Question

In Michaelis-Menten equation when \( K_m = C \):

• A. \( \text{The rate of process is equal to half of maximum rate} \)
• B. \( \text{Indicates zero-order process} \)
• C. \( \text{The rate process occurs at a constant rate} \)
• D. \( \text{Equation becomes identical to first-order elimination of drug} \)

Hint:

Michaelis-Menten kinetics describe the relationship between substrate concentration and reaction rate. Remember: - \( K_m \) is a key parameter indicating enzyme affinity. - \( [S] = K_m \) results in half the maximum reaction rate.

Answer 1

The Correct Option is A

Step 1: Write the Michaelis-Menten equation.
The rate of an enzyme-catalyzed reaction follows \[ v = \frac{V_{max} \cdot [S]}{K_m + [S]} \] where \( [S] \) is the substrate concentration and \( V_{max} \) is the top speed of the reaction.

Step 2: Express the rate as a fraction of \( V_{max} \).
Dividing both sides by \( V_{max} \) gives the fraction of maximum rate the reaction is running at: \[ \frac{v}{V_{max}} = \frac{[S]}{K_m + [S]} \]

Step 3: Substitute \( [S] = K_m = C \).
Putting \( [S] = K_m \) into this fraction gives \[ \frac{v}{V_{max}} = \frac{K_m}{K_m + K_m} = \frac{K_m}{2K_m} = \frac{1}{2} \] so the reaction is running at exactly half of its maximum rate the moment substrate concentration equals \( K_m \).

Final Answer:
When \( K_m = C \), the enzyme reaction runs at half its maximum rate, which is also the defining property of the Michaelis constant. \[ \boxed{v = \frac{V_{max}}{2}} \]