IB DP Biology: Enzymes and metabolism: active sites, rates, inhibition (DP Biology) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise IB Diploma Programme Biology section C1.1 (Enzymes and metabolism) from the IB Biology guide, first assessment 2025. Questions 1 to 9 are SL and HL. Questions 10 to 12 are HL only.

Learn the content first from the study guide and the revision notes. Course hub: IB DP Biology. Checklist: printable checklist. Diagnostics: all free 10-minute diagnostics.

Questions

1. Distinguish between anabolism and catabolism, giving one example of each. [4]

2. An enzyme’s active site is made of only a few amino acids. Explain how the rest of the enzyme helps the active site to catalyse a reaction. [3]

3. Describe induced-fit binding between an enzyme and its substrate. [3]

4. Explain why a substrate molecule and an active site must collide for catalysis, and give one situation in which the enzyme or the substrate is immobilised. [3]

5. Explain, using the structure of the active site, why a protease does not digest starch, and why heating the protease strongly stops it working. [4]

6. A bacterium from a hot spring makes an enzyme. The mass of product formed per minute (mg min⁻¹) was recorded at different temperatures.

Temperature / °C	40	50	60	70	80	90
Rate / mg min⁻¹	2.0	4.1	8.0	12.4	3.1	0.2

(a) State the range of temperatures that contains the optimum. [1] (b) Calculate the percentage change in rate between 80 and 90 °C. [2] (c) Explain the rise in rate from 40 to 70 °C. [2] (d) Explain the fall in rate above 70 °C. [2]

7. (a) Explain why an enzyme’s rate falls when pH moves far from its optimum. [3] (b) Explain why the rate stops rising when substrate concentration is very high. [2]

8. Oxygen released by an enzyme from hydrogen peroxide was measured as the total volume of gas collected.

Time / s	0	15	30	45	60	75	90
Volume / cm³	0	12	21	27	31	33	34

(a) Calculate the initial rate over the first 30 s, with units. [2] (b) Calculate the mean rate between 60 and 90 s. [2] (c) Explain the difference between the two rates. [2]

9. Energy values for a reaction are: reactants 55, products 70, peak of the uncatalysed pathway 120, peak of the catalysed pathway 78 (all kJ mol⁻¹). (a) Calculate the activation energy with and without the enzyme. [2] (b) Calculate the overall energy change and state whether the enzyme alters it. [2] (c) Explain how the enzyme lowers the activation energy. [2]

10. (HL only) (a) Distinguish between intracellular and extracellular enzyme-catalysed reactions, with one example of each. [3] (b) Explain why metabolism always generates heat, and state one use of this heat in mammals. [2] (c) State one cyclical metabolic pathway. [1]

11. (HL only) Rates of an enzyme reaction (arbitrary units) with two reversible inhibitors, X and Y, are shown.

Substrate / mM	2	4	8	16	32
No inhibitor	25	40	57	73	84
With X	10	18	31	47	64
With Y	15	24	34	44	51

(a) Calculate the percentage reduction in rate caused by X and by Y at 2 mM and at 32 mM. [4] (b) Identify the type of inhibition shown by X and by Y. [2] (c) Explain the difference using the way each inhibitor binds. [3]

12. (HL only) (a) Explain how isoleucine regulates its own synthesis from threonine. [4] (b) Penicillin is a mechanism-based inhibitor of transpeptidase. Explain how penicillin affects a bacterium, and how a change to transpeptidase gives resistance. [4]

Answers

1. Anabolism builds larger molecules from smaller ones, using condensation [1]; example: protein synthesis, glycogen formation or photosynthesis [1]. Catabolism breaks larger molecules into smaller ones [1]; example: hydrolysis in digestion or oxidation in respiration [1]. [4] Examiner insight: A “distinguish” answer should state both sides; a definition of only one is usually not enough, and each example must match its type.

2. Interactions (such as hydrogen bonds, ionic bonds and hydrophobic interactions) between amino acids throughout the protein hold the three-dimensional structure [1]. This keeps the few active-site amino acids in the correct positions [1], giving the active site the shape and chemical properties needed for catalysis [1]. [3] Examiner insight: Say that the whole structure positions the active-site amino acids; “the enzyme is a protein” alone does not answer the question.

3. Substrate binds to the active site [1]. The enzyme changes shape slightly as it binds [1], and the substrate changes shape too, which strains its bonds and eases the reaction [1]. [3] Examiner insight: Mentioning that the substrate also changes shape separates induced fit from a rigid lock-and-key description.

4. Molecules move randomly in solution, and a reaction needs the substrate to collide with the active site in the right orientation [1]. Without collision there is no binding [1]. Example: an enzyme embedded in a membrane, or a large immobilised substrate [1]. [3] Examiner insight: The example must be one where something is fixed in place; a free enzyme in solution does not earn the mark.

5. The active site of the protease has a shape and charge that match protein substrates, not starch [1], so starch cannot bind [1]. Heat breaks bonds holding the tertiary structure [1], so the active site is distorted and the enzyme is denatured [1]. [4] Examiner insight: Link the loss of activity to the active site changing shape; “the enzyme dies” gains nothing.

6. (a) Between 60 and 80 °C (accept 70 °C as the highest reading) [1] (b) (0.2 − 3.1) / 3.1 × 100 [1] = −94% (a 94% fall) [1] (c) Higher temperature gives more kinetic energy, so more frequent collisions between substrate and active sites [1], and more collisions have enough energy to react [1] (d) Bonds holding the enzyme’s shape break [1], the active site is distorted, so substrate cannot bind [1] [7] Examiner insight: A percentage-change answer needs the sign or the word fall; the optimum is a range because no reading lies between 70 and 80 °C.

7. (a) Away from the optimum, R group charges change [1], which disrupts ionic and hydrogen bonds [1], so the active site changes shape and the enzyme may be denatured [1] (b) Every active site is already occupied nearly all the time [1], so extra substrate cannot increase the rate [1] [5] Examiner insight: A plateau is explained by saturated active sites; “enzyme runs out” is a common wrong explanation.

8. (a) 21 / 30 [1] = 0.70 cm³ s⁻¹ [1] (b) (34 − 31) / 30 [1] = 0.10 cm³ s⁻¹ [1] (c) The rate is lower later because substrate has been consumed [1], so collisions with active sites are rarer [1] [6] Examiner insight: A rate is a change divided by time, with units; give the units even when the question does not repeat them.

9. (a) Without enzyme: 120 − 55 = 65 kJ mol⁻¹ [1]; with enzyme: 78 − 55 = 23 kJ mol⁻¹ [1] (b) 70 − 55 = +15 kJ mol⁻¹ (energy is taken in overall) [1]; the enzyme does not alter it [1] (c) It binds the substrate and strains its bonds [1], so less energy is needed to reach the transition state [1] [6] Examiner insight: Activation energy is measured from the reactants, not from zero, and the sign of the overall change should be stated.

10. (a) Intracellular reactions occur inside the cell producing the enzyme [1], for example glycolysis in the cytoplasm [1]. Extracellular reactions occur outside it, after secretion, for example digestion in the gut [1] (b) Metabolic reactions are not 100% efficient at transferring energy [1]; heat helps to maintain a constant body temperature [1] (c) The Krebs cycle (or the Calvin cycle) [1] [6] Examiner insight: Pair each type with a correct example; glycolysis is intracellular and linear, so it does not answer part (c).

11. (a) X at 2 mM: (25 − 10) / 25 = 60% [1]; at 32 mM: (84 − 64) / 84 = 24% [1]. Y at 2 mM: (25 − 15) / 25 = 40% [1]; at 32 mM: (84 − 51) / 84 = 39% [1] (b) X is competitive [1]; Y is non-competitive [1] (c) X resembles the substrate and binds reversibly to the active site [1], so more substrate outcompetes it [1]. Y binds reversibly to an allosteric site and changes the active site, so extra substrate does not remove the effect [1] [9] Examiner insight: Cite the figures that show the reduction shrinking for one inhibitor and staying near 40% for the other; naming the type without data is weaker.

12. (a) Isoleucine is the end product of the pathway [1]. It binds to and inhibits an early enzyme, threonine deaminase [1]. When isoleucine is plentiful the pathway slows [1]; when it falls, inhibition lifts and synthesis speeds up [1] (b) Penicillin binds irreversibly to the active site of transpeptidase [1], so cross-links in the cell wall are not formed and the wall is weakened [1]. A changed transpeptidase cannot bind penicillin [1], so those bacteria still make cell walls, survive and reproduce [1] [8] Examiner insight: Name the enzyme inhibited by isoleucine and describe the loop in both directions; for penicillin, “irreversible” is the key word.

Where marks are usually lost

• Saying enzymes are killed, rather than denatured.
• Stating that a substrate “does not fit” without naming the change in active site shape.
• Explaining a plateau as the enzyme being used up.
• Omitting units on a rate, or reading a rate from the wrong part of a curve.
• Giving percentage change without the sign or without dividing by the original value.
• Measuring activation energy from zero instead of from the reactants.
• Naming an inhibitor type without using the data or the binding site to justify it.
• Saying competitive inhibition cannot be overcome, or non-competitive inhibition can.

Next steps

• Revise with the revision notes and the study guide.
• Course hub: IB DP Biology, and the printable checklist.
• Find gaps with the all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. This page is a paraphrase aligned to section C1.1.