Question

The maximum velocity of an enzymatic reaction is 0.4 mole/sec. At 5 mM concentration of the substrate, the reaction velocity was found to be 0.2 mole/sec. If the enzyme shows standard Michaelis-Menten kinetics, the rate of the reaction at 10 mM substrate concentration in mole/sec is ............ (decimal digits up to 3 places)

Hint:

For Michaelis-Menten kinetics, use the equation \( v = \frac{V_{\text{max}} [S]}{K_m + [S]} \) to calculate reaction velocities at different substrate concentrations.

Answer 1

Correct Answer: 0.261

Step 1: Use the Michaelis-Menten equation.
The Michaelis-Menten equation for enzyme kinetics is: \[ v = \frac{V_{\text{max}} [S]}{K_m + [S]}, \] where \( v \) is the reaction velocity, \( V_{\text{max}} \) is the maximum velocity, \( [S] \) is the substrate concentration, and \( K_m \) is the Michaelis constant.

Step 2: Solve for \( K_m \).
At a substrate concentration of 5 mM, the reaction velocity is 0.2 mole/sec. We can substitute these values into the equation: \[ 0.2 = \frac{0.4 \times 5}{K_m + 5}. \] Solving for \( K_m \): \[ 0.2(K_m + 5) = 2 $\Rightarrow$ 0.2K_m + 1 = 2 $\Rightarrow$ 0.2K_m = 1 $\Rightarrow$ K_m = 5. \]

Step 3: Use the value of \( K_m \) to calculate the rate at 10 mM substrate concentration.
Now, substitute \( K_m = 5 \) and \( [S] = 10 \) mM into the Michaelis-Menten equation: \[ v = \frac{0.4 \times 10}{5 + 10} = \frac{4}{15} = 0.2667 \, \text{mole/sec}. \]

Step 4: Conclusion.
The reaction rate at 10 mM substrate concentration is 0.267 mole/sec.