Question

If two enzyme E1 and E2 have Km value for a substrate 'A' as 0.1 mM and 0.01 mM respectively, and for another substrate 'B' as 0.5 mM and 0.05 mM respectively, then

• A. E1 is more specific for substrate 'A' and 'B'
• B. E2 is more specific for substrate 'A' and 'B'
• C. E1 is mor specific for substrate 'A' and E2 is more specific for substrate 'B'
• D. E2 is more specific for substrate 'A' and E1 is more specific for substrate 'B'

Hint:

Enzyme Kinetics: Lower $K_m \implies$ Stronger substrate binding affinity and higher catalytic specificity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Michaelis constant ($K_m$) is inversely related to an enzyme's apparent substrate binding affinity (lower $K_m$ means higher affinity and specificity).
Key Formula or Approach:
\[ \text{Substrate Affinity} \propto \frac{1}{K_m} \]

Step 2: Detailed Explanation:
Comparing $K_m$ values for both enzymes:
1. For Substrate 'A':
- $K_m(\text{E1}) = 0.1\text{ mM}$
- $K_m(\text{E2}) = 0.01\text{ mM}$
Since $K_m(\text{E2}) < K_m(\text{E1})$, E2 requires a 10-fold lower substrate concentration to reach half-maximal velocity ($V_{\max}/2$), meaning E2 has higher affinity/specificity for A.
2. For Substrate 'B':
- $K_m(\text{E1}) = 0.5\text{ mM}$
- $K_m(\text{E2}) = 0.05\text{ mM}$
Since $K_m(\text{E2}) < K_m(\text{E1})$, E2 has higher affinity/specificity for B.
Therefore, E2 is more specific for both substrates 'A' and 'B'.

Step 3: Final Answer:
Therefore, E2 is more specific for substrate 'A' and 'B', matching option (B).