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Steady state. Michaelis and Menten assumed that the overall reaction for an enzy

ID: 490572 • Letter: S

Question

Steady state. Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as: Using this reaction, what expression can be used to describe the rate of breakdown of the enzyme-substrate complex? What does the "steady state assumption", as applied to enzyme kinetics, imply? Write the expression of the steady state assumption. When k_2 is very small compared to k_1 and k_-1, what process is rate determining in product formation? Why is k_2 is very small compared to k_1 and k_-1?

Explanation / Answer

ES is formed from reaction-1, decompose back to E and S throuh reaction-2 and also decomposed from reaction-3

Hence rate of ES = d[ES)/dt= K1[E][S]-K-1[ES]- K2[ES]

the net rate of formation of intermediate =0 is known as steady state assumption

when K2 is small, d(ES/dt)= K1[E][S]-K1-[ES] =0

or [ES] = (K1/K-1) [E][S]

when K2 is small, there is equilibrium between formation of ES from E and S and decompositino of ES back to E and S

since ES------>E+P is the rate limiting step, K2 is small.