A mechanism for enzyme-catalyzed reactions that leads to the typical rate equation for these reactions can be described.
A variety of rate equations are required to portray the rates of enzymes catalyzed reagents and physical conditions that are encountered. The rate equation of however, is a guide to many of these variations, and the mechanisms of this section, often called the Michaelis-Menten mechanism, is likewise a base for other variations.
The mechanism that accounts for the rate equation is similar to those dealt in with.
With S representing substrate, E the enzyme, and E. S and enzyme substrate complex, the mechanism is presumed to be adequately represented by
E + S E. S
E. S E + products
The steady state assumption, which, however, is not always clearly applicable in these reactions, leads to
k_{1}[E][S] = k^{-1}[E. S] + k_{2}[E. S]
And [E. S] = k_{1}/k^{-1} + k_{2} [E][S]
To bring these expressions to a form that can be compared with the empirical rate equation, we must recognize that only [E_{tot}] = [E] + [E. S], and not [E], is generally known. Often, in fact, only a quantity proportional to [Etot], and not even values of [E_{tot}], is available.
Replacement of [E] in equation by [E] = [E_{tot}] - [E. S] leads to
[E. S] = k_{1}[E_{tot}][S]/(k^{-1} + k_{2}) + k_{1}[S]
Now this expression for the intermediate E. S can be inserted into the expression for the rate of the net reaction. This rate can be based on the formation of products in the second mechanism step. We have
-d[S]/dt = R = k_{2}[E. S] = k_{1}k_{2}[E_{tot}][S]/k^{-1} + k_{2} + k_{1}[S]
= k_{2}[E_{tot}][S]/k^{-1} + k_{2})/k_{1} + [S]
It is customary for the term (k^{-1} + k_{2})/k_{1} to be obtained by the new symbol K_{M}, that is,
K_{M} = k^{-1} + k_{2}/k_{1} to give the rate equation result of this mechanism as
R = k_{2}[E_{tot}][S]/K_{M} + [S]
We have come at this stage to the form of the empirical rate equation obtained, we are now in a position to intercept the values of the parameters KM and k_{2}[E_{tot}] in terms of their roles in the roles in the steps of the mechanism.
Reference to equation shows that, as the reaction is proceeding
[E][S]/[E. S] = K_{M}
Thus K_{M} is related to species concentrations, as is the dissociation equilibrium constant for the species [E. S]. the value of K_{M}, however, is given by (k...1 + k_{2})/k_{1}, and this equal to the value of the dissociation constant for [E. S] only to the extent that k_{2} is small and can be neglected compared with k...1. Thus when the breakup of the E. S complex to form original E and S species dominates the process whereby the complex forms products, the value of KM approaches the dissociation constant for the E. S complex.
What, now, is the significance of the term k_{2}[E_{tot}]? One first notes that the rate of the overall reaction is
R = k_{2}[E. S]
It follows that k_{2}[E_{tot}] is the rate that the reaction would have if all the enzyme were in the form of the enzyme-substrate complex. Thus k_{2}[E_{tot}] is the maximum rate for a given value of [E_{tot}]. The turnover rate of an enzyme in a particular enzyme-catalyzed reaction is the rate per mole of enzyme, i.e. the turnover rate is equal to the value of k_{2}, and this can be calculated fromk_{2}[Etot] if the total enzyme concentration is known.