Q.44 An enzyme of 40 kDa is added to a substrate solution in a molar ratio of 1:3. The
concentration of the enzyme in the mixture is 12 mg/ml. What would be the corresponding
substrate concentration?
(A) 0.4 mM (B) 0.12 mM (C) 0.9 mM (D) 0.3 mM
An enzyme of 40 kDa at 12 mg/ml in a 1:3 enzyme-to-substrate molar ratio results in a substrate concentration of 0.9 mM.
Correct Answer: (C) 0.9 mM
The molar ratio [E]:[S] = 1:3 means [S] = 3 × [E]. Enzyme concentration converts to 0.3 mM, so substrate is 0.9 mM.
Step-by-Step Calculation
Convert enzyme mass concentration to molarity, then apply ratio:
- Enzyme MW = 40 kDa = 40,000 g/mol
- [E] = 12 mg/ml = 12 g/L
- Molarity [E] = 12/40,000 mol/L = 0.0003 M = 0.3 mM
- Since [S]/[E] = 3, [S] = 3 × 0.3 mM = 0.9 mM
Options Comparison
| Option | Concentration (mM) | Why Correct/Incorrect |
|---|---|---|
| (A) | 0.4 | ❌ Wrong—ignores ratio or miscalculates as 1:1 adjusted (~0.4 mM total) |
| (B) | 0.12 | ❌ Wrong—treats 12 mg/ml as molar without MW conversion |
| (C) | 0.9 | ✅ Correct—proper MW to molarity and ×3 ratio |
| (D) | 0.3 | ❌ Wrong—equals [E], forgets to multiply by 3 |
Quick Formula Reference
Molarity (M) = Concentration (g/L)/Molecular Weight (g/mol)
Substrate [S] = Ratio × Enzyme [E]
Key Takeaway: Always convert mass concentration (mg/ml → g/L) to molarity using MW before applying molar ratios in enzyme kinetics problems.
