Question

For a Michaelis-Menten reaction, k1=5*107 M-1S-1, k-1=2*104 S-1 and k2=4*102

S-1. Calculate the Km and Ks for this reaction. Does the substrate achieve equilibrium? How would inspection of the rate constant immediately have given you the same answer

Answer 1

Michaelis-Menten reaction:

with k₁, and k_-1 are forward, and backward rate constants for the first reaction where as k₂ is the forward rate constant for the second step of the enzyme catalytic reaction.

The rate of the above enzymatic reaction is given by reaction rate v related to [S], the concentration of a substrate S as follows:

where V_max represents the maximum rate achieved by the system, at maximum (saturating) substrate concentrations and Michaelis constant K_M is the substrate concentration at which the reaction rate is half of V_max.

Now, from the question:

For a Michaelis-Menten reaction, k₁=5*10⁷ M^-1s^-1, k_-1=2*10⁴ s^-1 and k₂=4*10² s^-1, the Michaelis constant K_M is given by the formula (refer for the derivation: https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics):

= (2*10⁴ s^-1 * 4*10² s^-1) / (5*10⁷ M^-1s^-1) = 0.16

In the Michaelis–Menten reaction, the subtrate is at equilibrium if it reaches equilibrium on a much faster time-scale than the product is formed or if the following holds true:

From question: = 4*10² / 2*10⁴ = 0.02 <<1. Thus, the substrate achieved equilibrium.

The inspection of the rate constant in the above calculation has given the answer.