Question

An enzyme, which follows Michaelis-Menten equation, catalyzes the reaction A\(\rightarrow\)B. When enzyme and substrate concentrations are 15 nM and 10 \(\mu\)M, respectively, the reaction velocity is 5 \(\mu\)M s\(^{-1}\). If \(K_m\) for the substrate A is 5 \(\mu\)M, the kinetic efficiency of the enzyme will be ______ \(\times 10^6\) M\(^{-1}\) s\(^{-1}\) (in integer).

Hint:

Kinetic efficiency is calculated using the ratio of the maximum velocity and the Michaelis constant. In the case of enzyme-substrate interaction, it gives the measure of enzyme efficiency.

Answer 1

To calculate the kinetic efficiency of the enzyme, we use the Michaelis-Menten equation for enzyme kinetics:
\[ v = \frac{V_{max} [S]}{K_m + [S]} \] Where:
- \( v \) is the reaction velocity,
- \( V_{max} \) is the maximum velocity,
- \( [S] \) is the substrate concentration,
- \( K_m \) is the Michaelis constant.
Given:
- The enzyme concentration is 15 nM,
- The substrate concentration is 10 \(\mu\)M = \(10 \times 10^{-6}\) M,
- The reaction velocity \(v = 5 \, \mu\)M/s = \(5 \times 10^{-6}\) M/s,
- \( K_m = 5 \, \mu\)M = \(5 \times 10^{-6}\) M.
The kinetic efficiency can be calculated as:
\[ {Kinetic efficiency} = \frac{k_{cat}}{K_m} \] Where \( k_{cat} \) is the catalytic rate constant, which can be related to the maximum velocity and enzyme concentration.
From the given reaction velocity, we know that the enzyme is working at a concentration of 15 nM. By plugging in the values into the Michaelis-Menten equation, we get:
\[ 5 \times 10^{-6} = \frac{V_{max} \times 10 \times 10^{-6}}{5 \times 10^{-6} + 10 \times 10^{-6}} \quad \Rightarrow \quad V_{max} = 10 \, \mu M/s \] Then, we calculate the kinetic efficiency:
\[ {Kinetic efficiency} = \frac{10 \, \mu M/s}{5 \times 10^{-6} M} = 100 \times 10^6 M^{-1} s^{-1} \] Thus, the kinetic efficiency of the enzyme is \( \mathbf{100 \times 10^6} \, M^{-1} s^{-1} \).