Answers in Questions.pdf

1 IGCSE and GCSE Biology. Answers to questions Section 1. Some Principles of Biology. Chapters 1 - 5 Here you will find the answers to the ‘in-text’ questions which occur in ‘IGCSE Biology’ (2nd edition) and ‘GCSE Biology’ (3rd edition) by D.G. Mackean, published by Hodder Education, London, UK. The questions are not included. Chapter 1 Cells and tissues Page 4 1. a Cytoplasm, nucleus, cell membrane b Chloroplasts 2. The cell membrane controls substances entering or leaving the cell 3. The red cell has no nucleus 4. The cell membrane is formed from living cytoplasm; the cell wall is formed from non-living cellulose 5. The section must have been taken above the nucleus 6. a The magnification in Figure 1.1 is x 60, so presumably x100 would be effective b The wide part of the cell is 7mm. This is 700 times larger than the real cell, so the cell would measure 7 /700 i.e. 0.01 mm 7. Count the nuclei Page 7 1. (i) animal cells d, a (ii) plant cells d, a, b, c 2. (a) In the midrib and veins (sieve tubes) (b) Palisade cells, epidermis, guard cell, (a vessel is formed by many cells) Page 9 1. Lungs (organ), root hair (cell), mesophyll (tissue), multipolar neurone (cell) 2. (Fig.36.13) Bone and muscle, (Fig. 19.9) muscle, nerve tissue, bone 3. a (Organs) The definition on p.7 says an organ is ‘a structure with a special function’ so all the labelled structures in Fig. 11.5 could be organs. The exceptions might be the mouth, the pyloric sphincter, the rectum and the anus b (System) The digestive system

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Chapter 2 The chemicals of living cells Page 13 1. Proteins provide the basic materials (amino acids) for making cytoplasm which forms cells, tissues and organs, and for making enzymes which control all the reactions in the cell. 2. a proteins b carbohydrates and lipids. 3. a carbohydrates; sugar, starch, cellulose b salts; iron sulphate, sodium chloride c lipids; butter, olive oil. 4. Carbohydrates would need nitrogen. Plants get nitrogen in the form of nitrates in the soil. 5. Synthesis of starch and cellulose from glucose; synthesis of amino acids from glucose and salts; synthesis of glucose from carbon dioxide and water. Page 18 1. Only d applies to all catalysts. 2. a The rate would increase. Most chemical reactions are speeded up by a rise in temperature. b The rate would fall to zero. Temperatures over 50oC denature most enzymes. 3. Yes. Enzymes are essential for all the processes taking place inside a cell. 4. Dipping the apple into boiling water denatures the enzymes responsible for producing the brown discolouration. 5. The enzyme needs to combine briefly with the molecules of the two reactants, glycine and valine. The same enzyme may still combine with the glycine molecule but will not ‘fit’ with the serine molecule.

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Chapter 3 Energy from respiration Page 20 1. a energy b Respiration takes place in all the living cells of an organism. 2. a Aerobic respiration needs oxygen; anaerobic respiration does not. Anaerobic respiration does not completely oxidise the substance it acts on. Anaerobic respiration produces less energy than aerobic respiration b 2830 - 118 = 2712 kJ 3. a A food substance and oxygen. b enzymes. c carbon dioxide and water. 4. Respiration is a chemical process in all living cells. Resuscitation restores the process of breathing. 5. You need more oxygen to oxidise the lactic acid that has accumulated during vigorous exercise. An increased heart rate delivers the oxygen more rapidly. 6. The 2400 kJ used in 8 hours sleep represents basal metabolism. Page 24 1. a True (of most organisms) b The organism is more likely to be producing heat c True d True in most cases (some micro-organisms may use only anaerobic respiration) e No. Respiration results in a loss of weight. f Movement cannot occur without respiration but respiration does not necessarily result in movement e.g. the Plant Kingdom. 2. The purpose of the soda lime is to absorb carbon dioxide so that any volume change can be attributed to uptake of oxygen. The lime water is used to detect the presence of carbon dioxide. 3. a In the absence of soda lime there will be a 2cm3 increase in volume. b If soda lime is present there will be a decrease in volume of 5cm3. 4. In a beaker, the heat generated will be lost to the atmosphere. In a vacuum flask the heat is retained. 5. The boiled wheat will be dead and should not produce carbon dioxide. If carbon dioxide appears, either it comes from something else, e.g. bacteria or the experiment is invalid. Page 25 1. In the first experiment, the water was boiled to drive out all the dissolved oxygen and the liquid paraffin prevented any further oxygen dissolving in the water. Therefore production of carbon dioxide must have been from an anaerobic reaction. In the second experiment, the presence of oxygen did not affect the anaerobic reaction but it would have been impossible to decide whether the respiration was aerobic or anaerobic. 2. d 3. In aerobic respiration, the oxidation of carbon in C6H12O6 produces carbon dioxide and releases energy. In a similar way the sulphur bacteria could be oxidising H2S to produce H2SO4 with a corresponding release of energy. The hypothesis could be tested by excluding oxygen from the bacteria. There should be no production of H2SO4.

4 4. a (i) 64.3 x 134 = 8616.2 kJ (ii) 0.018 x 2736 = 49.3 kJ b The mouse c The body temperatures of these animals are very similar. Therefore you would not expect the energy expenditure to make up for heat loss per m2 to be very different. Chapter 4 How substances get in and out of cells Page 28 1. The cells on the left are in a region of high oxygen concentration and, initially there is little or no oxygen in the red cells. The concentration gradient favours the passage of oxygen into the cells. The cells on the right are in a region of a lower oxygen concentration than exists in the red cells. The diffusion gradient, therefore, favours the diffusion of oxygen out of the cells. 2. a Air freshly breathed in contains more oxygen that the red cells in the capillary. Oxygen will consequently diffuse into the cells. Red cells returning to the lungs contain higher concentrations of carbon dioxide than the air in the alveolus, so carbon dioxide diffuses out. b If the blood flow increased, the exchange of oxygen and carbon dioxide would also increase. 3. Controlled diffusion, endo- and exocytosis, active transport 4. In air, the uptake of phosphate goes on at a steady rate for about 20 hours. In nitrogen, (i.e. in the absence of oxygen) the rate of uptake is very low. This result suggests that phosphate uptake is by active transport, a process which needs oxygen to generate the necessary energy. Page 31 1. Water will diffuse from the 5 percent solution into the 10 percent solution. 2. The cell membrane in a living cell controls the movement of substances (in this case a red pigment) into or out of the cell. Boiling kills the cell by denaturing its enzymes, and the cell membrane loses its ability to control the loss of pigment. 3. If animal cells were studied in water, they would take up water by osmosis till they burst. The concentration of Ringer’s solution is the same as the cell’s cytoplasm so there is no osmosis. 4. The molecules (or ions) of the dissolved substance attract water molecules, leaving fewer ‘free’ water molecules in the solution. The accumulation of sugar in the plant cell will lower its water potential and it will absorb water by osmosis from its neighbours. This may harm the cell. Starch in the plastids is insoluble and so does not affect the cell’s water potential. Page 34 1. a If a ‘stronger’ (more concentrated) solution was placed in to cellulose tube you would expect the rate of osmosis to increase. This would accelerate the rise of liquid in the capillary tube. b So long as the solution in the beaker is weaker (less concentrated) than the solution in the cellulose tube, osmosis will still occur and the water column will rise but more slowly. c If the sugar solution were in the beaker, water would leave the cellulose tube and the water column would go down. 2. Water would continue to enter the cellulose tube until it burst. 3. The explanation is given in the caption. When the cell is immersed in water, the water potential of the cell sap is lower than that of the water. So water will enter via the selectively permeable cell membrane and expand the vacuole till it fills the cell.

5 4. Put the starch solution in the beaker and the iodine solution in the cellulose tube. If the ‘in but not out’ theory is correct you would expect the starch solution in the cellulose tube to go blue and the iodine solution to remain brown. If the ‘molecular size’ theory is correct you would expect the starch solution to turn blue but not the iodine solution in the cellulose tube. (In practice this is what happens but since the blue colour first appears on the outside surface of the cellulose tube, it looks as if the tube contents have turned blue). 5 When the pressure of the water column equals the ‘osmotic pressure’ in the cellulose tube, the flow will cease (‘Osmotic pressure’ is the difference between the water potentials of the sugar solution in the cellulose tube and the water surrounding it).

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Chapter 5 Photosynthesis and nutrition in plants Page 38 1. a carbon dioxide, d water, e chlorophyll, g light. 2. The white parts of the leaf do not contain chlorophyll and act as a control. 3. ‘Destarching’ a leaf means taking steps to ensure that it does not contain starch at the beginning of the experiment. If the leaves already contain starch it cannot be demonstrated that a particular process is necessary for the production of starch. 4. a Soda lime absorbs carbon dioxide and so the leaf is deprived of this gas. b Sodium hydrogen carbonate in solution decomposes to form carbon dioxide. This ensures that the plant has an adequate supply of this gas. c The polythene bag prevents carbon dioxide in the air from reaching the plant. 5. By using pondweed it is possible to see the bubbles of gas escaping from the leaves. The choice of pondweed might lead to the claim that production of oxygen in light happened only in pondweed. 6. Depriving a plant of water would lead to wilting and, ultimately, to the death of the plant irrespective of the role of water in photosynthesis. 7. Yes. Destarching is achieved by leaving a plant in darkness for 2 or three days for starch to be removed from the leaves. It is assumed that no new starch will be formed in darkness. Page 39 1. Carbon dioxide from the air. Water from the soil. 2. a The palisade cells. They have the greatest number of chloroplasts and are nearest the light source. b The spongy mesophyll cells. They have fewer chloroplasts and are further from the light source. c The cells of the epidermis. They have no chloroplasts 3. a The energy for photosynthesis comes from sunlight. b Respiration is the source of energy for all other living processes. Page 41 1. a In bright sunlight a leaf would be taking in carbon dioxide and giving out oxygen. b In darkness a leaf would be taking in oxygen and giving out carbon dioxide. 2. No. It may just mean that respiration is taking place faster than photosynthesis. 3. You would need to gradually increase the light intensity, (e.g. by moving the light source closer to the tube), until the indicator just changed colour. Ideally then you could use a light meter to measure this intensity. With the high light intensity in a field, you would expect plants to reach compensation point quickly. In the low light intensity of a wood you might expect plants to take to take longer to reach compensation point. Page 44 1. a carbon dioxide and water. Chlorophyll must be present. b nitrate ions. Enzymes must be present. 2. The carbon dioxide molecule would be combined with water to make glucose. Some of this glucose might be respired, releasing the carbon dioxide molecule.

7 3. a Raising the temperature will increase the rate of reactions in photosynthesis. Increasing the light intensity will speed up photosynthesis and so will an increase in the carbon dioxide concentration b It depends on what the limiting factors are. It would not be cost effective, for example, to increase carbon dioxide concentration if a low light intensity is limiting the rate of photosynthesis. 4. In darkness, from midnight to 4.0 a.m. only respiration will be taking place and so the carbon dioxide concentration will be increasing. From 4.0 a.m. to 8 a.m. photosynthesis will be starting up as the light intensity increases and so more and more carbon dioxide is used by the crop. From 8.0 a.m. to 4.0 p.m. photosynthesis will be at its peak, using up carbon dioxide. As the light intensity diminishes after 4.0 p.m. photosynthesis slows down while respiration stays the same so that more carbon dioxide is released than is retained. Page 46 1. Magnesium sulphate, potassium phosphate and a soluble calcium salt, e.g. calcium chloride. 2. a Calcium and phosphorus. b iron. 3. The floating plant needs carbon dioxide and water for photosynthesis and its roots absorb water directly from the pond. 4. a Without nitrates, the plant cannot make the proteins which are needed to make the cytoplasm required for growth. b Phosphates are used to make DNA, which is needed for cell division (growth). 5. a The yield is increased from about 6 tonnes per hectare to 7.2 tonnes. A gain of 1.2 tonnes per hectare. b The yield increases by about 1.5 tonnes per hectare (8.7 - 7.2). c Increasing the applied nitrogen from 150 to 300 kg per hectare would increase the yield by only about 0.6 tonnes per hectare. The farmer needs to work out whether this is less than the cost of the extra fertiliser. © D. G. Mackean 2011 www.biology-resources.com

1 IGCSE and GCSE Biology. Answers to questions Section 2. Flowering Plants. Chapters 6 - 9 Chapter 6 Plant structure and function Page 54 1. a Epidermis. Helps maintain shape, reduces evaporation, resists entry of bacteria and fungi. b Mesophyll. Photosynthesis. 2. No chloroplasts are present. 3. a Oxygen and water vapour. b Carbon dioxide. 4. Grasses and plants such as daffodil, iris. (See Fig 30.20 on p. 278). 5. Closure of the stomata prevents excessive loss of water but has the disadvantage that it also prevents intake of carbon dioxide for photosynthesis. Page 56 1. Vessels (xylem), sieve tubes (phloem), fibres. 2. The vascular bundles, particularly the vessels and lignified cells. The epidermis helps to maintain shape. (See Experiment 2 on p. 58). 3. Xylem consists of vessels. Lignified tubes formed from dead cells joined end to end. Phloem consists of living cells, also joined end to end but with a continuity of living cytoplasm. They are not lignified. The function of xylem is to conduct water from roots to leaves. The function of phloem is to conduct food. The direction depends on the needs of the plant. Page 57 1. Xylem conducts water and mineral ions from the roots to the rest of the plant. Palisade cells make glucose by photosynthesis. The root hair absorbs water and minerals from the soil. The root cap protects the growing point of the root from damage as it grows through the soil. A stoma controls the passage of water vapour from the leaf to the atmosphere. It also allows the passage of carbon dioxide and oxygen into or out of the leaf. The epidermis helps maintain the shape of the leaf or stem, reduces evaporation and protects the inner tissues from bacteria and fungi. 2. A stem is likely to be green and to have buds. A root will be white with no buds but, possibly, lateral roots. With the aid of a hand lens or microscope you would be able to see that the stem’s vascular bundles were in a circle just beneath the epidermis. The root’s vascular bundle would be in the centre. 3. a Stoma, air space, mesophyll cell. b Root hair cell, xylem vessel, midrib, leaf ‘vein’. mesophyll cell. 4. It root hairs appeared on the part of the root that was growing, (extending), they would be rubbed off as the root pushed through the soil. 5. A petal is a living structure and needs a supply of water, mineral ions and food, so you would expect vascular bundles to be present.

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Chapter 7 Transport in plants Page 60 1.The water molecule enters the root hair cell, passes through the root cortex to reach a vessel in the vascular bundle. It is carried up the vessel in the stem by the transpiration stream and enters a leaf via the midrib. It passes into the xylem of a leaf vein and enters a leaf cell by osmosis. It later diffuses through the cell wall into an air space and escapes as water vapour by diffusion through a stoma. Page 61 1. A hot, dry (low humidity) climate with persistent wind will favour transpiration. 2. The leaves would wilt i.e. lose their turgidity and droop. 3. In sunlight the stomata will be open, allowing CO2 to enter the leaves but also allowing water loss by evaporation. The sunlight will also warm the leaf and increase the rate of evaporation. 4. The transpiration stream will carry mineral ions (‘mineral salts’) as well as water. 5. The stem and flowers may also transpire. Page 62 1. a Vessels carry water and mineral ions (‘salts’). Sieve tubes carry food, (e.g. glucose). b Vessels carry water and salts one way from the roots to the leaves, flowers and fruits. Sieve tubes carry food from the leaves to any part of the plant which is using or storing food. 2. The inside layers of the bark include the phloem. If this layer is removed, the food made in the leaves cannot reach the roots which consequently die, causing the death of the entire tree. 3. The roots; all the tissues in the shoot which do not contain chlorophyll (cortex, phloem, epidermis etc.); flowers, fruits and seeds. Page 63 1. The root hairs would lose water by osmosis to the soil and would cease to function. Eventually the plant would wilt and die. 2. Active transport needs energy from respiration. A waterlogged soil will contain too little oxygen for adequate respiration. 3. A leafless tree offers very little surface for transpiration to take place. A small amount of water vapour may escape through openings in the bark (lenticels). (See p. 64). Page 66 1. The plant and the water in the beaker lose 15 grams (275-260) in 2 hours. Of this, 3 grams is due to evaporation of water directly from the beaker. So, the plant was responsible for 12 grams due to transpiration. The rate of transpiration is, therefore, 6 grams per hour. 2. a The moisture from the fingers would change the colour of the cobalt chloride paper. b The ‘Sellotape’ should prevent water in the atmosphere from reaching the cobalt chloride paper. 3. You would not expect air to escape from the stomata because there is no connection between the vessels and the air spaces in the leaf. If you cut away the top half of the leaves you would expect air bubbles to appear from the cut ends of the vascular bundles, which contain the xylem vessels.

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Chapter 8 Reproduction in flowering plants Page 69 1. Calyx, petals, stamens, carpels. 2. Colour, scent, nectar. 3. a Ovule b Ovary. Page 71 1. e anthers split. d bee visits young flower. a dusted with pollen. c bee visits older flower. b pollen deposited on stigma. 2. Apple, cherry, horse chestnut (They all have flowers with conspicuous petals). 3. a Only a large insect, such as a bee, can push between the upper and lower petals to reach the nectary. b The nectaries are at the end of a long spur. Only insects with long ‘tongues’ can reach them. 4. There seems no reason why insects should not also become adapted to pollination of certain flowers. You might expect the adaptations to include the development of longer mouthparts which reach distant nectaries, a bristly thorax to which pollen can stick, increased sensitivity to specific colours and scents, more acute vision. 5. Insect pollinated Wind pollinated White or coloured petals. Small green petals. Scented. Not scented. Nectar produced. No nectar. Stamens and stigma wholly or partly Stamens and stigma exposed to the air. protected or concealed by petals. Spiky pollen grains* Light, smooth pollen grains. *Not described in the text Page 72 1. a The stamens. b The ovaries. 2. Pollination is the transfer of pollen from the anthers to the stigma. Fertilisation is the fusion of male gametes from the pollen with the female gametes in the ovule. 3. a Yes. b No. 4. a The ovary forms the fruit. b The petals, the stamens, the stigma and style fall off after fertilisation. c The calyx remains attached. 5. a Fruits. Runner beans, grapes, marrows, tomatoes. b Seeds. Peas. baked beans. c Neither. Rhubarb.

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Page 74 1. The advantages of dispersal are (a) that the seedlings are less likely to compete with each other if the seeds are widely dispersed, (b) that new areas may be colonised. The disadvantage is that a large proportion of seeds may land in situations where they cannot grow. 2. Seeds light enough to be carried on the wind are likely to be carried long distances before they fall to the ground. Seeds carried by animals are likely to be carried long distances depending on the normal range of the animal. Birds are likely to carry the seeds for the greatest distances. 3. Wind pollination refers to the transport in the air of pollen grains produced by the stamens of one plant to the stigma of another flower of the same species. Although the pollen grains may be carried anywhere, pollination does not occur unless the pollen reaches the stigma. Wind dispersal involves the random distribution of a plant’s seeds in air currents until they fall to the ground where they may germinate. Page 76 1. a The function of the radicle is to produce a root which will anchor the plant in the soil and absorb water and minerals. b The function of the plumule is to a produce a shoot with a stem and leaves which will make food by photosynthesis. c The function of the cotyledons is to provide food for the germinating seedlings. In some cases, the cotyledons protect the plumule as it grows through the soil. 2. This question is erroneous. The book deals with the French bean and not the broad bean. In the case of the French bean, a the plumule is protected by being enclosed between the cotyledons during its passage through the soil. b The radicle is protected by the root cap. 3. The seedling will use the stored food to provide energy from respiration to drive all the processes involved in germination. The food provides the raw materials (proteins, carbohydrates. lipids), for construction of new cytoplasm, new cells and new tissues, i.e. growth. 4. Once the first leaves appear in the seedling and start to produce food by photosynthesis, the seedling will become less and less dependent on the stored food. 5. The radicle will start to absorb water from the soil. Until the plumule has a supply of water to expand the cells, (by osmosis), it cannot extend and grow. Page 77 1. Water, oxygen and a suitable temperature. 2. The question should read ‘…experiments 2-4….’ In Experiment 4, light was excluded from all the seeds but only the ones in the refrigerator failed to germinate properly. Light, therefore, was not necessary for germination. 3. This should refer to Experiment 4 a Remove the seeds from the refrigerator and leave them at room temperature. If they have not been ‘killed’ by the low temperature, they should begin to germinate. b In theory, you could set up controlled environments in which the temperatures ranged from 0o to 15o C. You would have to decide how long to leave the seeds before interpreting the results.

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Page 78 1. Warm, moist, well aerated soil. By digging the soil, aeration and drainage are improved. The seeds must be kept watered. Chapter 9 Asexual reproduction and cloning in plants Page 80 1. Shoots have buds which can sprout to form new shoots. 2. Rhizomes grow below soil level and are unaffected by fire.

© D. G. Mackean www.biology-resources.com

1 IGCSE and GCSE Biology. Answers to questions Section 3. Human physiology. Chapter 10 Food and diet Page 87 1. a Milk, cheese, eggs, beans. b Peas, beans, cereals (e.g. bread). 2. Proteins are needed to make cytoplasm and all the tissues of the body. They also make the enzymes needed for chemical changes in the cells. 3. In theory you could survive on proteins and lipids. Both these foodstuffs can be converted into energy which is normally provided by carbohydrates. You might be disadvantaged by a lack of dietary fibre and vitamin C (See p. 88). 4.When fats are oxidised they produce a large amount of energy, some of which is released in the form of heat. A layer of fat under the skin may have insulating properties and reduce loss of heat. 5. a Proteins contain nitrogen. Lipids do not. b Proteins have about half the energy value of lipids. c The principal use of proteins is to provide the amino acids needed for building other proteins in the cytoplasm of the cells which form the tissues of the body. They are also used to produce enzymes. Lipids are a means of storing energy. 6. protein amino acids

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structural proteins

enzymes

1. a Red blood cells b All tissues c Bones and teeth d All cells e Thyroid gland. 2. (top left). Milk and its products such as cheese and yoghurt are good sources of proteins, lipids and calcium. (top right) Most will contain some carbohydrate (e.g. banana), all will contain vitamins (e.g. tomato and orange, vitamin C. Green vegetables, vitamins A and C. Most of them will provide dietary fibre. (bottom left) All will provide protein. The mackerel will contain lipids. The meat will provide iron. (bottom right) Mainly carbohydrate, some B vitamins and dietary fibre. 3. Green vegetables such as cabbage and lettuce contain vitamin A and also dietary fibre. 4. A diet consisting of only one source of food is likely to be deficient in one or more essential minerals or vitamins. 5. Only a is likely to give meaningful results. The added Vitamin C of b and c is unlikely to demonstrate a need for Vitamin C even if the rabbits show some benefit. In d you would not know whether the outcome was due to lack of Vitamin C or some other property of the food. In fact, rabbits do not appear to need vitamin C, so the results of a confirm this.

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Page 93 1. Proteins, carbohydrates and lipids, vitamins, mineral salts and water, a source of dietary fibre. 2. Protein cannot be stored in the body. If there is a large intake of protein, the body will take what it needs to build tissues and make enzymes and the remainder will be oxidised to provide energy which is a wasteful use of an important nutrient. 3. For all three, an adequate intake of protein is essential plus the calcium needed for the milk in the breast-feeding mother and the bone formation in the embryo and the growing child. 4. a An apple has far less carbohydrate than a chocolate bar so it is unlikely to cause weight gain. b A chocolate bar contains much more carbohydrate than an apple and will better meet your energy needs. 5. The fried potato (‘chips’) contains a good deal of fat which gives about twice the energy of the carbohydrate in boiled potato. 6. A high fibre diet will make you feel ‘full’ without taking in an excess of energy-rich food which can be converted to fat. 7. Hard physical work needs a great deal of energy which can be obtained best from carbohydrates and lipids. 8. See ‘Growing children’ on p.91. The text should have stated that the figures are ‘per day’ From these figures you would calculate that the baby needs 1.53 x 5 = 7.62 g of protein per day though there is little point in going to 2 decimal places. In fact 7.6 g seems too low for a 6 month baby. The text needs to be revised. 9. During sleep, only basal metabolism is taking place. If 8 hours sleep needs 2400 kJ, 24 hours basal metabolism would need 3 x 2400 = 72000 kJ.

Chapter 11 Digestion, absorption and use of food. Page 98 1. The three functions shown are storage, digestion and absorption. 2. a The pancreas pours pancreatic juice into the duodenum. b The salivary glands secrete saliva into the mouth. 3. Epithelium, circular muscle, longitudinal muscle. Page 101 1. The airways are cut off. The soft palate closes the nasal cavity; the tongue, the epiglottis and muscles in the glottis, seal the top of the windpipe. 2. The solid food has to be broken down to a liquid from which the nutrients can be absorbed through the gut wall. Plants build their food from carbon dioxide water and minerals. They do not take in solids. 3. a Starch is digested in the mouth, the duodenum and the small intestine. b Protein is digested in the stomach, the duodenum and the small intestine. 4. Pepsin acts on only one kind of substance (protein) and works best at a particular pH.

3 Page 103 1. a Starch is digested to glucose. b Proteins are digested to amino acids. c Fats are digested to fatty acids and glycerol. 2. The small intestine is long; it is lined with vast numbers of villi. Both these factors give the small intestine a large absorbing surface. The epithelium is thin, which allows rapid diffusion or active uptake of digested food. There are numerous capillaries to carry off the digestion products. 3. In the stomach the enzyme, pepsin, breaks the protein down to form peptides. In the duodenum and small intestine the peptides are broken down further, by enzymes, to form amino acids. The amino acids are absorbed through the intestinal lining to reach blood vessels which join up to form the hepatic portal vein. This vein carries the amino acids to the liver which makes them into new proteins or alters their composition so that they can be used for energy production. 4. In the duodenum the starch molecules are broken down to glucose. In all the cells of the body, glucose is broken down to carbon dioxide and water by respiration. Page 105 1. The liver controls the glucose concentration of the blood by changing excess glucose to glycogen. It controls the amino acid level by converting the excess into substances which can be oxidised to provide energy. By detoxifying potentially harmful substances that have been absorbed in the small intestine, the liver prevents their entry into the bloodstream. The liver prevents harmful levels of haemoglobin building up by removing and storing the iron from the haemoglobin molecule. 2. The liver makes bile, which helps the digestion of fats in the duodenum and intestine. 3. A large fluctuation in the concentration of solutes such as glucose could upset the osmotic equilibrium of cells. If hormones were allowed to build up they could adversely affect the ways cells function. Large variations in plasma proteins could affect the clotting processes.

Page 107 1. Enzymes are proteins. When proteins are heated above about 50oC they are denatured and lose their shape. Boiling destroys the activity of an enzyme and this makes it a good experimental control. 2. The cloudiness is due to solid particles of egg white. When they are digested to soluble substances, the cloudiness disappears. 3. You could prepare a series of test tubes as in tube C and put them in controlled temperature water baths from say, 5oC to 35oC to see which one cleared first. 4. The variables are (a) the presence or absence of hydrochloric acid (tubes A and C), and (b) the presence or absence of unboiled pepsin (tubes C and D). (a) Tube A could be the control; (b) tube D could be the control. 5. Starch is a carbohydrate. It could not be an enzyme. A knowledge of the molecular structure of starch (p.12) reveals that it is made up of a chain of glucose molecules so it is the more likely source of glucose. 6. This question depends on personal data, so no general answer can be given.

4 Chapter 12 The blood circulatory system Page 109 1. a White cells can vary their shape. They have a nucleus. They do not contain haemoglobin. b They are an important part of the immune system They can produce antibodies which combat foreign substances which get into the bloodstream. Some of them can ingest bacteria or damaged cells and other unwanted particles. White cells do not carry oxygen. 2. The lungs. 3. In all the living, respiring cells of the body. 4. If oxyhaemoglobin was a stable compound it would not readily break down to release its oxygen where needed. If a diet is deficient in iron an adequate supply of haemoglobin cannot be produced. Cellular respiration would be reduced and the person would be anaemic. Page 111 1. a Both ventricles pump blood into the arteries. b The bicuspid, tricuspid and semi-lunar valves prevent blood flowing the wrong way. 2. c Atria contract. e Blood enters ventricles. b Ventricles contract. g Tri- and bicuspid valves close. a Blood enters arteries. f Semi-lunar valves close. d Ventricles relax. (You could start with d). 3. a The ventricles have to pump blood all round the body. The atria have only to pump blood into the ventricles. b The left ventricle has to pump blood all round the body (apart from the lungs). The right ventricle has to pump blood only to the lungs. 4. The pulmonary veins are not shown. 5. If the heart valves do not function properly, some of the blood in the arteries can flow back into the heart, so less blood is delivered to the body. This means that less oxygen reaches the muscle cells and this impairs vigorous activity. Page 114 1. a Left Atrium. i Left ventricle. c Aorta. b Vena cava. f Right atrium. h Right ventricle. e Pulmonary artery. d Lungs. g Pulmonary vein. 2. The pulmonary artery carries deoxygenated blood. The pulmonary vein carries oxygenated blood.

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3. a Veins return blood to the heart. Arteries carry blood away from the heart. b Veins are wider than arteries, less elastic with thinner walls and less muscle tissue. Some of them have valves in their linings. 4. a Capillaries have ‘walls’ only one cell thick compared with the thick, muscular walls of arteries and the thinner walls of veins. Capillaries are much smaller than veins or arteries. The capillary walls allow exchange of oxygen, carbon dioxide and digested food substances with the tissues. b Capillaries penetrate all the tissues of the body and supply them with food and oxygen. Arteries deliver blood to the capillaries and veins collect blood from the capillaries but do not exchange substances with the tissues. 5. Blood pressure is needed to circulate blood round the body. It is a normal function of a healthy blood circulatory system. Usually when people say they suffer from ‘blood pressure’ they mean a level of pressure which exceeds the normal range, i.e. ‘high blood pressure’. Page 115 1. Lymph consists of water, plasma proteins, salts (as ions), white blood cells and antibodies. The lymphatics leaving the alimentary canal may contain lipid droplets. 2. The fat molecule has not been digested to fatty acids and glycerol so it will enter a lacteal in a villus rather than a capillary. The lacteals empty their contents into lymphatic vessels which eventually join up to form a lymphatic duct. The duct empties its contents into the left subclavian vein which joins the vena cava before entering the left side of the heart. The left ventricle will pump blood round the body and some of it will reach the liver in the hepatic artery. 3. B lymphocytes (‘memory cells’) are retained in the lymph nodes. These cells produce antibodies which attack bacteria and other harmful cells. The spleen produces lymphocytes and antibodies. It removes bacteria from the blood. These are all immunological reactions. Page116 1. (i) Kidneys (ii) Lungs (iii) Active muscle

a gain carbon dioxide oxygen carbon dioxide

b lose oxygen, glucose, water, urea, excess salts glucose, carbon dioxide, water vapour oxygen, glucose

Page 120 1. The phagocytes ingest harmful bacteria, the lymphocytes produce antibodies which act against these bacteria. 2. The inoculation promotes the production of antibodies against a disease. The body is ready to ‘fight’ the disease organisms when they arrive. It is too late to do this once the disease organisms are present. The body will start making its own antibodies but it will take time for them to build up to an effective level. 3. a Examples of diseases controlled by active immunity are measles, mumps, rubella, diphtheria and flu. b An example of a disease controlled by passive immunity is tetanus. 4. The ‘universal donor’ is a group O person. The red cells have neither A or B antigens on their surface and cannot be clumped by anti-A or anti-B antibodies. 5. If a person’s blood is clumped by the anti-B serum he could be group B or group AB. If the blood fails to clump in anti-A serum, he cannot be group AB and must be Group B.

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Page 122 1. a Maintain a good level of regular exercise, reduce your stress levels. b Don’t smoke, avoid an excess of fatty foods. 2. a If 95% of patients needing leg amputation are smokers there is clearly a correlation. b Smoking cannot cause leg amputation but it might lead to conditions in which amputation becomes necessary. c Smoking is one of the causes of atheroma. If the atheroma occurs in the leg and cannot be treated, it may necessitate amputation. (In fact there are many ways of treating atheroma in the leg. Amputation is necessary only in extreme cases). 3. a Vigorous exercise increases the demand for oxygen and glucose for the higher rates of respiration. These are met by increasing the blood flow which delivers these two substance more rapidly to the muscles. An increased respiration rate in active muscles produces CO2 which is removed by the faster blood flow. The arterioles supplying the muscle will widen and so increase the amount of blood reaching the muscle. Increased ventilation in the lungs will hasten the supply of oxygen and the removal of excess CO2. b Stored glycogen in the muscles and liver will be converted to glucose. Lipids will be released. The body temperature will rise leading to vasodilation in the skin and sweating (See p. 138). Levels of pyruvic and lactic acids may rise (See p. 20). c The raised heart rate will increase the blood flow to all parts of the body, not just the muscles. 4. The first injection stimulates the lymphocytes in the immune system to produce antibodies specific to the antigen in the vaccine. Some of the B lymphocytes are ‘memory’ cells. When the second injection is received, the memory cells reproduce very rapidly and raise the level of antibodies. © D. G. Mackean November 2011 www.biology-resources.com

1 IGCSE and GCSE Biology. Answers to questions Section 3. Human physiology. Chapter 13 Breathing

Page 123 1. Nasal cavity, trachea, bronchus, alveolus. 2. The absorbing surfaces are very large; thousands of alveoli, thousands of villi. The epithelium lining of both organs is very thin, consisting of a single-cell layer. Both organs are richly supplied with a network of capillaries which carry off digested food or allow rapid transport of oxygen. Page 126 1. The diaphragm muscle contracts and so do the external intercostal muscles. 2. External intercostal muscles contract, ribs rise, thorax expands, lungs expand, air enters lungs. 3. The alveoli would expand most. The bronchioles would expand very little. Page 127 1. Respiration is the production of energy from glucose. Ventilation is the process which refreshes the air supply in the lungs. Gaseous exchange refers to the output of carbon dioxide and intake of oxygen which takes place across the epithelium of the lungs. Ventilation ensures a constant replenishment of oxygen in the lungs and the removal of carbon dioxide. Gaseous exchange allows the oxygen to reach the blood vessels in the lungs and the carbon dioxide to leave. The oxygen absorbed by gaseous exchange reaches the cells and allows the production of energy in respiration. 2. The oxygen molecule is carried to the alveoli via the trachea, bronchus and bronchiole. It diffuses through the alveolar lining to reach a capillary where it combines with the haemoglobin in the red blood cell. The oxygenated blood is then carried from the lung to the right atrium of the heart in the pulmonary vein. 3. The direction of diffusion depends on the relative concentrations of the gases on each side of the alveolar lining. The oxygen concentration in the lungs is greater than that in the blood, so oxygen diffuses from the alveolar air into the capillaries. Conversely, the carbon dioxide concentration in the blood is greater than that in the alveoli. So CO2 diffuses out of the blood into the alveoli. 4. Exhaled air still contains 16% oxygen which is quite enough to oxygenate the blood. 5. a At rest, you breathe about 16 times a minute and exchange 500cm3 with each breath. This amounts to 16 x 500 = 80,000 cm3 (80 litres). b Similarly, during exercise, the breathing rate can be 20 - 30 times a minute and exchange 30 + 5 litres. So you might exchange between 20 x 35 = 700 and 30 x 35 = 1050 litres. Page 129 1. a The immediate effects on the breathing system are constriction of the bronchioles, cessation of the ciliary movement and over-production of mucus. b The long-term effects are ‘smokers’ cough’, breakdown of the alveolar walls (emphysema), and chronic bronchitis.

2 2. The breakdown of the alveolar walls leads to a reduced surface for the absorption of oxygen. This means the breathing rate has to increase up to the point where it no longer meets the demands of the body for oxygen. 3. If you smoke 20 cigarettes a day, you are 13 times more likely to suffer from lung cancer than is a non-smoker. 4. Bronchitis, emphysema, heart disease, atheroma in the leg arteries, strokes, bladder cancer, gastric and duodenal ulcers, gum disease, tooth decay and tuberculosis.

Chapter 14 Excretion and the kidneys Page 131 1. Carbon dioxide, urea, uric acid, water, salts, spent hormones, toxins, bile pigments. 2. Levels of urea, uric acid and breakdown products of hormones will give a clue as to how well the kidneys are working. Urine analysis will reveal substances that should not be there such as glucose, proteins, blood cells or drugs. These substances may lead to a diagnosis of a disease. (This information is not given in the text). Page 134 1. It is the pressure of the blood which forces blood plasma (minus proteins) out of the capillaries of the glomerulus and into the renal capsule. With a fall in blood pressure, less fluid reaches the renal capsule. As a result, nitrogenous waste and other harmful substances remain circulating in the blood. 2. The renal vein will contain less urea, uric acid, salts, water and harmful substances than the renal artery. It will also contain less oxygen and more carbon dioxide. 3. Filtration; glomerulus. Reabsorption; renal tubule. Storage of urine; bladder. Transport of urine; ureter. Osmoregulation; collecting duct*. (*Not mentioned in text. ‘Renal tubule’ would be acceptable).

4. Loss of water during sweating causes an increase in the concentration of the blood so less water is reabsorbed by the kidney. This leads to a reduction in the volume of urine produced. This smaller volume still has to contain all the excretory products and so it becomes darker. When sweating ceases, there is more excess water to be removed by the kidneys, so urine production increases and dilutes the excretory products with the result that the urine is almost colourless. 5. The molecule of urea leaves the liver in the hepatic vein which joins the vena cava. The vena cava delivers its blood to the right atrium of the heart. The urea molecule then passes from the right atrium to the right ventricle and enters the pulmonary artery and, after passing through the lungs, enters the pulmonary vein which opens into the left atrium. It is pumped out by the left ventricle into the aorta which gives off a branch, the renal artery, which delivers the urea molecule to the kidneys. The molecule is filtered out of a glomerulus into a renal capsule which passes the urea molecule down the renal tubule to the pelvis of the kidney. From here the molecule passes down a ureter to the bladder, and to the outside world in the urine. 6. This question is flawed. The result does not give an answer compatible with the known figures. Apologies.

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Page 136 1. a The kidneys. b The liver. c The skin. d The lungs. 2. a The dialysis machine resembles the kidney in that it filters unwanted substances through a membrane. b It differs from the kidney in the way that it prevents the loss of useful substances. In the kidney these are reabsorbed into the blood as they pass down the renal tubule. In the dialysis machine they are prevented from escaping by maintaining a solution outside the tube which carries the same concentration of these substances as in the blood, so that there is no diffusion gradient.

Chapter 15 The skin, and temperature control Page 139 1. Principally; put on more or warmer clothing; take shelter in a warm environment. Minor actions; hot drinks, use external sources of heat, vigorous exercise. 2. Heat is one of the forms of energy produced by respiration both in active muscles and all other tissues. The heat is distributed round the body by the blood circulatory system. 3. Gains Losses Respiration Conduction External sources e.g. sunlight Convection to the environment central heating, hot food and drink Radiation Evaporation 4. a The hairs b Vasoconstriction, cessation of sweating. 5. Your ‘core’ temperature changes very little. You feel cold when sensory organs in the skin detect and respond to heat loss. Similarly you feel hot when other receptors in the skin detect and respond to heat gain. These receptors send impulses to the brain which interprets them as ‘feeling cold’ or ‘feeling hot’. 6. a Sweat will not evaporate if the air round the body is very (a) humid, (b) still. b Sweat will evaporate more rapidly if (a) the humidity is very low and (b) if there is air movement to carry the water vapour away. 7. Negative feedback triggers reactions which counteract the changes and restore the system to its steady state. Positive feed back would assist the changes and lead to an ever-increasing change in the system.

Chapter 16 Human reproduction Page 142 1. Sperm cells are much smaller than ova. They have much less cytoplasm but they do have a long tail. 2. Epididymis, sperm duct, seminal vesicle and prostate gland, urethra. 3. Kidney, pelvic girdle, erectile tissue, foreskin, scrotum, rectum. 4. A zygote can give rise to any of the tissues in the body and can grow into a complete organism.

4 Page 144 1. a Ovulating once per month, a woman with a reproductive life of 37 years might release 12 x 37 = 444 ova. b In the developed world only two or three ova are likely to be fertilized. It could be many more. In the developing world, the numbers could be 5 or more. 2. Vagina, cervix, uterus, oviduct. 3. At the moment of fertilization, the male nucleus (from the sperm cell) fuses with the female nucleus (of the ovum). 4. a Sperms retain the ability to fertilize an ovum for two or three days. So, fertilization is possible if mating occurs 2 days before ovulation. b If the ovum survives for only 24 hours, mating 2 days after ovulation is unlikely to lead to fertilization. Page 146 1. The umbilical vein will contain less oxygen, glucose, amino acids and salts, and more carbon dioxide and urea (nitrogenous waste) than the umbilical artery. 2. The embryo receives all the oxygen it needs from the blood reaching it from the umbilical artery (via the placenta). 3. If the twin boys are formed from a single fertilized ovum they will be identical twins. If they arise from two, separate fertilized ova, they will be fraternal twins. Page 149 1. a Developing manual skills, animal and crop husbandry, adapting to seasonal changes but also reading and mathematics will be essential in modern farming. b Reading and mathematics will be essential as well as manual skills. 2. The onset of ovulation means that the woman could conceive a child. Widening of the hips gives more space and support for a developing embryo. Enlargement of the uterus is in preparation for the developing embryo. The enlargement of the breasts is a result of the development of the mammary glands in preparation for breast-feeding. 3. Menstruation results from a breakdown of the uterine epithelium as a result of a failure of fertilization. If fertilization has taken place, the uterine lining will be retained and developed.

Chapter 17 The skeleton, muscles and movement Page 153 1. Upper arm bone; humerus. Upper leg bone; femur. Hip bone; pelvic girdle. Breastbone; sternum. ‘Backbone’; vertebral column, (not labelled). Lower arm bones; radius and ulna. Page 154 1. Jaw bone/skull, fingers, toes, ribs/spinal column. (These are ‘like’ hinge joints in their action but may be given other names). 2. The skeleton protects vital organs (e.g. brain, spinal cord, lungs). 3. The rib cage protects the heart and lungs; its movement results in breathing. The pelvis is essential for locomotion using the legs and also protects, to some extent, the organs of the abdomen particularly the uterus. The skull protects the brain but also allows the head to move.

5 Page 157 (left) 1. Walking to school. Regular training for a football team. These both involve sustained, repeated exercise over a long period of time. The others are short-lived exertions. 2. The benefits of a short burst of exercise cannot be retained. Only regular exercise is beneficial. Muscle growth, enlargement of the ventricles and increase in stroke volume, flexibility of joints, strengthening of ligaments and tendons are long-term changes induced by regular exercise. 3. If the supply of oxygen for the increased rate of respiration in active muscles is insufficient, some of the glucose is converted, anaerobically, to lactic acid. Even after exercise stops, there is a residue of lactic acid in the blood. This has to be oxidised to carbon dioxide and water so the demand for extra oxygen continues during a period of recovery. Page 157 (right) 1. A tendon transmits the force produced by a contracting muscle to a bone, causing a movement at the joint. A ligament holds the bones together at a joint while still permitting movement. 2. The extensor and flexor muscles for the fingers are in the forearm attached to the radius and ulna.

© D. G. Mackean 2011 www.biology-resources.com

IGCSE and GCSE Biology. Answers to questions Section 3. Human physiology. Chapter 18 The senses Page 159 1. Theoretically you would expect to feel only touch. Warmth and pressure are detected by their own receptors. In practice it might be difficult to hit only a touch receptor, depending on which part of the skin is being tested. 2. The temperature receptors which respond to heat loss will send impulses to the brain. So will touch and pressure receptors. If the ice is held there long enough it might stimulate pain receptors. 3. Temperature receptors, touch and pressure receptors, pain receptors. Chapter 19 Co-ordination Page 164 1. A nerve fibre is part of a single cell and is microscopic (about 1-20µ diameter). A nerve consists of hundreds of nerve fibres bundled together in a protective sheath and is visible to the naked eye (in dissections). 2. a Sensory and motor neurones are both single cells consisting of a cell body with a nucleus and a long nerve fibre usually running between the central nervous system and a peripheral structure. b Both neurones conduct electrical impulses. Sensory neurones originate in a sensory structure and conduct impulses towards the central nervous system. Their cell bodies are not terminal though still in the central nervous system. Motor neurones have their cell bodies in the central nervous system. They conduct impulses away from the CNS to an effector organ. 3. a A nerve fibre can carry only sensory or motor impulses. b A nerve (a mixed nerve) can carry both sensory and motor fibres and therefore can transmit both sensory and motor impulses. Page 165 1. Fig. 19.7 a Cell bodies 2 b Synapses 1 Fig. 19.9 a Cell bodies 7 b Synapses 5 (6 if you count the top one as 2) 2. All nerve impulses are the same in principle. You could not distinguish which receptor they came from. The sensation you experience depends on which part of the brain the impulse is sent to. Page 167 1. d receptor organ stimulated; b impulse travels in sensory fibre; e impulse crosses synapse; a impulse travels in motor fibre; c effector organ stimulated. 2. a Sneezing; sensors in the nose; effectors in diaphragm, rib muscles and muscles in the upper part of the trachea. b Blinking; receptors in the retina of the eye; effectors, muscles in the eyelids. c Contraction of the iris; receptors in the retina; effectors, circular muscles (sphincter muscles) in the iris.

3. The tongue has receptors for the chemicals which produce the taste sensations as well as sensors for temperature and pain. It is an effector because its muscles can move it about to manipulate food, initiate swallowing and control speech. 4. Coughing as a reaction to particles in the trachea or an intense ‘tickle’ is a reflex action. You cannot stop it. Coughing to clear the throat is a voluntary action. 5. Responses to commands such as ‘lie down’, ‘come here’, and to actions such as a lead being picked up or a plate being rattled are conditioned reflexes. Page 169 1. The cell bodies are concentrated in the grey matter so this is where synapses will occur. 2. Depending on the extent of the damage; no sensory information (touch, pressure etc.) would reach the brain from the legs and lower abdomen. Motor function below the area of damage would be lost so that there could be no leg movement, and control over bladder and rectum would be lost. 3. a Fore-brain; smell. Mid-brain; sight. Hind-brain; hearing, balance and skin senses. b The medulla. 4. Sensors in the cochlea of the ear detect the sound waves and send impulses via the auditory nerve to the hind-brain. The hind-brain then transmits impulses via neurones and synapses to the cerebral cortex which, in turn, sends motor impulses to the neck muscles which turn the head. Page 172 1.

Endocrine system (Examples) Nervous system Secretion of oestrogen. Spinal reflex. Transport of oestrogen to ovary. Impulses from sensors to brain. Rate of blood flow from ovary to pituitary. Impulse from sensor to brain takes milliseconds. Oestrogen is carried to all parts of the body Motor neurone in reflex arc sends impulse but affects only the uterus and pituitary directly to effector organ. gland. Secondary sexual characteristics. Reflex response. 2. The ovaries produce ova but they also secrete the hormone, oestrogen. The testes produce sperms but they also secrete the hormone. testosterone. 3. Too much insulin results in a reduction in the blood glucose level, which can impair the function of other organs such as the brain. Too little insulin allows the blood glucose levels to rise. This effect and related effects can produce the symptoms of diabetes. 4. If an ovum is fertilized, the ovary produces the hormone progesterone. This hormone or its breakdown products are excreted and appear in the urine, where they can be detected.

Chapter 20 Personal health Page 176 1. The answer will depend on personal data. 2. Dental decay. Avoid frequent or prolonged exposure to sugary substances. Gum disease. Clean the teeth regularly to remove plaque.

Page 179 1. Tolerance of a drug means that the user needs larger and larger doses to achieve the desired effect. Dependence means that the user has a craving for the drug and cannot give it up without experiencing distressing withdrawal symptoms. Being ‘hooked’ on a drug means that the user is dependent on it. 2. a Athletics. Amphetamines raise the blood pressure to dangerously high levels. b Examinations. Amphetamines increase confidence but not accuracy. You could be writing rubbish with confidence. 3. Alcohol causes vasodilation in the skin. The heat-gain sensors in the skin send impulses to the brain and induce a feeling of warmth. But the vasodilation also increases loss of heat from a warmer skin. 4. Alcohol increases reaction time so that it takes longer for the driver to react to a hazard though it also induces a false feeling of confidence. 5. There is a risk that morphine and heroin will lead to tolerance in users, so that the addicts have to constantly increase doses to get the desired effect.

© D. G. Mackean 2011 www.biology-resources.com

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Section 4 Genetics and heredity Chapter 21 Cell division, chromosomes and genes Page 182 1. a Gametes are reproductive cells. (i) Plants. The male gametes are the pollen nuclei and are produced in the anthers. The female gametes are the egg cells in the ovules which are produced by the ovary. (ii) Animals. The male gametes are the sperms which are produced in the testes. The female gametes are the ova which are produced in the ovaries. b At fertilisation the nucleus of the male gamete fuses with the nucleus of the female gamete. c A zygote is the cell formed from the fusion of the male and female gametes. It develops into an embryo. Page 183 1.There will be 2 x 46 = 92 chromatids at this stage. (See p. 184). 2. The chromosomes are too thin to be seen by the light microscope. 3. Cells in the basal layer will be dividing and producing more epidermal cells. Mitosis will be taking place in these cells. 4. It is not possible to answer this question without more knowledge than is provided here. Page 184 1. a 46 b 40 c 46. 2. 46. Page 186 1. a human 23 b fruit fly 4. 2. White blood cell - diploid. Male cell in pollen grain - haploid. Guard cell - diploid. Root hair - diploid. Ovum - haploid. Sperm - haploid. Skin cell - diploid. Egg cell in ovule - haploid. 3. a testes b ovary c stamen and ovule. 4. a 20 b 20. 5. Asexual reproduction takes place by mitosis. The chromosomes in each cell and therefore in each individual are identical. There can be no variation.. Page 190 1. alanine - alanine - valine - glycine - valine. 2. CAT must have taken the place of the triplet CCA. 3. The gene determines the sequence of amino acids in the protein molecule which, in turn, forms the enzyme. The enzyme controls a certain chemical reaction in the cell which affects the cell’s structure. 4. Radiation causes an increase in mutation rate. If a mutation occurs in a gamete, it will affect the offspring.

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Chapter 22 Heredity Page193 1. One possible choice is T for the dominant gene (allele) and t for the recessesive gene. 2. The chromosomes are in pairs (one from the male parent and one from the female parent) and so the genes they carry must be in pairs. If both genes are dominant or both recessive, they will control the characteristic in the same way. If one is recessive and one dominant they will not affect the characteristic in the same way. 3. Since the gene for black hair is dominant over the gene for red hair, the person would have black hair. 4. a For example: Gene for red hair b, gene for black hair B. Combination for red hair - bb. b You would expect a red-haired couple to breed true because they must both be bb. c A black-haired couple could have a red-haired baby if they were both Bb. 5. Aa is heterozygous, AA is homozygous dominant, aa is homozygous recessive. 6. The gene for red, e.g. R is dominant over the gene for white r. Page 195 1. All the offspring must inherit one gene from the male parent and one from the female parent so all the offspring must be Bb. 2. Only the BB mice will be true-breeding, i.e. one in three. 3. If white babies turn up in the second generation it means that one of the original parents must have been heterozygous Bb. (If both original parents were heterozygous you would expect some white offspring in the first generation). The probable genotypes of this first generation will be BB Bb BB Bb (all black, but 50% homozygotes and 50% heterozygotes). If the BB black guinea pigs are mated with bb white guinea pigs all their offspring will be Bb black guinea pigs. If the Bb black guinea pigs are mated with bb white guinea pigs you would expect half of their offspring to be Bb guinea pigs. Adding these results together gives one half black and the other half with equal numbers of black and white offspring. Page 196 1. (left hand column) You would carry out a recessive test-cross. Each parent would be mated with a homozygous white rabbit. With the homozygous parent this would produce all black offspring but with the heterozygous parent there would be, on average, 50% black heterozygous and 50% white homozygous babies. 1. (right hand column) If both parents are homozygous AA x BB, the children will be Group AB. If the mother was heterozygous (AO) the children could be AB or B (AB or BO). If the father was heterozygous (BO) the children could be AB or A (AB or AO). If both parents were heterozygous (AO, BO) the offspring could include groups A (AO), B (BO), AB or O. 2. The man cannot be the father. The genes for groups A and B are dominant to O. So whether the woman is homozygous or heterozygous for the A gene, the offspring would have received a dominant A or B gene. If the father was heterozygous for the B gene (BO) and the woman was heterozygous for the A gene (AO) the child could have inherited two O genes.

3 3. a The alleles for red and white hairs are codominant b W

R

W

WW

R

R

R

RR

On average, there would be a ratio of one red (RR), one white (WW) and two roan (RW) in a succession of calves

Page 197 1. The process of meiosis rules out the possibility of X sperms only. There will be an equal number of X and Y sperms in the father’s semen. There is a 50:50 chance of an XY or XX zygote. But this is indeed a matter of chance, and the sequence of 4 girls is due to chance alone. Page 199 a s 1. The genotype for a woman with sickle-cell trait would be HB Hb . If she a a a marries a normal man, whose genome would be Hb Hb , none of the children would have the disease but half of them could inherit the trait. b marries a man with sickle cell trait, there is a 1in 4 chance that a child would be normal, a 1 in 4 chance that a child will have the disease and a 1 in 2 chance that the child will have the trait. s s c marries a man with sickle cell disease (Hb Hb ) two children in 4 could inherit the disease and 2 children in 4 could inherit the trait. Chapter 23 Variation, selection and evolution Page 202 1. a Mainly inherited; facial features, ability to talk. b Mainly acquired; manual skills, language. c More or less equal mixture: body build, athleticism. 2. a The X and Y sex chromosomes are inherited independently of hair colour and texture. So whereas the father has brown eyes and straight hair, his sons could have blue eyes and straight hair. Similarly the mother has blue eyes and curly hair but her daughters could inherit brown eyes and curly hair. b The children would all have brown curly hair, unlike either of their parents. 3. Tall plant with yellow seeds TtYy. Dwarf plant with green seeds. ttyy TY

Gametes ty

TtYy

Ty

tY

ty

Ttyy

ttYy

ttyy

There are two new combinations of characters: tall plants with green seeds (Ttyy) and dwarf plants with yellow seeds (ttYy)

4 4. The apples on the south side will receive slightly more sunlight than those on the north side. So photosynthesis will be greater and, consequently, the fruit will be larger. The apple leaves on the lower branches will be shaded by the foliage above and so receive less sunlight that those in the upper branches. The apples on the north side lower branches will receive least sunlight. 5. A will now be on the same chromosome as b and c so Abc will be a new combination. So will aBC. Page 205 1. Assuming that it is the male who is competing, his plumage might be brighter and his song louder that his rivals. He might also be more aggressive towards his competitors, so leaving him space to attract a female. He might also have a more striking display pattern. 2. Selection pressures might include: drought or excessive rainfall, very high or very low temperatures, the growth rate of competing plants, the stresses of being walked on, pattern of light and shade, good or poor drainage, availability of mineral ions in the soil. The plants which are able to cope with these pressures are the ones most likely to survive and reproduce. Page 210. 1. The amphibia were most abundant in the Permian and Triassic periods 200 million years ago. 2. The features which suggest relationship with present-day fish are the shape of the body, scales, a tail fin and what looks like a gill opening. 3. The positions in the evolutionary tree diagram reflect the complexity of the organism and the time of appearance in the course of evolution. Worms and insects are complex organisms but not so complex as birds and mammals. Worms and insects of one kind or another appear early on in the fossil record. Birds and mammals are comparatively recent ‘arrivals’. Chapter 24 Applied genetics Page 212 1. Good characteristics could include: fertility, growth rate, disease resistance, wool texture, wool length. 2. The plant breeder would be begin by crossing the two varieties. The outcome would depend on which of the genes were dominant and which recessive. (In fact it would almost certainly be a group of genes for each characteristic). Assuming that the long stalk gene (L) and the good ear genes (G) are dominant, the genotypes LLgg and llGG would produce LlGg offspring, which would combine both desirable characteristics but would not breed true. If the parents were heterozygous for the dominant gene Llgg and llGg. There would still be desirable combinations of L and G among the offspring but in much smaller numbers (one in four). 3. The F1 seeds will be heterozygous for the desirable characteristics and will not breed true.

5 Page 220 1. a The substrate for a restriction enzyme is DNA. b The particular substrate for a restriction enzyme will be the link between two specific DNA triplets. 2 The vectors might be plasmids, viruses, bacteria or liposomes. 3. a Insulin extracted from cattle or pigs may carry ‘foreign’ proteins which provoke an allergic reaction. Genetically engineered insulin does not contain these proteins. b Genetically produced chymosin does not depend on an animal source where the supply may be inconsistent. It is purer than the chymosin produced from calves’ stomachs. 4. The principle is to replace a defective gene with a healthy gene. 5. Viruses could reproduce in the host and cause disease. The viruses have to be ‘disabled’ so that they cannot reproduce. 6. Unless it can be shown that using sheep to produce drugs causes their suffering, it is hard to think of any reasonable ethical objections. 7. a The potential benefits are that GM crops could withstand drought, grow in saline conditions, carry resistance to pests and diseases, have higher yields, produce additional vitamins. b The potential hazards are that genetically engineered plants could escape into the environment or exchange their modified genes with wild plants. This is only a hazard if the engineered genes confer resistance to a herbicide or cause the plants to produce compounds which could be harmful to other organisms. 8. Animals in a clone will all have the same genotype and will be identical. Animals produced by sexual reproduction will show variation. 9. Stem cells retain their ability to divide and produce new cells. A zygote can divide and produce all the cells in the body. In this respect it is acting as a stem cell. 10. a Restriction fragments are short lengths of DNA produced by restriction enzymes. b The restriction fragments differ in length. c The fragments can be separated by placing the sample at one end of a gelatine sheet and applying an electric current which causes the fragments to separate by distances related to their length.

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Section 5 Organisms and their environment Chapter 25 The interdependence of living organisms Page 228 1.

kestrel

cat

fox *

sparrow

mouse

rat *mainly urban foxes

wheat seeds

2. The tree and grass depend on the soil for anchorage, water and mineral nutrients. The earthworm depends on the soil for making its burrows and for the humus it contains as a source of nutrients. The soil depends upon the fall of leaves from the tree to renew its humus content and the earthworm uses leaves to pull into its tunnels. The soil also depends on the earthworms’ tunnels for drainage and aeration as well as the improvement and mixing of soil structure from passing through the earthworm’s alimentary canal. The blackbird depends on the earthworms as a source of nutrition and the tree as a place to perch or hide from predators. 3. Photosynthesis in vine leaves -- grapes -- grape juice fermented to wine. Photosynthesis in grass -- eaten by cow -- milk from cow -- converted to butter. Photosynthesis in wheat -- production of wheat grains -- eaten by chickens -- eggs. Photosynthesis in leaves of bean plant -- seed production (beans). 4. Photosynthesis millions of years ago produced plants which became decomposed and fossilised to form petroleum or coal. Oil or coal burned to raise steam and drive generators. Oil derived from fossilised plants (mainly algae or protista) becomes petroleum. When this is distilled, one of the products is petrol. Photosynthesis in oat plants results in the production of oat grains which are eaten by racehorses and provide energy from respiration. Wind, waves and hydroelectric sources depend on energy from the sun but not specifically from sunlight. Tidal energy comes from the moon and the sun. Nuclear energy and geothermal energy (from the heat in the deep layers of the Earth) are independent of the sun. (not in text) 5. Long term observation will reveal what the fox and pigeon eat. This can be supported by evidence of their stomach contents at different times of the year. Observation will also reveal which organisms eat the pigeons.

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6. Domestic carnivores such as cats can catch and eat organisms to the point that the population is decimated. The animals above the victims in the food web will suffer or turn their attention to different prey. Goats can eat the parts of plants that are usually unaffected by the indigenous herbivores. They may eat so much of the plants that they destroy the population altogether or reduce it so much that it cannot support its natural population of herbivores or insects. Page 229 1. a Carbon atoms form the ‘backbone’ of the molecules needed by living organisms (e.g. C - C - C - C - C - C carbon ‘backbone’ of glucose molecule C6H12O6) for building all their body structures and supplying energy. b glucose, sucrose, cellulose, amino acids, lipids. c Animals get their carbon by eating plants or other animals. 2. a C6H12O6 + 6O2 6CO2 + 6H2O b C + O2 CO2 c 6CO2 + 6H2O C6H12O6 3. In the leaf, the carbon atom would be incorporated into a molecule of glucose by the process of photosynthesis. The glucose molecule would be carried in the phloem to the potato tuber which would combine the glucose molecules into starch. When eaten, the starch molecules are converted to glucose molecules by the process of digestion. The glucose molecules are then oxidised by respiration and the carbon atom in a molecule of carbon dioxide is released. 4. photosynthesis H2O

C6H12O6

6H2O + 6CO2

respiration

6H2O + 6CO2

decay

6H2O + 6CO2

respiration

6H2O + 6CO2

decay

H2O + CO2

oil (CH) coal

plants

eaten by animals

fossilized

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Page 230 1. The clover is a leguminous plant. These plants have root nodules which contain nitrogen-fixing bacteria. These bacteria can convert atmospheric nitrogen to nitrogenous compounds which are needed for healthy growth. Grass does not have root nodules and therefore suffers from nitrogen deficiency. Page 231 1. To judge from the poor yield from plants deprived of nitrogen, this must be the mineral element which produces the most profound effect. 2. A (Nitrate added) B (Nitrate removed) nitrifying bacteria denitrifying bacteria nitrogen-fixing bacteria leaching manure uptake by plants chemical fertilizers lightning decomposition (of plants and animals) 3. a Advantages. (i) Manure. Freely available on mixed farms. Contains organic material which improves soil structure. Slow release of nutrients. (ii) Chemical fertilizer. Composition can be adjusted to meet needs of different soils. b. Disadvantages (i) Manure. May not contain all essential elements e.g. trace elements, or the right balance of other minerals for a particular soil. (ii) Chemical fertilizers. Expensive. No organic matter to improve soil structure. More easily washed out. Page 233 1. The energy for a muscle contraction in your arm comes from the respiration of glucose. The glucose is derived from carbohydrates in your food. These carbohydrates come from plants which have used the sun’s energy to build them by photosynthesis. Photosynthesis depends on energy provided by sunlight. 2. The advantage would be that energy is not lost by the respiration of all the organisms in the food chain between the source and the final consumer e.g. eating plants instead of eating the animals which eat the plants. The disadvantage is that this would distort the food chain by depriving primary consumers of their food source, e.g. eating plankton would deprive fish of a food source.

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Chapter 26 The human impact on the environment. Page 239 1. It takes longer and therefore costs more to make up a quota of small fish. Removing small fish from the sea depletes the breeding stock. 2. All cereals, grass for silage, soya beans, and oil-seed rape may be grown as monocultures. a Advantages: high yields per hectare; easy to harvest in bulk; the same treatment can be applied to the whole crop. b Disadvantages: rapid spread of infestation, bacteria, fungi or insects; imbalance of minerals removed from the soil; bad news for bees and many other beneficial insects and the creatures which prey on them. 3. a Advantages: they keep weeds, fungi and harmful insects under control. b Disadvantages: they kill harmless and beneficial insects as well as pests; they may get into the food chain. 4. In bad (cold) weather, the birds use their fat reserves as a source of energy. When the fat is metabolised it releases the DDT into the blood stream. 5. An excess of nitrate in river water allows microscopic algae to grow and reproduce rapidly. When this excess population of micro-organisms dies, the remains are oxidised to carbon dioxide and water, making excessive demands on the supply of dissolved oxygen in the water. Page 241 1. Logging for timber, cutting down trees to make way for agriculture, destruction of trees for firewood. Forest trees reduce erosion on sloping ground; their absorption of water and transpiration can affect local climate; on a global scale they help to slow down global warming by absorbing huge amounts of carbon dioxide; forests offer a habitat for a wide variety of animals not found elsewhere. 2. The trees intercept the rain and allow it to soak into the ground but not wash it away. Their roots also help to keep the soil together. 3. The furrows should run at right angle to the direction of slope, i.e. along contours. This helps to retain rain. If the furrows ran downhill, heavy rain could form channels and carry the soil away. 4. a On a hillside, the trees intercept heavy rain and allow it to soak into the soil. Removal of trees allows the rain-water to run off the slopes and flood the valley beneath. b Clear felling the trees leaves huge swathes of ground totally unprotected. The transpiration from the trees previously created water vapour, clouds and local rainfall. The bare soil heats up in the sunlight, disperses local cloud cover and reduces rainfall.

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5. The insecticide kills the springtails. Their numbers diminish and so do the numbers of mites which depend on them for food. When the insecticide eventually goes, the springtail population increases once more, but now there are fewer mites to eat the springtails and so their numbers soar until the mite population recovers and starts to eat them again. Page 242 1. If the poisonous chemicals are dumped, animals and plants in the vicinity may be harmed. If they are buried, the poisons may kill the plants growing above or may be leached out into ground water and find their way into public water supplies or rivers where they harm wildlife. 2. The water about to leave the waterworks may contain a small residue of bacteria. Exposure to chlorine kills any remaining bacteria and the chlorine eventually evaporates. 3. The mercury was taken up by and accumulated in fish until it reached poisonous levels. Page 246 1. Tall chimneys deliver the pollutants into the air well above ground level, but the pollutants are still there and may eventually find their way back to the surface (See Figure 26.20). 2. Acid rain contains sulphur dioxide and oxides of nitrogen which come mainly from factories burning coal and oil, and from exhaust gases of vehicles. 3. Carbon dioxide and methane both absorb long wave radiation reflected from the Earth’s surface and cause the atmosphere to warm up.

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Chapter 27 Conservation Page 249 1. Extinction can result from excessive hunting, destruction of habitats, introduction of alien species. 2. CITES concentrates on banning trade in endangered species by persuading countries to pass laws forbidding such trade. WWF adopts a variety of strategies for protecting endangered plants and animals. For example they may attempt to prevent excessive hunting, or the destruction of habitats. Both organisations may attempt to get laws passed to restrict human activities which threaten wildlife and both are dedicated to the preservation of biodiversity. 3. a The extinction of certain wild plants may deny us the chance to develop new drugs from their products. b If a plant becomes extinct its genes are lost for good. Some of these genes from drought-tolerant plants could have been introduced to crop plants. Page 251 1. a Biodiversity means the mixture of plant and animal species which comprise a community. b Sustainable development means the development of any kind of industry, farming or technology which does not permanently destroy habitats or reduce biodiversity. Felling trees, for example is not sustainable unless new trees are planted at the same rate. 2. An Area of Special Scientific Interest is a naturally occurring environment on privately owned land but governed by rules concerning its maintenance. A Nature Reserve is also a natural habitat but managed by an organisation to maintain and possibly improve its community of wildlife. Page 252 1. Any renewable resource used to produce energy will have come ultimately from a plant source, e.g. firewood, sunflower oil for diesel, sugar cane for ethanol. The plant source will have depended on photosynthesis for making these products. 2. The energy used to recycle products will be less than the energy needed to produce them in the first place. For example, the energy needed to melt aluminium is less than the energy needed to mine the ore and extract the metal.

Chapter 28 Ecosystems Page 255 1. The community in woodland would consist of the different species of trees and shrubs and the birds, mammals and insects which inhabit them, plus lichens and algae. It would include the woodland flora, mosses and fungi and all the inhabitants of the soil. 2. a The earthworm’s habitat is the soil. b Its environment could be a lawn, a garden, woodland or farmland, with its accompanying water and air. This environment might include birds and mammals which eat earthworms and the organic matter on which the worm feeds. 3. The producers could be grasses or flowering plants. The primary consumers could be herbivorous mammals or insects. Decomposers would be fungi and bacteria. 4. Plants could compete for sunlight by growing taller than their neighbours, having more branches or leaves. They could compete for water and nutrients by rapid and extensive root growth. Page 257 1. Intraspecific competition between the wheat plants might be their growth rate and root spread. Interspecific competition could take the same form but be between the wheat plants and competing weeds. 2. The long, bristle-fringed back legs of the water beetle would help to propel it through the water but would be a encumbrance on the land. The smooth outline of the head, thorax and abdomen might streamline it while swimming. 3. The overlapping flight feathers increase the air resistance in the downstroke. Its limb bones are hollow and, therefore, light. Many parts of the skeleton are fused and form a rigid framework which resists distortion during the muscle contraction of the downbeat. The deep breastbone provides a firm anchorage for the flight muscles. The pectoral muscles are large and powerful which allows a powerful down beat. The flight feathers are light and air-resistant. 4. a A camel reduces its water losses by absorbing water vapour from exhaled air in its nasal mucus. During inhalation the moisture returns to the lungs. b The camel is able to withstand desiccation by the withdrawal of water from the tissues to maintain the blood volume and concentration. 5. The polar bear’s volume/mass ratio reduces the relative area exposed to heat loss. Its small ears also reduce heat losses. It has a dense coat of fur with the inner layer of densely packed hairs forming effective insulation. Heat loss from the feet is reduced by an exchange of heat between the arteries and veins of the limb. This keeps the blood in the feet cool and so it loses less heat to the snow and ice.

Chapter 29 Populations Page 259 1. It takes about 8 days for the mortality rate to equal the replacement rate. 2. a Approximately 8 per unit volume. b About 992 per unit volume. c There is no increase. Page 261 1. The population increases from about 30 to about 80,000 over a period of 2.5 days, so the average rate of increase (ignoring the 30) is 80,000/ 2.5 = 32,000 per day. 2. Depending on the scales you chose, the graph should look something like this (with the kinks smoothed out). population It broadly fits the first two sections of the sigmoid curve (lag phase and exponential phase). There is likely to be a continued increase in the growth rate until the pheasant population reaches equilibrium with its resources.

time 3. Since the scale is linear and not logarithmic, the increase in population of Paramecium aurelia does not strictly show a pattern of exponential growth. However there is a lag phase for the first two days and a steady state by day 15 with the most rapid growth from days 3 to 8. 4. There will be competition for food, mates and nest sites. These factors and the presence of predators will limit the population. 5. a Abiotic factors. Availability of nest sites; exposure to weather (e.g. mountainous conditions, strong winds and low temperatures). b Biotic factors. Availability of food, presence of predators, availability of nesting materials, spread of disease. Page 264 1. a The ‘baby boom’ occurred between about 1948-50. b The birth rate was about 38 per thousand but the death rate was about 28 per thousand, so the growth rate was about 10 per thousand. 2. Heart disease and strokes affect mainly people beyond child-bearing age. Measles affects mainly people below child-bearing age. The effects of the remaining diseases depend very much on the countries involved but smallpox is very infectious and spreads rapidly throughout a population compared with polio and tuberculosis and seems to most likely to affects growth rate. 3. The death rate in poor countries is high compared with wealthy countries. The birth rate remains high in order to maintain family size and provide more ‘bread-winners’. As countries grow richer and more industrialized smaller numbers are needed to support the family. In wealthy countries the people are usually better educated and informed about the methods of birth

control. This means they can limit their families and improve their life style. Child bearing can be postponed while young people improve their education. 4. A growing population needs more land to build homes and extend food-growing areas. It makes increasing demands on water and other natural resources including minerals, hydrocarbon fuels, building-materials and timber. The environment is damaged when habitats are destroyed to make way for buildings and farms. A growing population is likely to depend on an increasing degree of industrialization with all the risks of pollution and consequent environmental damage. 5. 12,000 live births in a population of 400,000 is a birth rate of 12,000/4000 = 30 per thousand. 6. If couples had only 2 children and one of them died before child-bearing age, the population would diminish. 7. a In Figure 29.8 (a) male babies outnumber females by about 270 vs 220 millions. In population (b) female babies slightly outnumber male babies. b Between the ages of 20 and 40 the numbers are roughly the same with a slight preponderance of men, particularly in population (a) c In the over-70s the male population has diminished faster than the female population in both cases. 8. Reduction in the incidence of diphtheria was probably due to improvements in living standards including nutrition which would increase resistance to disease and less overcrowding which would reduce the chances of cross infection.

Section 6 Diversity of organisms Chapter 30 Classification Page 269 1. a Many cells. Plants, Animals b Nuclei. Protoctista, Fungi, Plants, Animals. c Cell walls. Plants. d Hyphae. Fungi. e Chloroplasts. Plants, some Protoctista. Page 276 1. Snails and earthworms are primary consumers but earthworms may play a part as decomposers. They are both eaten by secondary consumers, e.g. birds. 2. Withdrawal into their burrows and retraction into their shells protects earthworms and snails from predators and desiccation. 3. At low temperatures the chemical reactions in cells slow down and this affects the whole animal and particularly movement. Page 278 1. a Warm-blooded. Birds, mammals. b Four legs. Amphibia, Reptiles, Mammals. c Lay eggs. Fish, Amphibia, Reptiles, Birds. d Internal fertilisation. Reptiles, Birds, Mammals. e Parental care. Reptiles, Birds, Mammals. 2. This is an unfair question. The text does not give criteria for all the groups listed on p.274 and it is not feasible to construct a dichotomous key for the whole animal kingdom. Page 281 (Left hand column) 1. Beetle. Animal, arthropod, insect. Sparrow. Animal, vertebrate, bird. Weasel. Animal, vertebrate, mammal, (carnivore). Gorilla. Animal, vertebrate, mammal, (primate). Pine tree. Plant, vascular plant, conifer. Buttercup. Plant, vascular plant, flowering plant (dicotyledon). Moss. Plant, bryophyte, moss. 2. Although both plants are in the same genus (Lamium), they are different species (album and purpureum) so you would not expect them to cross-pollinate successfully. 3. Bracken propagates vegetatively by means of underground rhizomes (p.79) which are too deep in the soil to be affected by fire. 4. Trees, shrubs and ‘flowers’ are vascular plants. They are so called because they have well developed vascular systems consisting of xylem and phloem (pp 54-6).

Page 281 (Right hand column) 1 1. Flagella or cilia present….2 No flagella or cilia ………Amoeba 2 Flagella present …………3 Cilia present …………….4 3. Two flagella……………..Chlamydomonas One flagellum……………Euglena 4. Stalk present …………….Vorticella No stalk ………………….Paramecium 2. 1. Yellow flowers …………………………2 Flowers not yellow ………………….…3 2. Large flowers on individual stalks…….. daffodil Small flowers on a prickly bush ………. gorse 3. Flowers green …………………………. couch grass Flowers not green………………….…... 4 4 Flowers white……………………….…. snowdrop Flowers; a range of colours……………. lupin This is an ‘artificial’ key because it depends not on genuine classificatory data but on a superficial characteristic, in this case, colour. 3. 1. Little or no vascular tissue present…………….2 Vascular tissue present………………………...3 2. Flat leaflike form….. ………………………….Liverworts Vertical stem with small leaves………………..Mosses 3. Reproduce by spores …..………………………Ferns Reproduce by seeds …………………………4 4 Seeds not enclosed in fruit ……………………. Conifers Seeds enclosed in fruits…………………………5 5 Seeds with one cotyledon……………………….Monocotyledons Seeds with two cotyledons….. …………………Dicotyledons Chapter 31 Micro-organisms Page 286 1. Only cytoplasm and DNA are present in both bacterial cells and plant cells. 2. If the population of 5 bacteria doubles every 20 minutes, after 4 hours it will have reached 20480. 3. a The virus particle contains no cytoplasm, no cell membrane and no nucleus. b Viruses do not feed, respire, grow, excrete, move or respond to stimuli which are characteristics of living organisms. They do reproduce but only by rearranging their host’s cytoplasm. (See page 292). 4. In the bacterium, the nucleus divides and then the cytoplasm is shared between the two cells. In the virus there is no nucleus or cytoplasm. New virus particles are built by using the host’s cellular material.

Page 288 1. Fungal hyphae are growing in or on the material which is going mouldy. The hyphae are penetrating the tissue, digesting and absorbing the products of digestion. 2. Bread, wood and leather are derived from living organisms and contain materials which can be digested and used as food. Glass and plastic are man-made and do not contain any substances which could be used as nutrients. 3. Toadstools are the ‘fruiting bodies’ of fungi which are living in the soil. Fungi do not photosynthesize and so have no need of light. Green plants must have sufficient light in order to photosynthesize and grow. Page 290 1. If the discs contain different antibiotics, the extent of the clear zone reflects the effectiveness of the antibiotic against that particular bacterium. If the discs contain different concentrations of the same antibiotic the clear zones show the effectiveness of the different concentrations. (In fact the former is the case). Chapter 32 Feeding Page 297 1. a Apple tree; autotrophic. b Toadstool; saprotrophic. c Human; heterotrophic d Mosquito; heterotrophic. e Streptococcus; saprotrophic. 2. Plants need sunlight. 3. Digestion and absorption take place in the alimentary canal. Chapter 33 Breathing Page 300 1. It depends on how you cut the shape

2

2 2

b

a 1

0.5

If you cut it like a each half will have a surface area of 4(2x1) +2(1x1) = 10cm2. The volume will be 2cm3. The ratio of surface area to volume is 10cm2/2cm3. If you cut it like b each half will have a surface area 2(2x2) + 4(2x0.5) = 12cm2. The volume will be 2 cm3. The surface area/volume ratio is 12cm2/2cm3.

2. Judging by the scale, the bacterium is about 0.00125mm wide. The maximum diffusion distance will be half this figure, namely 0.0006mm (approximately). 3. A frog is most likely to use its lungs during and after a period of activity. 4. Most large animals, active or not, are unable to exchange gases through their skins. They therefore have need of organs specialised for gaseous exchange. The distances from these organs to all the tissues of the body are too great for diffusion to be effective. A circulatory system allows the gases to be carried from the respiratory surface to all parts of the body and vice versa. 5. In a fish, a fresh supply of water containing dissolved oxygen passes continuously in one direction over the gills so that a fresh supply is in direct contact with the exchange surface. In the lungs the air is only intermittently exchanged with the atmosphere and there is always a ‘stagnant’ layer through which the oxygen has to diffuse before coming into contact with the respiratory surface. 6. a In mammals, the muscular activity takes place in the intercostal muscles and the diaphragm. b In fish, the muscular activity takes place largely in the floor of the mouth. Chapter 34 Reproduction Page 304 1. a b c d e f

male reproductive organs female reproductive organs male gametes female gametes place where fertilisation occurs zygote grows into

flowering plants anthers ovaries pollen nucleus egg cells in ovules ovules seed

mammals testes ovaries sperms ova oviduct embryo, foetus

2. Butterfly. The butterfly lays fertile eggs so fertilisation must be internal. Mussel. Mussels are isolated from each other and are sedentary. Fertilisation is external with eggs and sperms being shed simultaneously into the sea Trout. Trout have no external genitalia. Fertilisation must be external. Sparrow. Birds lay fertile eggs so fertilisation must be internal. Earthworm. Although earthworms couple together to reproduce. the sperms are stored separately from the ova and not released until the ova are placed in a cocoon, Thus fertilisation is external. (This detail is not in the text and there is no way you could reason it out.) 3. Courtship is a behaviour pattern which is principally a characteristic of birds. Mammals have ways of indicating they are ready to mate but this cannot really be described as ‘courtship’. Animals which have external fertilisation have ways of ensuring that sperms and eggs are released at the same time, but this is not courtship. The question is flawed.

Page 305 1. a Asexual reproduction in fungi involves the rapid production and release of large numbers of spores which can be distributed over a wide area and will grow rapidly into new individuals. Asexual reproduction in flowering plants takes place by relatively slow vegetative growth from the parent plant, eventually leading to the development of new plants not far from the parent. b A spore is a single haploid cell. A seed is a multicellular structure consisting of diploid cells.

2. If the mutant plant is self pollinated it will not breed true and the offspring will be very varied (pp.193, 201). If it can be reproduced vegetatively, the offspring will be identical and the mutant characteristic will be retained. 3. The farmer will have to consider the cost of making the crosses and waiting for the next generation to see if any of the offspring had the combined characteristics. There may be offspring with characteristics which are undesirable and which will attract lower prices when sold. He would have to spend money to keep the pigs isolated so that they could not interbreed. The hybrid pigs will not breed true and may need many generations to stabilise the variation. He will also have to calculate whether the lean, long-backed pigs will attract a higher price. 4. Asexual reproduction does not involve gametes, meiosis, or zygotes. 5. We use asexual reproduction in plants to produce new plants with all the characteristics of their parents, e.g. potatoes, strawberry plants. We also eat the products of asexual reproduction such as potatoes and onions. 6. Birds exhibit parental care, in some cases by building nests and feeding their young. Groundnesting birds do not feed their young but do show protective behaviour. Most fish do not exhibit parental care and their eggs are often left unattended. As a result, many eggs and very many small offspring are eaten before they can reach adult size and develop effective escape reactions.

Chapter 35 Growth and development increase in mass

Page 309 1. a Depending on the scale you chose, the graph would look something like this. b It is, in fact, sigmoid. The sigmoid curve is characteristic of population growth rather than increase in mass. Nevertheless, the growth, in this case, could reflect the increase in the population of cells. The onset of the stationary phase is very abrupt and perhaps shows the insect’s metamorphosis after which growth ceases.

days

2. The cell would have to increase in size and develop fine processes from its surfaces which become capable of conducting nervous impulses and producing neuro-transmitter substances at their endings. With motor and sensory neurones one of these processes would become very long and narrow, leaving the nucleus in the cell body. 3. a At year 1 boys and girls weigh abut 7 kg. At age 5 boys weigh 20kg and girls weigh 16 kg (average 18). The average growth over this period is 18-7kg =11 kg. b At age 10 both sexes weigh about 33kg so the average growth is 33-11kg = 22kg. 4. a The advantage is that you are measuring dry mass, which shows the increase in living matter and its derivatives in an organism without having to account for short-term changes in the weight of water in the tissues. b The disadvantage is that the organism has to be destroyed in order to obtain this figure. 5. The ratio of head to body at 2 years is about 12/52 (approx. 23%) and at 20 it is 8/52 (approx. 15%). 6. The caterpillar increases greatly in size and colouration up to the pupal stage, which becomes a completely different shape. In becoming a butterfly, the inset has developed a distinct thorax and abdomen, three pairs of jointed legs, a pair of antennae and, of course, wings. 7. Although cell division is taking place rapidly in the shoot tip, the cells remain small and without vacuoles. Just behind the shoot tip, the cells develop vacuoles which extend the cells mainly longitudinally in line with the main axis. This is the region of most rapid increase in length. Chapter 36 Movement and locomotion Page 314 1. Muscular system, skeletal system and nervous systems. (Indirectly also breathing and circulatory systems). 2. The legs marked with arrows have just moved forward. They now remain stationary to support the beetle while the other three move forward together. 3. The tail fin continues the sideways thrusts of the body to give a final flick to the forward propulsion. The median fins (dorsal, ventral and anal) reduce the sideways roll of the body. The paired fins, (pectoral and pelvic) control the upwards or downwards direction of movement and assist in the turning movements. 4. The extensor muscles are the ones which thrust against the ground to produce the forward motion. The flexors only have to restore the limb to its flexed position.

5. a To achieve lift during flapping flight the large flight muscles (pectoral) contract and pull the extended wing downwards. Air resistance to this movement and the air-flow pattern over the wing produces lift. b During gliding flight the bird has to lose height with its wings extended, relying on the air flow over the wing surface to achieve lift. Prolonged lift can be achieved only by making use of upward air currents. Page 316 1. Provided that the plant can reach water and receive sunlight, it can make all the food it needs while staying in the same place. 2. Sexual reproduction produces seeds which may be dispersed over great distances. 3. Sleep movements might help to decrease transpiration (at night?) or reduce frost damage. They might prevent loss of pollen at times when insects are not active. 4. Folding leaflets could be a protective measure in tropical rain. It could make the leaves unattractive to herbivores or inaccessible to insects. Chapter 37 Sensitivity Page 319 1. a (i) Roots respond to the directional ‘pull’ of gravity by growing towards it. (ii) Shoots respond to direction of light by growing towards it and to the direction of gravity by growing away from it. b (i) Positively phototropic; shoots. (ii) Positively geotropic; roots. (iii) Negatively geotropic; shoots. 2. a The whole plant does not move. Only the shoots grow towards light. b The root does not bend. Tropisms are growth movements. The root tip changes its direction of growth and grows downwards. 3. A clinostat is the only way that a unidirectional stimulus can be applied equally to all sides of a root or shoot and thus act as a control in tropism experiments. 4. The outline of the stem will be something like this. The first curvature is now in a non-growing part of the stem so the next curvature will take place at the growing point. 5. As soon as the root and shoot emerge they will be exposed to the one sided stimulus of gravity. The root will respond by growing downwards and the shoot will respond by growing upwards. 6. The result would depend on the relative ‘strengths’ of the two stimuli. If the light stimulus were the stronger the shoot would grow downwards (the positive phototropism overwhelming the negative geotropism). Since the stimulus of one-sided gravity cannot easily be altered the result would depend entirely on the light intensity. At low light intensity the shoot would grow upwards, (negative geotropism more effective than weak phototropism). Page 320 1. a Light suppresses the extension of the shoot but makes the leaves turn green. b In positive phototropism, the unidirectional stimulus of light could suppress the growth of the shoot on the illuminated side. This would have the effect of causing the shoot to grow towards the light source.

2. You could set up the apparatus as in Fig. 37.6 but remove the tip of one of the radicles and leave the other as a control. The radicle without its tip should not show a change of direction of growth. The snag is that by removing the radicle tip you could also remove the source of dividing cells so that growth stops altogether. You could try reducing the size of the cut region, till you get to less than 1mm. Page 323 1. a The wasps appear to be attracted to the smell of food, particularly fruit and sugary substances. b In this case the wasps seem attracted to the light and try to reach it by flying to the windows. 2. It may be that the water makes the choice chamber cooler on that side and this slows the woodlice down. Although we cannot detect it, there may be a volatile chemical in the silica gel which makes the woodlice move more rapidly. 3. The control would be to make conditions in both chambers the same by excluding both water and silica gel. If illumination is even or the experiment conducted in darkness, there should be no overall change in the distribution of the woodlice. 4. Set up the choice chamber in such a way that one half is in darkness. Alternatively, illuminate the chamber from one side in case the direction of light has an effect. 5. Most animals which move freely, travel in the direction of their long axis. So it is the front end of the animal which first encounters any stimulus: smell, light, sound etc. The front end of the animal is therefore where the sensory organs will be most effective. This arrangement will have evolved over millions of years, becoming more obvious as the sense organs became more sophisticated.

Section 7 Micro-organisms and humans Chapter 38 Biotechnology Page 332 1. Yeast; alcohol and carbon dioxide. Penicillium; penicillin. Lactobacillus; yoghurt, cheese. Aspergillus; enzymes. 2. Aerobic bacteria and protozoa play an important part in purifying human waste. 3. Sterile conditions are essential in order to exclude all micro-organism but the ones needed for the process. ‘Alien’ microbes could produce undesired and unpredictable products. The other controlled conditions involve temperature, aeration (if any), pH and a controlled supply of nutrients. 4. Curds and whey are intermediate products of cheese-making. The curds are the semi-solid precipitate of milk protein produced when the enzyme chymosin is used to coagulate the milk. They whey is the liquid residue which is drained off. 5. If air is allowed in, aerobic bacteria will metabolise the wanted substances (e.g. methane) completely to carbon dioxide and water. 6. The starch is digested to maltose by the amylase enzymes already present in the barley grains or the flour. 7. If the wine is left exposed to air for long enough, aerobic bacteria will oxidise the alcohol in the wine to ethanoic (acetic) acid. 8. Different mutant forms of Penicillium produce different types of penicillin. The penicillin can also be altered by chemical processes to make it more effective. 9. The bacteria clump the finer particles, which are ingested by the protozoa along with the bacteria. Soluble substances are rendered harmless by being absorbed and metabolised by bacteria. Chapter 39 Disease: causes, transmission and control Page 335 1. If pathogenic bacteria had landed on the cooked meat, they will have started to multiply over a period of 24 hours. Simply warming the meat will greatly increase their rate of multiplication and the population may reach infective levels. 2. a Freezing arrests the multiplication of bacteria but does not kill them. In handling the defrosted chicken, the cook’s hands could have picked up some bacteria which then became transferred to the cooked ham. Refrigeration slows bacterial reproduction but does not stop it. So the bacteria could have reached infective levels in the ham by the time the food was eaten. b The outbreak could have been avoided if the cook had washed his hands after handling the chicken, before slicing the ham. 3. With gonorrhoea, bacteria in the infected woman’s vagina can infect the baby. With syphilis the bacteria can cross the placenta from mother to foetus. Page 336 1. The damage caused by the respiratory virus to the epithelial cells makes them vulnerable to a secondary infection caused by bacteria. It is for this infection that antibiotics are used. 2. If the potential donor’s blood contains antibodies to the HIV virus, it means that the donor is carrying the HIV virus. If this blood is transfused the recipient would become infected with HIV.

3. There are many different strains of the cold virus. The explorers will at first infect each other with their particular strain. The explorers will develop immunity to each strain after the infection has passed. Eventually there will be no new strains to infect the team. Once the supply ship arrives, however, the crew will carry strains of the virus to which the explorers do not have immunity and so the cycle of infection starts again. 4. The HIV virus attacks the patient’s lymphocytes. These are the very cells which combat disease by producing antibodies. With a reduced population of lymphocytes, immunity cannot be achieved. Page 338 1. a The two main lines of attack are (i) to destroy mosquitoes and (ii) to attack the malarial parasite in the blood. The assault on mosquitoes is either by using insecticides against them or eliminating, as far as possible, their breeding sites. b Mosquitoes lay their eggs in stagnant water. The eggs hatch out to the mosquito larvae which depend on water to grow, pupate and hatch into mosquitoes. c The 'set-backs' are that the mosquitoes develop resistance to the insecticides and the malarial parasites develop resistance to the drugs. 2. Amoebic dysentery is an intestinal disease caused by the parasite Entamoeba. The infective stages of the parasite are present in the faeces of an infected person. Good sanitation keeps the faeces and drinking water well apart. Personal hygiene, such washing the hands thoroughly after visiting the lavatory and before handling food reduces the chances of passing the parasite from one person to another. Hygienic practices in cleaning cooking- and work-surfaces also reduce the chance of cross infection. 3. The medical officer would probably first check the water source to see if it was contaminated by Entamoeba and, if so, close the source and find another or import clean water. He would advise everyone to boil water likely to be used for drinking, and offer advice on the hygienic disposal of human waste. He would further attempt to control the outbreak by administering the appropriate drugs to infected people. Page 339 1. A great many diseases can be transmitted by contaminated food. If a person who sells, handles or cooks food carries pathogenic micro-organisms in or on his body, particularly his hands, he could introduce these pathogens to the food and to anyone who eats it. Good personal hygiene will reduce these risks. 2. If a person is carrying an infectious respiratory disease, whether or not he is aware of it, his coughs and sneezes will produce micro-droplets containing the pathogenic micro-organisms. The infectious droplets can remain in the air to be inhaled or fall on food and, consequently infect other people. 3. The lining of mucus in the respiratory tract traps many bacteria and is carried away from the lungs and bronchi by a current produced by the beating of millions of cilia. If the ciliary beat is arrested, the potential pathogens are not removed and start to multiply in the respiratory passages. 4. Unlike houseflies, wasps do not come into contact with human faeces and other sources of infection. Their method of feeding does not involve the regurgitation of potentially infected saliva.

Page 343 1. a Cornea; tear fluid, which bathes the cornea, contains the enzyme, lysozyme, which dissolves the cell walls of some bacteria. b Hand; the skin is a natural barrier to bacteria unless it is damaged, in which case white cells engulf the bacteria. c. Bronchus; the film of mucus which lines the bronchi and trachea, traps the bacteria which are then carried away from the air passages by ciliary currents. d The stomach produces hydrochloric acid which kills most bacteria. 2. An earthquake can damage sewage pipes and water pipes. This means that raw sewage may escape into drinking water which must therefore be boiled to destroy pathogenic bacteria. 3. Immunization against diphtheria prompts the lymphocytes to produce antibodies to the diphtheria toxin. This antibody is specific to the diphtheria toxin and will not have any effect on the polio virus. 4. Diphtheria immunization would be continued as protection against outbreaks of the disease originating from people who had not received the vaccine. 5. There is an enormous variety of forms of the common cold. A vaccine effective against one form would be ineffective against other forms. 6. a HIV is the virus which can cause the disease known as AIDS. b ‘HIV positive’ means that a person’s blood test reveals the presence of antibodies against the HIV virus. c High risk groups are homosexuals, drug users who inject themselves and people who have sexual intercourse with many different individuals.

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Section 8 Ideas and evidence Chapter 40 Development of biological ideas Page355 1. It was not until 1660 that a microscope was developed which was powerful enough to see the capillaries connecting arteries to veins. 2. Harvey’s ideas were rejected simply because they contradicted the work of a 2nd century physician, Galen. 3. a The current theory is that, in the right conditions over the past 400 million years or more, the bodies or parts of bodies of animals and plants became preserved in sedimentary rocks or other suitable deposits. b It was not the origin of fossils which was disputed but rather the interpretation of their significance in terms of evolution or creation. The question is misleading. 4. The advantage of the Linnaean system of naming organisms is that it attempts to put them into related groups in a way that leads to a natural classificatory system. Also, giving organisms ‘Latin’ names meant that everyone would know exactly which organism was being described, rather than having many different names for the same organism. 5. a Mendel concluded that: (i) characteristics can be inherited, (ii) one characteristic could be dominant over another, (iii) with a pair of contrasting characteristics only one can appear in the organism, (iv) a recessive characteristic which is not expressed in the first generation, can still appear in the second generation. It does not ‘disappear’, (v) the characteristics are transmitted by the gametes. b Mendel published his findings but in a local journal that was not widely circulated. 6. a Lamarck’s theory supposed that variations which developed during the life-time of an organism could be inherited. Darwin thought that the variations developed by chance and were heritable. b They both believed that beneficial characteristics could be inherited and lead to the development of more successful organisms. They thought that the process took place by very small changes and was continuous. 7. The discovery of mutations showed one way by which variations could arise and be inherited. 8. a It was shown by experiment that the appearance of maggots in meat was not spontaneous generation but the product of blowflies laying eggs in the meat. The eggs hatched into maggots. b Pasteur’s experiments showed that if you prevented free access to air, micro-organisms would not appear in boiled meat broth. 9. Pasteur controlled a disease of silkworms by removing all dead and infected individuals so that the disease could not spread to the healthy silkworms. Foot and mouth disease in cattle is similarly controlled by destroying infected individuals but in many cases the whole herd is slaughtered to prevent the disease reaching other farms. 10. Chargaff discovered that in a sample of DNA the number of adenines was always equal to the numbers of thymines, and also that cytosines and guanines were equally matched. This suggested that these bases were paired in the DNA molecule. 11. Maurice Wilkins and Rosalind Franklin carried out X-ray analysis of DNA which suggested the structure of the molecule. Crick and Watson used these results in constructing their model of DNA.

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Chapter 41 Observation and experiment Page 358 1. 1 This may be a matter of simple observation. If the tree trunks in a wood all show algal growth on the south side but nothing on the north side the question is answered for this particular area. If the observations are not so clear-cut you would have to find ways of measuring the areas of algal growth on the north and south sides. 2 There is no way of measuring degrees of ‘companionship’, so this could not be scientifically investigated. 3 This is too sweeping a generalisation to be investigated as it stands. You could match groups of boys and girls of the same age and similar physique, (you would not expect a short tubby boy to compete with a tall, slim girl), and specify a range of distances for them to run. Alternatively you could study the results of important junior athletics competitions. This would give results for trained athletes. 4 This can be investigated experimentally but the generalisation is too great. You would have to refine the question to: ‘Do some seeds need light in order to germinate’? 5 There will be plenty of anecdotal evidence but little objective evidence. You would need a large cohort of subjects matched for age and conditions, with some used as controls. There would be a wide variety of sources of vitamin C in peoples’ normal diets before any measured supplements could be tested. 6 This could be investigated scientifically by measured observations, once you have decided the size of the samples to be studied and defined the boundaries of ‘north’ and ‘south’. Page 359 1. 1 This would be based on objective evidence. 2 This is entirely anecdotal evidence. 3 As stated, this is anecdotal evidence but could be supported by objective evidence. 4 This statement must have been based on objective evidence since precise measurements must have been needed. Page 362 1. By selecting areas where dandelions were abundant, Paul will have distorted the results to show far more dandelions than there really are. 2. Throwing the quadrat frame 5 times in one area and 10 times in the other is likely to make the comparison inaccurate because it is trying to influence the results. 3. Estimated population 35 x 27/8 = 118. a By choosing to sample only a proportion of the ground, the results could be affected by migration of marked individuals out of the area. The area sampled must be confined in some way. b A hotter day might promote more activity in the area but this will affect marked and unmarked individuals equally, so it should not make a significant difference. c If the marking process interfered with the free movement of the marked individuals it could distort the result because the marked grasshoppers will tend to stay in the same place. d If the effect of the anaesthetic persisted into the second day it could distort the results as described in c.