Question

$K_m$ and $V_{max}$ of an enzyme preparation are 5 µM and 30 $µM min^{-1}$ respectively. Considering, $K_i$ value of competitive inhibitor is 60 µM, the velocity ($V_0$) of this enzyme-catalyzed reaction in the presence of 200 µM of substrate and 600 µM of competitive inhibitor is _______ $ µM min^{-1}$ (rounded off to two decimal places).

Answer 1

Correct Answer: 23.5

To solve this problem, we determine the reaction velocity ($V_0$) using the formula for enzyme kinetics in the presence of a competitive inhibitor. The modified Michaelis-Menten equation for competitive inhibition is:

V₀ = (V_max * [S]) / (K_m * (1 + [I]/K_i) + [S])

 

Given:

• $V_{max} = 30 \mu M \ min^{-1}$
• $K_m = 5 \mu M$
• $K_i = 60 \mu M$
• [S] = 200 µM
• [I] = 600 µM

First, calculate the factor by which $K_m$ is increased due to the inhibitor:

K_m,app = K_m * (1 + [I]/K_i) = 5 * (1 + 600/60) = 5 * (1 + 10) = 55 µM

 

Now use the modified Michaelis-Menten equation:

V₀ = (30 * 200) / (55 + 200)

 

= 6000 / 255

 

≈ 23.53 µM min^-1

 

The calculated velocity is 23.53 µM min^-1.