In the presence of a noncompetitive inhibitor, which is true about Km?

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Multiple Choice

In the presence of a noncompetitive inhibitor, which is true about Km?

Explanation:
Noncompetitive inhibition lowers the reaction’s maximum rate without changing how tightly the substrate binds. The inhibitor attaches to a site other than the active site and can bind to either the free enzyme or the enzyme-substrate complex. This reduces the number of active enzyme molecules, so Vmax drops, but the ability of the enzyme to bind substrate—the affinity, reflected by Km—remains the same. In the Michaelis-Menten view, V = (Vmax [S])/(Km + [S]), and when a noncompetitive inhibitor is present, Vmax decreases while Km stays unchanged, meaning half-maximal velocity occurs at the same substrate concentration as before.

Noncompetitive inhibition lowers the reaction’s maximum rate without changing how tightly the substrate binds. The inhibitor attaches to a site other than the active site and can bind to either the free enzyme or the enzyme-substrate complex. This reduces the number of active enzyme molecules, so Vmax drops, but the ability of the enzyme to bind substrate—the affinity, reflected by Km—remains the same. In the Michaelis-Menten view, V = (Vmax [S])/(Km + [S]), and when a noncompetitive inhibitor is present, Vmax decreases while Km stays unchanged, meaning half-maximal velocity occurs at the same substrate concentration as before.

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