Enzyme Kinetics with a Competitive Inhibitor in a Crude Lysate
A lysate contains an enzyme with Km = 0.2 mM for substrate S and Vmax = 100 nmol/min per mg protein. A contaminating protease cleaves 30% of the enzyme, removing a regulatory domain but leaving the catalytic site intact; the truncated form has Km = 0.05 mM and Vmax = 140 nmol/min per mg. An investigator adds a reversible inhibitor at 1 mM with Ki = 0.5 mM and observes that at [S] = 1 mM the velocity is 60 nmol/min per mg in the mixed sample. The inhibitor is later shown to bind only the free enzyme (competitive) and not the ES complex. The assay is run at 25°C, but the enzyme's optimum is 37°C.
1. Determine whether the observed velocity is consistent with pure competitive inhibition of the full-length enzyme alone, and explain the discrepancy.
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For competitive inhibition, v = Vmax[S]/(Km(1+[I]/Ki) + [S]). With [I]=1 mM, Ki=0.5 mM, α = 1 + 2 = 3. Using full-length values: v = 100×1/(0.2×3 + 1) = 100/1.6 = 62.5 nmol/min/mg. The observed 60 is close to this, which looks consistent-but the sample is a mixture. The truncated enzyme (30% of protein) has Km 0.05 and Vmax 140; if the inhibitor acts on it too, its contribution is 140×1/(0.05×3 + 1) = 140/1.15 = 121.7 per mg of that fraction. Weighted: 0.7×62.5 + 0.3×121.7 = 43.8 + 36.5 = 80.3 nmol/min/mg, higher than observed 60. The observed value is lower than the mixture prediction, so the inhibitor must act more strongly on the truncated form (or the truncation reduces inhibitor affinity assumptions). Judgment: a single-enzyme Michaelis-Menten fit to a mixed lysate gives a misleading Ki; the observed velocity alone cannot validate the model without resolving the two species.
2. The temperature is raised to 37°C and velocity rises to 90 nmol/min per mg. What can and cannot be concluded about Vmax and Km?
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Raising temperature increases kcat (and thus Vmax) and can change Km because substrate binding and conformational equilibria are temperature-dependent. A rise from 60 to 90 at a single [S] = 1 mM is consistent with either a higher Vmax at similar Km or a lower Km at similar Vmax-one data point cannot separate them. Also, at 37°C the protease may be more active, further shifting the full-length/truncated ratio, so the apparent change confounds temperature effects on kinetics with changes in enzyme composition. Conclusion: you can say the enzyme operates closer to its optimum and apparent velocity increased, but you cannot extract Vmax or Km without a substrate titration at the new temperature, and the mixed-species problem persists.