Skip to content
MarkSchemeMarkScheme
Sign up — no card

9700 · 3.1

Mode of action of enzymes — practice questions

Practice and worked examples for 9700 Mode of action of enzymes. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Outline the mechanism by which enzymes accelerate biochemical reactions.

Show solution outline
  1. Enzyme-Substrate Interaction: Enzymes are globular proteins with a unique, three-dimensional active site which is specifically shaped and chemically configured to bind a particular reactant molecule, known as the substrate.
  2. Enzyme-Substrate Complex Formation: When the substrate binds to the active site, an enzyme-substrate complex (ESC) is formed. This binding is temporary and involves weak non-covalent interactions.
  3. Induced Fit: The binding of the substrate often induces a slight conformational change in the enzyme's active site (the induced-fit hypothesis). This optimises the fit, places strain on the substrate's bonds, and correctly orientates it for the reaction.
  4. Lowering Activation Energy: The formation of the ESC facilitates the reaction by providing an alternative reaction pathway that requires significantly less energy. This process effectively lowers the activation energy – the minimum energy required for the reaction to proceed.
  5. Product Release and Enzyme Regeneration: Once the reaction occurs, converting the substrate into products, these products are released from the active site. The enzyme remains unchanged and is then free to bind to another substrate molecule, ready to catalyse the same reaction again.

Worked example 2

A solution contains 2.0 x 10⁻⁶ mol dm⁻³ of the enzyme carbonic anhydrase. When saturated with its substrate (CO₂), the enzyme is found to convert it to product at a maximum rate (V_max) of 1.2 mol dm⁻³ s⁻¹. Calculate the turnover number (k_cat) for carbonic anhydrase.

Show solution outline
  1. Recall the formula: The turnover number (k_cat) is calculated by dividing the maximum rate of reaction (V_max) by the total enzyme concentration ([E]t).
    • Formula: k_cat = V_max / [E]t
  2. Identify the given values:
    • V_max = 1.2 mol dm⁻³ s⁻¹
    • [E]t = 2.0 x 10⁻⁶ mol dm⁻³
  3. Substitute the values into the formula:
    • k_cat = (1.2 mol dm⁻³ s⁻¹) / (2.0 x 10⁻⁶ mol dm⁻³)
  4. Perform the calculation:
    • k_cat = 0.6 x 10⁶ s⁻¹
    • k_cat = 600,000 s⁻¹
  5. State the final answer with units: The turnover number for carbonic anhydrase is 600,000 s⁻¹. This means a single molecule of carbonic anhydrase can catalyse the conversion of 600,000 substrate molecules per second at its maximum speed.
Mark a 9700 question on this topicFull lessonOpen in course studio