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.
- C1.1.1SL / HL Enzymes as catalysts
- C1.1.2SL / HL Role of enzymes in metabolism
- C1.1.3SL / HL Anabolic and catabolic reactions
- C1.1.4SL / HL Enzymes as globular proteins with an active site for catalysis
- C1.1.5SL / HL Interactions between substrate and active site to allow induced-fit binding
- C1.1.6SL / HL Role of molecular motion and substrate-active site collisions in enzyme catalysis
- C1.1.7SL / HL Relationships between the structure of the active site, enzyme–substrate specificity and denaturation
- C1.1.8SL / HL Effects of temperature, pH and substrate concentration on the rate of enzyme activity
- C1.1.9SL / HL Measurements in enzyme-catalysed reactions
- C1.1.10SL / HL Effect of enzymes on activation energy
- C1.1.11HL Intracellular and extracellular enzyme-catalysed reactions
- C1.1.12HL Generation of heat energy by the reactions of metabolism
- C1.1.13HL Cyclical and linear pathways in metabolism
- C1.1.14HL Allosteric sites and non-competitive inhibition
- C1.1.15HL Competitive inhibition as a consequence of an inhibitor binding reversibly to an active site
- C1.1.16HL Regulation of metabolic pathways by feedback inhibition
- 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
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The practice paper
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Which row correctly classifies the metabolic reactions?
| Row | Protein synthesis from amino acids | Hydrolysis of starch to glucose |
|---|---|---|
| A. | anabolic | catabolic |
| B. | catabolic | anabolic |
| C. | anabolic | anabolic |
| D. | catabolic | catabolic |
- Row A
- Row B
- Row D
- Row C
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?
- The reaction proceeds at the maximum rate because the enzyme is at its optimum temperature
- The reaction is faster than normal because heating activated the enzyme
- The reaction proceeds slowly because some substrate was hydrolysed during heating
- Little or no reaction occurs because denaturation is (usually) irreversible
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?
- As the time taken to reach the end point, in s
- As the concentration of amylase divided by the time taken
- As 1 ÷ time taken, so that a shorter time gives a higher rate
- As the mass of starch added at the start of the reaction
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?
| Condition at 25 °C | Time for half of the substrate molecules to react |
|---|---|
| no enzyme present | 78 million years |
| bound in the active site of the enzyme | 0.018 s |
- The enzyme increases the amount of product that forms rather than the rate at which it forms
- An enzyme makes a reaction that is far too slow at 25 °C fast enough to be of use to the cell
- Raising the temperature of the cell by a few degrees would give a similar increase in rate
- Without the enzyme present, the reaction cannot happen at all, however long it is left
Outline how enzymes lower the activation energy of the reactions they catalyse.
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 / mm | Rate of ammonia production, not stirred / µmol min⁻¹ | Rate of ammonia production, stirred / µmol min⁻¹ |
|---|---|---|
| 2 | 8.4 | 12.6 |
| 4 | 5.1 | 7.9 |
| 6 | 3.6 | 5.4 |
Describe what happens when an enzyme catalyses a reaction, from substrate binding to product release.
Enzymes from thermophilic bacteria are widely used in industrial processes. Suggest reasons for this.
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.
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- A: anabolism builds larger molecules (condensation); catabolism breaks them down (hydrolysis/oxidation);
- D: heat-induced changes to tertiary structure are not reversed on cooling, so the active site remains non-functional;
- 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;
- 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).
- 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;
- (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;
- 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);
- 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.
- (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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The inhibition of an enzyme is reduced when the substrate concentration is increased. What can be concluded?
- The inhibitor binds irreversibly to the enzyme's active site
- The inhibitor competes with the substrate for the active site
- The inhibitor binds to an allosteric site
- The enzyme has been denatured by the inhibitor
In bacteria, isoleucine is synthesized from threonine by a pathway of enzyme-catalysed steps. A mutation alters threonine deaminase, an enzyme of this pathway, so that isoleucine can no longer bind to it, although its catalytic activity is unchanged. What is the most likely effect of the mutation?
- No isoleucine would be made, as threonine could no longer be converted
- The pathway would slow down sooner as isoleucine accumulated in the cell
- Threonine deaminase would be permanently inhibited, so threonine accumulates
- Isoleucine would be over-produced, even when the cell is not using it
Which statements about enzymes are correct?
- I. They are globular proteins.
- II. They are used up during the reactions they catalyse.
- III. They lower the activation energy of reactions.
- I and II only
- I and III only
- II and III only
- I, II and III
Which row correctly classifies the metabolic pathways?
| Row | Linear pathway | Cyclical pathway |
|---|---|---|
| A. | Krebs cycle | glycolysis |
| B. | Calvin cycle | Krebs cycle |
| C. | Krebs cycle | Calvin cycle |
| D. | glycolysis | Krebs cycle |
- Row A
- Row B
- Row C
- Row D
Metabolism is a network of interdependent chemical reactions. Which statement explains why a cell can exert control over its metabolism through its enzymes?
- Each reaction in the network depends on a specific enzyme, so changing that enzyme's activity changes the rate of the reaction
- Enzymes supply the energy that drives each metabolic reaction, so producing more enzyme makes more energy available
- Any enzyme can convert a substrate into every possible product, so the cell simply chooses which enzyme to make active at any time
- Enzymes are used up as they catalyse reactions, so the quantity of enzyme present sets how much product can be formed
Penicillin is a mechanism-based (irreversible) enzyme inhibitor. Which enzyme does it inhibit, and how does resistance commonly arise?
- Ribosomal peptidyl transferase; resistance by ribosome methylation
- DNA gyrase; resistance by over-expression of the gene
- Transpeptidase; resistance by a modified enzyme or secretion of β-lactamase
- β-lactamase; resistance by a mutant lactamase
What does the peak of each curve represent?
- The point at which bonds in the substrate are being broken
- The energy of the enzyme–substrate complex once the products have formed
- The energy released as new bonds form in the products
- The difference in energy between the substrates and the products
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?
| Condition at 25 °C | Time for half of the substrate molecules to react |
|---|---|
| no enzyme present | 78 million years |
| bound in the active site of the enzyme | 0.018 s |
- Without the enzyme present, the reaction cannot happen at all, however long it is left
- The enzyme increases the amount of product that forms rather than the rate at which it forms
- An enzyme makes a reaction that is far too slow at 25 °C fast enough to be of use to the cell
- Raising the temperature of the cell by a few degrees would give a similar increase in rate
Enzymes from thermophilic bacteria are widely used in industrial processes. Suggest reasons for this.
Draw an annotated graph to show the effect of substrate concentration on the rate of an enzyme-catalysed reaction, with enzyme concentration constant.
The rate of reaction of a wild-type enzyme and of variant G, which differs from the wild type by a single amino acid substitution in its active site, was measured over a range of substrate concentrations at constant temperature and pH. Vmax is the maximum rate, reached when the enzyme is saturated with substrate. The table shows Vmax and the substrate concentration at which the rate is half of Vmax for each form (mean ± standard error).
| Enzyme form | Vmax / μmol min⁻¹ (mean ± SE) | Substrate concentration at ½Vmax / mmol dm⁻³ (mean ± SE) |
|---|---|---|
| Wild-type | 48.0 ± 1.2 | 2.1 ± 0.2 |
| Variant G | 47.2 ± 1.5 | 6.8 ± 0.3 |
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-EAEAAAgAABQAt9aL
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.
- B: competitive inhibition is overcome at high substrate concentration;
- D: isoleucine normally binds an allosteric site on threonine deaminase, the first enzyme of the pathway, slowing the pathway as isoleucine accumulates (end-product inhibition); without that binding the pathway keeps running whatever the isoleucine concentration, so isoleucine is over-produced;
- B: enzymes are globular protein catalysts that lower activation energy and are not consumed;
- D: glycolysis is a linear chain of reactions from glucose to pyruvate, whereas the Krebs cycle (like the Calvin cycle) regenerates its starting compound;
- A: enzyme specificity means a separate enzyme is required for each reaction, so regulating the amount or activity of one enzyme regulates one step of the network; enzymes are not used up (A), they lower activation energy rather than supplying energy (B), and each enzyme catalyses only its own reaction (D);
- C: penicillin reacts covalently in transpeptidase's active site, permanently disabling wall cross-linking; resistant bacteria have altered transpeptidases or secrete β-lactamase, which destroys penicillin;
- A: the peak is the transition state, reached only when enough energy has been supplied to break bonds within the substrate; the drop after it is the energy yield from bond formation;
- C: 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).
- 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.
- axes labelled: rate of reaction (y) against substrate concentration (x);
- rate rises (steeply) at low substrate concentrations;
- curve levels off / plateaus at high substrate concentration, annotated as enzyme/active sites saturated;
Award marks for correct shape even if freehand; both annotations and axis labels required for full marks.
- (a) [2]
- variant G reaches half of Vmax only at a much higher substrate concentration (6.8 vs 2.1 mmol dm⁻³), so it binds substrate less tightly / has a lower affinity;
- the two Vmax values are almost equal (~47–48), so the maximum catalytic rate is essentially unchanged, OWTTE;
- (b) [3]
- the Vmax ranges overlap (≈46.8–49.2 vs 45.7–48.7), so that difference is not significant / could be due to chance;
- the ½Vmax ranges do not overlap (≈1.9–2.3 vs 6.5–7.1), so the difference in substrate binding is (likely) significant;
- a t-test (comparing two means) would test whether the difference is statistically significant;
- (c) [2 max]
- the different R-group alters the shape/chemistry of the active site, so it is less complementary to the substrate / binds it less readily;
- (so) fewer collisions lead to binding and a higher substrate concentration is needed to occupy half of the active sites, OWTTE;
- at very high substrate concentration the active sites are still all occupied and, once bound, the substrate is still converted to product, so Vmax is (almost) unchanged, OWTTE;
More in Theme C · Interaction and interdependence
- C1.2 Cell respiration 69
- C1.3 Photosynthesis 68
- C2.1 Chemical signalling 68
- C2.2 Neural signalling 67
- C3.1 Integration of body systems 74
- C3.2 Defence against disease 65
- C4.1 Populations and communities 67
- C4.2 Transfers of energy and matter 83
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