Question

How does the effectiveness of a competitive inhibitor gets affected under high physiological concentration of the substrate?

• A. The inhibitor becomes more effective.
• B. The effectiveness of the inhibitor gets reduced.
• C. The inhibitor remains equally effective
• D. The inhibitor gets irreversibly inactivated

Hint:

In competitive inhibition:
The apparent $\text{K}_m$ increases, but $\text{V}_{\max}$ remains unchanged.
This is because high substrate concentrations outcompete the inhibitor, restoring maximum enzymatic activity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Enzyme inhibitors can bind to enzymes to reduce their catalytic activity.
In competitive inhibition, the inhibitor structurally resembles the substrate and competes with it for binding to the active site of the free enzyme.

Step 2: Detailed Explanation:
The binding of a competitive inhibitor to the active site is a reversible equilibrium process:
\[ \text{E} + \text{I} \rightleftharpoons \text{EI} \]
Because the substrate ($\text{S}$) and the inhibitor ($\text{I}$) bind to the same active site, they compete directly with one another.
At low substrate concentrations, the inhibitor successfully binds to a fraction of the enzyme molecules, forming the inactive $\text{EI}$ complex and reducing the reaction rate.
However, if the physiological substrate concentration is increased to high levels, the substrate outcompetes the inhibitor for binding to the active site.
As the substrate concentration approaches infinity, virtually all enzyme molecules exist as the active $\text{ES}$ complex rather than the $\text{EI}$ complex.
This high substrate concentration reverses the effect of the competitive inhibitor, allowing the reaction velocity to approach its maximum rate ($\text{V}_{\max}$).
Therefore, the effectiveness of a competitive inhibitor is reduced under high substrate concentrations.

Step 3: Final Answer:
The effectiveness of the competitive inhibitor gets reduced under high substrate concentrations, which corresponds to Option (B).