These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. No constants are needed. A calculator is assumed. Keep full calculator values until the last step. All data sets are fictional.
Question 1 (multiple choice · foundation)
An enzyme’s rate of reaction falls sharply when the temperature is raised above its optimum. Which statement best explains this?
- (A) Hydrogen bonds and other weak interactions in the enzyme break, so the active site changes shape and binds substrate less well.
- (B) Peptide bonds in the enzyme break, so it falls apart into amino acids.
- (C) Molecules move more slowly at high temperatures, so there are fewer collisions.
- (D) The enzyme is used up faster at higher temperatures.
Answer and explanation
Answer: (A). Above the optimum, heat breaks the weak interactions that hold the fold. The active site loses its shape, so fewer enzyme–substrate complexes form.
- (B) Denaturation does not break peptide bonds; the sequence stays intact.
- (C) Molecules move faster at higher temperatures. Collision frequency rises, not falls.
- (D) Enzymes are not used up at any temperature; they are reused.
Question 2 (multiple choice · core)
The initial rate of an enzyme reaction is 18.0 µmol min⁻¹ at 25 °C and 31.5 µmol min⁻¹ at 35 °C. What is the percentage increase?
- (A) 13.5%
- (B) 42.9%
- (C) 75.0%
- (D) 175%
Answer and explanation
Answer: (C). (31.5 − 18.0) ÷ 18.0 × 100 = 13.5 ÷ 18.0 × 100 = 75.0%.
- (A) gives the difference in rate (13.5 µmol min⁻¹), not a percentage.
- (B) divides by the new value: 13.5 ÷ 31.5 × 100 = 42.9%.
- (D) gives the new rate as a percentage of the old (31.5 ÷ 18.0 × 100 = 175%), not the increase.
Question 3 (multiple choice · core)
A drug reduces an enzyme’s rate by about 60% at every substrate concentration tested, even very high ones. The drug’s shape is quite unlike the substrate’s. Which conclusion is best supported?
- (A) It is a competitive inhibitor that binds the active site.
- (B) It is a noncompetitive inhibitor that binds an allosteric site and changes the shape of the active site.
- (C) It denatures the enzyme permanently by breaking its peptide bonds.
- (D) It reacts with the substrate so that less substrate is available.
Answer and explanation
Answer: (B). Extra substrate does not overcome the inhibition, and the drug does not resemble the substrate. Both fit binding at an allosteric site, which changes the active site.
- (A) A competitive inhibitor’s effect shrinks at high substrate concentration, and it usually resembles the substrate.
- (C) Denaturation does not break peptide bonds, and nothing in the data shows the effect is permanent.
- (D) Using up substrate would matter less at very high substrate concentrations, so the inhibition would not stay at 60%.
Question 4 (data analysis · core)
The initial rate of enzyme Q, from a plant, is measured at different pH values at 25 °C.
| pH | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|
| Rate / µmol min⁻¹ | 1.0 | 4.2 | 10.5 | 14.0 | 11.2 | 5.6 | 1.4 |
Two samples are then held for 10 minutes at extreme pH and returned to pH 6 for testing: sample M (held at pH 3) gives 12.6 µmol min⁻¹; sample N (held at pH 10) gives 1.4 µmol min⁻¹.
(a) State the optimum pH and calculate the rate at pH 8 as a percentage of the rate at the optimum. (b) Explain why the rate at pH 8 is lower than at the optimum. (c) Calculate the activity of M and N as a percentage of the rate at pH 6, and interpret each result. (d) Suggest one improvement to the method for finding the optimum.
Worked solution and suggested Marlbridge rubric
(a) Optimum pH 6 (14.0 µmol min⁻¹). At pH 8: 5.6 ÷ 14.0 × 100 = 40%.
(b) At pH 8 there are fewer H⁺ ions. Some R groups lose H⁺ and change charge. This disrupts hydrogen bonds and ionic interactions, so the enzyme’s shape and the charge pattern of the active site change, and fewer enzyme–substrate complexes form.
(c) M: 12.6 ÷ 14.0 × 100 = 90%. Most activity returned at pH 6, so most of the change at pH 3 was reversible: the enzyme refolded once normal conditions returned. N: 1.4 ÷ 14.0 × 100 = 10%. Very little activity returned, so most of the enzyme was permanently denatured at pH 10.
(d) Any one: test at intervals of 0.5 pH between pH 5 and 7; repeat each pH at least three times and calculate a mean; use buffers to hold each pH constant during the test.
| Point | What earns it |
|---|---|
| 1 | Optimum pH 6 and 40% |
| 1 | pH 8 explained through changed R-group charges disrupting hydrogen bonds/ionic interactions |
| 1 | Both percentages (90% and 10%) |
| 1 | M interpreted as largely reversible and N as largely permanent |
| 1 | A valid improvement |
Question 5 (constructed response · core)
In a cell, enzyme 1 converts substance A to substance B, and enzyme 2 converts B to C. Enzyme 1’s reaction is reversible. In a test tube, with A fixed at 10 mmol L⁻¹, the net rate of enzyme 1 is measured with different starting concentrations of B:
| B / mmol L⁻¹ | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Net rate of A → B / µmol min⁻¹ | 20 | 15 | 10 | 5 | 0 |
(a) Describe the trend, using numbers. (b) Explain the trend in terms of the relative concentrations of substrate and product. (c) A drug blocks enzyme 2 in the cell. Predict the effect on the concentration of B and on the net rate of enzyme 1, and justify your prediction. (d) Explain why the net rate is zero at 8 mmol L⁻¹ of B even though enzyme 1 is fully active.
Model answer and suggested Marlbridge rubric
(a) As B rises from 0 to 8 mmol L⁻¹, the net rate falls steadily from 20 to 0 µmol min⁻¹, by 5 µmol min⁻¹ for every 2 mmol L⁻¹ of B. At 4 mmol L⁻¹ it is half the starting rate.
(b) The higher the product (B) concentration relative to the substrate (A), the more often enzyme 1 binds B and catalyses the reverse reaction. The reverse reaction cancels part of the forward reaction, so the net forward rate falls.
(c) B would build up, because it is no longer being removed by enzyme 2. As B accumulates, the reverse reaction increases and the net rate of enzyme 1 falls, possibly to zero. Blocking one enzyme slows the step before it.
(d) At these concentrations, the forward reaction (A → B) and the reverse reaction (B → A) happen at the same rate. The enzyme is still catalysing both, but there is no net change.
| Point | What earns it |
|---|---|
| 1 | Trend described with values (20 to 0, or 5 per 2 mmol L⁻¹) |
| 1 | Product build-up increases the reverse reaction, lowering the net rate |
| 1 | Correct prediction for B and for enzyme 1, linked to B no longer being removed |
| 1 | Zero net rate explained by equal forward and reverse rates, not by inactive enzyme |
Question 6 (constructed response · stretch)
Compound K has a shape similar to the substrate of a bacterial enzyme. Researchers think K is a competitive inhibitor.
(a) Predict the effect of K on the rate at low and at high substrate concentrations, and explain each prediction. (b) Describe an experiment to test whether K is competitive. Identify the independent variable(s), the dependent variable and an appropriate control. (c) Describe the results that would support the claim, and the results that would suggest K is noncompetitive instead. (d) Explain why it matters that K binds reversibly if it is to be used as a medicine that can be stopped.
Model answer and suggested Marlbridge rubric
(a) Low substrate: a large fall in rate, because K often occupies the active site and few substrate molecules compete with it. High substrate: a small fall, because substrate molecules greatly outnumber K and win the active site most of the time.
(b) Measure the initial rate at a range of substrate concentrations (for example, five from low to high), each with and without a fixed concentration of K. Independent variables: substrate concentration; presence of K. Dependent variable: initial rate (product formed per minute). Control: the same series without K (enzyme, substrate and buffer only), giving the uninhibited rate at each substrate concentration. Keep temperature, pH, enzyme concentration and volumes constant; repeat each condition.
(c) Supports competitive: the rate with K is much lower at low substrate but approaches the uninhibited rate at high substrate, so the percentage of normal rises with substrate. Suggests noncompetitive: the rate with K is reduced by about the same proportion at every substrate concentration and levels off at a lower value.
(d) Reversible binding means K attaches and detaches. When the medicine is stopped and K is cleared from the body, it leaves the active sites and normal enzyme activity returns. An irreversible inhibitor would leave the enzyme permanently blocked.
| Point | What earns it |
|---|---|
| 1 | Both predictions in (a), each linked to competition for the active site |
| 1 | Design varies substrate concentration with and without K, with a correct dependent variable |
| 1 | No-K control identified and justified, plus one controlled variable |
| 1 | Correct expected pattern for competitive and for noncompetitive |
| 1 | Reversible binding linked to recovery of activity when K is removed |
Question 7 (calculation · stretch)
The initial rate of an enzyme is measured at different substrate concentrations with a fixed amount of enzyme.
| Substrate / mmol L⁻¹ | 1 | 2 | 4 | 8 | 16 | 32 |
|---|---|---|---|---|---|---|
| Rate / µmol min⁻¹ | 20.0 | 30.0 | 40.0 | 48.0 | 53.3 | 56.5 |
Under these conditions the maximum possible rate is 60 µmol min⁻¹.
(a) Calculate the increase in rate per mmol L⁻¹ of substrate between 1 and 2 mmol L⁻¹, and between 16 and 32 mmol L⁻¹. (b) Calculate the rate at 32 mmol L⁻¹ as a percentage of the maximum, and explain why adding substrate has so little effect there. (c) The amount of enzyme is doubled. Predict the new maximum rate and the new rate at 8 mmol L⁻¹, assuming all other conditions are unchanged. (d) Instead, a noncompetitive inhibitor is added that effectively inactivates 40% of the enzyme molecules. Predict the new maximum rate and the rate at 32 mmol L⁻¹.
Worked solution
(a) 1 → 2 mmol L⁻¹: (30.0 − 20.0) ÷ (2 − 1) = 10.0 µmol min⁻¹ per mmol L⁻¹. 16 → 32 mmol L⁻¹: (56.5 − 53.3) ÷ (32 − 16) = 3.2 ÷ 16 = 0.20 µmol min⁻¹ per mmol L⁻¹, fifty times smaller.
(b) 56.5 ÷ 60 × 100 = 94%. Almost every active site is already occupied (the enzyme is nearly saturated), so extra substrate has few empty active sites to bind.
(c) Doubling the enzyme doubles the number of active sites, so the maximum becomes 120 µmol min⁻¹ and the rate at 8 mmol L⁻¹ becomes 2 × 48.0 = 96.0 µmol min⁻¹. (This assumes substrate is not used up noticeably faster during the initial-rate measurement.)
(d) Only 60% of the enzyme molecules work, so the maximum becomes 0.60 × 60 = 36 µmol min⁻¹ and the rate at 32 mmol L⁻¹ about 0.60 × 56.5 = 33.9 µmol min⁻¹. Extra substrate cannot restore the lost 40%, because the inhibitor is not competing for the active site.
Suggested mark points (4): 1 for both gradients; 1 for 94% with saturation explained; 1 for 120 and 96.0; 1 for 36 and 33.9 with the reason that substrate does not overcome noncompetitive inhibition.
Common error: in (a), using the change in rate without dividing by the change in concentration (3.2 instead of 0.20) hides how flat the curve has become.
How did you do?
- Q1 or Q4 wrong: re-read “Temperature” and “pH: changing the charges” in the study guide, with Figures 1 and 2.
- Q2 wrong: rework Worked example 1 (a); divide by the old value.
- Q4 (c) wrong: re-read “Can denaturation be reversed?” and Worked example 1 (d).
- Q5 or Q7 wrong: re-read “Substrate and product concentrations”.
- Q3 or Q6 wrong: rework Worked example 2 and look again at Figures 3 and 4.
Then tick off the topic checklist.