Question
Asked Nov 26, 2019
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For a Michaelis-Menten reaction, k1 = 5 x 107 M-1s-1, k-1 = 2 x 104s-1, and k2 = 4 x 102s-1. Calculate KS and KM for this reaction. Does substrate binding achieve equilibrium or the steady state?

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Expert Answer

Step 1

The enzyme substrate binding reaction is shown in the equation (1). In this equation, Enzyme (E) bonds with substrate (S) in the first step to form ES complex and in the second step, the ES complex breakdown to form E and P.

ES EP
E+S
(1)
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ES EP E+S (1)

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Step 2

According to the above equation, the rate of formation of ES is k1[E][S] and the rate of breakdown of ES is (k-1 + k2)[ES].

Consider the steady state approximation in the first case, when the rates of formation of ES is equal to the rate of breakdown of ES. In this case, the factor [E][S]/[ES] is equivalent to the Michaelis constant (KM) in the expression (2).

k, ES=(k. +k,)ES]
ES(k+k)
ES]
(k_k)
.(2)
К,
м
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k, ES=(k. +k,)ES] ES(k+k) ES] (k_k) .(2) К, м

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Step 3

Consider the state of equilibrium in which the value of k-1 is mech greater than the value of k2 due to which (k-1 + k2) is approximately equal to k-1. In this case, ...

ESk
ES]
(3)
Kg
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ESk ES] (3) Kg

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