ESAT Biology: Enzymes and animal physiology, set 3

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   An enzyme's active site can be blocked by a molecule with a shape similar to its normal substrate, called a competitive inhibitor. In an experiment, a fixed, low concentration of a competitive inhibitor is added to a reaction mixture together with the enzyme, and the substrate concentration is then increased steadily from a very low starting value while everything else (temperature, pH, enzyme concentration, inhibitor concentration) is kept constant. Which statement correctly predicts what happens to the rate of reaction as substrate concentration keeps increasing?

   1. A The rate cannot approach the rate obtained with no inhibitor present, however high the substrate concentration is raised, because the inhibitor blocks a fixed proportion of enzyme molecules for good.
   2. B The rate stays constant at whatever value it had at the start of the experiment, since it is the inhibitor concentration, not the substrate concentration, that fixes the rate of a competitively inhibited reaction.
   3. C The rate rises only until the substrate concentration numerically equals the inhibitor concentration, and then plateaus completely, because the two are treated as reaching a fixed one-to-one balance at that point.
   4. D The rate rises back towards the maximum rate seen with no inhibitor present at all, because at a high enough substrate concentration, substrate out-competes the fixed number of inhibitor molecules for the active site.
2. 21 mark

   A protease enzyme is exposed to a temperature well above its optimum, and separately, a fresh sample of the same enzyme is exposed to a pH far outside its optimum range. In both cases the enzyme loses almost all of its catalytic activity. Which statement correctly explains what has happened to the enzyme's structure in both cases?

   1. A Both extreme heat and extreme pH break the strong covalent peptide bonds joining amino acids together, cutting the protein chain into several shorter fragments, none of which can any longer fold into an active enzyme.
   2. B Both extreme heat and extreme pH disrupt the weak bonds holding the enzyme's tertiary structure in place, distorting the active site so it is no longer complementary to the substrate.
   3. C In both cases it is the enzyme's substrate that is chemically altered, so it is no longer recognised by the active site, even though the enzyme's own three-dimensional shape is left completely unaffected by either agent.
   4. D Extreme heat permanently destroys the enzyme molecule so it can never regain activity, but extreme pH only ever produces a temporary change that is fully and immediately reversed the moment normal pH is restored.
3. 31 mark

   An enzyme-catalysed reaction is being studied within its normal working range, below its optimum temperature. Across this range, the rate of reaction obeys the rule that it doubles for every 10 degrees C rise in temperature. At 15 degrees C the rate of reaction is 3 units. Using this rule, what is the rate of reaction at 45 degrees C?

   1. A 24 units
   2. B 12 units
   3. C 48 units
   4. D 33 units
4. 41 mark

   Bile is produced by the liver, stored in the gall bladder, and released into the small intestine, where it acts on large droplets of fat. Which statement correctly distinguishes the role of bile from the role of lipase in fat digestion?

   1. A Bile is itself an enzyme that catalyses the breakdown of fat into fatty acids and glycerol, working alongside lipase and speeding up fat digestion by directly attacking the same chemical bonds that lipase attacks.
   2. B Lipase physically breaks large fat droplets into many smaller droplets to increase the surface area available, while bile is the enzyme that then chemically digests the fat into fatty acids and glycerol.
   3. C Bile emulsifies large fat droplets into many smaller ones, increasing the surface area exposed to lipase, but bile is not an enzyme itself; lipase is the enzyme that hydrolyses the fat.
   4. D Bile neutralises the acidic contents arriving from the stomach so that lipase can begin working properly, and bile has no effect at all on the size of the fat droplets themselves.
5. 51 mark

   A trained runner completes an intense 400 metre sprint. During the sprint, oxygen cannot be delivered to the leg muscles fast enough to meet demand, so the muscle cells respire anaerobically for much of the race. For several minutes after crossing the finish line, the runner continues to breathe heavily and consume oxygen at a rate well above their normal resting rate. Which statement correctly explains this continued high oxygen consumption after the race has finished?

   1. A The extra oxygen is needed because anaerobic respiration continues in the leg muscles for several minutes after the race finishes, only slowing down once the runner's legs eventually stop moving altogether.
   2. B The extra oxygen is needed to replace the glucose used up during the race, since both aerobic and anaerobic respiration convert the oxygen breathed in directly into new glucose molecules for storage.
   3. C The extra oxygen is needed because the heart and breathing muscles, unlike the leg muscles during a sprint, can only ever respire aerobically, so they alone create the extra oxygen demand once exercise stops.
   4. D The extra oxygen consumed after the race is used to oxidise the lactic acid that built up in the muscles from anaerobic respiration during the sprint, which is what repays the runner's oxygen debt.
6. 61 mark

   At a synapse, a nerve impulse arriving at the end of one neurone is always passed on to the next neurone in only one direction, never backwards. Which statement correctly explains why this is the case?

   1. A Neurotransmitter is stored in, and released only from, vesicles in the presynaptic membrane, while the matching receptors sit only on the postsynaptic membrane, so a signal can only cross the gap in one direction.
   2. B The electrical impulse jumps directly across the synaptic gap as an electrical current, and it always travels in the same direction because the gap is too narrow to allow current to flow backwards.
   3. C The myelin sheath surrounding the presynaptic neurone insulates it so completely that an electrical impulse can only ever leave through the synaptic gap, and can never re-enter the neurone from that side.
   4. D Both neurones release the same neurotransmitter simultaneously, but only the postsynaptic neurone's neurotransmitter diffuses fast enough to have an effect, giving the appearance of one-way transmission.
7. 71 mark

   The rate of diffusion of oxygen from an alveolus into the blood, and of carbon dioxide from the blood into an alveolus, depends on there being a concentration gradient across the alveolar membrane. Alveolar surface area, membrane thickness and moistness all stay the same throughout a period of quiet breathing. Which of the following, if it stopped completely while the person kept breathing in and out through their mouth exactly as before, would most directly cause gas exchange at the alveoli to slow down and eventually stop?

   1. A Mucus production by cells lining the airways.
   2. B Blood flow through the capillaries surrounding the alveoli.
   3. C Production of new red blood cells in the bone marrow.
   4. D Surfactant secretion by cells lining the alveoli.
8. 81 mark

   In the human heart, the left ventricle has a much thicker, more muscular wall than the right ventricle, even though the two chambers pump the same volume of blood with each heartbeat. Which statement correctly explains this difference in wall thickness?

   1. A The left ventricle handles oxygenated blood, and oxygenated blood requires a thicker chamber wall to keep it fully separate from the deoxygenated blood flowing on the right side of the heart.
   2. B The right ventricle pumps a smaller volume of blood with each beat than the left ventricle does, since less blood is needed to fill the shorter pulmonary circuit, so it can get by with a thinner wall.
   3. C The right ventricle's wall is thinner because it only contracts during vigorous exercise, when the pulmonary circuit needs extra flow, whereas the left ventricle contracts with full force even at rest.
   4. D The left ventricle must generate far higher pressure than the right, because it pumps blood all the way around the long, high-resistance systemic circulation, unlike the short, low-resistance pulmonary circuit.
9. 91 mark

   Mammalian red blood cells lack a nucleus and are shaped as biconcave discs (flattened discs with a depression on each face, rather than a simple sphere). Which statement correctly explains how each of these two features helps the red blood cell carry out its function of transporting oxygen?

   1. A Lacking a nucleus stops the red blood cell producing carbon dioxide as it respires, so less carbon dioxide needs to be removed from the blood; the biconcave shape increases the total volume enclosed by the cell so that more haemoglobin can fit inside.
   2. B Lacking a nucleus allows the red blood cell to divide more often, increasing the total number of red blood cells available to carry oxygen; the biconcave shape allows red blood cells to stack together more easily as they flow through vessels.
   3. C Lacking a nucleus leaves more internal space for haemoglobin, increasing the oxygen each cell can carry; the biconcave shape gives a larger surface area relative to volume than a sphere would, speeding diffusion into and out of the cell.
   4. D Lacking a nucleus shortens the diffusion distance oxygen must travel to reach the centre of the cell; the biconcave disc shape maximises the volume enclosed within a given surface area, allowing each cell to carry the greatest possible amount of haemoglobin.
10. 101 mark

    A student writes in their notes: 'Egestion and excretion are just two different words for the same process: getting rid of waste from the body.' Which statement correctly evaluates this claim?

    1. A The claim is incorrect. Excretion removes waste made by the body's own metabolism, such as carbon dioxide or urea, whereas egestion removes faeces, material that was never absorbed into the body's cells at all.
    2. B The claim is correct, because both processes remove material that passes through the digestive system and out through the anus, so they describe exactly the same event happening at the same point in the body.
    3. C The claim is incorrect, but for the opposite reason: excretion refers only to material leaving via the anus, while egestion refers to metabolic waste such as urea and carbon dioxide leaving via the kidneys and lungs.
    4. D The claim is correct for humans, but incorrect for every other animal, because it is only in the human body that undigested food is ever able to leave separately from any metabolic waste product.
11. 111 mark

    As fluid passes along a healthy nephron, glucose is present in the filtrate at the start of the proximal convoluted tubule but has completely disappeared from the fluid by the end of it. Over the same stretch of the nephron, the concentration of urea in the fluid actually rises, even though no extra urea is added to the fluid at any point along the proximal convoluted tubule. Which statement correctly explains both observations?

    1. A Both glucose and urea are actively reabsorbed by the same transporter proteins, but the transporters have a much higher affinity for glucose, so urea is left behind at a higher concentration by comparison.
    2. B Glucose is actively reabsorbed from the filtrate by specific transporters, and in a healthy kidney this is complete; urea is not actively reabsorbed, so as water leaves the same fluid the remaining urea becomes more concentrated.
    3. C Glucose is filtered out of the blood only in a healthy kidney, so it never enters the nephron at all, while urea is filtered normally; this explains why glucose disappears without the need for any reabsorption.
    4. D Urea is chemically converted into a different, undetectable substance as it passes along the proximal convoluted tubule, which is why its concentration appears to rise even though the actual amount of urea has not changed.
12. 121 mark

    Two people, one with untreated type 1 diabetes and one with untreated type 2 diabetes, each eat an identical meal containing a large amount of carbohydrate. Blood glucose concentration is measured every 30 minutes for three hours afterwards. In a healthy person, blood glucose would rise after the meal and then be brought back down to its normal range within about two hours. Which statement correctly predicts and explains the two patients' results, compared to the healthy pattern?

    1. A The type 1 patient's blood glucose rises higher and stays high for longer than the healthy pattern, but the type 2 patient's pattern is indistinguishable from healthy, because type 2 diabetes only affects regulation once blood glucose is already very high.
    2. B Both patients' blood glucose rises and falls in the same pattern as a healthy person, because insulin is not actually needed to bring blood glucose back down after a meal; the liver alone removes excess glucose from the blood regardless of insulin.
    3. C Both patients show a higher, more prolonged rise in blood glucose than the healthy pattern; in type 1 diabetes this is because the pancreas produces little or no insulin, while in type 2 diabetes insulin is present but body cells have become resistant to it.
    4. D The type 2 patient's blood glucose rises higher and for longer than the healthy pattern because their pancreas produces no insulin at all, exactly as in type 1 diabetes; the two conditions differ only in how each is treated.
13. 131 mark

    A combined oral contraceptive pill contains synthetic versions of oestrogen and progesterone. A copper intrauterine device (IUD) is a small device fitted inside the uterus that releases no hormones at all. Which statement correctly distinguishes how each of these two methods of contraception prevents pregnancy?

    1. A The pill works by physically blocking sperm from ever reaching the uterus, exactly as the copper IUD does, but the pill's synthetic hormones make this physical barrier still more effective at stopping fertilisation.
    2. B The copper IUD releases a steady, low dose of oestrogen directly into the uterus, which is what makes it effective, while the pill instead works by physically blocking the entrance to the uterus.
    3. C Both methods work in the same way, by suppressing the release of FSH and LH from the pituitary gland to prevent ovulation; they differ only in how the hormone is delivered, as a tablet, or from a device in the uterus.
    4. D The pill's hormones inhibit the pituitary gland's release of FSH and LH, preventing ovulation in the first place; the copper IUD releases no hormones and instead acts locally, where copper ions are toxic to sperm and disrupt implantation.
14. 141 mark

    A pharmaceutical company has identified a new chemical compound that might be an effective treatment for a disease, and wants to bring it to market as a licensed medicine. Which sequence correctly orders the main stages the compound must go through, from earliest to latest, before it can be prescribed to patients generally?

    1. A Pre-clinical laboratory and animal studies first, to check basic toxicity and effectiveness; then a small trial in healthy volunteers to confirm safety and dose; then a larger, placebo-controlled trial in patients with the disease, to establish effectiveness.
    2. B Clinical testing on patients who have the disease first, to see if it seems to work at all, followed by pre-clinical laboratory and animal testing to confirm the mechanism, and finally testing on healthy volunteers to confirm it is safe for the general population.
    3. C Pre-clinical testing carried out directly in human patients who already have the disease, followed immediately by release of the drug to the whole affected population at large, with safety questions only monitored afterwards.
    4. D Testing on healthy volunteers only, since patients who already have the disease cannot ethically be given a drug that has not yet been proven to work; larger-scale testing and comparison against a placebo is not needed once safety has been shown.
15. 151 mark

    A patient with cardiovascular disease is prescribed three different medications: a statin, an anticoagulant, and an antihypertensive drug. Which statement correctly distinguishes what each of these three types of medication does?

    1. A All three drugs work in essentially the same way, by thinning the blood so it can flow more easily through arteries that have narrowed; they differ only in their dose and in how often each has to be taken.
    2. B A statin reduces the blood's ability to clot; an anticoagulant reduces the concentration of cholesterol in the blood; an antihypertensive drug widens the arteries directly to reduce cholesterol build-up.
    3. C A statin reduces blood cholesterol, slowing fatty deposits in artery walls; an anticoagulant reduces the blood's ability to clot; an antihypertensive drug lowers blood pressure, easing strain on the heart and arteries.
    4. D A statin lowers blood pressure directly by relaxing the artery walls; an anticoagulant reduces cholesterol build-up within artery walls; an antihypertensive drug instead reduces the blood's ability to clot.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: D

   1. A competitive inhibitor has a shape similar enough to the substrate to occupy the enzyme's active site, but unlike the substrate it is not converted to product, so while it occupies a site that enzyme molecule cannot catalyse a reaction.
   2. Because the inhibitor competes directly with the substrate for the same, unchanged active site, whether an active site is occupied by substrate or by inhibitor at any moment depends on their relative concentrations and how often each collides with the site.
   3. As substrate concentration is increased while inhibitor concentration stays fixed, substrate molecules collide with and occupy active sites increasingly more often than the unchanging number of inhibitor molecules do.
   4. At a sufficiently high substrate concentration, substrate can therefore out-compete the inhibitor almost completely, so the rate of reaction rises back towards the same maximum rate that would be reached with no inhibitor present at all: this reversibility by excess substrate is what separates a competitive inhibitor from a non-competitive one, so D is correct.
   • Why not A: Describes a non-competitive (allosteric) inhibitor, which acts at a separate site and is not overcome by adding more substrate. A competitive inhibitor competes for the same active site, so it can be out-competed.
   • Why not B: Ignores that the substrate and inhibitor compete directly for the same active site; as substrate concentration rises it collides with, and occupies, active sites more often, so the rate does change with substrate concentration.
   • Why not C: Invents a threshold rule with no basis in how competitive inhibition works; there is no special point at which substrate and inhibitor concentrations being numerically equal, or reaching some fixed one-to-one balance, changes the behaviour of the reaction.

2. Question 2Answer: B

   1. An enzyme's function depends on its precise three-dimensional shape, in particular the shape of its active site, which is held in place mainly by weak interactions (ionic and hydrogen bonds) between different parts of the folded amino acid chain.
   2. Both high temperature, through increased vibration of atoms in the molecule, and extreme pH, through disruption of the charges that hold ionic bonds together, can break these weak interactions without breaking the strong covalent peptide bonds that link amino acids into the chain itself.
   3. Once these weak bonds are disrupted, the tertiary structure unfolds or distorts, changing the shape of the active site so it is no longer complementary to the substrate; the enzyme is then denatured and can no longer bind its substrate effectively, however much substrate is present.
   4. Because the primary structure, the amino acid sequence itself, is unaffected by either agent, denaturation is fundamentally a change of shape rather than a chemical breakdown of the protein chain, whether the cause was extreme heat or extreme pH, so B is correct.
   • Why not A: Confuses the weak bonds that maintain tertiary structure with the strong covalent peptide bonds that link amino acids into the chain itself; neither heat nor pH within the ranges that cause denaturation breaks peptide bonds or cuts the chain.
   • Why not C: Misattributes the structural change to the substrate rather than the enzyme; it is the enzyme's own shape, specifically its active site, that is altered by denaturation, not the substrate molecule.
   • Why not D: Wrongly asserts an absolute permanent-versus-reversible split tied strictly to the cause; in reality whether denaturation can be reversed depends on how extreme and prolonged the exposure was, for both heat and pH, not on which of the two caused it.

3. Question 3Answer: A

   1. From 15 degrees C to 45 degrees C is a rise of 30 degrees C.
   2. Applying the doubling rule, a 30 degree C rise corresponds to 30 / 10 = 3 separate 10 degree C steps, so the rate doubles three times, not two or four times.
   3. Starting at 3 units: after the first doubling (to 25 degrees C) the rate is 6 units; after the second doubling (to 35 degrees C) it is 12 units; after the third doubling (to 45 degrees C) it is 24 units.
   4. So the rate at 45 degrees C is 3 x 2^3 = 3 x 8 = 24 units, matching option A.
   • Why not B: Under-counts the number of 10 degree C steps between 15 and 45 degrees C as two instead of three, so only two doublings are applied (3 to 6 to 12) instead of three.
   • Why not C: Over-counts the number of doublings by counting the four temperature values 15, 25, 35 and 45 as four steps, instead of counting the three 10 degree C gaps between them, giving one doubling too many (3 to 6 to 12 to 24 to 48).
   • Why not D: Misapplies the rule as an additive one, adding 10 units for each 10 degree C rise instead of doubling the rate, giving 3 + 10 + 10 + 10 = 33 rather than repeated multiplication by 2.

4. Question 4Answer: C

   1. Bile contains bile salts, which are not enzymes; they act as an emulsifier, physically breaking large fat droplets into many smaller droplets (emulsification) without breaking any chemical bonds within the fat molecules themselves.
   2. This emulsification greatly increases the total surface area of fat exposed to the surrounding fluid, which allows lipase to act on far more fat molecules per second than it could act on a small number of large droplets.
   3. Lipase is the enzyme responsible for the actual chemical digestion: it catalyses the hydrolysis of fat (triglyceride) molecules into fatty acids and glycerol.
   4. Bile also has a genuine, separate role in neutralising the acidic chyme arriving from the stomach, but this does not mean it has no effect on droplet size; bile performs both roles, so C, not D, correctly distinguishes bile's role from lipase's.
   • Why not A: Bile is not an enzyme; it is a mixture of salts and other substances that physically emulsifies fat, and it does not catalyse any chemical reaction itself.
   • Why not B: Swaps the two roles exactly: it is bile, not lipase, that physically breaks up fat droplets (emulsification), and lipase, not bile, is the enzyme that chemically hydrolyses fat.
   • Why not D: States a real function of bile (neutralising stomach acid) but wrongly denies its other, equally real function of emulsifying fat droplets, which does change droplet size and increases the surface area available to lipase.

5. Question 5Answer: D

   1. During intense exercise, the demand for energy in the leg muscles outstrips the rate at which oxygen can be delivered by the blood, so muscle cells switch, in part, to anaerobic respiration, which does not require oxygen: glucose to lactic acid.
   2. Anaerobic respiration releases much less energy per glucose molecule than aerobic respiration because the glucose is only partially broken down, and it produces lactic acid as a waste product, which builds up in the muscles and diffuses into the blood.
   3. After the race, once oxygen delivery is no longer limiting, the body needs extra oxygen to convert this accumulated lactic acid back into a form that can be used, largely by breaking it down aerobically; this extra oxygen use above the resting rate is often called repaying the oxygen debt.
   4. So the continued heavy breathing after the race reflects the oxidation of lactic acid produced during the anaerobic phase of the sprint, not ongoing anaerobic respiration itself, which is exactly what option D describes.
   • Why not A: Reverses the timing: anaerobic respiration in the muscles happens during the sprint itself because oxygen delivery is limiting there; once the race ends and oxygen delivery is no longer limiting, the extra oxygen used afterwards is for oxidising lactic acid, not for continued anaerobic respiration.
   • Why not B: Garbles the biochemistry: oxygen is not converted directly into glucose, and this is not what either aerobic or anaerobic respiration does with oxygen; aerobic respiration uses oxygen to help break down glucose, releasing carbon dioxide and water.
   • Why not C: Wrongly claims only the heart and breathing muscles can respire aerobically and that they alone create the extra demand; all cells, including leg muscle cells, can respire aerobically, and the main source of the extra oxygen demand is specifically the lactic acid built up in the leg muscles during the sprint.

6. Question 6Answer: A

   1. A nerve impulse is carried electrically along the length of a neurone, but neurones are not physically continuous with each other; they are separated by a tiny gap called the synapse.
   2. To cross this gap, the presynaptic neurone releases a chemical neurotransmitter from vesicles stored in its own membrane, and this neurotransmitter then diffuses across the gap.
   3. Receptor molecules that bind the neurotransmitter and trigger a new impulse are located only on the postsynaptic neurone's membrane, not on the presynaptic side.
   4. Because neurotransmitter release only happens on one side of the gap, and the matching receptors exist only on the other side, an impulse arriving at a synapse can only ever be passed on in the presynaptic-to-postsynaptic direction, never in reverse, so A is correct.
   • Why not B: Treats synaptic transmission as a direct electrical jump across the gap, when in fact the signal is carried across the synaptic gap chemically, by a neurotransmitter, not by electrical current flowing through the gap.
   • Why not C: Misapplies the role of the myelin sheath, which speeds up impulse conduction along the length of an axon (by saltatory conduction) and plays no part in enforcing the direction of transmission at a synapse.
   • Why not D: Invents a mechanism in which both neurones release neurotransmitter, when only the presynaptic neurone does so; the true explanation is the one-sided arrangement of vesicles and receptors, not a race between two simultaneous releases.

7. Question 7Answer: B

   1. Diffusion of a gas across the alveolar membrane depends on there being a difference in concentration (partial pressure) of that gas between the air in the alveolus and the blood in the surrounding capillaries.
   2. Ventilation continuously refreshes the air in the alveoli with air from outside, keeping alveolar oxygen concentration high and carbon dioxide concentration low, but this only maintains one side of the gradient.
   3. The other side of the gradient depends on blood flow: a continuous supply of relatively deoxygenated, carbon-dioxide-rich blood must keep arriving at, and then leaving, the capillaries around each alveolus, so that the blood there never reaches equilibrium with the air.
   4. If blood flow through these capillaries stopped, the small volume of blood remaining in contact with an alveolus would quickly equilibrate with the alveolar air, removing the concentration gradient and stopping net diffusion, even though breathing continued exactly as before, so B is correct.
   • Why not A: Mucus traps particles and pathogens in the airway lining; it plays no part in maintaining the concentration gradient that drives gas diffusion at the alveoli.
   • Why not C: Would reduce the blood's oxygen-carrying capacity only gradually, as existing red blood cells have a lifespan of months; it does not remove the diffusion gradient at the alveoli in the way that halting blood flow does.
   • Why not D: Surfactant reduces surface tension, preventing the alveoli from collapsing structurally; its absence would cause structural problems over time rather than directly and immediately removing the concentration gradient that drives diffusion.

8. Question 8Answer: D

   1. The heart's two sides work as a double circulatory system in series: the right side pumps blood only to the lungs and back (pulmonary circulation), while the left side pumps blood all the way around the rest of the body and back (systemic circulation).
   2. Over any period of time, both ventricles must pump the same volume of blood, since whatever leaves one side of the heart eventually arrives back at the other, so the difference between them cannot be about the volume pumped.
   3. What does differ is the pressure needed: the systemic circulation is a much longer circuit with far more total resistance than the short pulmonary circuit, so the left ventricle must contract with much greater force to generate the higher pressure needed.
   4. A thicker, more muscular wall can generate greater force, and therefore pressure, per contraction, which is why the left ventricle's wall is much thicker than the right ventricle's even though both pump equal volumes, exactly as option D describes.
   • Why not A: It is the septum, not wall thickness, that physically keeps oxygenated and deoxygenated blood apart; wall thickness is related to the pressure a chamber must generate, not to keeping the two blood supplies separate.
   • Why not B: In a double circulatory system in series, whatever volume leaves one side of the heart must eventually arrive back at the other, so both ventricles pump equal volumes over any period of time; a shorter circuit does not need a smaller volume of blood to fill it, so the difference between the ventricles is not one of volume at all.
   • Why not C: Both ventricles contract on every single heartbeat, at rest and during exercise; this invents a false distinction in when each chamber contracts.

9. Question 9Answer: C

   1. Haemoglobin, the oxygen-carrying protein, is packed into the cytoplasm of a red blood cell; because a mature mammalian red blood cell has no nucleus, that space is available to be filled with additional haemoglobin, increasing how much oxygen each cell can carry.
   2. A nucleus is not the site of gas exchange or of oxygen transport, so removing it does not change how much carbon dioxide the cell's metabolism produces, and it does not enable the cell to divide; a cell without a nucleus cannot undergo mitosis at all.
   3. For a fixed surface area, a sphere encloses the largest possible volume of any shape; a biconcave disc has a smaller volume than a sphere with the same surface area, which is precisely why it has a higher surface area to volume ratio than a sphere of the same size, letting oxygen diffuse into and out of the cell more quickly.
   4. So the correct pairing is: no nucleus frees up space for more haemoglobin, and the biconcave disc shape increases surface area relative to volume, not by maximising enclosed volume as option D wrongly claims, matching option C.
   • Why not A: The nucleus is not the site of cellular respiration, so removing it does not stop the cell's metabolism producing carbon dioxide; and a biconcave shape does not increase enclosed volume compared with other shapes of the same surface area, it reduces it.
   • Why not B: A nucleus is required for a cell to divide; lacking one actually prevents mitosis, the opposite of this claim, which is why red blood cells must be continually replaced from nucleus-containing precursor cells in the bone marrow rather than dividing themselves.
   • Why not D: The first half is a genuine effect, but the second half reverses the real geometry: for a fixed surface area, a sphere, not a flattened disc, encloses the greatest possible volume, so a biconcave disc has a smaller volume than a sphere of the same surface area, which is exactly why it has a higher surface area to volume ratio.

10. Question 10Answer: A

    1. Excretion is the removal from the body of waste products that result from the body's own metabolic reactions: carbon dioxide, a waste product of respiration, removed via the lungs, and urea, a waste product of the breakdown of excess amino acids in the liver, removed via the kidneys in urine, are both examples.
    2. Egestion is the removal of faeces from the body via the anus. Faeces are made up mainly of food that could not be digested and absorbed, together with dead gut bacteria and cells shed from the gut lining; this material was never taken up into the body's cells or used in any metabolic reaction.
    3. Because faeces never entered the body's metabolism in the first place, egestion is a fundamentally different process from excretion, even though both result in waste material leaving the body.
    4. So the student's claim is incorrect, and specifically because excretion concerns metabolic waste while egestion concerns undigested food residue, as option A states, not the reversed definitions or the false human exception given in the other options.
    • Why not B: Excretion includes waste that leaves via routes other than the anus, such as carbon dioxide exhaled from the lungs and urea removed in urine via the kidneys, so the two processes cannot be describing the same event at the same point in the body.
    • Why not C: Swaps the two definitions exactly: it is excretion, not egestion, that concerns metabolic waste such as urea and carbon dioxide, and egestion, not excretion, that concerns undigested material leaving via the anus.
    • Why not D: Invents a false human-specific exception; the distinction between egestion and excretion is based on where the waste material originates (metabolism versus undigested food), which applies to animals generally, not only to humans.

11. Question 11Answer: B

    1. At the glomerulus, small molecules including water, glucose and urea are all filtered out of the blood into the nephron's tubule, regardless of whether the body needs to keep or lose them; this is a non-selective process based only on molecular size.
    2. Along the proximal convoluted tubule, useful small molecules are then selectively reabsorbed back into the blood; glucose is actively reabsorbed by specific transporter proteins, and in a healthy kidney, where the amount of glucose does not exceed the transporters' capacity, all of it is normally reabsorbed, so none remains in the fluid by the end of this section.
    3. Urea, in contrast, is a waste product the body needs to excrete, so it is not actively reabsorbed; however, water is also being reabsorbed from the same fluid at this point, which reduces the total volume of fluid remaining.
    4. Because the same, unchanged amount of urea is now dissolved in a smaller volume of fluid, its concentration rises, even without any urea being added; this combination, complete active reabsorption for glucose and passive concentration by water loss for urea, is exactly what option B describes.
    • Why not A: Invents a shared transport mechanism between glucose and urea; urea is not actively reabsorbed at all, so its rising concentration is due to the loss of water from the fluid, not to competition for a shared transporter with glucose.
    • Why not C: Glucose, being a small molecule, is normally filtered into the nephron at the glomerulus along with water and other small solutes, exactly like urea; the reason it disappears is complete reabsorption further along the tubule, not exclusion from filtration in the first place.
    • Why not D: Invents a chemical conversion of urea that does not occur; urea passes along the nephron largely unchanged, and its rising concentration is simply due to water being removed from the same fluid, reducing the fluid's volume.

12. Question 12Answer: C

    1. In a healthy person, a rise in blood glucose after a meal triggers the pancreas to release insulin, which causes body cells, especially liver and muscle cells, to take up glucose from the blood and, in the liver, to store it as glycogen; this returns blood glucose to its normal range.
    2. In type 1 diabetes, the pancreas produces little or no insulin at all, often because the insulin-producing cells have been destroyed, so this signal to take up glucose is largely absent, and blood glucose remains elevated for longer after a meal than in a healthy person.
    3. In type 2 diabetes, the pancreas can still produce insulin, but the body's cells have become less responsive to it, so even though insulin is present, it is less effective at causing cells to take up glucose, again leaving blood glucose elevated for longer than normal.
    4. So although the underlying cause differs between the two conditions, no insulin production versus insulin present but ineffective, both produce a similar overall pattern compared to a healthy person: a higher peak in blood glucose and a slower return to the normal range, which is what option C describes.
    • Why not A: Understates type 2 diabetes: insulin resistance still impairs glucose uptake by cells after a normal meal, producing an elevated blood glucose pattern too, just through a different mechanism from type 1, rather than no abnormality at all.
    • Why not B: Insulin is precisely the hormone that signals cells, including liver and muscle cells, to take up and store glucose after a meal; without effective insulin action, glucose uptake by cells is impaired, contrary to this claim that insulin plays no role.
    • Why not D: Misdescribes type 2 diabetes as involving no insulin production at all, which is a description of type 1 diabetes; type 2 diabetes typically still involves insulin production, with the defect instead being reduced sensitivity of body cells to that insulin.

13. Question 13Answer: D

    1. The menstrual cycle is controlled by a sequence of hormones: FSH stimulates egg maturation, a surge in LH triggers ovulation, and oestrogen and progesterone regulate the build-up and breakdown of the uterus lining.
    2. The combined pill supplies synthetic oestrogen and progesterone at levels that feed back on the pituitary gland to suppress its normal release of FSH and LH, so no egg matures and no ovulation occurs; this is a hormonal method.
    3. A copper IUD, by contrast, releases no hormones and has no effect on FSH, LH or the ovarian cycle at all; instead, the copper ions it releases locally are toxic to sperm, reducing the chance of fertilisation, and also alter the uterus lining in a way that makes implantation of a fertilised egg unlikely.
    4. So the two methods prevent pregnancy by genuinely different routes, one hormonal and one non-hormonal, which is exactly the distinction option D draws, rather than the reversed or conflated mechanisms given in the other options.
    • Why not A: The pill's mechanism is hormonal suppression of ovulation, not a physical barrier to sperm; this option conflates the pill's mechanism with the copper IUD's local, physical/chemical action.
    • Why not B: Reverses the two mechanisms: a copper IUD is defined by releasing no hormones and working through copper ions, not oestrogen, while the pill's mechanism is hormonal suppression of ovulation, not a physical block of the uterus.
    • Why not C: Incorrectly attributes a hormonal, ovulation-suppressing mechanism to the copper IUD, when by definition this device releases no hormones and has no effect on the pituitary gland or the menstrual cycle's hormones at all.

14. Question 14Answer: A

    1. Before any human is exposed to a new compound, it must first go through pre-clinical testing: laboratory studies and animal studies, to establish basic information about toxicity, dosage, and whether the compound appears to have the intended biological effect at all.
    2. If pre-clinical testing is promising, clinical testing in humans can begin, starting with a small number of healthy volunteers; this stage is designed to check that the compound is safe in humans and to establish a safe range of doses, not yet to test whether it treats the disease.
    3. Only after safety has been established in healthy volunteers does testing move to a larger number of patients who actually have the disease; at this stage the drug's effectiveness is properly tested, often by comparing it against a placebo or an existing treatment, frequently using a double-blind method so that neither patient nor assessing clinician knows who received which treatment, reducing bias.
    4. Only once a compound has passed through all of these stages, showing it to be both safe and effective, can it normally be licensed for prescription to patients generally, matching the order given in option A.
    • Why not B: Reverses the necessary order; pre-clinical testing must always come first to establish basic safety before any human being, healthy or with the disease, is exposed to a completely untested compound.
    • Why not C: Misuses the term pre-clinical, which specifically refers to testing before any human trials (laboratory and animal studies), not testing in patients, and skips both the healthy-volunteer safety stage and the placebo-controlled patient trials stage entirely.
    • Why not D: Correctly starts with healthy volunteers for safety after pre-clinical work, but wrongly claims that testing in actual patients and placebo comparison are unnecessary once basic safety is shown; establishing that a drug is actually effective, not just safe, specifically requires trials in patients with the disease.

15. Question 15Answer: C

    1. A statin acts mainly on the liver to reduce the amount of cholesterol produced and circulating in the blood, which over time slows the build-up of fatty deposits inside artery walls, a major cause of coronary heart disease.
    2. An anticoagulant acts on the blood's clotting process itself, making the blood less likely to form a clot; this matters because a clot forming in a narrowed coronary artery, or breaking off and travelling elsewhere, can block blood flow and cause a heart attack or stroke.
    3. An antihypertensive drug acts to lower blood pressure, by mechanisms such as relaxing blood vessels or reducing the volume of fluid the heart has to pump, which reduces the mechanical strain on the heart muscle and on the walls of already-narrowed arteries.
    4. Each of the three drug types therefore targets a genuinely different risk factor or mechanism, cholesterol build-up, clot formation, and blood pressure respectively, rather than all three doing the same job or having their mechanisms shuffled between them, matching option C.
    • Why not A: Conflates three quite different mechanisms, cholesterol-lowering, anti-clotting, and blood-pressure-lowering, under one vague description ('blood thinning'), losing the real distinction between what each drug does.
    • Why not B: Swaps the statin and anticoagulant mechanisms with each other, and wrongly attributes cholesterol reduction to the antihypertensive drug's mechanism, when antihypertensives act on blood pressure, not cholesterol.
    • Why not D: Rotates all three mechanisms one place along, attaching each drug to a mechanism that actually belongs to one of the other two, rather than to its own real mode of action.