Question

The half-maximal velocity of an enzyme catalyzed reaction was found at a substrate concentration of \(0.5 \times 10^{-6} \, \text{M}\). This enzyme follows Michaelis-Menten kinetics. In the presence of a competitive inhibitor, the half-maximal velocity was found at a substrate concentration of \(1.5 \times 10^{-6} \, \text{M}\). Given that the enzyme-inhibitor pair has a dissociation constant of \(2 \times 10^{-7} \, \text{M}\), the concentration of the competitive inhibitor in \(\mu M\), rounded off to one place of decimal, was ______.

Hint:

For competitive inhibition, the presence of the inhibitor increases the substrate concentration required to reach half-maximal velocity. The concentration of the inhibitor can be determined using Michaelis-Menten kinetics.

Answer 1

Correct Answer: 0.4

The Michaelis-Menten equation for competitive inhibition is given by: \[ V = \frac{V_{\text{max}} [S]}{K_m(1 + [I]/K_i) + [S]} \] where:
- \( V \) is the observed velocity,
- \( V_{\text{max}} \) is the maximum velocity,
- \( [S] \) is the substrate concentration,
- \( K_m \) is the Michaelis constant,
- \( [I] \) is the concentration of the inhibitor,
- \( K_i \) is the dissociation constant of the enzyme-inhibitor pair.
For half-maximal velocity, \( V = \frac{V_{\text{max}}}{2} \), so the equation simplifies to: \[ \frac{V_{\text{max}}}{2} = \frac{V_{\text{max}} [S]}{K_m(1 + [I]/K_i) + [S]} \] Simplifying further: \[ 1 = \frac{[S]}{K_m(1 + [I]/K_i) + [S]} \] We know that the half-maximal velocity occurs at a substrate concentration of \( [S] = 0.5 \times 10^{-6} \, \text{M} \) without the inhibitor, and at \( [S] = 1.5 \times 10^{-6} \, \text{M} \) with the inhibitor. Hence, for the condition with the inhibitor, we use the fact that the half-maximal velocity corresponds to: \[ 1 = \frac{1.5 \times 10^{-6}}{0.5 \times 10^{-6}(1 + [I]/K_i) + 1.5 \times 10^{-6}} \] Given \( K_i = 2 \times 10^{-7} \, \text{M} \), solving for \( [I] \), we get: \[ [I] \approx 0.4 \, \mu M \] Thus, the concentration of the competitive inhibitor is approximately \( 0.4 \, \mu M \).