Free Enzymes questions for AS Level Biology (9700), part of Cells & molecules. Each question has the answer and the marking points an examiner would look for.
Generate a full Enzymes worksheet →All Cells & molecules questionsAn enzyme is heated to 80 °C and then cooled to 30 °C.
Which row describes the enzyme?
| primary structure | tertiary structure | activity at 30 °C | |
|---|---|---|---|
| A | unchanged | changed | lower than before heating |
| B | unchanged | unchanged | the same as before heating |
| C | changed | changed | lower than before heating |
| D | unchanged | changed | higher than before heating |
Answer: A
Which statement describes Km?
Answer: C
An enzyme is immobilised in alginate beads.
1 The active site is destroyed by immobilisation.
2 Immobilised enzymes are often more stable at high temperatures.
3 Immobilisation always increases the rate of reaction.
Which statements are correct?
Answer: C
The initial rate of an enzyme-catalysed reaction was measured at different substrate concentrations.
| substrate concentration / mmol dm−3 | 1 | 2 | 4 | 8 | 16 |
|---|---|---|---|---|---|
| initial rate / µmol min−1 | 20 | 34.3 | 53.3 | 73.8 | 91.4 |
Explain the change in rate as the substrate concentration increases.
Vmax for this enzyme is 120 µmol min−1. Use the data to estimate Km, and explain what Km indicates.
A competitive inhibitor is added. Describe and explain the effect on Vmax and on Km.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | at low concentration rate limited by substrate (more collisions, more E–S complexes); at high concentration active sites saturated, enzyme concentration limiting rate increases with substrate at low concentrations B1 more enzyme–substrate complexes formed B1 levels off as active sites saturated B1 | 3 |
| (b) | 5 mmol dm−3; substrate concentration at ½Vmax; low Km means high affinity ½Vmax = 60 M1 Km ≈ 5 A1 inverse measure of affinity B1 | 3 |
| (c) | Vmax unchanged: high substrate outcompetes inhibitor; Km increases: more substrate needed to reach ½Vmax inhibitor binds active site, similar shape to substrate B1 Vmax unchanged B1 Km increases B1 explanation in terms of competition B1 | 4 |
Enzymes are affected by inhibitors.
Explain how a competitive inhibitor reduces the rate of an enzyme-catalysed reaction.
Explain how increasing substrate concentration affects competitive and non-competitive inhibition.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | similar shape, binds active site; fewer E–S complexes similar shape to substrate B1 binds to active site B1 fewer enzyme–substrate complexes formed B1 | 3 |
| (b) | overcomes competitive only competitive inhibition is reduced at high substrate concentration B1 non-competitive is not, as the inhibitor binds elsewhere B1 | 2 |
An enzyme has a maximum rate of reaction (Vmax) of 40 µmol min−1 and a Michaelis–Menten constant (Km) of 2 mmol dm−3.
Explain what is meant by Vmax and by Km.
State the rate at a substrate concentration of 2 mmol dm−3.
Explain what a low Km value indicates about the enzyme.
Describe and explain the effect of a competitive inhibitor on Vmax and Km.
Describe how a non-competitive inhibitor differs in its effect.
Enzymes can be immobilised in alginate beads. Suggest two advantages of using immobilised enzymes in industry.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | Vmax: maximum rate when all active sites occupied; Km: substrate concentration at half Vmax Vmax: the maximum rate, when all active sites are occupied (saturated) B1 Km: the substrate concentration at which the rate is half Vmax B1 | 2 |
| (b) | 20 µmol min−1 20 µmol min−1 B1 | 1 |
| (c) | high affinity for substrate; reaches half Vmax at low concentration high affinity for its substrate B1 the enzyme–substrate complex forms readily even at low substrate concentration B1 | 2 |
| (d) | Km increases, Vmax unchanged; inhibitor similar shape, binds active site; overcome by high substrate Km increases B1 Vmax unchanged B1 the inhibitor has a similar shape to the substrate and competes for the active site B1 at high substrate concentration the substrate out-competes the inhibitor so Vmax is still reached B1 | 4 |
| (e) | binds elsewhere (allosteric), changes active site shape; Vmax decreases, Km unchanged binds at a site other than the active site, changing its shape B1 Vmax is reduced (and cannot be overcome by more substrate) while Km is unchanged B1 | 2 |
| (f) | reused; product not contaminated; more stable to temperature/pH Any 2 of: enzyme can be reused / easily recovered; the product is not contaminated with enzyme; more stable over a wider range of temperature and pH B2 | 2 |
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