Biology SL

1. 1.

   Explain how enzymes lower the activation energy of a reaction, using the concept of the enzyme-substrate complex.

   [3 marks] · no calculator

   Worked explanation

   1. The substrate binds in the active site, forming a temporary enzyme-substrate complex rather than instantly turning into product.
   2. Binding positions reacting groups and can strain bonds or stabilise the transition state. The pathway therefore has a lower activation-energy barrier.
   3. More collisions can then lead to reaction at the same temperature. Product leaves and the enzyme can catalyse another cycle; the overall reaction energy difference is unchanged.

   Common mistake: Lowering activation energy is not supplying energy or making the reaction's overall energy change more negative.

   Marking points

   • States that the substrate binds to the enzyme's active site, forming an enzyme-substrate complex.
   • States that this binding brings reacting molecules into close proximity and correct orientation, or strains bonds within the substrate, making the reaction more likely to proceed.
   • States that this reduces the activation energy needed for the reaction, increasing the rate at which the reaction proceeds without being consumed itself.

   Examiner tip: Enzymes speed up reactions by lowering activation energy, not by changing the overall energy difference (ΔG) between reactants and products — they do not make an unfavourable reaction favourable.

2. 2.

   Marking analysis: A learner attempts the following task: “Explain how enzymes lower the activation energy of a reaction, using the concept of the enzyme-substrate complex.” Their response addresses only this point: “States that the substrate binds to the enzyme's active site, forming an enzyme-substrate complex.” 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 the substrate binds to the enzyme's active site, forming an enzyme-substrate complex.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this binding brings reacting molecules into close proximity and correct orientation, or strains bonds within the substrate, making the reaction more likely to proceed.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this reduces the activation energy needed for the reaction, increasing the rate at which the reaction proceeds without being consumed itself.

   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. 3.

   Explain the effect of increasing temperature above the optimum on the rate of an enzyme-catalysed reaction.

   [3 marks] · no calculator

   Worked explanation

   1. Beyond the optimum, thermal disruption of the interactions holding the protein in its active conformation becomes increasingly important.
   2. An altered active site binds or catalyses less effectively, reducing productive complexes despite faster molecular movement.
   3. As more molecules lose their active conformation, the measured rate falls sharply; low-temperature slowing instead usually preserves structure.

   Common mistake: Heat denaturation generally changes folding rather than breaking the peptide backbone into amino acids.

   Marking points

   • States that above the optimum temperature, the rate of reaction decreases sharply.
   • States that excess kinetic energy breaks the hydrogen bonds and other interactions maintaining the enzyme's tertiary structure, denaturing it.
   • States that denaturation changes the shape of the active site so the substrate can no longer bind effectively, reducing (or eliminating) enzyme activity.

   Examiner tip: Distinguish loss of the active conformation during denaturation from the reversible slowing of molecular motion at low temperature. Severe heating often causes irreversible loss of activity, but denaturation is not inherently irreversible in every protein.

4. 4.

   Marking analysis: A learner attempts the following task: “Explain the effect of increasing temperature above the optimum on the rate of an enzyme-catalysed reaction.” Their response addresses only this point: “States that above the optimum temperature, the rate of reaction decreases sharply.” 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 above the optimum temperature, the rate of reaction decreases sharply.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that excess kinetic energy breaks the hydrogen bonds and other interactions maintaining the enzyme's tertiary structure, denaturing it.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that denaturation changes the shape of the active site so the substrate can no longer bind effectively, reducing (or eliminating) enzyme activity.

   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. 5.

   Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the difference between anaerobic cell respiration in yeast (alcoholic fermentation) and in human muscle cells (lactic acid fermentation).

   [3 marks] · no calculator

   Course scope: Optional enrichment

   Worked explanation

   1. Both pathways obtain a small net ATP yield through glycolysis, converting glucose to pyruvate without requiring oxygen.
   2. Yeast converts pyruvate to ethanol and carbon dioxide; human muscle reduces pyruvate to lactate without carbon dioxide release in that step.
   3. Each route regenerates NAD+ so glycolysis can continue. Much chemical energy remains in the organic end products rather than being recovered through aerobic oxidative phosphorylation.

   Common mistake: Lactate production in muscle does not release the carbon dioxide that makes yeast dough rise.

   Marking points

   • States that both processes begin with glycolysis, producing pyruvate and a small yield of ATP, without requiring oxygen.
   • States that in yeast, pyruvate is converted to ethanol and carbon dioxide (alcoholic fermentation).
   • States that in human muscle cells, pyruvate is converted to lactate/lactic acid (lactic acid fermentation), with no gas produced.

   Examiner tip: Extension practice: Anaerobic-versus-aerobic comparison in humans is shared content (C1.2.5); glycolysis, pyruvate, NAD regeneration and yeast fermentation extend into AHL. Guide references: C1.2.7-C1.2.10. This item is enrichment, not evidence of prescribed SL exam coverage. Both fermentation pathways serve the same purpose: regenerating NAD+ from NADH so that glycolysis can continue in the absence of oxygen.

6. 6.

   Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the difference between anaerobic cell respiration in yeast (alcoholic fermentation) and in human muscle cells (lactic acid fermentation).” Their response addresses only this point: “States that both processes begin with glycolysis, producing pyruvate and a small yield of ATP, without requiring oxygen.” 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 both processes begin with glycolysis, producing pyruvate and a small yield of ATP, without requiring oxygen.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that in yeast, pyruvate is converted to ethanol and carbon dioxide (alcoholic fermentation).
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that in human muscle cells, pyruvate is converted to lactate/lactic acid (lactic acid fermentation), with no gas produced.

   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. 7.

   Extension practice (includes HL content beyond the prescribed SL syllabus): Compare the net ATP yield per glucose molecule of fermentation in yeast or human muscle with aerobic cell respiration, and explain the reason for the difference.

   [3 marks] · no calculator

   Course scope: Optional enrichment

   Worked explanation

   1. Fermentation supplies about 2 net ATP per glucose from glycolysis; the fermentation reactions primarily regenerate NAD+ rather than providing the large aerobic yield.
   2. Aerobic oxidation continues through the link reaction and Krebs cycle, transferring electrons to the respiratory chain. Oxygen accepts the electrons at its end.
   3. The resulting proton gradient drives much more ATP synthesis, while glucose carbon is fully oxidised to carbon dioxide. Fermentation leaves energy-rich ethanol or lactate, explaining its smaller yield without requiring one universal aerobic ATP total.

   Common mistake: Oxygen is not an ATP ingredient; it enables electron flow and the large oxidative-phosphorylation contribution.

   Marking points

   • States that anaerobic fermentation yields a small net amount of ATP (glycolysis only, approximately 2 ATP per glucose).
   • States that aerobic respiration yields substantially more ATP per glucose, since it continues through the link reaction, Krebs cycle and oxidative phosphorylation.
   • Explains that aerobic respiration oxidises glucose completely to carbon dioxide, using oxygen as the final electron acceptor and forming water, whereas fermentation leaves much of the chemical energy in ethanol or lactate and regenerates NAD+ so glycolysis can continue.

   Examiner tip: Extension practice: Relative ATP yield is shared content (C1.2.5); the required glycolysis, link reaction, Krebs cycle, oxidative phosphorylation and electron-acceptor explanation is AHL. Guide references: C1.2.7-C1.2.16. This item is enrichment, not evidence of prescribed SL exam coverage. The large ATP yield difference exists because oxidative phosphorylation (which requires oxygen) accounts for the vast majority of ATP produced from one glucose molecule.

8. 8.

   Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Compare the net ATP yield per glucose molecule of fermentation in yeast or human muscle with aerobic cell respiration, and explain the reason for the difference.” Their response addresses only this point: “States that anaerobic fermentation yields a small net amount of ATP (glycolysis only, approximately 2 ATP per glucose).” 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 anaerobic fermentation yields a small net amount of ATP (glycolysis only, approximately 2 ATP per glucose).
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that aerobic respiration yields substantially more ATP per glucose, since it continues through the link reaction, Krebs cycle and oxidative phosphorylation.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that aerobic respiration oxidises glucose completely to carbon dioxide, using oxygen as the final electron acceptor and forming water, whereas fermentation leaves much of the chemical energy in ethanol or lactate and regenerates NAD+ so glycolysis can continue.

   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. 9.

   Annual net production is 48 000 kJ m^-2 yr^-1 for producers and 7200 kJ m^-2 yr^-1 for primary consumers. Calculate the percentage energy transfer and explain three causes of energy loss between these trophic levels.

   [5 marks]

   Worked explanation

   1. Compare annual production on the same area/time basis: transfer = (7200/48000) x 100 = 15%. Units cancel in the percentage.
   2. Some producer material is never eaten; some eaten material is not assimilated and passes to waste/detritus instead of consumer production.
   3. Of assimilated energy, respiration transfers energy into cellular work and ultimately heat, so it is unavailable as new consumer biomass. These are three different pathways, not three names for heat loss.

   Common mistake: Dividing producer energy by consumer energy reverses the efficiency fraction and can give an impossible value above 100%.

   Marking points

   • Uses percentage transfer = 7200/48 000 × 100.
   • Obtains 15%.
   • Energy is released as heat during cellular respiration.
   • Not all producer biomass is consumed.
   • Some consumed material is not assimilated and leaves as waste.

   Examiner tip: Quantify the transfer before explaining distinct biological loss pathways.

10. 10.

    Marking analysis: A learner attempts the following task: “Annual net production is 48 000 kJ m^-2 yr^-1 for producers and 7200 kJ m^-2 yr^-1 for primary consumers. Calculate the percentage energy transfer and explain three causes of energy loss between these trophic levels.” Their response addresses only this point: “Uses percentage transfer = 7200/48 000 × 100.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Marking points

    • Recognises credit for the stated point: Uses percentage transfer = 7200/48 000 × 100.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Obtains 15%.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Energy is released as heat during cellular respiration.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Not all producer biomass is consumed.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Some consumed material is not assimilated and leaves as waste.

    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. 11.

    Distinguish between a population, a community and an ecosystem.

    [3 marks] · no calculator

    Worked explanation

    1. All rabbits in one meadow at one time form a population: one species with a defined place and time.
    2. Rabbits, grasses, foxes and microbes there together form a community: multiple interacting populations.
    3. Adding soil, water, temperature and other abiotic surroundings gives the ecosystem, because energy and matter move between organisms and their physical environment.

    Common mistake: A community includes several species, while an ecosystem additionally includes abiotic components.

    Marking points

    • Defines a population as organisms of the same species living in the same area at the same time.
    • Defines a community as all the populations of different species living and interacting in the same area.
    • Defines an ecosystem as a community together with the abiotic (non-living) physical environment with which it interacts.

    Examiner tip: These three terms form a hierarchy of increasing scale: population → community → ecosystem — questions often test whether you can place a described scenario at the correct level.

12. 12.

    Marking analysis: A learner attempts the following task: “Distinguish between a population, a community and an ecosystem.” Their response addresses only this point: “Defines a population as organisms of the same species living in the same area at the same time.” 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: Defines a population as organisms of the same species living in the same area at the same time.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Defines a community as all the populations of different species living and interacting in the same area.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Defines an ecosystem as a community together with the abiotic (non-living) physical environment with which it interacts.

    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. 13.

    Explain the role of decomposers in recycling nutrients within an ecosystem.

    [3 marks] · no calculator

    Worked explanation

    1. Fungi and bacteria secrete enzymes that break dead material and waste into substances they can absorb.
    2. Their metabolism mineralises organic nutrients, returning forms such as ammonium to soil and carbon dioxide to air or water.
    3. Producers take up available inorganic nutrients again, connecting decomposition to renewed biomass production. Matter cycles; energy used during decomposition ultimately dissipates as heat.

    Common mistake: Decomposers recycle atoms and nutrients, not the heat energy lost from the food web.

    Marking points

    • States that decomposers (e.g. bacteria and fungi) break down dead organic matter and waste products.
    • States that this process releases inorganic nutrients (e.g. nitrogen and carbon compounds) back into the soil, water or air.
    • States that these recycled inorganic nutrients can then be taken up again by producers, allowing the nutrient cycle to continue.

    Examiner tip: Decomposers make nutrients in dead organic matter available again through mineralisation. Without them, biological nutrient recycling would be greatly reduced; do not claim that every nutrient would remain permanently locked, since abiotic processes also break down organic matter.

14. 14.

    Marking analysis: A learner attempts the following task: “Explain the role of decomposers in recycling nutrients within an ecosystem.” Their response addresses only this point: “States that decomposers (e.g. bacteria and fungi) break down dead organic matter and waste products.” 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 decomposers (e.g. bacteria and fungi) break down dead organic matter and waste products.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this process releases inorganic nutrients (e.g. nitrogen and carbon compounds) back into the soil, water or air.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that these recycled inorganic nutrients can then be taken up again by producers, allowing the nutrient cycle to continue.

    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. 15.

    Outline the processes by which carbon enters and leaves the atmosphere as part of the carbon cycle.

    [4 marks] · no calculator

    Worked explanation

    1. Photosynthesis fixes atmospheric carbon dioxide into organic compounds; respiration by producers, consumers and decomposers returns some organic carbon as carbon dioxide.
    2. Burning biomass or fossil fuels also returns carbon dioxide. Rapid combustion can release carbon that had remained in a reservoir for millions of years.
    3. Carbon dioxide dissolves in surface water and can outgas again. Deep-ocean transport or burial can retain it much longer, so surface dissolution alone does not imply permanent atmospheric removal.

    Common mistake: A fossil fuel's ancient origin does not make its combustion a slow transfer; reservoir age and transfer speed differ.

    Marking points

    • States that photosynthesis removes carbon dioxide from the atmosphere, fixing carbon into organic compounds in producers.
    • States that cellular respiration (in producers, consumers and decomposers) releases carbon dioxide back into the atmosphere.
    • States that combustion of fossil fuels or biomass rapidly releases stored carbon as carbon dioxide; the fossil carbon accumulated over geological time.
    • States that carbon dioxide also exchanges with the ocean by dissolution and outgassing; deep-ocean storage and burial in sediments or fossil reservoirs can remove carbon from rapid cycling for much longer periods.

    Examiner tip: Distinguish the age of a carbon reservoir from the speed of a transfer. Combustion can release ancient fossil carbon rapidly, and surface-ocean dissolution is not automatically permanent storage.

16. 16.

    Marking analysis: A learner attempts the following task: “Outline the processes by which carbon enters and leaves the atmosphere as part of the carbon cycle.” Their response addresses only this point: “States that photosynthesis removes carbon dioxide from the atmosphere, fixing carbon into organic compounds in producers.” 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 photosynthesis removes carbon dioxide from the atmosphere, fixing carbon into organic compounds in producers.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that cellular respiration (in producers, consumers and decomposers) releases carbon dioxide back into the atmosphere.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that combustion of fossil fuels or biomass rapidly releases stored carbon as carbon dioxide; the fossil carbon accumulated over geological time.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that carbon dioxide also exchanges with the ocean by dissolution and outgassing; deep-ocean storage and burial in sediments or fossil reservoirs can remove carbon from rapid cycling for much longer periods.

    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. 17.

    A food chain is: grass → grasshopper → frog → snake. Identify the trophic level of the frog and explain why food chains rarely extend beyond four or five trophic levels.

    [3 marks] · no calculator

    Worked explanation

    1. Grass is trophic level 1, grasshopper level 2 and frog level 3. The frog is therefore a secondary consumer, not the second trophic level.
    2. At each transfer, uneaten and unassimilated material bypasses the next consumer, while respiration dissipates energy as heat.
    3. Repeated losses leave less production available to support higher consumers, helping limit chain length. Four or five levels is a common pattern, not an absolute biological ceiling.

    Common mistake: Consumer number and trophic-level number differ because producers count as the first trophic level.

    Marking points

    • Identifies the frog as a secondary consumer (third trophic level).
    • States that only a fraction of energy passes to the next trophic level because of respiration, unconsumed biomass and unassimilated material; about 10% is a useful illustration, not a fixed efficiency.
    • Explains that repeated losses reduce the energy available to higher consumers, limiting the populations and number of trophic levels that can be supported.

    Examiner tip: The 10% rule is an approximation, but it explains both why food chains are short and why there are usually far fewer top predators than producers in an ecosystem.

18. 18.

    Marking analysis: A learner attempts the following task: “A food chain is: grass → grasshopper → frog → snake. Identify the trophic level of the frog and explain why food chains rarely extend beyond four or five trophic levels.” Their response addresses only this point: “Identifies the frog as a secondary consumer (third trophic level).” 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: Identifies the frog as a secondary consumer (third trophic level).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that only a fraction of energy passes to the next trophic level because of respiration, unconsumed biomass and unassimilated material; about 10% is a useful illustration, not a fixed efficiency.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that repeated losses reduce the energy available to higher consumers, limiting the populations and number of trophic levels that can be supported.

    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. 19.

    Outline how a pyramid of energy differs from a pyramid of numbers, and state one advantage of using a pyramid of energy.

    [2 marks] · no calculator

    Worked explanation

    1. A numbers pyramid counts organisms, so one large tree supporting many insects can create a narrow producer bar and broad consumer bar.
    2. An energy pyramid compares energy flow per area per time, for example kJ m^-2 yr^-1, rather than individual counts or a biomass snapshot.
    3. Transfers lose usable energy, so matched energy-flow bars decrease upward. This captures transfer efficiency without the distortion caused by very different organism sizes.

    Common mistake: An inverted numbers pyramid does not show energy being created; one tree can supply many small consumers.

    Marking points

    • States that a pyramid of numbers represents the number of individual organisms at each trophic level, while a pyramid of energy represents the rate of energy flow (e.g. kJ m⁻² yr⁻¹) through each trophic level.
    • States that a pyramid of numbers can sometimes give an inverted or unusual shape (e.g. many insects feeding on one large tree), whereas a pyramid of energy is always a true pyramid shape, since energy transfer between levels is always less than 100%.

    Examiner tip: A pyramid of energy is always upright because it directly reflects the law of energy loss at each trophic transfer, unlike pyramids of numbers or biomass, which can be distorted by organism size or timing.

20. 20.

    Marking analysis: A learner attempts the following task: “Outline how a pyramid of energy differs from a pyramid of numbers, and state one advantage of using a pyramid of energy.” Their response addresses only this point: “States that a pyramid of numbers represents the number of individual organisms at each trophic level, while a pyramid of energy represents the rate of energy flow (e.g. kJ m⁻² yr⁻¹) through each trophic level.” 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 a pyramid of numbers represents the number of individual organisms at each trophic level, while a pyramid of energy represents the rate of energy flow (e.g. kJ m⁻² yr⁻¹) through each trophic level.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a pyramid of numbers can sometimes give an inverted or unusual shape (e.g. many insects feeding on one large tree), whereas a pyramid of energy is always a true pyramid shape, since energy transfer between levels is always less than 100%.

    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. 21.

    Explain how competition for a limited resource can affect the carrying capacity of a population in an ecosystem.

    [3 marks] · no calculator

    Worked explanation

    1. Carrying capacity is the population size the current habitat can sustain with its resources and other constraints.
    2. As density rises, less food or space is available per individual. Competition reduces reproduction and/or increases deaths.
    3. Net growth consequently slows near carrying capacity: density feeds back on growth. If habitat resources change, the supported population size can change too.

    Common mistake: Carrying capacity is not an immutable species property; it depends on the environment and interactions.

    Marking points

    • Defines carrying capacity as the maximum population size that an environment can sustainably support given its available resources.
    • States that as a population approaches carrying capacity, competition for limited resources (e.g. food, space) intensifies, reducing individual survival and reproduction.
    • States that this negative feedback (density-dependent limitation) causes population growth rate to slow and stabilize near the carrying capacity.

    Examiner tip: Carrying capacity is not fixed — it can change if resource availability changes (e.g. due to climate, habitat loss, or a new competing species).

22. 22.

    Marking analysis: A learner attempts the following task: “Explain how competition for a limited resource can affect the carrying capacity of a population in an ecosystem.” Their response addresses only this point: “Defines carrying capacity as the maximum population size that an environment can sustainably support given its available resources.” 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: Defines carrying capacity as the maximum population size that an environment can sustainably support given its available resources.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that as a population approaches carrying capacity, competition for limited resources (e.g. food, space) intensifies, reducing individual survival and reproduction.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this negative feedback (density-dependent limitation) causes population growth rate to slow and stabilize 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.

23. 23.

    Extension practice (HL content beyond the prescribed SL syllabus): Outline the process of primary succession, from bare rock to a more established community.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Worked explanation

    1. Newly exposed rock has no developed soil. Tolerant pioneers such as lichens colonise where suitable propagules arrive.
    2. Weathering and accumulated dead organic matter help develop soil, allowing plants with different requirements to establish.
    3. Changing shade, nutrients and competition alter later communities; grasses, shrubs or trees may appear if local conditions permit. Disturbance and climate mean there is no guaranteed universal climax sequence.

    Common mistake: A field regrowing after fire already has soil and illustrates secondary, not primary, succession.

    Marking points

    • States that primary succession begins on newly exposed, lifeless substrate (e.g. bare rock) with no existing soil.
    • States that pioneer species (e.g. lichens or mosses) colonize first, tolerating harsh conditions and beginning to break down rock and accumulate organic matter.
    • States that soil development can allow later communities, such as grasses, shrubs or trees where conditions permit, to replace pioneers. The sequence and eventual community depend on climate, colonisation and disturbance rather than an inevitable progression to a single climax.

    Examiner tip: Extension practice: Primary succession and its relationship to later communities are AHL. Guide references: D4.2.12-D4.2.15. This item is enrichment, not evidence of prescribed SL exam coverage. Distinguish primary succession (starting with no soil, e.g. new volcanic rock) from secondary succession (starting with existing soil after a disturbance, e.g. fire or clearing), which proceeds faster.

24. 24.

    Marking analysis: A learner attempts the following task: “Extension practice (HL content beyond the prescribed SL syllabus): Outline the process of primary succession, from bare rock to a more established community.” Their response addresses only this point: “States that primary succession begins on newly exposed, lifeless substrate (e.g. bare rock) with no existing soil.” 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 primary succession begins on newly exposed, lifeless substrate (e.g. bare rock) with no existing soil.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that pioneer species (e.g. lichens or mosses) colonize first, tolerating harsh conditions and beginning to break down rock and accumulate organic matter.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that soil development can allow later communities, such as grasses, shrubs or trees where conditions permit, to replace pioneers. The sequence and eventual community depend on climate, colonisation and disturbance rather than an inevitable progression to a single climax.

    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. 25.

    Outline one method that estimates animal population size from directly sampled individuals and one method that uses indirect signs to estimate relative abundance in the field.

    [2 marks] · no calculator

    Worked explanation

    1. Capture, harmlessly mark and release M animals; after mixing, catch C animals and count R marked recaptures. If R/C estimates M/N, then N is approximately MC/R.
    2. This estimate assumes a suitably closed population, retained recognisable marks and representative/equal recapture chances; trap-shyness can bias the result.
    3. For an indirect method, count droppings along standardised transects. Repeated comparable counts index relative abundance, but absolute population size requires information about dropping production, persistence and detection.

    Common mistake: Twenty droppings do not mean twenty animals; signs must be calibrated before conversion into population numbers.

    Marking points

    • Outlines a direct method, e.g. the capture-mark-release-recapture technique, where a sample is captured, marked, released, and a second sample is later captured to estimate total population from the proportion of marked individuals recaptured.
    • Outlines an indirect method, e.g. counting indirect signs of presence such as droppings, tracks, nests or burrows, to estimate relative abundance without directly capturing individuals.

    Examiner tip: Capture-recapture estimates population size subject to assumptions such as a closed population and representative recapture. Tracks, droppings or nests usually provide an abundance index; estimating absolute numbers requires calibration to sign production, persistence and detection.

26. 26.

    Marking analysis: A learner attempts the following task: “Outline one method that estimates animal population size from directly sampled individuals and one method that uses indirect signs to estimate relative abundance in the field.” Their response addresses only this point: “Outlines a direct method, e.g. the capture-mark-release-recapture technique, where a sample is captured, marked, released, and a second sample is later captured to estimate total population from the proportion of marked individuals recaptured.” 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: Outlines a direct method, e.g. the capture-mark-release-recapture technique, where a sample is captured, marked, released, and a second sample is later captured to estimate total population from the proportion of marked individuals recaptured.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Outlines an indirect method, e.g. counting indirect signs of presence such as droppings, tracks, nests or burrows, to estimate relative abundance without directly capturing individuals.

    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. 27.

    Outline the process by which nerve impulses are transmitted across a chemical synapse.

    [4 marks] · no calculator

    Worked explanation

    1. Arrival of the impulse depolarises the presynaptic terminal and opens calcium channels. Calcium entry triggers vesicle fusion and neurotransmitter release.
    2. The transmitter diffuses across the cleft and binds matching postsynaptic receptors, changing ion-channel activity.
    3. The resulting postsynaptic electrical change may bring the cell to threshold for an impulse; release does not guarantee firing. Removal or breakdown of transmitter limits the signal.

    Common mistake: The electrical impulse does not jump unchanged across a chemical synapse; neurotransmitter carries the signal across the cleft.

    Marking points

    • States that an arriving nerve impulse causes calcium ion channels to open, allowing Ca²⁺ to enter the presynaptic neuron.
    • States that this causes synaptic vesicles containing neurotransmitter to fuse with the presynaptic membrane and release neurotransmitter into the synaptic cleft (exocytosis).
    • States that neurotransmitter diffuses across the synaptic cleft and binds to specific receptors on the postsynaptic membrane.
    • States that this binding opens ion channels in the postsynaptic membrane, potentially generating a new nerve impulse if a threshold is reached.

    Examiner tip: In a typical fast chemical synapse, vesicular release is presynaptic and the transmitted signal acts on postsynaptic receptors. Presynaptic autoreceptors also exist, so receptor presence is not exclusive to the postsynaptic membrane.

28. 28.

    Marking analysis: A learner attempts the following task: “Outline the process by which nerve impulses are transmitted across a chemical synapse.” Their response addresses only this point: “States that an arriving nerve impulse causes calcium ion channels to open, allowing Ca²⁺ to enter the presynaptic neuron.” 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 an arriving nerve impulse causes calcium ion channels to open, allowing Ca²⁺ to enter the presynaptic neuron.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this causes synaptic vesicles containing neurotransmitter to fuse with the presynaptic membrane and release neurotransmitter into the synaptic cleft (exocytosis).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that neurotransmitter diffuses across the synaptic cleft and binds to specific receptors on the postsynaptic membrane.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this binding opens ion channels in the postsynaptic membrane, potentially generating a new nerve impulse if a threshold is reached.

    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. 29.

    Outline how the antibody-mediated (humoral) immune response leads to the production of antibodies against a specific pathogen.

    [4 marks] · no calculator

    Worked explanation

    1. An antigen selects a B-cell whose receptor already matches it; the pathogen does not design a new antibody shape.
    2. With appropriate activation signals, often from helper T-cells, that cell proliferates into a clone sharing antigen specificity.
    3. Plasma cells from the clone secrete antibodies, while memory B-cells persist. A later encounter with the same antigen can recruit that expanded memory pool more rapidly.

    Common mistake: Antibodies are secreted by plasma cells, not by the pathogen; memory cells do not provide instant protection against every different antigen.

    Marking points

    • States that a B-lymphocyte with a receptor complementary to a specific antigen on the pathogen binds to that antigen.
    • States that antigen binding, often with helper T-cell signals, activates the selected B-lymphocyte, which proliferates by mitosis (clonal expansion).
    • States that most of these clones differentiate into plasma cells, which secrete large quantities of antibodies specific to that antigen.
    • States that a smaller number of clones become memory cells, which persist and allow a faster, stronger response if the same pathogen is encountered again.

    Examiner tip: The presence of memory cells after a primary immune response is exactly why the secondary response to the same pathogen is faster and stronger — the basis of both natural immunity and vaccination.

30. 30.

    Marking analysis: A learner attempts the following task: “Outline how the antibody-mediated (humoral) immune response leads to the production of antibodies against a specific pathogen.” Their response addresses only this point: “States that a B-lymphocyte with a receptor complementary to a specific antigen on the pathogen binds to that antigen.” 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 a B-lymphocyte with a receptor complementary to a specific antigen on the pathogen binds to that antigen.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that antigen binding, often with helper T-cell signals, activates the selected B-lymphocyte, which proliferates by mitosis (clonal expansion).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that most of these clones differentiate into plasma cells, which secrete large quantities of antibodies specific to that antigen.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a smaller number of clones become memory cells, which persist and allow a faster, stronger response if the same pathogen is encountered again.

    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. 31.

    A food chain transfers energy from producers storing 10,000 kJ m^-2 year^-1. Assuming 10% is transferred at each step, calculate the energy available to (a) primary consumers, (b) secondary consumers and (c) tertiary consumers. Explain how the decline in available energy limits food-chain length.

    [4 marks]

    Worked explanation

    1. Primary consumers receive 10000 x 0.10 = 1000 kJ m^-2 year^-1.
    2. Secondary consumers receive 1000 x 0.10 = 100 kJ m^-2 year^-1; tertiary consumers receive 100 x 0.10 = 10 kJ m^-2 year^-1.
    3. At just 0.1% of producer production by the tertiary level, little remains to support another consumer population. This illustrates energetic limits on chain length; the 10% efficiency is a stated model assumption.

    Common mistake: Giving every consumer 1000 applies 10% repeatedly to the producer rather than to the preceding trophic level.

    Marking points

    • Calculates energy available to primary consumers as 10,000 × 0.10 = 1,000 kJ m⁻² year⁻¹.
    • Calculates energy available to secondary consumers as 1,000 × 0.10 = 100 kJ m⁻² year⁻¹.
    • Calculates energy available to tertiary consumers as 100 × 0.10 = 10 kJ m⁻² year⁻¹.
    • Notes that this rapid decline in available energy at each trophic level explains why food chains rarely extend beyond four or five trophic levels.

    Examiner tip: Apply the 10% rule successively at each trophic level, always multiplying by the previous level's result rather than the original producer value — each step compounds the energy loss.

32. 32.

    Marking analysis: A learner attempts the following task: “A food chain transfers energy from producers storing 10,000 kJ m^-2 year^-1. Assuming 10% is transferred at each step, calculate the energy available to (a) primary consumers, (b) secondary consumers and (c) tertiary consumers. Explain how the decline in available energy limits food-chain length.” Their response addresses only this point: “Calculates energy available to primary consumers as 10,000 × 0.10 = 1,000 kJ m⁻² year⁻¹.” 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: Calculates energy available to primary consumers as 10,000 × 0.10 = 1,000 kJ m⁻² year⁻¹.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates energy available to secondary consumers as 1,000 × 0.10 = 100 kJ m⁻² year⁻¹.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates energy available to tertiary consumers as 100 × 0.10 = 10 kJ m⁻² year⁻¹.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Notes that this rapid decline in available energy at each trophic level explains why food chains rarely extend beyond four or five trophic levels.

    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. 33.

    A population of deer introduced to an island shows exponential growth initially, but the growth rate slows and eventually levels off as the population approaches the island's carrying capacity. Explain two factors that could limit further population growth as the population nears carrying capacity.

    [3 marks] · no calculator

    Worked explanation

    1. More deer share the island's vegetation, so food per deer falls. Poor nutrition can reduce fertility and juvenile survival, slowing births relative to deaths.
    2. Crowding also increases contacts through which pathogens spread, potentially raising mortality. This second factor has a different mechanism from food competition.
    3. Both effects strengthen with density, explaining the slowdown near carrying capacity. A storm can limit numbers too, but does not by itself explain a specifically density-dependent slowdown.

    Common mistake: Simply naming a disaster does not explain why the limiting effect intensifies as deer numbers approach the island's capacity.

    Marking points

    • Explains a valid density-dependent limiting factor, such as food/resource availability becoming scarce as more deer compete for a fixed supply of vegetation, reducing survival and reproduction rates.
    • Explains a second valid limiting factor, such as the build-up of waste/disease spreading more easily at higher population density, or increased predation pressure if predator populations also grow in response to more available prey.
    • Notes that these are density-dependent factors, meaning their limiting effect intensifies as population density increases, which is why population growth slows specifically as the population approaches carrying capacity rather than remaining constant throughout.

    Examiner tip: Density-dependent limiting factors (competition, disease, predation) intensify as population size increases, unlike density-independent factors (e.g. a natural disaster), which affect a population regardless of its size.

34. 34.

    Marking analysis: A learner attempts the following task: “A population of deer introduced to an island shows exponential growth initially, but the growth rate slows and eventually levels off as the population approaches the island's carrying capacity. Explain two factors that could limit further population growth as the population nears carrying capacity.” Their response addresses only this point: “Explains a valid density-dependent limiting factor, such as food/resource availability becoming scarce as more deer compete for a fixed supply of vegetation, reducing survival and reproduction rates.” 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: Explains a valid density-dependent limiting factor, such as food/resource availability becoming scarce as more deer compete for a fixed supply of vegetation, reducing survival and reproduction rates.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains a second valid limiting factor, such as the build-up of waste/disease spreading more easily at higher population density, or increased predation pressure if predator populations also grow in response to more available prey.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Notes that these are density-dependent factors, meaning their limiting effect intensifies as population density increases, which is why population growth slows specifically as the population approaches carrying capacity rather than remaining constant throughout.

    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. 35.

    Explain how carbon moves from the atmosphere into living organisms, and then back into the atmosphere, as part of the carbon cycle.

    [3 marks] · no calculator

    Worked explanation

    1. Producers fix atmospheric carbon dioxide through photosynthesis into organic molecules such as sugars.
    2. Feeding transfers those carbon atoms into consumer biomass; not every atom immediately returns to the air.
    3. Respiration by plants, animals and decomposers oxidises organic carbon and releases carbon dioxide, closing a route back to the atmospheric reservoir.

    Common mistake: Plants also respire: they are not exclusively carbon-dioxide absorbers, and decomposers do not need photosynthesis to release it.

    Marking points

    • Explains that producers (plants, algae) remove carbon dioxide from the atmosphere through photosynthesis, converting it into organic carbon compounds (e.g. glucose).
    • Explains that this organic carbon is passed along the food chain as consumers feed on producers and on each other.
    • Explains that carbon is returned to the atmosphere as carbon dioxide through the cellular respiration of producers, consumers, and decomposers, which break down organic matter.

    Examiner tip: Photosynthesis and respiration are the two opposing processes that together keep carbon cycling between its inorganic (atmospheric CO₂) and organic (living biomass) forms — always name both directions of the exchange.

36. 36.

    Marking analysis: A learner attempts the following task: “Explain how carbon moves from the atmosphere into living organisms, and then back into the atmosphere, as part of the carbon cycle.” Their response addresses only this point: “Explains that producers (plants, algae) remove carbon dioxide from the atmosphere through photosynthesis, converting it into organic carbon compounds (e.g. glucose).” 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: Explains that producers (plants, algae) remove carbon dioxide from the atmosphere through photosynthesis, converting it into organic carbon compounds (e.g. glucose).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that this organic carbon is passed along the food chain as consumers feed on producers and on each other.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that carbon is returned to the atmosphere as carbon dioxide through the cellular respiration of producers, consumers, and decomposers, which break down organic matter.

    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.

37. 37.

    A student counts a moss species in five equal quadrats beneath trees and five equal quadrats in an open area. Counts beneath trees are 12, 9, 15, 10 and 14; open-area counts are 3, 5, 4, 2 and 6. (a) Calculate the mean count for each habitat. (b) State the pattern shown. (c) Identify one limitation of the investigation. (d) Suggest one improvement that directly addresses that limitation.

    [4 marks] · no calculator

    Worked explanation

    1. Tree-site counts sum to 60, so the mean is 60/5 = 12 moss plants per quadrat. Open counts sum to 20, giving 20/5 = 4 per quadrat.
    2. Observed mean abundance is three times higher beneath trees, and the supplied ranges (9-15 versus 2-6) do not overlap. This describes this sample, not a significance test or proof that shade causes the difference.
    3. Only five quadrats per habitat limits precision and representation. Increase the number of randomly positioned equal-area quadrats, ideally across several sites, to address that limitation directly.

    Common mistake: A higher count beneath trees does not isolate light as the cause; moisture, substrate and placement can also differ.

    Marking points

    • Calculates a mean of 12 moss plants per quadrat beneath trees.
    • Calculates a mean of 4 moss plants per quadrat in the open area and states abundance is higher beneath trees.
    • Identifies a valid limitation, such as the small sample size, non-random placement, one site, or an uncontrolled environmental variable.
    • Suggests a matched improvement, such as more randomly placed quadrats across several sites or measuring the identified environmental variable.

    Examiner tip: An improvement earns credit only when it responds to the limitation you identified; avoid generic statements such as 'be more accurate'.

38. 38.

    Marking analysis: A learner attempts the following task: “A student counts a moss species in five equal quadrats beneath trees and five equal quadrats in an open area. Counts beneath trees are 12, 9, 15, 10 and 14; open-area counts are 3, 5, 4, 2 and 6. (a) Calculate the mean count for each habitat. (b) State the pattern shown. (c) Identify one limitation of the investigation. (d) Suggest one improvement that directly addresses that limitation.” Their response addresses only this point: “Calculates a mean of 12 moss plants per quadrat beneath trees.” 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: Calculates a mean of 12 moss plants per quadrat beneath trees.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates a mean of 4 moss plants per quadrat in the open area and states abundance is higher beneath trees.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Identifies a valid limitation, such as the small sample size, non-random placement, one site, or an uncontrolled environmental variable.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Suggests a matched improvement, such as more randomly placed quadrats across several sites or measuring the identified environmental variable.

    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.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.