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IB Biology 2025 · Theme C · Interaction and interdependence

C1.1 Enzymes and metabolism. Practice questions with markscheme.

67 original IB-style questions on C1.1, written from the 2025 guide: 31 multiple-choice, 24 short-answer, 8 data-based, 2 extended-response parts, 1 drawing, 1 labelling. Below are two ready-made practice papers built from them, one at standard level (21 marks) and one at higher level (20 marks); switch between them above the paper. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

10 statements at SL and HL, 7 additional higher level.

  1. C1.1.1SL / HL Enzymes as catalysts
  2. C1.1.2SL / HL Role of enzymes in metabolism
  3. C1.1.3SL / HL Anabolic and catabolic reactions
  4. C1.1.4SL / HL Enzymes as globular proteins with an active site for catalysis
  5. C1.1.5SL / HL Interactions between substrate and active site to allow induced-fit binding
  6. C1.1.6SL / HL Role of molecular motion and substrate-active site collisions in enzyme catalysis
  7. C1.1.7SL / HL Relationships between the structure of the active site, enzyme–substrate specificity and denaturation
  8. C1.1.8SL / HL Effects of temperature, pH and substrate concentration on the rate of enzyme activity
  9. C1.1.9SL / HL Measurements in enzyme-catalysed reactions
  10. C1.1.10SL / HL Effect of enzymes on activation energy
  11. C1.1.11HL Intracellular and extracellular enzyme-catalysed reactions
  12. C1.1.12HL Generation of heat energy by the reactions of metabolism
  13. C1.1.13HL Cyclical and linear pathways in metabolism
  14. C1.1.14HL Allosteric sites and non-competitive inhibition
  15. C1.1.15HL Competitive inhibition as a consequence of an inhibitor binding reversibly to an active site
  16. C1.1.16HL Regulation of metabolic pathways by feedback inhibition
  17. C1.1.17HL Mechanism-based inhibition as a consequence of chemical changes to the active site caused by the irreversible binding of an inhibitor

In the bank for C1.1

  • 31 multiple-choice
  • 24 short-answer
  • 8 data-based
  • 2 extended-response parts
  • 1 drawing
  • 1 labelling
  • 27 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

Make your own

The practice paper

21 marks at SL · 20 at HL · each has its own code
Take it online → Build a fresh paper Paper code BbB-EAAAAAgAABQAOzdd
Biology · topic quiz
Standard level · topic practice, not an exam format
30 minutes21 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers C1.1 Enzymes and metabolism · SL
Name:
1.

Which row correctly classifies the metabolic reactions?

[1]
RowProtein synthesis from amino acidsHydrolysis of starch to glucose
A.anaboliccatabolic
B.catabolicanabolic
C.anabolicanabolic
D.cataboliccatabolic
  1. Row A
  2. Row B
  3. Row D
  4. Row C
2.

A solution of an enzyme is heated to 80 °C for ten minutes and then cooled to its optimum temperature. Its substrate is then added. What is the expected result?

[1]
0 5 10 15 0 20 40 60 80 100 time / min temperature of the enzyme solution / °C substrate added held at 80 °C at the optimum
  1. The reaction proceeds at the maximum rate because the enzyme is at its optimum temperature
  2. The reaction is faster than normal because heating activated the enzyme
  3. The reaction proceeds slowly because some substrate was hydrolysed during heating
  4. Little or no reaction occurs because denaturation is (usually) irreversible
3.

Amylase digests starch. Samples of a starch–amylase mixture were tested with iodine solution at intervals, and the time taken for the blue-black colour to stop appearing was recorded at several temperatures. How should the rate of reaction be calculated from these results?

[1]
  1. As the time taken to reach the end point, in s
  2. As the concentration of amylase divided by the time taken
  3. As 1 ÷ time taken, so that a shorter time gives a higher rate
  4. As the mass of starch added at the start of the reaction
4.

The table shows the time taken for half of the substrate molecules of one reaction in nucleotide synthesis to react at 25 °C, with no enzyme present and when bound in the active site of the enzyme that catalyses the reaction. What do these data show about the role of enzymes in cells?

[1]
Condition at 25 °CTime for half of the substrate molecules to react
no enzyme present78 million years
bound in the active site of the enzyme0.018 s
  1. The enzyme increases the amount of product that forms rather than the rate at which it forms
  2. An enzyme makes a reaction that is far too slow at 25 °C fast enough to be of use to the cell
  3. Raising the temperature of the cell by a few degrees would give a similar increase in rate
  4. Without the enzyme present, the reaction cannot happen at all, however long it is left
5.

Outline how enzymes lower the activation energy of the reactions they catalyse.

[2]
6.

Urease, an enzyme that hydrolyses urea, was immobilized by trapping it in beads of alginate gel. Beads of three diameters were made from the same urease–alginate mixture. For each size, 10 cm³ of beads was placed in 50 cm³ of urea solution at 25 °C, and the rate of ammonia production was measured with the solution either left still or stirred continuously. The table shows the mean rates.

Bead diameter / mmRate of ammonia production, not stirred / µmol min⁻¹Rate of ammonia production, stirred / µmol min⁻¹
28.412.6
45.17.9
63.65.4
(a)Calculate the percentage increase in the rate of reaction caused by stirring, for the 4 mm beads.[1]
(b)Explain why the smaller beads gave a higher rate of reaction.[2]
(c)The enzyme molecules in the beads cannot move. Explain how stirring increased the rate of reaction.[2]
(d)Even when stirred, the 6 mm beads gave the lowest rate. Suggest why.[1]
7.

Describe what happens when an enzyme catalyses a reaction, from substrate binding to product release.

[2]
8.

Enzymes from thermophilic bacteria are widely used in industrial processes. Suggest reasons for this.

[2]
9.

The graph shows the effect of temperature on the activity of two enzymes: enzyme A from a human cell and enzyme B from a thermophilic bacterium living in hot springs.

0 20 40 60 80 Temperature / °C Rate / arbitrary units enzyme A enzyme B
(a)State the optimum temperature of enzyme B.[1]
(b)Compare and contrast the activity of enzyme A and enzyme B.[2]
(c)Explain the decrease in activity of enzyme A above 40 °C.[2]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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Markscheme BbB-EAAAAAgAABQAOzdd

One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. C1.1.3 [1]
  • A: anabolism builds larger molecules (condensation); catabolism breaks them down (hydrolysis/oxidation);
2. C1.1.8 [1]
  • D: heat-induced changes to tertiary structure are not reversed on cooling, so the active site remains non-functional;
3. C1.1.9 [1]
  • C: rate is inversely related to the time taken to reach the end point, so 1/time gives a value that increases as the reaction speeds up;
4. C1.1.1 [1]
  • B: the uncatalysed reaction does happen, but with a half-time of 78 million years it is far too slow to supply the cell; bound in the enzyme's active site the same reaction takes 0.018 s at the same temperature (about 10¹⁷ times faster), so enzymes allow the reactions of metabolism to run fast enough at the temperature of the cell;

B is refuted by the table itself (the uncatalysed reaction has a finite half-time); C tests scale (a few degrees raises rate by a small factor, not 10¹⁷); D confuses rate with yield. Data after Radzicka and Wolfenden (1995), orotidine 5'-phosphate decarboxylase: uncatalysed half-time 78 million years at 25 °C, kcat 39 s⁻¹ (half-time ln 2 ÷ 39 = 0.018 s).

5. C1.1.10 [2 max]
  • binding (in the active site) strains/distorts bonds in the substrate / holds substrates in the correct orientation;
  • (this) stabilizes the transition state, so less energy is needed to reach it;
  • local conditions in the active site (charged/acidic R-groups) assist the reaction, OWTTE;
6. C1.1.6
  • (a) [1]
    • (7.9 − 5.1) ÷ 5.1 × 100 = 55 %;

    Accept 54.9 %.

  • (b) [2]
    • the same volume of small beads has a larger total surface area, so more of the enzyme lies close to the surface;
    • urea molecules have a shorter distance to move into a small bead to reach the enzyme;
    • so collisions between urea molecules and active sites are more frequent / more enzyme–substrate complexes form per unit time, OWTTE;
  • (c) [2]
    • because the enzyme is immobilized, substrate molecules must move to the active sites for collisions to occur;
    • without stirring, urea close to the beads is used up faster than random movement replaces it / a layer of solution low in substrate forms round each bead;
    • stirring brings fresh urea solution to the surface of the beads / carries product away;
    • so collisions between substrate molecules and active sites are more frequent, OWTTE;
  • (d) [1]
    • stirring moves only the solution outside the beads, so urea must still move slowly through the gel to reach enzyme deep inside a large bead;
    • enzyme near the centre of a large bead meets little substrate / a smaller proportion of the enzyme is close to the surface, OWTTE;
7. C1.1.4-C1.1.5 [2 max]
  • substrate collides with / binds to the active site, forming an enzyme–substrate complex;
  • the active site changes shape (slightly) to fit the substrate more closely / induced fit;
  • the reaction occurs and products are released, leaving the enzyme unchanged (and reusable);
8. C1.1.7 [2 max]
  • they remain active / are not denatured at high temperatures;
  • reactions can run faster at higher temperatures (more collisions) without loss of enzyme;
  • less cooling of reaction vessels needed / lower contamination risk at high temperature / longer working life, OWTTE;

Any 2. Accept a named example such as Taq polymerase.

9. C1.1.7, C1.1.8
  • (a) [1]
    • 75 °C;

    Accept 73–77 °C. Units required.

  • (b) [2 max]
    • both show an optimum with activity falling either side / both reach a similar maximum rate;
    • enzyme B has a higher optimum than enzyme A / B remains active at temperatures that denature A;
    • A's activity falls to zero by ~45 °C, whereas B is still active (and rising towards its optimum), OWTTE;

    At least one similarity AND one difference required for full marks.

  • (c) [2 max]
    • heat breaks (hydrogen/ionic) bonds maintaining the tertiary structure;
    • enzyme is denatured / active site changes shape;
    • substrate no longer binds / fewer enzyme–substrate complexes form;

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