Question

In the presence of a competitive inhibitor

• A. $V_{\max }$ increases and $K_M$ decreases
• B. $V_{\max }$ remains same and $K_M$ increases
• C. $V_{\max }$ decreases and $K_M$ remains the same
• D. $V_{\max }$ decreases and $K_M$ increases

Answer 1

The correct option is A $V_{\max }$ increases and $K_M$ decreases

$K_M$ stands for Michaelis-Menten constant. It is the substrate concentration at which the velocity of the reaction is half of the maximum velocity that can be achieved.

It is the indirect measure of the affinity of the enzyme and the substrate. There is an indirect relationship between the value of $K_M$ and the affinity between substrate and enzyme molecules.

Higher $K_M$=lower affinity; lower $K_M$=higher affinity.
$V_{\max }$ stands for the maximum velocity or rate that a reaction can achieve.

A competitive inhibitor resembles the substrate very closely structurally and competes with the substrate for binding to the active site of the enzyme. It binds to the active site and forms the enzyme-inhibitor complex which doesn’t give products.

The inhibition is reversible. It can be reversed by increasing the substrate concentration. Increasing the substrate concentration increases the probability of substrate to bind with the enzymes and form products. With increase in a substrate concentration, $V_{\max }$ can be achieved. So, $V_{\max }$ remains the same but $K_M$ increases because the reaction is able to reach half of its $V_{\max }$ at an increased substrate concentration.