions 1 and 2 help explain why, at maximum contraction, muscle measures 70% of its extended (uncontracted) length? During contraction, the actin filaments are pulled toward the center of the sarcomere by myosin heads on the thick filaments. At maximum contraction, the sarcomere width is equal to the length of the thick filaments. In the relaxed sarcomere, the length of the thick filaments is about 70% of the sarcomere width from Z line to Z line. You can check this by measuring the widths of the relaxed sarcomere and the contracted sarcomere in Figure 50.26. 4. Review Figure 50.29 and the associated text in Biology, 8th edition. Then answer the next questions. a. Acetylcholine, an excitatory neurotransmitter, is the neurotransmitter released into the neuromuscular junction. This release can trigger an action potential along the length of the muscle membrane. Describe the process involved. A motor neuron synapses with a muscle fiber at a neuromuscular junction. If an action potential is generated in the neuron, the motor end plate of the neuron releases acetylcholine. This triggers an action potential in the muscle cell, which travels into the T tubule system. As noted in question 2, this causes the release of Ca2 from the sarcoplasmic reticulum, which then triggers contraction of the sarcomere. b. If acetylcholine always produces an excitatory response in the neuromuscular junction, how can we regulate which muscles (in the arm, for example) are contracted and which are extended at any given time? For example, how can we bend the arm only partially? Muscles are arranged into motor units. The contraction of each motor unit is controlled by separate neurons. The degree of response depends on how many motor units have been stimulated to contract. c. Severe calcium deficiency can lead to a reduction in bone mass. It can also have serious effects on the functioning of the nervous system and on the action of muscles. Explain what role(s) calcium plays in the activity of muscles. As noted, it is the release of Ca2 from the sarcoplasmic reticulum that ultimately causes the muscles to contract. In cases of severe calcium deficiency, muscles can become weak, or incapable of strong contractions.
50.1 Test Your Understanding The level of musculature in various organs is directly related to the amount of work done by that organ or organ part.
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A few examples are noted below. a. Give two examples of different parts of organ systems where this is evident. (Use two different organ systems.)
b. For each example, explain the advantage of the difference in musculature for each system.
Examples in the circulatory system include: • Heart—Atria are much less muscular than ventricles. • Heart—Right ventricle is much less muscular than left ventricle. • Arteries contain more muscle than veins.
This is advantageous because: • The atria pump blood a very short distance into the ventricles. The ventricles must pump blood greater distances. • The right ventricle pumps blood to the lungs, a shorter distance than the left ventricle, which pumps blood through the body. • Arteries carry blood under higher pressure than veins.
An example in the digestive system: • The stomach wall is more muscular than the esophagus.
This is advantageous because the stomach acts to churn the ingested food, which assists in breaking it into smaller pieces (aided by action of stomach acid). The primary functions of the esophagus—to undergo peristalsis and transfer food to the stomach—require much less musculature. (The peristaltic activity of the esophagus explains why you can eat upside down.)
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Activity 50.2 What would happen if you modified particular aspects of muscle function? Test your understanding of muscle function by asking yourself what would happen if a specific part was damaged. What would the system still be able to do? What would it be unable to do? In each of the following situations, indicate whether the proposed answers are true or false. If false, indicate why it is false. 1. A toxin in a newly discovered bacterial strain causes irreversible inactivation of the acetylcholine receptor. If you were infected with this organism, the symptoms would include: T/F a. Uncontrollable muscular contractions Explain: False—If the acetylcholine receptors were inactivated, no action potentials would be generated in muscle cells or T tubule systems. Therefore, no contractions would occur. T/F
b. Muscular paralysis Explain: True—As noted in part a, inactivation of the acetylcholine receptors would prevent muscle contractions.
T/F
c. Loss of membrane potential of muscle fibers Explain: False—The sodium-potassium pumps are responsible for maintaining the membrane potential in muscle fibers. Poisoning the acetylcholine receptors should not affect these pumps.
2. In an experiment, a physiologist was able to destroy some but not all of the motor neurons to a specific skeletal muscle. Given what you know about motor units, what is the result of this action? T/F a. There is no effect because motor neurons are redundant; multiple neurons innervate each muscle fiber as backup. Explain: False—Each motor unit is innervated by a specific motor neuron (and its branches). As a result, if you destroyed its motor neuron, the motor unit would be unable to respond. Activity 50.2
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T/F
b. The muscle will contract faster if the physiologist destroyed inhibitory motor neurons. Explain: False—No inhibitory motor neurons innervate motor units in skeletal muscle.
T/F
c. The muscle will in general, contract more weakly because fewer fibers are capable of contracting. Explain: True.
T/F d. The ability of the muscle to contract will not change because the electrical potential will spread from stimulated fibers to adjacent fibers. Explain: False—In skeletal muscle, each muscle cell in a motor unit receives separate innervation from the branches of the motor neuron associated with that motor unit.
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Notes to Instructors Chapter 51 Animal Behavior What is the focus of this activity? The morphology, physiology, and behavior of organisms are all tightly integrated or interconnected. In addition, all these characteristics of organisms can affect both an organism’s survival and its reproductive success. In other words, an organism’s behavior occurs directly (or proximally) as a response to current stimuli and indirectly (or ultimately) as the result of the process of natural selection or evolution of the species as a whole.
What is this particular activity designed to do? Activity 51.1 What determines behavior? In this activity, students are given specific examples of types of behavior. Then they are asked to propose: • what questions they would need to ask to determine proximate causation for each behavior, • what questions they would need to ask to determine ultimate causation for each behavior, and • what kinds of experiments or investigations they would propose to answer the questions.
Answers Activity 51.1 What determines behavior? Note: A wide range of questions can be asked to determine proximate versus ultimate causation in each of the examples. A few representative questions are provided below. Similarly, a wide range of experiments or investigations can be proposed. 1. Some plant species (for example, many orchids) rely on a single species of insect for pollination. If the insect species dies out, so will the plant species. In a through c, refer only to the behavior of the insect species. a. What questions would you need to ask to determine proximate causation for this species-specific insect behavior? What are the specific characteristics of the orchid and the insect? Notes to Instructors
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Does some feature(s) of the morphology of the insect limit the types of flowers that it can pollinate? Is this specific insect species unable to pollinate other flowers? What factors (color pattern, fragrance, type of nectar, and so on) draw the insect species to only this type of orchid? b. What questions would you need to ask to determine ultimate causation for this insect behavior? Why would this type of tight symbiotic relationship between orchid and pollinator evolve? c. What kinds of experiment(s) or investigation(s) you would propose to answer at least one of the questions in parts a and b? What factor(s) draw the insect species to the flower? You would need to test for each factor separately. You could place individual insects in test chambers. To test for nectar preference, put small samples of nectar from the orchid and from several other orchid species into the chamber. Then observe the insects and record their responses to the different types of nectar. You can determine whether or not the insects are attracted by nectar from only the one species of orchid and how often (per unit time) they sample each type of nectar. To test for fragrance preference, remove the nectar-containing parts of various flowers, including the orchid. Then put the orchid into a sealed box that has a small opening in one side. Do the same for another species of flower. Connect one end of a Y tube to one box and the other to the second box. Place the insect in the vertical arm of the Y tube and record how often it moves to one box or the other. Repeat the procedure many times. You can randomize which box (right side or left side of Y tube) contains the orchid. In some experiments, use no orchid. 2. Many bird species that are common in the northern United States during spring and summer fly south in the fall to overwinter and feed in Central or South America. In the spring, they return to states in the northern United States to breed. a. What questions would you need to ask to determine proximate causation for this behavior? How do birds “know” when to fly south versus north? For example: What environmental factors (for example, day length or temperature) stimulate the birds to flock and fly south in the fall? What environmental factors stimulate the birds to fly north in the spring? What, if any, physiological changes occur to trigger these flights? How do birds navigate (find their way) on these long flights? 338
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b. What questions would you need to ask to determine ultimate causation for this behavior? Why do birds fly south? Why would this type of behavior evolve? c. What kinds of experiment(s) or investigation(s) you would propose to answer at least one of the questions in parts a and b? What environmental factors stimulate flocking behavior for the flight south? You could test the effects of changes in day length on bird behavior. For example, maintain different flocks of birds in separate artificial habitats that allow you to regulate day length. In one habitat, maintain the flock on normal day-length cycles. In others, alter the day-length cycles by speeding up or slowing down the approach of shorter day lengths. You can then observe any differences in the behaviors of the different flocks of birds. 3. Many species of animals engage in complex courtship rituals. Among these species is the bower bird of Australia. Male bower birds construct elaborate structures or bowers from twigs, leaves and moss and decorate them with colorful objects, such as berries and shells. The bowers and the dances the males perform are designed to attract female bower birds for mating. After mating, the females fly away to build a nest and raise the offspring. The males remain at their bowers and try to attract additional mates. These birds can live for up to 17 years. Males are territorial and build their bowers in the same location each year. In studying their behavior, researchers have noted that about 25% of the females “shop around” going from one bower to another before deciding on a mate. The other 75% appear to go directly to a single bower to mate. These behaviors cannot be observed in captivity. a. What questions would you need to ask to determine proximate causation for the bower building behavior? What characteristics of the environment (etc.) trigger the bower building response in the male? b. What questions would you need to ask to determine the proximate causation for the female choice? What characteristics of the bower or the male’s dance trigger the mating response in the female? c. What questions would you need to ask to determine ultimate causation for these behaviors? Why did this elaborate mating ritual develop?
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d. What kinds of experiment(s) or investigation(s) would you propose to answer at least one of the questions in either parts a or b? What characteristics of the environment trigger the male bower building behavior? Because these birds cannot be studied in captivity, you would need to do a detailed analysis of the environmental characteristics in the days/weeks leading up to the behavior and compare these from year to year to determine key factors. To determine what characteristics of the bower attract females you could record the characteristics of the bowers that attract the most females and do an analysis of similarities among these. You would need to do the same to compare the characteristics and dance patterns of the males that were most successful. There are obviously many more questions that students could ask. Some would have to do with the life span of the birds and their ability to learn over time not only which males are considered the best mates but also the location of these males, which could explain why many females fly directly to specific males’ bowers (etc.)
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Notes to Instructors Chapter 52 An Introduction to Ecology and the Biosphere What is the focus of this activity? Ecology is the scientific study of the interactions of organisms and their environments. These interactions can be studied at a variety of levels, ranging from the individual organism to populations, communities, ecosystems, landscapes, biomes, and ultimately the entire biosphere. As a result, studying ecology requires that we integrate our understanding of the biology of organisms (for example, cell structure and function, genetics, evolution, physiology) with our understanding of the physical and chemical factors that affect the world and its climates.
What is this particular activity designed to do? Activity 52.1 What factors determine climate? In this activity, students apply their understanding of the factors that produce the biomes and climates on Earth to propose how these factors would affect environmental conditions on a hypothetical continent on Earth.
Answers Activity 52.1 What factors determine climate? The map on page 343 shows a hypothetical continent on Earth. Assume that biomes and climates on this continent are produced by the same factors that produce biomes and climates on Earth’s real continents. Use this map to answer the questions in this activity. Where needed, draw the required features directly on the map. 1. a. On the map of the hypothetical continent, indicate the location(s) of the biomes listed in the table below. To do this, draw approximate boundary lines to delimit each biome type, and then label each delimited area with the type of biome it contains. Use the information in Figures 52.10, 52.19, and 52.20 of Biology, 8th edition, to answer these questions.
Notes to Instructors
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b. In the table indicate the annual temperature and precipitation ranges for each biome. Biome type
Annual temperature and precipitation ranges
a. Tropical rain forest(s)
In general, tropical rain forests lie between the equator and about 25° north and south latitudes, where the annual mean precipitation exceeds 150 cm and the annual mean temperature exceeds about 23°C.
b. Temperate grasslands Temperate grasslands tend to lie in the interior of continents or on the east coast in South America, between 25° and 45° north or south latitudes. Annual precipitation is between 30 and 100 cm, with periodic droughts common. Temperature ranges between 10°C and 30°C.
c. Chaparral
The chaparral biome tends to lie between 25° and 45° north and south latitudes, where the general direction of the trade winds is away from the coast and the annual mean precipitation averages 30 to 50 cm. Average winter temperatures range between 10 and 12°C and average summer temperatures can reach 30°C with a maximum of 40°C.
d. Temperate forest(s)
Temperate forests tend to lie between 25° and 60° north and south latitudes, where the annual mean precipitation ranges from 70 to 200 cm and the annual temperature range is from below zero to 30°C.
e. Tundra
Tundra lies above 60° north and south latitude (Note: On maps of the Earth, there is no tundra in the southern hemisphere because there are no land masses between 60° and 90° south latitude. Antarctica lies above 90° south latitude for the most part). Annual precipitation ranges from 20 to 60 cm, and temperatures range from below 30°C up to about 10°C in summer.
2. Atmospheric circulation is driven primarily by differential heating of Earth’s surface. More heat is delivered near the equator than near the poles. This seems to explain the northward and southward flows of air. What introduces the eastward and westward components into air movement? (Hint: Review Figure 52.10, Global Air Circulation and Precipitation Patterns and Global Wind Patterns, in Biology, 8th edition.) As indicated in Figure 52.10, “as Earth rotates on its axis, land near the equator moves faster than that at the poles, deflecting the winds from the vertical paths” introduced by differential heating. This deflection creates more easterly flow patterns between the equator and 30° north and south latitudes and more westerly flow patterns between 30° and 60° north and south latitudes. 342
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Temperate grassland
Tundra
60ºN
A
Temperate forest
X1
W1 B Chaparral E
30ºN
O2
C
X2 Temperate D Rainforest
0º
W2 Chaparral 30ºS F
O3
Temperate forest
X3
Temperate Grass
60ºS W3 90ºS = mountains A hypothetical continent
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3. Use your understanding of global air circulation and wind patterns to draw arrows on the map of the hypothetical continent indicating: a. The direction of prevailing winds at points W1, W2, and W3 b. The direction of flow of surface currents in the ocean at points O1, O2, and O3 (Hint: Note on the map in Figure 52.10 that surface currents in the ocean follow the major wind systems at the surface.) 4. Are the surface winds at the given points warming or cooling as they move? Explain. Are the winds warming or cooling as they move?
Point on the map a. X1
Explanation
The winds are warming The general wind pattern in this area is from the southwest (westerlies). These winds tend to pick up moisture as they move from 30° to 60° N latitude.
b. X2
The winds in this region tend to be cooling.
This region tends to be under the influence of cooling easterly winds. These winds tend to pick up moisture as they flow from 30° to 0° N latitude.
c. X3
The winds are warming The general wind pattern in this region is westerly. These winds tend to pick up moisture as they flow from 30° to 60° S latitude.
5. What biomes or vegetation types would most likely be found at the given points? (Assume all are at sea level or low altitudes.) Point on the map
Most likely type of biome or vegetation
A
This region would most likely contain coniferous forest.
B
This region would most likely be a temperate grassland that will graduate into temperate forest as it moves east.
C
This would most likely be tropical forest, but could be savanna depending on local water conditions.
D
This is most likely a region of tropical rainforest.
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6. Would the climate at point E be relatively wet or dry? Explain. This region would be relatively dry. It lies in the rain shadow of the mountains. In addition it lies at about 30° N latitude, a region of very drying winds. 7. What would the direction of the prevailing winds be at Earth’s surface at point X3? The prevailing winds at this point should be westerly. 8. In the United States, temperate forest extends from the East Coast westward for about 1,200 miles (to the Mississippi River Valley). From there, the forest begins to thin out toward the west into oak savannas (or temperate woodland), and it finally gives way to open grassland (Great Plains). The grasslands extend 1,000 miles westward to the foothills of the Rocky Mountains. a. Why does grassland replace forest west of the Mississippi River? This area is under the influence of the westerlies (wind flow pattern), which carry moist air from the Pacific Ocean onto the North American continent. When these winds reach the Rocky Mountains, they rise and cool. This causes the moisture in the air to condense and fall as rain on the western slopes of the Rockies. The wind that reaches the eastern slopes is dry, so much of the land in this “rain shadow area” is desert. As the winds continue overland, they pick up some water; however, the area west of the Mississippi receives less rainfall than areas farther east. Grasslands are found in such regions where the mean annual precipitation is between about 30 and 80 cm. b. What is the rain shadow effect? As noted in part a, the rain shadow effect occurs on the leeward side of mountain ranges. When the air currents encounter the mountains, they rise. As air rises, it cools and the moisture in it condenses and falls as rain. As a result, the air that reaches the other side of the mountain is dry, so the leeward side of the mountain is dry. c. Draw a rain shadow somewhere in the southern hemisphere of the map. The rain shadow is always on the leeward side of the mountain. Therefore, in the southern hemisphere, the rain shadow should be on the western side of the mountains if the site is between 0° and 30° south latitude but on the eastern side if the site is between 30° and 60° south latitude. 9. What biome would exist at point F? Point F is at the transition between temperate grassland and temperate forest. It is difficult to determine which biome would exist in this exact location, and therefore it is likely a transition zone.
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10. How are the general characteristics of plants (for example, morphology) influenced by climate? In other words, explain what effects climate has on the types of plants that grow in an area. As indicated in Figure 52.20, the type of biome is strongly influenced by both the annual mean temperature of the region and the annual mean precipitation. For example, when both are relatively high, you find temperate rain forests. Plants in these regions do not experience extremes in temperature or water over the year. When mean annual temperature is high and mean annual precipitation is low, deserts are formed and the plants are those that can survive in this type of climate. Refer to Figure 52.21 for examples of the types of plants that would be found in the various types of biomes. 11. Refer to Figure 52.20 in Biology, 8th edition. In general, how is the distribution of major ecosystems or biomes related to climate? If you know the mean annual precipitation and the mean annual temperature of an area, would you be able to accurately predict the type of biome that could exist there? Explain. As noted above, the type of biome or ecosystem is determined by the mean annual temperature and the mean annual precipitation in an area. However, how the precipitation and temperature are distributed over the year also has a significant impact on the type of biome. For example, is the temperature relatively constant over the course of the year or does it range between high summer temperatures and very low winter temperatures? Other factors—for example, type of soil—can also affect the type of biome. 12. Why isn’t Earth’s climate uniform? To answer this, summarize the major factors that can produce differences in climate from place to place. The primary factors affecting the type of climate are the annual amount of solar energy received and the annual amount of water received. The amount of solar energy received per year in a given area is directly related to its location on the globe and the fact that the earth is tilted 23 degrees on its axis. As a result, the equator receives approximately the same amount of solar energy throughout the year. On the other hand, more northern areas in the United States will experience different seasons over the course of the year. The major wind patterns are set up by differential heating at the equator. The major precipitation patterns are a function of these wind patterns.
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Name _______________________
Course/Section_______________________
52.1 Test Your Understanding
1. A large asteroid hits the Earth, and alters its axis tilt from 23.5° to 10°. a. What effect(s) would this have on seasons in the northern hemisphere? With less tilt, there would be less seasonality. Overall, the seasons would be warmer and less distinctly different from one another. The sun’s rays over the course of the year would be more direct, and the day length would tend to be less variable in summer versus winter. Days in the summer would still be longer than days in the winter, but the difference in average day length would not be as great. b. What effect(s) would this have on biome distribution in the Northern Hemisphere? (Refer to Figure 52.20 in Biology, 8th edition.) As the annual mean temperature increased, we would see much less tundra and boreal/coniferous forest. Over time, the land that was previously occupied by the tundra and boreal/coniferous forest would become temperate forest, grassland, or desert, depending on location and precipitation. A catastrophic event such as this would likely lead to mass extinction events and it would take a very long time for the biomes and flora/fauna to reposition/establish themselves on the Earth. 2. If you travel across the United States on a line from North Carolina to Southern California, you will find that deciduous forest extends from the East Coast westward for approximately 1,200 miles to about the valley of the Mississippi River. There it begins to thin out toward the west into oak savannas and finally gives way entirely to open grassland (Great Plains), which extends another 1,000 miles westward up into the foothills of the Rocky Mountains. If you cross over the mountains you find chaparral in Southern California. T/F The single most important factor affecting the type of biome present in each of these major regions of the United States is rainfall. True 3 to 6.
T/F
The type of vegetation that will grow in a particular region of the earth is strongly affected by:
3. Extremes of temperature (high and low) experienced over the year in that region. True
T/F 4. Altitude of that region. True T/F 5. Seasonal availability of water. True T/F 6. Longitudinal location of that region. False Activity 52.1
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Notes to Instructors Chapter 53 Population Ecology What is the focus of these activities? To study ecology, we often use mathematical models and statistical analysis to estimate population size and distribution. We have also used mathematical models to make predictions about what may happen to the population in the future. Students should keep in mind, however, that any model is limited by the assumptions used in building it. At best, models provide us with a generalized idea of what could happen if the specific assumptions of the model are met.
What are the particular activities designed to do? Activity 53.1 What methods can you use to determine population density and distribution? The questions used in this activity are designed to help students review and understand some of the basic methods used in determining population density and distribution. Activity 53.2 What models can you use to calculate how quickly a population can grow? The questions are designed to help students review and understand how various mathematical models can be used in studying populations.
Answers Activity 53.1 What methods can you use to determine population density and distribution? 1. To measure the population density of chipmunks occupying a particular park, you sample several quadrats and capture 50 chipmunks. You mark each of them with a small dot of red paint on their backs, and then release them. The next day, another 50 chipmunks are captured. Among the 50, you find 10 that are marked. a. Using the mark–recapture formula below, how many chipmunks do you estimate the population contains? Number of recaptures in second catch Total number in second catch
Number marked in first catch Total population N
If 10/50 = 50/N, then N = 250 chipmunks. 348
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b. What effect would each of the following discoveries have on your estimate? i. You later discover that you sampled the one area of the park that was most favored by the chipmunks. If the area was favored by the chipmunks, you most likely captured more chipmunks in this area than you would have caught in other areas of the park. As a result, your estimate of the population size would be smaller than if you had sampled more of the park. ii. You later discover that the chipmunks were licking the mark off of each others’ backs. Some of the recaptured chipmunks that were counted as unmarked could have been among those that had their marks licked off. As a result, your estimate of population size would be higher than if the marks had not been licked off. For example, if 10 of those captured had licked off their marks, the actual recapture ratio would be 20/50. Then the population estimate would change to 125 (rather than 250) chipmunks in the population. iii. You later discover that the marked chipmunks are easier to see and therefore more susceptible to predation. If more marked chipmunks are captured by predators, you will recapture fewer marked chipmunks than you should. For example, if you mark 100 and 50 are eaten, the probability of your recapturing any of the marked chipmunks is reduced by 50%. Therefore, the population size you calculate will also be reduced by up to 50%. As a result, you will not be able to accurately quantify the population size. c. How could you modify your sampling program to ensure that you make more accurate estimates of population size? At a minimum, you should: i. sample several areas of the park. ii. use a marking system that cannot be removed easily. iii. use a marking system that is not easily detected by predators.
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2. Refer to the two proposals for the distribution of a tree species below.
Proposed distribution 1
Proposed distribution 2
a. What type of distribution is represented in each of the proposals? Distribution 1: = clumped Distribution 2: = random b. Given these two possible distributions, what factors do you need to consider in setting up a sampling plan for the area? Propose sampling strategies and the results you would get if organisms were distributed as in 1 vs. 2 above. For each sampling strategy proposed, indicate how you will know if you have chosen both an appropriate quadrat size and number of quadrates to provide you a good representation of both the size of the population and the actual distribution of organisms within the sampling area. The answer to this question will depend on the size of the total area relative to the size of the sample quadrats and the number of quadrats sampled per distribution. The following gives an overview of the results you would get for each type of distribution and different sampling strategies. Assume that you divide each area into four quadrats and then sample only quadrat D in both areas.
A
B
C
D
If you count the total number of trees in quadrat D in both distributions 1 and 2, you discover that quadrat D in distribution 1 contains 10 trees, while quadrat D in distribution 2 contains 5 trees. These differences would indicate that the density of trees in distribution 1 may be higher than the density in distribution 2. It doesn’t help you determine how the trees are distributed, however. 350
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Now, assume that you divide each area into 16 quadrats.
A1
A2
B1
B2
A3
A4
B3
B4
C1
C2
D1
D2
C3
C4
D3
D4
If you sample the four quadrats in section D in each area (1 and 2), your data would look something like this: Quadrat number:
Number of trees Area 1 Area 2 1 2 6 1 3 1 0 1
D1 D2 D3 D4
These data indicate that areas 1 and 2 contain not only different densities but different distributions of trees as well. If you divide the areas further into 64 quadrats, you would get a better idea of the exact distribution of the trees. However, if your goal is to determine relative distributions rather than absolute distributions, you don’t need this much detail. In other words, the size and number of quadrats that need to be sampled depend on the goals of your study.
3. The following table provides the numbers of deaths per acre per year resulting from two different agents of mortality applied to grasshopper populations of different densities.
Activity 53.1
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a. Fill in the mortality rates for agents A and B in the table below. Grasshopper population density (individuals/acre)
Deaths per year per acre Agent A Agent B
Mortality rate (%) Agent A Agent B
100
4
0
4%
0%
1,000
40
25
4%
2.5%
10,000
400
500
4%
5%
100,000
4,000
50,000
4%
50%
b. Graph the data below.
Mortality Rate (%)
60 50 40 Agent A
30
Agent B
20 10 0 100
1,000
10,000 100,000
Population Density/Acre
c. Which of the two agents of mortality (A or B) is operating in a densityindependent manner? Explain your answer. Regardless of the population size per acre, agent A kills only 4% of the population. In other words, it is working in a density-independent manner. Agent B is acting in a density-dependent manner because as the density changes, the percentage of individuals killed by agent B also changes. In this example, as the density of grasshoppers increases, the percentage of grasshoppers killed per acre also increases. d. Which of the two agents of mortality (A or B) is likely to act as a factor stabilizing the size of the grasshopper population? Explain your answer. No matter how large the population gets, agent A removes only 4%. In contrast, agent B removes an increasing percentage of the grasshoppers as the population size increases. Therefore, agent B is more likely to act to stabilize the size of the population over time. 352
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53.1 Test Your Understanding A researcher has recently discovered three species of parasites (A, B, and C) that infect developing salamanders. He suspects that one or more of these species cause fatalities during salamander development and that the death rate varies with salamander population density. To test his hypothesis, the researcher sets up a series of experiments. In each experiment, he varies the density of the salamander populations. At the start of each experiment, he infects 5% of each test population with a single parasite species and then measures the mortality/death rate after four weeks. To establish a baseline mortality rate, he sets up a control experiment that differs only in that no parasite is introduced at any density. The data table and graph below relate the mortality rate of salamanders (caused by the three different parasites) to the original density of the developing salamander population. a. Given the data presented in the graph, indicate whether each of the parasite species (A, B, and C) is acting in a density-dependent or density-independent fashion. Explain your answers. Exp. 1 Density
Exp. 3
Control
Control
A Mortality B Mortality C Mortality Mortality Density Density Density Rate Rate Rate Rate 0.1 20 0.82 20 0.4 20 0.1 0.2 40 0.74 40 0.25 40 0.096 0.4 80 0.76 80 0.49 80 0.12 0.6 160 0.8 160 0.35 160 0.15 0.85 320 0.76 320 0.36 320 0.092
Mortality rate
20 40 80 160 320
Exp. 2
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 20
40
80 Salamander density
A
B
160
C
320
Control
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Parasite A is acting in a density-dependent fashion, because parasite A causes progressively higher mortality rates as salamander density increases. Parasite B produces greater mortality than the control. However, the percent mortality is fairly consistent over the range of population densities. As a result, it is acting in a density-independent fashion. Parasite C, while it produces greater mortality than that shown in the control, also does not appear to act in a density-dependent fashion. Line D is the control, showing approximately 10% mortality regardless of population density. b. Looking at the data for parasites A and B, develop an argument to indicate which is more likely to cause extinction of the salamander population. Explain your reasoning. Because of its overall high mortality rate at all salamander densities, parasite B may decrease the population size such that it cannot rebound. In contrast, parasite A has lower mortality at lower population densities. As a result you are likely to see an oscillation in population size and parasite mortality over time. Only at very high population densities (>320) is parasite A likely to cause extinction.
Activity 53.2 What models can you use to calculate how quickly a population can grow? 1. In the simplest population growth model (dN/dt = rN). a. What do each of the terms stand for? Term
Stands for
dN
Change in the number of individuals
dt
Time interval during which the change occurred
r
Per capita population growth rate = per capita birth rate minus per capita death rate
N
Initial population size
b. What type of population growth does this equation describe? Exponential growth of a population (J-shaped curve) c. What assumptions are made to develop this equation? Unlimited space (habitat) and resources 354
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2. Population growth may also be represented by the model, dN/dt = MmaxN[(K – N)/K ]. a. What is K? K is the carrying capacity or the number of individuals that can be sustained over time. K is a function of the environment. b. If N = K, then what is dN/dt? When N = K, dN/dt = 0; that is, there is no change in the population size over time. c. Describe in words how dN/dt changes from when N is very small to when N is large relative to K. When N is very small, the population is growing exponentially. The actual number of new individuals added is relatively small, however, because N is so small. As N nears the value of K/2, dN/dt approaches its maximum. As N approaches K, dN/dt decreases. d. What assumptions are made to develop this equation? The logistic model assumes that resources and space limit the growth of a population and that these factors determine the maximum number of individuals that can be sustained over time. 3. You and your friends have monitored two populations of wild lupine for one entire reproductive cycle (June year 1 to June year 2). By carefully mapping, tagging, and censusing the plants throughout this period, you obtain the data listed in the chart. Parameter
Population A
Population B
500
300
100
30
20
100
Initial number of plants Number of new seedlings established Number of the initial plants that died
a. Calculate the following parameters for each population. Parameter
Population A
Population B
B (births during time interval)
100
30
D (deaths during time interval)
20
100
b (per capita birth rate)
100/5000.20
30/3000.10
m (per capita death rate)
20/5000.04
100/3000.33
0.200.040.16
0.100.330.23
r (per capita rate of increase) Activity 53.2
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b. Given the initial population size and assuming that the population is experiencing exponential growth at growth rate r, what will the number of plants be in each population in 5 years? (Use the initial population size as time 0, and compute to time 5.) Population A: 1,050 Population B: 81 4. Using the exponential growth formula, you can determine the amount of time it will take for a population to double in size if you know rmax or r. Doubling time is equal to: log102 / log10 (1 r) Alternatively this value can be estimated by using the formula: 70 divided by the percent increase per unit time (as a whole number) = doubling time per unit time or 70/r = approximate doubling time. Using either of these formulas—the exponential growth formula or the approximate doubling rate formula—calculate the following. a. If the population of a country is growing at 2% per year, how many years will it take for the population to double? Doubling time = 70/2 per year = 35 years. b. If your bank account is growing at a rate of 1% per year, how many years will it take for it to double? Doubling time = 70/1 per year = 70 years. 5. You are studying the growth of a particular strain of bacteria. You begin with a tiny colony on a Petri plate. One day later, you determine that the colony grew and exactly doubled in size. A calculation showed that if the colony continued to grow at the same (constant) rate, it would cover the entire plate in 30 days. (Assume that colony size is directly proportional to the number of individual bacteria.) a. What is value of r? See question 4. You can calculate r using the formula log10 2/log10 (1 + r). Alternatively, r can be estimated by using the formula: 70 divided by the percent increase per unit time (as a whole number) equals the doubling time per unit time. Therefore, if 70/r = 1 day, then r must equal approximately 0.70, or 70%.
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b. On what day would the bacteria cover half the plate? If the bacteria colony doubles in size every day and completely covers the plate on day 30, then it must cover half the plate on day 29. 6. You collect data on birth and mortality in three populations of grasshoppers, and you calculate the following birth and death rates for these populations. Both populations are experiencing exponential growth: b 0.90 0.45 0.15
Population A Population B Population C
d 0.80 0.35 0.05
Are the following statements true or false? F a. T b. F c. T d.
Population A is growing at the fastest rate. Population C has the lowest death rate. Population C is growing at the slowest rate. All populations are growing at the same rate.
7. In a herd of bison, the number of calves born in 1992, 1993 and 1994 was 55, 80, and 70, respectively. In which year was the birth rate greatest? You cannot answer this question unless you also know the population size for each of those years. 8. A population of pigeons on the west side of town has a per capita annual growth rate of 0.07. A separate population of pigeons on the east side of town has a per capita annual growth rate of 0.10. If the populations are both growing exponentially and both are censused the following year, in which of the populations will dN/dt be greatest? Again, you cannot answer this question because you don’t know how large the populations are initially. If the two population sizes are equal, or if the population on the east side is larger than the population on the west side, then dN/dt will be greater on the east side. However, if the population on the west side is large and the population on the east side is much smaller, then dN/dt of the west-side population could be the greater of the two. 9. Suppose you have a “farm” on which you grow harvest, and sell edible freshwater fish. The growth of the fish population is logistic. You want to manage your harvest to maintain maximum yields (that is, the maximum rate of production) from your farm over a number of years. As a fisheries manager, you are responsible for deciding how many walleye can be harvested without destabilizing the population. Activity 53.2
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a. Below is a data table showing the walleye population in a typical pond on your fish farm over 24 weeks. Draw a graph showing how population size in the pond changes through time. Time (weeks)
1
Population Size
3
4
5
6
7
8
9
10
11
12
100 101 102 103 104 106 110 115 125 140 155 172
Time (weeks)
13
Population Size
Pooulation size
2 14
15
16
17
18
19
20
21
22
23
24
188 201 209 217 221 225 229 233 235 237 238 239
250 225 200 175 150 125 100 75
E
D C B A 1
9
5
13 Time
17
21
b. How large should you let the population get before you harvest? Identify the point on your graph and explain why. The rate of production will be fastest at the point on the curve where the slope is the steepest. This would be at about point C (the midpoint) on the graph. dN/dt is the measure of the rate of change (that is, production) of individuals in the population. When you graph N versus t, the slope corresponds to dN/dt. When the population size (N) is very small, then [(K ⫺ N)/K ] is large. However, there are so few individuals in the population that the gain in population numbers, dN/dt, is also relatively small. When the population size (N) is very large, then there are many individuals but [(K – N)/K ] is small. The rate of population growth (production measured as dN/dt) is at a maximum when N is equal to 1/2 K. c. Assume the carrying capacity for your pond is 250 individuals. Check your answer in part b by using the data in the chart and computing the change in the population size (dN/dt) when the population is at several different levels relative to its carrying capacity. Use K = 250 and rmax = 0.20.
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Population size (N)
(K – N)/K
25 (low) 50 (moderately low) 125 (half K) 200 (moderately high) 250 (high)
dN/dt
0.1 0.8 0.5 0.2 0
0.5 8 12.5 8 0
10. A rabbit population has the following life table.
Age class 0–1
Number of survivors 100
Number of deaths 10
Number of offspring per reproducing pair Mortality rate 0.10 0
1–2
90
30
0.33
1.5
2–3
60
30
0.50
2.0
3–4
30
24
0.80
2.5
4–5
6
6
1.0
0
a. Fill in the missing data in the table. b. Owing to a good food supply and small predator population, the rabbit population is growing by leaps and bounds. The rabbits call a meeting to discuss population control measures. Two strategies are proposed: • Delay all rabbit marriages until age class 2–3 (rabbits never breed until after marriage). • Sterilize all rabbits in age class 3–4. Which of the proposed strategies will be more effective in slowing population growth? Explain your reasoning and show your calculations. In the first two years of their lives, each reproducing pair of rabbits will have an average of 1.5 offspring. For a cohort of 100 rabbits born in year 0–1, 45 pairs of rabbits (the 90 survivors) will each have 1.5 offspring during year 1–2; they will add about 67.5 rabbits to the population. When this same cohort reaches age class 3–4, it will consist of 15 pairs, each of which will produce an average of 2.5 offspring for a total of 37.5 rabbits. Therefore, the population size will be more limited by delaying rabbit marriages until age class 2–3.
Activity 53.2
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Notes to Instructors Chapter 54 Community Ecology What is the focus of this activity? Biology, 8th edition, defines community ecology as “the study of how interactions between species affect community structure and organization.” Biologists recognize that the organisms in communities are continuously evolving in association with one another. This implies that community structure and organization are also continuously evolving. This activity focuses on some of the key principles and factors that can affect the structure and organization of communities.
What is the particular activity designed to do? Activity 54.1 What do you need to consider when analyzing communities of organisms? Activity 54.2 What effects can disturbance have on a community? These activities are designed to help students recognize that studying community ecology requires the consideration and integration of a number of ecological principles. Activity 54.3 How can distance from the mainland and island size affect species richness? This activity allows students to explore the theory of island biogeography and the effects distance from the mainland and island size have on species numbers. All of these activities give students practice in data analysis and graphical representation and interpretation.
Answers Activity 54.1 What do you need to consider when analyzing communities of organisms? Understanding problems in community ecology most often requires integration of a number of ecological principles. For questions 1 to 5, analyze the situations described. Then explain which of the following ecological principles could be active in each particular situation.
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Ecological principles: coevolution realized niche fundamental niche tolerance competitive exclusion resource partitioning
character displacement top-down vs. Bottom-up controls trophic structure or trophic levels keystone species competition disturbance
1. A small clan of hyenas killed an antelope. While they were feeding on the carcass, two female lions approached, growled at the hyenas, and chased them away from the carcass. The kill is an example of predation by the hyenas. The action of the lionesses is an example of competition. One species is actively keeping another away from a resource. 2. Two species of closely related swallows live in England. The black swallow lives in coniferous forests, and the yellow swallow lives in deciduous forests. In Ireland, where the black swallow has never been introduced, only the yellow swallow is present and it lives in both coniferous and deciduous forests. This is an example of how a population’s realized (actual) niche can differ from its fundamental or ecological niche. When the black swallow is not present, the yellow swallow is capable of living in both deciduous and coniferous forests. In England, where both the black swallow and the yellow swallow are found, the yellow swallow is limited to deciduous forests. You would need more information to determine whether this is also a case of competitive exclusion or resource partitioning. For example, the birds’ distribution in England could be the result of resource partitioning if black sparrows are known to live in both deciduous and coniferous forests when yellow sparrows are not present. The distribution could be the result of competitive exclusion if black swallows live only in coniferous forests. 3. In a woodland community, three species of rodents coexist: voles, field mice and shrews. All three species eat seeds and nuts. Each species has a preference for seeds of the most appropriate sizes for their teeth and mouths. However, all three species compete for the same kinds of nuts. An owl species also lives in this woodland community. The owl preys on all three rodent species. During one particular year, a parasite that causes pneumonia in birds is introduced into the community. This dramatically reduces the owl population, which remains low for several years as a result. Following the initial reduction in the owl population, there is a dramatic increase in the population of field mice and a dramatic decrease in the populations of both voles and shrews.
Activity 54.1
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The owl is a keystone species. Because the vole and shrew populations decline and the mouse population increases following removal of the owls, the mice are obviously better competitors for the available resources. 4. In 1962, five mute swans escaped from captivity and began a breeding population in Chesapeake Bay. Today, there are over 4,000 mute swans living in the bay. Each year they eat approximately 10.5 million pounds of aquatic grasses. These grasses provide habitat for waterfowl and crustaceans, improve water quality, decrease erosion, and increase dissolved oxygen concentrations in the bay. The birds are also aggressively territorial, and have been known to trample nests of other birds (e.g., least terns and black skimmers) and drive away native birds like tundra swans and black ducks from feeding and roosting areas. This is an example of competition for territory and resources as well as an example of disturbance, in this case caused by the cascading effects of non-native species introduction. Disturbance of the trophic structure is occurring as the primary producers are decimated, and cannot perform their other ecological functions (oxygenation, cover, sediment stabilization). This could also be considered a form of top-down control gone wrong. Here, introduction of the mute swans caused loss of the aquatic grasses that supported the system. Their loss led to the destruction of the system. 5. Known as the “Hawaiian woodpecker,” the `akiapola`au (aki-a-pul-a-ow) is found only in montane mesic old-growth koa/`ohi` forests, and only on the Big Island (Hawaii). It has a distinctive beak that is like a multiple-use tool.
The short straight lower mandible is used to peck holes in the wood and the long curved upper mandible is used to probe for insects and larvae. Males have larger beaks than females and feed on the trunks of trees. Females feed higher on branches and twigs. ‘Akiapola`au are thought to have the lowest reproduction rate for a small bird—only one chick per year, which is cared for by the parents for 6 months or more. The decline in their numbers appears to correspond with the introduction of rats, cats and logging on the island. 362
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This species has a fairly specialized niche in old-growth forest. As a result, it is suffering loss of niche from logging. Introduced species are preying on the birds and may also be competing for habitat. Significant reduction in numbers implies the birds have a limited tolerance range. Character displacement (beak size) in males and females exists; this allows resource partitioning (feeding on different parts of trees). Relating this example back to previous chapters, the parental care aspect of this bird indicates that it is K-selected, which is unusual for animals of such small size. 6. A researcher collected data on an experiment she conducted on two desert islands. The islands were of similar size, climate, and species composition and richness, and were the same distance from the mainland. Originally, the same species of snake was present on both islands (A and B). In her experiment, the researcher removed the species of snake from island A. For comparison, the species of snake was NOT removed from island B. She then recorded the number of animal species on each island over a period of 24 months. Her data are presented in the table below. a. Graph the data. # Species Time (months) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
A
B
36 38 35 33 31 32 29 29 26 24 20 18 18 13 11 10 9 10 10 8 9 10 9 8
36 37 34 35 36 38 40 36 38 39 36 37 35 34 36 38 40 38 36 38 39 36 37 35
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45 40 35 30 25 20 15 10 5 0
A B
1
4
7
10 13 16 Time (months)
19
22
b. Construct a hypothesis that explains the difference between the numbers of species present on island A versus island B over the 24 month period. One reasonable hypothesis would be that the snake was acting as a keystone species on both islands. When it was removed from island A, another competitor/predator dominated and drove many of the other species on the island to extinction. c. Propose an experimental design to test your hypothesis. Explain the reasoning behind your design. One possible design would be to remove the snake from island B, reintroduce these snakes to island A, and again collect data on species diversity over time. You would ideally collect data on not only species diversity, but also population size and species richness. This would allow you to better understand which competitor species is being held in check by the snake. d. What would you expect to find as a result of your experiment? Describe your expected results and draw them on the graph below. Over time you would expect to see the numbers of species on island A increase and the number of species on island B decrease.
Number of species
Possible graph: 40 35 30 25 20 15 10 5 0
B A
1
4
7
10 13 16 Time (months)
19
22
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Activity 54.2 What effects can a disturbance have on a community? Disturbances rarely affect only one species of a community. More often disturbances that have direct effects on a single species produce a cascade of additional effects on other species. For questions 1 and 2, analyze the ecological situations described. Then answer the questions that follow each situation. 1. A particularly popular island vacation site is home to many species of orchids. The primary income generating businesses on the island are tourism and orchid sales. Disturbance: To make the island more attractive to visitors, a local politician suggested that the island be periodically “fogged” with insecticides. Response: This fogging reduced the insect numbers. It also appeared to reduce the number of birds and new growths of orchids. (Hint: Examine the ideas of limits of tolerance, coevolution, predation, and bioaccumulation.) a. Is it likely that the disturbance directly caused the response? Fogging with insecticides is likely to have directly reduced the number of insects. However, the reduction in the number of birds and new growths of orchids may not have been directly caused by the pesticides. b. What other factors might be involved? Many pesticides are fat-soluble. The amount of pesticide directly encountered by the birds may have been within their limits of tolerance. As insect eaters, however, they may have accumulated much more of the pesticide over time. Any one insect contains only a small amount of the pesticide, but over time, the birds consume many insects and the pesticide in each accumulates in the birds’ adipose (fat) tissues. Most orchids are pollinated by insects. General fogging for insect pests would also kill the insects that pollinate the orchids and result in less new growth in the orchid population. Many plants have coevolved with specific insect species. If their pollinator dies out, the orchid species dies out as well.
Activity 54.2
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c. How could you test different factors for their effect on the response? For example, what experiments could you set up? Different types of experiments could be set up. One possible experiment would test the effect of direct spraying on a population of birds that had no prior exposure to the pesticides. Groups of birds would need to be maintained in a facility that allows you to control what they eat and to observe their behavior. At the beginning of the experiment, all the birds would be directly sprayed with pesticides at a level consistent with possible exposure in the wild. For the remainder of the experiment, half of the birds would be fed insects that contained no pesticides. The other half would be fed insects sprayed with pesticide. Immediately before spraying and immediately after spraying, you would measure the level of pesticides in specific tissues in the birds. You would continue to measure the levels of pesticide in these tissues at set time intervals. The behavior and levels of pesticides in both sets of captive birds could be compared to each other and to those of birds in the wild. d. What would you expect to find if the other factors you proposed affected the response? If direct spraying caused the deaths in the birds, you would expect to see equal death rates in both populations of captive birds. If bioaccumulation caused the deaths, you would expect to see higher numbers of deaths in the birds fed insects sprayed with pesticide. You would also expect to see a relatively constant level of pesticide in the tissues of the birds fed unsprayed insects and increasing levels of pesticide in the tissues of birds fed sprayed insects. When you compared the levels of pesticides in the tissues, you would expect to see a strong correlation between the levels of pesticides in the birds fed sprayed insects and in the birds from the wild. 2. Australia and New Zealand are home to a very wide range of marsupials (for example, kangaroos and other pouched mammals). Until colonization by foreign traders and other developments, placental mammals were not found in these areas. Disturbance: Following colonization, the rabbit was introduced to Australia. Response: The rabbit multiplied rapidly and ultimately became a pest species, doing considerable damage to both crop and natural plants. (Hint: Examine the ideas of competition, competitive exclusion, predation, and coevolution.) a. Is it likely that the disturbance directly caused the response? The introduction of the rabbit is the most likely disturbance. The response is the considerable damage to crops and natural plants caused by the uncontrolled rabbit populations. There is a direct cause-and-effect relationship between the introduction of the rabbit and the rabbits’ effects on crops and other plants. 366
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b. What other factors might be involved? In areas where the rabbit is a native species, it is generally not a pest because the rabbit has coevolved with its predators (including disease organisms), its food organisms, and its competitors. The situation is different when a species is artificially introduced into a new habitat. In the new habitat, there may be no predators or disease organisms to keep the introduced species’ population under control. The introduced species may be a better competitor for available food sources, so it is likely to become a pest. c. How could you test different factors for their effect on the response? For example, what experiments could you set up? The government of Australia tested the hypothesis that lack of disease organisms was a central cause for the explosion in the rabbit population. In 1950, the government imported rabbit myxovirus and infected the wild populations. Significant decreases in the rabbit population were reported. As a result, there were significant increases in the populations of not only native plants but also native animals such as the kangaroo that depend on the native plants. (More recently, rabbit calcivirus, RCD, has been used to keep some of the more resistant rabbit populations under control.) d. What would you expect to find if the other factors you proposed affected the response? If lack of disease organisms was a factor in the overpopulation of rabbits, you would expect that the introduction of disease organisms would significantly reduce the population (which did occur). 3. In the boreal forest of Canada, wildfires are important disturbance factors. A single wild fire seldom burns the whole forest. Instead it burns large patches or stands and leaves others untouched. Following a wildfire in a black spruce forest, there is usually a predictable regrowth of the vegetation, starting with ground lichens and small spruce seedlings. As the spruce trees grow and form a closed-crown canopy, feather mosses (Bryophytes) are found in an increasing proportion on the forest floor. In some cases, the peat moss Sphagnum outcompetes the feather mosses and eventually dominates the ground cover. Because wildfires occur naturally about every 10 years, a forest stand can sometimes burn before Sphagnum dominance is reached, and the whole process repeats. a. Graph “Tree Biomass vs. Time” over a 100-year period: i. at the stand scale (the stand is a particular part of the forest that burned) and ii. at the landscape scale (composed of many forest stands). • Assume, at the stand scale, that when a stand burns, all trees in that stand die. • Assume, at the landscape scale, that there is one fire every 10 years, and that each fire burns a different stand. Activity 54.1
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Tree biomass
Tree biomass
The graphs should look something like this (X-axix is approx 100 yrs):
Landscape scale
Stand scale
b. Explain how (and why) the graphs differ. In the scenario presented, even though there are large variations in tree biomass at the stand scale, the variations are damped at the landscape scale, and at the landscape scale the overall biomass remains fairly constant.
Activity 54.3 How can distance from the mainland and island size affect species richness? 1. Island biogeography theory attempts to explain the patterns of species richness and turnover on islands as a function of the size of the island and its distance from the mainland. a. What are the basic tenets of this theory? (Refer to Figure 54.27 in Biology, 8th edition.) The theory states that species richness (number of species) on islands is determined by both the size of the island and its distance from a mainland (source of species). Among the basic tenets are the following: Larger islands tend to have higher immigration rates and lower extinction rates than smaller islands. As a result, larger islands have more species than smaller islands. If two islands are the same size, they have equal rates of extinction. However, the one closer to the mainland has a higher rate of immigration and therefore a larger number of species. b. In the graph below, curves A and A’ represent the historic immigration and extinction rates, respectively, for an island off the coast of South America. Given these data, what is the equilibrium number of species for this island?
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The equilibrium species number would be approximately 8. Island Biogeography 18 Rate of immigration or extinction
16 14 A A' B C D E
12 10 8 6 4 2 0 1
2
3
4
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8
9 10 11 12 13 14 15 16 17 18
Number of species on island
c. If a land bridge or an isthmus forms and connects the island to the mainland, which two curves would best represent the resulting immigration and extinction rates for the island? Line C would represent the new rate of immigration and line A’ would remain the rate of extinction. Rate of extinction is a factor of island size (and therefore of habitat availability); as a result it is generally not affected by distance from the mainland. d. What would be the new equilibrium number of species? The new equilibrium species number would be between 10 and 11. 2. A research scientist comes back from a study of a group of islands in the South Pacific. He says that his data show that in some cases, the smaller islands he investigated had higher numbers of lizard species on them than did the larger islands. He suggests these data indicate that we should reexamine the whole theory of island biogeography. a. What additional information would you like to have about the group of islands he studied?
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Some of the questions that could be asked include the following. • How far are the islands from the mainland? • What types of habitat are available on the islands? • How did the scientist determine how many species each island had? b. Is there any way his data could be accounted for using the current theory of island biogeography? Explain your reasoning. Given the theory of island biogeography, it is possible that a smaller island may have more species than a larger island if the smaller island is much closer to a mainland than the larger island is.
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Notes to Instructors Chapter 55 Ecosystems What is the focus of this activity? In Activity 53.2, students compared exponential growth and logistic growth models. On one hand, the exponential growth model assumes no limits on the availability of resources. On the other hand, the logistic growth model assumes that each population has a carrying capacity, an upper limit on its population size that is determined by its environment. In Activity 55.1, students investigate various factors that can affect a population’s carrying capacity.
What is this particular activity designed to do? Activity 55.1 What limits do available solar radiation and nutrients place on carrying capacities? This activity is designed to help students understand basic energy relationships in ecosystems, including how the availability of energy can limit population growth.
Answers Activity 55.1 What limits do available solar radiation and nutrients place on carrying capacities? 1. Energy transformations in a community can be diagrammatically represented as a trophic structure. a. Diagram a simple trophic structure for a grassland community. Figure 55.10 on page 1,229 of Biology, 8th edition, shows an idealized pyramid of net production and may be used as an example of the trophic structure for a grassland community. b. At what trophic level(s) would humans belong in your diagram? Explain. Humans, in general, have an omnivorous diet. Much like the mouse in Figure 55.10, humans consume both other animals and plants. Humans who are strictly vegetarians and eat only food derived from plants would be classified as primary consumers. However, if part of their diet is from plants and part from animals, which are primary consumers, then humans are classified as secondary consumers. (This demonstrates one of the difficulties associated Notes to Instructors
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with using very simple models. Not all organisms fit neatly into only one of the categories.) 2. Energy is lost at every step in the trophic structure where energy is transferred—for example, from primary producers to primary consumers. It is estimated that only 1% of the solar energy striking plants is converted to chemical energy, only 50–90% of gross primary productivity (GPP) becomes net primary productivity (NPP), and only about 10% of the energy available at each tropic level transfers to the next level (as biomass). a. Restate the preceding information in your own words. Be sure to explain what GPP and NPP are and how they relate to the percentage of energy transfers from one trophic level to another. Gross primary productivity (GPP) is a measure of the total amount of light energy used in photosynthesis per unit time. It is usually expressed in joules per square meter per year (J/m2/yr). Net primary productivity (NPP) equals GPP minus the energy the photosynthetic organisms use for cellular respiration. NPP can similarly be expressed in J/m2/yr, or it can be expressed as the total biomass of primary producers added per year to an area. Biomass is usually expressed in g/m2/yr. (The assumption is that each gram of biomass contains a given amount of energy, which can be released by oxidation.) The information given in the question states that, of the sunlight that reaches plants, only 1% is converted to chemical energy. This is a measure of GPP. Only 50 to 90% of this 1%—that is, 0.5 to 0.9%—of the energy that reaches plants becomes actual biomass, or NPP. If all the NPP were consumed by primary consumers, you would expect to see 10% of it end up in biomass of secondary consumers. That is, 0.05 to 0.09% of the sunlight that reaches plants is actually converted into biomass of primary consumers. b. Where does the “lost energy” go? Is it correct to say that energy is lost in these transfers? If not, what would be a better way of expressing this? The energy that is “lost” in these transfers is energy used by the organisms in their own metabolism. For example, a primary consumer needs to eat 10 g of plant food to gain 1 g of biomass. The other 9 g of plant food are oxidized (via cellular respiration) to produce the ATP needed to build more biomass and to maintain the life functions of the organism. 3. Biology, 8th edition, indicates that each day the earth is “bombarded by about 1022 joules of solar radiation. (1 J = 0.239 cal).” It has also been estimated that primary producers on Earth collectively create about 170 billion tons of organic material per year. Do the following calculation to determine how these two values compare. 372
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a. Convert 1022 joules/day into kilocalories per year. 1022 J/day 0.239 cal/J = 0.239 1022 cal/day = 2.39 1021 cal/day = 2.39 1018 kcal/day 365 days/yr = 8.7 1023 kcal/yr b. Convert 170 billion tons of organic material into kilocalories. To do this, assume all of the 170 billion tons is glucose. Also, assume tons are metric tons. (Note: 180 grams of glucose when burned in a calorimeter gives off 686 Kcal of energy.) 170 109 metric tons/yr 1,000 g/kg 1,000 kg/metric ton 686 kcal/mole = 170 1015 g/yr of biomass 1 mole of glucose/180 g = 9.4 1014 moles/yr 686 kcal/mole = 6.5 1017 kcal/yr c. Given your answers to parts a and b, what percentage of the total incoming solar energy is captured as biomass (glucose)? 6.5 1017 kcal/yr = 7 10 5 = 0.00007 = 0.007% 8.7 1023 kcal/yr d. Measurements of photosynthetic conversion of the energy in sunlight to biomass indicate that at best, plants can convert about 75% of the energy they absorb as sunlight into biomass. The other 25% is used to support metabolism. What could account for the apparently low efficiency of photosynthesis relative to total incoming solar energy that you calculated in part c? Much of Earth is covered by water. In fact, only about 29% of Earth is land. Of that land, only about 11% is arable (suitable for agriculture). Some of the arable lands have been paved over (as streets, for example) or covered with buildings and are not available for agriculture. In addition, most of the arable land is in the temperate regions, where crops can grow only part of the year. This alone reduces the amount of biomass produced by at least 50%. Other factors that can affect productivity include variations in weather patterns from year to year, which can produce droughts or floods that cause crop loss and therefore loss of the energy stored in them. 4. If we can calculate the earth’s total primary productivity, we can use this value to develop an estimate of the total number of humans Earth can support.
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a. How much energy in kilocalories per year would it take to support the Earth’s current human population if all individuals weighed about 75 kg and each required 2,000 kcal per day? Assume 6 109 humans currently inhabit Earth. 2,000 kcal/day/human 365 days/yr = 7.3 105 kcal/yr/human 6 109 humans on Earth = 4.4 1015 kcal/yr consumed by the human population b. How does the estimate of the energy contained in the 170 billion tons of organic material (from part b in question 3) compare to the amount of energy required to support the current human population (from part a in question 4)? If we accept the assumptions made in part a, then: 4.4 1015 kcal/yr = 6.8 101 6.5 1017 kcal/yr or 68% of the total biomass is consumed by humans. c. Is it reasonable to assume that all of the primary productivity on Earth is available to support humans? If not, what else do you need to consider? The plant biomass on Earth ultimately supports or nourishes all the animals, the fungi, and many of the prokaryotes. If all the humans alive today were consuming 2,000 kcal per day, only 32% of the total biomass would be available for the other organisms. 5. You have been monitoring the net primary productivity (NPP) of a grassland area for several years. Over the years, NPP increased initially and then leveled off. You suspect that the availability of a nutrient is limiting productivity. a. Design an experiment to determine whether there is a limiting nutrient. To test whether a particular nutrient is limiting NPP, you could subdivide the area into a number of smaller plots. Treat some plots by adding more of the nutrient, and leave others untreated. Then measure the NPP produced on each type of plot. If the nutrient is limiting, you should find higher NPPs on treated plots than on untreated plots. You can repeat the experiment with a number of different nutrients. b. Given your design, what results would you expect if only one of the nutrients you test is limiting? If your test plots are each set up to test addition of a single nutrient, you would expect to see an increase in growth only on the plot supplemented with the limiting nutrient.
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c. What results would you expect if more than one of the nutrients are limiting? Would this affect your design in any way? If your test plots are each set up to test addition of a single nutrient, you would likely see no difference in growth across all plots if more than one nutrient is limiting. At this point you could start testing combinations of nutrients or other factors (see d below). d. What factor(s) other than nutrient limitation might cause NPP to level off? Among other things, plant growth can be limited by water availability, physical or chemical conditions related to the density of plants per unit area, and the presence or absence of symbiotic mycorrhizal fungi.
55.1 Test Your Understanding 1 to 3. Two species of hawks are found at various times of the year in the United States. Hawk species A eats mice and squirrels (both of which eat seeds). Hawk species B eats snakes that feed on mice and squirrels. Given what you know about trophic structures, if other characteristics of the two hawk species are similar, which of the following are likely to be true? Explain your reasoning. A ⫽ Likely to be true. B ⫽ Not likely to be true. 1. For a given feeding area, the number of hawk species A that could be supported is likely to be smaller than the number of hawk species B that could be supported. B—Not likely to be true. Species A, feeding on primary consumers, will have on average a 10-fold greater food base available than species B, which feeds on snakes, which are predators and thus secondary consumers. 2. Reduction in the available primary productivity in the feeding area would tend to have a lesser effect on the hawk species A than on hawk species B. B—Not likely to be true. Assume primary productivity is reduced from 100 kg of biomass to 50 kg. Prior to the loss, the feeding area would support 10 kg of primary consumers (e.g., mice and squirrels). That 10 kg would be used to support 1 kg of secondary consumers (e.g., snakes and hawk A). Therefore, less than 0.1 kg of hawk B would be supported. The reduction of primary productivity biomass to 50 kg would support 5 kg of primary consumers (on average) and 0.5 kg of snakes and hawk A. As a result, less than 0.05 kg of hawk B could be supported. In other words, the percent reduction for both populations would be about the same.
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3. If the seeds consumed by the mice and squirrels in this feeding area were contaminated with heavy metals, you would expect to find much lower concentrations of heavy metals in hawk species A than in hawk species B. A—Likely to be true. Hawk A would need to eat 10 g of mice and squirrels for every g of hawk mass added. Hawk B would have to eat 10 g of snake that ate 100 g of mice and squirrels for every g of hawk mass added. As a result, heavy metals, which are not metabolized but accumulate through the food chain, would be found in much higher concentrations in hawk species B.
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Notes to Instructors Chapter 56 Conservation Biology and Restoration Ecology What is the focus of this activity? Many species of both plants and animals are rapidly going extinct as a result of habitat reduction or loss. Efforts to conserve individual species and species diversity require an understanding of the species’ physiology and behavior as well as its community ecology, evolution, and population genetics.
What is this particular activity designed to do? Activity 56.1 What factors can affect the survival of a species or community? Students are asked to consider and integrate a wide range of principles to address questions in conservation biology.
Answers Activity 56.1 What factors can affect the survival of a species or community? Making decisions to preserve communities requires an understanding and integration of many factors. Assume you work for the U.S. government and manage a large national forest. You are told that to maintain the economy in the area, the government has agreed to allow foresters to remove half a million acres of trees from a million-acre parcel. This parcel is almost square. You have asked your staff to do an analysis of two possible methods for implementing the plan: Proposal I: Split the million acres into two parcels of a half million acres each and allow the foresters to harvest all trees on one of these parcels. Proposal II: Divide the million-acre tract into 50 parcels of 10,000 acres each and allow the foresters to cut half the trees in each parcel. 1. List some of the ecological advantages and disadvantages of each proposal. (Note: Each proposal has many possible advantages and disadvantages. The following are just a few of these.)
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Proposal I Advantages
Proposal II
Disadvantages
Advantages
Disadvantages
• Fewer roads would • The regeneration • All cut areas would • There would be be needed. rate for trees would be closer to seed more roads and • There would be be slower because sources, and therefore more less edge habitat seed sources are natural edge and more relative to interior more distant. regeneration would disturbance. habitat. • More soil nutrients be faster. • The result may be • Interior species would be lost from • Fewer soil invasion of more may be favored the land. nutrients would be exotic species. more. • Complete loss of lost due to runoff. • Replanting efforts • Harvesting and some communities • Disturbed animal would be more replanting would could result if species may be diffuse and less be more focused animal or plant able to move to efficient. and therefore could communities were adjacent uncut occur more distributed areas. efficiently with heterogeneously. • Species that roam less disturbance to over a wide the other halfgeographic range million acres. may benefit.
2. Given the following characteristics of various animal species described below, which of the forestry-cutting proposals would be more likely to ensure continued success of each animal species? Explain your answers. a. C. arnivora is a secondary and occasionally a tertiary consumer or carnivore. Behaviorally, it roams over about 20 square miles of “home range” in search of food—for example, rabbits (herbivores) and foxes (carnivores). One square mile contains 2.8 107 square feet; 20 square miles contains 56 107 square feet; 1 acre contains 2.6 105 square feet. Therefore, a 1 million-acre forest contains 26 1010 square feet, or 26 1010/2.8 107, or about 9.3 103 square miles. As a result, this area could support as many as 4.6 102 C. arnivora (if these were the only carnivores). Half of the forest (proposal I) contains about 4.6 103 square miles. This area could support a maximum of 2.3 102 C. arnivora. The half of the forest left in proposal II is patchy. Although patches would be contiguous, the patchiness of the environment may have an effect on its desirability to predators.
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In general, the effects of reducing the forest size on this species depend not only on the absolute area remaining but also on the size of the existing population, the minimum number of individuals required to maintain the population long term, the response of the carnivore to disturbance, and other factors. b. R. odentia is a small rodentlike herbivore that exists in small numbers in the forest. Its preferred food and habitat are found along the edges of the forest and are composed primarily of herbaceous (nonwoody) annual plants which produce tender shoots in the spring and plentiful seeds later in the year. Proposal II would produce more edge habitat than proposal I. If the success of this species is directly related to the amount of edge habitat, then it would be more successful if proposal II were adopted. c. P. redatoria is a predatory bird which feeds on small rodents and occasionally on snakes and other reptiles. It can range over large distances looking for food. It nests in hollows that occur naturally in a particular species of tree. These hollows are not found in trees under 20 years of age but are commonly found in trees 40 years and older. If we assume that 40-year-old trees are randomly distributed throughout the forest, then both proposals would eliminate about half of these trees. In other words, both proposals would reduce the available nesting sites for this species by half. The immediate effect on the population would depend on the existing population size. If the number of P. redatoria in the population and the number of available nest sites are currently about equal, then half of the P. redatoria would lose their nest sites. They would not be able to reproduce and would eventually die without leaving offspring. If the P. redatoria population is much smaller than the number of available nesting sites, then there may be little immediate effect on the population. However, if this plan is repeated in 20 years and the other half of the forest is cut, the P. redatoria species would have no nesting sites left and would disappear in this region. 3. Conservation biologists have debated extensively which is better: many, small reserves or a few large ones. a. What types of factors should be considered in making judgments about size and location of reserves? What is better will vary from species to species. Some species roam over wide ranges and may do better if many small reserves are connected into one large one. Some species can survive in only old-growth forest areas that have appropriate nesting sites. Other species are much more successful in open areas or edge habitat. In other words, the requirements of the species to be preserved
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should be of primary importance in making decisions about the size and location of reserves. b. Some ecologists argue that we should be concerned about preserving the largest number of species. Others argue that we should be most concerned with saving those species judged to be of unusual importance. Develop an argument to support one of these viewpoints. Your argument should be based on our existing understanding of biology in general and ecology in particular. Many logical arguments can be proposed. In developing your argument, you should consider the habitat and energy (food) requirements of the plant and animal species as well as any behavioral requirements of the species. 4. Based on your answers to questions 1–3, which of the proposals for removing half a million acres from the national forest in question 1 would you recommend? Explain your reasoning. Again, a number of possible arguments can be developed to support either proposal I or proposal II. Your argument should include a comparison of the advantages and disadvantages of each proposal as they relate to both the plant and animal communities in the forest.
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Appendix A An Introduction to Data Analysis and Graphing for a PC This appendix provides a quick introduction to the use of some of the data analysis and graphing capabilities of one of the more commonly used computer programs, Excel. This and other similar software systems make it much easier to graph and analyze large data sets. 1. TO SET UP EXCEL FOR DATA ANALYSIS AND GRAPHING • Open Excel and activate the “Data Analysis” tool. 䊊 Under the “Tools” pull-down menu, look for “Data Analysis.” 䊊 If you don’t see “Data Analysis,” look for “Add-Ins.” Click on “Data Analysis” or “Add-Ins” and select “Analysis ToolPack”; then click “OK.” 䊊 If you did a custom install instead of a complete install of Excel, you may not find “Analysis ToolPack” under “Add-Ins.” If this is the case, go to your original installation CD to add this feature. • Enter the data to be analyzed or graphed in columns on the Excel spreadsheet. 䊊 If you want to graph height versus arm length, enter the data in two columns with the data you want to appear on the y-axis (independent variable) of a graph in the right-hand column. 䊊 The data to appear on the x-axis (dependent variable) should be in the adjacent left-hand column. 䊊 An example is on the next page. Note: If you are using Excel 2007 to find and activate Add-ins: • Click on the “Office” button in the upper left-hand corner. • Then click on the “Excel options” button at the bottom of the pop-up window. • Select “Add-ins” from the left-hand column. • Select/highlight “Analysis ToolPak” in the list of Inactive Application Add-ins, then click “Go.” • Check the box in front of “Analysis Tool Pak” in the window that appears and click “OK.” • Now click on the “Data” tab at the top of the screen. You should see a new “Data Analysis” pull-down menu at the right top of the screen.
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A Gender
B Arm Length (cm)
C Height (cm)
f m m f f m m m
73.7 76.7 71.1 66 68.6 76.7 78.7 81.3
170.2 177.8 170.2 165.1 172.7 172.7 177.8 182.9
2. TO SORT THE DATA • In the upper left-hand corner of the Excel spreadsheet, click the box between Row 1 and Column A. The whole table should be highlighted when you do this. • Pull down the “Data” menu at the top of the screen and click “Sort.” • If you have headers or titles on your columns, click the 䊊 in front of “Header row.” • Select the column you want to sort by (e.g., Gender). • Select either “Ascending” if you want the sort to go from low to high (or A to Z) or “Descending” for high to low (or Z to A). • Click “OK.” 3. TO GET DESCRIPTIVE/SUMMARY STATISTICS FOR THE DATA a. Pull down the “Tools” menu at the top of the Excel spreadsheet and click on “Data Analysis.” b. Click on “Descriptive Statistics” and then “OK.” • In the window that appears, for input range click on the icon to the right of the “Input Range:” box. • Highlight (select) the data in the Excel spreadsheet that you want analyzed. For example, select all of the arm length data (males plus females). • Click on the chart icon next to “Input Range” again to return to the Descriptive Statistics window. • To indicate where to place the output, click on the 䊊 in front of “Output Range:” and then click on the chart icon to the right of the “Output Range:” box. 䊊 To select the location for the data to appear, click on any open box in the Excel spreadsheet. 䊊 Click the chart icon next to “Output Range” again to return to the Descriptive Statistics window. • In the Descriptive Statistics window select: 䊐 Summary Statistics 䊐 Confidence Level for Mean • Click “OK.” A box containing the summary statistics will appear. For example: 382
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Definitions:
Arm Length Mean
74.1
Standard Error
The average of all the values.
1.849421
An estimate of the standard deviation of the sample mean.
Median
75.2
The number that lies in the middle of the set of numbers.
Mode
76.7
The most frequent number in the set.
Standard Deviation
5.230952
The square root of the variance.
Sample Variance
27.36286
A measure of how far from the average the data points are.
Kurtosis
–0.99518
A measure of how peaked or flat the distribution is.
Skewness
–0.29163
A measure of the degree of symmetry of the distribution.
Range
15.3
Minimum
66
Maximum
81.3
Sum
592.8
Count
8
The difference between the lowest and highest values in the set. The smallest value in the set. The largest value in the set. The total of all values in the set. The total number of values in the set.
Two data sets may have the same mean but very different amounts of variability. To indicate the amount of variability that exists in a given set of data, the arithmetic mean for those data is often expressed as the mean ⫾ the standard deviation. Here, the smaller the standard deviation is, the smaller the variance is in the treatment group. The smaller the variance is, the more confidence we have that the data in our sample set represent a good estimate of the values for the larger population. 4. TO GRAPH THE DATA a. To make a linear graph of x vs. y values: • Click on the “Chart Wizard” icon at the top of the Excel spreadsheet. • Click on “XY (Scatter).” • Click “Next.” • Click the chart icon to the left of “Data range:” 䊊 Select all of the data in the two columns (Arm length [cm] and Height [cm]). Appendix A
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Note: If your data are not in side-by-side columns with the y data in the right-hand column, you can click on the “Series” tab instead of the “Data range:” tab. Click “Add” and then enter the data separately for the x- and y-axes by highlighting the columns separately here. Click “Next” and select the “Titles” tab to add titles for the chart and the x- and y-axes. Click “Next” and select where you want the chart to appear (on the same sheet or a new sheet). Then click “Finish.” On the graph that appears, click on one of the data points to highlight it, then right click (on PC computers) and select “Add trendline.” 䊊 Under the “Type” tab, select “Linear.” 䊊 Under the “Options” tab, select: 䊐 “Display equation on chart” and “Display R-squared value on chart.” 䊐 Click “OK.” For the data in the previous table you should get the following graph. 䊊
• • •
•
Arm Length vs Height y = 0.9409x 103.95 R2 = 0.7753
185
Height (cm)
180 175
Height (cm) Linear (Height (cm))
170 165 160 0
20
40
60
80
100
Arm Length (cm)
The formula y ⫽ 0.9409x ⫹ 103.95 is in the format y ⫽ mx ⫹ b. When the value of m is positive it indicates that there is a direct relationship between x and y; in this case it indicates that as x increases, y increases. (If m were negative, it would indicate an inverse relationship—as x increases, y decreases.) For this specific graph, if it were reasonable to extrapolate the line to a point where x ⫽ 0, the line would intercept the y axis at b, or 103.95 cm. With this as the zero point for x on the line, the y values would increase from this point by m, or 0.9409 for every unit change in x. The R2 value has no units and a range of 0 to 1. An R2 value of zero means there is no linear relationship between x and y values. On the other hand, high values of R2 indicate a strong correlation between the x and y values. High R2 values are also a good indicator of statistical significance. 384
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b. To make a histogram of the data: • Select a set of data to use (e.g., heights for males only). (See previous section “To Sort the Data.”) • Make a new column of data containing heights increasing by 5 cm each—for example, 160, 165, 170, 175, 180, 185 cm. (This is called your Bin Range.) • Go to the “Tools” pull-down menu and click on “Data Analysis,” then on “Histogram.” Then click “OK.” 䊊 Click on the chart icon to the right of the “Input Range:” box. 䊐 Select/highlight the height data for males only (include a column heading). 䊐 Click the chart icon again to return to the Histogram window. 䊊 Click on the chart icon to the right of the “Bin Range:” box. 䊐 Select/highlight the column containing heights increasing by 5 cm each. 䊐 Click on the chart icon again to return to the Histogram window. 䊊 Click on the 䊊 in front of “Output Range:” and then click on the chart icon to the right of the “Output Range:” box. 䊐 To select the location for the data to appear, click on any open box in the Excel spreadsheet. 䊐 Click the chart icon again to return to the Histogram window. 䊊 In the Histogram window select: 䊐 Chart Output 䊊 Click “OK.” The frequency data and histogram should appear.
Frequency
Histogram 3 2 1 0 160 165 170 175 180 185 More
Frequency
Histograms are often developed to help determine whether or not the data appear to be distributed normally. In this case, only five values (the five heights for males) are included—enough to determine how the heights among these five individuals are distributed but not enough to make any general statements about the total population of males. Note: If the graph of the data approximates a bell-shaped curve or normal distribution, the standard deviation takes on additional meaning. For normal data distributions, about 68% of the data points will fall within one standard deviation (on either side) of the mean. Ninety-five percent of the data will fall within two standard deviations of the mean; and 99% will fall within three standard deviations of the mean. Appendix A
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5. TO DETERMINE STATISTICAL SIGNIFICANCE The statistical significance of the difference in means can be determined by using a statistical test. The null hypothesis (H0) is generally assumed for such tests. A simple null hypothesis would state that there is no difference in x (some characteristic) between two populations (e.g., there is no difference in arm length between males and females). This null hypothesis further assumes that any differences observed between the two populations are the result of chance variation or sampling error alone. An alternative hypothesis (Ha) would be that there is a difference or relationship between the populations. The alternative hypothesis is supported when the null hypothesis is rejected. The Data Analysis function of Excel allows you to perform a number of different statistical tests, including t tests and analysis of variance (ANOVA). These tests are designed to show whether the observed difference in means, between groups or between control and test groups, is due to random chance/variability. A significance level of 0.05 or 5% means that statistically there is only a 5% chance that the difference observed is due to chance alone. A significance level of 0.05 is usually accepted as evidence that the treatment produced a significant effect. A significance level of 0.01, or 1%, is considered highly significant. It would mean that only 1% of the time would such a difference be likely to occur as a result of chance alone. In these cases, the null hypothesis is rejected.
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Appendix B An Introduction to Data Analysis and Graphing for a Mac This appendix provides a quick introduction to the use of some of the data analysis and graphing capabilities of one of the more commonly used computer programs, Excel. This and other similar software systems make it much easier to graph and analyze large data sets. 1. TO SET UP EXCEL FOR DATA ANALYSIS AND GRAPHING • Open Excel and activate the “Data Analysis” tool. 䊊 Under the “Tools” pull-down menu, look for “Data Analysis.” 䊊 If you don’t see “Data Analysis,” look for “Add-Ins.” Click on “Data Analysis” or “Add-Ins” and select “Analysis ToolPack”; then click “OK.” 䊊 If you did a custom install instead of a complete install of Excel, you may not find “Analysis ToolPack” under “Add-Ins.” If this is the case, go to your original installation CD to add this feature. • Enter the data to be analyzed or graphed in columns on the Excel spreadsheet. 䊊 If you want to graph height versus arm length, enter the data in two columns with the data you want to appear on the y-axis (independent variable) of a graph in the right-hand column. 䊊 The data to appear on the x-axis (dependent variable) should be in the adjacent left-hand column. 䊊 For example: A
B
C
Gender
Arm Length (cm)
Height (cm)
3
f
73.7
170.2
4
m
76.7
177.8
5
m
71.1
170.2
6
f
66
165.1
7
f
68.6
172.7
8
m
76.7
172.7
9
m
78.7
177.8
10
m
81.3
182.9
1 2
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2. TO SORT THE DATA • In the upper left-hand corner of the Excel spreadsheet, click the box between Row 1 and Column A. The whole table should be highlighted when you do this. • Pull down the “Data” menu at the top of the screen and click “Sort.” • If you have headers or titles on your columns, click the 䊊 in front of “Header row.” • Select the column you want to sort by (e.g., Gender). • Select either “Ascending” if you want the sort to go from low to high (or A to Z) or “Descending” for high to low (or Z to A). • Click “OK.” 3. TO GET DESCRIPTIVE/SUMMARY STATISTICS FOR THE DATA a. Pull down the “Tools” menu at the top of the Excel spreadsheet and click on “Data Analysis.” b. Click on “Descriptive Statistics” and then “OK.” • In the window that appears, for input range click on the icon to the right of the “Input Range:” box. • Highlight (select) the data in the Excel spreadsheet that you want analyzed. For example, select all of the arm length data (males plus females). • Click on the icon to the right of “Input Range” again to return to the Descriptive Statistics window. • To indicate where to place the output, click on the 䊊 in front of “Output Range:” and then click on the icon to the right of the “Output Range:” box. 䊊 To select the location for the data to appear, click on any open box in the Excel spreadsheet. 䊊 Click the icon next to “Output Range” again to return to the Descriptive Statistics window. • In the Descriptive Statistics window select: 䊐 Summary Statistics 䊐 Confidence Level for Mean • Click “OK.” A box containing the summary statistics will appear. For example:
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Definitions:
Arm Length Mean
74.1
The average of all the values.
1.849421
An estimate of the standard deviation of the sample mean.
Median
75.2
The number that lies in the middle of the set of numbers.
Mode
76.7
The most frequent number in the set.
Standard Error
Standard Deviation
5.230952
The square root of the variance.
Sample Variance
27.36286
Kurtosis
–0.99518
Skewness
–0.29163
A measure of how far from the average the data points are. A measure of how peaked or flat the distribution is. A measure of the degree of symmetry of the distribution. The difference between the lowest and highest values in the set.
Range
15.3
Minimum
66
Maximum
81.3
Sum
592.8
Count
8
The smallest value in the set. The largest value in the set. The total of all values in the set. The total number of values in the set.
Two data sets may have the same mean but very different amounts of variability. To indicate the amount of variability that exists in a given set of data, the arithmetic mean for those data is often expressed as the mean ⫾ the standard deviation. Here, the smaller the standard deviation is, the smaller the variance is in our set of results or data. The smaller the variance is, the more confidence we have that the data in our sample set represent a good estimate of the values for the larger population. 4. TO GRAPH THE DATA a. To make a linear graph of x vs. y values: • Click on the “Chart Wizard” icon at the top of the Excel spreadsheet. • Click on “XY (Scatter).” • Click “Next.” • Click the icon to the right of “Data range:” 䊊 Select all of the data in the two columns (Arm length [cm] and Height [cm]). 䊊 Note: If your data are not in side-by-side columns with the y data in the right-hand column, you can click on the “Series” tab at the top of the screen Appendix B
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that pops up (instead of the “Data range” tab). Click “Add” and then enter the data separately for the x- and y-axes by highlighting the columns separately here. Click “Next” and select the “Titles” tab to add titles for the chart and the x- and y-axes. Click “Next” and select where you want the chart to appear (on the same sheet or a new sheet). Then click “Finish.” On the graph that appears, click on one of the data points to highlight it, then click on the “Chart” pull-down menu at the top of the screen and select “Add trendline.” 䊊 Under the “Type” tab, select “Linear.” 䊊 Under the “Options” tab, select: 䊐 “Display equation on chart” and “Display R-squared value on chart.” 䊐 Click “OK.” For the data in the previous table you should get the following graph. Arm Length vs Height y = 0.9409x 103.95 R2 = 0.7753
185
Height (cm)
180 175
Height (cm) Linear (Height (cm)
170 165 160 0
20
40
60
80
100
Arm Length (cm)
The formula y ⫽ 0.9409x ⫹ 103.95 is in the format y ⫽ mx ⫹ b. When the value of m is positive it indicates that there is a direct relationship between x and y; in this case it indicates that as x increases, y increases. (If m were negative, it would indicate an inverse relationship—as x increases, y decreases.) For this specific graph, if it were reasonable to extrapolate the line to a point where x ⫽ 0, the line would intercept the y-axis at b, or 103.95 cm. With this as the zero point for x on the line, the y values would increase from this point by m, or 0.9409 for every unit change in x. The R2 value has no units and a range of 0 to 1. An R2 value of zero means there is no linear relationship between x and y values. On the other hand, high values of R2 indicate a strong correlation between the x and y values. High R2 values are also a good indicator of statistical significance.
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b. To make a histogram of the data: • Select a set of data to use (e.g., heights for males only). (See the previous section “To Sort the Data.”) • Make a new column of data containing heights increasing by 5 cm each for example, 160, 165, 170, 175, 180, 185 cm. (This is called your Bin Range.) • Go to the “Tools” pull-down menu and click on “Data Analysis,” then on “Histogram.” Then click “OK.” 䊊 Click on the icon to the right of the “Input Range:” box 䊐 Select/highlight the height data for males only (include a column heading). 䊐 Click the chart icon again to return to the Histogram window. 䊊 Click on the icon to the right of the “Bin Range:” box. 䊐 Select/highlight the column containing heights increasing by 5 cm each. 䊐 Click on the icon again to return to the Histogram window. 䊊 Click on the 䊊 in front of “Output Range:” and then click on the icon to the right of the “Output Range:” box. 䊐 To select the location for the data to appear, click on any open box in the Excel spreadsheet. 䊐 Click the icon again to return to the Histogram window. 䊊 In the Histogram window select: 䊐 Chart Output 䊊 Click “OK.” The frequency data and histogram should appear.
Frequency
Histogram 3 2 1 0 160 165 170 175 180 185 More
Frequency
Histograms are often developed to help determine whether or not the data appear to be distributed normally. In this case, only five values (the five heights for males) are included— enough to determine how the heights among these five individuals are distributed but not enough to make any general statements about the total population of males. Note: If the graph of a large data set approximates a bell-shaped curve or normal distribution, the standard deviation takes on additional meaning. For normal data distributions, about 68% of the data points will fall within one standard deviation (on either side) of the mean. Ninety-five percent of the data will fall within two standard deviations of the mean; and 99% will fall within three standard deviations of the mean.
Appendix B
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5. TO DETERMINE STATISTICAL SIGNIFICANCE The statistical significance of the difference in means can be determined by using a statistical test. The null hypothesis (H0) is generally assumed for such tests. A simple null hypothesis would state that there is no difference in x (some characteristic) between two populations (e.g., there is no difference in arm length between males and females). This null hypothesis further assumes that any differences observed between the two populations are the result of chance variation or sampling error alone. An alternative hypothesis (Ha) would be that there is a difference in arm length (or other characteristic) between the populations. The alternative hypothesis is supported when the null hypothesis is rejected. The Data Analysis function of Excel allows you to perform a number of different statistical tests, including t tests and analysis of variance (ANOVA). These tests are designed to show whether the observed difference in means, between groups or between control and test groups, is due to random chance/variability. A significance level of 0.05 or 5% means that statistically there is only a 5% chance that the difference observed is due to chance alone. A significance level of 0.05 is usually accepted as evidence that the treatment produced a significant effect. A significance level of 0.01, or 1%, is considered highly significant. It would mean that only 1% of the time would such a difference be likely to occur as a result of chance alone. In these cases, the null hypothesis is rejected.
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