Question

Which of the following is true for competitive inhibition?

• A. $K_m$ decreases but $V_{max}$ increases
• B. $K_m$ increases but $V_{max}$ decreases
• C. $K_m$ increases but $V_{max}$ remains the same
• D. $K_m$ decreases but $V_{max}$ remains the same

Answer 1

The correct option is C $K_m$ increases but $V_{max}$ remains the same

In an enzymatic chemical reaction, the substrate binds at the active site of the enzyme and forms the product. The enzyme is always substrate-specific. But when an inhibitor with similar characteristics of the substrate binds at the active site of the enzyme, like the substrate, then there will be no product formation. This type of inhibition is known as competitive inhibition.


The above-given graph represents the change in the rate of reaction with respect to change in the substrate concentration. From the above graph, it can be inferred that as the substrate concentration increases, the rate of reaction increases and becomes constant at a point. The substrate concentration at which the rate of reaction is half the maximum rate $\left(V_{max}\right)$ is represented by the Michaelis-Menten constant $\left(K_m\right)$.
The $K_m$ value changes in the presence of a competitive inhibitor. The Km value increases when there is a competitive inhibitor.
To achieve a required concentration of product, a larger amount of substrate is used that leads to an increase in $K_m$. Whereas $V_{max}$ remains unchanged in the presence or absence of an inhibitor.
So, the correct answer is - Km increases and Vmax remains same.