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9700 · 3.2

Factors that affect enzyme action — practice questions

Practice and worked examples for 9700 Factors that affect enzyme action. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A student investigates the effect of temperature on the activity of amylase. They record the time taken for starch to be completely hydrolysed at various temperatures. Describe and explain the expected trend in reaction rate as the temperature increases from 10°C to 70°C.

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Description of Trend: As temperature increases from 10°C up to an optimum (typically around 37-40°C for amylase), the rate of starch hydrolysis will increase. Beyond the optimum, as temperature continues to rise to 70°C, the rate of reaction will rapidly decrease.

Explanation:

  1. 10°C to Optimum: Increasing temperature provides substrate and enzyme molecules with more kinetic energy. This leads to more frequent and energetic collisions, increasing the chances of successful enzyme-substrate complex formation. Consequently, the rate of starch breakdown increases.
  2. At Optimum: This is the temperature at which the enzyme exhibits its highest activity, as the balance between collision frequency and enzyme stability is ideal.
  3. Above Optimum (to 70°C): Beyond the optimum temperature, the excessive kinetic energy causes vibrations within the amylase molecule. These vibrations break the weak hydrogen and ionic bonds maintaining the active site's specific three-dimensional shape. The active site permanently changes shape, a process known as denaturation. A denatured enzyme can no longer bind effectively with starch, leading to a significant and irreversible loss of catalytic activity, and thus a sharp decrease in the rate of hydrolysis.

Worked example 2

An experiment measures the initial rate of an enzyme-catalysed reaction at different substrate concentrations. The data shows that Vmax is 80 µmol min⁻¹. The rate of reaction is found to be 40 µmol min⁻¹ when the substrate concentration is 5 mmol dm⁻³. Calculate the Michaelis-Menten constant (Km) for this enzyme.

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Step 1: Understand the definitions

  • Vmax (Maximum Velocity): The maximum rate of reaction when the enzyme is fully saturated with substrate. Given as 80 µmol min⁻¹.
  • Km (Michaelis-Menten Constant): The substrate concentration at which the reaction rate is exactly half of Vmax (½ Vmax).

Step 2: Calculate ½ Vmax First, we need to find the rate that is half of the maximum velocity. ½ Vmax = Vmax / 2 ½ Vmax = 80 µmol min⁻¹ / 2 ½ Vmax = 40 µmol min⁻¹

Step 3: Determine Km from the data The definition of Km is the substrate concentration that gives a rate of ½ Vmax. From the problem statement, we are told that the reaction rate is 40 µmol min⁻¹ when the substrate concentration is 5 mmol dm⁻³. Since the rate of 40 µmol min⁻¹ is equal to our calculated ½ Vmax, the corresponding substrate concentration is the Km.

Step 4: Final Answer Therefore, the Michaelis-Menten constant (Km) for this enzyme is 5 mmol dm⁻³.

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