The enzyme-catalysed reaction shown below follows Michaelis- Menten kinetics.

​$$\mathrm{E} + \mathrm{S} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{ES} \overset{k_2}{\longrightarrow} \mathrm{E} + \mathrm{P}$$

$$k_1 = 1 \times 10^8\ \mathrm{M}^{-1}\mathrm{s}^{-1},\quad k_{-1} = 4 \times 10^4\ \mathrm{s}^{-1},\quad k_2 = 8 \times 10^2\ \mathrm{s}^{-1}$$

From the information given above, calculate $$K_m$$ and $$K_s$$.​

Correct option is D

To solve this question, we need to calculate Kₘ and Kₛ (Michaelis constant and the rate constant for the enzyme-substrate complex dissociation, respectively) using the given values of the rate constants.

Given :   $$k_1 = 1 \times 10^8 \, M^{-1}s^{-1} \, \text{(rate constant for binding)}$$

$$k_{-1} = 4 \times 10^4 \, s^{-1} \, \text{(rate constant for dissociation)}$$

$$k_2 = 8 \times 10^2 \, s^{-1} \, \text{(rate constant for conversion of ES to E + P)}$$

Step 1: Calculate Kₛ (rate constant for dissociation of enzyme-substrate complex):

The formula for K_s   is given by:

​$$K_s = \frac{k_{-1}}{k_1}$$

Substitute the given values:

$$K_s = \frac{4 \times 10^4 \, s^{-1}}{1 \times 10^8 \, M^{-1}s^{-1}} = 4 \times 10^{-4} \, M = 400 \, \mu M$$​​​

​So, Kₛ = 400 μM.

Step 2: Calculate Kₘ (Michaelis constant):

The Michaelis constant K_m is given by:

​$$K_m = \frac{k_{-1} + k_2}{k_1}$$​​

​Substitute the given values:

​$$K_m = \frac{4 \times 10^4 \, s^{-1} + 8 \times 10^2 \, s^{-1}}{1 \times 10^8 \, M^{-1}s^{-1}} = \frac{4.08 \times 10^4 \, s^{-1}}{1 \times 10^8 \, M^{-1}s^{-1}} = 4.08 \times 10^{-4} \, M = 408 \, \mu M$$

So, Kₘ = 408 μM.

Answer:

The correct combination of Kₛ and Kₘ is:

Kₛ = 400 μM, Kₘ = 408 μM.

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