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Biology for Cambridge O Level Answer Book Phil Bradfield I Published by PEAK PUBLISHING LIMITED London UK © Peak Publishing Limited 2016 Text © Phil Bradfield 2016 First published 2016 ISBN: 1 845222 16 4 All rights reserved. No part of this publication may be reproduced, copied or transmitted save with written permission or in accordance with the provisions of the Copyright, Design and Patents Act 1988, or under the terms of any license permitting limited copying issued by the Copyright Licensing Agency, 90 Tottenham Court Road, London W1P 9HE. Cambridge International Examinations bears no responsibility for the example answers to questions which are contained in this publication. The answers to exam-style questions and to Cambridge past paper questions in this book have been written by the author. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. Printed in Pakistan II Contents 1 Cell structure and organisation ....................................................................2 2 Diffusion and osmosis ..................................................................................3 3 Enzymes .......................................................................................................5 4 Plant Nutrition ..............................................................................................6 5 Animal Nutrition .......................................................................................... 8 6 Transport in Flowering Plants ..................................................................... 9 7 Transport in humans ..................................................................................10 8 Respiration .................................................................................................12 9 Excretion .................................................................................................... 14 10 Homeostasis ...............................................................................................15 11 Coordination and response ........................................................................16 12 Support, movement and locomotion .........................................................19 13 The use and abuse of drugs .......................................................................20 14 Microorganisms and biotechnology .......................................................... 21 15 The relationships of organisms with one another and with the environment .............................................................................................. 23 16 Development of organisms and the continuity of life ................................25 17 Inheritance .................................................................................................27 1 Chapter 1: Cell structure and organisation Self – Assessment Answers 1. b 2.a 3. d 4.B 5.d 6. C 7. a) b) c) A = cytoplasm B = nucleus C = chloroplast D = sap vacuole E = cell wall A: living material of the cell (many functions) B: controls the activities of the cytoplasm C: absorbs light during photosynthesis D: store of solutes / sugars, ions etc. E: maintains the shape of the cell / allows cell to be turgid Structure Root hair cell Xylem vessel Red blood cell A B C D E 8. a) Carbohydrate in plant cell walls. Maintains the shape of the cell. b) Green pigment in chloroplasts. Absorbs light during photosynthesis. c) Woody material in some plant cell walls/xylem vessels. Supports the vessel/prevents collapse/supports the plant. d) Red pigment in red blood cells. Transports oxygen from lungs to tissues. 9. a) Breaking down food and absorbing the (soluble) products into the blood. b) Any three, e.g. oesophagus, stomach, liver, pancreas, intestine. c) Any two of: respiratory, circulatory, excretory, nervous, endocrine, reproductive. d) Correct function and two organs, e.g. (Circulatory system) transport of materials around body. Heart/arteries/veins. (Nervous system) coordination/sending messages. Brain/spinal cord/nerves. 10. a) Clear, clean drawing lines Drawing is reasonable shape and proportions Labels: cell wall, cytoplasm, nucleus (cell membrane and sap vacuole not visible). b) Correct calculation, from drawing size/photo size. c) Iodine (solution) / methylene blue (solution). d) Cut open an onion and separate some of the leaves. Make a small square cut in the inner surface of one of the leaves. Using forceps peel off the thin inner epidermis of the leaf. Place the epidermis on a drop of water on a clean microscope slide. Add a 2 drop of stain. Gently lower a cover slip onto the specimen. Use a piece of filter paper to soak up any excess solution. Chapter 2: Diffusion and osmosis Self – Assessment Answers 1.b 2.c 3. d 4.A 5.a 6.C 7.c 8.b 9. Feature Diffusion Osmosis Active transport Movement is due to kinetic energy of the particles Movement requires energy from cell respiration Movement occurs down a concentration gradient 10. a) b) c) A: No cells visible B: Red cells with normal (biconcave disc) appearance C: Red cells shrunken, with crinkly edges (crenated) Solution B. The water potential of solution A is higher than the water potential inside the cells. The cells absorb water by osmosis, and burst. The water potential of solution B is the same as the water potential inside the cells. There is no net movement of water in or out of the cells by osmosis, so they maintain their normal appearance. The water potential of solution C is lower than the water potential inside the cells. The cells lose water by osmosis, and shrink. d) There are no intact cells left in tube A. The clear solution is red from the contents of the cells after they have burst (due to the haemoglobin released). Tubes B and C both contain intact cells. 11. a) From X to Y b)Osmosis c) Drawing should show a label line clearly ending on the position of the cell membrane, just beneath the cell wall. 3 Alternatively – the membrane around the sap vacuole (tonoplast) is also correct. d)Diffusion 12. a) At X, movement of the sodium ions is from low to high concentrations, i.e. against a concentration gradient. The method must be active transport. b) At Y, movement of the sodium ions is from high to low concentrations, i.e. down a concentration gradient. The movement is by diffusion. c) The folds increase the surface area, which enables a faster rate of diffusion. 13.a) 10.0 × × × 8.0 6.0 % change in mass 4.0 × 2.0 0.0 −2.0 −4.0 0 0.1 0.20.30.40.50.6 Concentration of sucrose solution/mol per dm3 × −6.0 −8.0 × −10.0 × b) 0.35 mol per dm3 (where the curve crosses the X-axis) c) Sucrose solution of concentration 0.5 mol per dm3 has a lower water potential than the contents of the potato cells. Water leaves the cells by osmosis, causing a decrease in mass of the tissue. d) Both these solutions are (very) hypotonic to the contents of the potato cells. The cells gain water by osmosis. When the cells become fully turgid, they can take in no more water, and their mass remains constant. e) There are several possibilities, e.g. ● Use several pieces of potato at each concentration of sucrose solution (find an average value of mass change at each concentration.) ● Repeat with intermediate concentrations of sucrose solution (0.25, 0.45 mol per dm3 etc.) ● Cut each piece of potato into smaller slices to increase the surface area for osmosis. 4 Chapter 3: Enzymes Self – Assessment Answers Rate of reaction / arbitrary units 1.d 2.b 3.a 4.a 5.c 6.a)70°C b) At 50°C the molecules of substrate and enzyme have more kinetic energy than at 20°C, so they move around faster and collide more often. Therefore enzyme-substrate collisions are more likely to result in a reaction. c) At 90°C the high temperature damages the structure of the enzyme molecule, including that of its active site, so that it can no longer bind with the substrate, decreasing the rate of reaction. This is called denaturing. 7. a) 40 cm3 b) 30 cm3 c) The curve at 40°C should be drawn to the left of the curve in Figure 3.11, with a steeper slope, but reaching the same maximum value. The higher temperature increases the rate of reaction, but the same amount of product (volume of gas) is produced, in a shorter time interval. 8. a) Iodine (solution) b) The colour changes from yellow to blue-black. c) 0.35 × 0.3 × 0.25 × × 0.2 0.15 × × 0.1 × 0.05 0 × 2 46 810 pH d) pH = 5 e) The optimum pH for most human enzymes is 7. This means that the two types of amylase must be chemically different, since they are differently affected by changes in pH. f) At this extreme of pH the very acidic conditions cause the enzyme to be denatured, so it does not catalyse the reaction. 5 g) Modifications should include: ● Repeat the experiment using different concentrations of amylase, such as 0.1%, 0.5%, 1%, 2%, 5% etc. ● Use 5 cm3 of pH = 5 buffer (i.e. the optimum pH) in all tubes ● Keep all other variables constant: temperature (20°C), concentration of starch suspension, volume of starch suspension (5 cm3) volume of amylase solution (1 cm3) Chapter 4: Plant Nutrition Self – Assessment Answers 1.a 2.d 3.A 4.c 5.B 6.c 7. d 8.a 9. a)Respiration b) With increasing light intensity the rate of photosynthesis increases, so more carbon dioxide is absorbed by the leaf. c) Another factor is limiting the rate of photosynthesis – carbon dioxide concentration or temperature. 10. a) Place the leaf in boiling water for 30 seconds. Transfer the leaf to a tube of ethanol and boil until the leaf is decolourised. Remove the leaf and wash in cold water. Spread the leaf out on a white tile and cover it with iodine solution. b) Yellow-brown to blue-black. c) Areas with starch should be those outside the area covered by the foil, plus the area of the triangular hole. d) Storage of carbohydrate. 11. a) Part of the leaf Function Epidermis Reduces water loss / protects against disease-causing organisms Stomata Allow entry and exit of carbon dioxide, oxygen and water vapour Palisade mesophyll Main site of photosynthesis Spongy mesophyll Photosynthesis and gas exchange between the cells and the air spaces Xylem Transport of water and mineral ions Phloem Transport of sugars and other products of photosynthesis 6 b) Many chloroplasts for photosynthesis / thin cell walls so light can easily enter them / vertical arrangement of cells (only one cell wall for light to cross). c) Spongy mesophyll cells have fewer chloroplasts / more rounded shape / less densely packed / not arranged vertically / surrounded by air spaces. 12. The first carbohydrate produced by photosynthesis is the sugar sucrose. This substance is a ‘single sugar’ or monosaccharide. It can be converted into other carbohydrates, such as the polysaccharide cellulose, which is the main component of plant cell walls. Carbohydrates consist of three elements: carbon, hydrogen and oxygen. Proteins contain two further elements nitrogen and sulfur, which are obtained from the soil. 13. a) Independent variable = light intensity, dependent variable = rate of photosynthesis. b) To supply carbon dioxide (for photosynthesis). c) 100 90 × Rate of photosynthesis / bubbles per minute 80 × × × 7 8 9 × 60 × 50 × 40 30 × 20 × 10 0 × 0 1 2 3 4 5 6 10 Light intensity / arbitrary units d) 51 bubbles per minute (approx.) e) Over the range 0-5 units of light intensity the rate of photosynthesis increases linearly with increasing light intensity, from 0 to 77 bubbles per minute. (The slope of the curve is constant). f) Above 5 units of light intensity the increase in the rate of photosynthesis starts to level off (the slope of the curve decreases). Above 7 units of light intensity the rate remains constant at about 90 bubbles per minute (the slope of the curve is zero). g) Between 0-5 units, light intensity is the limiting factor affecting the rate of photosynthesis. It is the factor, which, if increased, affects the rate. Above 5 units light intensity is no longer a limiting factor – some other factor is limiting the rate of photosynthesis. h) Between 0 and 5 units of light intensity there would be no difference in the curve. Above 5 units the curve would level off at a higher rate of photosynthesis (above 90 bubbles per minute). i) Bubbles may vary in size. The gas in the bubbles may not be pure oxygen (so bubble rate may not be a good indicator of the rate of photosynthesis). 7 j) There are several possibilities. The more obvious ones are: ● Collect the gas and measure the volume produced per minute ● Analyse the gas for its oxygen content ● Carry out replicates at each light intensity ● Carry out bubble counts at intermediate levels of light intensity. Chapter 5: Animal Nutrition Self – Assessment Answers 1.c 2.B 3.a 4.c 5.d 6.b 7.b 8. c 9. d 10.c 11.a 12.D 13. Enzyme Substrate Products amylase starch maltose maltase maltose glucose pepsin protein peptides lipase lipid fatty acids and glycerol 14.a) b) Starch – add iodine solution, colour change from yellow-brown to blue-black. Protein – add biuret solution. Violet (mauve) solution formed. Lipid – dissolve in ethanol and pour into water. White emulsion formed. i) Saliva is secreted in the mouth, containing the enzyme amylase. Amylase breaks down starch in the bread into maltose. Mechanical digestion takes place – chewing by the teeth. Saliva also moistens the food. ii) In the stomach the enzyme pepsin is secreted, which breaks down protein in the bread into peptides (short chains of amino acids). iii) In the duodenum bile is secreted from the liver. Bile emulsifies lipids in the bread. Pancreatic juice is added, which contains three enzymes: amylase (same reaction as in the mouth); lipase, which digests lipids into fatty acids and glycerol; and trypsin, which breaks down proteins into peptides. 15. (Diagram similar to Figure 5.20.) When the circular muscles contract and the longitudinal muscles relax, the diameter of 8 the gut becomes narrower. If this happens behind the food it will squeeze the food along the lumen. Rhythmic contractions and relaxations produce the wave known as peristalsis, which pushes the food along the length of the gut. 16. a) (Diagram similar to the villus shown in Figure 5.23.) b) i) The epithelial cells on the outside of the villus produce various enzymes that complete the digestion of food. ii) Glucose is absorbed through the epithelium and into the network of capillaries in the middle of the villus. Absorption is by active transport. iii) Fatty acids and glycerol enter the lacteal in the middle of the villus and are transported through the lymphatic system into the blood. 17. a) 37°C is body temperature, which is likely to be the optimum temperature for the enzyme. b) Pepsin digested the albumen, producing peptides (short chains of amino acids). c) A control is a part of the experiment that is set up to minimise the effects of variables other than the one under test, which in this case is the action of the pepsin. The conditions in tube 2 are exactly the same as in tube 1, except that there is no pepsin in the second tube. d) The lower temperature in tube 3 meant that the reaction took longer, due to the lower kinetic energy of the reactant and enzyme molecules. e) The optimum pH for pepsin is low (about 2) so in alkaline conditions there is little or no pepsin activity. f) The acid is present to kill any bacteria in the food, preventing food poisoning. g) Acid is neutralised by alkaline juices from the pancreas juice and bile. Chapter 6: Transport in Flowering Plants Self – Assessment Answers 1. c 2.C 3. b 4.A 5.D 6. A 7.d 8.a 9. At the leaves, water evaporates from the mesophyll cells, in the process called transpiration. This draws water up through the plant, in a movement called the transpiration stream. Water travels through the xylem in specialised cells called vessels which have cell walls thickened with lignin. Xylem forms a continuous pathway through the plant, from the roots to the leaves. The other transport tissue is phloem, which together with xylem forms structures in the stem known as vascular bundles. This second transport tissue contains living cells called sieve tubes, which carry the products of photosynthesis to different parts of the plant, in a process known as translocation. 9 10. a) A = epidermis, B = root hair, C = cortex, D = xylem, E = phloem b) The root hair cell contains a higher concentration of solutes than the soil water, so it has a lower water potential. Water therefore enters the root hair from the soil by osmosis. c) The cortex cells actively transport ions into the xylem, lowering the water potential in the xylem vessels. Water follows by osmosis. d) Diagram should show vascular bundles arranged in a peripheral ring around the stem. Labels to include xylem, phloem and vascular bundle. e) (Description of the method in Activity 6.1.) 11. a) Palisade mesophyll b) Any suitable examples, such as roots, stems, flowers, buds, young leaves, tubers, bulbs etc. c) Water evaporates from the mesophyll cells in the leaf into the air spaces in the spongy mesophyll. It then diffuses out through the stomata in the lower epidermis. 12. a) The diagram should be large, with clear lines and correct proportions. Labels should include three from: epidermal cell, stoma, guard cell, chloroplast. b) The magnification depends on the size of the drawing, and is calculated from the equation: length of guard cell (X) in drawing / μm Magnification = 25 μm c) Loss of water vapour is controlled by the opening and closing of stomata, which is brought about by changes in shape of the guard cells. On the side of each cell next to the stoma, the guard cell wall is thickened and less flexible. When water enters the guard cells by osmosis, the cells become turgid and bend outwards in a C-shape, opening up the stoma between the cells. Conversely, when water leaves the guard cells and they become flaccid, they straighten and close the stoma. Chapter 7: Transport in humans Self – Assessment Answers 1.b 2.d 3.c 4. B 5.A 6.C 7.b 8. a 9.d 10.b 11. a) i)muscle ii) coronary artery b) K – vena cava, L – aorta, M – pulmonary vein 10 c) i) Valves labelled correctly – semilunar valves at bases of aorta and pulmonary artery, bicuspid valve between left atrium and left ventricle, tricuspid valve between right atrium and right ventricle. ii) Arrows drawn correctly – arrow shown from left atrium to left ventricle and out through the aorta, second arrow from right atrium to right ventricle and out of the heart via the pulmonary artery. d)a → c → d → b 12. a)i) Red blood cells contain the pigment haemoglobin. In the lungs where the concentration of oxygen is high the haemoglobin combines with the oxygen, forming oxyhaemoglobin. In the tissues, where oxygen is at a low concentration, oxyhaemoglobin dissociates and gives up its oxygen, supplying the tissues with oxygen for respiration. Carbon dioxide is produced by respiring tissues, and passes into the blood by diffusion. Most carbon dioxide is carried as hydrogencarbonate ions in the plasma, with smaller amounts as carbon dioxide gas in solution, or combined with haemoglobin in the red cells. The blood carries the carbon dioxide to the lungs, where it is eliminated from the body. ii) There are two main ways the blood defends the body against pathogens – phagocytosis and antibody production. Phagocytes produce pseudopodia, which surround a bacterium and engulf it, enclosing the microorganism in a vacuole inside the cell. The phagocyte then secretes digestive enzymes into the vacuole, which break down the bacterium. Lymphocytes detect foreign antigens on a bacterium or virus and make antibodies, which are secreted into the blood plasma. The antibodies stick to the surface antigens on the pathogen, and destroy it in a variety of ways, such as neutralising bacterial toxins, causing a bacterium to burst, ‘labelling’ bacteria or clumping them together so that they are more easily dealt with by phagocytes. Some lymphocytes develop into memory cells, which remain in the body for many years and provide immunity to a pathogen if a second infection occurs. b) Fats from animal sources such as meat, cheese and other dairy products are particularly high in saturated fat. A diet that is rich in fat, especially saturated fat, raises blood cholesterol. Cholesterol is a lipid substance that combines with other fatty substances in the blood and can form a blockage in a coronary artery, called an atheroma, reducing blood flow to the heart muscle. Atheromas can result in various heart complaints, such as angina and heart attacks. 13. a) (Any five differences) Arteries Veins Small lumen Thick wall Wall contains much muscle and elastic tissue No valves, except at start of pulmonary artery and aorta Carry blood away from heart Blood pressure high Contain oxygenated blood (except pulmonary artery) Large lumen Thin wall Wall contains less muscle and elastic tissue Valves present throughout veins Carry blood towards heart Blood pressure low Contain deoxygenated blood (except pulmonary vein) 11 b) The walls of a capillary are made of a single layer of flattened epithelial cells. The walls of the capillaries are leaky, so that blood pressure forces fluid from the blood plasma out of the capillaries, forming the tissue fluid. Nutrients such as glucose and amino acids, as well as oxygen, can diffuse out of the blood in the capillaries into the tissue fluid, and from there into the body cells. Excretory waste products produced by the cells such as urea and carbon dioxide can diffuse in the opposite direction, from the cells into the blood in the capillaries. 14. a) 25.0 ÷ 5.0 = 5 times. b) An increase in heart rate and an increase in stroke volume. c) At rest: 0.8 × 100 = 16% 5.0 During exercise: 20.0 × 100 = 80% 25.0 d) The walls of blood vessels (arterioles) leading to these organs can constrict or dilate to vary the volume of blood passing through them. e) The brain needs a constant supply of blood to provide oxygen and nutrients for the brain cells to function correctly. 15. a) P = white blood cell / phagocyte; Q = red blood cell; X = plasma. b) Platelets; blood clotting. c) white blood cell phagocyte of drawing d)Magnification = width width of cell (both widths must be in the same units) 30 mm For example, if drawing width is 30 mm, magnification = 0.015 = x 2000 mm e) The biconcave disc shape of the red blood cell provides a large surface area to volume ratio, which allows more oxygen to diffuse into the cell. The flattened disc shape means that there is only a short distance for the oxygen to diffuse to the middle of the cell. Red cells have no nucleus, and contain haemoglobin, which combines with oxygen and transports it around the body. Chapter 8: Respiration Self – Assessment Answers 1.b 2.a 3.c 4.c 5.D 6.d 7.A 12 8.c 9. a) A unicellular fungus. b) glucose → ethanol + carbon dioxide (+ some energy) C6H12O → 2C6H5OH + 2CO2 (+ some energy) c) Manufacture of bread and alcoholic drinks (wine/beer). 10. a) P = larynx, Q = bronchus, R = intercostal muscles, S = diaphragm. b) Air passing over the larynx causes the vocal cords to vibrate, producing sound. c) R and S. d) Cells lining the trachea secrete a sticky fluid called mucus, which traps bacteria. Other cells are covered with cilia, which beat backwards and forwards in a coordinated way, pushing the mucus and bacteria up the airway in the direction of the mouth, where it is swallowed. 11. Inhalation Exhalation external intercostal muscles contract relax internal intercostal muscles relax contract ribs move down and inwards move down and inwards diaphragm contracts and flattens volume of thoracic cavity increases relaxes and assumes a dome shape decreases pressure in thoracic cavity decreases increases volume of air in the lungs increases decreases 12. a) ● Alveoli provide a large surface area for efficient diffusion of oxygen and carbon dioxide ● There is a short distance through the thin alveolus wall and the surrounding capillary wall for the rapid diffusion of respiratory gases between the air in the alveoli and the blood ● There is an intimate capillary network surrounding the alveolus, with a continuous blood supply to maintain diffusion gradients for the gases b) Carbon dioxide is mainly present as hydrogencarbonate ions in deoxygenated blood. The hydrogencarbonate ions are converted into carbon dioxide, which diffuses out through the capillary walls, across the alveolar walls and into the fluid lining the alveolus. From there it diffuses into the space in the alveolus, and expelled from the lungs during exhalation. c) (Description/diagram of apparatus shown in Figure 8.3). Gently breathe in and out through the rubber tube. Air is drawn in through the indicator solution in one boiling tube and pass out through the solution in the second tube. The indicator that has exhaled air passing through it will change colour first. 13. a) Tidal volume = 500 cm3. b) Six complete breaths are shown in 30 seconds, which = 12 breaths / minute. c) Total volume of air breathed per minute = tidal volume x number of breaths = (500 x 12) cm3 = 6000 cm3 = 6 cm3. 13 d) i) The volume increased (by 500 cm3). ii) The external intercostal muscles contracted, moving the ribs upwards and outwards. The diaphragm muscles contracted, lowering the diaphragm. The volume of the chest increased, lowering the pressure inside the chest cavity and drawing air into the lungs. e) The line should show an increase in the tidal volume and an increase in the frequency of breaths compared with that at rest 14. a) glucose+ oxygen → carbon dioxide + water (+ energy) C6H12O6+ 6O2 →6CO2 +6H2O (+ energy) b) Soda lime (or sodium hydroxide / potassium hydroxide). c) Carbon dioxide produced by the maggots is absorbed by the soda lime. The maggots use up oxygen in respiration, which decreases the volume of gas in the respirometer and draws the coloured liquid along the capillary tube towards the seeds. The rate of movement of the liquid is a measure of the rate of respiration of the seeds. d) An identical respirometer but with glass beads in place of the maggots. Chapter 9: Excretion Self – Assessment Answers 1. c 2.a 3.b 4.d 5.b 6.c 7. a) Excretion is the removal of the waste products of metabolism from the body. b) In the capillaries surrounding the alveoli of the lungs, hydrogencarbonate ions in the plasma are converted back into carbon dioxide, which then diffuses across the capillary wall and through the wall of the alveolus. The carbon dioxide diffuses into the air space in the middle of the alveolus, and is expelled from the lungs during exhalation. c) In light plants carry out photosynthesis and oxygen is a waste product of this process. 8. a) A = kidney, B = ureter, C = urethra. b) X = aorta, Y = (right) renal vein. c) Blood in Y (renal vein) contains less urea (and other nitrogenous waste products), more carbon dioxide and less oxygen than the blood in X (aorta). (It may also have different concentrations of water and salts, depending on the levels in the blood.) d) In the wall of the urethra close to the bladder is a sphincter muscle. This is a ring-like muscle, which contracts to close off the outlet from the bladder. 9. a) A urine sample would contain no protein, no glucose, more urea and more ammonia than the sample of plasma. b) The process is deamination. Amino acids are broken down into a carbohydrate and ammonia. The ammonia is combined with carbon dioxide to make urea. 14 10. The kidney machine, or renal dialysis machine, filters the patient’s blood to remove urea and other waste products, along with excess water and salts. Arterial blood is taken from the patient’s arm and pumped through the dialyser. The filtered blood is returned to a vein. The filter in a dialyser consists of a dialysis membrane. This is partially permeable – it has holes that are large enough to allow small molecules to pass through, but will not allow the passage of larger molecules such as proteins, or the passage of blood cells. Modern kidney machines use dialysers containing many narrow tubes made of dialysis membrane. Blood flows through the tubes and dialysing fluid flows around the outside of the tubes, in the opposite direction. Unwanted solutes pass through the membrane from the blood to the dialysis fluid. Dialysis fluid contains the same concentration of glucose and salts as blood, but no urea. As the patient’s blood flows past the membrane, urea, along with any excess salts and water, diffuses through the holes in the membrane into the dialysis fluid, while cells and proteins are retained in the blood. There is urea in the blood but not in the fresh dialysis fluid, so there is a steep concentration gradient for diffusion of the urea. However there is no concentration gradient for diffusion of glucose, which remains in the blood. The dialysis fluid is constantly replaced with fresh solution and the ‘used’ fluid removed, so that after several hours the patient’s blood is ‘cleaned’ of waste and the correct water and salt balance achieved. Chapter 10: Homeostasis Self – Assessment Answers 1.c 2.b 3.d 4.d 5.c 6.a 7. a) P = sweat gland; Q = hair erector muscle; R = capillaries; X = epidermis. b) When the body temperature rises, the sweat glands increase their secretion of sweat. The Sweat passes out of the sweat gland through a pore onto the surface of the skin, where it evaporates. The heat to evaporate the sweat is supplied by the body, cooling the surface. When the body temperature falls, little sweat is produced, so there is no cooling effect. c) Melanin absorbs ultraviolet radiation from the sun, preventing it from causing mutations to DNA. 15 8. Hot environment Blood flow Sweating Hairs in skin Shivering Metabolism Cold environment Vasodilation increases blood flow (Vasoconstriction reduces blood to surface capillaries, so that more flow to surface capillaries, so that heat is lost from the skin surface less heat is lost from the skin surface) Secretion of sweat increases. Secretion of sweat decreases. More heat is taken from the skin in Less heat is taken from the skin in evaporating the sweat evaporating the sweat Erector muscles relax, allowing hairs to lie flat, so there is no insulating layer of air next to the skin Shivering (rapid contraction and relaxation of muscle) does not occur Erector muscles contract, pulling hairs upright. Hairs trap an insulating layer of air next to the skin Shivering (rapid contraction and relaxation of muscle) occurs, generating heat The rate of metabolism in organs such as the liver and muscles decreases, generating less heat The rate of metabolism in organs such as the liver and muscles increases, generating more heat 9. a) Cover one beaker with a layer of cotton wool and secure it in place using the rubber bands. Using the measuring cylinder, measure out the same volume of hot water from a kettle into the two beakers. Cover each beaker with a lid. Measure the starting temperature of the water in each beaker using the thermometer. Repeat the measurements of temperature at five-minute intervals. b) The temperature of the water in the insulated beaker will fall more slowly than the water in the beaker without insulation. c) i) Repeat the investigation and find an average of the results. ii) Use two thermometers, leave one in each beaker throughout the investigation. d) i) Body temperature is 37°C, lower than that of the hot water from the kettle. ii) The model is different in many respects – insulation in the skin is fat, not cotton wool, the body is not shaped like a beaker (any sensible difference). Chapter 11: Coordination and response Self – Assessment Answers 1.A 2.c 3. b 4.A 5.D 6.c 7.b 16 8.B 9.d 10.c 11. a)i) medulla ii)cerebellum iii) cerebrum b) Patient A: the motor area of the cerebrum Patient B: the part of the cerebrum concerned with memory 12. a) i) Light intensity is detected by nerve endings in the retina, and transmitted to the brain via the optic nerve. In the brain impulses are passed to motor neurones, which pass out of the brain via a different nerve to the iris muscles. In bright light, circular muscles of the iris contract, constricting the pupil and reducing the amount of light entering the eye. In dim light, radial muscles in the iris contract, dilating the pupil and allowing more light to enter the eye. ii) The response is a reflex because it is involuntary – not initiated by the brain. iii) The reflex allows the correct intensity of light to fall on the retina in order to produce an image, and too much light would damage the retina. b) Light entering the eye is refracted by the cornea and passes through the pupil. It is further refracted by the lens, which focuses the light rays onto the retina. In the retina, light-sensitive cells called rods and cones convert the light into nerve impulses, which pass out via the optic nerve to the brain. c) Clouding of the lens means that less light will be able to pass through it and reach the retina. The image will be blurred, and the person will have difficulty in seeing clearly (particularly in dim light). Intensity of colours will be decreased. 13. a) A = sensory neurone, B = relay neurone, C = motor neurone. b) The sensory neurone carries impulses from the pain receptor to the spinal cord. The relay neurone receives impulses (via synapses) from the sensory neurone and passes them to the motor neurone. The motor neurone sends impulses out to the muscle fibres, causing them to contract. c) P = grey matter, Q = white matter. d) i) The person would neither move the arm nor be able to feel the pain (since both the route to the motor neurone and to the brain are blocked). ii) The person could feel the pain, but the reflex would not work to move their arm (the route to the motor neurone is blocked, but not neurones in the spinal cord leading to the brain). 17 14. Nervous system Endocrine system Form in which information is carried Route by which information is transmitted Speed of response (fast or slow) Response short-lived or longer-lasting Effect localised or widespread electrical signals (also chemical across synapses) By neurones In the blood Response voluntary or involuntary Voluntary or involuntary (Usually) fast (Usually) slow Short-lived Mostly longer-lasting (Usually) localised Sometimes affects one target organ, but often widespread Involuntary 15. a) 80 mg per 100 cm3 of blood (approximately). b) Approximately 1.5 hours (after ingesting the glucose). c) Insulin stimulates the uptake of glucose into cells from the blood. It stimulates cells of the liver and muscles to convert glucose into glycogen, and increases the oxidation of glucose by tissue respiration. d) The concentration of glucose in the blood of patient Q is always higher than that in the normal person, for example it starts at 180 mg per 100 cm3 of blood, compared with the 80 mg per 100 cm3 in the normal person. Patient Q’s blood glucose rises more rapidly, reaching a maximum of 360 mg per 100 cm3 after 1 hour. After this time, the level in patient Q falls more slowly than in the normal patient, and is still 220 mg per 100 at 6.5 hours, while that of the normal patient has returned to 80 mg per 100 after 3 hours. e) Patient Q may be producing some insulin (but less than normal). She will excrete some glucose in her urine, and some will be oxidised by cell respiration. 16. a) Student number of correct responses finger tip back of hand palm of hand back of neck A 10 8 9 5 B 10 8 7 6 C 9 7 8 4 D 10 8 9 7 E 10 7 8 4 mean 9.8 7.6 8.2 5.2 18 b) Ten replicate tests allow the calculation of a mean value and increases the reliability of the results. c) i) Most sensitive – finger tip. Least sensitive – back of neck. ii) Some areas of the skin (e.g. the finger tips) have a higher concentration of touch receptors. d) Any suitable suggestions, such as: ● Test the skin in other areas of the body ● Repeat using the points at different distances apart (2 mm, 10 mm etc.) ● Increase the sample size. Chapter 12: Support, movement and locomotion Self – Assessment Answers 1.c 2.d 3.a 4.b 5.A 6. Joint Bones meeting at the joint Type of joint Number of possible planes of movement elbow humerus meets radius and ulna hinge one shoulder scapula meets humerus ball and socket three hip pelvis meets femur ball and socket three 7.a)A = scapula, B = humerus, C = triceps, D = tendon, E = ligaments, F = biceps, G = radius, H = ulna. b) i) The tendon (D) connects the (triceps) muscle to the (ulna) bone. The tendon is strong and inelastic, so that it doesn’t stretch when the muscle contracts to pull on the bone. ii) The ligaments (E) run from one bone to another across the elbow joint. They hold the bones together at the joint, preventing them from dislocating. c) Antagonistic muscles are pairs of muscles that work together to move a bone. When one muscle contracts, the other relaxes, and vice versa. d) The triceps muscle contracts and the biceps relaxes. Contraction of the triceps pulls on the ulna, moving the arm to the extended position. 19 Chapter 13: The use and abuse of drugs Self – Assessment Answers 1.d 2.c 3. c 4.a 5.b 6.D 7. a) i) A drug is any externally administered substance that affects chemical reactions in the body. ii) Drug abuse means using a drug for a purpose other than that for which it is normally prescribed. b) A narcotic drug results in a powerful feeling of contentment and euphoria. A depressant acts on the nervous system to slow down nervous communication and reaction times. Tolerance means that the drug user needs to take more of the drug to have the same effects. Addiction means that the user becomes psychologically dependent on taking the drug, despite the adverse consequences. c) Heroin addicts are unable to live without regularly taking the drug, and become preoccupied with obtaining and using heroin, often turning to crime to fund the habit. 8. a) The airways leading to the lungs are protected by the secretion of mucus, which traps dirt particles and bacteria. The mucus is removed by the beating of tiny hairlike structures called cilia. Tar in cigarette smoke destroys these structures, so that contaminants are not removed from the lungs. Long-term irritation of the airways produces a disease called bronchitis, where the smoker develops a persistent cough and difficulty in breathing. Tar also contains carcinogens, which can cause lung cancer. When this happens, cells in the lungs divide out of control, producing a growth called a tumour. b) Many people now regard smoking as being no longer socially acceptable, because of the unpleasant smell and irritation caused by tobacco smoke. There is also strong evidence that passive smoking is harmful to the non-smoker, and can even increase their risk of developing lung cancer. 9. a) An antibiotic is a substance that is produced by a microorganism, which kills or reduces the growth of other microorganisms. It is used in medicine to treat bacterial infections. b) Penicillin. It prevents the proper synthesis of the bacterial cell wall. This weakens the wall, so that water enters the bacterial cells by osmosis and they burst. (Alternatively – tetracycline). It kills bacteria by preventing them from synthesising proteins, including enzymes. 10. a) The short-term effects of alcohol abuse include decreased mental functioning, and an increase in reaction times; loss of muscular control, coordination and balance, slurred speech and blurred vision, nausea and vomiting, and reduced self-control. Long-term abuse leads to cirrhosis of the liver, and other diseases such as liver cancer, stomach ulcers and cardiovascular (heart) disease. Alcoholism also causes a wide range of psychological problems, including anxiety, depression and personality disorders. 20 b) Alcohol abuse is associated with domestic violence, and crimes such as burglary and assault. It can lead to the alcoholic becoming isolated from family and friends, and to marriage problems and divorce. Chapter 14: Microorganisms and biotechnology Self – Assessment Answers 1.b 2.d 3.a 4.c 5.d 6. d 7. c 8. Feature Type of organism mould yeast bacterium virus Composed of hyphae Reproduce inside living cells Use saprotrophic nutrition Have a cell wall Form reproductive spores 9. a) A diagram similar to Figure 14.7 is needed, with the labels asked for in the question. b) Plasmids are small circular rings of DNA, present in the cytoplasm of a bacterium. These carry some of the cell’s genes. 10. a) A cell grows until its length has doubled. In the cytoplasm, the single chromosome is copied by a process called replication, and the two copies separate to opposite ends of the cell. The cell membrane now folds inwards, forming a double layer across the middle of the cell. Two new cell walls are made, in between the two membranes. The two daughter cells then separate. b) Time / minutes 0 20 40 60 80 100 120 140 Number of cells 1 2 4 8 16 32 64 128 21 160 180 200 220 240 c) 256 512 1024 2048 4096 After 4 hours there are 4096 cells. 5000 × Number of cells 4000 3000 × 2000 × 1000 × × × × × 0× × × × × 050 100150200250 Time / hours 11. The graph shows a steep increase in the gradient of the curve (exponential). Milk is pasteurised by heating to 71.7°C for 15-20 seconds. A starter culture containing lactic acid bacteria is added to the milk. The bacteria convert lactose sugar into lactic acid, which lowers the pH of the milk. The solids are separated and allowed to ripen and mature into the finished cheese. A substance called rennet is added, which contains an enzyme that coagulates proteins in the milk. The milk separates into solids called curds and a liquid called whey. 22 12. a) Very hot steam at a high pressure is passed through the fermenter to sterilise it. b) The fermenter tank is surrounded by a water jacket. Cold water is passed through the jacket, keeping the contents of the fermenter at the optimum temperature. c) Sugars, a nitrogen source such ammonia or amino acids, and mineral ions (e.g. magnesium, potassium, sulfate and phosphate). d) The air supplies oxygen for the (aerobic) respiration of the Penicillium. The air is filtered to prevent other microorganisms entering the fermenter from the outside and contaminating the broth. e) If the paddles stopped, the broth would not be stirred. The Penicillium would not be mixed with the nutrients so well, so growth of the mould would be reduced. f) The penicillin is extracted from the broth using a series of organic solvents. It is then re-dissolved back into a solution of sodium hydrogencarbonate, and allowed to crystallise out as the sodium salt of penicillin. Chapter 15: The relationships of organisms with one another and with the environment Self – Assessment Answers 1.C 2.b 3.a 4.b 5.d 6.d 7.B 8.a 9.a 10.c 11.b 12.d 13. a) large fish humans small fish crabs zooplankton worms phytoplankton mangrove leaves 23 b) Any food chain with 4 trophic levels, e.g. phytoplankton → zooplankton → small fish → large fish c) Dead organisms are food for decomposers (fungi and bacteria). The decomposers produce carbon dioxide by respiration, which can be used by the producers for photosynthesis. 14. a)phytoplankton → copepods → krill → Antarctic cod → squid → Weddell seal b) primary consumer and secondary consumer c) Energy is lost by respiration of the copepods, and in excretory products and egestion. d) In general the efficiency decreases – the % energy transfer is lower near the ends of the food chains. e) The producers in marine food chains (phytoplankton) have a higher productivity than producers in terrestrial food chains. They are more efficient at trapping the Sun’s energy than terrestrial plants, so there is more energy available at the start of the food chain. 15. a) P = ammonium ions, Q = nitrate b) X = decomposers / fungi and bacteria, Y = nitrifying bacteria c) Uptake via root hairs, by active transport d) By excretion (e.g. as urea) and by death and decomposition by decomposers, forming ammonium ions. 16. a) adult mosquito (lives on land) eggs (laid on water) pupa (lives in water) larva (lives in water) b) A vector is any organism that transmits pathogens from one host to another (in this case the mosquito transmits the malaria parasite to humans). c) Any three methods from: ● Spraying the insecticide on the water where the larvae live ● Draining marshes, ponds, lakes or any standing water ● Stocking lakes and ponds animals that feed on mosquito larvae ● Spraying a thin layer of oil onto the water where mosquitoes breed 17.a) carbon dioxide, methane, water vapour (or any other correct greenhouse gas) b) (Answer should include a diagram similar to Figure 15.25.) Short wavelength infrared 24 c) d) 18. a) b) c) radiation from the Sun passes through the atmosphere, and is absorbed by the surface of the Earth. The Earth radiates a longer wavelength infrared radiation, which is absorbed by greenhouse gases in the atmosphere. They re-emit the radiation, warming the atmosphere and the surface of the Earth. Without the greenhouse effect the Earth would be 33°C cooler – too low to sustain life. If there are more greenhouse gases in the atmosphere, they absorb and re-radiate more infrared radiation, increasing the surface temperature of the Earth. The probable cause is extra carbon dioxide produced by the combustion of fossil fuels. An ecosystem is a community of living organisms, together with their non-living environment. The insecticide was probably applied to crops on land nearby, and entered the lake by leaching or run-off. The phytoplankton absorbed the insecticide from the lake water. When the crustaceans fed on the phytoplankton, the insecticide was not broken down in their bodies or excreted, but became concentrated in their tissues. When the fish ate the crustaceans this process was repeated, and again when the bird ate the fish, so that the insecticide built up in concentration along the food chain. Chapter 16: Development of organisms and the continuity of life Self – Assessment Answers 1. D 2. c 3.c 4.a 5. b 6. d 7.b 8.c 9.B 10. b 11.a 12. d 13. a) Tubers are formed by one parent plant and do not involve the fusion of gametes from two parents. b) Tubers are a result of cell division by mitosis, which results in cells that are genetically identical. c) The plant has flowers, which are the sex organs that produce gametes. d) Asexual reproduction produces genetically identical offspring. This is an advantage when the environment is stable and the organism is well adapted. Sexual reproduction produces offspring that show variation. If the environment changes, this variation may allow new forms of the plant to be able to survive better. 25 14. a) b) c) d) P = stigma, Q = anther, R = ovule. Brightly coloured petals, nectar, scent. Answer should be a diagram similar to Figure 16.19. Any four from: ● flower is small and green, with no brightly coloured petals ● no nectaries ● stamens dangle loosely outside the flower instead of being stiff and enclosed within the flower ● feathery stigma, exposed outside the flower ● large amounts of small, smooth light pollen grains, instead of smaller amounts of large, sticky or spiky pollen grains 15. a) Both fruits are wind-dispersed. One has hairs forming a ‘parachute’, the other is shaped like a wing or helicopter blade. Both adaptations allow the fruits to carry their seeds away from the parent plant. b) Any two of the following methods of animal dispersal: ● Animals remove the fruits and carry them some distance away from the parent plant. They eat the fruit and drop the seeds. ● Small seeds are eaten with the fruit, pass undigested through an animal’s gut and out with the faeces. ● Seeds have dry pericarps with hooks, hairs or spikes that catch in an animal’s fur. ● Seeds are carried in mud on the feet of animals or birds. (Alternatively methods of self-dispersal or dispersal by water may be described.) c) If seeds all germinated and grew near the parent plant they would compete for resources, such as water, light and minerals. Seed dispersal prevents overcrowding and reduces competition for resources, as well as allowing plants to colonise new habitats. d) The offspring would be different from the parent plants, because they are formed by sexual reproduction. Sexual reproduction involves meiosis to produce gametes, and fertilisation. Both these processes result in genetic variation, which would produce variation in the appearance of the offspring. 16. a) A = sperm duct, B = penis, C = urethra, D = prostate gland, E = testis, F = scrotum. b) A carries sperm from the testes to the urethra during sexual intercourse. D secretes a nutritive fluid that is added to the sperm, forming semen. F is a sac that holds the testes outside the main body cavity. This maintains the temperature of the testes a few degrees lower than the core body temperature, which is necessary for sperm development. c)E 17.a)oestrogen b) day 14 c) Thickening prepares the lining of the uterus for implantation of a fertilised egg. d)progesterone e) The fetus is inside the amniotic sac, which secretes amniotic fluid. The fluid fills the space between the amnion and the embryo. The fluid gives mechanical protection to the developing embryo, reducing friction and acting as a shock absorber. 26 18.a) b) 19.a) b) c) d) e) Any three from: ● sexual intercourse with an infected person ● blood-to-blood contact, e.g. when drug users share an infected needle or syringe ● across the placenta from an HIV-positive woman to the fetus, or at birth ● by breastfeeding, from an HIV-positive woman to her baby. ● Using condoms – this barrier method of contraception greatly reduces the risk of transmission of the virus. When condoms are used by a couple in which one person is HIV-positive, the rate of HIV infection is less than 1% per year. ● Reducing the sharing of needles by drug users – drug users can be encouraged to use sterile needles and syringes, e.g. by supplying these through needle-exchange programmes. ● Supplying anti-viral drugs to pregnant women who are HIV-positive – treating the mother with anti-viral medicines before the birth (and after birth giving these medicines to the child) reduces the risk of transmission by up to 99%. ● Encouraging HIV-positive mothers to bottle-feed their babies – if bottle-feeding is possible, it can prevent transmission of the virus through breast milk. Alternatively, the mothers may be given anti-viral medicines during breastfeeding. ● Education programmes – educating people about HIV/AIDS, and promoting the idea that they should keep to one sexual partner, as well as avoiding ‘high risk’ groups, such as prostitutes. Boiling removes oxygen from the water. The oil prevents oxygen entering the water from the air. A = water, B = oxygen, C = a suitable temperature. A control tube would contain peas on wet cotton wool, and the tube would be maintained at room temperature (20°C). Tube A = no germination, tube B = no germination, Tube C = no germination (or less than in the control tube). Chapter 17: Inheritance Self – Assessment Answers 1. d 2.a 3.c 4.b 5. c 6.d 7.A 8.a 9. b 10.c 27 11. a) A gene is a unit of inheritance, consisting of a length of DNA in a chromosome. Each gene determines a particular feature of an organism. Alleles are different forms of the gene. b) Each member of a homologous pair of chromosomes is identical in size (length). Each carries genes in a fixed order, with genes for certain characteristics at the same position on each of the pair. The alleles at a particular gene locus may be the same or different in each member of the pair. c) Each gene consists of a length of DNA that contains a set of instructions for making a particular protein. (The code consists of a sequence of base pairs in the DNA molecule). 12. a) Three. b) Individual 4 has CF, yet neither of his parents is affected. If the allele for CF was dominant, at least one of individual 4’s parents would have the allele, and suffer from CF. c) 1 = Aa, 4 = aa, 5 = Aa, 7 = Aa, 10 = aa d) Parents’ genotypes: father (4) = aa, mother (5) = Aa Probability that next child is affected = 0.5 (50%) father’s gametes mother’s gametes (a) (a) (A) Aa normal Aa normal (a) aa affected aa affected e) i) Gene mutation is the mutation that affects a single gene, so that a new allele is produced. ii) CFTR is made of proteins. iii) Experiments are carried out by using viruses as vectors. 13. Cross Ratio of genotypes in F1 Ratio of phenotypes in F1 TT x TT all TT all tall TT x Tt 1TT : 1Tt all tall TT x tt all Tt all tall Tt x Tt 1TT : 2Tt : 1tt 3 tall : 1 dwarf Tt x tt 1Tt : 1tt 1 tall : 1 dwarf tt x tt all tt all dwarf 28 14.a) A codominant gene is one in which both alleles are expressed in the phenotype of a heterozygote. b) Let allele for red = R and allele for white = W i) Parents’ genotypes: RR x WW gametes (R) (R) (W) RW roan RW roan RW roan RW roan gametes (R) (R) (R) RR red RW roan RR red RW roan gametes (R) (W) (R) RR red RW roan (W) RW roan WW white (W) ii) Parents’ genotypes: RR x RW (W) iii) Parents’ genotypes: RW x RW 15.a) Parents genotypes: man = HbSHbS woman = HbAHbS eggs sperm (HbA) (HbS) (HbS) HbAHbS carrier HbSHbS sickle cell anaemia (HbS) HbAHbS carrier HbSHbS sickle cell anaemia b) The parasite that causes malaria spends part of its life cycle inside red blood cells. The red blood cells of carriers are mostly normal in appearance, but because they contain about 40% abnormal haemoglobin, they are more delicate than normal red blood cells, and are easily damaged. When the malaria parasite enters these cells they burst, and the parasite is killed before it can develop any further. This interrupts the life cycle of the parasite, so the person is less likely to develop the full signs and symptoms of malaria. 29 c) In areas where malaria is endemic, the heterozygote has an increased resistance to the disease, so carriers of the allele are more likely to survive to reproduce, passing on the allele to their offspring. This is called a selective advantage – natural selection acts in favour of the phenotype, and maintains the sickle cell allele in the population. A selection pressure is an environmental factor that produces differential mortality. In this case the selection pressure is the presence of malaria. d) The genotype homozygous for the normal allele (HbAHbA). 16. When myxomatosis was first used to kill the rabbits it was very effective, because they had no gene for resistance to the disease. Soon strains of rabbits appeared that were less likely to die from the disease. These carried a gene mutation that gave them resistance to myxomatosis. The resistant rabbits had a strong selective advantage and survived to breed, while the non-resistant rabbits were at a selective disadvantage, and more of these were killed. The selection pressure is the presence of myxomatosis, and resistant rabbits are better adapted to the environment (fitter) when the disease is present. This is an example of survival of the fittest. 17. a) Step 1 = restriction enzyme (restriction endonuclease); step 2 = DNA ligase. b) The plasmid acts as a vector to transfer the human insulin gene into bacteria. c) The transgenic bacteria are cultured in an industrial fermenter, where they can be grown to produce insulin on a commercial scale. d) The insulin is used to treat people with diabetes. This avoids the need to use insulin from animal tissues, which is slightly different in structure from human insulin, and does not give such good control of blood glucose levels in humans. 30 Biology for Cambridge O Level Biology for Cambridge O Level has been written to provide up-to-date coverage of the material required for the latest Cambridge International Examinations O Level Biology syllabus. Key features include: • A comprehensive text covering the material needed for Cambridge O Level Biology • Full colour illustrations, engaging photographs and clear diagrams • Learning objectives at the start of each chapter and chapter summaries to draw together key points and help students review their learning • Margin boxes containing extension material and additional points of interest to develop students’ knowledge and understanding of the topics • In-text boxes containing definitions and explaining key terms • A full range of practical activities covering the requirements of the syllabus • Self-assessment questions throughout the chapters to allow students to check their understanding as they go along • Cambridge O Level style practice questions at the end of each chapter, to support revision • A separate section of past Cambridge O Level questions Dr Phil Bradfield is a highly experienced biology teacher and author of many leading texts. 31