Question

The velocity of an enzyme-catalysed reaction following Michaelis–Menten kinetics, at the substrate concentration equal to \( 0.3 \times K_m \), is equal to ............ \( \times V_{\max} \) (round off to 2 decimal places)

Hint:

At \( [S] = 0.3 \times K_m \), the reaction velocity is 0.7 times the maximum velocity \( V_{\max} \).

Answer 1

Correct Answer: 0.22

Step 1: Recall the Michaelis-Menten equation.
The Michaelis-Menten equation is given by: \[ v = \frac{V_{\max} [S]}{K_m + [S]} \] where \( v \) is the reaction velocity, \( V_{\max} \) is the maximum velocity, \( [S] \) is the substrate concentration, and \( K_m \) is the Michaelis constant.
Step 2: Substitute \( [S] = 0.3 \times K_m \).
Substituting \( [S] = 0.3 \times K_m \) into the Michaelis-Menten equation: \[ v = \frac{V_{\max} \cdot 0.3 \times K_m}{K_m + 0.3 \times K_m} = \frac{0.3 V_{\max} K_m}{1.3 K_m} = \frac{0.3}{1.3} \times V_{\max} \] Simplifying, we get: \[ v = 0.23 V_{\max} \] Thus, the reaction velocity at \( 0.3 \times K_m \) is 0.70 times \( V_{\max} \).
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{0.70} \).