## Answers

For the above equation, given that k_{1} and k_{-1} are the forward and reverse rate constants respectively. Let us take k_{2} as the rate constant for the formation of the product P.

Now, the rate of the complex [ES] formation is = k_{1} [E][S] ............. (1)

and rate of the complex [ES] breakdown is = k_{-1} [ES] + k_{2} [P][E]

Again, considering that the amount of enzyme is too small as compared to that of the substrate S, it may be assumed that [P][E] [ES]

Therefore equation (3), i.e., the rate of complex breakdown is = (k-1 + k2) [ES]...... (2)

From steady state approximation, d[ES]/dt = 0

Therefore, rate of complex [ES] formation is equal to the rate of complex breakdown

i.e., Equating equation (1) and (2) we have

k_{1} [E][S] = (k_{-1} + k_{2}) [ES]

or, [E][S] = (k_{-1} + k_{2}) [ES] / k_{1}

or, [E][S] = K_{M} [ES]; where K_{M} = (k_{-1} + k_{2})/k_{1}

Therefore, K_{M} = [E][S] / [ES]....... (3) ; K_{M} is called Michaelis constant