We can show that the mechanism above
leads to the given initial
velocity versus substrate concentration equation. One relatively simple
way to do this is to make the assumption that step (1), formation of
the ES complex, is significantly faster than step (2), production
of products from ES, *i.e*. the mechanism can be represented as
(1) E + S = ES (fast)
(2) ES → E + P (slow)
This
is not true for all enzymes, but it is true enough for some, and
those cases where this assumption does not hold can be dealt with
using an alternate assumption that leads ultimately to the same result.
Now if step (2) is the much slower step, it is the rate-limiting
step, and its rate determines the overall reaction rate. Writing
the rate law for this elementary step, we have
*V*_{0} = *k*_{2} [ES]
where *k*_{2} is the first-order rate constant for
step (2) in the forward direction. Since ES is an intermediate
and the rate law for a reaction is expressed in terms of a rate constant
and *reactant* concentrations,
we must find an expression for [ES] in terms of [S] (strictly
speaking, we mean [S]_{0}, the initial substrate concentration).
In this case, we will also make use of the total enzyme concentration,
[E]_{T},
since this is a quantity that we can control in the experimental
setting. We can see that at high substrate concentrations, essentially
all enzyme at any instant has substrate bound, and therefore
[ES] = [E]_{T},
and the enzyme is being driven as fast as it can possibly go,
given its limited amount. Under such conditions, we say that
the enzyme is *saturated* with substrate, and we have
*V*_{max} = *k*_{2} [E]_{T}
The parameter *V*_{max} is thus set by the catalytic
rate constant and total enzyme concentration. To get [ES] at less
than saturating substrate concentrations, we use both the relation
[ES] =* *[E]_{T} – [E], where [E] is the concentration
of free enzyme, and the relation we get from assuming that step
(1) is at equilibrium:
*k*_{1}/* k*_{–1} = [ES]
/ [E][S]
where *k*_{1} and *k*_{–1} are
the forward and reverse rate constants, respectively, for step (1).
Solving the latter for [E] and substituting into the above expression
for [ES], we obtain
[ES] = * *[E]_{T} – [E] = [E]_{T} – (*k*_{–}_{1}/* k*_{1})(
[ES] / [S] )
Solving this for [ES], we have
[ES] = * *[E]_{T} – [E] = [E]_{T} /{(*k*_{–}_{1}/*k*_{1})(1
/ [S]) + 1} = [E]_{T} [S] /{(*k*_{–}_{1}/*k*_{1})
+ [S]}
If we now define *K*_{M} = *k*_{–}_{1}/*k*_{1}
(*i.e*. *K*_{M} is the equilibrium constant
for the *reverse* of step
1, ES → E + S) and combine the expression for [ES] we just
obtained with the rate equation *V*_{0} = *k*_{2} [ES],
we have
*V*_{0} = *k*_{2} [ES]
= *k*_{2}[E]_{T} [S] /{*K*_{M} +
[S]} = *V*_{max} [S] /{*K*_{M} +
[S]}
where we have substituted *V*_{max} = *k*_{2} [E]_{T} to
obtain the last equality, which gives us the equation we were after.
This shows that the simple mechanism we proposed above as a model
for at least some enzyme-catalyzed reactions does indeed give us
an initial rate vs. substrate concentration curve consistent with
what is typically observed in enzyme kinetics experiments. |