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: Michaelis-Menten equation. The Michaelis-Menten equation is given by: \[ v = \frac{V_{\text{max}} \cdot [S]}{K_m + [S]} \] where: - \( v \) is the reaction rate, - \( V_{\text{max}} \) is the maximum rate, - \( [S] \) is the substrate concentration, - \( K_m \) is the Michaelis constant (substrate concentration at which the reaction rate is half of \( V_{\text{max}} \)).

 Step 2: Condition when \( K_m = C \). When \( [S] = K_m \): \[ v = \frac{V_{\text{max}} \cdot K_m}{K_m + K_m} = \frac{V_{\text{max}}}{2} \] This shows that the rate of the process is equal to half of the maximum rate (\( V_{\text{max}} \)) when the substrate concentration equals the Michaelis constant (\( K_m \)). 

Step 3: Comparison with other options. - Option \( (B) \): Zero-order kinetics occur when \( [S] \gg K_m \). 
- Option \( (C) \): Constant rate occurs in zero-order kinetics. 
- Option \( (D) \): First-order elimination occurs when \( [S] \ll K_m \). 

Conclusion: The correct answer is \( (A) \).

Answer 2

The Michaelis constant \( K_m \) has a built-in physical meaning: it is defined as the substrate concentration at which the reaction runs at exactly half its maximum possible rate. The question is asking what happens when the substrate concentration \( C \) equals \( K_m \), so let's check each option against that definition.

1. The rate of process is equal to half of maximum rate: This is the textbook definition of \( K_m \) itself, the substrate concentration at which the enzyme is working at exactly half of its top speed \( V_{max} \). Setting the substrate concentration equal to \( K_m \) is, by definition, the condition this statement describes.
2. Indicates zero-order process: A zero-order process happens when substrate concentration is far higher than \( K_m \), so the enzyme is fully saturated and the rate no longer depends on how much more substrate is added. Setting \( [S] = K_m \) is nowhere near this saturating condition.
3. The rate process occurs at a constant rate: A constant rate independent of substrate concentration is another way of describing zero-order behavior, which again only happens at substrate concentrations well above \( K_m \), not when they are equal.
4. Equation becomes identical to first-order elimination of drug: First-order behavior, where rate is directly proportional to substrate concentration, shows up when substrate concentration is much lower than \( K_m \), the opposite extreme from setting them equal.

Since \( K_m \) is defined as the substrate concentration giving half-maximal rate, the condition \( [S] = K_m \) directly matches the first option, while zero-order and first-order behavior only appear at the two opposite extremes, far above or far below \( K_m \).

So the correct answer is The rate of process is equal to half of maximum rate.