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DSE Biology | Enzymes and Factors Practice Paper 1 (Basics and Graphs)

DSE Biology | Enzymes and Factors Practice Paper 1 (Basics and Graphs)

  • 2026-11-04

DSE Biology | Enzymes and Factors Practice Paper 1 (Basics and Graphs)

From Q10, average rate and activation energy through substrate saturation, enzyme amount, pH and competitive inhibition, with every figure checkable.

Attempt first, mark afterwards Allow about thirty minutes; write the formula before substituting, and for any temperature item ask first whether the optimum has been passed.

01 Paper details (check answers afterwards)

Item Detail
Questions 10 items
Suggested time About 30 minutes
Difficulty mix Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter Chapter: DSE Biology | Enzymes and Affecting Factors
Level S4 (Secondary 4) Biology

02 Questions (10 items)

  1. FoundationAn enzyme makes 18 mg of product per minute at 20°C and 36 mg per minute at 30°C. What is its Q10?
  2. FoundationIf Q10 stays at 2 and the rate is 36 mg/min at 35°C, what is the theoretical rate at 45°C?
  3. FoundationIn 40 minutes the product rises from 0 to 96 mg. What is the average rate?
  4. FoundationAn enzyme lowers activation energy from 60 to 25 kJ/mol. By how many kJ/mol and what percentage?
  5. IntermediateRaising substrate from 2 to 4 mmol/L lifts the rate from 15 to 24, and from 4 to 8 only to 27. What are the two percentage rises and what does it show?
  6. IntermediateWith plenty of substrate, raising the enzyme from 0.5 to 1.5 mg lifts the rate from 8 to 24 units. How many times faster, and what relationship is shown?
  7. IntermediatePepsin works best near pH 2 and keeps only 12%% of that activity at pH 7. By how many percentage points does activity fall?
  8. ChallengeWith Q10 = 2 and a rate of 10 units at 25°C, 45°C should give 40 units but the measured value is only 8. Explain the difference.
  9. ChallengeAn enzyme has Vmax 120 units and Km 5 mmol/L. A competitive inhibitor raises Km to 20 while Vmax stays 120. At 20 mmol/L substrate, what is the rate before and after, and the fall in percent?
  10. Error redoError drill: a student writes that boiling only slows an enzyme down and cooling restores it fully. What is correct?

03 Answers and one-line notes

# Answer One-line note
1 Q10 = 2 Q10 = rate at 30°C divided by rate at 20°C = 36 over 18 = 2, meaning the rate doubles for every 10°C rise.
2 72 mg/min Multiply by Q10 once per 10°C step: 36 × 2 = 72.
3 2.4 mg/min Rate = increase in product divided by time = 96 over 40 = 2.4.
4 By 35 kJ/mol, which is 58.3% The fall is 60 minus 25 = 35, which is 58.3% of the original 60.
5 Up 60.0% then only 12.5%, showing the active sites are near saturation Doubling the substrate lifts the rate by 60.0% and then by just 12.5%, the classic saturation curve.
6 3 times, directly proportional to enzyme amount 24 over 8 = 3; triple the enzyme gives triple the rate when substrate is not limiting.
7 It falls by 88.0 percentage points 100% minus 12% = 88.0 points; extreme pH alters the charges and shape of the active site.
8 The rate rises to 20 by 35°C, then denaturation sets in, so the measured 8 stays below the theoretical 40 at 45°C Q10 holds only below the denaturation point; past the optimum the tertiary structure breaks down and the active site fails.
9 From 96.0 down to 60.0, a fall of 37.5% Michaelis-Menten v = Vmax × S over (Km + S): 120×20 over 25 = 96.0, and 120×20 over 40 = 60.0 with the inhibitor.
10 Boiling denatures the enzyme, destroying the tertiary structure irreversibly, so cooling cannot restore activity Denaturation damages the protein shape rather than slowing it temporarily, and an enzyme does not die.

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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