A level-readyAll subjects · Biology · WBI11 (Unit 1)
Core practical 4: factors affecting enzyme activity
Core practical 4 investigates how temperature, pH, enzyme concentration and substrate concentration affect the initial rate of an enzyme-catalysed reaction.
A typical method
Example: catalase breaking down hydrogen peroxide, collecting oxygen in an upturned measuring cylinder or gas syringe. Or trypsin digesting the protein in milk, timing how long the cloudy milk takes to clear.
1. Equilibrate enzyme and substrate separately in a water bath at the chosen temperature.
2. Mix and start timing; record product (or loss of substrate) at regular intervals.
3. Repeat for each value of the independent variable, at least three times.
Control: volumes and concentrations, temperature (water bath), pH (buffer), time.
Initial rate
The rate is fastest at the start, before substrate is used up. To find the initial rate, plot product against time and draw a tangent at t = 0; its gradient is the initial rate. A simpler measure is 1/time to reach an end point.
Expected results
- Temperature: rate rises as molecules gain kinetic energy and collide more often, up to an optimum; above it, bonds holding the tertiary structure break, the active site changes shape and the enzyme denatures.
- pH: each enzyme has an optimum; away from it, H+ ions disrupt ionic and hydrogen bonds, changing the active site.
- Enzyme concentration: more active sites, rate rises (if substrate is in excess).
- Substrate concentration: rate rises, then levels off when all active sites are occupied (enzyme concentration becomes limiting).
Worked example
24 cm3 of oxygen is collected in the first 30 s. Calculate the initial rate.
- Rate = volume / time = 24 / 30
Answer: 0.8 cm3 s-1
Check yourself
- 18 cm3 of oxygen is released in the first 40 s. Calculate the initial rate in cm3 s-1.
Show answer
0.45 cm3 s-1. 18 / 40 = 0.45 cm3 s-1.
- Milk takes 50 s to clear with trypsin. Calculate the rate as 1/time in s-1.
Show answer
0.02 s-1. 1 / 50 = 0.02 s-1.
- How is the initial rate found from a graph of product against time?
- Read the final value of product
- Draw a tangent at time zero and find its gradient
- Find the area under the curve
- Divide final product by total time
Show answer
Draw a tangent at time zero and find its gradient. The gradient at t = 0 gives the rate before substrate is used up.
- Why does the rate fall above the optimum temperature?
- Molecules move more slowly
- There is less substrate
- Bonds holding the tertiary structure break, so the active site changes shape
- The enzyme is used up
Show answer
Bonds holding the tertiary structure break, so the active site changes shape. The enzyme denatures, so substrates no longer fit.
- Why does the rate level off at high substrate concentration?
- All active sites are occupied, so enzyme concentration is limiting
- The enzyme has denatured
- Substrate is used up
- The temperature is too low
Show answer
All active sites are occupied, so enzyme concentration is limiting. With every active site in use, adding substrate cannot increase the rate.
- Why is a buffer added when investigating temperature?
- To change the temperature
- To denature the enzyme
- To keep the pH constant
- To act as a substrate
Show answer
To keep the pH constant. pH is a control variable here.
- Why are enzyme and substrate placed separately in the water bath before mixing?
- So enzyme and substrate are at the test temperature before the reaction starts
- To denature the enzyme
- To increase the volume
- To remove oxygen
Show answer
So enzyme and substrate are at the test temperature before the reaction starts. The reaction must take place at the chosen temperature from the start.
Sources
Last updated 7 November 2026. Spotted a mistake? Tell us.