Question

why does Vmax not change in the presence or absence of a competitive inhibitor? conversely, Vmax will decrease in the presence of either non-competitive or uncompetitive inhibitor. explain how binding inhibitor leads to an apparent decrease in Vmax.

Answer 1

In case of competitive inhibition the Vmax does not change because increasing amounts of substrate can saturate (swamp) the inhibitor which is present in fixed concentration thus allowing the enzyme to effectively not see the inhibitor at high substrate concentrations. On the other hand in case of non-competitive or uncompetitive inhibitor, they do not bind to the active site of the enzyme does not resemble the substrate in this case. Therefore, further addition of substrate will not eliminate the effect of the inhibitor. As a result, there is always a fixed amount of enzyme inactive in non-competitive inhibition. So when you change the amount of enzyme, Vmax changes. Thus, in the presence of a non-competitive inhibitor Vmax decreases.