An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 HL (TZ1) IB Diploma Programme Biology paper. Not affiliated with or reproduced from IB.
Paper 1
Answer all 40 multiple-choice questions. No calculator is allowed.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
During intense physical exercise, the human body must coordinate cardiac output and ventilation rate to meet increased metabolic demands. Which of the following correctly describes the direct stimulus detected by chemoreceptors and the primary integration centre in the brain responsible for coordinating this response?
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Worked solution
During intense exercise, cell respiration increases in muscle tissues, releasing more carbon dioxide (\(CO_2\)) into the blood. \(CO_2\) reacts with water to form carbonic acid, which dissociates into hydrogen ions (\(H^+\)) and hydrogen carbonate ions, thereby lowering the blood pH. Central and peripheral chemoreceptors detect this decrease in blood pH and send signals to the cardiovascular and respiratory control centres located in the medulla oblongata of the brain. The medulla oblongata then coordinates an increase in ventilation rate and heart rate to restore homeostasis.
Marking scheme
Award [1] for the correct answer (B). Reject A because increased exercise decreases blood pH rather than increasing it, and the cerebellum controls motor coordination, not automatic autonomic reflexes. Reject C because oxygen concentration is not the primary direct stimulus for daily respiratory regulation, and the hypothalamus is not the primary coordinator of cardiac/ventilation rhythm during exercise. Reject D because blood carbon dioxide increases rather than decreases, and the cerebrum controls conscious actions rather than autonomic cardiorespiratory regulation.
Question 2 · multiple-choice
1 marks
How do the sympathetic and parasympathetic nervous systems integrate to regulate peristalsis in the human digestive tract?
A.Sympathetic and parasympathetic nerves synapse directly onto smooth muscle cells, completely bypassing the enteric nervous system.
B.Sympathetic stimulation increases peristalsis by activating enteric neurons, whereas parasympathetic stimulation decreases it.
C.Parasympathetic stimulation increases peristalsis by releasing acetylcholine, whereas sympathetic stimulation decreases it by releasing norepinephrine.
D.Both sympathetic and parasympathetic stimulation increase peristalsis to accelerate nutrient absorption during a flight-or-fight response.
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Worked solution
The autonomic nervous system regulates digestion through opposing actions of its two divisions. The parasympathetic division ('rest and digest') increases digestive activity, including peristalsis, primarily via the release of acetylcholine, which excites enteric neurons and smooth muscles. In contrast, the sympathetic division ('fight or flight') decreases digestive activity and redirects blood flow to skeletal muscles, primarily via the release of norepinephrine, which inhibits enteric pathways and relaxes digestive smooth muscle.
Marking scheme
Award [1] for the correct answer (C). Reject A because autonomic nerves synapse extensively with the enteric nervous system to modulate its activity, rather than bypassing it completely. Reject B because the roles are reversed; parasympathetic increases motility and sympathetic decreases it. Reject D because sympathetic stimulation decreases peristalsis during a flight-or-flight response.
Question 3 · multiple-choice
1 marks
During aerobic cell respiration, specific coenzymes act as electron carriers. Which of the following options correctly identifies the net yield of reduced coenzymes produced per molecule of glucose specifically within the citric acid (Krebs) cycle?
A.\(6\text{ NADH} + 2\text{ FADH}_2\)
B.\(2\text{ NADH} + 2\text{ FADH}_2\)
C.\(8\text{ NADH} + 2\text{ FADH}_2\)
D.\(10\text{ NADH} + 4\text{ FADH}_2\)
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Worked solution
For each molecule of glucose that enters glycolysis, two molecules of pyruvate are produced. These two pyruvates are converted into two molecules of acetyl-CoA during the link reaction. Consequently, the citric acid (Krebs) cycle turns twice per glucose molecule. Each turn of the citric acid cycle yields \(3\text{ NADH}\) and \(1\text{ FADH}_2\) (along with 1 ATP/GTP and \(2\text{ CO}_2\)). Therefore, two turns of the cycle yield a net total of \(6\text{ NADH}\) and \(2\text{ FADH}_2\).
Marking scheme
Award [1] for the correct answer (A). Reject B because this corresponds to only a single turn of the cycle (plus one extra NADH) or is mathematically incorrect. Reject C because it incorrectly includes NADH from the link reaction or glycolysis. Reject D because it overestimates the yield of both NADH and FADH2.
Question 4 · multiple-choice
1 marks
In mitochondria, what is the immediate consequence if the inner mitochondrial membrane becomes highly permeable to protons (\(H^+\)) due to the presence of an uncoupling protein?
A.The rate of the link reaction decreases because NAD+ can no longer be regenerated.
B.Protons flow back into the matrix without passing through ATP synthase, reducing ATP synthesis.
C.Oxygen consumption decreases because electrons can no longer be transferred to oxygen.
D.The pH of the intermembrane space decreases significantly.
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Worked solution
An uncoupling protein provides an alternative pathway for protons to diffuse down their electrochemical gradient from the intermembrane space into the mitochondrial matrix. Because they bypass ATP synthase, the energy stored in the proton gradient is dissipated as heat instead of driving the phosphorylation of ADP to ATP. Consequently, ATP synthesis decreases, even though the electron transport chain continues to function (often at an accelerated rate).
Marking scheme
Award [1] for the correct answer (B). Reject A because the electron transport chain continues to oxidize NADH to NAD+, meaning NAD+ regeneration is not inhibited. Reject C because oxygen consumption actually increases as the ETC works harder to try to re-establish the lost proton gradient. Reject D because protons leaving the intermembrane space would increase its pH (make it less acidic), not decrease it.
Question 5 · multiple-choice
1 marks
A person becomes dehydrated after a long run in hot weather. Which of the following sequences correctly describes how the nervous and endocrine systems coordinate with the kidneys to restore water balance?
A.Osmoreceptors in the medulla oblongata detect decreased solute concentration; the posterior pituitary releases more ADH; the collecting duct becomes less permeable to water.
B.Osmoreceptors in the hypothalamus detect decreased solute concentration; the posterior pituitary releases less ADH; the collecting duct becomes less permeable to water.
C.Osmoreceptors in the hypothalamus detect increased solute concentration; the posterior pituitary releases more ADH; the collecting duct becomes more permeable to water.
D.Osmoreceptors in the hypothalamus detect increased solute concentration; the anterior pituitary releases less ADH; the collecting duct becomes more permeable to water.
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Worked solution
Dehydration leads to a loss of water from the blood, which increases blood osmolarity (solute concentration). This change is detected by osmoreceptors in the hypothalamus. In response, the hypothalamus stimulates the posterior pituitary gland to secrete antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it binds to receptors on the collecting ducts, initiating a cascade that inserts aquaporins into the apical membranes. This increases the water permeability of the collecting ducts, allowing more water to be reabsorbed back into the blood, producing concentrated urine and restoring water balance.
Marking scheme
Award [1] for the correct answer (C). Reject A because the integration center is the hypothalamus, not the medulla, and increased ADH increases (not decreases) collecting duct permeability. Reject B because dehydration increases solute concentration, and less ADH would lead to more water loss (diuresis). Reject D because ADH is released by the posterior pituitary, not the anterior pituitary, and more (not less) ADH is released.
Question 6 · multiple-choice
1 marks
Which of the following processes occurs at the source during translocation in the phloem of a vascular plant?
A.Active transport of protons out of companion cells, which drives the co-transport of sucrose into the phloem, causing water to enter by osmosis.
B.Active transport of sucrose out of the phloem into sink cells, causing water to leave the phloem by osmosis.
C.Passive diffusion of sucrose into sieve tube elements, which decreases the hydrostatic pressure at the source.
D.Active transport of water into companion cells, which increases the hydrostatic pressure to push sap downwards.
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Worked solution
At the source (e.g., photosynthesizing leaves), sucrose must be actively loaded into the phloem. Companion cells use proton pumps (\(H^+\)-ATPases) to actively pump hydrogen ions out of the cell into the cell wall space, creating a proton gradient. Protons then diffuse back down their gradient through a co-transporter protein, carrying sucrose with them into the companion cell-sieve tube complex. This high concentration of sucrose lowers the water potential inside the sieve tube, causing water to flow into the phloem from the adjacent xylem by osmosis, which increases the hydrostatic pressure at the source.
Marking scheme
Award [1] for the correct answer (A). Reject B because unloading at the sink is described, not loading at the source. Reject C because sucrose loading is an active process and increases, rather than decreases, the hydrostatic pressure. Reject D because water movement into the phloem is passive (via osmosis), not active.
Question 7 · multiple-choice
1 marks
A population of wild rabbits is monitored over several decades. Which of the following is a density-dependent factor that would limit the growth of this population as it approaches its carrying capacity (\(K\))?
A.An increase in the transmission rate of a contagious viral disease within the warren.
B.A sudden, severe winter freeze that kills off young offspring.
C.A forest fire that destroys a large portion of the grassland habitat.
D.An increase in the concentration of toxic heavy metals in the local groundwater.
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Worked solution
Density-dependent factors are environmental factors whose impact on a population varies depending on the density of the population itself. As the rabbit population increases and approaches carrying capacity, individuals live in closer proximity. This higher density facilitates the rapid transmission of pathogens and contagious diseases (such as myxomatosis) through the population, increasing mortality rates. In contrast, severe weather (B), forest fires (C), and toxic chemical pollution (D) are density-independent factors because their occurrence and lethality do not depend on how many rabbits are present in the area.
Marking scheme
Award [1] for the correct answer (A). Reject B, C, and D because these represent density-independent factors whose impacts are independent of population density.
Question 8 · multiple-choice
1 marks
During the follicular phase of the human menstrual cycle, what trigger causes the sudden surge in luteinizing hormone (LH) that induces ovulation?
A.Progesterone levels drop to a critical minimum, releasing the hypothalamus from negative feedback.
B.High levels of follicle-stimulating hormone (FSH) directly convert into LH within the ovarian follicles.
C.Human chorionic gonadotropin (hCG) stimulates the corpus luteum to produce LH.
D.Estrogen levels rise continuously and reach a high threshold, switching from negative feedback to positive feedback on the hypothalamus and anterior pituitary.
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Worked solution
During the early and mid-follicular phase, estrogen secreted by the developing follicles exerts negative feedback on the hypothalamus and anterior pituitary, keeping LH and FSH levels relatively low. However, as the dominant follicle matures, it secretes increasingly high levels of estrogen. When estrogen levels rise above a specific threshold and remain elevated for about 36 hours, the feedback mechanism switches from negative to positive. This positive feedback stimulates a massive, sudden release of GnRH and a subsequent surge of LH from the anterior pituitary, which triggers ovulation.
Marking scheme
Award [1] for the correct answer (D). Reject A because progesterone is very low during the follicular phase and only rises after ovulation when the corpus luteum forms. Reject B because hormones are separate proteins/glycoproteins and do not convert into one another. Reject C because hCG is only secreted by an embryo/placenta during pregnancy to maintain the corpus luteum; it does not trigger ovulation.
Question 9 · multiple_choice
1 marks
During intense physical exercise, the body coordinates several physiological responses to meet the metabolic demands of skeletal muscle. Which of the following correctly describes the integrated response of the nervous system and the cardiovascular system?
A.Increased sympathetic stimulation releases acetylcholine, leading to vasoconstriction in active skeletal muscles.
B.Decreased parasympathetic activity and increased sympathetic activity result in an increased heart rate and increased cardiac output.
C.Baroreceptors detect an increase in blood pressure, triggering the cardiovascular center in the cerebellum to increase heart rate.
D.Epinephrine released by the anterior pituitary gland stimulates general vasodilation in all vascular beds.
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Worked solution
During physical exercise, the body needs to deliver oxygen and nutrients to active skeletal muscles more rapidly. The cardiovascular control center in the medulla oblongata coordinates a decrease in parasympathetic nervous activity (which normally slows the heart) and an increase in sympathetic nervous activity. This increases both heart rate and stroke volume, resulting in an increased cardiac output. Sympathetic stimulation also leads to selective vasoconstriction of vessels feeding non-essential organs (like the digestive tract) and vasodilation of vessels feeding active skeletal muscles.
Marking scheme
Award 1 mark for the correct answer (B). Reject all other options as they contain anatomical or physiological inaccuracies: option A incorrectly states sympathetic fibers release acetylcholine to cause vasoconstriction; option C incorrectly states the cardiovascular center is in the cerebellum (it is in the medulla); option D incorrectly states epinephrine is released by the anterior pituitary (it is released by the adrenal medulla).
Question 10 · multiple_choice
1 marks
During exercise, blood carbon dioxide levels increase. Which pathway correctly outlines the homeostatic response to return blood pH and carbon dioxide levels back to normal?
A.Chemoreceptors in the medulla detect a decrease in pH \(\rightarrow\) respiratory center in the medulla oblongata \(\rightarrow\) increased stimulation of the diaphragm and external intercostal muscles \(\rightarrow\) increased ventilation rate.
B.Baroreceptors in the carotid sinus detect an increase in carbon dioxide \(\rightarrow\) cardiovascular center in the pons \(\rightarrow\) decreased stimulation of the external intercostal muscles \(\rightarrow\) decreased tidal volume.
C.Chemoreceptors in the lungs detect a decrease in oxygen \(\rightarrow\) anterior pituitary gland \(\rightarrow\) secretion of erythropoietin \(\rightarrow\) immediate increase in breathing depth.
D.Thermoreceptors in the hypothalamus detect increased temperature \(\rightarrow\) motor cortex \(\rightarrow\) voluntary control of alveolar ventilation \(\rightarrow\) decreased respiration rate.
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Worked solution
Increased cellular respiration during exercise releases more carbon dioxide into the blood, which reacts with water to form carbonic acid, lowering blood pH. Central chemoreceptors in the medulla oblongata (and peripheral chemoreceptors in the carotid and aortic bodies) detect the resulting decrease in pH. They send signals to the respiratory center in the medulla oblongata, which increases the frequency of nerve impulses sent via the phrenic and intercostal nerves to the diaphragm and external intercostal muscles. This increases ventilation rate and depth, accelerating the excretion of carbon dioxide and restoring normal pH.
Marking scheme
Award 1 mark for the correct pathway (A). Reject B because baroreceptors detect pressure (not carbon dioxide) and the respiratory control center is in the medulla, not the pons. Reject C because erythropoietin is a hormone regulating red blood cell production released by the kidneys, not an immediate ventilator reflex regulator. Reject D because thermoreception and voluntary control are not the primary homeostatic pathways for blood carbon dioxide regulation.
Question 11 · multiple_choice
1 marks
If a eukaryotic cell is exposed to an inhibitor that specifically blocks the active transport of pyruvate across the mitochondrial membrane, what would be the immediate effect on cell respiration?
A.Glycolysis will stop because of the lack of oxidized NAD+ in the cytoplasm.
B.The rate of lactate or ethanol production in the cytoplasm will increase to regenerate NAD+.
C.The Krebs cycle will continue at a normal rate using acetyl-CoA derived directly from cytoplasmic glucose.
D.Oxygen consumption by the electron transport chain will increase to compensate for the lack of NADH.
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Worked solution
Pyruvate is produced by glycolysis in the cytoplasm. To enter the link reaction and the Krebs cycle, it must be actively transported across the mitochondrial outer and inner membranes. If this transport is blocked, pyruvate accumulates in the cytoplasm. Because the mitochondria cannot accept pyruvate, aerobic respiration stops. To keep glycolysis running and generating ATP, the cell must regenerate oxidized NAD+ in the cytoplasm by converting pyruvate to lactate (in animal cells) or ethanol and carbon dioxide (in yeast), thereby increasing the rate of anaerobic fermentation.
Marking scheme
Award 1 mark for the correct answer (B). Reject A because glycolysis does not stop immediately; rather, anaerobic pathways are up-regulated to regenerate NAD+. Reject C because the Krebs cycle cannot run without pyruvate entering the mitochondria to form acetyl-CoA. Reject D because without pyruvate entry, NADH and FADH2 production in the matrix drops, meaning the electron transport chain will have fewer electron donors and oxygen consumption will decrease, not increase.
Question 12 · multiple_choice
1 marks
Brown adipose tissue contains high levels of uncoupling protein 1 (UCP1) in its mitochondrial inner membranes. UCP1 allows protons (\(H^+\)) to leak down their electrochemical gradient into the matrix without passing through ATP synthase. What is the metabolic consequence of this leak?
A.More ATP is produced per molecule of glucose oxidized.
B.The electron transport chain stops operating because the proton gradient is lost.
C.Energy from the proton gradient is dissipated as heat instead of being captured as ATP.
D.Oxygen is no longer reduced to water at the end of the electron transport chain.
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Worked solution
In typical mitochondria, protons pumped into the intermembrane space by the electron transport chain can only flow back into the matrix through ATP synthase, coupling the proton motive force to the phosphorylation of ADP to ATP. Uncoupling protein 1 (UCP1) provides an alternative pathway for protons to leak back into the matrix. This collapses the proton gradient without producing ATP, dissipating the stored electrochemical energy directly as heat (non-shivering thermogenesis).
Marking scheme
Award 1 mark for the correct answer (C). Reject A because ATP production per glucose decreases, not increases. Reject B because the electron transport chain actually runs faster as it attempts to restore the collapsed proton gradient. Reject D because oxygen continues to act as the terminal electron acceptor and is reduced to water.
Question 13 · multiple_choice
1 marks
A person is severely dehydrated. How does the body respond to restore osmotic balance, and what is the direct cellular effect on the kidneys?
A.Osmoreceptors in the hypothalamus detect high solute concentration \(\rightarrow\) posterior pituitary releases ADH \(\rightarrow\) aquaporins are inserted into the collecting duct membranes \(\rightarrow\) water reabsorption increases.
B.Osmoreceptors in the medulla detect low blood volume \(\rightarrow\) anterior pituitary releases less ADH \(\rightarrow\) aquaporins are removed from the collecting duct membranes \(\rightarrow\) water reabsorption decreases.
C.Baroreceptors in the aorta detect high blood pressure \(\rightarrow\) adrenal cortex secretes aldosterone \(\rightarrow\) the loop of Henle becomes impermeable to water \(\rightarrow\) water excretion increases.
D.Chemoreceptors in the carotid body detect high urea levels \(\rightarrow\) hypothalamus secretes ADH \(\rightarrow\) proximal convoluted tubule actively transports water into blood capillaries.
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Worked solution
Dehydration increases the solute concentration (osmolarity) of the blood. Osmoreceptors in the hypothalamus detect this change and stimulate the posterior pituitary gland to release antidiuretic hormone (ADH) into the blood. ADH travels to the kidneys, where it binds to receptors on the cells of the collecting ducts. This triggers a secondary messenger cascade causing vesicles containing aquaporin channels to fuse with the apical membrane, increasing water permeability and allowing more water to be reabsorbed back into the bloodstream.
Marking scheme
Award 1 mark for the correct homeostatic pathway and response (A). Reject B because the posterior pituitary, not the anterior pituitary, stores and releases ADH, and dehydration leads to more ADH release, not less. Reject C because aldosterone is involved in sodium balance and its primary action is not making the loop of Henle impermeable. Reject D because water reabsorption in response to ADH occurs in the collecting ducts, not the proximal tubule, and travels via osmosis (passive), not active transport.
Question 14 · multiple_choice
1 marks
According to the pressure-flow hypothesis of translocation in phloem, how is sucrose loaded into sieve tube members at the source, and what is the physical consequence of this loading?
A.Sucrose is actively transported out of the phloem, causing the water potential inside the sieve tube to increase and lowering the hydrostatic pressure.
B.Sucrose is loaded by active transport, which decreases water potential inside the sieve tube, causing water to enter by osmosis and increasing hydrostatic pressure.
C.Sucrose diffuses passively down its concentration gradient, decreasing water potential and causing water to leave the phloem, which increases hydrostatic pressure.
D.Sucrose is actively pumped along with water through plasmodesmata, keeping water potential constant and maintaining equal hydrostatic pressure from source to sink.
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Worked solution
At the source (e.g., photosynthesizing leaves), sucrose is actively loaded into sieve tube members. This active transport is usually driven by a proton-sucrose symport mechanism, which uses energy. The accumulation of sucrose inside the sieve tube lowers the water potential (makes it more negative). Consequently, water from the adjacent xylem moves into the sieve tube by osmosis. The entry of water increases the hydrostatic pressure at the source end, driving the bulk flow of sap toward the sink.
Marking scheme
Award 1 mark for the correct option (B). Reject A because active transport moves sucrose into, not out of, the phloem at the source. Reject C because sucrose loading is an active process against its concentration gradient, not passive diffusion. Reject D because plasmodesmata facilitate symplastic transport, but active loading is needed to build up the pressure gradient, which is unequal between source and sink.
Question 15 · multiple_choice
1 marks
In a classic sigmoid population growth curve (S-curve), which phase is characterized by the population size approaching the carrying capacity (K) of the environment, and what is the main driver of this phase?
A.Exponential phase, driven by abundant resources and a lack of environmental resistance.
B.Transitional phase, driven by increasing density-dependent factors such as resource depletion and competition.
C.Plateau phase, driven by a birth rate that greatly exceeds the death rate of the population.
D.Lag phase, driven by the organisms adapting to a brand-new, hostile environment.
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Worked solution
A sigmoid population growth curve consists of the lag phase, exponential phase, transitional phase, and plateau phase. During the transitional phase, resources start to become limited as the population density rises. Density-dependent factors such as increased competition for food, space, and increased predation or disease begin to slow down the rate of population growth, causing the curve to transition from exponential growth to a plateau near the carrying capacity (K).
Marking scheme
Award 1 mark for the correct phase and description (B). Reject A because the exponential phase is where growth is at its maximum and the population is far below carrying capacity. Reject C because during the plateau phase, the birth rate equals the death rate (it does not greatly exceed it). Reject D because the lag phase occurs at the very beginning of population colonization when numbers are low, not when approaching carrying capacity.
Question 16 · multiple_choice
1 marks
In the female menstrual cycle, what is the primary role of the surge in Luteinizing Hormone (LH) around day 14, and what event triggers this sudden surge?
A.It stimulates the development of primary follicles, and is triggered by high levels of progesterone.
B.It causes the shedding of the endometrium, and is triggered by a sudden drop in estrogen.
C.It triggers ovulation and the formation of the corpus luteum, and is stimulated by high estrogen levels from the dominant follicle.
D.It inhibits FSH secretion by the anterior pituitary, and is triggered by the release of human chorionic gonadotropin.
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Worked solution
Around day 14 of the menstrual cycle, the dominant follicle (Graafian follicle) releases high and sustained levels of estrogen. This high concentration of estrogen exerts positive feedback on the hypothalamus and anterior pituitary gland, triggering a massive, sudden release (surge) of Luteinizing Hormone (LH). This LH surge stimulates the completion of meiosis I in the oocyte, triggers ovulation (rupture of the follicle to release the secondary oocyte), and causes the remaining follicular cells to develop into the corpus luteum.
Marking scheme
Award 1 mark for the correct option (C). Reject A because FSH (follicle stimulating hormone) is responsible for follicle development and is inhibited, not triggered, by progesterone. Reject B because shedding of the endometrium (menstruation) is triggered by a drop in progesterone and estrogen at the end of the cycle, not by LH. Reject D because hCG is produced by an embryo after implantation to maintain the corpus luteum, not to trigger ovulation or inhibit FSH during the normal cycle.
Question 17 · Multiple Choice
1 marks
During vigorous exercise, multiple body systems must coordinate their functions to maintain homeostasis. Which of the following correctly describes a physiological mechanism occurring during heavy exercise?
A.Epinephrine causes bronchodilation in the lungs and vasoconstriction in arterioles supplying skeletal muscles.
B.An increase in blood carbon dioxide concentration decreases blood pH, which is detected by chemoreceptors that stimulate the ventilation center in the medulla oblongata to increase breathing rate.
C.A decrease in blood pH shifts the oxygen-hemoglobin dissociation curve to the left, ensuring oxygen is more tightly bound to hemoglobin in active tissues.
D.The cardiovascular center in the medulla oblongata sends parasympathetic signals via the vagus nerve to increase cardiac outputExternal Link.
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Worked solution
During heavy exercise, increased cell respiration produces more carbon dioxide (\(CO_2\)). This dissolved \(CO_2\) forms carbonic acid, decreasing blood pH. Chemoreceptors in the medulla oblongata, aorta, and carotid arteries detect this change and signal the respiratory center in the medulla oblongata to increase the rate and depth of ventilation. Option A is incorrect because epinephrine causes vasodilation (not vasoconstriction) in arterioles supplying skeletal muscles to increase blood flow. Option C is incorrect because a decrease in pH shifts the curve to the right (Bohr effect), decreasing hemoglobin's affinity for oxygen to facilitate oxygen release to active muscles. Option D is incorrect because increasing cardiac output during exercise is achieved via sympathetic signals, while parasympathetic signals via the vagus nerve decrease heart rate.
Marking scheme
Award [1] mark for the correct choice B. Award [0] marks for any other choice.
Question 18 · Multiple Choice
1 marks
Which of the following correctly describes the products formed during the link reaction of aerobic cell respiration per molecule of pyruvate, and their immediate fates?
A.One molecule of \(CO_2\) is released as waste, one molecule of \(FADH_2\) is sent to the electron transport chain, and one acetyl group is transferred to coenzyme A.
B.One molecule of \(CO_2\) is released as waste, one molecule of \(NADH + H^+\) is sent to the electron transport chain, and one acetyl group is transferred to coenzyme A.
C.Two molecules of \(CO_2\) are released as waste, two molecules of \(NADH + H^+\) are sent to the electron transport chain, and one acetyl group is transferred to coenzyme A.
D.One molecule of \(ATP\) is produced by substrate-level phosphorylation, one molecule of \(NADH + H^+\) is sent to the electron transport chain, and one acetyl group is transferred to coenzyme A.
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Worked solution
The link reaction decarboxylates and oxidizes pyruvate. For each molecule of pyruvate entering the mitochondrial matrix: 1. One carbon atom is removed as carbon dioxide (\(CO_2\)), which diffuses out of the cell as waste. 2. Pyruvate is oxidized by transferring hydrogen atoms to \(NAD^+\), forming one molecule of \(NADH + H^+\), which goes to the electron transport chain. 3. The remaining 2-carbon acetyl group is attached to coenzyme A, forming acetyl-CoA, which enters the Krebs cycle. Option A is incorrect because \(FADH_2\) is produced in the Krebs cycle, not the link reaction. Option C is incorrect because it describes the products per glucose molecule (two pyruvates), not per single pyruvate molecule. Option D is incorrect because no ATP is produced during the link reaction.
Marking scheme
Award [1] mark for the correct choice B. Award [0] marks for any other choice.
Question 19 · Multiple Choice
1 marks
A person consumes a large volume of water over a short period. Which of the following endocrine and physiological responses will occur in the body to restore osmotic balance?
A.Osmoreceptors in the hypothalamus detect a decrease in blood solute concentration, leading to decreased secretion of ADH from the posterior pituitary gland and decreased water reabsorption in the collecting ducts.
B.Osmoreceptors in the hypothalamus detect an increase in blood solute concentration, leading to increased secretion of ADH from the posterior pituitary gland and increased water reabsorption in the collecting ducts.
C.Osmoreceptors in the medulla oblongata detect a decrease in blood solute concentration, leading to decreased secretion of ADH from the anterior pituitary gland and decreased water reabsorption in the loop of Henle.
D.Osmoreceptors in the hypothalamus detect a decrease in blood solute concentration, leading to increased secretion of ADH from the posterior pituitary gland and increased water reabsorption in the distal convoluted tubule.
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Worked solution
When a large volume of water is consumed, the blood becomes more dilute (decreased solute concentration or osmolarity). Osmoreceptors in the hypothalamus detect this decrease. Consequently, the hypothalamus signals the posterior pituitary gland to decrease the secretion of Antidiuretic Hormone (ADH). Lower levels of ADH make the walls of the collecting ducts less permeable to water (fewer aquaporins are inserted into the membrane), resulting in decreased water reabsorption into the blood and the production of a large volume of dilute urine. Option B describes the response to dehydration. Option C is incorrect because osmoreceptors are in the hypothalamus (not medulla oblongata), ADH is released from the posterior (not anterior) pituitary, and ADH acts mainly on the collecting ducts (not loop of Henle). Option D is incorrect because ADH secretion is decreased, not increased, when the blood is dilute.
Marking scheme
Award [1] mark for the correct choice A. Award [0] marks for any other choice.
Question 20 · Multiple Choice
1 marks
Which of the following best describes the active mechanism that occurs during phloem loading in source tissues, such as leaves?
A.Sucrose is actively transported directly into the sieve tube elements by a sodium-potassium pump.
B.Hydrogen ions are actively pumped out of companion cells, creating a proton gradient that drives the co-transport of sucrose into the companion cells.
C.Water is actively pumped into the phloem from the xylem, generating a high hydrostatic pressure that pushes sucrose toward the sink.
D.Sucrose is actively transported out of the phloem into the companion cells via facilitated diffusion channel proteins.
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Worked solution
Phloem loading involves active transport of sucrose into companion cells. Proton pumps (\(H^+\)-ATPases) in the plasma membrane of companion cells actively pump hydrogen ions out of the cell into the cell wall space, creating a high concentration of protons outside the cell. Protons then flow back down their concentration gradient into the companion cells through a co-transporter protein, which simultaneously transports sucrose against its concentration gradient (cotransport). Sucrose then moves from the companion cells into the sieve tube elements via plasmodesmata. Option A is incorrect because it uses a proton gradient, not sodium-potassium. Option C is incorrect because water moves passively by osmosis, not active pumping. Option D is incorrect because sucrose is loaded into the phloem, not out, and loading is an active process, not passive facilitated diffusion.
Marking scheme
Award [1] mark for the correct choice B. Award [0] marks for any other choice.
Question 21 · Multiple Choice
1 marks
A population of yeast is grown in a closed laboratory culture vessel containing a fixed amount of nutrient broth. Which of the following describes the phase of population growth where the natality rate is approximately equal to the mortality rate, and the primary reason for this?
A.Exponential phase, because resources are abundant and waste products have not yet accumulated.
B.Transitional phase, because the rate of population growth is slowing down as resources begin to limit survival.
C.Plateau phase, because the population has reached the carrying capacity of the environment due to limiting factors such as nutrient depletion and toxic waste accumulation.
D.Lag phase, because the organisms are adapting to their new environment and synthesizing necessary enzymes.
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Worked solution
In a sigmoid population growth curve (which occurs in closed environments with limited resources), the plateau phase represents the stage where population growth slows to zero. At this stage, the natality (birth/division) rate equals the mortality (death) rate (and since it is a closed culture, there is no immigration or emigration). This occurs because the population has reached the carrying capacity (\(K\)) due to density-dependent limiting factors like the depletion of glucose (nutrients) and the accumulation of toxic metabolic wastes (such as ethanol in yeast). Option A describes the phase of rapid growth where natality greatly exceeds mortality. Option B describes the phase where growth is slowing but natality is still greater than mortality. Option D is the initial phase where there is little growth as the cells adjust to the medium.
Marking scheme
Award [1] mark for the correct choice C. Award [0] marks for any other choice.
Question 22 · Multiple Choice
1 marks
During the human menstrual cycle, hormones secreted by the pituitary gland and the ovaries interact via positive and negative feedback loops. Which of the following events is triggered by the peak concentration of estrogen secreted by the developing follicle?
A.Negative feedback on the hypothalamus, leading to an immediate decrease in LH levels.
B.Positive feedback on the pituitary gland, leading to a surge in LH secretion that induces ovulation.
C.Inhibition of progesterone production by the corpus luteum, causing menstruation.
D.Direct stimulation of the endometrium to break down and shed.
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Worked solution
During the follicular phase, estrogen is secreted by the growing follicle. When estrogen levels rise and reach a peak, they exert a positive feedback effect on the pituitary gland (and hypothalamus), causing a massive surge in luteinizing hormone (LH) secretion. This LH surge is the direct trigger for ovulation (the release of the secondary oocyte from the mature follicle). Option A is incorrect because the peak level of estrogen triggers positive feedback, not negative feedback, leading to an LH surge, not a decrease. Option C is incorrect because the corpus luteum is only formed after ovulation, and menstruation is caused by a decline in progesterone and estrogen when the corpus luteum degenerates. Option D is incorrect because estrogen stimulates the thickening (proliferation) of the endometrium, not its breakdown.
Marking scheme
Award [1] mark for the correct choice B. Award [0] marks for any other choice.
Question 23 · Multiple Choice
1 marks
The secretion of gastric juice in the stomach is regulated by both nervous and hormonal mechanisms. Which of the following correctly describes the events during the gastric phase of digestion?
A.The sight and smell of food trigger sympathetic pathways to release gastrin from the stomach lining, which inhibits pepsinogen secretion.
B.The stretching of the stomach wall by food triggers impulses to the brain, which sends parasympathetic signals via the vagus nerve to stimulate gastric glands to secrete gastric juice and gastrin.
C.High acidity in the duodenum triggers the release of secretin and somatostatin, which stimulate the gastric glands to increase hydrochloric acid production.
D.Food entering the stomach stimulates the release of acetylcholine, which acts on the pituitary gland to release antidiuretic hormone to conserve water for digestion.
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Worked solution
The gastric phase of digestion begins when food enters the stomach. The physical stretching (distension) of the stomach wall activates stretch receptors, which send nervous impulses to the brainstem (medulla oblongata). The brain sends parasympathetic signals back via the vagus nerve to stimulate the gastric glands to release gastric juice (hydrochloric acid and pepsinogen) and stimulates G-cells to release the hormone gastrin into the blood, which further enhances secretion. Option A describes the cephalic phase (sight and smell), which is parasympathetic, not sympathetic, and gastrin stimulates (not inhibits) pepsinogen. Option C describes the intestinal phase, where high acidity and hormones like secretin inhibit (not stimulate) gastric secretion. Option D is incorrect because acetylcholine acts locally on the gastric glands, and antidiuretic hormone (ADH) from the pituitary is not involved in stimulating gastric secretion.
Marking scheme
Award [1] mark for the correct choice B. Award [0] marks for any other choice.
Question 24 · Multiple Choice
1 marks
In aerobic cell respiration, what is the direct source of energy used by ATP synthase to phosphorylate ADP to ATP during oxidative phosphorylation?
A.The transfer of electrons from reduced coenzymes (NADH and \(FADH_2\)) directly to ADP.
B.The movement of protons (\(H^+\)) down their electrochemical gradient from the intermembrane space to the matrix.
C.The hydrolysis of water molecules in the intermembrane space to release oxygen and high-energy protons.
D.The active transport of protons across the outer mitochondrial membrane into the cytoplasm.
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Worked solution
During oxidative phosphorylation, the electron transport chain pumps protons (\(H^+\)) from the matrix into the intermembrane space, creating a high concentration of protons (an electrochemical proton gradient). The direct source of energy used by ATP synthase is the passive movement of these protons down their electrochemical gradient back into the matrix through the ATP synthase channel. This movement of protons (chemiosmosis) drives the rotation of the enzyme, providing the mechanical energy to synthesize ATP from ADP and inorganic phosphate. Option A is incorrect because the transfer of electrons provides energy to pump protons, not directly to ADP. Option C describes photolysis of water in photosynthesis, whereas in respiration, water is formed as a product, not hydrolyzed. Option D is incorrect because protons are pumped across the inner (not outer) mitochondrial membrane, and their flow through ATP synthase is passive down a gradient, not active transport.
Marking scheme
Award [1] mark for the correct choice B. Award [0] marks for any other choice.
Question 25 · multiple-choice
1 marks
During exercise, cellular respiration increases, leading to a rise in carbon dioxide levels in the blood. Which of the following correctly identifies the location of the chemoreceptors that detect this change and the brain region that coordinates the homeostatic response to increase ventilation?
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Worked solution
Carbon dioxide reacts with water to form carbonic acid, which dissociates and lowers blood pH. This change is detected by central chemoreceptors in the medulla oblongata and peripheral chemoreceptors in the aorta and carotid arteries. The respiratory control center, also located in the medulla oblongata, coordinates the autonomic response to increase ventilation rate and depth to expel the excess carbon dioxide.
Marking scheme
Award 1 mark for the correct option B. Reject all other options because they misidentify either the location of the respiratory chemoreceptors or the region of the brain coordinating the ventilation response.
Question 26 · multiple-choice
1 marks
Which comparison correctly describes the blood in the hepatic portal vein compared to the blood in the hepatic vein two hours after a meal rich in carbohydrates?
A.The hepatic portal vein has a lower concentration of glucose because the liver has already stored it as glycogen.
B.The hepatic portal vein has a higher concentration of glucose and a lower concentration of urea.
C.The hepatic vein has a higher concentration of glucose because the liver is actively breaking down glycogen.
D.The hepatic vein has a higher concentration of amino acids because the liver is synthesizing plasma proteins.
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Worked solution
Two hours after a carbohydrate-rich meal, glucose is actively absorbed from the small intestine into the capillaries that drain into the hepatic portal vein, making its glucose concentration higher than in the hepatic vein (as the liver hepatocytes absorb and store excess glucose as glycogen). Additionally, the liver produces urea via deamination of excess amino acids and releases it into the hepatic vein, meaning the blood entering via the hepatic portal vein has a lower urea concentration than the blood leaving via the hepatic vein.
Marking scheme
Award 1 mark for the correct option B. Reject option A and C because the hepatic portal vein carries glucose-rich blood from the gut directly to the liver before glycogen synthesis occurs. Reject option D because the liver synthesizes plasma proteins but does not release free amino acids in higher concentrations into the hepatic vein.
Question 27 · multiple-choice
1 marks
During aerobic cell respiration, which stages directly produce carbon dioxide as a product?
A.Glycolysis and Link reaction only
B.Link reaction and Krebs cycle only
C.Krebs cycle and Oxidative phosphorylation only
D.Glycolysis, Link reaction, and Krebs cycle
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Worked solution
Carbon dioxide is produced during decarboxylation reactions. These occur in the link reaction (where the three-carbon pyruvate is decarboxylated to a two-carbon acetyl group) and in the Krebs cycle (during the decarboxylation steps of six-carbon and five-carbon intermediates). Glycolysis does not involve decarboxylation, and oxidative phosphorylation uses oxygen as a terminal electron acceptor to form water, without producing carbon dioxide.
Marking scheme
Award 1 mark for the correct option B. Reject any options containing glycolysis or oxidative phosphorylation as steps that produce carbon dioxide.
Question 28 · multiple-choice
1 marks
Cyanide is a potent poison that binds irreversibly to cytochrome c oxidase (complex IV) in the electron transport chain. What is the immediate effect of this inhibition inside the mitochondrion?
A.The pH of the intermembrane space increases as proton pumping ceases, stopping ATP synthesis.
B.The pH of the matrix increases because protons are pumped at an accelerated rate.
C.Oxygen consumption increases as the electron transport chain attempts to overcome the block.
D.NADH oxidation increases to compensate for the lost ATP production.
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Worked solution
Inhibition of complex IV prevents the transport of electrons along the electron transport chain to oxygen. Consequently, proton pumping from the matrix into the intermembrane space ceases. Since protons are no longer actively pumped, the proton concentration in the intermembrane space decreases, meaning its pH increases. Without this proton concentration gradient, ATP synthase cannot function, and ATP synthesis stops.
Marking scheme
Award 1 mark for the correct option A. Reject option B because matrix pH would decrease (become more acidic) as protons fail to be pumped out of it. Reject option C and D because electron transport is completely blocked, preventing oxygen reduction and NADH oxidation.
Question 29 · multiple-choice
1 marks
During a prolonged run on a hot day, a runner becomes dehydrated. Which of the following describes the correct homeostatic response to restore water balance?
A.Hypothalamus detects low blood solute concentration; pituitary gland releases less ADH; collecting duct becomes less permeable to water.
B.Hypothalamus detects high blood solute concentration; pituitary gland releases more ADH; collecting duct becomes more permeable to water.
C.Hypothalamus detects high blood solute concentration; pituitary gland releases less ADH; collecting duct becomes more permeable to water.
D.Hypothalamus detects low blood solute concentration; pituitary gland releases more ADH; collecting duct becomes less permeable to water.
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Worked solution
Dehydration leads to an increase in blood solute concentration (high osmolarity). This change is detected by osmoreceptors in the hypothalamus, which stimulates the posterior pituitary gland to secrete more antidiuretic hormone (ADH). ADH travels in the bloodstream to the kidneys, where it increases the water permeability of the collecting duct cells by promoting the insertion of aquaporins, allowing more water to be reabsorbed back into the blood.
Marking scheme
Award 1 mark for the correct option B. Reject other options because they misstate either the initial stimulus detection (dehydration causes high, not low, solute concentration) or the resulting ADH levels and collecting duct permeability changes.
Question 30 · multiple-choice
1 marks
Active transport is used to load sucrose into the sieve tube elements of the phloem at a source. What is the direct physical consequence of this high sucrose concentration?
A.Water potential increases, causing water to diffuse out of the phloem into adjacent xylem vessels.
B.Water potential decreases, causing water to enter the phloem from the xylem by osmosis, which increases hydrostatic pressure.
C.Hydrostatic pressure decreases, drawing organic solutes along the phloem toward the sink.
D.Water is actively transported into the phloem to dilute the sucrose concentration.
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Worked solution
The active loading of sucrose into the sieve tube elements at the source decreases the solute potential and thus decreases the overall water potential inside the phloem. Consequently, water moves from the adjacent xylem vessels (where the water potential is higher) into the phloem by osmosis. This influx of water increases the hydrostatic pressure inside the sieve tubes at the source, driving the bulk flow of sap toward the sink.
Marking scheme
Award 1 mark for the correct option B. Reject option A because water potential decreases, not increases. Reject option C because hydrostatic pressure increases rather than decreases. Reject option D because water movement is passive (osmosis), not active transport.
Question 31 · multiple-choice
1 marks
A biologist uses the Lincoln index to estimate a population of land snails in a woodland. They capture, mark, and release 40 snails. One week later, they capture 50 snails from the same area, of which 10 are marked. What is the estimated population size?
A.80
B.150
C.200
D.400
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Worked solution
Using the Lincoln index formula: \(N = \frac{n_1 \times n_2}{m_2}\) where \(n_1 = 40\) (the number of individuals marked in the first sample), \(n_2 = 50\) (the total number of individuals caught in the second sample), and \(m_2 = 10\) (the number of marked individuals recaptured in the second sample). Substituting these values: \(N = \frac{40 \times 50}{10} = 200\).
Marking scheme
Award 1 mark for the correct option C. Reject other options which result from incorrect application or rearrangement of the Lincoln index formula.
Question 32 · multiple-choice
1 marks
During the human menstrual cycle, estrogen levels rise significantly during the late follicular phase. What is the primary homeostatic effect of this high concentration of estrogen just before ovulation?
A.It exerts negative feedback on FSH secretion, immediately causing the degeneration of the dominant follicle.
B.It exerts positive feedback on the anterior pituitary, triggering a surge in luteinizing hormone (LH) secretion.
C.It exerts negative feedback on the hypothalamus, inhibiting any further secretion of luteinizing hormone (LH).
D.It directly stimulates the corpus luteum to begin secreting large quantities of progesterone.
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Worked solution
During most of the follicular phase, estrogen exerts negative feedback on the pituitary. However, when estrogen levels reach a high threshold concentration near the end of the follicular phase, they switch to exerting positive feedback on both the hypothalamus and the anterior pituitary. This stimulates a massive, rapid surge in luteinizing hormone (LH) secretion, which triggers ovulation of the mature follicle.
Marking scheme
Award 1 mark for the correct option B. Reject option A and C because high estrogen levels at this specific point stimulate rather than inhibit LH secretion. Reject option D because the corpus luteum has not yet formed before ovulation has taken place.
Question 33 · multiple-choice
1 marks
During intense exercise, several physiological changes occur to meet the increased oxygen demands of skeletal muscle. Which of the following correctly describes a coordinated response between the cardiovascular and respiratory systems during exercise?
A.An increase in blood pH stimulates the ventilation center in the medulla oblongata, decreasing cardiac output.
B.Increased carbon dioxide production lowers blood pH, which is detected by chemoreceptors, leading to an increased heart rate and increased ventilation rate.
C.Local vasodilation in active muscles increases total peripheral resistance, causing a reflex decrease in ventilation depth.
D.Decreased stretch receptor activity in the lungs directly stimulates the sinoatrial node to increase heart rate.
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Worked solution
During exercise, increased aerobic respiration in muscles produces more carbon dioxide, which reacts with water to form carbonic acid, lowering blood pH. This is detected by central and peripheral chemoreceptors, which send signals to the medulla oblongata to increase both ventilation rate and heart rate (via the sympathetic nervous system) to enhance oxygen delivery and carbon dioxide removal.
Marking scheme
Award [1] for the correct answer (B). Reject alternative options: A is incorrect because blood pH decreases, not increases, and cardiac output increases. C is incorrect because vasodilation decreases peripheral resistance, and ventilation increases. D is incorrect because stretch receptors in the lungs inhibit inhalation and do not directly stimulate the SA node in this manner.
Question 34 · multiple-choice
1 marks
Which of the following processes represents an endocrine integration of the digestive system response to food entering the duodenum from the stomach?
A.Vagus nerve stimulation causes the release of gastrin, which increases gastric acid secretion.
B.The acidic chyme stimulates the release of secretin from the duodenum, which triggers the pancreas to release bicarbonate ions.
C.Bile salts emulsify lipids in the small intestine through direct neural reflexes from the enteric nervous system.
D.Low pH in the duodenum inhibits the secretion of cholecystokinin (CCK), preventing pancreatic enzyme release.
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Worked solution
Acidic chyme entering the duodenum stimulates endocrine cells in the duodenal mucosa to release the hormone secretin. Secretin travels via the bloodstream (endocrine pathway) to the pancreas, stimulating the secretion of bicarbonate-rich pancreatic juice to neutralize the acid.
Marking scheme
Award [1] for the correct answer (B). Reject A as it represents a nervous/endocrine interaction of the gastric phase. Reject C because emulsification is a physical process, not endocrine. Reject D because low pH and lipids stimulate, rather than inhibit, CCK release.
Question 35 · multiple-choice
1 marks
If a cell is treated with a chemical that blocks the transfer of electrons from cytochrome c oxidase to oxygen, what is the immediate effect on the link reaction and the Krebs cycle?
A.Both will continue at an increased rate to compensate for the loss of ATP production.
B.The link reaction will stop, but the Krebs cycle will continue using ATP instead of \(NAD^+\).
C.Both will stop because \(NAD^+\) and \(FAD\) are not regenerated by the electron transport chain.
D.Only the link reaction will stop, while the Krebs cycle will run in reverse.
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Worked solution
Oxygen is the final electron acceptor in the electron transport chain. If it is blocked, electrons cannot flow, and NADH and FADH2 cannot be oxidized back to \(NAD^+\) and \(FAD\). Without these oxidized coenzymes, the link reaction and Krebs cycle cannot proceed, as they require \(NAD^+\) and \(FAD\) as cofactors.
Marking scheme
Award [1] for the correct answer (C). Reject A because without \(NAD^+\) these processes cannot run. Reject B because the Krebs cycle cannot use ATP in place of \(NAD^+\). Reject D because neither can run in reverse to produce ATP.
Question 36 · multiple-choice
1 marks
During glycolysis, glucose is phosphorylated twice. What is the primary biological significance of these phosphorylation events?
A.To decrease the potential energy of the sugar and make it more stable.
B.To prevent the glucose from diffusing out of the cell and to activate it for subsequent cleavage.
C.To generate ATP directly via oxidative phosphorylation.
D.To allow the glucose molecule to bind to the mitochondrial outer membrane.
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Worked solution
Phosphorylation adds negatively charged phosphate groups to glucose, preventing it from passing back through the hydrophobic plasma membrane, thereby trapping it in the cell. It also destabilizes (activates) the molecule, preparing it for cleavage into two three-carbon sugars.
Marking scheme
Award [1] for the correct answer (B). Reject A because phosphorylation increases potential energy and makes the molecule less stable. Reject C because these steps consume ATP rather than generating it, and glycolysis does not involve oxidative phosphorylation. Reject D because glycolysis occurs in the cytoplasm and does not require binding to the mitochondrial membrane.
Question 37 · multiple-choice
1 marks
A person is severely dehydrated. Which sequence of physiological events correctly describes the homeostatic response to restore water balance?
A.Decreased blood solute concentration -> detected by hypothalamic osmoreceptors -> decreased ADH release -> decreased water reabsorption in collecting ducts.
D.Decreased blood solute concentration -> detected by pituitary gland -> increased aldosterone release -> excretion of excess salt.
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Worked solution
Dehydration causes an increase in blood solute concentration (osmolarity). Osmoreceptors in the hypothalamus detect this change and stimulate the posterior pituitary to release antidiuretic hormone (ADH). ADH binds to receptors on the collecting duct cells of the kidney, causing the insertion of aquaporin channels, which increases water reabsorption back into the blood.
Marking scheme
Award [1] for the correct answer (B). Reject A as it describes the response to overhydration. Reject C because dehydration decreases blood volume, and baroreceptors detect pressure/volume, not solute concentration. Reject D because dehydration increases blood solute concentration, and aldosterone is not the primary hormone for direct water reabsorption.
Question 38 · multiple-choice
1 marks
Which of the following best explains the mechanism of phloem translocation from source to sink in plants?
A.Active transport of sucrose into the phloem at the source lowers the water potential, causing water to enter from the xylem and creating high hydrostatic pressure.
B.Passive diffusion of sucrose into the phloem at the source increases water potential, causing water to leave the phloem and creating low hydrostatic pressure.
C.Transpiration pull draws water and sucrose upward through the sieve tubes under negative tension.
D.Gravity pulls the dense sucrose solution downward through the sieve tubes without the need for water movement.
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Worked solution
Active transport of sucrose into the companion cells and sieve tube elements (phloem loading) at the source decreases water potential. Water then enters the phloem from the adjacent xylem by osmosis. This accumulation of water creates high hydrostatic pressure at the source, driving the mass flow of sap towards the sink (where pressure is lower).
Marking scheme
Award [1] for the correct answer (A). Reject B because active transport, not passive diffusion, is used, and it lowers (not increases) water potential. Reject C because it describes xylem transport (transpiration pull), not phloem translocation. Reject D because translocation is driven by hydrostatic pressure gradients, not gravity, and can occur in any direction.
Question 39 · multiple-choice
1 marks
A population of rabbits is introduced to an island with abundant resources and no predators. After an initial period of rapid growth, the growth rate slows down and the population stabilizes. Which of the following is a density-dependent factor most likely responsible for this stabilization?
A.A sudden volcanic eruption that covers half of the island in ash.
B.An unusually cold winter that causes widespread hypothermia.
C.Increased competition for nesting sites and food resources as population density rises.
D.A change in the island's daylight hours due to seasonal shifts.
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Worked solution
Density-dependent factors are those whose effects on the size or growth of the population vary with the population density. Competition for resources (food, nesting sites) increases as the population density increases, leading to a decreased birth rate or increased death rate, stabilizing the population at carrying capacity. A, B, and D are density-independent factors.
Marking scheme
Award [1] for the correct answer (C). Reject A, B, and D because they are abiotic, density-independent factors that affect the population regardless of its density.
Question 40 · multiple-choice
1 marks
During the follicular phase of the menstrual cycle, what is the primary role of LH (luteinizing hormone) and how does it interact with estrogen?
A.LH stimulates follicle development, which leads to a gradual decline in estrogen levels to trigger menstruation.
B.LH is inhibited by low levels of estrogen but stimulated by sustained high levels of estrogen, leading to a surge that triggers ovulation.
C.LH directly stimulates the corpus luteum to secrete large amounts of progesterone before ovulation occurs.
D.LH causes the breakdown of the endometrium, which stimulates a positive feedback loop increasing FSH.
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Worked solution
During the follicular phase, developing follicles secrete estrogen. Low levels of estrogen exert negative feedback on LH, but as a follicle matures, it secretes very high levels of estrogen. Sustained high estrogen levels switch to positive feedback, causing a rapid surge in LH secretion from the pituitary, which triggers ovulation.
Marking scheme
Award [1] for the correct answer (B). Reject A because follicle development increases estrogen levels and LH surge triggers ovulation, not menstruation. Reject C because the corpus luteum is only formed after ovulation. Reject D because progesterone and estrogen maintain the endometrium, and LH does not cause its breakdown.
Paper 2 Section A
Answer all questions. Calculators are permitted. Focuses on data evaluation and short-answer theory.
5 Question · 40 marks
Question 1 · Data-based & Short Answer
8 marks
An investigation was conducted into the cardiovascular and respiratory responses of two groups of young adults, an **Athletic Group** and a **Sedentary Group**, during graded treadmill exercise up to their maximum capacity. The table below shows the cardiac output (\(\text{L~min}^{-1}\)) measured at different running speeds (\(\text{km~h}^{-1}\)).
(a) State the relationship between running speed and cardiac output for both groups. [1] (b) Calculate the percentage difference in cardiac output between the Athletic and Sedentary groups at 12 \(\text{km~h}^{-1}\). Show your working. [2] (c) Explain how the nervous system and endocrine system coordinate to increase cardiac output during exercise. [3] (d) Outline why the athletic group can sustain higher running speeds without reaching fatigue compared to the sedentary group. [2]
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Worked solution
(a) For both groups, there is a positive relationship between running speed and cardiac output. The sedentary group plateaus at 14-16 \(\text{km~h}^{-1}\), while the athletic group continues to increase up to 28 \(\text{L~min}^{-1}\).
(c) Proprioceptors in muscles/joints detect movement, and chemoreceptors in the aorta/carotid arteries detect drops in pH due to rising \(\text{CO}_2\). Impulses are sent to the cardiovascular center in the medulla oblongata. The sympathetic nerve stimulates the sinoatrial node (SA node) via noradrenaline to increase heart rate. Concurrently, the endocrine system releases adrenaline from the adrenal glands into the blood stream, which also acts directly on the SA node to increase both rate and force of contraction (stroke volume).
(d) The athletic group has adapted to training with an increased stroke volume (larger left ventricle) and increased capillary density in skeletal muscles. This allows for a much higher rate of oxygen delivery to muscles, allowing them to remain in aerobic respiration at high speeds, delaying lactic acid production and subsequent fatigue.
Marking scheme
(a) [1 max] - Cardiac output increases as running speed increases (for both groups) OR positive correlation / relationship.
(b) [2 max] - Correct calculation of difference (5.0) and substitution into percentage formula: \(\frac{22 - 17}{17} \times 100\) OR \(\frac{22-17}{22} \times 100\) [1 mark] - Correct final value: 29.4% (or 29%) OR 22.7% (or 23%) [1 mark]
(c) [3 max] - Chemoreceptors detect low blood pH / high \(\text{CO}_2\) levels (or proprioceptors detect movement) [1 mark] - Medulla oblongata / cardiovascular center sends impulses down sympathetic nerve to release noradrenaline at the SA node [1 mark] - Adrenal glands secrete adrenaline into blood, which increases heart rate / stroke volume [1 mark]
(d) [2 max] - Athletes have larger stroke volume / higher maximum cardiac output / more capillaries [1 mark] - Allows for sustained aerobic respiration / less reliance on anaerobic respiration / less lactate accumulation at high speeds [1 mark]
Question 2 · Data-based & Short Answer
8 marks
The rate of cell respiration in yeast (*Saccharomyces cerevisiae*) was investigated using three different sugars (glucose, maltose, and lactose) at various temperatures. The rate of respiration was measured by the volume of carbon dioxide released (\(\text{cm}^3\text{~min}^{-1}\)).
(a) Identify the sugar that yields the lowest rate of respiration and suggest a biological reason for this. [2] (b) Describe the effect of temperature on the rate of respiration using glucose. [2] (c) Distinguish between aerobic and anaerobic respiration in yeast in terms of products and energy yield. [2] (d) Explain the sudden decrease in carbon dioxide production at \(50^{\circ}\text{C}\). [2]
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Worked solution
(a) Lactose is the sugar with the lowest respiration rate. This is because yeast (*S. cerevisiae*) does not possess the gene for the enzyme lactase (beta-galactosidase) and therefore cannot break down lactose (a disaccharide) into monosaccharides (glucose and galactose) for glycolysis.
(b) For glucose, the respiration rate increases with temperature from 20 °C (1.2 \(\text{cm}^3\text{~min}^{-1}\)) to an optimum at 40 °C (4.5 \(\text{cm}^3\text{~min}^{-1}\)). Above 40 °C, the rate drops rapidly, reaching 0.8 \(\text{cm}^3\text{~min}^{-1}\) at 50 °C.
(c) Aerobic respiration in yeast yields carbon dioxide and water as products, with a very high ATP yield (~32 ATP molecules per glucose molecule). Anaerobic respiration (ethanol fermentation) yields carbon dioxide and ethanol, with a very low ATP yield (only 2 ATP molecules per glucose molecule from glycolysis).
(d) At 50 °C, the temperature is high enough to disrupt hydrogen bonds, ionic bonds, and hydrophobic interactions within the tertiary structure of the yeast's respiratory enzymes. This causes the enzymes to denature, changing the conformation of their active sites so that the substrates can no longer bind, thus halting the metabolic pathways of respiration.
Marking scheme
(a) [2 max] - Lactose [1 mark] - Yeast lacks the lactase / beta-galactosidase enzyme to hydrolyze/break down lactose into fermentable monosaccharides [1 mark]
(b) [2 max] - Respiration rate increases as temperature rises from 20 to 40 °C [1 mark] - Drops rapidly / significantly at 50 °C / peak/optimum is at 40 °C [1 mark]
(c) [2 max] - Aerobic produces carbon dioxide and water, while anaerobic produces carbon dioxide and ethanol [1 mark] - Aerobic has a high ATP yield (~30-38 ATP per glucose) whereas anaerobic has a low ATP yield (2 ATP per glucose) [1 mark]
(d) [2 max] - High temperature (50 °C) causes denaturation of enzymes / proteins involved in cell respiration [1 mark] - Destroys active sites / tertiary structure due to broken hydrogen bonds, preventing substrate binding / enzyme-substrate complexes from forming [1 mark]
Question 3 · Data-based & Short Answer
8 marks
The physiological responses of human volunteers exposed to rising ambient temperatures in a climate-controlled room were monitored. Core body temperature and sweating rate were recorded.
(a) Describe the trend in sweating rate as ambient temperature increases from \(20^{\circ}\text{C}\) to \(45^{\circ}\text{C}\). [2] (b) Explain the role of the hypothalamus in maintaining core body temperature when environmental temperature rises. [3] (c) Outline the physiological mechanism of vasodilation and how it aids in heat loss. [3]
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Worked solution
(a) Sweating rate increases with increasing ambient temperature. The increase is non-linear/exponential; it increases slowly from 20 °C to 25 °C (from 50 to 60 \(\text{g~m}^{-2}\text{h}^{-1}\)), but increases very rapidly above 25 °C up to 45 °C (reaching 600 \(\text{g~m}^{-2}\text{h}^{-1}\)).
(b) The hypothalamus contains thermoreceptors that monitor blood temperature and receives input from peripheral thermoreceptors in the skin. When the temperature rises, the hypothalamus coordinates a negative feedback response. It sends nerve impulses to the sweat glands to increase secretion (sweat evaporates, drawing latent heat from the skin) and to skin arterioles to dilate, allowing more heat to be lost. It also suppresses shivering and metabolic heat production.
(c) Vasodilation involves the relaxation of the circular smooth muscle in the walls of arterioles leading to the skin capillaries. This decreases resistance, allowing more blood to flow through capillaries close to the skin surface. As a result, heat from the blood is lost to the surroundings through radiation, conduction, and convection, cooling the body.
Marking scheme
(a) [2 max] - Sweating rate increases as ambient temperature increases [1 mark] - Rise is non-linear / increases slowly below 25 °C and rapidly / exponentially above 25 °C (must refer to data values to support) [1 mark]
(b) [3 max] - Hypothalamus acts as a thermostat / monitors blood temperature / receives nerve signals from skin thermoreceptors [1 mark] - Coordinates negative feedback mechanisms to bring core temperature back to set point [1 mark] - Sends nerve impulses to effectors: sweat glands (stimulate sweating) and arterioles (stimulate vasodilation) [1 mark]
(c) [3 max] - Smooth muscle in skin arterioles relaxes [1 mark] - Causes lumen of arterioles to widen / dilate, increasing blood flow to skin capillaries [1 mark] - Heat is lost from blood to the air via radiation / conduction / convection [1 mark]
Question 4 · Data-based & Short Answer
8 marks
Transpiration rates of bean plants (*Phaseolus vulgaris*) were measured at various wind speeds (\(\text{m~s}^{-1}\)) under two different relative humidity levels: Low Humidity (40%) and High Humidity (80%). The table shows the transpiration rate (\(\text{g~dm}^{-2}\text{h}^{-1}\)).
(a) Compare the effect of wind speed on transpiration rate in high and low humidity conditions. [3] (b) Explain how stomatal closure acts as a regulatory mechanism during high wind speeds or dry conditions. [2] (c) Outline the structure and function of xylem vessels that adapt them for the transport of water under tension. [3]
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Worked solution
(a) Similarities: For both humidity levels, transpiration rate increases with wind speed. For both, the rate of increase is greater between 0 and 5 \(\text{m~s}^{-1}\) than between 5 and 10 \(\text{m~s}^{-1}\) (where it begins to plateau). Differences: The transpiration rate is consistently higher under low humidity (40%) than under high humidity (80%) at every wind speed. For instance, at 10 \(\text{m~s}^{-1}\), the transpiration rate under low humidity is 5.0 \(\text{g~dm}^{-2}\text{h}^{-1}\), while under high humidity it is only 2.0 \(\text{g~dm}^{-2}\text{h}^{-1}\).
(b) Stomatal closure is triggered under dry or extremely windy conditions, often mediated by the plant hormone abscisic acid (ABA). Guard cells pump out potassium ions, causing water to leave the guard cells by osmosis. The guard cells become flaccid, which closes the stomatal pore, reducing transpiration and preventing dehydration.
(c) Xylem vessels are adapted in several ways: 1. Dead cells: There is no cytoplasm or plasma membrane, leaving a hollow lumen that offers minimal resistance to the mass flow of water. 2. Lignified walls: The deposition of lignin provides immense tensile strength, preventing the vessels from collapsing inward under the high tension (negative pressure) generated by transpiration pull. 3. Cohesion/adhesion support: The narrow lumen increases the capillary effect and surface area for adhesion of water molecules to the hydrophilic cell walls, helping maintain a continuous water column.
Marking scheme
(a) [3 max] - Both show an increase in transpiration rate with wind speed [1 mark] - Both show a plateau/slowing of the rate of increase at higher wind speeds (5 to 10 \(\text{m~s}^{-1}\)) [1 mark] - Low humidity (40%) consistently has a higher transpiration rate than high humidity (80%) across all tested wind speeds [1 mark]
(b) [2 max] - High wind/drought causes guard cells to lose water / become flaccid (mediated by hormone ABA) [1 mark] - Flaccid guard cells close the stomatal pore, preventing further diffusion of water vapor out of the leaf / conserving water [1 mark]
(c) [3 max] - Dead/hollow cells without cytoplasm provide an unobstructed pathway for water flow [1 mark] - Lignin in cell walls provides strength to withstand tension/negative pressure [1 mark] - Pits/pores in walls allow lateral movement of water between vessels [1 mark]
Question 5 · Data-based & Short Answer
8 marks
An experiment studied the growth of two protozoan species, *Paramecium aurelia* and *Paramecium caudatum*, when grown alone (monoculture) and together (mixed culture) in a liquid medium. The table shows population density (cells/mL) over 16 days.
(a) State the ecological principle demonstrated by the survival of *P. aurelia* and the decline of *P. caudatum* in the mixed culture. [1] (b) Using the data, compare the growth of *P. caudatum* in monoculture and in mixed culture. [3] (c) Distinguish between density-dependent and density-independent factors that limit population growth, providing an example of each. [4]
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Worked solution
(a) The ecological principle demonstrated is the **Competitive Exclusion Principle** (also known as Gause's Principle), which states that two species competing for the exact same limiting resource cannot coexist indefinitely if their ecological niches overlap completely.
(b) Compare growth of *P. caudatum* in monoculture vs. mixed culture: - Both show growth during the initial phase (Days 0 to 8). - In monoculture, *P. caudatum* continuous to grow and reaches a stable plateau (carrying capacity) of approximately 110 cells/mL by Day 16. - In mixed culture, the population of *P. caudatum* peaks early at Day 8 (at only 45 cells/mL, which is lower than the 80 cells/mL in monoculture), and then rapidly declines to extinction (0 cells/mL) by Day 16.
(c) Distinguishing between limiting factors: - **Density-dependent factors**: Factors whose effects on the size or growth of the population vary with the population density. As density increases, the mortality rate increases or birth rate decreases. Examples: Competition for resources (food, space), disease/parasitism, predation, accumulation of toxic waste. - **Density-independent factors**: Factors that affect population size regardless of the density of the population. Their impact is the same whether the population is large or small. Examples: Natural disasters (floods, fires), weather/climatic events (sudden freeze, drought), volcanic eruptions.
(b) [3 max] - Both show initial growth up to day 8 [1 mark] - In monoculture, *P. caudatum* reaches a carrying capacity/plateau of ~110 cells/mL, whereas in mixed culture it peaks at 45 cells/mL and then declines [1 mark] - In mixed culture, *P. caudatum* goes extinct/reaches 0 cells/mL by day 16, whereas it survives indefinitely in monoculture [1 mark]
(c) [4 max] - Density-dependent factors: effect increases as population density increases (or vice-versa) [1 mark] - Example of density-dependent: competition for food / nesting sites / spread of infectious disease / predation [1 mark] - Density-independent factors: effect is constant regardless of population density [1 mark] - Example of density-independent: forest fire / flood / extreme cold weather / volcanic eruption [1 mark]
Paper 2 Section B
Answer two out of three extended-response questions. One additional mark is available for the quality of construction of each response.
2 Question · 32 marks
Question 1 · extended response
16 marks
a. Explain the pathway and products of anaerobic cell respiration in both yeast and humans, including practical applications of these pathways. [4]
b. Describe the processes that occur in the link reaction and the Krebs cycle during aerobic respiration in eukaryotic cells. [5]
c. Explain how the structure of the mitochondrion is adapted to its function of generating ATP via chemiosmosis. [6]
(Plus 1 mark for quality of communication)
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Worked solution
### Part a: Anaerobic Cell Respiration * Anaerobic respiration occurs in the cytoplasm of cells in the absence of oxygen. * In yeast, glucose is converted to pyruvate through glycolysis, which is then decarboxylated and reduced to form ethanol and carbon dioxide. * This process in yeast is utilized in baking (where CO2 causes dough to rise) and brewing (for ethanol production in alcoholic beverages). * In humans, glucose is converted to pyruvate, which is then reduced to lactate (lactic acid) with no production of carbon dioxide. * This pathway allows human muscle cells to continue producing ATP at a rapid rate during intense exercise when oxygen supply is insufficient. * Both pathways yield only 2 ATP molecules per molecule of glucose (derived from glycolysis).
### Part b: Link Reaction and Krebs Cycle * **Link Reaction**: * Pyruvate enters the mitochondrial matrix from the cytoplasm via active transport. * Pyruvate is decarboxylated (releasing CO2) and oxidized (reducing NAD+ to NADH + H+). * The remaining 2-carbon acetyl group combines with Coenzyme A to form acetyl-CoA. * **Krebs Cycle**: * Acetyl-CoA transfers the acetyl group to a 4-carbon compound (oxaloacetate) to form a 6-carbon compound (citrate). * Citrate undergoes a series of decarboxylation reactions, releasing two molecules of CO2. * Citrate also undergoes oxidation reactions, transferring hydrogen/electrons to reduce three molecules of NAD+ to NADH and one molecule of FAD to FADH2. * One molecule of ATP is produced per cycle via substrate-level phosphorylation. * The 4-carbon compound is regenerated at the end of the cycle to react with another acetyl-CoA.
### Part c: Chemiosmosis and Mitochondrial Adaptations * NADH and FADH2 donate high-energy electrons to the electron transport chain (ETC) located on the inner mitochondrial membrane. * As electrons pass through the carriers of the ETC, energy is released and used to pump protons (H+ ions) from the matrix into the intermembrane space. * This active transport establishes a steep proton gradient (electrochemical gradient) across the inner membrane. * Protons flow down their electrochemical gradient back into the matrix through the channel enzyme ATP synthase. * The kinetic energy of this proton flow drives the phosphorylation of ADP and inorganic phosphate to form ATP. * **Adaptations of Mitochondrion Structure**: * **Cristae** (highly folded inner membrane): Dramatically increases the surface area available for the electron transport chain proteins and ATP synthase complexes. * **Intermembrane space**: Has a very small/narrow volume, allowing a rapid build-up of a high concentration of protons to establish the gradient quickly. * **Matrix**: Semi-fluid environment containing appropriate enzymes and a suitable pH for the link reaction and Krebs cycle. * **Inner membrane**: Impermeable to protons except through ATP synthase, preventing the proton gradient from leaking away.
Marking scheme
### Part a (Max [4 marks]) * Anaerobic respiration occurs in cytoplasm / without oxygen; [1] * In yeast, glucose is converted to pyruvate, then to ethanol and carbon dioxide; [1] * Yeast fermentation is used in baking (CO2 for rising) and brewing (ethanol production); [1] * In humans, glucose is converted to pyruvate, then to lactate; [1] * Human anaerobic pathway provides rapid ATP during short, intense exercise when oxygen is scarce; [1] * Low ATP yield compared to aerobic respiration (2 ATP vs 30+ ATP); [1]
### Part b (Max [5 marks]) * Pyruvate enters the mitochondrial matrix from cytoplasm; [1] * Link reaction: Pyruvate is decarboxylated (loses CO2) and oxidized (NAD+ to NADH), then bound to Coenzyme A to form acetyl-CoA; [1] * Krebs cycle: Acetyl-CoA (2C) combines with oxaloacetate (4C) to form citrate (6C); [1] * Citrate is decarboxylated to release two CO2 molecules; [1] * Citrate is oxidized, reducing 3 NAD+ to 3 NADH and 1 FAD to 1 FADH2 per turn; [1] * Substrate-level phosphorylation produces 1 ATP per turn; [1] * Oxaloacetate is regenerated at the end of the cycle; [1]
### Part c (Max [6 marks]) * NADH/FADH2 donate electrons to the electron transport chain (ETC) on the inner membrane/cristae; [1] * Electron transfer releases energy used to pump protons (H+ ions) into the intermembrane space; [1] * Creates a high concentration of protons / electrochemical gradient; [1] * Protons diffuse back into the matrix through ATP synthase; [1] * ATP synthase uses proton motive force / kinetic energy of proton flow to synthesize ATP from ADP + Pi; [1] * Cristae folding increases surface area for ETC and ATP synthase; [1] * Small volume of intermembrane space allows rapid proton buildup; [1] * Inner membrane is impermeable to protons except through ATP synthase; [1]
### Quality of Communication [1 mark] * 1 mark awarded if the essay is structured logically, presents clear transitions between parts, and uses accurate biology nomenclature throughout.
Question 2 · extended response
16 marks
a. Outline the concept of homeostasis and the role of negative feedback mechanisms. [3]
b. Explain the homeostatic control of blood glucose concentration in humans, including the roles of the pancreas and liver. [6]
c. Explain how the human body responds to a decrease in core body temperature. [6]
(Plus 1 mark for quality of communication)
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Worked solution
### Part a: Homeostasis and Negative Feedback * **Homeostasis** is the maintenance of a constant internal environment (e.g., pH, temperature, water potential) within narrow limits, despite changes in the external environment. * **Negative feedback** is a control mechanism where a change in a physiological variable triggers a response that counteracts or reverses the direction of that change. * This system requires: * A **sensor/receptor** to detect the deviation from the set point. * A **control center** (often the brain/hypothalamus or endocrine glands) to process the signal. * An **effector** (muscles or glands) to carry out the corrective action and return the system to its optimal set point.
### Part b: Control of Blood Glucose * Blood glucose concentration is monitored by specialized endocrine cells in the islets of Langerhans within the **pancreas**. * **When blood glucose levels rise (hyperglycemia, e.g., after eating)**: * **Beta (\(\beta\)) cells** in the pancreas detect the increase and secrete the hormone **insulin** into the bloodstream. * Insulin binds to specific receptors on target cells (mostly liver, muscle, and adipose cells). * This stimulates increased uptake of glucose from the blood into cells. * In hepatocytes (liver cells) and muscle cells, insulin activates enzymes that convert glucose into glycogen (**glycogenesis**) for storage. * This decreases blood glucose concentration back to the normal set point. * **When blood glucose levels fall (hypoglycemia, e.g., during exercise or fasting)**: * **Alpha (\(\alpha\)) cells** in the pancreas detect the decrease and secrete **glucagon** into the bloodstream. * Glucagon binds to receptors on liver cells (hepatocytes). * This stimulates the breakdown of stored glycogen into glucose (**glycogenolysis**). * It can also promote **gluconeogenesis** (the synthesis of glucose from non-carbohydrate sources like amino acids or glycerol). * Glucose is released into the blood, raising blood glucose concentration back to the normal set point.
### Part c: Thermoregulation in Cold Environments * A decrease in core body temperature is detected by **thermoreceptors** in the skin and the **hypothalamus** in the brain. * The hypothalamus acts as the control center, coordinating several physiological and behavioral responses to reduce heat loss and increase heat production: * **Vasoconstriction**: * Arterioles supplying the skin capillaries constrict. * This shunts blood away from the skin surface to deeper tissues, minimizing heat loss via radiation and conduction. * **Shivering**: * Involuntary, rapid contraction and relaxation of skeletal muscles. * This muscle activity increases the rate of cellular respiration, generating metabolic heat. * **Piloerection (Goosebumps)**: * Arrector pili muscles at the base of hair follicles contract, causing hairs to stand on end. * This traps a layer of warm insulating air close to the skin. * **Metabolic Rate Increase**: * The hypothalamus stimulates the release of hormones (such as adrenaline and thyroxin) which increase the basal metabolic rate to produce more heat. * **Behavioral Adaptations**: * The brain promotes conscious actions such as putting on warmer clothes, huddling, or moving closer to heat sources to reduce heat loss.
Marking scheme
### Part a (Max [3 marks]) * Homeostasis: maintenance of a constant internal environment within narrow limits; [1] * Negative feedback reverses/counteracts a change from the set point; [1] * Requires a detector/receptor, a control center, and an effector; [1] * Examples of homeostatic variables include blood pH, blood glucose, temperature, and water potential; [1]
### Part b (Max [6 marks]) * Blood glucose is monitored by the pancreas / Islets of Langerhans; [1] * High blood glucose stimulates beta (\(\beta\)) cells to secrete insulin; [1] * Insulin stimulates body/muscle/liver cells to take up glucose from blood; [1] * Insulin promotes glycogenesis (conversion of glucose to glycogen) in the liver/muscles; [1] * Low blood glucose stimulates alpha (\(\alpha\)) cells to secrete glucagon; [1] * Glucagon stimulates glycogenolysis (breakdown of glycogen to glucose) in the liver; [1] * Glucagon can stimulate gluconeogenesis (synthesis of glucose from non-carbohydrates); [1] * Negative feedback returns glucose levels to the normal range; [1]
### Part c (Max [6 marks]) * Lower temperature detected by thermoreceptors in skin and hypothalamus; [1] * Hypothalamus coordinates corrective actions as the control center; [1] * Vasoconstriction of skin-bound arterioles reduces blood flow to the surface, reducing heat loss; [1] * Shivering occurs due to involuntary muscle contractions, generating metabolic heat; [1] * Piloerection (hairs standing on end) traps a layer of insulating air; [1] * Metabolic rate increases via hormone secretion (e.g., thyroxin/adrenaline) to produce metabolic heat; [1] * Behavioral responses (e.g., huddling, adding clothes) reduce exposure to cold; [1]
### Quality of Communication [1 mark] * 1 mark awarded if the essay is structured logically, presents clear transitions between parts, and uses accurate biology nomenclature throughout.
Paper 3 Section A
Answer all compulsory questions based on experimental work and skills.
3 Question · 15 marks
Question 1 · Data-based Experimental
5 marks
Students investigated the rate of respiration in germinating mung beans (Vigna radiata) using a respirometer at three temperatures. Potassium hydroxide (KOH) was placed in the chamber to absorb carbon dioxide. The displacement of a colored fluid index in a capillary tube of 1.0 mm internal diameter was measured over 15 minutes.
Data:
At 15°C: 12.0 mm
At 25°C: 28.5 mm
At 35°C: 45.0 mm
Questions:
(a) Calculate the volume of oxygen consumed per minute by the seeds at 25°C. Show your working and state the units. (Use \(\pi \approx 3.1416\); volume of cylinder = \(\pi r^2 h\)). [2]
(b) Explain the role of potassium hydroxide (KOH) in this experimental design. [1]
(c) Predict and explain the displacement of the fluid if germinating seeds were replaced with boiled, dead seeds of the same mass. [2]
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Worked solution
(a) Step 1: Find the radius of the capillary tube: \(r = 1.0\text{ mm} / 2 = 0.5\text{ mm}\). Step 2: Calculate total volume consumed: \(V = \pi r^2 h = 3.1416 \times (0.5)^2 \times 28.5 = 22.384\text{ mm}^3\). Step 3: Calculate rate per minute: \(22.384 / 15\text{ minutes} = 1.49\text{ mm}^3\text{ min}^{-1}\) (or \(1.5\text{ mm}^3\text{ min}^{-1}\)). (b) Respiration consumes oxygen and produces carbon dioxide. Absorbing the carbon dioxide with KOH ensures the net gas volume decreases as oxygen is consumed, drawing the fluid along the tube. (c) Boiling denatures the enzymes involved in cellular respiration. Therefore, dead seeds cannot carry out cellular respiration, no oxygen is consumed, and the fluid index remains stationary.
Marking scheme
(a) [2 marks max] • \(1.49\text{ mm}^3\text{ min}^{-1}\) or \(1.5\text{ mm}^3\text{ min}^{-1}\) (correct value with unit) [1] • Correct working showing radius as 0.5 mm and division by 15 [1] Note: Accept correct calculation if diameter was used as radius by mistake (giving 5.97 mm3 min-1) for 1 mark max. (b) [1 mark max] • Absorbs carbon dioxide produced by respiration so that pressure changes reflect only oxygen uptake [1] (c) [2 marks max] • Fluid would not move / remains at 0 mm [1] • Boiled seeds are dead / enzymes are denatured so cellular respiration does not occur [1]
Question 2 · Data-based Experimental
5 marks
A potometer was used to investigate the effect of wind speed on the rate of transpiration in a leafy shoot of Geranium. A hair dryer simulated three wind speeds: Low (2.0 m/s), Medium (4.5 m/s), and High (7.0 m/s). The distance moved by the air bubble (in mm) was recorded every 2 minutes for 10 minutes.
Data:
At Low (2.0 m/s): 4, 8, 13, 17, 21 mm
At Medium (4.5 m/s): 9, 18, 27, 36, 44 mm
At High (7.0 m/s): 11, 22, 31, 39, 45 mm
Questions:
(a) Calculate the rate of transpiration at Medium wind speed over the 10-minute period. State the units. [1]
(b) Describe the effect of wind speed on the rate of transpiration between 2.0 m/s and 7.0 m/s, using the provided data. [2]
(c) Explain the rate of transpiration at High wind speed (7.0 m/s) between 6 and 10 minutes. [2]
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Worked solution
(a) The total distance moved at Medium wind speed at 10 minutes is 44 mm. The rate is calculated as: \(\text{Rate} = 44\text{ mm} / 10\text{ min} = 4.4\text{ mm min}^{-1}\). (b) Increasing wind speed from 2.0 m/s to 4.5 m/s significantly increases the rate of transpiration (from 2.1 mm min\(^{-1}\) to 4.4 mm min\(^{-1}\)). However, from 4.5 m/s to 7.0 m/s, the rate remains almost constant / plateaus (44 mm versus 45 mm total displacement). (c) At very high wind speeds (7.0 m/s), extreme transpiration causes the leaves to lose water faster than the roots can absorb it, causing water stress. This triggers stomatal closure (mediated by plant hormones like abscisic acid) to conserve water, causing the transpiration rate to flatten out in the final minutes of the trial (only 6 mm moved between 8 and 10 minutes compared to 11 mm between 0 and 2 minutes).
Marking scheme
(a) [1 mark max] • \(4.4\text{ mm min}^{-1}\) (must include correct unit) [1] (b) [2 marks max] • As wind speed increases, transpiration rate increases [1] • Rate plateaus/shows minimal increase between 4.5 m/s and 7.0 m/s (supported by data, e.g., 44 mm vs 45 mm at 10 mins) [1] (c) [2 marks max] • Rapid water loss causes guard cells to lose turgor / causes water stress [1] • Leads to stomatal closure / reduces transpiration rate to conserve water [1]
Question 3 · Data-based Experimental
5 marks
An experiment was conducted to study blood homeostasis during strenuous exercise. The blood pH and mean ventilation rate of healthy volunteers were measured at rest, after 5 and 10 minutes of intense running, and after 5 minutes of recovery.
(a) State the relationship between blood pH and ventilation rate during exercise. [1]
(b) Explain the physiological mechanism that links the decrease in blood pH to the increase in ventilation rate. [3]
(c) Identify one variable that must be controlled in the volunteers to ensure the validity of this study. [1]
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Worked solution
(a) There is a negative correlation / inverse relationship between blood pH and ventilation rate during exercise (as pH falls, ventilation rate rises). (b) During intense exercise, increased cell respiration in muscles produces more carbon dioxide (\(\text{CO}_2\)). \(\text{CO}_2\) reacts with water in blood plasma to form carbonic acid (\(\text{H}_2\text{CO}_3\)), which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogen carbonate ions (\(\text{HCO}_3^-\)), lowering pH. Chemoreceptors in the medulla oblongata and major blood vessels (aorta and carotid arteries) detect this decrease in pH. They send nervous signals to the respiratory control center in the medulla oblongata, which increases the rate and depth of ventilation by sending motor impulses to the diaphragm and external intercostal muscles. (c) Variables to control include: age of the volunteers, baseline physical fitness/athletic training level, ambient room temperature, relative humidity, or type of exercise performed (e.g., constant treadmill speed).
Marking scheme
(a) [1 mark max] • Inverse/negative relationship (as pH decreases, ventilation rate increases) [1] (b) [3 marks max] • Increased cellular respiration produces \(\text{CO}_2\), lowering blood pH / increasing \(\text{H}^+\) concentration [1] • Chemoreceptors in carotid arteries/aorta/medulla detect the drop in pH [1] • Signals sent to ventilation center in medulla oblongata, which stimulates intercostal muscles/diaphragm to contract faster [1] (c) [1 mark max] • Accept any one valid control variable: age of subjects, physical fitness level, environmental temperature/humidity, diet/hydration status before testing, or exercise intensity [1]
Paper 3 Section B
Answer all questions from one selected Option (e.g., Option D - Human Physiology).
6 Question · 30 marks
Question 1 · Short Response
5 marks
During sudden physical exertion, the human body must rapidly adjust its cardiac output to meet increased metabolic demands. Explain how the nervous system and the endocrine system coordinate to increase the heart rate during exercise.
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Worked solution
First, increased cellular respiration during exercise releases more carbon dioxide, which lowers blood pH. Chemoreceptors detect this change and signal the cardiovascular center in the medulla oblongata. The medulla sends nerve impulses down the sympathetic nerve to the sinoatrial (SA) node, releasing noradrenaline to increase the rate of contraction. Concurrently, the sympathetic nervous system stimulates the adrenal glands to secrete adrenaline into the bloodstream. Adrenaline travels to the heart, binding to receptors on the SA node to sustain and further increase the heart rate.
Marking scheme
Award [1] for each of the following, up to [5 max]: [1] Increased blood CO2 levels or decreased blood pH is detected by chemoreceptors. [2] Impulses are sent to the cardiovascular center in the medulla oblongata. [3] Sympathetic nerves send impulses to the sinoatrial (SA) node. [4] Noradrenaline is released at the SA node to increase heart rate. [5] Adrenal glands (adrenal medulla) are stimulated to secrete adrenaline (epinephrine). [6] Adrenaline travels in the blood to the SA node to accelerate the heart rate. [7] Parasympathetic (vagus nerve) activity is reduced.
Question 2 · Short Response
5 marks
Describe how the nervous system and endocrine system cooperate to regulate gastric juice secretion during the cephalic and gastric phases of digestion.
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Worked solution
During the cephalic phase, the sight, smell, or taste of food activates sensory receptors, sending signals to the brain. The vagus nerve (parasympathetic nervous system) carries impulses to the stomach, stimulating gastric glands to secrete hydrochloric acid and pepsinogen, and stimulating G-cells to secrete the hormone gastrin. During the gastric phase, food entry distends the stomach, activating mechanoreceptors (stretch receptors). These trigger local enteric nervous reflexes to sustain gastric juice secretion. Concurrently, peptides in the food stimulate further gastrin secretion, which travels in the blood to target parietal and chief cells, promoting maximum gastric secretion.
Marking scheme
Award [1] for each of the following, up to [5 max]: [1] Cephalic phase is initiated by sensory stimuli (sight/smell/taste of food). [2] Vagus nerve (parasympathetic) sends impulses from the brain to the stomach. [3] Gastric glands are stimulated to secrete acid/pepsinogen and G-cells to secrete gastrin. [4] Gastric phase is initiated by food entering the stomach, causing distension. [5] Stretch receptors (mechanoreceptors) activate local neural reflexes. [6] Gastrin (hormone) is released into the bloodstream to stimulate parietal/chief cells to increase gastric secretions. [7] High pH or presence of peptides also stimulates gastrin secretion.
Question 3 · Short Response
5 marks
The inner mitochondrial membrane is highly folded into cristae. Explain how the structure of this membrane and the generation of a proton gradient are essential for ATP synthesis during aerobic cell respiration.
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Worked solution
The folding of the inner mitochondrial membrane into cristae increases the surface area available for the electron transport chain (ETC) and ATP synthase proteins. High-energy electrons carried by NADH and FADH2 are passed along the ETC, releasing energy. This energy is utilized by proton pumps to actively transport protons (\(H^+\)) from the matrix across the inner membrane. Because the inner membrane is impermeable to protons, they accumulate in the extremely narrow intermembrane space, quickly building a steep electrochemical gradient (proton motive force). Finally, protons flow down this gradient back into the matrix through ATP synthase, driving the mechanical rotation of the enzyme which phosphorylates ADP to ATP.
Marking scheme
Award [1] for each of the following, up to [5 max]: [1] Cristae increase the surface area of the inner mitochondrial membrane for ETC and ATP synthase. [2] Electron transport chain carriers accept electrons from reduced coenzymes (NADH/FADH2). [3] Energy from electron transport is used to pump protons (\(H^+\)) from the matrix into the intermembrane space. [4] Inner membrane is impermeable to protons, preventing direct diffusion back to the matrix. [5] Narrow intermembrane space allows rapid accumulation of protons, creating a steep electrochemical/proton gradient. [6] Protons flow back into the matrix through ATP synthase (chemiosmosis). [7] Flow of protons provides energy for ATP synthase to phosphorylate ADP to ATP.
Question 4 · Short Response
5 marks
Dehydration triggers a homeostatic pathway to conserve water. Explain the roles of the hypothalamus, posterior pituitary gland, and ADH (antidiuretic hormone) in regulating blood solute concentration during dehydration.
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Worked solution
Dehydration increases the solute concentration (osmolarity) of the blood. Osmoreceptors in the hypothalamus detect this change. In response, the hypothalamus produces antidiuretic hormone (ADH), which is transported along axons to the posterior pituitary gland where it is stored. The posterior pituitary releases ADH into the bloodstream. ADH travels to the kidneys, where it binds to receptors on the cells of the collecting ducts. This triggers the insertion of water channels called aquaporins into the luminal membranes. As a result, water is reabsorbed by osmosis into the hypertonic medulla and then the blood, lowering blood osmolarity back to normal levels.
Marking scheme
Award [1] for each of the following, up to [5 max]: [1] Osmoreceptors in the hypothalamus detect high blood solute concentration / high blood osmolarity. [2] Hypothalamus synthesizes ADH (antidiuretic hormone). [3] ADH is transported to and released by the posterior pituitary gland. [4] ADH travels via the bloodstream to the kidneys. [5] ADH binds to receptors on the collecting duct (and/or distal convoluted tubule) cells. [6] ADH stimulates the insertion of aquaporins (water channels) into the membranes. [7] Water permeability increases, causing more water to be reabsorbed into the blood, producing concentrated urine.
Question 5 · Short Response
5 marks
Translocation in the phloem is an active process that moves organic solutes throughout a plant. Explain how a hydrostatic pressure gradient is generated and maintained to drive phloem transport from source to sink.
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Worked solution
At the source, organic solutes such as sucrose are actively loaded into the sieve tubes of the phloem using energy (often via proton-sucrose symport). This high concentration of solutes lowers the water potential within the sieve tube. Consequently, water moves from the adjacent xylem into the phloem by osmosis. Because the cell walls of sieve tubes are rigid, this influx of water causes a rapid increase in turgor or hydrostatic pressure at the source. At the sink, sucrose is actively or passively unloaded from the sieve tube, which increases the water potential inside the phloem. Water then exits the phloem by osmosis, decreasing the hydrostatic pressure. This difference in hydrostatic pressure between the source and the sink drives the mass flow of phloem sap along the hydrostatic gradient.
Marking scheme
Award [1] for each of the following, up to [5 max]: [1] Active loading of sucrose/solutes into phloem sieve tubes at the source. [2] High solute concentration lowers water potential in the sieve tube. [3] Water enters the phloem from the xylem by osmosis. [4] Influx of water creates high hydrostatic/turgor pressure at the source. [5] Unloading of sucrose at the sink increases water potential. [6] Water leaves the phloem at the sink, reducing hydrostatic pressure. [7] The resulting hydrostatic pressure gradient drives mass flow from source to sink.
Question 6 · Short Response
5 marks
According to the competitive exclusion principle, two species cannot occupy the exact same ecological niche indefinitely. Explain how niche partitioning can allow two competing species to coexist within the same community.
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Worked solution
When two species compete for the same limiting resources, one will eventually outcompete the other, leading to extinction or exclusion. However, natural selection favors individuals that minimize competition. This leads to niche partitioning, where species divide resources. In spatial partitioning, species use different physical zones of the same ecosystem. In temporal partitioning, they feed or reproduce at different times of the day or year. In morphological partitioning, natural selection drives character displacement, leading to physical adaptations (such as different beak sizes) that allow them to exploit different resources, thereby narrowing their realized niches and enabling stable coexistence.
Marking scheme
Award [1] for each of the following, up to [5 max]: [1] Complete niche overlap leads to competitive exclusion where one species is eliminated. [2] Natural selection favors individuals that avoid direct competition. [3] Niche partitioning is the division of resources to reduce competition. [4] Spatial partitioning: utilizing different physical areas or microhabitats (e.g., canopy vs forest floor). [5] Temporal partitioning: being active or feeding at different times of day/year. [6] Morphological partitioning / character displacement: evolution of different traits to exploit different resources. [7] Allows species to occupy distinct realized niches within the same habitat.
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