Practise 37 exam-style questions on Aerobic and anaerobic respiration for AQA GCSE Biology. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
37 questions
2–4 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
3 marks
Explain why germinating seeds have a higher respiration rate than dry seeds.
Model answer
Germinating seeds are actively growing and need energy for cell division and biosynthesis, so they respire more (higher metabolic rate). Dry seeds are dormant with low metabolic activity and low energy demand.
Explanation
Germination involves breaking dormancy and growth processes that require ATP produced by respiration, so rates are higher than in dormant (dry) seeds.
Mark scheme
1 mark
germinating seeds are metabolically active/ growing/ need energy for growth
Any statement linking germination to active growth acceptable.
1 mark
therefore higher respiration rate (higher O2 uptake / more CO2 produced)
Link to gas exchange acceptable.
1 mark
dry seeds are dormant so have low metabolic rate / low energy demand and low respiration
Reference to dormancy acceptable.
Question 2
3 marks
Explain why pure aerobic respiration cannot occur in completely anoxic (no oxygen) environments.
Model answer
Aerobic respiration requires oxygen as the final electron acceptor in the electron transport chain. Without oxygen the chain cannot operate, oxidative phosphorylation cannot occur and cells must rely on anaerobic pathways which do not require oxygen.
Explanation
Oxygen is indispensable for the electron transport chain and large-scale ATP production in aerobic respiration; in its absence organisms use anaerobic mechanisms that yield less energy.
Mark scheme
1 mark
aerobic respiration requires oxygen (final electron acceptor)
Essential point.
1 mark
without oxygen electron transport chain stops so oxidative phosphorylation cannot run
Mechanistic link needed.
1 mark
therefore anaerobic pathways must be used instead
Conclusion about alternative metabolism.
Question 3
4 marks
A student measures oxygen uptake of seeds at different temperatures and obtains a graph that peaks then falls at high temperature. Explain the likely reasons for the increase with temperature and the drop at very high temperature.
Model answer
Increasing temperature raises enzyme activity and kinetic energy, so respiration rate increases. At very high temperatures enzymes denature, metabolic reactions slow or stop, so respiration rate falls.
Explanation
Between low and optimum temperatures reaction rates increase (more collisions and enzyme activity). Beyond the optimum, heat damages protein structure (denaturation), lowering respiration.
Mark scheme
1 mark
increase due to increased enzyme activity / reaction rates as temperature rises
Link to kinetic energy acceptable.
1 mark
higher temperature increases kinetic energy of molecules so more collisions and faster reactions
Alternative phrasing allowed.
1 mark
drop at high temperature due to enzyme denaturation
Must mention denaturation or irreversible damage to enzymes.
1 mark
denaturation reduces rate of respiration as metabolic pathways fail
Explain effect on metabolic pathways.
Question 4
4 marks
A factory uses yeast fermentation to make ethanol. Name two variables they must control to maximise yield and explain briefly why for each.
Model answer
Temperature: must be kept near optimum for yeast enzymes to maximise rate without denaturation. pH: maintain optimum pH so enzymes function. Sugar concentration: sufficient substrate to make ethanol but not too high to cause osmotic stress. Oxygen level: keep anaerobic to favour ethanol production (not respiration).
Explanation
Industrial fermentation requires tight control of environmental conditions and substrate to maintain yeast activity and maximally direct metabolism toward ethanol rather than respiration or stress responses.
Mark scheme
1 mark
temperature: control near yeast optimum to maximise enzyme activity
Requirement and reason needed.
1 mark
pH: maintain optimum pH so enzymes function efficiently
Accept any other valid variable with reason: substrate, contamination control, oxygen levels.
1 mark
sugar concentration: adequate substrate increases yield but too high causes osmotic stress
Reason must be given.
1 mark
oxygen: keep anaerobic to prevent yeast respiring aerobically which reduces ethanol yield
Either oxygen or contamination acceptable with explanation.
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