aqa a level biology topic 3 exam fully solved & updated

In large cells of U. marinum, most mitochondria are found close to the cell-surface membrane. In smaller cells, the mitochondria are distributed evenly throughout the cytoplasm. Mitochondria use oxygen during aerobic respiration. Use this information and your knowledge of surface area to volume ratios to suggest an explanation for the position of mitochondria in large U. marinum cells. - answer-1. Large(r) cells have small(er) surface area to volume ratio; 2. (Takes) longer for oxygen to diffuse (to mitochondria) OR Less/no oxygen diffuses (to mitochondria) OR Diffusion distance/pathway is long(er); Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake (2 marks) - answer-1. Large(r) organisms have a small(er) surface area:volume (ratio); OR Small(er) organisms have a large(r) surface area:volume (ratio); 2. Overcomes long diffusion pathway OR Faster diffusion; Figure 1 shows two models of oxygen uptake found in animals. Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B. (2 marks) - answer-1. Water has low(er) oxygen partial pressure/concentration (than air); 2. So (system on outside) gives large surface area (in contact with water) OR So (system on outside) reduces diffusion distance (between water and blood); 3. Water is dense(r) (than air); 4. (So) water supports the systems/gills; The table below shows features of two mammals. Bats are flying mammals; shrews are ground-living mammals. Mammal Mean body mass / kg Mean lung volume / cm3 Bat 0.096 12.48 Shrew 0.024 0.72 Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew. Give your answer to an appropriate number of significant figures. Give one suggestion to explain this difference. - answer-1. and 2. Correct answer for 2 marks, 4.3 (times greater);; Accept for 1 mark, 4. (correct answer not given to 2 significant figures) OR Evidence of 130 (cm3 kg-1) and 30 (cm3 kg-1) 3. Provides more oxygen for respiration A scientist investigated the affinity for oxygen of horse haemoglobin and mouse haemoglobin. Some of their results are shown in the table. Animal Partial pressure of oxygen when haemoglobin is 50% saturated / kPa Partial pressure of oxygen when haemoglobin is 25% saturated / kPa Body mass of one animal / g Horse 3.2 1.9 550 000 Mouse 6.5 3.3 23 (b) The following equation can be used to estimate the metabolic rate of an animal. Metabolic rate = 63 × BM−0.27 BM = body mass in grams Use this equation to calculate how many times faster the metabolic rate of a mouse is than the metabolic rate of a horse. - answer-Correct answer of 15 (times faster) = 2marks 23−0.27 divided by 550 000−0.27 Mammals such as a mouse and a horse are able to maintain a constant body temperature. Use your knowledge of surface area to volume ratio to explain the higher metabolic rate of a mouse compared to a horse. (3 marks) - answer-Mouse 1. (Smaller so) larger surface area to volume ratio; 2. More/faster heat loss (per gram/in relation to body size); 3. (Faster rate of) respiration/metabolism releases heat; Describe the relationship between size and surface area to volume ratio of organisms (1 mark) - answer-As size increases, ratio (of surface area to volume) decreases; A scientist calculated the surface area of a large number of frog eggs. He found that the mean surface area was 9.73 mm2. Frog eggs are spherical. The surface area of a sphere is calculated using this equation Surface area = 4πr2 where r is the radius of a sphere π = 3.14 Use this equation to calculate the mean diameter of a frog egg. Show your working - answer-Two marks for correct answer in range of 1.75 to 1.76032;; Accept for 1 mark, incorrect answer using radius 0.87 / 0.88 / 0.880 / 0.8802 / 0.88015; OR Accept for 1 mark, incorrect answer with correct rearranged equation, e.g., Radius r2 = surface area ÷ 4 π OR r2 = 9.73 ÷ 12.56 OR r2 = 0.77 / 0.774 / 0.775 The scientist calculated the ratio of surface area to mass for eggs, tadpoles and frogs. He also determined the mean rate of oxygen uptake by tadpoles and frogs. His results are shown in the table. Stage of frog development Ratio of surface area to mass Mean rate of oxygen uptake / μmol g-1 h-1 Egg 2904 : 1 no information Tadpole 336 : 1 5.7 Adult 166 : 1 1.3 (c) The scientist used units of μmol g-1 h-1 for the rate of oxygen uptake. Suggest why he used μmol in these units. - answer-(Measures) small uptake / amount / quantity / volume / concentration / rate (of oxygen uptake); OR Avoids use of powers of ten / standard form / many decimal places; The scientist decided to use the ratio of surface area to mass, rather than the ratio of surface area to volume. He made this decision for practical reasons. Suggest one practical advantage of measuring the masses of frog eggs, tadpoles and adults, compared with measuring their volumes. - answer-More accurate / less error (in measuring mass); OR Causes less distress / damage to animal (to measure mass); OR Easier / quicker (to find mass) because irregular shapes; OR Fewer measurements / calculations; Explain why oxygen uptake is a measure of metabolic rate in organisms (1 mark) - answer-(Oxygen used in) respiration, which provides energy / ATP; OR (Oxygen is used in) respiration, which is a metabolic process / chemical reaction; A student who looked at these results said that they could not make a conclusion about the relationship between stage of development and metabolic rate. Use information in the table to explain reasons why they were unable to make a conclusion. (3 marks) - answer-1. No information about egg; 2. So cannot compare all stages (in Table 2); 3. No statistical information / test / t-test / comparison of standard deviations; OR No measure of significant differences; Tubifex worms are small, thin animals that live in water. They have no specialised gas exchange or circulatory system. The figure below shows a tubifex worm. Name the process by which oxygen reaches the cells inside the body of a tubifex worm. (1 mark) - answer-(Simple) diffusion Using the information provided, explain how two features of the body of the tubifex worm allow efficient gas exchange (2 marks) - answer-1. Thin/small so short diffusion pathway; 2. Flat/long/small/thin so large surface area to volume ratio/surface area : volume; Weddell seals are diving mammals that live in cold environments. A Weddell seal is shown in Figure 1. (i) Explain how the body shape of a Weddell seal is an adaptation to living in a cold environment. - answer-1. Small SA:VOL; 2. (So) reduces heat loss / (more) heat retained; Weddell seals can remain underwater for long periods of time. Figure 2 shows the rate of blood flow to different organs of a Weddell seal before a dive and during a long dive. Describe and explain the changes in the rate of blood flow to the different organs during a long dive. (3 marks) - answer-1. Brain is the same, others fall; 2. Brain controls other organs / remains active / needs constant supply of oxygen; 3. Lungs not used / are used less / seal is not breathing / heart rate decreases / heart pumps less / blood diverted to muscles; Describe how oxygen in the air reaches capillaries surrounding alveoli in the lungs. Details of breathing are not required. (4 marks) - answer-1. Trachea and bronchi and bronchioles; 2. Down pressure gradient; 3. Down diffusion gradient; 4. Across alveolar epithelium. 5. Across capillary endothelium / epithelium. Breathing out as hard as you can is called forced expiration. Describe and explain the mechanism that causes forced expiration. (4 marks) - answer-1. Contraction of internal intercostal muscles; 2. Relaxation of diaphragm muscles / of external intercostal muscles; 3. Causes decrease in volume of chest / thoracic cavity; 4. Air pushed down pressure gradient. Describe and explain one feature of the alveolar epithelium that makes the epithelium well adapted as a surface for gas exchange. Do not refer to surface area or moisture in your answer. (2 marks) - answer-1. Flattened cells OR Single layer of cells; 2. Reduces diffusion distance/pathway; 3. Permeable; 4. Allows diffusion of oxygen/carbon dioxide; Doctors measure the health of lungs by calculating the FEV1:FVC ratio. • FEV1 is the maximum volume of air exhaled in one second. • FVC is the maximum volume of air exhaled in one breath. The minimum FEV1:FVC ratio of healthy lungs is 0.7:1 A man with the lung disease emphysema inflated his lungs fully. He then exhaled as much of this air as quickly as possible in one breath. The figure below shows how the volume of exhaled air changed during this breath. Use the information provided to determine the FEV1:FVC ratio of this man's lungs. Go on to determine how many times greater the minimum ratio of healthy lungs is than his ratio. - answer-Correct answer for 2 marks = 1.10-1.15;; Accept for 1 mark, 0.6(1) : 1 (correct FEV1 : FEC ratio) Tidal volume is the volume of air inhaled and exhaled during a single breath when a person is resting. The tidal volume in a person with emphysema is reduced compared with the tidal volume in a healthy person. Suggest and explain how a reduced tidal volume affects the exchange of carbon dioxide between the blood and the alveoli. - answer-1. Less carbon dioxide exhaled/moves out (of lung) OR More carbon dioxide remains (in lung); 2. (So) reduced diffusion/concentration gradient (between blood and alveoli); 3. Less/slower movement of carbon dioxide out of blood OR More carbon dioxide stays in blood; Figure 2 shows changes in concentration of oxygen in two gas exchange systems. A student studied Figure 2 and concluded that the fish gas exchange system is more efficient than the human gas exchange system. Use Figure 2 to justify this conclusion. - answer-1. In fish, blood leaving (V) has more oxygen than water leaving (E); 2. (But) in humans, blood leaving (V) has less oxygen than air leaving (E); 3. Difference in oxygen (concentration) between artery and vein is greater in fish than in humans; 4. (So) fish remove a greater proportion from the oxygen they take in; Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system (2 marks) - answer-1. Blood and water flow in opposite directions; 2. Diffusion/concentration gradient (maintained) along (length of) lamella/filament; Describe and explain the mechanism that causes lungs to fill with air. (3 marks) - answer-1. Diaphragm (muscle) contracts and external intercostal muscles contract; 2. (Causes volume increase and) pressure decrease; 3. Air moves down a pressure gradient Ignore along OR Air enters from higher atmospheric pressure; Two solutions often used to stain tissues are haematoxylin solution and iodine solution. • Haematoxylin solution stains DNA a blue colour. • Iodine solution stains starch a blue-black colour. The scientist used haematoxylin solution and not iodine solution to stain the lung tissue. Suggest why. (2 marks) - answer-1. This/animal/lung tissue does not contain starch; 2. (Makes) nucleus visible; OR Nucleus contains DNA; Particulate matter is solid particles and liquid particles suspended in air. Polluted air contains more particulate matter than clean air. A high concentration of particulate matter results in the death of some alveolar epithelium cells. If alveolar epithelium cells die inside the human body they are replaced by non-specialised, thickened tissue. Explain why death of alveolar epithelium cells reduces gas exchange in human lungs. (3 marks) - answer-1. Reduced surface area; 2. Increased distance for diffusion; 3. Reduced rate of gas exchange; Describe and explain the advantage of the counter-current principle in gas exchange across a fish gill. (3 marks) - answer-1. Water and blood flow in opposite directions; 2. Maintains diffusion/concentration gradient of oxygen OR Oxygen concentration always higher (in water); 3. (Diffusion) along length of lamellae/filament/gill/capillary; The water potential of leaf cells is affected by the water content of the soil. Scientists grew sunflower plants. They supplied different plants with different volumes of water. After two days, they determined the water potential in the leaf cells by using an instrument that gave a voltage reading. The scientists generated a calibration curve to convert the voltage readings to water potential. Figure 1 shows their calibration curve. Figure 1. Sunflowers are not xerophytic plants. The scientists repeated the experiment with xerophytic plants. Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 2. - answer-1. Low/slow growth; 2. Due to smaller number/area of stomata (for gas exchange); OR 3. Growth may continue at lower water potentials; 4. (Due to) adaptations in enzymes involved in photosynthesis/metabolic reactions; Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow only slowly. (2 marks) - answer-1. Stomata close; 2. Less carbon dioxide (uptake) for less photosynthesis/glucose production; Describe the pathway taken by an oxygen molecule from an alveolus to the blood. (3 marks) - answer-1. (Across) alveolar epithelium; 2. Endothelium / epithelium of capillary; Explain how one feature of an alveolus allows efficient gas exchange to occur. (2 marks) - answer-1. (The alveolar epithelium) is one cell thick; 2. Creating a short diffusion pathway / reduces the diffusion distance; Carbon monoxide is a poisonous gas that is present in cigarette smoke. This carbon monoxide can be absorbed into the blood where it binds with haemoglobin. Scientists investigated the concentration of carbon monoxide in cars in which people were smoking or not smoking. They measured the concentration with the car windows open and closed. The graph shows the scientists' results as they presented them. A value of ± 2 standard deviations from the mean includes over 95% of the data. In England, in October 2015, a law was introduced making it illegal to smoke in a car carrying someone who is under the age of 18. Following the introduction of the law, a politician stated: 'It is dangerous to smoke when a child is in the car. Higher levels of deadly toxins can build up, even on short journeys, and children breathe faster than adults, meaning they inhale more of the deadly toxins.' Use the information provided and the dat - answer-For 1. Significantly higher concentrations of CO (compared with no smoking) with closed window (as no overlap in 2 × SD); 2. Any increase in CO could be dangerous; OR CO causes less oxygen to be carried / provided (which could be deadly in children); 3. (significantly) higher levels after (just) 5 minutes (with closed windows supporting short journey statement); Against 4. No idea if (roughly) 5ppm is 'deadly'; 5. No significant difference with open window (as 2 × SD overlaps); 6. No data on child breathing rates; OR Idea that children breathe faster but have smaller lung volume, so overall volume of CO inhaled could be similar; Describe the gross structure of the human gas exchange system and how we breathe in and out. (6 marks) - answer-1. Named structures - trachea, bronchi, bronchioles, alveoli; 2. Above structures named in correct order OR Above structures labelled in correct positions on a diagram; 3. Breathing in - diaphragm contracts and external intercostal muscles contract; 4. (Causes) volume increase and pressure decrease in thoracic cavity (to below atmospheric, resulting in air moving in); 5. Breathing out - Diaphragm relaxes and internal intercostal muscles contract; 6. (Causes) volume decrease and pressure increase in thoracic cavity (to above atmospheric, resulting in air moving out); Figure 1 shows the stages of development of an insect called a damselfly. The adult damselfly uses a tracheal system for gas exchange. Explain three ways in which an insect's tracheal system is adapted for efficient gas exchange. (3 marks) - answer-1. Tracheoles have thin walls so short diffusion distance to cells; 2. Highly branched / large number of tracheoles so short diffusion distance to cells; 3. Highly branched / large number of tracheoles so large surface area (for gas exchange); 4. Tracheae provide tubes full of air so fast diffusion (into insect tissues); 5. Fluid in the end of the tracheoles that moves out (into tissues) during exercise so faster diffusion through the air to the gas exchange surface; OR Fluid in the end of the tracheoles that moves out (into tissues) during exercise so larger surface area (for gas exchange); 6. Body can be moved (by muscles) to move air so maintains diffusion / concentration gradient for oxygen / carbon dioxide; The damselfly larva is a carnivore that actively hunts prey. It has gills to obtain oxygen from water. Some other species of insect have larvae that are a similar size and shape to damselfly larvae and also live in water. These larvae do not actively hunt prey and do not have gills. Explain how the presence of gills adapts the damselfly to its way of life. (2 marks) - answer-1. Damselfly larvae has high(er) metabolic / respiratory (rate); 2. (So) uses more oxygen (per unit time / per unit mass); A scientist measured the size of each gill lamella of the gills of 40 damselfly larvae. His results are shown in the table. Mean width / mm (± uncertainty / mm) 1.61 (± 0.19) Mean width / mm (± uncertainty / mm) 6.12 (± 0.41) Calculate the mean surface area of one side of one gill lamella. Assume that a gill lamella is rectangular and give your answer to an appropriate number of significant figures. Include the percentage error (uncertainty) of surface area in your answer. Show your working. - answer-Mean SA = 9.85 mm2 / 9.9 mm2 ; Percentage uncertainty of SA = 18.5 / 18.7 / 19; Explain two ways in which the structure of fish gills is adapted for efficient gas exchange (2 marks) - answer-1. Many lamellae / filaments so large surface area; 2. Thin (surface) so short diffusion pathway; Explain how the counter current mechanism in fish gills ensures the maximum amount of the oxygen passes into the blood flowing through the gills. (3 marks) - answer-1. Water and blood flow in opposite directions; 2. Blood always passing water with a higher oxygen concentration; 3. Diffusion gradient maintained throughout length (of gill) OR Diffusion occurs throughout length of gill OR If water and blood flowed in same direction equilibrium would be reached; Name the process by which oxygen passes from an alveolus in the lungs into the blood. (1 mark) - answer-(Simple) diffusion; The photograph shows a fire-breather creating a ball of fire. Fire-breathers do this by blowing a fine mist of paraffin oil onto a flame. Some of this mist can be inhaled and may eventually lead to fibrosis. People who have been fire-breathers for many years often find they cannot breathe out properly. Explain why. (2 marks) - answer-1. Loss of elasticity / elastic tissue / increase in scar tissue; 2. Less recoil; Describe and explain how the counter current system leads to efficient gas exchange across the gills of a fish. (3 marks) - answer-1. Water and blood flow in opposite directions; 2. Maintains concentration / diffusion gradient / equilibrium not reached / water always next to blood with a lower concentration of oxygen; 3. Along whole / length of gill / lamellae; Amoebic gill disease (AGD) is caused by a parasite that lives on the gills of some species of fish. The disease causes the lamellae to become thicker and to fuse together. AGD reduces the efficiency of gas exchange in fish. Give two reasons why. (2 marks) - answer-1. (Thicker lamellae so) greater / longer diffusion distance / pathway; 2. (Lamellae fuse so) reduced surface area; The table below shows some features of gas exchange of a fish at rest. Volume of oxygen absorbed by the gills from each dm3 of water / cm3 7 Mass of fish / kg 0.4 Oxygen required by fish / cm3 kg-1 hour -1 90 (i) Calculate the volume of water that would have to pass over the gills each hour to supply the oxygen required by the fish. Show your working. (2 marks) - answer-Correct answer of 5.1 or 5.14(2857) (dm3) = 2 marks;; The volume of water passing over the gills increases if the temperature of the water increases. Suggest why. (1 mark) - answer-1. Increased metabolism / respiration / enzyme activity; 2. Less oxygen (dissolved in water); Two groups of people volunteered to take part in an experiment. • People in group A were healthy. • People in group B were recovering from an asthma attack. Each person breathed in as deeply as they could. They then breathed out by forced expiration. A scientist measured the volume of air breathed out during forced expiration by each person. The graph below shows the results. Forced expiration volume (FEV) is the volume of air a person can breathe out in 1 second. Using data from the first second of forced expiration, calculate the percentage decrease in the FEV for group B compared with group A. (1 mark) - answer-19(%); The people in group B were recovering from an asthma attack. Explain how an asthma attack caused the drop in the mean FEV shown in the figure below. (4 marks) - answer-1. Muscle walls of bronchi / bronchioles contract; 2. Walls of bronchi / bronchioles secrete more mucus; 3. Diameter of airways reduced; 4. (Therefore) flow of air reduced. A biologist investigated the effect of water temperature on the rate of ventilation of gills in a species of fish. She kept four fish in a thermostatically controlled aquarium and measured the mean ventilation rate by counting movements of their gill covers. Her results are shown in Figure 1.In this investigation, the biologist also monitored the concentration of oxygen in the water in the aquarium. The concentration of oxygen in water changes with temperature of the water. Figure 2 shows how it changes. Suggest a difficulty of counting movements of gill covers as a method of measuring rate of ventilation in fish. (1 mark) - answer-Fish keep moving / swimming / movement of gill covers too fast to count (at higher temperatures). The biologist concluded that there was a correlation between rate of ventilation of the gills and temperature of the water. A scatter diagram can be used to look for a correlation but, in this investigation, it was not the appropriate graph for her data. Explain why. (1 mark) - answer-1. There is only one dependent variable / there are not two dependent variables / water temperature is the independent variable / breathing rate is dependent on water temperature; 2. Water temperature plus breathing rate are not both properties of fish OR water temperature plus breathing rate are not both properties of water. Describe the relationship between temperature of water, oxygen in water and rate of ventilation. (1 mark) - answer-As (water) temperature increases, oxygen (concentration / solubility) falls and ventilation rate increases. Use Figure 1 and Figure 2 to explain the advantage to the fish of the change in its rate of ventilation. (3 marks) - answer-1. As concentration / solubility of oxygen falls less oxygen flows over gills / less oxygen enters gills / less oxygen enters fish; 2. (As a result) blood oxygen (concentration) falls / is lower; 3. An increase in ventilation rate increases / maintains the flow of oxygen / carbon dioxide across gills / into (or out of) fish; 4. Maintains diffusion / concentration gradient(s) (in gills); 5. To maintain oxygen supply to cells / tissues / organs / to maintain respiration. Q6.A scientist used grasshoppers to investigate the effect of composition of air on breathing rate in insects. He changed the composition of air they breathed in by varying the concentrations of oxygen and carbon dioxide. The scientist collected 20 mature grasshoppers from a meadow. He placed the grasshoppers in a small chamber where he could adjust and control the composition of air surrounding them. The small chamber restricted the movement of the grasshoppers. His results for three of the grasshoppers are shown in the table below in the form in which he presented them. The percentages of oxygen and carbon dioxide in Column A do not add up to 100% but in columns C and D they do. Suggest two reasons for this difference. (2 marks) - answer-1. Other gases / nitrogen / water vapour in atmosphere / A; 2. Only oxygen and carbon dioxide in gas mixtures / C and D; 3. Composition of / gases in A not controlled / composition of gas mixtures/C and D controlled Use all the data to describe the effect of concentration of carbon dioxide on the breathing rate of grasshoppers. (3 marks) - answer-1. Breathing rate lowest when no carbon dioxide / in (pure) oxygen / B; 2. (Generally) presence of carbon dioxide increases breathing rate / as concentration of carbon dioxide increases breathing rate increases / there is a positive correlation; 3. Breathing rate increases when (carbon dioxide) higher than 0.1% / concentration in atmosphere / A; 4. Breathing rate of grasshopper 3 falls in D / 16% / gas mixture 2 (whereas others increase). One of the different types of air was similar to the air in the meadow where the grasshoppers were collected. It provides data that might be used to calculate a mean breathing rate for grasshoppers in the meadow. (i) Use the data to estimate the mean breathing rate of the three grasshoppers in the meadow. Show your working. (2 marks) - answer-54; The estimate does not provide a reliable value for the mean breathing rate of all insect species in the meadow. Other than being an estimate, suggest and explain three reasons why this value would not be reliable. (3 marks) - answer-1. Small sample / only 3 (grasshoppers) so may not be representative (of all grasshoppers / insects); 2. Grasshoppers are not the only insects / species; so genetic / behavioural / metabolic differences; 3. (Insects) not all mature / are at different stages of development / different sizes; so different metabolic rates; 4. Movement not restricted / not at rest in meadow; so (rate of) respiration higher; 5. (Naturally-occurring) carbon dioxide concentration lower in meadow; so breathing rate lower; Scientists studied the rate of carbon dioxide uptake by grape plant leaves. Grape leaves have stomata on the lower surface but no stomata on the upper surface. The scientists recorded the carbon dioxide uptake by grape leaves with three different treatments: Treatment 1 − No air-sealing grease was applied to either surface of the leaf. Treatment 2 − The lower surface of the leaf was covered in air-sealing grease that prevents gas exchange. Treatment 3 − Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange. The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in light and then for 20 minutes in the dark. The scientists' results are shown in the diagram below. (a) Suggest the purpose of each of the three leaf treatments. - answer-1. (No grease) means stomata are open/ allows normal CO2 uptake. 2. (Grease on lower surface) seals stomata/stops CO2 uptake through stomata 3. (Grease on both surfaces) shows sealing is effective/stops all CO2 uptake. Describe the results shown for Treatment 1. (2 marks) - answer-1. (Mean rate of) carbon dioxide uptake was constant and fell after the light turned off; 2. Uptake fell from 4.5 to 0 / uptake started to fall at 60 minutes and reached lowest at 80 minutes / uptake fell over period of 20 minutes; The stomata close when the light is turned off. Explain the advantage of this to the plant. (2 marks) - answer-1. (Because) water is lost through stomata; 2. (Closure) prevents / reduces water loss; 3. Maintain water content of cells. Treatment 2 shows that even when the lower surface of the leaf is sealed there is still some uptake of carbon dioxide. Suggest how this uptake of carbon dioxide continues. (1 mark) - answer-(Carbon dioxide uptake) through the upper surface of the leaf / through cuticle. In both Treatment 1 and Treatment 2, the uptake of carbon dioxide falls to zero when the light is turned off. Explain why. (2 marks) - answer-1. No use of carbon dioxide in photosynthesis (in the dark); 2. No diffusion gradient (maintained) for carbon dioxide into leaf / there is now a diffusion gradient for carbon dioxide out of leaf (due to respiration). Scientists studied three species of plant. They selected fully grown leaves from five different plants of each species. For each leaf they measured: • leaf surface area • leaf thickness • the number of stomata per mm2 . The scientists' results are shown in the table below. How did the scientists ensure they could make a valid comparison between leaves from different species? (1 mark) - answer-(Scientists) used fully grown leaves / used five plants of each (species). Describe a method you could use to find the surface area of a leaf. (3 marks) - answer-1. Draw around leaf on graph paper; 2. Count squares (however described); 3. Multiply by 2 (for upper and lower leaf surface); OR 4. Draw around a leaf on paper of known mass (per unit area); 5. Cut out and weigh; 6. Multiply by 2 (for upper and lower leaf surface). Which species, A or B, would you predict grew in a drier environment? Explain one feature that caused you to choose this species. (1 mark) - answer-Species B (no mark) 1. Smaller surface area so less evaporation / less heat absorbed; 2. Thicker leaves so greater diffusion distance (for water); 3. Fewer stomata / lower stomatal density so less diffusion / evaporation (of water); 4. Smaller surface area to volume ratio so less evaporation. Other than the features of leaves in the table above, give two features of leaves of xerophytes. For each feature explain how it reduces water loss. (2 marks) - answer-1. Thick(er) cuticle so increase in diffusion distance / slower (rate of) diffusion; 2. Hairs on leaves so reduction in air movements / increase in humidity / decrease in water potential gradient; 3. Curled leaves so reduction in air movements / increase in humidity / decrease in water potential gradient; 4. Sunken stomata so reduction in air movements / increase in humidity / decrease in water potential gradient. Species C has a high number of stomata per mm2 Despite this it loses a small amount of water. Use the data to explain why. (1 mark) - answer-Small leaves / surface area so (total) number of stomata is low. Describe the processes involved in the absorption and transport of digested lipid molecules from the ileum into lymph vessels (5 marks) - answer-1. Micelles contain bile salts and fatty acids/monoglycerides; 2. Make fatty acids/monoglycerides (more) soluble (in water) OR Bring/release/carry fatty acids/monoglycerides to cell/lining (of the iluem) OR Maintain high(er) concentration of fatty acids/monoglycerides to cell/lining (of the ileum); 3. Fatty acids/monoglycerides absorbed by diffusion; 4. Triglycerides (re)formed (in cells); 5. Vesicles move to cell membrane; Figure 1 shows a cell from the lining of the ileum specialised for absorption of products of digestion. SGLT1 is a carrier protein found in the cell-surface membrane of this cell, it transports glucose and sodium ions (Na+ ) into the cell. Figure 1 (a) The action of the carrier protein X in Figure 1 is linked to a membrane-bound ATP hydrolase enzyme. Explain the function of this ATP hydrolase. (2 marks) - answer-1. (ATP to ADP + Pi ) Releases energy; 2. (energy) allows ions to be moved against a concentration gradient OR (energy) allows active transport of ions; The movement of Na+ out of the cell allows the absorption of glucose into the cell lining the ileum. Explain how. (2 marks) - answer-1. (Maintains/generates) a concentration/diffusion gradient for Na+ (from ileum into cell); 2. Na+ moving (in) by facilitated diffusion, brings glucose with it OR Na+ moving (in) by co-transport, brings glucose with it; To study lipid digestion, a scientist placed a tube into the gut of a healthy 20-year-old man. The end of the tube passed through the stomach but did not reach as far as the ileum. The scientist fed the man a meal containing triglycerides through the tube. The scientist also used the tube to remove samples from the man's gut at intervals after the meal. The scientist measured the type of lipid found in the samples. Some of her results are shown in the table below. Sample Time of collection after meal / min Concentration of fatty acids / mg cm-3 Concentration of triglycerides / mg cm-3 A 45 2.7 0.6 B 75 3.3 0.0 (a) Use your knowledge of lipid digestion to explain the differences in the results for samples A and B shown in the table above. You should assume that no absorption had occurred. (3 marks) - answer-1. Triglycerides decrease because of the action of lipase OR Fatty acids increase because of the action of lipase; 2. Triglycerides decrease because of hydrolysis (of triglycerides) OR Fatty acids increase because of hydrolysis (of triglycerides); 3. Triglycerides decrease because of digestion of ester bonds (between fatty acid and glycerol) OR Fatty acids increase because of digestion of ester bonds (between fatty acid and glycerol); After collecting the samples, the scientist immediately heated them to 70 °C for 10 minutes. Explain why. (2 marks) - answer-1. To denature the enzymes/lipase; 2. So no further digestion/hydrolysis/catalysis occurred; Describe the role of micelles in the absorption of fats into the cells lining the ileum. (3 marks) - answer-1. Micelles include bile salts and fatty acids; 2. Make the fatty acids (more) soluble in water; 3. Bring/release/carry fatty acids to cell/lining (of the ileum); 4. Maintain high(er) concentration of fatty acids to cell/lining (of the ileum); 5. Fatty acids (absorbed) by diffusion; Describe the role of enzymes in the digestion of proteins in a mammal. (4 marks) - answer-1. (Reference to) hydrolysis of peptide bonds; 2. Endopeptidase act in the middle of protein/polypeptide OR Endopeptidase produces short(er) polypeptides/ increase number of ends; 3. Exopeptidases act at end of protein/polypeptide OR Exopeptidase produces dipeptides/amino acids; 4. Dipeptidase acts on dipeptide/between two amino acids OR Dipeptidase produces (single) amino acids; Scientists investigated how the diet of rabbits affected their digestion and absorption of protein. The scientists fed rabbits an identical mass of food but varied the percentage of protein in the food. The scientists measured the mean mass of protein fed to the rabbits that was absorbed, which they then expressed as a percentage value. The scientists' results are shown in Figure 1. The error bars show ± 2 standard deviations. ± 2 standard deviations cover 95% of the data. What can you conclude about the absorption of the products of protein digestion as the percentage of protein increased in the rabbits' food? (3 marks) - answer-1. No significant difference (in protein absorption); 2. (because ± 2) SDs overlap; 4. Amount of protein (in diet) is not a limiting factor OR Something else is limiting factor e.g. amount of protease; 5. (But) small range of protein in diet OR (Should) Investigate wider range; The digestive system of a rabbit is shown in Figure 2. The food eaten by a rabbit is digested mainly by microorganisms in its caecum. The caecum is a section of intestine attached between the ileum and the large intestine. The resulting semi-digested material leaves the anus of a rabbit as soft, caecal droppings. The rabbit then eats these caecal droppings. Use this information and Figure 2 to suggest how eating its own caecal droppings helps a rabbit's digestion and absorption of dietary protein. (3 marks) - answer-1. More/remaining/undigested (protein) broken down; 2. (So more) amino acids absorbed; 3. (Because) protein/food passes again through stomach/ileum; The diagram outlines the digestion and absorption of lipids. Tick () the box by the name of the process by which fatty acids and glycerol enter the intestinal epithelial cell. - answer-Diffusion Explain the advantages of lipid droplet and micelle formation. (3 marks) - answer-1. Droplets increase surface areas (for lipase / enzyme action); 2. (So) faster hydrolysis / digestion (of triglycerides / lipids); 3. Micelles carry fatty acids and glycerol / monoglycerides to / through membrane / to (intestinal epithelial) cell; Name structure Q in the diagram above and suggest how it is involved in the absorption of lipids. (4 marks) - answer-1. Golgi (apparatus); 2. Modifies / processes triglycerides; 3. Combines triglycerides with proteins; 4. Packaged for release / exocytosis OR Forms vesicles; Cells lining the ileum of mammals absorb the monosaccharide glucose by co-transport with sodium ions. Explain how. (3 marks) - answer-1. Sodium ions actively transported from ileum cell to blood; 2. Maintains / forms diffusion gradient for sodium to enter cells from gut (and with it, glucose); 3. Glucose enters by facilitated diffusion with sodium ions; A student set up the experiment shown in the diagram below. The material from which Visking tubing is made is partially permeable. After 15 minutes, the student removed samples from the liquid in the beaker and from the liquid inside the Visking tubing. She carried out biochemical tests on these samples. She drew the table below to record her results. Complete the table by placing a tick () in each box that you expect to have shown a positive result. Justify your answers to part (b) - answer-1. Biuret: protein molecules too large to pass through tubing; 2. Iodine in potassium iodide solution: starch molecules too large to pass through tubing; 3. Benedict's: starch hydrolysed to maltose, which is able to pass through tubing. Scientists investigated the relationship between the percentage of fat in the diet and the death rate from breast cancer in 24 different countries. They plotted the data from each country on the graph below. Describe the information given by point A on the graph. (1 mark) - answer-In one country where the percentage of fat (in the diet) is 35%, the death rate (from breast cancer) is 20 per 100 000; Describe how the scientists calculated the death rate from breast cancer for each country. (1 mark) - answer-1. No. of deaths from breast cancer divided by total population × 100 000; 2. No. of deaths from breast cancer divided by all deaths × 100 000; 3. Sample and count deaths from breast cancer in 100 000 people; Some people have used the graph to conclude that a high percentage of fat in the diet causes breast cancer. Evaluate this conclusion. (3 marks) - answer-1. Positive correlation; 2. But correlation does not show causation / some other (named) factor may be involved; 3. Evidence against positive correlation e.g. different death rates at same % fat / similar death rates at different % fat / some countries with higher death rate have lower fat intake; The concentration of glucose in the blood rises after eating a meal containing carbohydrates. The rise is slower if the carbohydrate is starch rather than sucrose. Explain why. (3 marks) - answer-1. Starch digested to maltose / by amylase; 2. Maltose digested to glucose / by maltase; 3. Digestion of sucrose is a single step / only one enzyme / sucrase; The glycaemic load (GL) of a diet is a measure of how much digestible carbohydrate it contains. The higher the GL of a diet the more quickly it raises the blood glucose concentration after a meal. A diet with a high GL also increases the concentration of harmful lipids in the blood. Scientists investigated the relationship between diets with different glycaemic loads and the risk of developing coronary heart disease (CHD) in women. The scientists determined the glycaemic loads of the diets of a large number of women. They then divided the women into 5 groups. Group 1 had diets with the lowest glycaemic load and group 5 had diets with the highest glycaemic load. The scientists determined the risk of developing CHD in each group. The graph shows their results. The scientists excluded women who smoked from the study. Explain why. (1 mark) - answer-Smoking increases risk of CHD / introduces another variable; What do these data show about the effect that glycaemic load of the diet has on the risk of developing CHD? (1 mark) - answer-1. No effect on risk with diet group 1 and 2 / lowest glycaemic load; 2. Above diet group 2 / in higher groups, risk increases as glycaemic load increases; Use the information provided to explain the effect that glycaemic load of the diet has on the risk of developing CHD. (2 marks) - answer-1. (Higher GL diets lead to) more (harmful) lipids (in blood), so greater risk of atheroma; 2. Atheroma leads to blockage of coronary artery / increased risk of blood clot in coronary artery; A student investigated the effect of chewing on the digestion of starch in cooked wheat. He devised a laboratory model of starch digestion in the human gut. This is the method he used. 1. Volunteers chewed cooked wheat for a set time. The wheat had been cooked in boiling water. 2. This chewed wheat was mixed with water, hydrochloric acid and a protein-digesting enzyme and left at 37 °C for 30 minutes. 3. A buffer was then added to bring the pH to 6.0 and pancreatic amylase was added. This mixture was then left at 37 °C for 120 minutes. 4. Samples of the mixture were removed at 0, 10, 20, 40, 60 and 120 minutes, and the concentration of reducing sugar in each sample was measured. 5. Control experiments were carried out using cooked wheat that had been chopped up in a blender, not chewed. (a) What reducing sugar, or sugars, would you expect to be produced during chewing? Give a reason for your answer. (2 marks) - answer-1. Maltose; 2. Salivary amylase breaks down starch. In this model of digestion in the human gut, what other enzyme is required for the complete digestion of starch? (1 mark) - answer-Maltase. What was the purpose of step 2, in which samples were mixed with water, hydrochloric acid and pepsin? (1 mark) - answer-(Mimics / reproduces) effect of stomach. In the control experiments, cooked wheat was chopped up to copy the effect of chewing. Suggest a more appropriate control experiment. Explain your suggestion. (2 marks) - answer-1. Add boiled saliva; 2. Everything same as experiment but salivary amylase denatured. Explain what these results suggest about the effect of chewing on the digestion of starch in wheat. (3 marks) - answer-1. Some starch already digested when chewing / in mouth; 2. Faster digestion of chewed starch; 3. Same amount of digestion without chewing at end. Describe the structure of proteins. (5 marks) - answer-1. Polymer of amino acids; 2. Joined by peptide bonds; 3. Formed by condensation; 4. Primary structure is order of amino acids; 5. Secondary structure is folding of polypeptide chain due to hydrogen bonding; 6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic / disulfide bonds; 7. Quaternary structure is two or more polypeptide chains. Describe how proteins are digested in the human gut. (4 marks) - answer-1. Hydrolysis of peptide bonds; 2. Endopeptidases break polypeptides into smaller peptide chains; 3. Exopeptidases remove terminal amino acids; 4. Dipeptidases hydrolyse / break down dipeptides into amino acids. Maltose is hydrolysed by the enzyme maltase. Explain why maltase catalyses only this reaction. (3 marks) - answer-1. Active site (of enzyme) has (specific) shape / tertiary structure / active site complementary to substrate / maltose; 2. (Only) maltose can bind / fit; 3. To form enzyme substrate complex. What is the function of a red blood cell? - answer-Red blood cells contain haemoglobin so they can transport oxygen in the blood. what is the structure of haemoglobin? - answer-Haemoglobins are groups of proteins found in different organisms. Haemoglobin is a protein with a quaternary structure. Haemoglobin contains 4 haem groups. Haemoglobin contains 4 polypeptide chains. what does the affinity of haemoglobin for oxygen mean? - answer-The ability of haemoglobin to attract/bind, oxygen. what does saturation of haemoglobin with oxygen mean? - answer-When haemoglobin in holding the maximum amount of oxygen it can bind. what does loading/association of haemoglobin mean? - answer-The binding of oxygen to haemoglobin. what does unloading/dissociation of haemoglobin mean? - answer-when oxygen detaches or unbinds from haemoglobin. what graph is used to describe the affinity of haemoglobin for oxygen? - answer-The oxyhaemoglobin dissociation curve. It has a sigmoid (s) shape. What does the oxyhaemoglobin dissociation curve show? - answer-Oxygen is loaded in regions with a high partial pressure of oxygen (e.g. at the alveoli in the lungs) and is unloaded in regions of low partial pressure of oxygen (e.g. respiring tissues). what is cooperative binding? - answer-The shape of the haemoglobin molecule makes it difficult for the first oxygen molecule to bind to one of the sites on its four polypeptide subunits because they are closely united. Therefore at low oxygen concentrations, little oxygen binds to haemoglobin. The gradient of the curve is shallow initially. However, the binding of this first oxygen molecule changes the quaternary structure of the haemoglobin molecule, causing it to change shape. This change makes it easier for the other subunits to bind to an oxygen molecule. The binding of the first oxygen molecule induces the other subunits to bind to an oxygen molecule. It therefore takes a smaller increase in the partial pressure of oxygen to bind the second oxygen molecule than it did to bind the first one. This is known as positive cooperativity because binding of the first molecule makes binding of the second easier and so on. The gradient of the curve steepens. After the binding of the third molecule. It is harder for haemoglobin to bind the fourth oxygen molecule because the majority of the binding sites are occupied, it is less likely that a single oxygen molecule will find an empty site to bind to. The gradient of the curve reduces and the graph flattens off. How does the Bohr effect affect haemoglobins affinity for oxygen? - answer-The Bohr affect is when a high concentration of carbon dioxide causes haemoglobins affinity for oxygen to decrease (because the low PH created by carbonic acid causes the shape of haemoglobin to change) so the oxyhaemoglobin curve shifts right. what happens when there is a low partial pressure of carbon dioxide? - answer-Curve shifts to the left because haemoglobin has a higher affinity for oxygen so oxygen is loaded more readily. (at lungs) what happens when there is a high partial pressure of carbon dioxide? - answer-Curve shifts to the right because haemoglobin has a lower/decreased affinity for oxygen so oxygen is unloaded more readily. (at respiring tissues) why do different animals have different haemoglobins? - answer-Animals have different types of haemoglobin which have different affinities for oxygen, which is an adaptation to their environments. The oxygen dissociation curve shifts to the right during vigorous exercise. Explain the advantage of this shift: - answer-1. Lower affinity for oxygen/oxygen unloads more readily 2. To tissues 3. for rapid respiration Explain how the body shape of a Weddell seal is an adaptation to living in a cold environment. - answer-1. small surface area to volume ratio 2. So reduces heat loss The oxygen dissociation curve of the foetus is to the left of that for its mother. Explain the advantage of this for the foetus. (2 marks) - answer-1. Higher affinity / loads more oxygen; 2. At low/same/high partial pressure/pO2; 3. Oxygen moves from mother/to foetus; After birth, foetal haemoglobin is replaced with adult haemoglobin. Use the graph to suggest the advantage of this to the baby. - answer-1. Low affinity/oxygen dissociates; 2. (Oxygen) to respiring tissues/muscles/cells; Hereditary persistence of foetal haemoglobin (HPFH) is a condition in which production of foetal haemoglobin continues into adulthood. Adult haemoglobin is also produced. People with HPFH do not usually show symptoms. Suggest why. - answer-Enough adult Hb produced / enough oxygen released /more red blood cells produced; Explain how oxygen is loaded, transported and unloaded in the blood (6) - answer-1. Haemoglobin carries oxygen/ has a high affinity for oxygen/ oxyhaemoglobin; 2. In red blood cells; 3. Loading/uptake/association in lungs; 4. at high p.02; 5. Unloads/ dissociates / releases to respiring cells/tissues; 6. at low p.02; 7. Unloading linked to higher carbon dioxide (concentration);