How to Solve IBO Probelms_2

January 27, 2018 | Author: martynapet | Category: Photosynthesis, Kidney, Physiology, Biochemistry, Physical Sciences
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How to Solve IBO Probelms_2...

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Let’s continue with past paper of IBO 2012 Part A.

To answer this question you need to remember the difference between C3 and C4 plants. C3 plants are not prosperous in hot dry conditions because their photosynthetic efficiency is affected by a process called photorespiration. This problem of photorespiration is overcome in C4 plants by a two-stage strategy which keeps high CO2 concentrations and low oxygen concentrations in the bundle sheath cells where the Calvin cycle takes place. Therefore, we can conclude that a. is correct because of high CO2 assimilation at high temperatures. b. is incorrect as we stated above that C3 plants are not adapted to high temperatures. c. must be correct because if we chose the first graph for C4 plants, then the second graph is for C3 plants and we see that at temperature 10o net assimilation of CO2 is higher as compared to the first graph’s curve at 10o. Answers:

Another question concerned with photosynthesis:

You should remember the main steps of photosynthesis which consists of light reactions and dark reaction. The processes that occur during the light reactions are depicted below:

The main products of light reaction are ATP and NADPH as well as oxygen. During dark reactions Kalvin cycle takes place during which carbon dioxide is fixed, then reduced by NADPH followed by regeneration of RuBP with G3P as a by-product.

a. b.

c. d.

Having all this in mind, we can start tackling the problems: is right because light reaction involves the generation of ATP. is incorrect because oxygen is just a by-product and the essential initial step is excitation of P680 electrons which are transferred to the primary electron acceptor. You can come to the answer by remembering about photorespiration in which oxygen interferes with photosynthesis and thus is unwanted product of photosynthesis. in photosynthesis ATP is synthesized in linear electron flow involving cytochrome complex. ATP synthase complex resides in the inner membrane of mitochondrion, no chloroplast. Therefore, the answer is wrong. the linear electron flow proceeds from photosystem II to photosystem I and cyclic electron flow involves only photosystem I. Take a look at the diagram below. Therefore the statement is incorrect.

e. is correct because as you know light reactions takes place on thylakoid membrane but Kalvin cycle takes place in the stroma.

For such types of question you need to know some concepts because it is hard to make guesses if you do not need the steps of photosynthesis. The answers:

Let’s take a look at another question:

18.1 The key point in this part is to realize that the smaller the surface area per unit volume ratio, the bigger the organism (remember that cells are really small so that they have a big surface are to volume ratio to maintain high levels of exchange between intracellular and extracellular environments). Bigger organisms have smaller surface area for exchange and thereby their metabolic rate is quite sluggish, they do not lose the heat that rapidly and therefore do not need to compensate for the loss by having high heart or breathing rates. On the contrary, smaller organisms have bigger surface area and thus lose heat quicker and in order to maintain a constant temperature they have high metabolic rate reflected by higher breathing rate and heart rate. Having this in mind, you can rank organisms as follows: A will have the highest ratio of surface area to volume, followed by C, then B and finally D. 18.2 Total volume of blood will be highest in biggest organisms in order to provide nutrients and oxygen to more cells and will be lowest in small organisms due to small size. This means that the answer will be the reverse of 18.1: first would be D, followed by B, then C and finally A.

Let’s look at the human anatomy:

You should tick a. as a right one if you remember the renin-angiotensin-aldosterone system which involves a change in permeability of some ion channels in the kidneys as well as constriction of vessel in the kidneys. Not to mention reabsorption of water and sodium ions taking place in the nephron. b. is also right because of the effects of ADH and aldosterone which both increase water reabsorption in the blood concurrently increase blood volume. The c. is incorrect because the loop of Henle contains only ion channels in the ascending part and aquaporins only in descending part. Try to remember the diagram showed below to remind yourself of kidney’s functional capabilities.

In d. you should remember that hyperosmotic urine means that urine is rich in solutes and low in water. As loop of Henle helps to concentrate the urine by regulating NaCl and water reabsorption, thus directly affecting concentration of urine. The longer the loops, the more concentrated the urine could be because of large area for reabsorption. Therefore the statement in question d. is incorrect. Some knowledge of chemistry can help you with question e. If you remember, CO2 dissolves in aqueous intracellular fluid of erythrocytes and forms a weak H2CO3 acid which almost immediately dissociates into hydrogen ion (which alters pH) and bicarbonate ion which diffuses into the plasma. When bicarbonate reaches the lung alveoli, it combines with hydrogen ions and produce carbon dioxide which we expire. This reaction helps regulate pH and complements regulation of blood pH by kidneys which secrete hydrogen ions and reabsorb bicarbonate ions. Thus, e. is correct. I would choose f. as a right one, however, the answers below shows that it should be wrong. From the diagram above we can clearly see that hydrogen ions are secreted and bicarbonate ions are reabsorbed. Kidneys regulate pH by reducing excretion of hydrogen ions and thereby acidifying overly alkaline urine or they may increase reabsorption of bicarbonate in order to neutralise overly acidic urine. As for question g. you should remember that all acids produced in the body are non-volatile apart from carbonic acid, which is the sole volatile acid because it can dissociate into carbon dioxide and water. If you look at kidney’s diagram above you will see that kidneys do not excrete carbon dioxide which is volatile, but instead reabsorbs bicarbonate ions. Therefore, the answer is wrong. To come to the answer of h. you need to remember the diagram of the nephron and that acidosis means a decrease in pH of blood. Ammonia (NH3) is secreted by the cells in the proximal tubule which acts as a buffer to trap hydrogen ions producing ammonium ions (NH4+). The more acidic the filtrate is in the proximal tubule, the more ammonia is produced. So the statement is correct. The statement i. is correct if you remember that glomerular filtration is influenced by high blood pressure resulting from differences between diameter of afferent and efferent arterioles. Also try to remember the effect of angiotensin II which involves constriction of kidney vessels and thus decreasing filtration rate thereby conserving water. j. is a hard nut if you are not acquainted with hormones but you should remember that antidiuretic hormone and oxytocin are two hormones secreted by posterior pituitary. Thus, statement j. is incorrect.

The answers are:

The following question is interesting because it does not require previous knowledge:

Before starting answering the question you should realize that chewing time is directly related to the type of food, that is vegetarian food is hard to chew because it contains insoluble cellulose whereas meat is quite easy to chew. Also you should take into account the size of animal and its energy requirements. Let’s begin by classifying those four animals into two groups: horse and cattle are herbivores whereas wolf and human are carnivores (but humans are more omnivores nowadays). Thus, as plant food is harder to chew, herbivores need more saliva to degrade the material then carnivores. And as cattle are bigger than horse, they need more food for energy and therefore you have one part of the chain which is b  c. I would place humans in the category of 0.75 to 1.5, that is, they should secrete less then herbivores but more that carnivores due to some cellulose-rich food in the human diet. The least amount of saliva, to my mind, should be secreted by wolf which takes relatively easily digestible diet. In my opinion, the answers should be a 45°, is by drawing a small picture with vectors and applying law of tangents.

The number 1.1 means that D is higher than L, the same as in the first part, thus we can conclude that organism b. will have a higher angle than 45 degrees.

Now let’s look at the graphs to make correct conclusions. a. statement is true because if you look at the first graph you should notice that there are no points of Draco volans log weight beyond point 1.25, whereas log weight of Ptychozoon kuhli extends to 1.5. If you evaluate the lines in the first graph of two organisms you can see that they quite coincide. b. statement is incorrect because when you look at the second graph and compare y values for Ptychozoon kuhli and Draco Volans when x value is the same for both species, you see that the line of Draco volans is significantly higher than that of Ptychozoon kuhli. c. Statement is incorrect because when you look at graph 3, you can see that values for Ptychozoon kuhli are higher as compared to Draco volans values. d. is correct because you have to look at graph 4 and the values quite neatly coincides. e. is correct because you compare graphs 2 and 3 and see that Draco volans; has bigger patagium area whereas Ptychozoon kuhli has bigger accessory area. The answers are provided below:

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