Designing a Fair Enzyme Practical
A student plans to investigate the effect of temperature on the activity of catalase from potato tissue. The method: cut potato cylinders of 'about the same size', place one in a test tube with 5 mL of 3% hydrogen peroxide, and measure the volume of oxygen collected in 60 seconds using a gas syringe. Temperatures are to be 10, 20, 30, 40, 50 and 60 °C, achieved by standing the test tube in a water bath for 2 minutes before adding the potato. The student notices that at 60 °C the gas volume is very low, and concludes that catalase is denatured at 60 °C. A classmate points out that the potato was added to the peroxide before the tube had reached 60 °C.
1. Identify the independent, dependent and at least three controlled variables, and explain why each control matters.
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Independent variable: temperature of the reaction mixture (10–60 °C). Dependent variable: volume of oxygen collected in 60 seconds (a measure of catalase activity).
Controlled variables and reasons:
- Volume and concentration of hydrogen peroxide (5 mL of 3%) - substrate concentration affects reaction rate, so it must be constant to attribute differences to temperature.
- Size/surface area of the potato cylinder - catalase is in the tissue; a larger piece provides more enzyme, increasing rate. Surface area also affects diffusion of substrate in.
- Time of collection (60 s) - rate is volume per unit time; varying time changes the measured volume.
- pH - enzymes have an optimum pH; changes would confound the temperature effect.
- Same potato and same batch - enzyme concentration varies between tubers.
- Equilibration time in the water bath - ensures the reaction starts at the intended temperature.
Each control removes an alternative explanation for any observed difference, so the conclusion can be attributed to temperature.
2. Evaluate the classmate's criticism and the student's conclusion. How should the method be changed?
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The classmate is correct. The potato was added before the tube reached 60 °C, so the reaction began at a lower temperature and warmed during the 60-second collection. The measured volume therefore reflects a mixture of temperatures, not a constant 60 °C, and the low volume cannot be attributed to denaturation at 60 °C. The student's conclusion is unsupported by the data as collected.
A better method: equilibrate both the peroxide and the potato separately in the water bath until the peroxide reaches the target temperature (check with a thermometer), then add the potato and start timing. Keep the tube in the bath throughout. Use a fresh potato cylinder and fresh peroxide for each trial, and repeat each temperature at least three times to calculate a mean and identify outliers.
To test denaturation specifically, the student could pre-heat the potato at 60 °C for several minutes, cool it back to room temperature, then assay activity. If activity does not recover, denaturation is supported; if it does, the low rate was due to temperature effects on rate rather than irreversible denaturation.
3. Predict the shape of the expected graph and explain the biological reason for each part of the curve.
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The graph of oxygen volume against temperature should rise from 10 °C to a peak around 30–40 °C (the optimum for potato catalase), then fall sharply at 50–60 °C.
Rising phase: increasing temperature increases the kinetic energy of enzyme and substrate molecules, so more frequent and more energetic collisions occur, and more enzyme–substrate complexes form per unit time. The rate increases roughly exponentially at first.
Peak: at the optimum temperature, the rate is maximal because collisions are frequent without disrupting the enzyme's shape.
Falling phase: above the optimum, increased thermal energy breaks the hydrogen bonds and other weak interactions holding the enzyme's tertiary structure. The active site loses its complementary shape, substrate can no longer bind efficiently, and the rate falls. At high enough temperatures this denaturation is irreversible, which is why activity does not return on cooling.
A steep fall rather than a gradual one is expected because denaturation is cooperative - once the structure begins to unfold, it rapidly loses function.