Question

In a plot of 1/V against 1/[S] for an enzyme-catalyzed reaction, the presense of a non-competitive inhibitor will:

Answer 1

Answer:

d) leave the intercept of the 1/[S] axis unchanged

Explanation:

A non-competitive inhibitor binds to the enzyme in an allosteric site, which is a different site than the active site where the substrate binds.

If you add more substrate molecules, the inhibitor will still be able to bind to the enzyme and inhibit its activity, so Vmax decreases but Km (the substrate concentration at which the rate of the reaction is half the Vmax) remains unchanged.

In a Lineweaver-Burk plot (1/V vs 1/[S]), the point where the line hits the y-axis is 1/Vmax and the point where the line hits the x-axis is -1/Km.

Because Vmax decreases the line intercepts the y-axis farther away from zero, and the x-axis in the same place because Km doesn't change.