Chapter 9: Reaction Mechanisms, Pathways, Bioreactions and Bioreactors

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Example R9.6-2: Derive an Initial Rate Law for Alcohol Dehydrogenates

Before considering the possibility of going directly from the apoenzyme to the holoenzyme, assume that the rate of dissociation of the complex (ADHNAD+),

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is irreversible, and show that the initial rate law for ethanol in the enzyme cofactor reaction sequence discussed earlier is of the form

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(RE7.4-2.1)
 



     
Solution
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Let E = ADH, S1 = CH3CHO, S2 = NADH, = NAD+, and P1 = CH3CH2OH. Then

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By adding the rate law for the rate of formation of ethanol (P 1),

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to the equation forIMAGE 07eq81.gif(below), we see that the rate law for ethanol can be written as

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(RE7.4-2.2)

Application of the PSSH to the holoenzyme (EmiddotS2) and the apoenzyme
(Emiddot)
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allows one to solve for the concentrations of the cofactor-enzyme complexes in terms of S2, S2*, and E.
   

image 07eq84.gif

(RE7.4-2.3)



(RE7.4-2.4)

The total concentration of bound and unbound enzyme is

image 07eq85.gif

(RE7.4-2.5)

Substituting Equations (RE7.4-2.3) and (RE7.4-2.4) into Equation (RE7.4-2.5), the unbound enzyme concentration is
       
   

(RE7.4-2.6)
       

Setting (P1) = 0, we obtain the initial rate law by combining Equations (RE7.4-2.2), (RE7.4-2.4), and (RE7.4-2.6):

IMAGE 07eq87.gif

(RE7.4-2.7)

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