Question

An enzymatic reaction exhibits Michaelis–Menten kinetics. What will happen if the concentration of enzyme is doubled keeping \( [S_0] \gg [E] \)?

• A. Both \( K_m \) and \( V_{max} \) will remain same
• B. Both \( K_m \) and \( V_{max} \) will increase
• C. \( V_{max} \) will increase; \( K_m \) will remain same
• D. \( K_m \) will increase; \( V_{max} \) will remain same

Hint:

In Michaelis–Menten kinetics, increasing enzyme concentration increases \( V_{max} \), but \( K_m \) remains constant as it is an intrinsic property of the enzyme-substrate complex.

Answer 1

The Correct Option is C

Step 1: Recall Michaelis–Menten equation:
\[ V = \frac{V_{max}[S]}{K_m + [S]} \] Here, \( V_{max} = k_{cat}[E] \), where \( [E] \) is the enzyme concentration and \( k_{cat} \) is the turnover number.
Step 2: Analyze what happens when enzyme concentration \([E]\) is doubled:
If enzyme concentration is doubled, \( V_{max} \) also doubles because it is directly proportional to \([E]\). However, \( K_m \) is a constant that depends on the enzyme-substrate interaction and is not affected by enzyme concentration.
Step 3: Use condition \( [S_0] \gg [E] \):
This ensures we are in the range where Michaelis–Menten assumptions apply and substrate saturation is sufficient for the reaction rate to approach \( V_{max} \).