Question

In the presence of a non 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 C $V_{\max }$ decreases and $K_M$ remains the same

$K_M$ 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 non competitive inhibitor binds at a site other than the active site. It can bind to the free enzyme or the ES complex. It changes the shape of the active site such that substrate is not able to bind to it. Since the inhibitor can bind to either the free enzyme or the ES complex, $V_{\max }$ is lowered. The $K_M$ however remains the same, because to reach half of the maximum attainable velocity, the substrate concentration remains the same.