Catalase Kinetics: Reading a Plateau and Diagnosing an Inhibitor
A student homogenised 20 g of potato tissue in cold buffer and used the supernatant, after centrifugation, as a source of catalase. She measured the volume of oxygen released in one minute when 2 cm3 of extract was added to 20 cm3 of hydrogen peroxide at pH 7.0 and 25 °C.
| H2O2 / mmol dm-3 | 10 | 20 | 40 | 80 | 160 |
|---|---|---|---|---|---|
| O2 / cm3 min-1 | 12 | 22 | 38 | 52 | 58 |
In a second run at 80 mmol dm-3 hydrogen peroxide she added copper(II) sulfate to give 1.0 mmol dm-3 Cu2+ in the mixture; the rate fell to 31 cm3 min-1. At 10 mmol dm-3 the rate with Cu2+ was 6 cm3 min-1. A boiled extract still gave 2 cm3 of gas in one minute.
1. Estimate the maximum rate (Vmax) from the data and explain, in terms of enzyme structure, why the curve flattens. Give one reason why this experiment cannot measure the true Vmax directly.
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Reading the trend. The rate rises steeply at first (+10, then +16 cm3 min-1 between 10 and 40 mmol dm-3) and then only slowly (+14, then +6). The last two values, 52 and 58, are close together, so the curve is approaching a plateau of about 60 cm3 min-1. Any value between 58 and 65 is a defensible estimate; the honest statement is "Vmax is approximately 60 cm3 min-1", not an exact figure.
Why it flattens. Catalase molecules have a fixed number of active sites. At low substrate concentration most sites are empty, so raising [H2O2] binds more substrate and the rate rises almost in proportion. As concentration rises, an increasing fraction of sites is occupied at any instant; once every site is occupied as soon as it becomes free, the enzyme is saturated and adding more substrate cannot increase the rate. The rate then depends on the turnover number of the enzyme, not on substrate concentration.
Why the true Vmax is not reached. At 160 mmol dm-3 the plateau may not yet be complete, and pushing the concentration higher risks substrate inhibition or denaturation of the protein by concentrated peroxide, so the enzyme may be destroyed before saturation is demonstrated. The true Vmax is therefore an extrapolation from the curve, not a measured point.
2. Using both Cu2+ data points, decide whether the inhibition is competitive or non-competitive. Justify your conclusion and state one further measurement that would make it firmer.
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Key evidence - the near-saturating point. At 80 mmol dm-3 the enzyme is close to saturation (52 of an estimated 60 cm3 min-1). With Cu2+ present the rate falls to 31 cm3 min-1. Competitive inhibitors bind the active site and are displaced by excess substrate, so at near-saturating substrate a competitive inhibitor should have little or no effect - the same Vmax is still reachable, just at a higher substrate concentration. Here the rate is reduced by about 40% even when substrate is plentiful, which rules out simple competitive inhibition.
Supporting evidence - the low-substrate point. At 10 mmol dm-3 the rate halves, from 12 to 6 cm3 min-1. A roughly constant fractional inhibition at both low and high substrate concentration is what non-competitive (or mixed) inhibition predicts: the inhibitor binds a site other than the active site and distorts the enzyme, so a proportion of enzyme molecules is effectively removed whatever the substrate concentration.
Conclusion. The data fit non-competitive inhibition better than competitive inhibition, although mixed inhibition is not excluded by two points alone.
A firmer measurement. Repeat over the full substrate range (10–320 mmol dm-3) with Cu2+ present and compare the plateau: a lowered plateau confirms non-competitive inhibition, whereas an unchanged plateau reached only at much higher [H2O2] would indicate competitive inhibition.
3. Explain the 2 cm3 of gas produced by the boiled extract and suggest one improvement that would make the control more reliable.
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Why boiling matters. Boiling denatures catalase: the hydrogen bonds and other weak interactions holding the tertiary structure are broken, the active site loses its shape, and the enzyme can no longer catalyse the decomposition of hydrogen peroxide. The 2 cm3 cannot, therefore, be enzyme-catalysed oxygen.
Where the gas comes from. Several non-biological sources are plausible:
- Air dissolved in the buffer or trapped in the apparatus expands as the mixture warms towards room temperature or as it is handled.
- Hydrogen peroxide decomposes slowly and non-enzymatically, especially in light or in the presence of traces of metal ions, releasing oxygen.
- The syringe or respirometer has dead space and the first reading drifts before the apparatus settles.
Because this blank is small relative to the experimental rates (2 against 12–58 cm3 min-1) it does not overturn the conclusions, but it does mean every rate is slightly overestimated.
Improvement. Equilibrate all solutions to 25 °C in a water bath before starting, and run a parallel control in which the extract is replaced by an equal volume of buffer only. Subtracting this control value from each reading corrects for non-enzymatic decomposition and for apparatus drift, so the rates reported reflect catalysis alone.