Home · A-Level Biology 9700 · Enzymes · Factors that affect enzyme action · Practice
AS · Practice questions · Factors that affect enzyme action

Two questions: meet and fit.

Six original Cambridge-style questions on the factors that change enzyme activity, from denaturation to reading Km off a graph.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Collisions, shape, sites.

01
Explain
[4 marks]

Explain the effect of increasing temperature from 20 °C to 60 °C on the rate of a reaction catalysed by a human enzyme.

  • From 20 °C to the optimum (about 37-40 °C) the rate increases, because the molecules have more kinetic energy. ✓
  • So there are more frequent collisions between substrate and active site, and more enzyme-substrate complexes form per second. ✓
  • Above the optimum, hydrogen and ionic bonds in the tertiary structure break, so the active site changes shape. ✓
  • The substrate no longer fits / is no longer complementary; the enzyme is denatured and the rate falls towards zero. ✓
02
Analysis
[3 marks]

Describe how you would investigate the effect of pH on the activity of catalase, identifying the independent variable and two variables to control.

  • Independent variable: pH, set with a range of buffer solutions (for example pH 4, 5, 6, 7, 8, 9) added to the hydrogen peroxide. ✓
  • Measure the volume of oxygen released in a fixed time (or the initial rate) with a gas syringe. ✓
  • Control two of: temperature (water bath), concentration and volume of hydrogen peroxide, concentration and volume of catalase. ✓
03
Data
[3 marks]

For an enzyme, Vmax is 24 μmol min−1. At a substrate concentration of 3.0 mmol dm−3 the rate is 12 μmol min−1. State Km and explain how a second enzyme with Km 0.8 mmol dm−3 for the same substrate compares.

  • 12 is half of 24, so Km = 3.0 mmol dm−3. ✓
  • The second enzyme has a lower Km, so it reaches half of its Vmax at a lower substrate concentration. ✓
  • It therefore has a higher affinity for the substrate. ✓
04
Explain
[4 marks]

Compare the effects of a competitive and a non-competitive reversible inhibitor on enzyme activity.

  • Competitive: similar shape to the substrate, binds to the active site, so blocks substrate binding. ✓
  • Its effect is reduced by increasing substrate concentration; Vmax is unchanged but Km increases. ✓
  • Non-competitive: binds to a site other than the active site, changing the shape of the active site. ✓
  • Increasing substrate does not overcome it; Vmax decreases (Km unchanged). Both bind reversibly by weak bonds. ✓
05
Explain
[2 marks]

Explain why increasing the substrate concentration has no further effect on the rate once Vmax is reached, and state what would raise the rate.

  • At Vmax all the active sites are occupied at any moment (saturated), so enzyme concentration is the limiting factor. ✓
  • Increasing the enzyme concentration (or the temperature, up to the optimum) would raise the rate. ✓
06
Analysis
[3 marks]

A student compares lactase immobilised in alginate beads with free lactase. Describe how the beads are made and give two advantages of immobilisation.

  • Mix lactase with sodium alginate solution and drip it into calcium chloride solution; calcium alginate beads form, trapping the enzyme. ✓
  • Advantage: the enzyme can be recovered and reused / allows a continuous process. ✓
  • Advantage: the product is not contaminated with enzyme; or the enzyme is more stable at higher temperatures and over a wider pH range. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.