Secondary Biology: Enzymes and Metabolic Reactions (Paper 2)
- 2026-10-31
Secondary Biology: Enzymes and Metabolic Reactions (Paper 2)
Ten original questions on rate against time, the effect of a ten degree rise, limiting factors of enzyme and substrate, and the common mistake that enzymes are used up.
Direct answer Rate is product divided by time, so halving the time doubles the rate. Extra enzyme only helps while enzyme is limiting.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: Secondary Biology: Enzymes and Metabolic Reactions |
| Level | 4 basic, 3 advanced, 2 challenge and 1 redo question |
02 Questions (10 items)
- FoundationTube A produces 30 cm³ of gas in 5 minutes. What is the rate?
- FoundationTube B produces 48 cm³ in 8 minutes. Which tube is faster?
- FoundationAt a steady 6.0 cm³ per minute, how long is needed for 60 cm³?
- FoundationDo all enzymes have the same optimum pH? Give an example.
- IntermediateThe rate goes from 3 to 6 when warmed from 20 to 30 °C, and to 12 at 40 °C. What is the factor per 10 °C?
- IntermediateWith plenty of substrate, doubling the enzyme takes the rate from 5 to 10 cm³ per minute. By what factor?
- IntermediateThe time for the reaction falls from 8 minutes to 4 minutes. By what factor does the rate rise?
- ChallengeThere is 8 g of substrate and the rate is 2 g per minute. When is it used up, and does extra enzyme help afterwards?
- ChallengeWhy does the rate fall sharply rather than gradually above the optimum temperature?
- Error redoA pupil writes that enzymes are used up in reactions and must be constantly replaced. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 6.0 cm³ per minute (30 divided by 5). | Rate is the amount of product divided by the time, with units stated. |
| 2 | Neither: both give 6.0 cm³ per minute (48 divided by 8). | Convert both readings to the same time unit before comparing. |
| 3 | 10 minutes (60 divided by 6.0). | Time equals the total amount divided by the rate. |
| 4 | No; pepsin works best near pH 2 and trypsin near pH 8. | Each enzyme is adapted to the pH of the place where it works. |
| 5 | About 2 times (6 divided by 3 and 12 divided by 6 are both 2). | This rule breaks down above the optimum because the enzyme starts to denature. |
| 6 | 2 times (10 divided by 5). | Adding enzyme helps only while enzyme is the limiting factor. |
| 7 | 2 times (8 divided by 4). | Halving the time doubles the rate; the two are inversely related. |
| 8 | After 4 minutes; extra enzyme does not help because the substrate has run out. | Once substrate is limiting, the amount of enzyme is no longer the key factor. |
| 9 | Heat denatures the enzyme, destroying the shape of the active site so the substrate cannot bind, and the change is irreversible. | Cooling only slows an enzyme and is reversible, whereas denaturation is not. |
| 10 | Enzymes are catalysts: they are unchanged at the end of the reaction and can be reused. | It is the substrate that is used up, not the enzyme. |
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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