AQA A-level Biology: Enzymes and Required Practical 1 (7402) – 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 section 3.1.4.2 and Required Practical 1 of the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015. Use the study guide and revision notes first. Course links: the AQA A-level Biology hub, the printable checklist and the free diagnostics.

Questions

1. (a) Calculate the pH of a solution with a hydrogen ion concentration of 5.0 × 10⁻⁵ mol dm⁻³. [1] (b) Calculate the hydrogen ion concentration of a buffer at pH 9.0. [1] (c) How many times greater is the hydrogen ion concentration at pH 5 than at pH 8? [1]

2. Describe how, according to the induced-fit model, an enzyme lowers the activation energy of a reaction. [4]

3. A mutation changes one amino acid in the active site of an enzyme and the enzyme stops working. Explain why. [3]

4. A protease from a bacterium that lives in hot springs was mixed with a cloudy protein suspension at different temperatures. The time taken for the suspension to clear was recorded.

Temperature (°C)	40	50	60	70	80
Time to clear (s)	300	150	75	50	120

(a) Calculate the rate at 70 °C as 1 ÷ time, in s⁻¹. [1] (b) Calculate the percentage change in rate between 70 °C and 80 °C. [2] (c) Explain the changes in rate from 40 °C to 70 °C and from 70 °C to 80 °C. [3]

5. One enzyme reaction was run three ways: without inhibitor, with inhibitor X and with inhibitor Y. Rates are in arbitrary units.

Substrate concentration	0.5	1	2	4	8	16
No inhibitor	1.6	2.7	4.0	5.3	6.4	7.1
Inhibitor X	0.6	1.1	2.0	3.2	4.6	5.8
Inhibitor Y	0.8	1.3	2.0	2.7	3.2	3.6

(a) Identify which inhibitor is competitive and which is non-competitive. Use calculations of percentage reduction in rate to support your answer. [4] (b) Explain how the non-competitive inhibitor reduces the rate. [2]

6. A student has a 3.0% enzyme stock and wants 20 cm³ of each diluted solution. (a) Calculate the volume of stock and the volume of water needed for a 1.5% solution. [2] (b) Explain why every solution must have the same final volume. [1] (c) With excess substrate, predict the shape of a graph of rate against enzyme concentration. [1]

7. A student followed the formation of a product at a fixed temperature.

Time (s)	0	15	30	45	60	90	120	180
Product (mmol dm⁻³)	0.00	0.23	0.39	0.51	0.59	0.69	0.74	0.79

A tangent drawn at time 0 passes through (0, 0.00) and (30, 0.53). (a) Calculate the initial rate, to 2 significant figures, with units. [2] (b) Calculate the mean rate over the first 60 s, to 2 significant figures. [1] (c) Explain why the mean rate is lower than the initial rate. [3]

8. (a) Compare the lock-and-key model with the induced-fit model of enzyme action. [3] (b) State whether each enzyme works inside or outside the cell that made it: (i) catalase in a liver cell; (ii) amylase in saliva. [2]

9. Plan an investigation of how pH affects the speed at which catalase breaks down hydrogen peroxide, giving the variable you would measure and how you would use your results. [7]

10. A student measured the volume of oxygen from a catalase reaction. Each syringe reading has an uncertainty of ±0.5 cm³, so a volume found by difference has an uncertainty of ±1.0 cm³. (a) The student collected 8.0 cm³. Calculate the percentage uncertainty. [2] (b) Calculate the percentage uncertainty if 20.0 cm³ were collected with the same syringe. [1] (c) Explain why the student should repeat each measurement. [2]

11. Pepsin, a protease secreted into the stomach, is fully active at pH 2.0. Tested at pH 6.8 it showed no activity. (a) Calculate the hydrogen ion concentration at pH 2.0 and at pH 6.8. [2] (b) Calculate how many times greater the concentration is at pH 2.0, to 2 significant figures. [1] (c) Explain, using tertiary structure, why pepsin has no activity at pH 6.8. [4] (d) Explain why pepsin is classified as an extracellular enzyme. [1]

Answers

1. (a) pH = -log₁₀(5.0 × 10⁻⁵) = 4.3 [1] (b) [H⁺] = 10⁻⁹·⁰ = 1.0 × 10⁻⁹ mol dm⁻³ [1] (c) Three pH units, so 10³ = 1000 times [1] Examiner insight: Show the substitution into the formula and give the unit for concentration, so that the working is easy to follow. [3]

2. The substrate attaches at the active site, giving an enzyme-substrate complex [1]. Before binding, the active site is not an exact fit [1]. It shifts slightly in shape during binding and wraps around the substrate [1]. This strains bonds in the substrate (or holds two substrates close together), so less energy is needed to start the reaction [1]. [4] Examiner insight: Answers gain credit for saying the active site changes shape, not just that substrate fits. State the effect on activation energy at the end.

3. The amino acid sequence (primary structure) determines the tertiary structure [1]. A different amino acid changes the shape or charge of the active site [1]. The substrate is no longer complementary, so an enzyme-substrate complex cannot form [1]. [3] Examiner insight: Link the sequence to tertiary structure, then to the active site. Denaturation is a different process, so writing “it was denatured” does not answer this question.

4. (a) 1 ÷ 50 = 0.020 s⁻¹ [1] (b) Rate at 80 °C = 1 ÷ 120 = 0.0083 s⁻¹ [1]. Change = (0.0083 - 0.020) ÷ 0.020 × 100 = -58% [1] (c) Up to 70 °C, more kinetic energy gives more collisions and more enzyme-substrate complexes per second [1]. Above 70 °C, bonds holding the tertiary structure break [1]. The active site changes shape, and some enzyme molecules are denatured, so the rate falls [1]. [6] Examiner insight: A percentage change needs a sign or the word fall. Explain both halves of a rise-then-fall curve, because a one-sided explanation can only earn part of the marks.

5. (a) X at substrate concentration 0.5: (1.6 - 0.6) ÷ 1.6 × 100 = 62.5% [1]. At 16: (7.1 - 5.8) ÷ 7.1 × 100 = 18% [1]. X has a smaller effect as substrate rises and is heading for the same maximum, so X is competitive [1]. Y: 50% at 0.5 and about 49% at 16, an almost constant percentage reduction with a lower plateau, so Y is non-competitive [1] (b) It binds to a site other than the active site [1]. This alters the tertiary structure and the shape of the active site, so fewer substrates can bind, and extra substrate cannot overcome this [1] [6] Examiner insight: Quote figures from the table for both inhibitors. Saying only “the curve is lower” gives no evidence for the type.

6. (a) Stock volume = (1.5 ÷ 3.0) × 20 = 10.0 cm³ [1]. Water = 20 - 10.0 = 10.0 cm³ [1] (b) The same final volume keeps the substrate concentration the same, so enzyme concentration is the only variable changed [1] (c) A straight line through the origin (rate proportional to enzyme concentration) [1] [4] Examiner insight: Both the stock volume and the water volume are asked for, so give both. A graph prediction should say where the line passes.

7. (a) Gradient = 0.53 ÷ 30 = 0.01767 [1]. Rate = 0.018 mmol dm⁻³ s⁻¹ [1] (b) Mean rate = 0.59 ÷ 60 = 0.0098 mmol dm⁻³ s⁻¹ [1] (c) The rate is greatest at time 0, because substrate is most plentiful then [1]. The rate falls as substrate is used up (and product builds up) [1]. The mean averages this decrease, so it is lower than the rate at time 0 [1] [6] Examiner insight: Give the unit and the stated number of significant figures. For the explanation, link the falling gradient to the changing substrate concentration.

8. (a) In lock and key, the active site is a rigid shape that is complementary to the substrate [1]. In induced fit, the active site changes shape as the substrate binds [1]. This puts strain on the substrate and further lowers activation energy, which lock and key does not explain [1] (b) (i) Inside (intracellular) [1] (ii) outside (extracellular), because it is secreted by the salivary glands [1] [5] Examiner insight: A comparison needs both models side by side, saying what each does.

9. Independent variable: pH, using at least five buffer solutions across a range of values [1]. Dependent variable: the volume of oxygen collected, using a gas syringe or an inverted measuring cylinder, recorded at set time intervals [1]. Control: the same volume and concentration of hydrogen peroxide, the same volume of catalase source, and a constant temperature in a water bath [1]. Repeat at each pH and calculate a mean [1]. Plot volume against time for each pH and draw a tangent at time 0 to find the initial rate [1]. Plot the initial rate against pH [1]. Safety: wear eye protection because hydrogen peroxide can irritate [1] [7] Examiner insight: A plan should say how each variable is kept constant. Naming a control variable without saying how it is controlled is less secure than giving the method.

10. (a) 1.0 ÷ 8.0 [1] × 100 = 12.5% [1] (b) 1.0 ÷ 20.0 × 100 = 5.0% [1] (c) Repeats show whether results are reliable and let an anomaly be identified and left out [1]. A mean reduces the effect of random error [1] [5] Examiner insight: Percentage uncertainty is absolute uncertainty divided by the measurement, not the other way round. Larger readings from the same apparatus give a smaller percentage uncertainty.

11. (a) [H⁺] at pH 2.0 = 1.0 × 10⁻² mol dm⁻³ [1]. At pH 6.8 = 1.6 × 10⁻⁷ mol dm⁻³ [1] (b) (1.0 × 10⁻²) ÷ (1.6 × 10⁻⁷) = 6.3 × 10⁴ times [1] (c) The pH is far from the optimum, so hydrogen ion concentration changes [1]. This disrupts ionic and hydrogen bonds in the enzyme [1]. The tertiary structure changes, so the active site changes shape [1]. The substrate is no longer complementary, so no enzyme-substrate complexes form [1] (d) It is secreted by cells and acts outside them, in the stomach [1] [8] Examiner insight: In pH explanations, name the bonds and the structure level affected. Saying “pH denatures it” without a mechanism leaves out the reasoning that earns marks.

Where marks are usually lost

• Using “killed” or “dead” for an enzyme in place of “denatured”.
• Saying the substrate “fits exactly” and ignoring induced fit.
• Explaining only the rising part of a temperature graph and not the fall.
• Forgetting units, or the number of significant figures asked for.
• Quoting a percentage change without a sign.
• Naming a control variable without saying how it is kept constant.
• Describing inhibitors without saying where each binds.
• Drawing a tangent at the wrong place instead of at time 0.

Next steps

• Re-read the revision notes and the study guide.
• Go to the AQA A-level Biology hub and the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Biology specification (7401, 7402), for first teaching 2015, published by AQA. Section 3.1.4.2 and Required Practical 1.