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Enzymes and Metabolism · Paper 1 — Free Biology SL Practice Questions

1FoundationMCQEnzyme kinetics and inhibition1 markPaper 1~2 min
In an experiment using catalase, the rate of oxygen production is measured at increasing hydrogen peroxide concentrations. At low concentrations, the rate increases proportionally, but above a certain concentration the rate remains constant regardless of further increases in substrate. Which of the following best explains this plateau?
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Solutions

2MasteryMCQEnzyme kinetics and inhibition1 markPaper 1~2 min
An enzyme has a Michaelis constant (KmK_m) of 2.0 mmol dm−32.0 \ \text{mmol dm}^{-3}. At a substrate concentration of 4.0 mmol dm−34.0 \ \text{mmol dm}^{-3}, the reaction velocity V0V_0 is approximately 67 percent of VmaxV_{max}. A student predicts that doubling the substrate concentration to 8.0 mmol dm−38.0 \ \text{mmol dm}^{-3} will raise V0V_0 to 90 percent of VmaxV_{max}. Which statement correctly evaluates this prediction using the Michaelis–Menten equation V0=Vmax[S]Km+[S]V_0 = \dfrac{V_{max}[S]}{K_m + [S]}?
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Solutions

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