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IB Biology: Enzymes and Reaction Kinetics (10 Error-Retry Items)

IB Biology: Enzymes and Reaction Kinetics (10 Error-Retry Items)

  • 2026-11-08

IB Biology: Enzymes and Reaction Kinetics (10 Error-Retry Items)

Ten error-retry items correcting rate units, relative-activity drops, Q10 estimates, Vmax with Km, immobilised-enzyme trends and the belief that boiling is reversible.

Focus Write the equation and unit before substituting, and state the error being corrected, not just the new answer.

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 (SL/HL)

02 Questions (10 items)

  1. FoundationAn enzyme experiment produces 36 cm³ of gas in 6 minutes. Find the mean rate.
  2. FoundationActivity at the optimum is set at 80 and measures 32 at another temperature. Find the relative activity and the fall.
  3. FoundationThe rate is 5 cm³/min at 10°C and 10 cm³/min at 20°C. Find Q10.
  4. Foundation20 cm³ of substrate is completely broken down in 5 minutes. Find the mean rate of use.
  5. IntermediateQ10 is 1.6 and the rate at 20°C is 12 cm³/min. Estimate the rate at 30°C, assuming no denaturation.
  6. IntermediateVmax is 60 cm³/min and Km is 2 mM. Find the rate at a substrate concentration of 2 mM.
  7. IntermediateKm is 4 mM and the substrate rises from 4 mM to 8 mM. Find the rate as a percentage of Vmax before and after.
  8. ChallengeAn immobilised enzyme reused 5 times shows activities of 100, 92, 84, 76 and 68 (%). Find the mean fall per reuse.
  9. ChallengeProduct rises from 0 to 18 cm³ in minute 1 and by another 9 cm³ in minute 2. Find the initial rate and the mean rate over two minutes.
  10. Error redoA student writes that a boiled enzyme solution regains full activity on cooling because the enzyme was not destroyed. What is wrong?

03 Answers and one-line notes

# Answer One-line note
1 36 divided by 6 gives 6 cm³/min. Rate is always amount of product divided by time; never invert the two.
2 Relative activity is 32 divided by 80, giving 40%, so the fall is 60 percentage points. Relative activity and percentage fall are different figures; answer the one asked.
3 10 divided by 5 gives 2, so Q10 equals 2. Q10 is the rate multiplier per 10°C rise, not a temperature difference.
4 20 divided by 5 gives 4 cm³/min. Using substrate used up also means dividing by time, and the unit must be given.
5 12 multiplied by 1.6 gives 19.2 cm³/min. Check that the temperature is still below the optimum before extrapolating, or the rate is overestimated.
6 v equals 60 times 2 divided by 4, giving 30 cm³/min, which is half of Vmax. When substrate equals Km the rate is exactly half of Vmax; that is the definition of Km.
7 Four over eight gives 50%, and eight over twelve gives about 66.7%, so doubling substrate only lifts the rate from half to two thirds. Enzyme kinetics is not proportional; saturation must be read from Vmax and Km.
8 100 minus 68 divided by 4 gives 8, so activity falls 8 percentage points per reuse. Five readings create only four intervals; dividing by five is the classic off-by-one error.
9 The initial rate is 18 cm³/min and the mean over two minutes is 27 divided by 2, giving 13.5 cm³/min. The two figures differ, so the time interval must be named in the answer.
10 Boiling denatures the enzyme: hydrogen and ionic bonds break, the tertiary structure and active site change permanently, and cooling cannot restore them. Only low-temperature slowing is reversible. Confusing low-temperature slowing with heat denaturation is the commonest conceptual error in enzyme work.

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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