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S4 Biology: Cellular Respiration Paper 2 (Application)
S4 Biology: Cellular Respiration Paper 2 (Application)
This paper mixes the equation, respiratory quotient, energy efficiency and inhibitors. Watch the units and percentages.
Quick answer 1. C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O | 2. larger inner membrane area | 3. glycolysis | 4. rate falls | 5. RQ 0.8, fat-leaning | 6. 9 cm3 per day | 7. 31.8% efficient | 8. glycolysis keeps working | 9. 15 times | 10. glycolysis is cytoplasmic
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Biology — Cellular Respiration |
| Level | Application |
02 Questions (10 items)
- FoundationWrite the balanced equation for the aerobic respiration of glucose.
- FoundationThe inner mitochondrial membrane is folded into many cristae. What advantage does this give?
- FoundationOf the three main stages of aerobic respiration, which one does not use oxygen directly?
- FoundationWhen temperature falls from 30 C to 5 C, how does the respiration rate of seeds change and why?
- IntermediateAn animal uses 500 cm3 of oxygen and releases 400 cm3 of carbon dioxide. What is the RQ and which substrate dominates?
- IntermediateA seed uses 45 cm3 of oxygen over 5 days. What is the average daily uptake?
- IntermediateComplete oxidation releases 2880 kJ per mole of glucose and 30 ATP store 30.5 kJ/mol each. What is the storage efficiency (1 d.p.)?
- ChallengeCyanide blocks the electron transport chain. Why can a cell still survive for a short time?
- ChallengeOne glucose gives 2 ATP anaerobically and 30 aerobically. How many times greater is the aerobic yield?
- Error redoA student says respiration happens only inside mitochondria. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O. | Carbon, hydrogen and oxygen atom counts balance on both sides. |
| 2 | It increases the surface area, holding more electron transport chains and ATP synthase. | A larger surface allows a higher rate of oxidative phosphorylation. |
| 3 | Glycolysis does not use oxygen. | Oxygen only acts as the final electron acceptor in the chain. |
| 4 | The rate falls because molecular kinetic energy and enzyme activity both drop. | Cold slows enzymes without normally denaturing them. |
| 5 | RQ = 0.8, pointing to fat. | Fat gives about 0.7 and carbohydrate about 1.0, so 0.8 is a fat-leaning mix. |
| 6 | About 9.0 cm3 per day. | Average rate = total divided by the number of days. |
| 7 | Storage is 915 kJ, so efficiency is 31.8%. | The remaining energy is lost as heat. |
| 8 | Glycolysis still supplies a net 2 ATP as an emergency source. | Blocking the chain stops aerobic output, yet glycolysis keeps running in the cytoplasm. |
| 9 | It is 15 times greater. | The gap comes from the Krebs cycle and the electron transport chain. |
| 10 | Glycolysis happens in the cytoplasm, and anaerobic respiration is completed there too. | Mitochondria only host the last two stages of aerobic respiration. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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