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, then round to the significant figures of the data. All data sets are fictional.
Question 1 (multiple choice · foundation)
Which statement best describes how an enzyme increases the rate of a reaction in a cell?
- (A) It supplies the energy that the substrate molecules need to react.
- (B) It lowers the activation energy, so a much larger fraction of substrate molecules can react in a given time.
- (C) It makes the free energy change of the reaction more negative, so more product forms.
- (D) It reacts with the substrate and is converted into one of the products.
Answer and explanation
Answer: (B). An enzyme provides a route with a lower energy barrier, so more encounters succeed each second.
- (A) Enzymes do not add energy; they lower the barrier.
- (C) ΔG depends only on reactants and products; the enzyme does not change it.
- (D) The enzyme is released unchanged and reused; it is not a reactant.
Question 2 (multiple choice · core)
An energy profile is drawn for a reaction in a cell. The values read from it are:
| Feature | Value / kJ mol⁻¹ |
|---|---|
| Activation energy without enzyme | 86 |
| Activation energy with enzyme | 34 |
| Free energy of products minus free energy of reactants (no enzyme) | −20 |
Which row is correct for the reaction with the enzyme?
- (A) Activation energy lowered by 52 kJ mol⁻¹; ΔG = −20 kJ mol⁻¹
- (B) Activation energy lowered by 52 kJ mol⁻¹; ΔG = −72 kJ mol⁻¹
- (C) Activation energy lowered by 120 kJ mol⁻¹; ΔG = −20 kJ mol⁻¹
- (D) Activation energy lowered by 34 kJ mol⁻¹; ΔG = −20 kJ mol⁻¹
Answer and explanation
Answer: (A). The barrier falls from 86 to 34 kJ mol⁻¹, a drop of 86 − 34 = 52 kJ mol⁻¹. The reactant and product levels do not move, so ΔG stays at −20 kJ mol⁻¹.
- (B) adds the drop in barrier to ΔG (−20 − 52 = −72). An enzyme does not change ΔG.
- (C) adds the two activation energies (86 + 34 = 120) instead of subtracting.
- (D) gives the new activation energy (34) as if it were the amount by which it was lowered.
Question 3 (multiple choice · core)
The active site of an enzyme is a narrow, curved groove. Two positively charged R groups sit at opposite ends of the groove, and its floor is lined with nonpolar R groups. Which molecule is most likely to bind and react?
- (A) A long, curved molecule with a negatively charged group at each end and a nonpolar middle section.
- (B) A long, curved molecule with a positively charged group at each end and a nonpolar middle section.
- (C) A long, curved molecule that is nonpolar along its whole length.
- (D) A large, branched molecule with a negatively charged group at each end and a nonpolar middle.
Answer and explanation
Answer: (A). It matches the groove in shape, its negative ends attract the positive R groups, and its nonpolar middle lies on the nonpolar floor.
- (B) has the right shape, but its positive ends would be repelled by the positive R groups.
- (C) has the right shape but no negative groups to be attracted to the positive R groups, so it is held only weakly and is less likely to bind than (A).
- (D) has the right charges but the wrong shape: a large, branched molecule cannot fit a narrow groove.
Question 4 (data analysis · core)
Lactase hydrolyses the milk sugar lactose into glucose and galactose. A student adds lactase to lactose solution at 30 °C and measures glucose every minute. A second tube gets water instead of lactase.
| Time / min | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Lactase: glucose / mmol L⁻¹ | 0.0 | 1.2 | 2.4 | 3.5 | 4.4 | 5.0 | 5.3 |
| No lactase: glucose / mmol L⁻¹ | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
(a) Calculate the initial rate of glucose production over the first 2 minutes. (b) Calculate the mean rate between 4 and 6 minutes, and express it as a percentage of the initial rate. (c) Explain why the rate falls. (d) A student says: “The rate falls because the lactase is being used up.” Evaluate this claim and describe a simple extra step that would test it. (e) State the purpose of the tube without lactase.
Worked solution and suggested Marlbridge rubric
(a) (2.4 − 0.0) mmol L⁻¹ ÷ 2 min = 1.2 mmol L⁻¹ min⁻¹.
(b) (5.3 − 4.4) ÷ 2 = 0.9 ÷ 2 = 0.45 mmol L⁻¹ min⁻¹. As a percentage: 0.45 ÷ 1.2 × 100 = 37.5% (about 38%) of the initial rate.
(c) As lactose is hydrolysed, its concentration falls. Fewer lactose molecules collide with active sites each minute, so fewer enzyme–substrate complexes form.
(d) The claim is not supported. Lactase is released unchanged after each reaction and reused; the falling substrate concentration explains the slowdown. Test: at 6 minutes, add more lactose. If the rate rises again, the lactase was still active.
(e) It is a negative control. Glucose stays at 0.0, so lactose does not break down by itself in this time; the glucose in the first tube comes from the lactase.
| Point | What earns it |
|---|---|
| 1 | Initial rate 1.2 mmol L⁻¹ min⁻¹ with units |
| 1 | 0.45 mmol L⁻¹ min⁻¹ and about 38% |
| 1 | Rate falls because substrate concentration falls, so fewer collisions with active sites |
| 1 | Claim rejected because the enzyme is reused, with a valid test (add more substrate) |
| 1 | No-lactase tube identified as a negative control that rules out breakdown without enzyme |
Do not award the third point for “the enzyme slows down” without naming substrate.
Question 5 (constructed response · core)
Fresh pineapple contains bromelain, a protease (an enzyme that digests proteins). Gelatin is a protein that sets into a gel. To test whether fresh pineapple juice digests gelatin, a student sets gelatin in four identical tubes, adds 2 cm³ of a different liquid to each, and after 24 hours at 20 °C measures the depth of gelatin turned to liquid.
| Tube | Liquid added | Depth of liquefied gelatin / mm |
|---|---|---|
| W | fresh pineapple juice | 12 |
| X | pineapple juice boiled for 5 min, then cooled | 0 |
| Y | distilled water | 0 |
| Z | a purchased protease solution | 15 |
(a) Identify the independent variable and the dependent variable. (b) Identify the negative control(s) and the positive control, and justify each one. (c) Explain what tube X adds that tube Y does not. (d) State two variables that should be kept the same.
Model answer and suggested Marlbridge rubric
(a) Independent variable: the liquid added. Dependent variable: depth of liquefied gelatin after 24 hours, in mm.
(b) Negative controls: Y (water) shows that gelatin does not liquefy just from having liquid on it for 24 hours. X (boiled juice) shows what happens when the juice is present but its proteins have lost their shape. Positive control: Z (known protease) shows that the method can detect protein digestion.
(c) Tube X keeps everything else in the juice (sugars, acids, other small molecules) but destroys protein shape. Because boiled juice has no effect, the digestion in W depends on a heat-sensitive protein, consistent with an enzyme rather than, for example, acid in the juice. Tube Y cannot show this.
(d) Any two: volume of liquid added; volume and concentration of gelatin; temperature; time; tube diameter.
| Point | What earns it |
|---|---|
| 1 | Both variables correct, with the dependent variable measurable (depth in mm) |
| 1 | Water identified as a negative control with a reason |
| 1 | Protease identified as a positive control with a reason (shows the test can work) |
| 1 | Boiled juice explained as ruling out non-protein substances in the juice |
| 1 | Two sensible controlled variables |
Question 6 (constructed response · stretch)
The substrate of a fictional bacterial enzyme carries a negatively charged phosphate group. In the active site, the R group of an arginine (positively charged) sits next to this phosphate when the substrate binds. Researchers make two mutants, each with one amino acid changed, and measure activity under identical conditions.
| Enzyme | Change at this position | Activity / µmol min⁻¹ mg⁻¹ |
|---|---|---|
| Wild type | none (arginine, positive) | 50.0 |
| Mutant 1 | leucine (nonpolar, uncharged) | 0.8 |
| Mutant 2 | lysine (positive, slightly different shape) | 31.5 |
(a) Calculate the activity of each mutant as a percentage of the wild type. (b) Explain the large fall in activity for Mutant 1. (c) Explain why Mutant 2 keeps much more activity than Mutant 1, but still less than the wild type. (d) A third mutation changes a nonpolar amino acid on the far side of the enzyme to another nonpolar amino acid. Predict its effect and justify your prediction.
Model answer and suggested Marlbridge rubric
(a) Mutant 1: 0.8 ÷ 50.0 × 100 = 1.6%. Mutant 2: 31.5 ÷ 50.0 × 100 = 63%.
(b) In the wild type, the positive arginine attracts the negative phosphate and holds the substrate in position. Leucine has no charge, so this attraction is lost. Far fewer enzyme–substrate complexes form, or the substrate is not held where the activation energy is lowered, so activity drops to 1.6%.
(c) Lysine is also positive, so the attraction to the phosphate is kept and the substrate still binds. But lysine’s R group has a different shape, so the fit is less exact: charge compatible, shape poorer, giving 63%.
(d) Little or no change. The change is far from the active site and swaps one nonpolar R group for another, so the hydrophobic interactions holding the fold are probably kept, and the active site’s shape and charge are unlikely to change.
| Point | What earns it |
|---|---|
| 1 | Both percentages correct |
| 1 | Mutant 1: loss of positive charge and loss of attraction to the negative phosphate |
| 1 | Consequence: fewer enzyme–substrate complexes / substrate not held in position to lower activation energy |
| 1 | Mutant 2: charge kept but shape fit poorer |
| 1 | Third mutation: little effect, justified by distance from active site and similar R group |
Question 7 (calculation · stretch)
In a test tube containing 2.0 nmol of an enzyme and plenty of substrate, 0.60 mmol of product forms at a steady rate in 5.0 minutes. Without the enzyme, the same reaction produces product at 4.0 × 10⁻⁸ mol min⁻¹. (1 mmol = 10⁻³ mol; 1 nmol = 10⁻⁹ mol.)
(a) Calculate the rate of the catalysed reaction in mmol min⁻¹. (b) Calculate how many substrate molecules each enzyme molecule converts per second, on average. (c) Calculate how many times faster the catalysed reaction is, and how long the uncatalysed reaction would take to make 0.60 mmol. (d) Use your answer to (b) to explain why cells need only small amounts of each enzyme.
Worked solution
(a) Rate = 0.60 mmol ÷ 5.0 min = 0.12 mmol min⁻¹ (1.2 × 10⁻⁴ mol min⁻¹).
(b)
- Moles of product per mole of enzyme: (0.60 × 10⁻³ mol) ÷ (2.0 × 10⁻⁹ mol) = 3.0 × 10⁵. Each enzyme molecule catalysed about 300 000 reactions.
- Time: 5.0 min = 300 s.
- Per second: 3.0 × 10⁵ ÷ 300 s = 1.0 × 10³ s⁻¹, about 1000 reactions per enzyme molecule per second.
As background, this number is called the turnover number.
(c) Ratio: (1.2 × 10⁻⁴ mol min⁻¹) ÷ (4.0 × 10⁻⁸ mol min⁻¹) = 3000 times faster. Time without enzyme: (0.60 × 10⁻³ mol) ÷ (4.0 × 10⁻⁸ mol min⁻¹) = 15 000 min = 250 h, about 10.4 days, instead of 5 minutes.
(d) Each enzyme molecule is released unchanged after every reaction and binds a new substrate. One molecule catalyses hundreds of thousands of reactions in minutes, so a few enzyme molecules process a lot of substrate.
Suggested mark points (4): 1 for 0.12 mmol min⁻¹; 1 for about 1000 s⁻¹ with the unit conversion shown; 1 for 3000 times and about 250 h (10 days); 1 for linking reuse of the unchanged enzyme to the large number in (b).
Common error: skipping the unit conversions gives 0.60 ÷ 2.0 ÷ 300 = 0.001 s⁻¹, a million times too small.
How did you do?
- Q1 or Q2 wrong: re-read “Activation energy: the barrier every reaction must cross” in the study guide and look again at Figure 1.
- Q3 or Q6 wrong: re-read “Enzyme structure: the active site” and Figure 2; use shape and charge.
- Q4 or Q7 calculations wrong: rework Worked example 1; check units and use the gradient.
- Q5 incomplete: re-read Worked example 2, especially the purpose of each control.
Then tick off the topic checklist.