Question

1. Please explain why “noncompetitive inhibition”- where it is observed experimentally- is more likely to be a coincidence than a reveal of something profound in the mechanism of an enzyme (in contrast to competitive and uncompetitive inhibition).

Answer 1

Sol :-

Noncompetitive inhibition lowers only Vmax and binds to a site other than catalytic (allosteric site) on the enzyme. By doing so, they change the structural configuration of the enzyme cleft (probably by disrupting Hydrogen bonding or hydrophobic interactions). This also makes enzyme-substrate binding impossible. It is a special case of mixed inhibition where the inhibitor can be bound to either the substrate occupied or the substrate free enzyme. If the binding constants to both of these states is equivalent, then the Michaelis constant, Kmm, is unaffected by the inhibitor, but pseudo-zero order rate constant Vmax,appmax,app becomes smaller.

Also the inhibitor in this case usually binds to place other than active site which leads to deformation of original active site where the real substrate would have bounded, thus doesn't matter even if we increase substrate concentration, it still won't bind effectively or at all. “there is no actual competition between substrate and inhibitor!” , thus substrate concentration will definitely have no effect on such inhibition.

In non-competitive inhibition Km does not change because the inhibitor binds the free enzyme and the enzyme-substrate complex with the same affinity (that is Ki = K’i, so α=α’). As a result because km = (k-1 + k2)/k1, the ratio does not change because k1 and k2 are reduced by the same amount. Vmax is affected but for Km is just the ratio of the different rate constants that matters.

Hence, it is more likely to be a coincidence than a reveal of something profound in the mechanism of an enzyme.