IB · BIOLOGY HL

Biology HL

Theme C: Interaction and interdependence (HL) — Theme C HL

Name: ____________________Date: October 25, 2026

1.
Optional kinetic-model extension, not claimed as prescribed IB HL depth: Explain the difference between reversible competitive inhibition in the standard active-site model and pure noncompetitive enzyme inhibition, referring to substrate binding and catalytic activity.
[4 marks] · no calculator
Course scope: Optional enrichment
Syllabus placement evidence: Biology guide, March 2024 update, C1.1.14-C1.1.15, printed p.66. The complete prescribed section covers inhibition mechanisms and effects of substrate concentration, but not recall of pure-model Km/Vmax or equal enzyme/enzyme-substrate affinity. Those additional kinetic distinctions are conservatively offered as enrichment, not a board exclusion ruling or prescribed HL coverage. Editorial placement judgment from the course guidance, not an exam-board ruling.
Worked explanation
1. In the standard competitive model, substrate and inhibitor cannot occupy the active site together. More substrate can increasingly outcompete inhibitor, so the same Vmax is approached with a higher apparent Km.
2. A pure noncompetitive inhibitor binds free enzyme and enzyme-substrate complex equally at another site. Substrate may still bind, but catalytic capacity falls.
3. Its lower Vmax is not restored by increasing substrate, while Km is unchanged in the pure model. A remote binding site alone could also produce mixed inhibition, so location is insufficient to establish the kinetic class.
Common mistake: Pure noncompetitive inhibition does not require preventing substrate binding; its defining effects concern catalytic capacity and equal affinity for E and ES.
Marking points
- States that in the standard competitive model the inhibitor and substrate bind mutually exclusively at the active site; the inhibitor need not be a close structural copy of the substrate.
- States that increasing substrate concentration can overcome competitive inhibition, since substrate and inhibitor compete for the same active site.
- States that a pure noncompetitive inhibitor binds free enzyme and enzyme-substrate complex with equal affinity at a site distinct from the substrate-binding site.
- Explains that it reduces catalytic capacity and Vmax without changing Km, so increasing substrate concentration cannot restore the uninhibited maximum rate; substrate binding is not necessarily prevented.
Examiner tip: Competitive inhibition increases apparent Km while leaving Vmax unchanged; pure noncompetitive inhibition lowers Vmax without changing Km. A remote binding site alone does not distinguish pure noncompetitive from mixed inhibition.
2.
Marking analysis: A learner attempts the following task: “Optional kinetic-model extension, not claimed as prescribed IB HL depth: Explain the difference between reversible competitive inhibition in the standard active-site model and pure noncompetitive enzyme inhibition, referring to substrate binding and catalytic activity.” Their response addresses only this point: “States that in the standard competitive model the inhibitor and substrate bind mutually exclusively at the active site; the inhibitor need not be a close structural copy of the substrate.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that in the standard competitive model the inhibitor and substrate bind mutually exclusively at the active site; the inhibitor need not be a close structural copy of the substrate.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that increasing substrate concentration can overcome competitive inhibition, since substrate and inhibitor compete for the same active site.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a pure noncompetitive inhibitor binds free enzyme and enzyme-substrate complex with equal affinity at a site distinct from the substrate-binding site.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that it reduces catalytic capacity and Vmax without changing Km, so increasing substrate concentration cannot restore the uninhibited maximum rate; substrate binding is not necessarily prevented.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
3.
Outline how the rate of an enzyme-catalysed reaction changes with increasing substrate concentration, at a fixed enzyme concentration, and explain why it eventually plateaus.
[3 marks] · no calculator
Worked explanation
1. At low substrate concentration many active sites are unoccupied, so additional substrate increases productive encounters and initial reaction rate.
2. At high substrate concentration the enzyme spends most of its time cycling with bound substrate. With enzyme quantity and conditions fixed, turnover capacity becomes limiting.
3. The rate approaches a plateau rather than rising indefinitely. More active enzyme can raise that plateau; more substrate alone cannot remove enzyme saturation.
Common mistake: The plateau does not mean the reaction has stopped; it continues near its maximal rate.
Marking points
- States that at low substrate concentration, the rate of reaction increases roughly proportionally with substrate concentration.
- States that at high substrate concentration, the rate of reaction plateaus and becomes independent of further increases in substrate concentration.
- Explains that the plateau occurs because all available active sites become saturated (occupied) with substrate, so the fixed amount of enzyme becomes the limiting factor.
Examiner tip: At fixed temperature, pH and other conditions, increasing substrate alone cannot raise the saturation plateau. More active enzyme raises Vmax; changing conditions can also alter catalytic activity, so enzyme quantity is not the only possible way to change Vmax.
4.
Marking analysis: A learner attempts the following task: “Outline how the rate of an enzyme-catalysed reaction changes with increasing substrate concentration, at a fixed enzyme concentration, and explain why it eventually plateaus.” Their response addresses only this point: “States that at low substrate concentration, the rate of reaction increases roughly proportionally with substrate concentration.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that at low substrate concentration, the rate of reaction increases roughly proportionally with substrate concentration.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that at high substrate concentration, the rate of reaction plateaus and becomes independent of further increases in substrate concentration.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that the plateau occurs because all available active sites become saturated (occupied) with substrate, so the fixed amount of enzyme becomes the limiting factor.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
5.
Outline the role of the electron transport chain and chemiosmosis in producing ATP during oxidative phosphorylation.
[4 marks] · no calculator
Worked explanation
1. NADH and FADH2 donate electrons to the inner-mitochondrial-membrane respiratory chain. Energy released by electron transfer powers proton pumping from matrix to intermembrane space.
2. The membrane separates a proton electrochemical gradient. Protons return to the matrix through ATP synthase, coupling downhill flow to ADP + phosphate forming ATP.
3. Oxygen accepts electrons at the chain's end and, with protons, forms water. Without that outlet, electron flow and maintenance of the gradient fail.
Common mistake: ATP synthase does not pump protons outward in this ATP-producing mode; their return drives synthesis.
Marking points
- States that electrons from NADH and FADH₂ are passed along a series of protein carriers (the electron transport chain) in the inner mitochondrial membrane.
- States that energy released as electrons move along the chain is used to actively pump H⁺ ions across the inner membrane, creating a proton gradient (higher H⁺ concentration in the intermembrane space).
- States that H⁺ ions flow back down this gradient through ATP synthase (chemiosmosis), and this flow drives the synthesis of ATP from ADP and inorganic phosphate.
- States that oxygen acts as the final electron acceptor at the end of the chain, combining with electrons and H⁺ to form water.
Examiner tip: Chemiosmosis is the general mechanism (a proton gradient driving ATP synthase) shared by both oxidative phosphorylation in mitochondria and photophosphorylation in chloroplasts.
6.
Marking analysis: A learner attempts the following task: “Outline the role of the electron transport chain and chemiosmosis in producing ATP during oxidative phosphorylation.” Their response addresses only this point: “States that electrons from NADH and FADH₂ are passed along a series of protein carriers (the electron transport chain) in the inner mitochondrial membrane.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that electrons from NADH and FADH₂ are passed along a series of protein carriers (the electron transport chain) in the inner mitochondrial membrane.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that energy released as electrons move along the chain is used to actively pump H⁺ ions across the inner membrane, creating a proton gradient (higher H⁺ concentration in the intermembrane space).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that H⁺ ions flow back down this gradient through ATP synthase (chemiosmosis), and this flow drives the synthesis of ATP from ADP and inorganic phosphate.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that oxygen acts as the final electron acceptor at the end of the chain, combining with electrons and H⁺ to form water.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
7.
Explain why oxygen is essential for aerobic cell respiration to continue, in terms of its role at the end of the electron transport chain.
[3 marks] · no calculator
Worked explanation
1. Oxygen receives electrons and protons at the respiratory chain's end, forming water and allowing upstream carriers to be reoxidised.
2. Without oxygen, carriers remain reduced and electron transfer/proton pumping cannot continue normally. The proton gradient is no longer sustained for most aerobic ATP production.
3. NADH and FADH2 are not efficiently reoxidised, restricting reactions that need NAD+ or FAD, including the link reaction and Krebs cycle. Fermentation can regenerate some NAD+ for glycolysis but does not replace oxidative phosphorylation.
Common mistake: Oxygen is not consumed directly by the Krebs cycle; its terminal-acceptor role enables the necessary upstream cofactor recycling.
Marking points
- States that oxygen acts as the final electron acceptor at the end of the electron transport chain, combining with electrons and H⁺ ions to form water.
- States that without oxygen to accept electrons, the electron transport chain becomes blocked (electrons have nowhere to go) and stops functioning.
- States that this in turn prevents regeneration of NAD+ and FAD from NADH and FADH₂ via the chain, which stops the Krebs cycle and the link reaction, halting most ATP production from aerobic respiration.
Examiner tip: Oxygen's role is indirect but critical: it does not directly participate in ATP synthesis itself, but without it as the final electron acceptor, the entire electron transport chain (and the ATP synthesis it drives) grinds to a halt.
8.
Marking analysis: A learner attempts the following task: “Explain why oxygen is essential for aerobic cell respiration to continue, in terms of its role at the end of the electron transport chain.” Their response addresses only this point: “States that oxygen acts as the final electron acceptor at the end of the electron transport chain, combining with electrons and H⁺ ions to form water.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that oxygen acts as the final electron acceptor at the end of the electron transport chain, combining with electrons and H⁺ ions to form water.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that without oxygen to accept electrons, the electron transport chain becomes blocked (electrons have nowhere to go) and stops functioning.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this in turn prevents regeneration of NAD+ and FAD from NADH and FADH₂ via the chain, which stops the Krebs cycle and the link reaction, halting most ATP production from aerobic respiration.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
9.
Outline the light-dependent reactions of photosynthesis, including the roles of water and the products formed.
[4 marks] · no calculator
Worked explanation
1. Light absorbed by thylakoid pigments excites electrons in photosystems. Water splitting at photosystem II replaces lost electrons and releases oxygen and protons.
2. Electron transfer supports proton accumulation in the thylakoid lumen; return to the stroma through ATP synthase produces ATP.
3. Further light excitation supports reduction of NADP+ to NADPH. ATP and NADPH then supply energy and reducing power for carbon fixation, while oxygen is a water-splitting by-product.
Common mistake: The released oxygen comes from water, not from splitting carbon dioxide in the Calvin cycle.
Marking points
- States that light energy is absorbed by chlorophyll and other pigments in the thylakoid membrane, exciting electrons.
- States that water molecules are split (photolysis) to replace the excited electrons lost from chlorophyll, releasing oxygen as a by-product.
- States that the excited electrons pass along an electron transport chain in the thylakoid membrane, driving chemiosmotic ATP synthesis.
- States that NADP+ is reduced to NADPH using the energized electrons, and both ATP and NADPH are produced as products used in the light-independent reactions.
Examiner tip: ATP and NADPH supply the Calvin cycle; oxygen is released by water splitting and is not a Calvin-cycle input. Do not extend this to saying oxygen has no other role in illuminated plant metabolism: respiration and photorespiration can consume it.
10.
Marking analysis: A learner attempts the following task: “Outline the light-dependent reactions of photosynthesis, including the roles of water and the products formed.” Their response addresses only this point: “States that light energy is absorbed by chlorophyll and other pigments in the thylakoid membrane, exciting electrons.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that light energy is absorbed by chlorophyll and other pigments in the thylakoid membrane, exciting electrons.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that water molecules are split (photolysis) to replace the excited electrons lost from chlorophyll, releasing oxygen as a by-product.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the excited electrons pass along an electron transport chain in the thylakoid membrane, driving chemiosmotic ATP synthesis.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that NADP+ is reduced to NADPH using the energized electrons, and both ATP and NADPH are produced as products used in the light-independent reactions.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
11.
Outline the light-independent reactions (Calvin cycle) of photosynthesis, including the roles of ATP, NADPH and carbon dioxide.
[3 marks] · no calculator
Worked explanation
1. Rubisco adds carbon dioxide to five-carbon RuBP; the unstable six-carbon product splits into three-carbon compounds.
2. ATP supplies energy and NADPH reducing power to convert these into triose phosphate.
3. Most triose phosphate is used to regenerate RuBP so fixation can continue; a smaller net output supplies sugars and other organic molecules. The cycle uses no photons directly but depends on light-reaction products.
Common mistake: Calling these dark reactions does not mean they normally run all night independently of ATP and NADPH supply.
Marking points
- States that carbon dioxide is fixed by combining with a five-carbon acceptor molecule (ribulose bisphosphate, RuBP), catalysed by the enzyme rubisco.
- States that ATP and NADPH (from the light-dependent reactions) are used to reduce the resulting three-carbon compound into a sugar (e.g. G3P/triose phosphate).
- States that some of the triose phosphate produced is used to regenerate RuBP, allowing the cycle to continue, while the rest is used to synthesize glucose and other organic molecules.
Examiner tip: The Calvin cycle depends entirely on the ATP and NADPH supplied by the light-dependent reactions; it cannot proceed in the dark for long, despite the older name 'dark reactions'.
12.
Marking analysis: A learner attempts the following task: “Outline the light-independent reactions (Calvin cycle) of photosynthesis, including the roles of ATP, NADPH and carbon dioxide.” Their response addresses only this point: “States that carbon dioxide is fixed by combining with a five-carbon acceptor molecule (ribulose bisphosphate, RuBP), catalysed by the enzyme rubisco.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that carbon dioxide is fixed by combining with a five-carbon acceptor molecule (ribulose bisphosphate, RuBP), catalysed by the enzyme rubisco.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that ATP and NADPH (from the light-dependent reactions) are used to reduce the resulting three-carbon compound into a sugar (e.g. G3P/triose phosphate).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that some of the triose phosphate produced is used to regenerate RuBP, allowing the cycle to continue, while the rest is used to synthesize glucose and other organic molecules.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
13.
Explain the effect of increasing light intensity on the rate of photosynthesis, and why the rate eventually plateaus even as light intensity continues to increase.
[3 marks] · no calculator
Worked explanation
1. On the rising part of the stated response, extra light increases photochemical energy supply, so light is limiting.
2. At the plateau, more light alone no longer raises rate. Carbon dioxide supply or temperature-dependent reaction capacity may now constrain carbon fixation.
3. To distinguish those possibilities, vary carbon dioxide at fixed high light and temperature, or vary temperature separately. The curve alone does not identify which alternative factor is limiting.
Common mistake: A plateau does not mean light has become unnecessary or that photosynthesis has stopped.
Marking points
- States that at low light intensity, the rate of photosynthesis increases as light intensity increases, since light is the limiting factor.
- States that at high light intensity, the rate plateaus and further increases in light intensity have no additional effect.
- Explains that at this point, another factor (e.g. carbon dioxide concentration or temperature) has become limiting instead of light, so light is no longer the factor restricting the rate.
Examiner tip: This is an application of the principle of limiting factors: the rate of a process is always constrained by whichever factor is currently in shortest supply relative to the others.
14.
Marking analysis: A learner attempts the following task: “Explain the effect of increasing light intensity on the rate of photosynthesis, and why the rate eventually plateaus even as light intensity continues to increase.” Their response addresses only this point: “States that at low light intensity, the rate of photosynthesis increases as light intensity increases, since light is the limiting factor.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that at low light intensity, the rate of photosynthesis increases as light intensity increases, since light is the limiting factor.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that at high light intensity, the rate plateaus and further increases in light intensity have no additional effect.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that at this point, another factor (e.g. carbon dioxide concentration or temperature) has become limiting instead of light, so light is no longer the factor restricting the rate.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
15.
Outline why fermentation in human muscle or yeast yields much less ATP per glucose than aerobic respiration, referring to the electron transport chain.
[3 marks] · no calculator
Worked explanation
1. Glycolysis generates a small ATP yield and reduces NAD+. Fermentation reoxidises NADH by reducing pyruvate or a derivative, allowing glycolysis to continue.
2. This does not send those electrons through a respiratory chain to drive oxidative phosphorylation, so that major aerobic ATP contribution is absent.
3. The net yield remains about 2 ATP per glucose, with much chemical energy retained in lactate or ethanol. This describes fermentation, not microbial anaerobic respiration using a different terminal electron acceptor.
Common mistake: Fermentation is not a low-efficiency oxygen-powered respiratory chain; it regenerates NAD+ without that chain.
Marking points
- States that fermentation regenerates NAD+ by reducing pyruvate or a derivative, allowing glycolysis to continue without using a respiratory electron transport chain.
- States that the oxidative-phosphorylation ATP yield associated with the aerobic respiratory chain is therefore absent.
- States that only the small amount of ATP produced directly during glycolysis (substrate-level phosphorylation) is obtained, giving a much lower net ATP yield overall.
Examiner tip: Here fermentation is meant, sometimes called anaerobic respiration in school terminology. Some microorganisms perform anaerobic respiration with an electron transport chain and a final electron acceptor other than oxygen; that is not the process described here.
16.
Marking analysis: A learner attempts the following task: “Outline why fermentation in human muscle or yeast yields much less ATP per glucose than aerobic respiration, referring to the electron transport chain.” Their response addresses only this point: “States that fermentation regenerates NAD+ by reducing pyruvate or a derivative, allowing glycolysis to continue without using a respiratory electron transport chain.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that fermentation regenerates NAD+ by reducing pyruvate or a derivative, allowing glycolysis to continue without using a respiratory electron transport chain.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the oxidative-phosphorylation ATP yield associated with the aerobic respiratory chain is therefore absent.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that only the small amount of ATP produced directly during glycolysis (substrate-level phosphorylation) is obtained, giving a much lower net ATP yield overall.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
17.
Explain why an increase in temperature, up to the optimum, increases both enzyme reaction rate and the rate of respiration.
[3 marks] · no calculator
Worked explanation
1. Warming within the functional range increases enzyme/substrate molecular motion and the frequency of effective collisions.
2. Respiration comprises enzyme-catalysed steps, so faster productive encounters can raise its overall rate when other requirements are available.
3. Above the observed optimum, structural damage and other heat stress can outweigh that gain. The optimum is specific to enzyme systems and conditions, not a universal temperature.
Common mistake: Heating does not increase the number of enzyme molecules in this controlled comparison; it changes encounters and eventually structure.
Marking points
- States that increasing temperature increases the kinetic energy of both enzyme and substrate molecules, causing them to move faster.
- States that this increases the frequency of successful collisions between enzyme active sites and substrate molecules.
- States that since respiration is controlled by a sequence of enzyme-catalysed reactions, an increased collision/reaction rate for these enzymes increases the overall rate of respiration, up to the point where the enzymes begin to denature.
Examiner tip: Beyond the optimum temperature, the rate of respiration falls sharply because the enzymes controlling its reactions denature — the same principle covered for general enzyme-substrate kinetics applies directly here.
18.
Marking analysis: A learner attempts the following task: “Explain why an increase in temperature, up to the optimum, increases both enzyme reaction rate and the rate of respiration.” Their response addresses only this point: “States that increasing temperature increases the kinetic energy of both enzyme and substrate molecules, causing them to move faster.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that increasing temperature increases the kinetic energy of both enzyme and substrate molecules, causing them to move faster.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this increases the frequency of successful collisions between enzyme active sites and substrate molecules.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that since respiration is controlled by a sequence of enzyme-catalysed reactions, an increased collision/reaction rate for these enzymes increases the overall rate of respiration, up to the point where the enzymes begin to denature.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
19.
Compare the location within the chloroplast of the light-dependent reactions and the light-independent reactions of photosynthesis.
[2 marks] · no calculator
Worked explanation
1. Light-dependent electron transport and photophosphorylation occur at the thylakoid membranes, which separate lumen from stroma for proton-gradient formation.
2. The Calvin cycle occurs in the stroma, where enzymes use ATP and NADPH to fix carbon dioxide and regenerate RuBP.
3. The two locations are functionally coupled: membrane products supply stromal reactions, rather than the two stages operating as unrelated processes.
Common mistake: The stroma is the fluid around thylakoids, not their internal lumen.
Marking points
- States that the light-dependent reactions occur on/in the thylakoid membranes.
- States that the light-independent reactions (Calvin cycle) occur in the stroma, the fluid surrounding the thylakoids.
Examiner tip: This spatial separation mirrors the mitochondrion, where the electron transport chain is membrane-bound (cristae) and the Krebs cycle occurs in the surrounding fluid (matrix).
20.
Marking analysis: A learner attempts the following task: “Compare the location within the chloroplast of the light-dependent reactions and the light-independent reactions of photosynthesis.” Their response addresses only this point: “States that the light-dependent reactions occur on/in the thylakoid membranes.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that the light-dependent reactions occur on/in the thylakoid membranes.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the light-independent reactions (Calvin cycle) occur in the stroma, the fluid surrounding the thylakoids.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
21.
Explain how auxin controls the phototropic response of a shoot growing towards a unidirectional light source.
[4 marks] · no calculator
Worked explanation
1. Unidirectional light leads to unequal auxin distribution near the shoot tip, with more auxin on the shaded side.
2. In shoots, that greater auxin signal promotes greater cell elongation on the shaded side.
3. Unequal growth makes the shaded side longer and bends the tip toward light. Equal elongation on both sides would lengthen the shoot without the phototropic bend.
Common mistake: The side away from light elongates more; saying the illuminated side grows fastest predicts bending in the wrong direction.
Marking points
- States that auxin is produced in the shoot tip and is transported laterally away from the light towards the shaded side.
- States that this creates a higher concentration of auxin on the shaded side of the shoot than the illuminated side.
- States that auxin stimulates cell elongation, so cells on the shaded side elongate more than cells on the illuminated side.
- States that this differential growth bends the shoot towards the light source (positive phototropism).
Examiner tip: The mechanism depends on unequal lateral distribution of auxin, not simply 'more auxin overall' — always frame the answer as a comparison between the two sides.
22.
Marking analysis: A learner attempts the following task: “Explain how auxin controls the phototropic response of a shoot growing towards a unidirectional light source.” Their response addresses only this point: “States that auxin is produced in the shoot tip and is transported laterally away from the light towards the shaded side.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that auxin is produced in the shoot tip and is transported laterally away from the light towards the shaded side.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this creates a higher concentration of auxin on the shaded side of the shoot than the illuminated side.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that auxin stimulates cell elongation, so cells on the shaded side elongate more than cells on the illuminated side.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this differential growth bends the shoot towards the light source (positive phototropism).
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
23.
Outline the role of gibberellin in breaking seed dormancy and initiating germination in cereal seeds.
[3 marks] · no calculator
Course scope: Optional enrichment
Syllabus placement evidence: C3.1.19-C3.1.23 (AHL), printed pp.77-78; D2.2.11, p.93. General phytohormones and a hormone effect on expression are prescribed. This prompt requires recall of the cereal gibberellin-aleurone-amylase mechanism without supplying it; that named mechanism exceeds the listed auxin/cytokinin/ethylene examples.
Worked explanation
1. Water uptake activates the cereal embryo, which releases gibberellin toward the aleurone layer.
2. The hormone induces expression and production of hydrolytic enzymes including amylase; it does not itself digest starch.
3. Amylase breaks down endosperm starch, making sugars available for embryo respiration and growth before photosynthetic leaves are established.
Common mistake: Gibberellin is a signal, not the enzyme catalysing starch hydrolysis.
Marking points
- States that water uptake by the seed triggers the embryo to release gibberellin.
- States that gibberellin diffuses to the aleurone layer and induces the synthesis of the enzyme amylase.
- States that amylase hydrolyses starch stored in the endosperm into sugars that are used by the embryo for growth during germination.
Examiner tip: Gibberellin acts on gene expression in the aleurone layer, not directly on the starch itself; keep the causal chain (gibberellin → amylase gene → amylase enzyme → starch breakdown) in the correct order.
24.
Marking analysis: A learner attempts the following task: “Outline the role of gibberellin in breaking seed dormancy and initiating germination in cereal seeds.” Their response addresses only this point: “States that water uptake by the seed triggers the embryo to release gibberellin.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that water uptake by the seed triggers the embryo to release gibberellin.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that gibberellin diffuses to the aleurone layer and induces the synthesis of the enzyme amylase.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that amylase hydrolyses starch stored in the endosperm into sugars that are used by the embryo for growth during germination.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
25.
Outline how the primary and secondary immune responses differ in speed and magnitude, and explain the biological reason for this difference.
[4 marks] · no calculator
Worked explanation
1. During first exposure, rare matching naive B-cells must be selected, activated and expanded before many plasma cells produce antibodies, giving a delay and a smaller initial response.
2. Some activated descendants persist as antigen-specific memory B-cells.
3. Re-exposure recruits this larger, experienced pool, leading to faster and usually greater, more sustained antibody production. Memory cells still need suitable recognition and activation signals.
Common mistake: A faster secondary response is specific to previously encountered antigens, not automatic immunity to all pathogens.
Marking points
- States that the primary response is slow to develop (taking days) and produces a relatively low concentration of antibody.
- States that the secondary response is much faster and produces a greater, more prolonged concentration of antibody.
- Explains that the primary response requires selection and activation of naive antigen-specific B-cells, followed by proliferation (clonal expansion) and differentiation.
- Explains that memory B-cells produced during the primary response persist and allow a much faster, larger clonal expansion upon a second exposure to the same antigen.
Examiner tip: Antigen-experienced memory B-cells respond more rapidly on re-exposure and generate a larger response. They still require appropriate recognition and activation signals; they do not simply skip activation altogether.
26.
Marking analysis: A learner attempts the following task: “Outline how the primary and secondary immune responses differ in speed and magnitude, and explain the biological reason for this difference.” Their response addresses only this point: “States that the primary response is slow to develop (taking days) and produces a relatively low concentration of antibody.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that the primary response is slow to develop (taking days) and produces a relatively low concentration of antibody.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the secondary response is much faster and produces a greater, more prolonged concentration of antibody.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that the primary response requires selection and activation of naive antigen-specific B-cells, followed by proliferation (clonal expansion) and differentiation.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that memory B-cells produced during the primary response persist and allow a much faster, larger clonal expansion upon a second exposure to the same antigen.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
27.
Explain how vaccination can stimulate protective immunity without the person having to experience the corresponding natural disease.
[3 marks] · no calculator
Worked explanation
1. A vaccine presents antigens or instructions for producing them without requiring the full natural infection, allowing immune recognition at much lower risk.
2. Activation and clonal expansion generate effector responses and antigen-specific B- and/or T-cell memory.
3. Later pathogen exposure can be controlled faster, reducing infection or severe-disease risk. Protection is not absolute and depends on vaccine, pathogen and recipient; this explanation is a mechanism, not individual treatment advice.
Common mistake: Vaccination stimulates active immune memory; it is not simply an injection of ready-made antibodies for passive immunity.
Marking points
- States that a vaccine supplies pathogen antigens, or instructions for making an antigen, in a form designed to stimulate immunity with much lower risk than natural infection.
- States that antigen recognition activates an adaptive response that can generate memory B-cells and/or T-cells.
- Explains that subsequent exposure can produce a faster, stronger response that reduces infection or disease risk and severity, without guaranteeing that all infection or symptoms are prevented.
Examiner tip: Do not claim vaccines always prevent symptoms or can never cause the target disease: rare live-vaccine-associated disease is possible, particularly with contraindicated use in immunocompromised people. This is an immunity mechanism question, not individual medical advice.
28.
Marking analysis: A learner attempts the following task: “Explain how vaccination can stimulate protective immunity without the person having to experience the corresponding natural disease.” Their response addresses only this point: “States that a vaccine supplies pathogen antigens, or instructions for making an antigen, in a form designed to stimulate immunity with much lower risk than natural infection.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that a vaccine supplies pathogen antigens, or instructions for making an antigen, in a form designed to stimulate immunity with much lower risk than natural infection.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that antigen recognition activates an adaptive response that can generate memory B-cells and/or T-cells.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that subsequent exposure can produce a faster, stronger response that reduces infection or disease risk and severity, without guaranteeing that all infection or symptoms are prevented.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
29.
Outline the roles of nitrogen-fixing bacteria, nitrifying bacteria, and denitrifying bacteria in the nitrogen cycle.
[3 marks] · no calculator
Course scope: Optional enrichment
Syllabus placement evidence: C4.2.1-C4.2.22, printed pp.83-85; C4.1.12, p.82. Prescribed matter cycling focuses on carbon; legume-nodule mutualism does not prescribe the separate fixation, nitrification and denitrification conversions required here. Full nitrogen-cycle recall is extra content.
Worked explanation
1. Nitrogen fixers convert atmospheric N2 into ammonia/ammonium, introducing biologically accessible nitrogen; Rhizobium in legume nodules is one example.
2. Nitrifying microorganisms oxidise ammonium to nitrite and then nitrate. Plants can assimilate nitrate and/or ammonium, so nitrate formation is not compulsory before every uptake event.
3. Denitrifiers reduce nitrate toward N2, commonly in low-oxygen conditions, returning nitrogen to the atmosphere. These are distinct conversions, not three names for decomposition.
Common mistake: Fixation is N2 to reduced nitrogen, not nitrate to N2; the latter is denitrification.
Marking points
- States that nitrogen-fixing bacteria (e.g. Rhizobium, often in root nodules of legumes, or free-living in soil) convert atmospheric nitrogen gas (N₂), which most organisms cannot use directly, into ammonia/ammonium (NH₃/NH₄⁺).
- States that nitrifying microorganisms oxidise ammonia/ammonium to nitrite and then nitrate; plants can absorb both nitrate and ammonium, with preferences varying among species and conditions.
- States that denitrifying bacteria convert nitrate back into atmospheric nitrogen gas (N₂), typically under anaerobic (low-oxygen) soil conditions, returning nitrogen to the atmosphere and completing the cycle.
Examiner tip: Distinguish fixation, nitrification and denitrification by their chemical conversions. Nitrogen can follow several routes, including direct ammonium uptake and assimilation into biomass; the cycle is not a compulsory one-way relay through every listed bacterial group.
30.
Marking analysis: A learner attempts the following task: “Outline the roles of nitrogen-fixing bacteria, nitrifying bacteria, and denitrifying bacteria in the nitrogen cycle.” Their response addresses only this point: “States that nitrogen-fixing bacteria (e.g. Rhizobium, often in root nodules of legumes, or free-living in soil) convert atmospheric nitrogen gas (N₂), which most organisms cannot use directly, into ammonia/ammonium (NH₃/NH₄⁺).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that nitrogen-fixing bacteria (e.g. Rhizobium, often in root nodules of legumes, or free-living in soil) convert atmospheric nitrogen gas (N₂), which most organisms cannot use directly, into ammonia/ammonium (NH₃/NH₄⁺).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that nitrifying microorganisms oxidise ammonia/ammonium to nitrite and then nitrate; plants can absorb both nitrate and ammonium, with preferences varying among species and conditions.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that denitrifying bacteria convert nitrate back into atmospheric nitrogen gas (N₂), typically under anaerobic (low-oxygen) soil conditions, returning nitrogen to the atmosphere and completing the cycle.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
31.
A population has an intrinsic growth rate (r) of 0.5 per year, a current size (N) of 50 individuals, and the habitat has a carrying capacity (K) of 200 individuals. (a) Using the logistic growth equation dN/dt = rN(1 − N/K), calculate the current rate of population growth. (b) Explain what happens to this growth rate as N approaches K.
[4 marks]
Syllabus placement evidence: C4.1.5-C4.1.8, printed pp.81-82; skills: mathematics, p.31. Carrying capacity and sigmoid growth modelling are prescribed. The equation and all parameters are supplied, requiring only substitution and limiting behaviour, not recall, differentiation or solution of a logistic differential equation. Core denotes supplied-model application, not a claim that the equation must be memorized.
Worked explanation
1. Use the supplied model: dN/dt = (0.5 year^-1)(50 individuals)(1 - 50/200). The dimensionless crowding term is 0.75.
2. The current rate is 25 x 0.75 = 18.75 individuals year^-1. This instantaneous model rate need not be a whole number of animals.
3. As N approaches 200, 1 - N/K approaches zero and net growth approaches zero. At K the model has balanced net additions, not a population of zero or cessation of all births.
Common mistake: 18.75 is a growth rate, not the population size after one full year; that would require solving or approximating the model over time.
Marking points
- Substitutes the given values into the logistic growth equation: dN/dt = 0.5 × 50 × (1 − 50/200).
- Simplifies 1 − 50/200 = 0.75.
- Calculates dN/dt = 0.5 × 50 × 0.75 = 18.75 individuals per year.
- Explains that as N approaches K, the term (1 − N/K) approaches zero, so the overall growth rate dN/dt also approaches zero, causing population growth to slow and level off near the carrying capacity.
Examiner tip: The term (1 − N/K) is the key 'braking' factor in the logistic equation — it equals nearly 1 (barely slowing growth) when N is much smaller than K, and shrinks toward 0 (strongly slowing growth) as N approaches K.
32.
Marking analysis: A learner attempts the following task: “A population has an intrinsic growth rate (r) of 0.5 per year, a current size (N) of 50 individuals, and the habitat has a carrying capacity (K) of 200 individuals. (a) Using the logistic growth equation dN/dt = rN(1 − N/K), calculate the current rate of population growth. (b) Explain what happens to this growth rate as N approaches K.” Their response addresses only this point: “Substitutes the given values into the logistic growth equation: dN/dt = 0.5 × 50 × (1 − 50/200).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Substitutes the given values into the logistic growth equation: dN/dt = 0.5 × 50 × (1 − 50/200).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Simplifies 1 − 50/200 = 0.75.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates dN/dt = 0.5 × 50 × 0.75 = 18.75 individuals per year.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that as N approaches K, the term (1 − N/K) approaches zero, so the overall growth rate dN/dt also approaches zero, causing population growth to slow and level off near the carrying capacity.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
33.
State the competitive exclusion principle, and explain how niche differentiation allows two similar species to coexist in the same habitat without one driving the other to local extinction.
[3 marks] · no calculator
Worked explanation
1. If two species use exactly the same limiting resources in the same stable setting, a persistent competitive advantage prevents indefinite coexistence in the simple exclusion model.
2. Using different feeding times, prey sizes or microhabitats reduces overlap. For example, birds feeding at different tree heights share less of the same food resource.
3. Sufficient partitioning can allow both populations to replace themselves, even while some competition remains. Coexistence does not require completely separate habitats or zero interaction.
Common mistake: Similar species can coexist; exclusion concerns effectively identical limiting niches, not merely similar appearance.
Marking points
- States the competitive exclusion principle: two species competing for exactly the same limited resource (occupying an identical ecological niche) cannot indefinitely coexist in the same habitat — one will eventually outcompete and exclude the other.
- Explains niche differentiation as a process by which similar species evolve or behaviourally shift to use slightly different resources, times, or locations, reducing direct competition between them.
- Explains that sufficiently different resource use reduces niche overlap and the intensity of competition enough to permit coexistence; competition need not disappear entirely.
Examiner tip: The competitive exclusion principle applies specifically to species occupying truly identical niches — niche differentiation is the evolutionary 'escape route' that allows similar species observed coexisting in nature to avoid this fate.
34.
Marking analysis: A learner attempts the following task: “State the competitive exclusion principle, and explain how niche differentiation allows two similar species to coexist in the same habitat without one driving the other to local extinction.” Their response addresses only this point: “States the competitive exclusion principle: two species competing for exactly the same limited resource (occupying an identical ecological niche) cannot indefinitely coexist in the same habitat — one will eventually outcompete and exclude the other.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States the competitive exclusion principle: two species competing for exactly the same limited resource (occupying an identical ecological niche) cannot indefinitely coexist in the same habitat — one will eventually outcompete and exclude the other.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains niche differentiation as a process by which similar species evolve or behaviourally shift to use slightly different resources, times, or locations, reducing direct competition between them.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that sufficiently different resource use reduces niche overlap and the intensity of competition enough to permit coexistence; competition need not disappear entirely.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
35.
An enzyme is tested at substrate concentrations of 1, 2, 4 and 8 mmol dm⁻³. Rates without inhibitor are 12, 20, 31 and 40 units; with inhibitor they are 6, 12, 24 and 38 units. (a) Compare the effect of the inhibitor at low and high substrate concentration. (b) Deduce the most likely type of inhibition. (c) Explain the molecular basis for this deduction. (d) Suggest additional data that would strengthen the conclusion.
[5 marks] · no calculator
Worked explanation
1. At 1 mmol dm^-3, inhibition reduces rate from 12 to 6 units, a 50% reduction; at 8 mmol dm^-3 it reduces 40 to 38, only 5%. The inhibition effect weakens as substrate rises.
2. This convergence is most consistent with competitive inhibition: abundant substrate increasingly occupies sites that inhibitor would otherwise block.
3. The data support a similar limiting maximum rate, but do not establish an exact shared Vmax from a single near-convergent pair.
4. Collect replicated initial rates over more concentrations, especially above 8 mmol dm^-3, and fit curves with uncertainty to compare Vmax and apparent Km. A common Vmax with increased apparent Km would strengthen the competitive interpretation.
Common mistake: Forty versus thirty-eight is not proof that Vmax is identical; neither dataset is demonstrably at its asymptote and uncertainty is absent.
Marking points
- States that the inhibitor substantially reduces rate at low substrate concentration.
- States that the rates become very similar at the highest substrate concentration (40 versus 38 units).
- Deduces that the pattern is most consistent with competitive inhibition.
- Explains that inhibitor and substrate compete for the active site, so abundant substrate increasingly outcompetes inhibitor and a similar maximum rate can be approached.
- Suggests collecting replicated rates over more, especially higher, substrate concentrations to estimate and compare Vmax and Km with uncertainty.
Examiner tip: Do not identify inhibition from one point; use the whole pattern and distinguish evidence for a shared Vmax from evidence about Km.
36.
Marking analysis: A learner attempts the following task: “An enzyme is tested at substrate concentrations of 1, 2, 4 and 8 mmol dm⁻³. Rates without inhibitor are 12, 20, 31 and 40 units; with inhibitor they are 6, 12, 24 and 38 units. (a) Compare the effect of the inhibitor at low and high substrate concentration. (b) Deduce the most likely type of inhibition. (c) Explain the molecular basis for this deduction. (d) Suggest additional data that would strengthen the conclusion.” Their response addresses only this point: “States that the inhibitor substantially reduces rate at low substrate concentration.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that the inhibitor substantially reduces rate at low substrate concentration.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the rates become very similar at the highest substrate concentration (40 versus 38 units).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Deduces that the pattern is most consistent with competitive inhibition.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that inhibitor and substrate compete for the active site, so abundant substrate increasingly outcompetes inhibitor and a similar maximum rate can be approached.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Suggests collecting replicated rates over more, especially higher, substrate concentrations to estimate and compare Vmax and Km with uncertainty.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.