In noncompetitive inhibition, which statement is true regarding Vmax and Km?

In noncompetitive inhibition, which statement is true regarding Vmax and Km?

• A. Vmax decreases, Km unchanged
• B. Vmax unchanged, Km increases
• C. Vmax decreases, Km decreases
• D. Vmax increases, Km unchanged

Answer: (A)

Noncompetitive inhibition reduces the overall amount of active enzyme available to convert substrate to product without changing how tightly the substrate binds. The inhibitor binds at a site other than the active site, and in pure noncompetitive inhibition it can bind to either the free enzyme or the enzyme-substrate complex with similar affinity. This lowers the maximum velocity you can achieve because some enzyme molecules are effectively out of commission, but it doesn’t change the substrate binding affinity, so Km remains the same. So you end up with a lower Vmax while Km stays unchanged.

Noncompetitive inhibition reduces the overall amount of active enzyme available to convert substrate to product without changing how tightly the substrate binds. The inhibitor binds at a site other than the active site, and in pure noncompetitive inhibition it can bind to either the free enzyme or the enzyme-substrate complex with similar affinity. This lowers the maximum velocity you can achieve because some enzyme molecules are effectively out of commission, but it doesn’t change the substrate binding affinity, so Km remains the same. So you end up with a lower Vmax while Km stays unchanged.