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IB Biology: Enzymes and Reaction Kinetics (10 Advanced Items)
IB Biology: Enzymes and Reaction Kinetics (10 Advanced Items)
Ten advanced items on Q10 and initial rate, activity units, denaturation versus reversible slowing, competitive and non-competitive inhibition, and reading Vmax with Km.
Focus Extrapolate with Q10 only below the optimum, and judge inhibition type from Vmax and Km together.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 30 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: IB Biology: Enzymes and Reaction Kinetics |
| Level | IB Biology (HL) |
02 Questions (10 items)
- FoundationAn enzyme produces 24 cm³ of gas in 4 minutes. Find the mean rate.
- FoundationWhat is Q10, and what value is typical for enzyme reactions?
- FoundationCan an enzyme boiled and then cooled regain its activity? Why?
- FoundationWhat is the usual definition of one unit (U) of enzyme activity?
- IntermediateFrom 0 to 2 minutes the product rises from 0 to 18 cm³. Find the mean rate and its relation to the initial rate.
- IntermediateRate is 5 cm³/min at 20°C and 10 cm³/min at 30°C. (a) Find Q10. (b) Predict the rate at 40°C with the same Q10, and state the limitation.
- IntermediateHow do competitive and non-competitive inhibitors differ in binding site and in whether more substrate can overcome them?
- ChallengeAn enzyme has Vmax = 120 cm³/min and Km = 4 mM. What is the rate when substrate concentration equals Km?
- ChallengeWith an inhibitor, Vmax stays at 120 cm³/min but Km rises from 4 to 8 mM. Which type of inhibition is this, and why?
- Error redoA student writes “90 cm³ produced in 30 minutes, so rate = 90 cm³/min”. Is this right? Why?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 24 ÷ 4 = 6 cm³/min. | Rate is always product divided by time, with the unit kept. |
| 2 | Q10 is the factor by which rate multiplies for each 10°C rise; for most enzyme reactions it is about 2. | Q10 only applies below the optimum temperature; above it the rate falls instead. |
| 3 | No. Boiling breaks hydrogen and ionic bonds, so the tertiary structure is permanently changed (denatured) and the active site cannot reform. | Denaturation is an irreversible change of structure, unlike the temporary slowing caused by cold. |
| 4 | One unit is the amount of enzyme that catalyses the conversion of 1 μmol of substrate per minute under stated conditions. | Activity units measure catalytic ability, not mass, so enzyme activity cannot be compared by grams. |
| 5 | 18 ÷ 2 = 9 cm³/min, which is the initial rate taken from the near-linear start of the curve. | Initial rate comes from the first, near-linear part of the curve; later the rate falls as substrate is used up. |
| 6 | (a) Q10 = 10 ÷ 5 = 2. (b) Predicted 10 × 2 = 20 cm³/min, but above the optimum the enzyme denatures, so the real rate will not double again. | Extrapolating with Q10 only works below the optimum; near or above it the prediction overestimates. |
| 7 | A competitive inhibitor binds the active site and is overcome by more substrate; a non-competitive inhibitor binds elsewhere and distorts the active site, so extra substrate does not help. | This is also how inhibition type is judged from data: unchanged Vmax with a rising Km means competitive inhibition. |
| 8 | At [S] = Km the rate is half of Vmax: 120 ÷ 2 = 60 cm³/min. | Km is defined as the substrate concentration giving half of Vmax, so the relation can be used directly. |
| 9 | Competitive inhibition: unchanged Vmax means the enzyme can still be fully saturated, and a higher Km means more substrate is needed for half-maximal rate. | Read both parameters together; judging from Km or Vmax alone easily misclassifies the inhibition. |
| 10 | No. Rate must be divided by time: 90 ÷ 30 = 3 cm³/min; treating the total as a rate makes it 30 times too large. | This is the commonest lost mark: total product, product at an instant and rate are three different quantities. |
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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