A propionate kinase enzyme utilizes two substrates, propionate and acetate, with Kₘ for propionate being half that of acetate. Which one of the following options about the rate of the reaction at very low substrate concentrations is correct?

Name: Adda247- Exam Preparation App
Rating: 4.8 (1546868 reviews)

Correct option is B

Information Booster

Michaelis constant ( $K_m$ ) represents the substrate concentration at which an enzyme operates at half its maximum velocity ( $V_{max}$ ).
A lower $K_m$ indicates higher affinity of the enzyme for the substrate.
At very low substrate concentrations, the reaction rate is inversely proportional to $K_m$ .
Substrates with lower $K_m$ are utilized faster at low concentrations.
Competitive inhibition affects substrate utilization by increasing the effective $K_m$ .
If $K_m$ of propionate is half that of acetate, its reaction rate will be doubled at very low concentrations.

Option 1 (Half the rate of acetate) – Incorrect:
- Since $K_m$ for propionate is lower, it is utilized faster, not slower.
Option 3 (Equal rate to acetate) – Incorrect:
- The reaction rate is different because the $K_m$ values are different.
Option 4 (Four times the rate of acetate) – Incorrect:
- The correct factor is 2× not 4×.