An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 HL (TZ2) IB Diploma Programme Biology paper. Not affiliated with or reproduced from IB.
Paper 1
Answer all 40 multiple-choice questions. No calculator is permitted.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
During a period of intense physical exercise, the autonomic nervous system coordinates several physiological responses to meet the metabolic demands of skeletal muscle. Which of the following correctly describes the response of the cardiovascular and respiratory systems to sympathetic stimulation during exercise?
A.Vasodilation of arterioles feeding the kidneys and digestive tract to maximize core metabolic rate.
B.Increased stroke volume of the heart and bronchodilation in the lungs.
C.Decreased ventilation rate and vasoconstriction of coronary arteries.
D.Activation of the vagus nerve leading to a positive inotropic effect on the myocardium.
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Worked solution
During exercise, sympathetic stimulation is activated. This leads to: 1. An increase in cardiac output by increasing both heart rate and stroke volume (force of contraction). 2. Bronchodilation of the airways in the lungs to reduce resistance and increase ventilation. 3. Vasoconstriction of arterioles supplying non-essential organs (such as the kidneys and digestive tract) to redirect blood flow to active skeletal muscles (making choice A incorrect). Vasodilation occurs in active muscles and coronary arteries. The vagus nerve is part of the parasympathetic nervous system, which decreases heart rate, not increases it (making choice D incorrect).
Marking scheme
Award 1 mark for the correct option (B). Reject A because sympathetic stimulation causes vasoconstriction, not vasodilation, of arterioles supplying the digestive tract and kidneys. Reject C because ventilation rate increases and coronary arteries dilate. Reject D because the vagus nerve is parasympathetic and slows heart activity.
Question 2 · multiple-choice
1 marks
Osmoreceptors in the hypothalamus monitor blood solute concentration. If a person becomes dehydrated, which physiological mechanism is activated to restore homeostasis?
A.The posterior pituitary gland releases less ADH, decreasing water reabsorption in the collecting ducts.
B.The posterior pituitary gland releases more ADH, causing aquaporins to be inserted into the apical membranes of collecting duct cells.
C.The adrenal cortex secretes less aldosterone, leading to increased sodium retention.
D.The glomerulus increases its glomerular filtration rate (GFR) to maximize water conservation.
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Worked solution
Dehydration leads to an increase in blood osmolarity (solute concentration), which is detected by hypothalamic osmoreceptors. This stimulates the release of Antidiuretic Hormone (ADH) from the posterior pituitary gland. ADH travels in the blood to the kidneys, where it binds to receptors on the collecting duct cells. This triggers a signaling cascade resulting in the insertion of aquaporin water channels into the apical membranes, increasing water reabsorption back into the blood and producing concentrated urine.
Marking scheme
Award 1 mark for the correct option (B). Reject A because dehydration increases ADH release, not decreases it. Reject C because decreased aldosterone would decrease sodium retention, and aldosterone is not the primary immediate response to high blood osmolarity. Reject D because increasing GFR would increase urine output, worsening dehydration.
Question 3 · multiple-choice
1 marks
In an ecosystem, two closely related species of seed-eating birds coexist in the same forest. Species A has a beak size ranging from 8 mm to 11 mm, while Species B has a beak size ranging from 12 mm to 15 mm. However, in regions where only Species A is found, its beak size ranges from 9 mm to 13 mm. Which ecological phenomenon best explains this observation?
A.Competitive exclusion
B.Resource partitioning via character displacement
C.Biomagnification
D.Primary succession
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Worked solution
Character displacement is the evolutionary divergence of traits in sympatric populations (coexisting in the same area) compared to allopatric populations (living in separate areas). When species A and B live together, competition for seeds of intermediate sizes drives natural selection for different beak sizes, allowing resource partitioning. When living alone, species A faces less interspecific competition and can utilize a broader range of seed sizes, resulting in a broader range of beak sizes.
Marking scheme
Award 1 mark for the correct option (B). Reject A because competitive exclusion would lead to the elimination of one species from the habitat. Reject C because biomagnification refers to the concentration of toxins up a food chain. Reject D because primary succession is the colonization of previously lifeless areas.
Question 4 · multiple-choice
1 marks
A pedigree analysis of a rare genetic condition in a family shows that affected fathers always pass the condition to all of their daughters, but none of their sons. Affected mothers can pass the condition to both sons and daughters. What is the most likely mode of inheritance for this condition?
A.Autosomal recessive
B.Y-linked
C.X-linked dominant
D.X-linked recessive
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Worked solution
In X-linked dominant inheritance: 1. An affected father has one X chromosome with the dominant disease allele (\(X^D Y\)). 2. He passes his Y chromosome to all of his sons (so they do not inherit the condition from him). 3. He passes his \(X^D\) chromosome to all of his daughters, so they must all inherit the dominant allele and exhibit the condition. 4. An affected mother can be heterozygous (\(X^D X^d\)) or homozygous (\(X^D X^D\)) and can pass the mutant X chromosome to either sons or daughters with equal probability.
Marking scheme
Award 1 mark for the correct option (C). Reject A because autosomal recessive traits would not show a pattern where 100% of daughters of affected fathers are affected while 0% of sons are affected. Reject B because Y-linked traits are only passed from father to son. Reject D because if it were X-linked recessive, an affected father (\(X^d Y\)) would only pass the carrier state to daughters, not the fully affected condition (unless the mother also carried the allele).
Question 5 · multiple-choice
1 marks
Phloem transport involves the active loading of organic solutes at the source. Which of the following best describes the mechanism of sucrose loading into the phloem sieve tubes of most herbaceous plants?
A.Sucrose is loaded passively down its concentration gradient through open plasmodesmata.
B.Protons are actively pumped out of the companion cells, creating an electrochemical gradient that drives sucrose entry via a sucrose-proton cotransporter.
C.Sodium-potassium pumps create an electrical potential that drives the facilitated diffusion of sucrose.
D.Sucrose is converted to starch inside the companion cells to maintain a low concentration of free sucrose.
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Worked solution
Apoplastic phloem loading is an active transport mechanism. Proton pumps (H+-ATPases) in the plasma membrane of companion cells actively transport hydrogen ions (\(H^+\)) out of the cell into the cell wall spaces. This builds up a high concentration of protons outside the cell. The protons then diffuse back down their electrochemical gradient into the companion cell through a sucrose-proton symporter (cotransport protein), bringing sucrose against its concentration gradient with them. Sucrose then diffuses into the sieve tube element via plasmodesmata.
Marking scheme
Award 1 mark for the correct option (B). Reject A because passive loading through plasmodesmata is symplastic and does not represent the active loading described. Reject C because animal cells use Na+/K+ pumps, but plants use H+ pumps for secondary active transport. Reject D because starch is insoluble and cannot be transported in phloem; sucrose must remain as a soluble sugar.
Question 6 · multiple-choice
1 marks
During the cardiac cycle, there is a brief period when the ventricles are contracting but all four heart valves (tricuspid, bicuspid, pulmonary, and aortic) are closed. What is this phase of the cardiac cycle called?
A.Isovolumetric ventricular contraction
B.Atrial systole
C.Ventricular diastole
D.Isovolumetric ventricular relaxation
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Worked solution
During early ventricular systole, the ventricles begin to contract, which rapidly increases the pressure of the blood inside them. This pressure causes the atrioventricular (AV) valves (tricuspid and bicuspid) to snap shut to prevent backflow. However, the pressure in the ventricles is not yet high enough to exceed the pressure in the aorta and pulmonary artery, so the semilunar valves remain closed. Because all four valves are closed, no blood can enter or leave the ventricles, resulting in 'isovolumetric' contraction.
Marking scheme
Award 1 mark for the correct option (A). Reject B because during atrial systole, the AV valves are open. Reject C because during ventricular diastole, the ventricles are relaxing. Reject D because during isovolumetric relaxation, the ventricles are relaxing (not contracting) after the semilunar valves close but before the AV valves open.
Question 7 · multiple-choice
1 marks
Thyroxin is a hormone secreted by the thyroid gland. What is its primary physiological role in homeostasis, and how is its secretion regulated?
A.It lowers blood glucose levels; regulated by insulin secretion from the pancreas.
B.It increases the basal metabolic rate; regulated by negative feedback involving thyroid-stimulating hormone (TSH) from the anterior pituitary.
C.It decreases body temperature by promoting sweating; regulated by the adrenal medulla.
D.It increases calcium deposition in bones; regulated by parathyroid hormone (PTH).
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Worked solution
Thyroxin (T4) is secreted by the thyroid gland in response to Thyroid Stimulating Hormone (TSH) from the anterior pituitary (which is itself stimulated by Thyrotropin-Releasing Hormone, TRH, from the hypothalamus). Thyroxin acts on almost all body cells to increase their basal metabolic rate, which also increases body temperature. High levels of thyroxin exert negative feedback on both the hypothalamus and pituitary gland to inhibit further release of TRH and TSH.
Marking scheme
Award 1 mark for the correct option (B). Reject A because insulin and glucagon regulate blood glucose. Reject C because thyroxin increases body temperature and metabolic rate, and is not regulated by the adrenal medulla. Reject D because calcitonin (also from the thyroid) and PTH regulate calcium homeostasis, not thyroxin.
Question 8 · multiple-choice
1 marks
A population of wild rabbits is growing in an environment with limited resources. As the population size (\(N\)) approaches the carrying capacity (\(K\)), which of the following statements correctly describes the dynamics of this population?
A.The per capita birth rate increases while the death rate decreases.
B.The population growth rate (\(dN/dt\)) approaches zero as the environmental resistance increases.
C.The carrying capacity (\(K\)) increases proportionally with the population size.
D.The population transitions from a logistic growth curve to an exponential growth curve.
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Worked solution
As a population reaches its carrying capacity (\(K\)), density-dependent limiting factors (like food shortages, disease, and predation) increase environmental resistance. In the logistic growth model, represented by the equation \(dN/dt = rN(K-N)/K\), as \(N\) approaches \(K\), the term \((K-N)/K\) approaches 0, which reduces the overall population growth rate (\(dN/dt\)) to zero, leading to a stable population size.
Marking scheme
Award 1 mark for the correct option (B). Reject A because birth rate decreases and/or death rate increases as resources become scarce. Reject C because carrying capacity (\(K\)) is determined by environmental factors, not by population size itself. Reject D because the population transitions from exponential growth to logistic growth as it approaches carrying capacity.
Question 9 · multiple-choice
1 marks
During high-intensity muscular exercise, the rate of cellular respiration increases, leading to changes in blood chemistry. How do the circulatory and respiratory systems integrate their responses to maintain homeostasis?
A.Increased carbon dioxide decreases blood pH, which is detected by chemoreceptors that signal the medulla oblongata to increase both ventilation rate and heart rate via the sympathetic nervous system.
B.Decreased oxygen levels are detected by baroreceptors in the carotid sinus, which directly trigger the sinoatrial node to increase heart rate while inhibiting ventilation to prevent respiratory alkalosis.
C.Increased lactate levels stimulate the pituitary gland to release epinephrine, causing systemic vasodilation and an immediate decrease in tidal volume to maximize oxygen retention.
D.Increased blood pH stimulates peripheral chemoreceptors to reduce heart rate and slow down blood flow, allowing more time for gas exchange in the pulmonary capillaries.
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Worked solution
During intense exercise, increased cell respiration releases more CO2, which dissolves in blood plasma to form carbonic acid, lowering blood pH. Chemoreceptors in the medulla oblongata, carotid bodies, and aortic bodies detect this decrease in pH and send nerve impulses to the control centres in the medulla oblongata. The medulla oblongata then increases the rate of ventilation and increases heart rate via sympathetic nerve stimulation of the sinoatrial (SA) node to rapidly deliver oxygen and remove carbon dioxide.
Marking scheme
Award [1] mark for identifying A as the correct option. Other options are incorrect because baroreceptors monitor blood pressure, epinephrine increases tidal volume, and exercise-induced carbon dioxide accumulation decreases rather than increases blood pH.
Question 10 · multiple-choice
1 marks
A healthy volunteer drinks 1.5 litres of pure water within a short period. How will the homeostatic mechanism respond to this change to maintain blood plasma osmolarity?
A.Plasma solute concentration increases, stimulating hypothalamic osmoreceptors to release more ADH, leading to increased water reabsorption in the collecting duct.
B.Plasma solute concentration decreases, inhibiting hypothalamic osmoreceptors, which reduces ADH release and results in the production of large volumes of dilute urine.
C.Plasma solute concentration decreases, stimulating the pituitary gland to release more aldosterone, which increases the permeability of the collecting duct to water.
D.Plasma solute concentration remains constant, but the increased blood volume is detected by stretch receptors in the bladder, triggering immediate secretion of ADH.
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Worked solution
Drinking a large volume of water dilutes the blood plasma, lowering its osmolarity. Hypothalamic osmoreceptors detect this decrease and reduce their signaling. Consequently, the posterior pituitary gland releases less antidiuretic hormone (ADH). Lower ADH levels reduce the water permeability of the collecting ducts in the kidneys, meaning less water is reabsorbed back into the blood, and a large volume of dilute urine is excreted.
Marking scheme
Award [1] mark for identifying B as the correct option. Option A is incorrect because water intake decreases plasma solute concentration. Option C is incorrect because aldosterone regulates sodium, and ADH controls water permeability. Option D is incorrect because ADH release would decrease.
Question 11 · multiple-choice
1 marks
In a forest, several species of insectivorous warblers feed on insects on the same species of spruce tree. However, the Cape May warbler feeds exclusively on the outermost tips of the upper branches, while the yellow-rumped warbler feeds on the lower branches near the trunk. What ecological principle is demonstrated by this behavior?
A.Competitive exclusion, resulting in the local extinction of the yellow-rumped warbler.
B.Intraspecific competition, where members of the same species divide resources to avoid conflict.
C.Resource partitioning, which minimizes interspecific competition and allows co-existence.
D.Primary succession, as new species colonize different microhabitats of a mature community.
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Worked solution
Resource partitioning occurs when species living in the same habitat use different resources or use the same resource in different ways. By foraging in different parts of the same tree, the warblers minimize niche overlap and reduce direct interspecific competition, enabling multiple species to co-exist in the same habitat.
Marking scheme
Award [1] mark for identifying C as the correct option. Reject A because both species continue to co-exist. Reject B because this is interaction between different species (interspecific), not within the same species. Reject D because succession refers to community changes over time, not spatial distribution.
Question 12 · multiple-choice
1 marks
In tomato plants, the allele for purple stems (P) is dominant to green stems (p), and the allele for cut leaves (C) is dominant to potato leaves (c). A tomato plant heterozygous for both traits (PpCc) is crossed with a testcross parent (ppcc). The offspring phenotypes are: Purple stem, cut leaves: 388; Green stem, potato leaves: 396; Purple stem, potato leaves: 112; Green stem, cut leaves: 104. What is the recombination frequency between these two gene loci?
A.11.2%
B.21.6%
C.50.0%
D.78.4%
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Worked solution
Identify the recombinant phenotypes, which are the non-parental phenotypes: Purple stem, potato leaves (112) and Green stem, cut leaves (104). The total number of offspring is 388 + 396 + 112 + 104 = 1000. The sum of the recombinants is 112 + 104 = 216. The recombination frequency is (216 / 1000) * 100 = 21.6%.
Marking scheme
Award [1] mark for identifying B as the correct option. Show working: Recombination Frequency = (Number of recombinant offspring / Total offspring) * 100 = (216 / 1000) * 100 = 21.6%.
Question 13 · multiple-choice
1 marks
During periods of high transpiration, water columns in the xylem vessels are subjected to immense tension (negative pressure). What property of water molecules prevents these continuous columns from breaking under such tension?
A.The high density of liquid water compared to water vapor, which creates an upward buoyant force.
B.Cohesive forces due to hydrogen bonding between water molecules, combined with adhesive forces to cellulose walls.
C.Active transport of mineral ions by xylem vessel elements, which maintains a positive root pressure.
D.The low specific heat capacity of water, which allows rapid temperature and pressure changes.
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Worked solution
Water columns in xylem are under high tension. The continuous flow depends on cohesion (hydrogen bonds between water molecules that hold them together, preventing the column from breaking) and adhesion (hydrogen bonds between water molecules and the hydrophilic cellulose components of the xylem cell wall, which helps support the column against gravity).
Marking scheme
Award [1] mark for identifying B as the correct option. Reject A because density differences do not provide tensile strength. Reject C because xylem vessel elements are dead at maturity and cannot perform active transport. Reject D because water has a high specific heat capacity.
Question 14 · multiple-choice
1 marks
An unknown unicellular organism is isolated from a hypersaline lake. Analysis reveals that the cell lacks membrane-bound organelles, its cell wall does not contain peptidoglycan, and its membrane lipids contain ether linkages instead of ester linkages. To which domain of life does this organism belong?
A.Eukaryota
B.Archaea
C.Bacteria
D.Protozoa
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Worked solution
Archaea are prokaryotic organisms. Unlike Bacteria, their cell walls lack peptidoglycan, and their cell membranes are composed of ether-linked lipids (often branched hydrocarbons) rather than the ester-linked lipids found in Bacteria and Eukaryota.
Marking scheme
Award [1] mark for identifying B as the correct option. Reject A and D because eukaryotes have membrane-bound organelles and ester-linked lipids. Reject C because bacteria have ester-linked lipids and peptidoglycan cell walls.
Question 15 · multiple-choice
1 marks
In yeast cells, when oxygen is unavailable, pyruvate is converted into ethanol and carbon dioxide during fermentation. What is the metabolic significance of this process?
A.It generates additional ATP through the reduction of pyruvate.
B.It regenerates oxidized NAD+ to allow glycolysis to continue producing a net yield of ATP.
C.It produces oxygen as a byproduct to kickstart aerobic respiration.
D.It converts toxic lactic acid into less harmful ethanol, preventing cellular acidosis.
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Worked solution
Glycolysis requires oxidized NAD+ to convert glyceraldehyde-3-phosphate into 1,3-bisphosphoglycerate, producing NADH. Under anaerobic conditions, there is no electron transport chain to oxidize NADH back to NAD+. Fermentation processes use pyruvate or acetaldehyde as an electron acceptor to oxidize NADH back to NAD+, enabling glycolysis (and thus ATP production) to continue.
Marking scheme
Award [1] mark for identifying B as the correct option. Reject A because fermentation itself does not produce any additional ATP. Reject C because oxygen is not produced. Reject D because yeast cells produce ethanol, not lactic acid.
Question 16 · multiple-choice
1 marks
An enzyme-catalyzed reaction is studied in the presence of an inhibitor. It is observed that as the substrate concentration is increased to very high levels, the rate of reaction eventually reaches the same maximum velocity (Vmax) as the uninhibited reaction. However, a higher concentration of substrate is required to reach half of this maximum velocity (1/2 Vmax). What can be concluded about the inhibitor?
A.It is a non-competitive inhibitor that binds to an allosteric site on the enzyme.
B.It is a competitive inhibitor that binds reversibly to the active site of the enzyme.
C.It is an irreversible inhibitor that permanently denatures the enzyme's tertiary structure.
D.It is an end-product inhibitor that stabilizes the inactive transition state of the enzyme.
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Worked solution
A competitive inhibitor competes with the substrate for binding to the active site. Because the inhibitor and substrate compete, increasing the substrate concentration to a very high level eventually overwhelms the inhibitor, allowing the reaction to reach the same maximum velocity (Vmax). However, because more substrate is needed to overcome the competition, the Michaelis constant (Km, which is the substrate concentration at 1/2 Vmax) increases.
Marking scheme
Award [1] mark for identifying B as the correct option. Reject A because non-competitive inhibitors decrease Vmax and cannot be overcome by high substrate concentrations. Reject C because irreversible inhibitors reduce active enzyme concentration, lowering Vmax.
Question 17 · multiple_choice
1 marks
During intense exercise, several physiological changes occur to meet the increased demand for oxygen in skeletal muscles. Which of the following correctly describes the integration of the nervous and endocrine systems in regulating heart rate during exercise?
A.The parasympathetic nervous system releases acetylcholine to increase heart rate, while the adrenal medulla secretes epinephrine to prolong this effect.
B.Ventilation rate increases, causing the carotid baroreceptors to send impulses to the motor cortex, which directly stimulates the sinoatrial (SA) node.
C.High blood carbon dioxide levels are detected by chemoreceptors, which send impulses to the medulla oblongata, increasing sympathetic stimulation to the sinoatrial (SA) node and stimulating epinephrine release.
D.Increased blood pressure during exercise is detected by stretch receptors in the vena cava, triggering a decrease in sympathetic signals to the AV node to prevent overfilling.
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Worked solution
During exercise, increased cellular respiration elevates blood carbon dioxide levels, which lowers blood pH. This is detected by chemoreceptors in the carotid and aortic bodies, as well as the medulla oblongata. Nerve impulses are sent to the cardiovascular centre in the medulla oblongata, which increases sympathetic nerve impulses to the sinoatrial (SA) node to increase heart rate, while simultaneously stimulating the adrenal glands to release epinephrine into the blood, reinforcing and prolonging the effect.
Marking scheme
Award 1 mark for the correct option (C). Award 0 marks for incorrect options. Reject option A because the parasympathetic nervous system decreases heart rate via acetylcholine. Reject option B because the motor cortex does not directly stimulate the SA node in response to baroreceptors. Reject option D because increased blood pressure would trigger negative feedback to decrease heart rate, not increase it.
Question 18 · multiple_choice
1 marks
The coordination of digestion involves both nervous and hormonal pathways. Which of the following events is a result of nervous control rather than hormonal control?
A.The release of gastrin by G cells in the stomach lining in response to the presence of peptides.
B.The secretion of hydrogen ions by parietal cells stimulated by the sight and smell of food.
C.The release of secretin by the duodenum in response to low pH of the chyme.
D.The secretion of pancreatic enzymes in response to cholecystokinin (CCK) in the blood.
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Worked solution
The sight and smell of food triggers the cephalic phase of digestion, which is mediated entirely by the nervous system (vagus nerve). This directly stimulates gastric secretion (including hydrochloric acid from parietal cells) before food even enters the stomach. In contrast, the other options involve the action of hormones (gastrin, secretin, and cholecystokinin) circulating in the blood.
Marking scheme
Award 1 mark for the correct option (B). Award 0 marks for incorrect options.
Question 19 · multiple_choice
1 marks
A patient is diagnosed with an endocrine disorder that prevents the secretion of thyroid-stimulating hormone (TSH). Assuming the negative feedback loop for thyroid hormone secretion is otherwise intact, what would be the expected concentrations of thyrotropin-releasing hormone (TRH) and thyroxine in this patient's blood compared to a healthy individual?
A.TRH concentration would be higher, and thyroxine concentration would be lower.
B.TRH concentration would be lower, and thyroxine concentration would be higher.
C.Both TRH and thyroxine concentrations would be lower.
D.Both TRH and thyroxine concentrations would be higher.
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Worked solution
Normally, TRH from the hypothalamus stimulates TSH from the pituitary, which stimulates thyroxine from the thyroid. Thyroxine exerts negative feedback on both. If the pituitary cannot secrete TSH, thyroxine levels will decrease significantly. Because thyroxine levels are very low, there is minimal negative feedback on the hypothalamus. Consequently, the hypothalamus will secrete high amounts of TRH in an attempt to stimulate the pituitary.
Marking scheme
Award 1 mark for the correct option (A). Award 0 marks for incorrect options.
Question 20 · multiple_choice
1 marks
The diagram below shows the feedback loop regulating blood solute concentration (osmolarity): [Hypothalamus detects high blood solute concentration] -> [Pituitary gland releases Hormone X] -> [Target cells in Kidney] -> [Response Y]. Which of the following correctly identifies Hormone X and Response Y?
A.Hormone X is Aldosterone; Response Y is decreased permeability of the distal convoluted tubule to sodium ions.
B.Hormone X is ADH; Response Y is insertion of aquaporins into the membranes of collecting duct cells.
C.Hormone X is ADH; Response Y is active transport of urea out of the loop of Henle.
D.Hormone X is Aldosterone; Response Y is increased active transport of water into the vasa recta.
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Worked solution
When osmoreceptors in the hypothalamus detect high blood solute concentration (dehydration), the posterior pituitary gland is stimulated to release ADH (antidiuretic hormone). ADH targets the cells of the collecting ducts in the kidney, causing the insertion of aquaporin water channels into their membranes, which increases water reabsorption.
Marking scheme
Award 1 mark for the correct option (B). Award 0 marks for incorrect options.
Question 21 · multiple_choice
1 marks
The population size of a species of yeast is grown in a closed culture vessel with a fixed amount of nutrient medium. After an initial lag phase and exponential growth phase, the population enters a stationary phase. During which phase does the mortality rate first become approximately equal to the natality rate, and what is the primary limiting factor in this closed culture?
A.Phase II; accumulation of toxic waste products.
B.Phase III; lack of nesting sites.
C.Phase IV; depletion of finite nutrient resources.
D.Phase IV; predation by secondary consumers.
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Worked solution
The stationary phase (Phase IV) is reached when the growth rate becomes zero, meaning the natality rate equals the mortality rate (in a closed culture with no migration). The primary limiting factor for a yeast population in a closed vessel is the depletion of the finite nutrient resources in the growth medium.
Marking scheme
Award 1 mark for the correct option (C). Award 0 marks for incorrect options.
Question 22 · multiple_choice
1 marks
Two species of protists, Paramecium caudatum and Paramecium aurelia, were grown together in a culture with a constant supply of food. After 16 days, P. caudatum died out, while P. aurelia reached its carrying capacity. When grown separately under the same conditions, both species survived and thrived. What ecological principle is best illustrated by this experiment?
A.Mutualism, where both species benefit from sharing the same ecological niche.
B.Competitive exclusion, because both species share the identical niche and compete for the same limiting resource.
C.Resource partitioning, which allows both species to coexist by utilizing different food sources.
D.Commensalism, because P. aurelia benefits while P. caudatum is unaffected.
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Worked solution
Gause's principle of competitive exclusion states that two species competing for the exact same limiting resource cannot coexist if other ecological factors remain constant. One species will always have a slight advantage, leading to the exclusion of the other.
Marking scheme
Award 1 mark for the correct option (B). Award 0 marks for incorrect options.
Question 23 · multiple_choice
1 marks
In tomato plants, the allele for tall stems (D) is dominant to dwarf stems (d), and the allele for hairy stems (H) is dominant to hairless stems (h). A dihybrid tall, hairy plant was test-crossed with a dwarf, hairless plant. The following offspring phenotypes were obtained: Tall, hairy: 412; Dwarf, hairless: 388; Tall, hairless: 104; Dwarf, hairy: 96. Which of the following is the most accurate conclusion from these results?
A.The genes assort independently, yielding a recombinant frequency of 50%.
B.The genes are linked on the same chromosome, with a recombination frequency of 20%.
C.The genes are linked on the same chromosome, with a recombination frequency of 80%.
D.The genes are sex-linked, meaning recombinants can only occur in female offspring.
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Worked solution
A test cross of a dihybrid (DdHh x ddhh) is expected to produce a 1:1:1:1 ratio if the genes assort independently. Here, the offspring show a high frequency of parental phenotypes (Tall, hairy and Dwarf, hairless) and a low frequency of recombinant phenotypes (Tall, hairless and Dwarf, hairy). This indicates gene linkage. The total number of offspring is 1000, and the number of recombinant offspring is 200 (104 + 96). The recombination frequency is (200 / 1000) * 100% = 20%.
Marking scheme
Award 1 mark for the correct option (B). Award 0 marks for incorrect options.
Question 24 · multiple_choice
1 marks
Active transport is essential for translocation in the phloem of plants. Which of the following statements correctly describes the process of phloem loading in companion cells?
A.Protons (H+) are actively pumped out of the companion cells, creating an electrochemical gradient that drives the co-transport of sucrose into the companion cells against its concentration gradient.
B.Sucrose is actively pumped out of the sieve tube elements, lowering the hydrostatic pressure and pulling water in from the xylem by osmosis.
C.Potassium ions (K+) are actively pumped into the companion cells, causing sucrose to diffuse down its concentration gradient through plasmodesmata.
D.Water is actively pumped into the source tissues, generating high turgor pressure that pushes the phloem sap towards the sink.
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Worked solution
In apoplastic phloem loading, protons (H+) are actively pumped out of the companion cells into the cell wall space using ATP. This builds an electrochemical gradient. Protons then diffuse back down their gradient into the companion cells through a co-transporter protein, which simultaneously transports sucrose against its concentration gradient into the companion cells.
Marking scheme
Award 1 mark for the correct option (A). Award 0 marks for incorrect options.
Question 25 · multiple-choice
1 marks
When blood pressure decreases during a sudden change in posture, which autonomic response is coordinated by the cardiovascular control center in the medulla oblongata to restore homeostasis?
A.Increased parasympathetic stimulation to the sinoatrial node.
B.Decreased sympathetic stimulation to the ventricles.
C.Increased sympathetic stimulation to the sinoatrial node and arterioles.
D.Increased parasympathetic stimulation to the arterioles causing vasodilation.
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Worked solution
A decrease in blood pressure is detected by baroreceptors (stretch receptors) in the carotid sinus and aortic arch. This information is sent to the cardiovascular control center in the medulla oblongata. In response, the medulla increases sympathetic nervous output and decreases parasympathetic output. The increased sympathetic stimulation acts on the sinoatrial (SA) node to increase heart rate, on the ventricles to increase contractility, and on smooth muscle in arterioles to cause vasoconstriction, thereby raising blood pressure back to normal levels.
Marking scheme
Award 1 mark for the correct option (C). Correct option identifies increased sympathetic stimulation to the SA node and vasoconstriction of arterioles as the correct homeostatic reflex to a decrease in blood pressure. Incorrect options describe parasympathetic actions or sympathetic inhibition, which would further lower blood pressure.
Question 26 · multiple-choice
1 marks
During the cephalic phase of digestion, which pathway leads to the initial secretion of gastric juice before food even enters the stomach?
A.Distension of the stomach wall stimulates mechanoreceptors, which signal via local enteric reflexes.
B.Sight or smell of food stimulates the brain, which sends impulses via the vagus nerve to stimulate gastric glands.
C.High pH in the duodenum triggers the release of secretin, which stimulates parietal cells.
D.Partially digested proteins in the duodenum stimulate the release of CCK, triggering the enteric nervous system.
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Worked solution
The cephalic phase of digestion is initiated by the sight, smell, taste, or thought of food. These sensory stimuli are processed by the brain, which sends impulses along the vagus nerve (parasympathetic nervous system) to the stomach. This nervous pathway stimulates gastric glands to secrete gastric juice (containing hydrochloric acid and pepsinogen) and triggers the release of gastrin, preparing the stomach for incoming food.
Marking scheme
Award 1 mark for the correct option (B). The cephalic phase is characterized by nervous system regulation (via the vagus nerve) triggered by sensory inputs, preceding the mechanical or chemical entry of food into the stomach.
Question 27 · multiple-choice
1 marks
An increase in blood solute concentration is detected by osmoreceptors in the hypothalamus. What is the correct sequence of events that follows to restore osmotic balance?
A.Decreased ADH release \(\rightarrow\) decreased aquaporins in the collecting duct \(\rightarrow\) production of concentrated urine.
B.Increased ADH release \(\rightarrow\) increased aquaporins in the collecting duct \(\rightarrow\) increased water reabsorption.
C.Decreased ADH release \(\rightarrow\) increased aquaporins in the collecting duct \(\rightarrow\) decreased water reabsorption.
D.Increased ADH release \(\rightarrow\) decreased aquaporins in the collecting duct \(\rightarrow\) production of dilute urine.
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Worked solution
An increase in blood solute concentration (high osmolarity) is detected by hypothalamic osmoreceptors. This triggers the posterior pituitary gland to release antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it binds to receptors on the cells of the collecting ducts, increasing the insertion of aquaporins (water channels) into their apical membranes. This increases water reabsorption back into the blood, producing a lower volume of highly concentrated urine and restoring blood osmolarity to normal.
Marking scheme
Award 1 mark for the correct option (B). The pathway correctly links high solute concentration with increased ADH release, increased aquaporin density, and increased water reabsorption in the collecting ducts.
Question 28 · multiple-choice
1 marks
When core body temperature drops below the set point, which homeostatic response is initiated by the hypothalamus?
A.Vasodilation of skin arterioles to increase heat radiation.
B.Stimulation of sweat glands to release sweat for evaporative cooling.
C.Vasoconstriction of skin arterioles and stimulation of skeletal muscles to shiver.
D.Decreased metabolic rate to prevent internal heat production.
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Worked solution
In response to cold stress, the thermoregulatory center in the hypothalamus coordinates physiological changes to conserve and generate heat. This includes vasoconstriction of cutaneous (skin) arterioles to divert blood away from the skin surface (minimizing heat loss) and the stimulation of rapid skeletal muscle contractions (shivering) to generate metabolic heat.
Marking scheme
Award 1 mark for the correct option (C). The correct response to cold involves vasoconstriction (to reduce heat loss) and shivering (to generate heat).
Question 29 · multiple-choice
1 marks
A population of yeast is grown in a closed batch culture with a limited amount of nutrient broth. Which factor represents a density-dependent limit on the population growth as it approaches the stationary phase?
A.Sudden decrease in incubation temperature.
B.Accumulation of ethanol to toxic levels.
C.Dilution of the nutrient medium by adding sterile water.
D.Sudden exposure of the culture to ultraviolet light.
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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. In a closed batch culture of yeast, as the population density increases, yeast cells consume glucose and produce ethanol as a byproduct of anaerobic respiration. The accumulation of ethanol to toxic concentrations is directly dependent on yeast density and acts as a limiting factor that stalls population growth, pushing it into the stationary phase.
Marking scheme
Award 1 mark for the correct option (B). Ethanol accumulation is a density-dependent metabolic byproduct. Temperature changes and UV light are density-independent factors, while dilution does not limit growth as a result of density changes.
Question 30 · multiple-choice
1 marks
In Drosophila, the gene for body color (gray \(G\) vs. black \(g\)) and wing length (long \(L\) vs. vestigial \(l\)) are linked on an autosome. A heterozygous gray, long-winged female (\(GgLl\)) is test-crossed with a black, vestigial-winged male (\(ggll\)). The offspring phenotypes are: Gray, long: 415; Black, vestigial: 385; Gray, vestigial: 105; Black, long: 95. What is the recombination frequency between these two gene loci?
A.10%
B.20%
C.40%
D.80%
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Worked solution
Recombination frequency is calculated as: \(\text{Recombination Frequency} = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} \times 100\). The parent genotypes are \(GgLl\) (heterozygous) and \(ggll\) (homozygous recessive test-cross partner). The parental-type offspring are Gray, long (415) and Black, vestigial (385), as these represent the linked alleles in their original configurations. The recombinant offspring are Gray, vestigial (105) and Black, long (95). Total offspring \(= 415 + 385 + 105 + 95 = 1000\). Total recombinants \(= 105 + 95 = 200\). Recombination frequency \(= \frac{200}{1000} \times 100 = 20\%\).
Marking scheme
Award 1 mark for the correct option (B). Method of calculation requires identifying the recombinants, summing them, and dividing by the total population.
Question 31 · multiple-choice
1 marks
According to the pressure-flow hypothesis of phloem translocation, what causes the movement of sap from the source to the sink?
A.Tension created by the transpiration of water from leaves.
B.Active transport of sucrose out of the sieve tubes at the source.
C.Hydrostatic pressure gradient generated by the active loading of sucrose at the source and subsequent entry of water by osmosis.
D.Capillary action within the narrow lumen of the dead companion cells.
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Worked solution
Active transport is used to load sucrose into the phloem sieve tubes at the source. This high concentration of solute decreases the water potential inside the sieve tubes, causing water to enter from the adjacent xylem by osmosis. This accumulation of water increases the hydrostatic pressure at the source. At the sink, sucrose is unloaded, water potential increases, water leaves the phloem, and hydrostatic pressure drops. Sap flows down this hydrostatic pressure gradient from source to sink.
Marking scheme
Award 1 mark for the correct option (C). The driving force for mass flow in phloem is the hydrostatic pressure gradient established by active loading of solutes and the resulting osmotic water movement.
Question 32 · multiple-choice
1 marks
Which structural feature of xylem vessels prevents them from collapsing inward when water is pulled upward under extremely high tension?
A.Highly permeable cell membranes that allow water to flow freely.
B.Helical or annular rings of lignin reinforcing the cellulose cell wall.
C.Active sodium-potassium pumps in the cell walls that maintain turgor pressure.
D.Sieve plates that divide the vessel into discrete, pressurized chambers.
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Worked solution
Water is pulled up the xylem under tension (negative pressure) due to transpiration. To prevent the vessels from collapsing inward under this substantial sucking force, xylem vessel walls are heavily reinforced with lignin. Lignin is a strong, woody polymer deposited in helical, annular, or reticulate patterns within the cellulose cell wall, providing high tensile strength and structural support.
Marking scheme
Award 1 mark for the correct option (B). Lignified cell walls are the specific evolutionary adaptation that prevents structural collapse of xylem vessels under negative pressure.
Question 33 · multiple-choice
1 marks
During exercise, the concentration of carbon dioxide in the blood increases. How does the body coordinate the respiratory and circulatory systems to restore homeostasis?
A.The medulla oblongata detects a decrease in blood pH and sends nerve impulses to increase both ventilation rate and heart rate.
B.The hypothalamus detects an increase in blood pH and sends hormone signals to increase ventilation rate and decrease heart rate.
C.The cerebellum detects a decrease in blood pH and sends nerve impulses to decrease ventilation rate and increase heart rate.
D.The cerebrum detects an increase in blood pH and sends hormone signals to decrease both ventilation rate and heart rate.
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Worked solution
Carbon dioxide dissolves in the blood to form carbonic acid, which lowers blood pH. Chemoreceptors in the medulla oblongata detect this decrease in pH. The medulla oblongata coordinates a response via the autonomic nervous system, sending nerve impulses to the sinoatrial (SA) node of the heart and the respiratory muscles to increase both heart rate and ventilation rate, thereby accelerating carbon dioxide removal.
Marking scheme
Award 1 mark for identifying the correct regulatory center (medulla oblongata), the stimulus (decreased pH), and the coordinated responses (increased ventilation and heart rate). Correct option is A.
Question 34 · multiple-choice
1 marks
What is the direct cellular response in the collecting duct cells of the kidney nephron when blood solute concentration increases?
A.Decreased secretion of ADH causes endocytosis of aquaporins from the luminal membrane.
B.Increased secretion of ADH causes exocytosis of aquaporins into the luminal membrane.
C.Decreased secretion of ADH causes active transport of sodium ions out of the filtrate.
D.Increased secretion of ADH causes the luminal membrane to become less permeable to water.
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Worked solution
An increase in blood solute concentration (high osmolarity) is detected by hypothalamic osmoreceptors, triggering the release of ADH (antidiuretic hormone) from the posterior pituitary. ADH binds to receptors on the collecting duct cells, initiating a cascade that leads to the exocytosis of vesicles containing aquaporins. These aquaporins insert into the luminal membrane, increasing its permeability to water and allowing more water to be reabsorbed back into the bloodstream.
Marking scheme
Award 1 mark for linking high blood solute concentration to increased ADH secretion, followed by the insertion of aquaporins into the luminal membrane via exocytosis to increase permeability. Correct option is B.
Question 35 · multiple-choice
1 marks
In a marine ecosystem, nutrient runoff stimulates phytoplankton growth, which subsequently increases the biomass of zooplankton and small fish. However, the population of top-predator sharks remains unchanged due to intensive commercial fishing. Which statement correctly describes the limiting factors acting on this community?
A.The zooplankton population is experiencing top-down control only.
B.The increase in phytoplankton biomass is an example of bottom-up control on zooplankton.
C.The shark population is primarily limited by bottom-up factors.
D.The small fish population is unaffected by density-dependent factors.
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Worked solution
Bottom-up control occurs when the population size of trophic levels is determined by the availability of resources (like nutrients and primary producers) at the base of the food chain. Here, nutrient runoff increases phytoplankton, which subsequently boosts zooplankton and fish, showing bottom-up control. The shark population is limited by top-down human exploitation (fishing), not bottom-up resources.
Marking scheme
Award 1 mark for correctly identifying that resource-driven population increases represent bottom-up control on zooplankton. Correct option is B.
Question 36 · multiple-choice
1 marks
A dihybrid test cross was performed between an organism heterozygous for two traits (AaBb) and a homozygous recessive individual (aabb). The offspring showed 420 parental phenotypes and 80 recombinant phenotypes. Which of the following is the correct recombination frequency and conclusion about the gene loci?
A.Recombination frequency is 16% and the genes are unlinked.
B.Recombination frequency is 16% and the genes are linked.
C.Recombination frequency is 19% and the genes are linked.
D.Recombination frequency is 19% and the genes are unlinked.
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Worked solution
The recombination frequency is calculated as (number of recombinants / total offspring) x 100. Total offspring = 420 + 80 = 500. Recombination frequency = (80 / 500) x 100 = 16%. Because the recombination frequency is less than 50%, the genes do not assort independently, meaning they are linked on the same chromosome.
Marking scheme
Award 1 mark for the correct calculation of 16% recombination frequency and the logical deduction that this indicates gene linkage. Correct option is B.
Question 37 · multiple-choice
1 marks
In the apoplastic loading of sucrose into phloem companion cells, how is the concentration gradient necessary for sucrose transport established?
A.Active transport of sucrose directly out of the companion cells into the sieve tube elements via carrier proteins.
B.Passive diffusion of hydrogen ions out of the companion cells, creating an electrical gradient.
C.Active transport of hydrogen ions out of the companion cells into the cell wall, generating a proton gradient.
D.Facilitated diffusion of sucrose into the companion cells from the mesophyll cell cytoplasm.
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Worked solution
During apoplastic phloem loading, proton pumps actively transport hydrogen ions (protons) out of the companion cells into the cell wall (apoplast) using ATP. This generates a high concentration of protons outside the cell. Protons then diffuse back down their electrochemical gradient into the companion cell through a co-transporter protein, which simultaneously carries sucrose against its concentration gradient.
Marking scheme
Award 1 mark for recognizing that proton pumps actively transport hydrogen ions into the cell wall to create the concentration gradient driving sucrose co-transport. Correct option is C.
Question 38 · multiple-choice
1 marks
During intense exercise, actively respiring muscle tissue produces large amounts of carbon dioxide and lactic acid. How does this affect oxygen transport and delivery to these tissues?
A.The local decrease in pH increases hemoglobin's affinity for oxygen, causing it to retain oxygen.
B.The local increase in pH decreases hemoglobin's affinity for oxygen, promoting oxygen release.
C.The local decrease in pH shifts the oxygen dissociation curve to the right, promoting oxygen release.
D.The local increase in pH shifts the oxygen dissociation curve to the left, promoting oxygen uptake.
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Worked solution
Increased carbon dioxide and lactic acid production lowers pH in actively respiring tissues. According to the Bohr effect, lower pH shifts the oxygen dissociation curve of hemoglobin to the right. This shift represents a decreased affinity of hemoglobin for oxygen at any given partial pressure, causing hemoglobin to release oxygen more readily to the tissues that need it.
Marking scheme
Award 1 mark for identifying that decreased pH causes a rightward shift of the oxygen dissociation curve, reducing affinity and increasing oxygen release. Correct option is C.
Question 39 · multiple-choice
1 marks
When body temperature rises above the set point, which of the following sequences represents the correct response coordinated by the hypothalamus to restore homeostasis?
A.Vasoconstriction of skin arterioles, activation of sweat glands, and shivering.
B.Vasodilation of skin arterioles, activation of sweat glands, and decrease in metabolic rate.
C.Vasodilation of skin arterioles, inhibition of sweat glands, and shivering.
D.Vasoconstriction of skin arterioles, inhibition of sweat glands, and increase in metabolic rate.
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Worked solution
To lower body temperature, the hypothalamus coordinates mechanisms to increase heat loss and decrease heat production. This includes: vasodilation of skin arterioles (bringing blood closer to the surface to radiate heat), activation of sweat glands (promoting evaporative cooling), and lowering general metabolic activity. Shivering and vasoconstriction are responses to cold.
Marking scheme
Award 1 mark for selecting the correct physiological mechanisms of heat loss (vasodilation, sweating, decreased metabolism). Correct option is B.
Question 40 · multiple-choice
1 marks
Two closely related species of beetle feed on the same pine tree. Species X feeds exclusively on the young needles at the tips of the branches, while Species Y feeds exclusively on the older needles closer to the trunk. What ecological concept does this scenario best demonstrate?
A.Competitive exclusion
B.Resource partitioning
C.Biomagnification
D.Primary succession
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Worked solution
Resource partitioning occurs when species divide a niche to avoid direct competition for resources. By feeding on different parts of the same tree (different ages of needles), the two beetle species minimize direct competition and can coexist within the same habitat.
Marking scheme
Award 1 mark for identifying resource partitioning as the division of a resource to reduce competition and allow coexistence. Correct option is B.
Paper 2 Section A
Answer all structured questions including the database question. Calculator permitted.
5 Question · 40 marks
Question 1 · Short Answer
8 marks
A runner starts a 400-meter sprint, causing a rapid shift in physiological states. (a) Describe how the medulla oblongata detects and coordinates the response to a high concentration of carbon dioxide in the blood during exercise. [3] (b) Explain the role of epinephrine (adrenaline) in mobilizing resources during exercise, linking the nervous and endocrine systems. [3] (c) State two changes in the cardiovascular system, other than increased heart rate, that occur during exercise to optimize oxygen delivery. [2]
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Worked solution
Detailed explanation of the autonomic control of heart and ventilation rate via medullary chemoreceptors (a), the hormonal action of epinephrine via the bloodstream to target tissues (b), and systemic blood redistribution and stroke volume increase during exercise (c).
Marking scheme
a) Award [1] for chemoreceptors in the medulla oblongata (or major arteries) detecting a drop in pH / rise in carbon dioxide concentration. Award [1] for sympathetic impulses sent via the cardiac accelerator nerve to the sinoatrial (SA) node to increase heart rate. Award [1] for impulses sent to the diaphragm / intercostal muscles to increase ventilation rate/depth. b) Award [1] for the sympathetic nervous system stimulating the adrenal glands/medulla to release epinephrine. Award [1] for epinephrine traveling through the blood to target tissues (linking systems). Award [1] for explaining a specific effect: stimulating glycogenolysis in the liver / causing bronchodilation in the lungs / increasing cardiac stroke volume. c) Award [1] for each correct cardiovascular change up to [2]: Vasodilation of arterioles supplying active skeletal muscles; Vasoconstriction of arterioles supplying non-essential organs (e.g., gut/kidneys); Increased stroke volume.
Question 2 · Short Answer
8 marks
Osmoregulation is crucial for terrestrial mammals to maintain internal solute concentrations. (a) Describe how the loop of Henle acts as a countercurrent multiplier to establish an osmotic gradient in the medulla. [3] (b) Explain the effect of high blood solute concentration on the cells of the collecting duct. [3] (c) Distinguish between osmoregulation in a desert mammal (such as a kangaroo rat) and a freshwater mammal (such as an otter) in terms of nephron structure. [2]
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Worked solution
Comprehensive description of the loop of Henle countercurrent multiplier system (a), the cellular mechanism of ADH action on aquaporin insertion in the collecting duct membranes (b), and structural adaptations of nephrons to different environmental water availabilities (c).
Marking scheme
a) Award [1] for stating that the descending limb is permeable to water but impermeable to solute (water leaves filtrate by osmosis). Award [1] for stating that the ascending limb actively transports sodium and chloride ions out into the medulla tissue fluid (and is impermeable to water). Award [1] for establishing a high solute concentration / hypertonic gradient in the medulla tissue. b) Award [1] for high blood solute concentration stimulating ADH release from the posterior pituitary. Award [1] for ADH binding to receptors on the collecting duct cells, initiating a signaling cascade. Award [1] for the insertion of aquaporins into the cell membranes, making the collecting duct highly permeable to water. c) Award [1] for desert mammals having longer loops of Henle / thicker medulla to allow greater water reabsorption. Award [1] for freshwater mammals having shorter loops of Henle / thinner medulla as water conservation is less critical.
Question 3 · Short Answer
8 marks
A population of rabbits is introduced to an isolated island with abundant resources. (a) Describe the changes in growth rate and birth/death rates during the three phases of a sigmoid population growth curve. [3] (b) Distinguish between density-dependent and density-independent factors that limit population growth, giving one example of each. [3] (c) Explain the concept of carrying capacity \(K\) and how overshoot can lead to a population crash. [2]
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Worked solution
Detailed analysis of the three phases of sigmoid growth in relation to birth and death rates (a), categorization and examples of limiting factors (b), and definition of carrying capacity with ecological consequences of overshoot (c).
Marking scheme
a) Award [1] for exponential phase: birth rate far exceeds death rate, leading to rapid increase. Award [1] for transitional phase: birth rate begins to fall and/or death rate begins to rise, slowing growth. Award [1] for plateau phase: birth rate equals death rate, and growth rate becomes zero. b) Award [1] for defining density-dependent factors as those whose effect depends on population density, with a valid example (e.g., competition for food, disease, predation). Award [1] for defining density-independent factors as those affecting the population regardless of its density, with a valid example (e.g., natural disasters, extreme temperatures). Award [1] for contrasting their mechanisms of action / impact relative to density. c) Award [1] for defining carrying capacity (\(K\)) as the maximum population size that an environment can sustainably support over time. Award [1] for explaining that overshoot leads to overexploitation of resources, reducing the environment's future carrying capacity and causing a rapid population crash/starvation.
Question 4 · Short Answer
8 marks
In tomato plants, red fruit (\(R\)) is dominant over yellow fruit (\(r\)), and tall stem (\(T\)) is dominant over dwarf stem (\(t\)). A heterozygous tall, red-fruited plant is crossed with a dwarf, yellow-fruited plant. (a) State the expected phenotypic ratio of the offspring if the two genes are unlinked and assort independently. [1] (b) In a trial cross of 400 offspring, the following phenotypes were observed: - Tall, red-fruited: 182 - Dwarf, yellow-fruited: 178 - Tall, yellow-fruited: 21 - Dwarf, red-fruited: 19 Explain these results in terms of gene linkage and crossing over. [4] (c) Calculate the recombination frequency for these two genes, showing your working. [2] (d) State the genotype of the recombinant offspring. [1]
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Worked solution
Step-by-step resolution of dihybrid test cross involving autosomal linkage: independent assortment ratio prediction (a), interpretation of deviation from expected ratio due to linkage and crossing-over (b), calculation of recombination frequency from experimental data (c), and determination of recombinant genotypes (d).
Marking scheme
a) Award [1] for 1:1:1:1 (or equal proportions of the four phenotypes). b) Award [1] for identifying that the genes are linked on the same chromosome. Award [1] for stating that the dominant alleles \(R\) and \(T\) (and recessive \(r\) and \(t\)) are located together on the parental chromosomes. Award [1] for identifying that the less frequent phenotypes (tall/yellow and dwarf/red) are recombinants. Award [1] for explaining that recombinants only arise from crossing over between non-sister chromatids during prophase I of meiosis. c) Award [1] for showing correct working: \(\frac{21 + 19}{400} \times 100\%\) (or \(\frac{40}{400}\)). Award [1] for the correct final answer of 10%. d) Award [1] for both correct genotypes: \(Ttrr\) and \(ttRr\) (accept equivalent representations).
Question 5 · Short Answer
8 marks
Plants must balance the uptake of carbon dioxide for photosynthesis with the loss of water vapor. (a) Explain how the properties of water and the structure of xylem vessels allow water to be transported under tension from the roots to the leaves. [4] (b) Describe how a potometer can be used to measure the rate of transpiration, including one essential precaution to ensure valid results. [3] (c) Outline one structural adaptation of a xerophyte that reduces water loss. [1]
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Worked solution
Detailed explanation of transpiration pull driven by cohesion-tension theory in xylem (a), methodology of using a potometer to estimate transpiration rate via water uptake, including controls and precautions (b), and adaptation of xerophytes to arid environments (c).
Marking scheme
a) Award [1] for cohesion where hydrogen bonding between water molecules maintains a continuous water column. Award [1] for adhesion where water molecules adhere to the hydrophilic cellulose walls of xylem vessels, resisting gravity/tension. Award [1] for xylem vessels having a hollow lumen / being dead / lacking end walls, reducing resistance to flow. Award [1] for lignified cell walls providing structural support to prevent the xylem from collapsing under negative pressure/tension. b) Award [1] for explaining that a potometer measures the rate of water uptake by tracking the movement of an air bubble in a capillary tube. Award [1] for an essential precaution: cutting the plant stem under water to prevent air bubbles from blocking xylem vessels / sealing all joints with petroleum jelly (Vaseline) to ensure the system is airtight. Award [1] for stating how the rate is calculated: distance moved by the bubble divided by time (or calculating volume of water using cylinder volume formula). c) Award [1] for any of the following: Thick waxy cuticle to block cuticular transpiration; Rolled leaves to trap humid air around stomata, reducing concentration gradient; Sunken stomata to create a microclimate of high humidity; Reduced leaves/spines to decrease total surface area for transpiration.
Paper 2 Section B
Answer two questions of your choice from a choice of three extended-response options. Clarity of communication is assessed.
2 Question · 32 marks
Question 1 · Extended Response
16 marks
Homeostasis is the maintenance of a constant internal environment within narrow physiological limits.
(a) Outline the principle of negative feedback, using the control of blood glucose concentration as an example. [4]
(b) Explain the role of the kidney and antidiuretic hormone (ADH) in maintaining water balance (osmoregulation) when the body is dehydrated. [6]
(c) Describe the physiological mechanisms used by endothermic homeotherms to increase core body temperature when exposed to cold conditions. [5]
[Plus 1 mark for quality of communication]
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Worked solution
**(a) Outline the principle of negative feedback, using the control of blood glucose concentration as an example. [4]** - Negative feedback is a regulatory mechanism where a change in a physiological variable triggers responses that counteract the change, restoring the variable back to its normal set point. - If blood glucose levels rise (e.g., after eating), beta cells in the pancreatic islets of Langerhans detect this change and secrete insulin into the bloodstream. - Insulin stimulates body cells (especially muscle and liver cells) to increase glucose uptake and triggers glycogenesis (conversion of glucose into glycogen for storage), lowering blood glucose levels. - If blood glucose levels drop (e.g., during exercise or fasting), alpha cells in the pancreatic islets detect this and secrete glucagon into the bloodstream. - Glucagon stimulates liver cells to undergo glycogenolysis (breakdown of glycogen to glucose) and release glucose into the blood, raising blood glucose levels back to normal.
**(b) Explain the role of the kidney and antidiuretic hormone (ADH) in maintaining water balance (osmoregulation) when the body is dehydrated. [6]** - Dehydration causes an increase in blood solute concentration (osmolarity) / a decrease in blood water potential. - Osmoreceptors in the hypothalamus detect this change in blood osmolarity. - The hypothalamus stimulates the posterior pituitary gland to release antidiuretic hormone (ADH) into the bloodstream. - ADH travels to the kidneys, where it binds to specific receptors on the cells lining the distal convoluted tubules and collecting ducts. - This binding activates a secondary messenger system that causes vesicles containing aquaporins (water channels) to fuse with the apical membrane of these cells. - The insertion of aquaporins significantly increases the permeability of the collecting duct to water. - As the filtrate passes through the collecting duct, water is reabsorbed by osmosis into the highly concentrated (hypertonic) interstitial fluid of the renal medulla and then back into the capillaries. - As a result, a smaller volume of highly concentrated (hypertonic) urine is excreted, conserving water in the body and returning blood osmolarity to normal.
**(c) Describe the physiological mechanisms used by endothermic homeotherms to increase core body temperature when exposed to cold conditions. [5]** - Thermoreceptors in the skin and the hypothalamus detect a decrease in external or core body temperature. - **Vasoconstriction:** Sympathetic nerves stimulate arterioles supplying the skin capillaries to constrict, while shunt vessels dilate. This reduces blood flow to the skin surface, minimizing heat loss via radiation, conduction, and convection. - **Shivering:** Rapid, involuntary cycles of skeletal muscle contraction and relaxation are initiated, converting metabolic energy into heat. - **Piloerection:** Hair erector muscles contract, pulling hairs upright to trap a layer of warm, still air close to the skin, which acts as an insulator (though this is more effective in furry mammals than humans). - **Metabolic rate increase:** Endocrine glands release hormones such as thyroxine (from the thyroid gland) and adrenaline (from the adrenal medulla), which increase cellular metabolism and generate heat.
Marking scheme
**(a) Negative feedback and blood glucose [Max 4 marks]:** - Award 1 mark for defining negative feedback as counteracting a deviation from a set point to restore homeostatic balance. - Award 1 mark for noting that high blood glucose stimulates insulin secretion from pancreatic beta cells. - Award 1 mark for noting that insulin promotes glucose uptake by cells/glycogenesis in the liver/muscle. - Award 1 mark for noting that low blood glucose stimulates glucagon secretion from pancreatic alpha cells. - Award 1 mark for noting that glucagon promotes glycogenolysis/release of glucose from the liver.
**(b) Osmoregulation and ADH [Max 6 marks]:** - Award 1 mark for stating that dehydration increases blood osmolarity/solute concentration (or decreases water potential). - Award 1 mark for identifying that osmoreceptors in the hypothalamus detect this change. - Award 1 mark for stating that ADH is released by the posterior pituitary gland. - Award 1 mark for noting that ADH increases the permeability of the collecting duct (and distal convoluted tubule) to water. - Award 1 mark for describing the mechanism: ADH causes aquaporins (water channels) to be inserted into the membrane. - Award 1 mark for explaining that water is reabsorbed by osmosis into the hypertonic medulla/blood. - Award 1 mark for noting that this results in a small volume of highly concentrated/hypertonic urine.
**(c) Response to cold [Max 5 marks]:** - Award 1 mark for stating that thermoreceptors in the skin/hypothalamus detect the temperature drop. - Award 1 mark for vasoconstriction: constriction of skin arterioles reducing blood flow and heat loss at the surface. - Award 1 mark for shivering: rapid muscle contractions generating metabolic heat. - Award 1 mark for piloerection: contraction of hair erector muscles trapping an insulating layer of air. - Award 1 mark for increased metabolic rate: release of thyroxine/adrenaline to raise cellular respiration.
**Quality of Communication [1 mark]:** - Award 1 mark if the overall response is written in a clear, structured, and coherent manner using precise biological terms (e.g., islet of Langerhans, aquaporins, vasoconstriction, medulla, etc.).
Question 2 · Extended Response
16 marks
Multicellular organisms require the integration of various organ systems to function effectively and respond to changing environments.
(a) Explain how the nervous and endocrine systems cooperate in the 'fight-or-flight' response when an animal perceives a threat. [4]
(b) Describe how the nervous, cardiovascular, and respiratory systems are integrated to meet the increased metabolic demands of active skeletal muscles during strenuous exercise. [6]
(c) Outline how the nervous and endocrine systems coordinate the secretion of digestive juices in the stomach. [5]
[Plus 1 mark for quality of communication]
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Worked solution
**(a) Explain how the nervous and endocrine systems cooperate in the 'fight-or-flight' response when an animal perceives a threat. [4]** - When a threat is perceived, sensory inputs are processed by the brain, activating the amygdala and then the hypothalamus. - The hypothalamus sends nerve impulses along sympathetic fibers of the autonomic nervous system directly to the adrenal medulla. - This nervous stimulation causes the adrenal medulla to rapidly secrete the hormones adrenaline (epinephrine) and noradrenaline into the blood. - This represents a direct nervous-to-endocrine relay (neuroendocrine integration). - Concurrently, the hypothalamus activates the slower HPA (hypothalamic-pituitary-adrenal) endocrine axis, releasing Corticotropin-Releasing Hormone (CRH), which stimulates the pituitary to release ACTH, which in turn prompts the adrenal cortex to release cortisol. - Together, these nervous and endocrine signals prepare the body by increasing heart rate, dilating airways, mobilizing glucose (via glycogen breakdown), and prioritizing blood flow to skeletal muscles.
**(b) Describe how the nervous, cardiovascular, and respiratory systems are integrated to meet the increased metabolic demands of active skeletal muscles during strenuous exercise. [6]** - During exercise, active skeletal muscles undergo high rates of cellular respiration, consuming oxygen (\(O_2\)) and generating carbon dioxide (\(CO_2\)) and hydrogen ions (\(H^+\) / lactic acid). - The accumulation of \(CO_2\) and \(H^+\) lowers blood pH. - Chemoreceptors in the medulla oblongata (central) and in the carotid and aortic bodies (peripheral) detect this decrease in pH and increase in \(CO_2\). - The respiratory center in the medulla oblongata sends more frequent nerve impulses (via the phrenic and intercostal nerves) to the diaphragm and external intercostal muscles, increasing ventilation rate and depth. This maximizes gas exchange in the lungs. - Simultaneously, the cardiovascular center in the medulla oblongata increases sympathetic nervous output to the sinoatrial (SA) node of the heart, while decreasing parasympathetic (vagus) output. - This increases both heart rate and stroke volume, boosting cardiac output to deliver blood faster to the muscles. - Local chemical changes (low \(O_2\), high \(CO_2\), heat) cause local vasodilation of arterioles feeding active muscles, while sympathetic signals cause vasoconstriction of arterioles in non-essential organs (such as the digestive tract), redistributing oxygenated blood where it is needed most.
**(c) Outline how the nervous and endocrine systems coordinate the secretion of digestive juices in the stomach. [5]** - The secretion of gastric juice is coordinated in phases involving both nervous and endocrine mechanisms. - **Cephalic Phase (Nervous):** The sight, smell, taste, or thought of food triggers impulses in the cerebral cortex and hypothalamus, which send signals down the vagus nerve (parasympathetic) to the stomach. - This nervous pathway directly stimulates gastric glands in the stomach mucosa to secrete hydrochloric acid (HCl) and pepsinogen. - The vagus nerve also stimulates endocrine G-cells in the stomach wall to secrete the hormone gastrin into the blood. - **Gastric Phase (Nervous and Endocrine):** Once food enters the stomach, physical distension (stretching) is detected by stretch receptors in the stomach wall. - This sends impulses via local nervous pathways (enteric nervous system) and the vagus nerve, further stimulating gastric secretions. - Chemical components of food (e.g., peptides) also stimulate G-cells directly to secrete more gastrin. - Gastrin circulates in the blood and binds to receptors on parietal cells, strongly stimulating sustained secretion of hydrochloric acid. - When the stomach contents become highly acidic (pH drops below 1.5–2), a negative feedback mechanism inhibits gastrin secretion, reducing further acid production.
Marking scheme
**(a) Fight-or-flight response [Max 4 marks]:** - Award 1 mark for stating that the hypothalamus processes the perception of threat. - Award 1 mark for noting that the sympathetic nervous system stimulates the adrenal medulla. - Award 1 mark for stating that the adrenal medulla secretes adrenaline/noradrenaline into the bloodstream. - Award 1 mark for outlining the HPA axis (hypothalamus -> pituitary -> adrenal cortex releasing cortisol) as a secondary endocrine pathway. - Award 1 mark for mentioning the combined physiological effect (e.g., increased heart rate, glycogen breakdown/mobilization of glucose).
**(b) Exercise integration [Max 6 marks]:** - Award 1 mark for stating that active muscle respiration increases \(CO_2\) and decreases blood pH. - Award 1 mark for noting that central/peripheral chemoreceptors detect these chemical changes. - Award 1 mark for describing the respiratory response: the medulla oblongata stimulates the respiratory muscles (diaphragm/intercostals) to increase rate and depth of breathing. - Award 1 mark for describing the cardiovascular response: the cardiovascular center in the medulla increases sympathetic input to the SA node. - Award 1 mark for explaining that this increases heart rate and cardiac output. - Award 1 mark for explaining vasodilation in active muscles and vasoconstriction in non-essential organs (blood shunting).
**(c) Gastric secretion [Max 5 marks]:** - Award 1 mark for describing the cephalic phase where sensory stimuli trigger vagus nerve (parasympathetic) signals. - Award 1 mark for stating that the vagus nerve stimulates gastric glands to secrete gastric juice/HCl/pepsinogen. - Award 1 mark for stating that the vagus nerve (or stomach stretch) stimulates G-cells to release the hormone gastrin. - Award 1 mark for describing the gastric phase: distension of the stomach activates stretch receptors leading to local nervous reflexes. - Award 1 mark for explaining that gastrin travels via blood to stimulate acid secretion by gastric glands. - Award 1 mark for mentioning the negative feedback loop where high acidity/low pH inhibits gastrin secretion.
**Quality of Communication [1 mark]:** - Award 1 mark if the overall response is written in a clear, structured, and coherent manner using precise biological terms (e.g., chemoreceptors, medulla oblongata, vagus nerve, gastrin, etc.).
Paper 3 Section A
Answer all questions based on experimental techniques and data.
3 Question · 15 marks
Question 1 · Short Answer & Calculation
5 marks
A group of students investigated the effect of light intensity on the rate of photosynthesis in *Chlorella* cells. They used a dissolved oxygen sensor to measure the rate of net oxygen production. The table below displays their results:
(a) Identify the rate of cellular respiration of *Chlorella* in this experiment, stating the correct units. [1] (b) Calculate the gross rate of photosynthesis at a light intensity of 200 \(\text{\mu mol m}^{-2}\text{s}^{-1}\). Show your working. [2] (c) Explain the relationship between light intensity and the rate of oxygen production between 400 and 800 \(\text{\mu mol m}^{-2}\text{s}^{-1}\). [2]
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Worked solution
(a) At a light intensity of 0 \(\text{\mu mol m}^{-2}\text{s}^{-1}\), no photosynthesis occurs. The net oxygen production rate is -2.1 \(\text{\mu mol O}_2\text{ dm}^{-3}\text{ min}^{-1}\). This value represents the rate of oxygen consumption due to cellular respiration, which is 2.1 \(\text{\mu mol O}_2\text{ dm}^{-3}\text{ min}^{-1}\).
(c) Between 400 and 800 \(\text{\mu mol m}^{-2}\text{s}^{-1}\), the rate increases only marginally from 8.5 to 8.6, indicating that the photosynthetic machinery is operating at maximum capacity under these conditions. Light intensity is no longer the factor that limits the rate of photosynthesis. Other environmental factors, such as ambient temperature or carbon dioxide concentration, are now the limiting factors.
Marking scheme
(a) Award [1] mark for the correct rate and units: - 2.1 \(\text{\mu mol O}_2\text{ dm}^{-3}\text{ min}^{-1}\) (accept -2.1).
(b) Award [2] marks in total: - [1] mark for correct working shown: \(4.8 + 2.1\) or \(4.8 - (-2.1)\). - [1] mark for correct final answer: 6.9 \(\text{\mu mol O}_2\text{ dm}^{-3}\text{ min}^{-1}\) (accept unit-free answer if unit was graded in part a, but reject if incorrect units are written here).
(c) Award [2] marks in total: - [1] mark for stating that the rate of oxygen production plateaus/reaches a maximum/remains virtually constant. - [1] mark for explaining that light intensity is no longer the limiting factor / another factor (such as carbon dioxide concentration or temperature) is now limiting.
Question 2 · Short Answer & Calculation
5 marks
An investigation was conducted to determine the osmolarity of potato tissue. Cylinders of potato were weighed and placed in sucrose solutions of various concentrations. After 24 hours, the cylinders were reweighed and the percentage change in mass was calculated:
(a) Estimate the sucrose concentration that is isotonic to the potato tissue cytoplasm, and explain how you determined this. [2] (b) Explain the change in mass of the potato cylinders in the 0.0 \(\text{mol dm}^{-3}\) sucrose solution. [2] (c) Suggest one standard procedure used to ensure the mass measurements of the potato cylinders were accurate and comparable before and after immersion. [1]
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Worked solution
(a) The isotonic concentration is where there is no net movement of water, which corresponds to 0% change in mass. Looking at the data, the mass increases by +5.1% at 0.2 \(\text{mol dm}^{-3}\) and decreases by -1.2% at 0.3 \(\text{mol dm}^{-3}\). By linear interpolation, 0% change occurs at approximately 0.28 \(\text{mol dm}^{-3}\) (accept any value in the range of 0.27 to 0.29 \(\text{mol dm}^{-3}\)).
(b) Pure water (0.0 \(\text{mol dm}^{-3}\)) has a higher water potential (less negative) than the cytoplasm of the potato cells. Consequently, water moves down the water potential gradient from the surrounding solution into the potato cells via osmosis. This entry of water increases the turgor and mass of the tissues.
(c) Before weighing, each potato cylinder must be gently blotted with a paper towel. This ensures that any excess liquid adhering to the external surface of the potato is removed, preventing it from artificially inflating the mass measurement.
Marking scheme
(a) Award [2] marks in total: - [1] mark for correct estimated concentration: 0.28 \(\text{mol dm}^{-3}\) (accept range 0.27 to 0.29 \(\text{mol dm}^{-3}\)). - [1] mark for explaining that this is the point where there is no net water movement / 0% change in mass.
(b) Award [2] marks in total: - [1] mark for stating that the surrounding solution (pure water) has a higher water potential than the potato cell cytoplasm. - [1] mark for stating that water moves into the cells by osmosis, increasing their mass.
(c) Award [1] mark for any of the following standard procedures: - Blotting the potato cylinders dry / removing surface water with a paper towel before weighing. - Using potato cylinders cut to the same initial length/diameter (surface area to volume ratio). - Ensuring the cylinders are completely submerged in the solution.
Question 3 · Short Answer & Calculation
5 marks
To estimate the population size of a species of beetle (*Tribolium castaneum*) in a grain storage facility, a biologist used the capture-mark-recapture method. - First sample size (\(n_1\)): 120 beetles were captured, marked with non-toxic paint, and released. - Second sample size (\(n_2\)): 150 beetles were captured after a 48-hour period. - Recaptured marked beetles (\(nm\)): 30 of the beetles in the second sample were found to have the mark.
(a) Calculate the estimated population size of the beetles. Show your working. [2] (b) Outline two conditions that must be met during the 48-hour period for this method to yield a valid population estimate. [2] (c) State how the reliability of this population estimate could be improved. [1]
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Worked solution
(a) The Lincoln Index formula is: \(N = \frac{n_1 \times n_2}{nm}\) Where: - \(n_1\) = 120 - \(n_2\) = 150 - \(nm\) = 30
(b) For the estimate to be valid, several assumptions must hold true over the 48 hours: 1. The population is closed: there is no significant birth, death, immigration, or emigration. 2. The marked beetles mix randomly with the unmarked individuals before the second sample is taken. 3. The marking technique does not affect the survival rate of the beetles (e.g., does not make them more visible to predators or toxic to them) or alter their behavior. 4. Marks are permanent and do not rub off.
(c) Reliability can be improved by repeating the capture-mark-recapture process multiple times to obtain a mean value, or by capturing larger sample sizes in both the first and second events to minimize sampling error.
Marking scheme
(a) Award [2] marks in total: - [1] mark for showing the correct working: \(\frac{120 \times 150}{30}\). - [1] mark for the correct final answer: 600 (beetles).
(b) Award [2] marks for any two of the following conditions outlined: - No migration (no immigration or emigration) / closed population. - No births or deaths during the study period. - The mark does not affect survival / does not increase predation risk. - The mark does not wear off or fade. - Marked individuals disperse/mix randomly back into the population. - All individuals have an equal chance of being caught in both samples.
(c) Award [1] mark for: - Repeating the entire procedure and calculating a mean. - Increasing the size of the samples captured (larger \(n_1\) and/or \(n_2\)).
Paper 3 Section B
Answer all questions from one Option of your choice (A, B, C, or D).
6 Question · 30 marks
Question 1 · extended answer
5 marks
Explain how the nervous system and endocrine system coordinate to increase cardiac output during strenuous physical exercise.
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Worked solution
1. **Chemoreceptor detection**: Chemoreceptors in the carotid and aortic bodies detect a drop in blood pH caused by increased \(CO_2\) concentration during exercise. 2. **Medulla oblongata integration**: Sensory impulses are sent to the cardiovascular control centre in the medulla oblongata. 3. **Sympathetic nervous system stimulation**: The cardiovascular centre sends impulses along sympathetic motor neurons (specifically, the accelerator nerve) to the sinoatrial (SA) node. 4. **Neurotransmitter release**: Noradrenaline (norepinephrine) is released at the SA node, increasing the rate of depolarization and thus the heart rate. 5. **Endocrine contribution**: Adrenaline (epinephrine) is released from the adrenal medulla into the bloodstream, acting directly on the SA node and cardiac muscle to increase both heart rate and stroke volume (force of contraction).
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - Detection of low blood pH / high carbon dioxide levels by chemoreceptors (in aorta/carotid arteries). [1] - Signals sent to the cardiovascular control centre in the medulla oblongata. [1] - Sympathetic nerves (accelerator nerve) send impulses to the sinoatrial (SA) node. [1] - Release of neurotransmitter (noradrenaline) increases the heart rate/frequency of SA node firing. [1] - Adrenal glands secrete epinephrine/adrenaline into the bloodstream. [1] - Epinephrine binds to receptors on the SA node/cardiac muscle to increase heart rate and force of contraction (stroke volume). [1]
Question 2 · extended answer
5 marks
Describe the physiological and behavioral responses coordinated by the hypothalamus to restore normal core body temperature when a human is exposed to extreme cold.
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Worked solution
1. **Detection**: Hypothalamus acts as a thermostat, detecting a fall in blood temperature and receiving sensory input from skin thermoreceptors. 2. **Vasoconstriction**: Sympathetic signals cause vasoconstriction of arterioles supplying the skin capillaries, diverting warm blood away from the body surface to reduce convective and radiative heat loss. 3. **Shivering**: Involuntary, rapid contractions of skeletal muscles (shivering) are triggered, converting chemical energy from ATP into thermal energy to generate heat. 4. **Metabolic increase**: The hypothalamus stimulates the release of thyroxine (via TSH) and adrenaline, which increase the basal metabolic rate, generating more heat. 5. **Behavioral adjustments**: The cerebral cortex is stimulated to drive behavioral actions (e.g., huddling, seeking shelter, or putting on additional clothing).
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - Hypothalamus acts as the control centre / receives input from skin thermoreceptors / monitors blood temperature. [1] - Vasoconstriction of arterioles supplying skin capillaries reduces heat loss by radiation/conduction. [1] - Shivering / involuntary muscle contraction is stimulated to generate heat through respiration. [1] - Increase in metabolic rate/cell respiration mediated by hormones (thyroxine/adrenaline). [1] - Piloerection (erector pili muscles contract) traps a layer of insulating air. [1] - Behavioral responses (e.g., putting on clothes/moving to warm areas) are triggered via cerebral cortex stimulation. [1]
Question 3 · short answer
5 marks
Distinguish between 'top-down' and 'bottom-up' control of populations in a community, using named examples to illustrate your answer.
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Worked solution
1. **Bottom-up definition**: Population sizes are limited by the availability of resources (e.g., nutrients, water, light) at the base of the food web. 2. **Bottom-up example**: In aquatic ecosystems, nitrogen or phosphorus runoff causes algal blooms, which increases zooplankton populations, and subsequently higher trophic levels. 3. **Top-down definition**: Populations at lower trophic levels are regulated by predation, herbivory, or disease from higher trophic levels. 4. **Top-down example**: In a kelp forest ecosystem, sea otters (top predators) control sea urchin populations, preventing them from overgrazing and destroying the kelp forest. 5. **Key difference**: Bottom-up affects trophic levels upwards through energy flow, whereas top-down control exerts pressure downwards from consumers to producers.
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - Bottom-up control is driven by the availability of resources/nutrients/primary producers. [1] - Example of bottom-up control: e.g., nutrient enrichment (nitrogen/phosphorus) increasing phytoplankton and subsequently fish abundance, OR water/nutrients limiting plant biomass in grasslands. [1] - Top-down control is driven by predation/herbivory from higher trophic levels. [1] - Example of top-down control: e.g., wolves controlling deer populations which allows vegetation to regenerate, OR sea otters preying on sea urchins to preserve kelp forests. [1] - Clear statement of contrast (e.g., bottom-up regulates from the base of the food chain upwards, whereas top-down regulates from the apex downwards). [1]
Question 4 · extended answer
5 marks
Explain how a researcher can use a chi-squared (\(\chi^2\)) test to determine whether two genes in a dihybrid cross are linked or assorting independently.
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Worked solution
1. **Null Hypothesis**: State the null hypothesis (\(H_0\)) that the two genes assort independently (i.e., they are not linked). 2. **Expected Ratio**: For a test cross of a dihybrid (e.g., \(AaBb \times aabb\)), the expected Mendelian ratio of phenotypes is \(1:1:1:1\) (or \(9:3:3:1\) for an \(F_2\) self-cross). 3. **Calculation**: Calculate expected values (\(E\)) based on this ratio, then use the formula \(\chi^2 = \sum \frac{(O - E)^2}{E}\) where \(O\) represents observed offspring counts. 4. **Degrees of Freedom**: Determine the degrees of freedom (\(df = \text{number of classes} - 1\)), which is 3 for a dihybrid test cross. 5. **Comparison and Conclusion**: Compare the calculated \(\chi^2\) value to the critical value at a significance level of \(p = 0.05\). If the calculated value exceeds the critical value, reject the null hypothesis, demonstrating that the genes are linked (located on the same chromosome).
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - State the null hypothesis (\(H_0\)): there is no link between the genes / they assort independently. [1] - Calculate the expected phenotypic ratio: 1:1:1:1 for a test cross (or 9:3:3:1 for selfing of double heterozygotes). [1] - Use of the chi-squared formula: \(\chi^2 = \sum \frac{(O-E)^2}{E}\) (with terms clearly defined). [1] - Identify correct degrees of freedom: \(df = 3\) for four phenotypic classes. [1] - Comparison with critical value: if calculated \(\chi^2 >\) critical value (at \(p = 0.05\)), reject the null hypothesis / conclude genes are linked. [1]
Question 5 · extended answer
5 marks
Explain the mechanism of translocation of organic solutes in the phloem of angiosperms, including the role of active transport and hydrostatic pressure.
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Worked solution
1. **Active Loading**: Sucrose is actively loaded into the companion cells and sieve tube elements at the source (e.g., leaves) against its concentration gradient using energy from ATP (specifically via a proton gradient and \(H^+\)/sucrose cotransporters). 2. **Osmosis at Source**: The high concentration of solute inside the sieve tube decreases the solute potential (and thus water potential), causing water to flow from the adjacent xylem into the phloem by osmosis. 3. **Hydrostatic Pressure Generation**: The entry of water into the non-elastic sieve tube walls increases hydrostatic pressure at the source end. 4. **Unloading and Osmosis at Sink**: At the sink (e.g., roots, fruits), sucrose is unloaded (actively or passively), which increases water potential in the phloem, causing water to exit back into the xylem. 5. **Mass Flow**: This creates a hydrostatic pressure gradient between the source (high pressure) and sink (low pressure), driving the bulk/mass flow of phloem sap along the sieve tube.
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - Active transport / active loading of sucrose/solutes into companion cells/sieve tubes at the source. [1] - Proton pumps create an \(H^+\) gradient to co-transport sucrose into phloem. [1] - High solute concentration lowers water potential, causing water to enter phloem from xylem by osmosis. [1] - Influx of water increases hydrostatic pressure at the source. [1] - At the sink, sucrose unloading increases water potential, causing water to leave the phloem (lowering hydrostatic pressure). [1] - A pressure gradient is established, causing mass flow/bulk flow of organic solutes from source to sink. [1]
Question 6 · extended answer
5 marks
Explain how the renin-angiotensin-aldosterone system (RAAS) responds to a sudden drop in blood pressure to restore cardiovascular homeostasis.
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Worked solution
1. **Renin release**: A drop in blood pressure or blood volume is detected by juxtaglomerular cells in the kidneys, which secrete the enzyme renin into the bloodstream. 2. **Angiotensin I formation**: Renin acts enzymatically on the circulating plasma protein angiotensinogen (produced by the liver) to convert it into angiotensin I. 3. **Angiotensin II conversion**: Angiotensin-converting enzyme (ACE) in the lungs converts angiotensin I to angiotensin II. 4. **Vasoconstriction**: Angiotensin II is a potent vasoconstrictor, causing systemic arterioles to constrict, which directly increases total peripheral resistance and blood pressure. 5. **Aldosterone action**: Angiotensin II also stimulates the adrenal cortex to secrete aldosterone, which increases \(Na^+\) reabsorption (and subsequent water reabsorption by osmosis) in the collecting ducts and distal tubules of the kidney, raising blood volume and pressure.
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - Low blood pressure/volume stimulates the kidneys to secrete renin. [1] - Renin converts angiotensinogen into angiotensin I. [1] - Angiotensin-converting enzyme (ACE) converts angiotensin I into angiotensin II (in the lungs). [1] - Angiotensin II causes vasoconstriction of arterioles, directly raising blood pressure. [1] - Angiotensin II stimulates the adrenal cortex to secrete aldosterone. [1] - Aldosterone increases sodium (and water) reabsorption in the distal tubules/collecting ducts, increasing blood volume. [1]
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