Question

Two different enzyme catalysis reactions I and II have identical Y-intercepts for the Lineweaver–Burke plots. The slope for reaction I is twice that of reaction II. If the initial concentrations of enzymes in I and II are same, the correct statement(s) is(are)
\[ \frac{1}{v}=\frac{1}{v_{\max}}+\frac{K_M}{v_{\max}}\frac{1}{[S]} \]

• A. Reactions I and II have same turn over number
• B. Michaelis–Menten constants for reactions I and II are identical
• C. Michaelis–Menten constant for reaction I is twice than that of reaction II
• D. The rates of the elementary steps for reactions I and II are identical

Hint:

In a Lineweaver–Burke plot: intercept \(=1/v_{\max}\), slope \(=K_M/v_{\max}\).
If intercepts are equal, differences in slope reflect differences in \(K_M\) only.

Answer 1

The Correct Option is A, C

Step 1: Use the Lineweaver–Burke form. The intercept equals \(1/v_{\max}\). Identical Y-intercepts \(\Rightarrow v_{\max}^{(I)}=v_{\max}^{(II)}\). With equal enzyme concentrations \([\mathrm{E}]_0\), the turnover number \(k_{\text{cat}}=v_{\max}/[\mathrm{E}]_0\) is the same for I and II \(\Rightarrow\) (A) true.
Step 2: Compare the slopes. Slope \(=K_M/v_{\max}\).
Since \(v_{\max}\) is the same and slope(I) \(=2\times\) slope(II), we get \(K_M^{(I)}=2K_M^{(II)}\) \(\Rightarrow\) (C) true; (B) false.
Step 3: Elementary steps. Equality of \(v_{\max}\) and the slope relation gives no information about the detailed microscopic steps; therefore (D) cannot be concluded (false).