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3.1.4Additional lessons

Enzyme Reaction Rates Skills — IB Biology SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 53 of 142

📚 IB Biology SL Hub / Topical Revision
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Part 1 of 6Progress Curves

3.1.4 Enzyme Reaction Rates Skills
1
A graph of product formed against time starts steeply because plenty of substrate is present.
2
The gradient decreases as substrate is used up and fewer enzyme molecules are occupied.
3
The curve finally levels off when all the substrate has been converted into product.
4
The initial rate is the gradient of the curve at time zero, before conditions change.
5
To find it, draw a tangent at the start and calculate change in y divided by change in x.

Part 2 of 6Calculating Rates

Rate formula
  • Rate equals the amount of product formed, or substrate used, divided by the time taken.
Units
  • Choose units that fit the measurement, such as cm³ s⁻¹ for gas volume or g min⁻¹ for mass.
Tangent
  • The gradient at one point of a curve is found from a line touching the curve at that point.
Mean rate
  • The average rate over a period is the total change divided by the total time.
Comparing
  • Initial rates should be compared, since the later rate is affected by changes in substrate concentration.

Part 3 of 6Interpreting Graphs

Interpreting Graphs
Substrate graph
  • Rate increases then plateaus because the enzymes become saturated with substrate.
Temperature graph
  • Rate rises to an optimum and then drops sharply as the enzyme denatures.
pH graph
  • A bell-shaped curve shows an optimum pH, with activity falling on either side.
Plateau
  • Reasons for a plateau include saturation of enzymes, limited substrate or a limiting enzyme concentration.

Part 4 of 6Using Rate Data

Using Rate Data
Comparing
  • Use the initial rates to compare conditions, since they are measured when the substrate has not yet run out.
Error
  • A tangent drawn by eye gives some error, so the gradient should be treated as an estimate.
Link to theory
  • A greater rate at higher temperature is explained by more frequent collisions with greater energy.

Part 5 of 6Worked Example

Question

  • Oxygen collected from catalase and hydrogen peroxide is shown in a graph.
  • The tangent at the origin passes through (0 s, 0 cm³) and (20 s, 10 cm³).
  • Calculate the initial rate.

  1. Gradient = change in volume ÷ change in time.
  2. Rate = (10 − 0) cm³ ÷ (20 − 0) s = 0.5 cm³ s⁻¹.

Answer The initial rate is 0.5 cm³ s⁻¹ (example data).

Part 6 of 6Exam Checklist & Tips

3.1.4 – Key Points for the Exam

  • Initial rate The rate of reaction at the start, found from the gradient at time zero.
  • Gradient The change in the y value divided by the change in the x value.
  • Saturation The state when all active sites of an enzyme are occupied.
  • Plateau A flat part of a graph where the value no longer changes.

Common Mistakes

  • Calculating the rate from the whole graph when the initial rate was requested.
  • Drawing a tangent that crosses the curve instead of touching it.
  • Leaving out units for the rate or giving the wrong units.
  • Explaining a plateau as enzyme denaturation, when it is usually due to the substrate being used up.

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