Cambridge International AS & A Level Biology 9700: Factors that affect enzyme action – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover subtopic 3.2, Factors that affect enzyme action, from the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. This is AS Level content, set on Papers 1 and 2 and assumed for Paper 4. The focus is data, Km and planning; for recall questions use the Enzymes practice questions. Learn the content first with the study guide and revision notes. Course hub: Cambridge A Level Biology; printable checklist.

Questions

1. A student times how long an amylase–starch mixture takes to stop giving a blue-black colour with iodine. The times are 40 s, 64 s and 125 s. Calculate the rate as 1000/t for each, to one decimal place. [2]

2. At a fixed, limited quantity of substrate, a graph of initial rate against enzyme concentration rises as a straight line and then levels off. Explain the shape of the graph. [3]

3. The table shows the time taken for amylase to clear a starch solution at different temperatures, all at pH 7.

Temperature / °C	20	30	40	50	60
Time / s	250	125	80	100	500

(a) Calculate the rate as 1000/t at each temperature. [2] (b) Calculate the percentage decrease in rate between 40 °C and 60 °C. [2] (c) Explain the change in rate from 20 °C to 40 °C and from 40 °C to 60 °C. [4]

4. A protease has an optimum pH of 8. Explain why it shows very little activity at pH 2. [3]

5. The initial rate of enzyme X was measured at different substrate concentrations. The curve levels off at a rate of 50 arbitrary units.

Substrate concentration / mmol dm⁻³	1	2	4	8	16	32	64
Initial rate / arbitrary units	10.0	16.7	25.0	33.3	40.0	44.4	47.1

(a) Use the data to find Km for enzyme X. Show how you used Vmax. [2] (b) Enzyme Y acts on the same substrate, has the same Vmax and has a Km of 12 mmol dm⁻³. At 4 mmol dm⁻³ substrate, enzyme Y has a rate of 12.5 units. Compare the affinities of X and Y for the substrate, using the data. [3]

6. The rate of an enzyme-catalysed reaction was measured with increasing concentrations of a reversible inhibitor.

Inhibitor concentration / mmol dm⁻³	0	0.5	1.0	2.0	4.0
Rate / arbitrary units	48	36	30	20	12

(a) Calculate the percentage inhibition at 2.0 mmol dm⁻³ inhibitor. [2] (b) Explain the trend in the data. [2]

7. The rate of an enzyme reaction was measured with no inhibitor, with inhibitor A and with inhibitor B, at a low and a high substrate concentration. Each inhibitor was used at the same fixed concentration.

Substrate concentration / mmol dm⁻³	No inhibitor	Inhibitor A	Inhibitor B
1	20.0	8.6	10.0
20	54.5	46.2	27.3

(a) Calculate the rate with each inhibitor as a percentage of the uninhibited rate at both substrate concentrations. [2] (b) Identify the type of inhibitor A and inhibitor B. Explain your answers. [4]

8. Describe and explain the effect of a competitive inhibitor and of a non-competitive inhibitor on Vmax and on Km. [4]

9. Equal quantities of an enzyme were used free in solution and immobilised in alginate beads. Initial rates were measured before and after both were kept at 60 °C for 10 minutes.

	Free enzyme	Immobilised enzyme
Rate before heating / arbitrary units	2.40	1.60
Rate after heating / arbitrary units	0.36	1.12

(a) Calculate the percentage of its original rate that each preparation kept after heating. [2] (b) Suggest why the immobilised enzyme had a lower rate before heating. [2] (c) Suggest why the immobilised enzyme kept more of its activity. [1] (d) State two other advantages of using immobilised enzymes. [2]

10. Plan an investigation into the effect of pH on the activity of a bacterial protease, using a cloudy suspension of milk protein that becomes clear as the protein is digested. [8]

11. Catalase was added to hydrogen peroxide and the volume of oxygen collected was plotted against time. A tangent drawn at t = 0 passes through (0 s, 0 cm³) and (20 s, 13.0 cm³). A tangent drawn at t = 90 s passes through (60 s, 31.0 cm³) and (120 s, 37.0 cm³). By 30 s, 17.5 cm³ had been collected.

(a) Calculate the initial rate of reaction. [2] (b) Calculate the rate at 90 s. [2] (c) Explain why the rate at 90 s is lower than the initial rate. [2] (d) The mean rate over the first 30 s is 0.58 cm³ s⁻¹. Explain why the initial rate, not this mean rate, should be used to compare different catalase concentrations. [1]

Answers

1. Uses 1000 ÷ time for each [1]; 25.0, 15.6 and 8.0 [1] Examiner insight: Rates in one column must share the same decimal places; mixing 25 and 15.625 loses the accuracy mark.

2. Initially, as enzyme concentration increases, more active sites are available, so more enzyme–substrate complexes form per unit time [1]; rate is proportional to enzyme concentration while substrate is in excess [1]; the line levels off because substrate becomes the limiting factor and some active sites are unoccupied [1] Examiner insight: “Enzyme is used up” or “enzyme is saturated” scores nothing here; the mark needs substrate named as the limiting factor.

3. (a) Method 1000 ÷ t [1]; 4.0, 8.0, 12.5, 10.0, 2.0 [1] (b) (12.5 − 2.0) ÷ 12.5 × 100 [1] = 84% [1] (c) 20–40 °C: more kinetic energy, so more frequent collisions between enzyme and substrate [1]; more enzyme–substrate complexes form per second [1]. 40–60 °C: hydrogen and ionic bonds in the tertiary structure break, so the active site changes shape [1]; substrate is no longer complementary, fewer enzyme–substrate complexes, enzyme denatured [1] Examiner insight: Each half of (c) carries its own marks; error carried forward is allowed in (b) from wrong rates in (a).

4. The high H⁺ concentration changes the charges on R groups in the enzyme [1]; ionic and hydrogen bonds holding the tertiary structure are disrupted, so the active site changes shape [1]; the substrate no longer fits or is no longer attracted to the active site, so few enzyme–substrate complexes form [1] Examiner insight: Mark schemes expect the link from H⁺ ions to R-group charges; “the pH is wrong for the enzyme” earns no credit.

5. (a) ½Vmax = 50 ÷ 2 = 25 units [1]; reading across to the data, Km = 4 mmol dm⁻³ [1] (b) X has a lower Km (4 vs 12 mmol dm⁻³), so X has the higher affinity [1]; X reaches half its maximum rate at a lower substrate concentration, so forms enzyme–substrate complexes more readily [1]; at 4 mmol dm⁻³, X works at 25 units, twice the rate of Y at 12.5 units [1] Examiner insight: “Use the data” means a quoted or manipulated figure is needed for full marks.

6. (a) (48 − 20) ÷ 48 × 100 [1] = 58% [1] (b) As inhibitor concentration increases, rate decreases [1]; more enzyme molecules have inhibitor bound at any moment, so fewer enzyme–substrate complexes form [1] Examiner insight: Describing the trend alone gets one mark; the second needs the reason in terms of enzyme–substrate complexes.

7. (a) A: 43% and 85%; B: 50% and 50% — one mark for each inhibitor’s pair of values [1] [1] (b) A is competitive [1]: its inhibition is much smaller at high substrate concentration, because substrate molecules out-compete A for the active site [1]. B is non-competitive [1]: it reduces the rate by the same proportion at both concentrations, because it binds away from the active site and more substrate cannot overcome it [1] Examiner insight: Naming each type is one mark and the evidence from the figures is a separate mark.

8. Competitive: Vmax unchanged, because at very high substrate concentration substrate out-competes the inhibitor [1]; Km increased, because more substrate is needed to reach half Vmax [1]. Non-competitive: Vmax lowered, because inhibited enzyme molecules cannot form products whatever the substrate concentration [1]; Km unchanged, because uninhibited enzyme molecules bind substrate normally [1] Examiner insight: Each mark needs the effect and its explanation; a bare list of effects scores little.

9. (a) Free: 0.36 ÷ 2.40 × 100 = 15% [1]; immobilised: 1.12 ÷ 1.60 × 100 = 70% [1] (b) Substrate has to diffuse into the bead and product out [1]; some active sites inside the gel are less accessible to substrate [1] (c) The alginate gel holds the enzyme molecules and limits unfolding, so fewer are denatured [1] (d) Any two: product not contaminated with enzyme [1]; enzyme recovered and reused [1]; allows a continuous process in a column Examiner insight: “Other” means stability will not be credited again in (d), and “cheaper” alone needs the reason (reuse) to score.

10. Independent variable: pH, using buffer solutions at five or more values, for example pH 4, 5, 6, 7, 8, 9, 10 [1]; same volume of buffer added to each tube [1]. Dependent variable: time for the milk suspension to clear, or change in absorbance measured with a colorimeter [1]. Standardised: temperature, using a thermostatically controlled water bath [1]; same concentration and volume of protease and of milk suspension [1]. Equilibrate enzyme and substrate separately at the test temperature before mixing [1]. Three or more replicates at each pH, calculating a mean and identifying anomalies [1]. Control with boiled protease to show clearing is due to enzyme activity; wear eye protection and gloves as enzymes can be irritants [1] Examiner insight: Planning marks need specific detail; “keep the temperature the same” without a water bath, or “different pHs” without buffers and values, scores nothing.

11. (a) Gradient = 13.0 ÷ 20 [1] = 0.65 cm³ s⁻¹ [1] (b) (37.0 − 31.0) ÷ (120 − 60) [1] = 0.10 cm³ s⁻¹ [1] (c) Hydrogen peroxide is being broken down, so its concentration falls [1]; fewer collisions between substrate and active sites, so fewer enzyme–substrate complexes per second [1] (d) At t = 0 substrate concentration is the same in every tube, so only enzyme concentration differs; the mean rate is affected by substrate being used up [1] Examiner insight: Units are required; a correct value without cm³ s⁻¹ loses the accuracy mark.

Where marks are usually lost

• Recording rates in a table to inconsistent decimal places.
• Saying the substrate or enzyme curve levels off “because the enzyme is used up”.
• Explaining a fall above the optimum without linking broken bonds to active-site shape.
• Explaining pH effects without mentioning H⁺ ions and R-group charges.
• Reading Km at half the highest measured rate instead of half of Vmax.
• Naming an inhibitor type without quoting evidence from the data.
• Leaving units off rates calculated from a tangent.
• Using a mean rate over a long time interval instead of the initial rate.

Next steps

• Revise with the Factors that affect enzyme action revision notes
• Re-read the Factors that affect enzyme action study guide
• Whole topic: Enzymes study guide and Enzymes revision notes
• Take the 9700 AS diagnostic
• Course hub: Cambridge A Level Biology and printable checklist
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, published by Cambridge International: Topic 3 Enzymes, subtopic 3.2 Factors that affect enzyme action, learning outcomes 1–4.