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Chapter 7

# Chapter 7 - C Derivation of the M M equation Assumption...

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C. Derivation of the M & M equation: Assumption 1 (assumption of equilibrium): Early in the reaction, little P has accumulated so k -2 can be ignored. Rate = P/time = ( o ) = k 2 [ES] ES must be formed in order to there to be product, therefore this governs the overall rate of the reaction k 1 represents enzyme substrate binding, while k 2 is the chemistry step which is the slower, rate determining step

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Eqtn. 1 Formation of ES: rate = k 1 [E] [S] Eqtn. 2 Breakdown of ES: rate = k -1 [ES] rate = k 2 [ES] This can breakdown two ways. Steady State Assumption: Once reaction gets started, the [ES] remains constant. As a result, the formation of ES must equal the Breakdown of ES: Eqtn. 3 Eqtn. 4 Now rewrite this expression in terms of ES: Eqtn. 5 Define K m = 3 k
Substituting K m into Eqtn. 5 gives: Eqtn. 6 Now we have an expression for [ES], but it is difficult to know [E] free enzyme at any given moment in the reaction. Hence, we need to rewrite the expression in terms of Total Enzyme [E o ] = [ES] + [E free ] Solving for [E] = [E o ] - [ES] If we plug this into Eqtn. 6 then: Eqtn. 7 Through a series of algebraic rearrangements (make sure you go through these steps on your own), we can solve for [ES] Eqtn. 8

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Let’s take a moment to remember where we started out. We said that for: the initial velocity ( o ) = k 2 [ES] Plugging [ES] from Eqtn. 8 above into ( o ) = k 2 [ES]: Eqtn. 9 *****Under saturating conditions of S (i.e. [S] = very high) then [E o ] = [ES]. Recall that Vmax is achieved when all E has S bound. Therefore, the Michaelis-Menten equation is: v = initial velocity at any substrate concentration, you can know the velocity this is where the idea of instantaneous velocity comes in so we can treat the [S] as a constant Eqtn. 10 SO WHO CARES?
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