Cambridge International AS & A Level Biology 9700: Mode of action of enzymes – 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, Mode of action of enzymes, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. This is AS Level content, examined in Papers 1 and 2. Questions 5 to 9 practise the practical skills tested in Paper 3, and Question 11 practises the planning skills tested in Paper 5 at A Level. Calculators are allowed in all 9700 papers. Give calculated answers to the same number of significant figures as the data, or one more.

Read the 3.1 study guide or the 3.1 revision notes first. These questions are different from those in the topic-wide Enzymes practice questions. Course pages: Cambridge A Level Biology hub and printable checklist.

Questions

1. (a) Explain why enzymes are described as globular proteins. [2] (b) State whether each enzyme is intracellular or extracellular: (i) catalase in mung bean cells; (ii) amylase in the small intestine; (iii) DNA polymerase. [3]

2. For one reaction, the activation energy is 75 kJ mol⁻¹ without an enzyme and 30 kJ mol⁻¹ with the enzyme. The products have 40 kJ mol⁻¹ less energy than the substrates.

(a) Calculate the percentage reduction in activation energy. [2] (b) State the overall energy change of the catalysed reaction and explain your answer. [2]

3. Amylase hydrolyses starch but not cellulose, although both are polymers of glucose. Explain why. [3]

4. Scientists compared the three-dimensional structure of an enzyme with and without its substrate. Without substrate, the active site was more open. With substrate bound, several R groups in the active site had moved closer around the substrate.

(a) State which hypothesis of enzyme action these results support, and explain why. [2] (b) Explain how this change could help the enzyme lower the activation energy. [2]

5. Catalase extract from sweet potato was added to hydrogen peroxide. Oxygen was collected in a gas syringe.

Time / s	0	30	60	90	120	150	180	210
Volume of O₂ / cm³	0	18	31	40	46	49	50	50

(a) Calculate the mean rate of oxygen production in the first 30 s. [2] (b) A tangent drawn to the curve at t = 0 passes through (0, 0) and (25, 20). Calculate the initial rate. [2] (c) Calculate the mean rate between 90 s and 150 s, and explain why it is lower than the initial rate. [3] (d) Explain why no more oxygen is collected after 180 s. [1]

6. A student followed starch hydrolysis by amylase, sampling the mixture into iodine solution every 15 s. In three repeats, the iodine first stayed orange-brown at 150 s, 165 s and 180 s.

(a) Calculate the mean rate of reaction as 1/time. [2] (b) Identify two limitations of this method and suggest an improvement for each. [4]

7. Outline how a colorimeter could be used to follow the disappearance of starch in the reaction in Question 6. [5]

8. Starch solutions of known concentration were mixed with iodine and their absorbance measured. The calibration line was straight: absorbance = 0.060 × starch concentration (g dm⁻³). Samples from an amylase reaction gave absorbances of 0.60 at 0 min, 0.42 at 3 min and 0.27 at 6 min.

(a) Calculate the starch concentration at 3 min and at 6 min. [2] (b) Calculate the mean rate of starch disappearance from 0 to 3 min and from 3 to 6 min. [2] (c) Explain the difference between the two rates. [1]

9. A student used sweet potato discs as a source of catalase.

(a) Explain why all the discs were cut to the same size. [2] (b) Explain why the student set up a tube with boiled discs. [2]

10. Describe and explain how catalase breaks down hydrogen peroxide, from the arrival of a substrate molecule to the release of products. Include enzyme specificity and the induced-fit hypothesis. [8]

11. Plan an investigation to compare the catalase activity of extracts of mung beans and potato by measuring the rate of formation of product. [8]

Answers

1. (a) Its polypeptide folds into a compact, roughly spherical tertiary structure [1] with hydrophilic R groups mostly on the outside, so it is soluble [1]. (b) (i) intracellular [1]; (ii) extracellular [1]; (iii) intracellular [1]. Examiner insight: One mark per item in (b). Catalase stays intracellular even when you extract it from tissue; classification is by where the enzyme normally acts.

2. (a) (75 − 30) ÷ 75 × 100 [1] = 60% [1]. (b) 40 kJ mol⁻¹ released, the same as without the enzyme [1]; the enzyme changes only the activation energy, not the energy of the substrates or products [1]. Examiner insight: A bare 60% earns both marks only if it is correct; show the subtraction so a slip still earns the method mark.

3. Amylase’s active site has a shape complementary to starch, which has α-1,4 glycosidic bonds [1]. Cellulose has β-1,4 glycosidic bonds, with alternate glucose units inverted, so its shape is different [1]. Cellulose cannot fit the active site, so no enzyme–substrate complex forms [1]. Examiner insight: “Same shape” instead of “complementary” usually loses the first mark.

4. (a) Induced fit [1]: the active site changed shape when the substrate bound, so it is not rigid as in lock-and-key [1]. (b) R groups moving around the substrate put strain on bonds in it [1], so less energy is needed to break them and the reaction proceeds more easily [1]. Examiner insight: Say that the active site changes shape. Answers saying the substrate changes shape to fit do not gain the mark.

5. (a) 18 ÷ 30 [1] = 0.60 cm³ s⁻¹ [1]. (b) 20 ÷ 25 [1] = 0.80 cm³ s⁻¹ [1]. (c) (49 − 40) ÷ 60 = 0.15 cm³ s⁻¹ [1]. Hydrogen peroxide is being used up, so substrate concentration falls [1] and fewer enzyme–substrate complexes form per second [1]. (d) All the hydrogen peroxide has been broken down; the volume is limited by the amount of substrate, not the enzyme [1]. Examiner insight: Units are required for the accuracy marks in (a) to (c). In (b), the gradient must come from the tangent, not from the data table.

6. (a) Mean time = (150 + 165 + 180) ÷ 3 = 165 s [1]; rate = 1 ÷ 165 = 0.0061 s⁻¹ (6.1 × 10⁻³ s⁻¹) [1]. (b) The end point is only known to within 15 s [1]; sample more often, for example every 5 s [1]. The colour is judged by eye, which is subjective [1]; use a colorimeter, or compare with a colour standard [1]. Also accept: temperature may not be constant; keep tubes in a thermostatically controlled water bath. Examiner insight: Each improvement must match its limitation. “Repeat more” does not fix sampling-interval uncertainty and usually gains no credit.

7. Use a red filter, complementary to blue-black [1]. Zero the colorimeter with a blank of iodine solution and water [1]. Make a calibration curve of absorbance against known starch concentrations [1]. Take samples at set time intervals, add each to the same volume of iodine solution and read the absorbance [1]. Convert absorbance to concentration using the curve and plot concentration against time [1]. Examiner insight: “Measure the colour” is too vague. Name what is read (absorbance) and how it becomes a concentration.

8. (a) 0.42 ÷ 0.060 = 7.0 g dm⁻³ [1]; 0.27 ÷ 0.060 = 4.5 g dm⁻³ [1]. (b) (10 − 7.0) ÷ 3 = 1.0 g dm⁻³ min⁻¹ [1]; (7.0 − 4.5) ÷ 3 = 0.83 g dm⁻³ min⁻¹ [1]. (c) Starch concentration is lower later, so there are fewer enzyme–substrate collisions per minute [1]. Examiner insight: Allow error carried forward in (b) from wrong values in (a). The 0 min value (10 g dm⁻³) comes from 0.60 ÷ 0.060.

9. (a) Same size gives the same surface area [1], so the same amount of enzyme is exposed to hydrogen peroxide [1]. (b) Boiling denatures catalase [1], showing that oxygen is produced because of the enzyme and not by hydrogen peroxide breaking down on its own [1]. Examiner insight: “Fair test” alone earns nothing. Say which variable is controlled and why it affects the result.

10. Catalase is a globular protein whose active site has a shape complementary to hydrogen peroxide [1]. Hydrogen peroxide collides with and binds to the active site, forming an enzyme–substrate complex held by temporary bonds [1]. Induced fit: the active site changes shape slightly as the substrate binds [1], straining bonds in the substrate [1]. This lowers the activation energy [1]. The products, water and oxygen, are released [1]; the enzyme is unchanged and can bind another substrate [1]. Specificity: other molecules are not complementary to the active site, so cannot form an enzyme–substrate complex [1]. Examiner insight: Real mark schemes often list more points than marks, with a maximum. Extended answers that stop at “the substrate fits the active site” cap well below full credit.

11. Blend equal masses of mung beans and potato with equal volumes of buffer, then filter [1]. Use the same volume and concentration of hydrogen peroxide and the same volume of each extract [1]. Keep temperature constant in a water bath [1] and pH constant with a buffer [1]. Collect oxygen in a gas syringe, reading the volume at regular intervals [1]. Plot volume against time and find the initial rate from a tangent at t = 0 [1]. Repeat at least three times and calculate means [1]. Control with boiled extract [1]. Examiner insight: Planning marks need operational detail: “same amount” earns less than “same mass of tissue in the same volume of buffer”.

Where marks are usually lost

• Writing “same shape” instead of complementary.
• Saying the substrate changes shape in induced fit.
• Claiming an enzyme changes the overall energy change of a reaction.
• Using a mean rate when the question asks for the gradient of a tangent.
• Leaving out units, or quoting 1/time without s⁻¹.
• Pairing a limitation with an improvement that does not address it.
• Saying a colorimeter “measures colour” without naming absorbance, the filter, the blank or the calibration curve.
• Writing “the enzyme is used up” when a reaction stops; the substrate is used up.

Next steps

• 3.1 revision notes
• 3.1 study guide
• Enzymes practice questions for the whole topic
• Cambridge A Level Biology hub
• Printable Biology checklist
• AS Level Biology 9700 diagnostic, or all free 10-minute diagnostics
• 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 Assessment International Education (Cambridge University Press & Assessment, September 2022). Topic 3, Enzymes, section 3.1 Mode of action of enzymes, learning outcomes 1 to 4.