Enzymes Practice Questions

Enzyme questions appear in almost every biology and chemistry exam, and they reward precise language: “active site”, “complementary”, “denatured”, “collision”. This set of 14 questions starts with the basics and builds up to kinetics calculations and graph analysis. Work through them with a pencil and calculator, then check the answer key. For background, see enzymes explained and enzyme kinetics.

Questions

Q1. Define the term enzyme and explain how enzymes speed up reactions, using the idea of activation energy.

Q2. Compare the lock-and-key model with the induced-fit model of enzyme action. Which is better supported by evidence, and why?

Q3. Explain why each enzyme usually catalyses only one reaction or type of reaction.

Q4. Pepsin works best at pH 2 and trypsin at pH 8. Explain why a change in pH away from the optimum reduces enzyme activity. Refer to specific bonds.

Q5. A student measures the time taken for an enzyme to digest a fixed amount of starch at different temperatures:

Temperature / °C	10	20	30	40	50	60
Time / s	300	150	80	50	120	no reaction

(a) Calculate the rate (1/time) at 20 °C and at 40 °C, in s⁻¹. (b) State the approximate optimum temperature. (c) Explain the result at 60 °C.

Q6. Between 10 °C and 20 °C, the rate in Q5 doubles. Calculate the temperature coefficient Q₁₀ for this interval and comment on its value.

Q7. Explain why increasing substrate concentration increases the rate of an enzyme reaction at first, but eventually has no further effect.

Q8. An enzyme has Vmax = 120 µmol min⁻¹ and Km = 2.0 mmol dm⁻³. Using the Michaelis–Menten equation, v = Vmax[S] / (Km + [S]), calculate the rate when [S] is: (a) 2.0 mmol dm⁻³, (b) 6.0 mmol dm⁻³, (c) 0.50 mmol dm⁻³.

Q9. Two enzymes act on the same substrate. Enzyme A has Km = 0.1 mmol dm⁻³; enzyme B has Km = 5 mmol dm⁻³. Which has the higher affinity for the substrate? Explain.

Q10. Copy and complete the table:

	Competitive inhibitor	Non-competitive inhibitor
Where it binds
Effect on Vmax
Effect on apparent Km
Overcome by more substrate?

Q11. Malonate, ⁻OOC–CH₂–COO⁻, inhibits the enzyme succinate dehydrogenase, whose substrate is succinate, ⁻OOC–CH₂–CH₂–COO⁻. Suggest the type of inhibition and explain using the structures.

Q12. In a Lineweaver–Burk plot (1/v against 1/[S]), the y-intercept is 1/Vmax and the x-intercept is −1/Km. A plot has a y-intercept of 0.020 min µmol⁻¹ and an x-intercept of −0.40 dm³ mmol⁻¹. Calculate Vmax and Km.

Q13. Catalase breaks down hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. A student collects 36 cm³ of oxygen in 60 s at room temperature and pressure, where 1 mol of gas occupies 24.0 dm³. (a) Calculate the moles of oxygen produced. (b) Calculate the moles of H₂O₂ broken down. (c) Calculate the mean rate of H₂O₂ breakdown in mol s⁻¹.

Q14. Design an experiment to investigate the effect of pH on the activity of catalase using potato discs and hydrogen peroxide. Include the independent, dependent and controlled variables, how you’d measure the rate, and one safety precaution.

Answer key

A1. An enzyme is a biological catalyst — usually a protein — that speeds up a reaction without being used up. It provides an alternative reaction pathway with a lower activation energy, for example by holding substrates in the right orientation, straining bonds, or providing groups that donate or accept protons. More collisions then have enough energy to react at the same temperature (see reaction rates and catalysts).

A2. Lock-and-key: the active site has a rigid shape exactly complementary to the substrate. Induced fit: the active site is flexible and changes shape slightly as the substrate binds, forming a tighter fit and straining bonds in the substrate. Induced fit is better supported — X-ray structures show many enzymes changing shape on binding (hexokinase closes around glucose, for example). See lock-and-key vs induced fit.

A3. The active site has a specific 3D shape and arrangement of charged, polar and non-polar groups, set by the enzyme’s tertiary structure. Only substrates with a complementary shape and chemistry can bind and form an enzyme–substrate complex.

A4. Changes in pH alter the charges on amino acid side chains (e.g. –COO⁻ becomes –COOH at low pH; –NH₃⁺ becomes –NH₂ at high pH). This disrupts ionic bonds and hydrogen bonds that hold the tertiary structure, changing the shape of the active site. It can also change charges on groups in the active site that bind the substrate or take part in catalysis, so the substrate binds less well. Extreme pH denatures the enzyme (see factors affecting enzymes).

A5. (a) 20 °C: 1/150 = 6.7 × 10⁻³ s⁻¹. 40 °C: 1/50 = 2.0 × 10⁻² s⁻¹. (b) About 40 °C (the fastest measured; the true optimum lies somewhere between 30 and 50 °C). (c) The enzyme is denatured: vibrations break bonds holding the tertiary structure, the active site loses its shape, and no enzyme–substrate complexes form.

A6. Rate at 10 °C = 1/300 = 3.33 × 10⁻³ s⁻¹; at 20 °C = 6.67 × 10⁻³ s⁻¹. Q₁₀ = rate(T + 10)/rate(T) = 6.67/3.33 = 2.0. This is typical of chemical reactions: a 10 °C rise roughly doubles the rate because more molecules have energy above the activation energy.

A7. At low [S], many active sites are free, so more substrate means more frequent collisions and more enzyme–substrate complexes — the rate rises. At high [S], nearly all active sites are occupied (the enzyme is saturated), so the rate is limited by how fast the enzyme can turn substrate into product. Adding more substrate has no further effect; the rate approaches Vmax.

A8. (a) v = 120 × 2.0 / (2.0 + 2.0) = 60 µmol min⁻¹ (at [S] = Km, v = Vmax/2 — a useful check). (b) v = 120 × 6.0 / 8.0 = 90 µmol min⁻¹. (c) v = 120 × 0.50 / 2.5 = 24 µmol min⁻¹.

A9. Enzyme A. Km is the substrate concentration giving half Vmax. A lower Km means the enzyme reaches half its maximum rate at a lower concentration, i.e. it binds substrate more readily — higher affinity.

A10.

	Competitive	Non-competitive
Where it binds	Active site	Another (allosteric) site
Effect on Vmax	Unchanged	Decreased
Effect on apparent Km	Increased	Unchanged (for pure non-competitive)
Overcome by more substrate?	Yes	No

See enzyme inhibition.

A11. Competitive inhibition. Malonate has the same two carboxylate groups as succinate, separated by one CH₂ instead of two, so it has a similar shape and charge distribution and fits the active site. But it can’t be dehydrogenated (it lacks the –CH₂–CH₂– unit that forms a C=C double bond), so it occupies the site without reacting. Succinate dehydrogenase is part of the Krebs cycle.

A12. Vmax = 1/0.020 = 50 µmol min⁻¹. −1/Km = −0.40, so Km = 1/0.40 = 2.5 mmol dm⁻³.

A13. (a) n(O₂) = 0.036 dm³ / 24.0 dm³ mol⁻¹ = 1.5 × 10⁻³ mol. (b) Ratio H₂O₂ : O₂ = 2 : 1, so n(H₂O₂) = 3.0 × 10⁻³ mol. (c) Rate = 3.0 × 10⁻³ / 60 = 5.0 × 10⁻⁵ mol s⁻¹.

A14. A good answer includes:

• Independent variable: pH, using buffer solutions (e.g. pH 4, 5, 6, 7, 8, 9) (see buffers explained).
• Dependent variable: rate of oxygen production — volume of gas collected in a gas syringe or upturned measuring cylinder in a fixed time (e.g. 60 s), or the time for a disc to float.
• Controlled variables: temperature (water bath), concentration and volume of H₂O₂, number, size and surface area of potato discs (cut with a cork borer to the same thickness), volume of buffer, same potato.
• Control: boiled potato discs to show that the enzyme causes the reaction.
• Repeats: at least three at each pH; calculate a mean and discard anomalies.
• Safety: wear eye protection — hydrogen peroxide is an irritant and oxidiser; take care with cork borers and scalpels.

See the enzyme experiment guide for a full method.

Key takeaways

• Use precise terms: active site, complementary, enzyme–substrate complex, denatured, saturated.
• Explain temperature effects with kinetic energy and collisions below the optimum, and denaturation above.
• Km measures affinity (lower = tighter); at [S] = Km, v = Vmax/2.
• Competitive inhibitors raise apparent Km; non-competitive inhibitors lower Vmax.