Deriving the Michaelis-Menten Equation Gale Rhodes
Chemistry Department University of Southern Maine rhodes@usm.maine.edu
Memorize this derivation as soon as your
encounter it in your text, and you will be able to read the remainder of the chapter with
far greater understanding. For other suggestions on how to make your study of biochemistry
easier, see Learning Strategies.
A simple model of enzyme action:
We would like to know how to recognize an enzyme that behaves according to this model. One
way is to look at the enzyme's kinetic behavior -- at how substrate concentration affects
its rate. So we want to know what rate law such an enzyme would obey. If a newly
discovered enzyme obeys that rate law, then we can assume that it acts according to this
model. Let's derive a rate law from this model.
For this model, let V0 be the initial velocity of the
reaction. Then V0 = kcat [ES]. (2)
The maximum velocity Vmaxoccurs when the enzyme
is saturated -- that is, when
all enzyme molecules are tied up with S, or
[ES] = [E]total .
So Vmax = kcat [E]total. (3)
We want to express V0 in terms of measurable
quantities, like [S] and [E]total , so we can
see how to test the mechanism by experiments in kinetics. So we must replace [ES]
in (2) with measurables.
During the initial phase of the reaction, as long as the reaction velocity
remains constant, the reaction is in a steady state, with ES being
formed and consumed at the same rate. During this phase, the rate of formation of
[ES] equals its rate of consumption. According to model (1), Rate of formation of
[ES] = k1[E][S].
Rate of consumption of [ES] = k-1[ES] + kcat [ES].
So in the steady state, k-1[ES] + kcat [ES] = k1[E][S].
(4)
Remember that we are trying to solve for [ES] in terms of measurables, so
that we can replace it in (2). First, collect the kinetic constants in (4): (k-1+kcat)[ES]=k1[E][S],
and (k-1 + kcat)/k1 = [E][S]/[ES]. (5)
To simplify (5), first group the kinetic constants by
defining them as Km : Km=(k-1+kcat)/k1(6)
and then express [E] in terms of [ES] and [E]total:
[E]=[E]total-[ES] (7) Substitute (6) and (7) into (5): Km=([E]total-[ES])
[S]/[ES] (8)
Solve (8) for [ES]: First multiply both sides by [ES]: [ES]
Km=[E]total[S]-[ES][S]Then collect
terms containing [ES] on the right: [ES]Km+[ES][S]=[E]total[S]
Factor [ES] from the left-hand terms: [ES](Km+[S])=[E]total[S]
and finally, divide both sides by (Km + [S]): [ES]=[E]total[S]/(Km+[S])
(9)
Substitute (9) into (2): V0 = kcat [E]total
[S]/(Km + [S]) (10)
Recalling (3), substitute Vmax into (10) for kcat
[E]total: V0=Vmax [S]/(Km+[S])
(11)
This equation expresses the initial rate of reaction in terms of a measurable
quantity, the initial substrate concentration. The two kinetic parameters, Vmaxand Km , will be different for every enzyme-substrate
pair.
Equation (11), the Michaelis-Menten equation, describes the kinetic behavior
of an enzyme that acts according to the simple model (1). Equation (11) is of the form
y = ax/(b + x) (does this look familiar?)
This is the equation of a rectangular hyperbola, just like the saturation
equation for the binding of dioxygen to myoglobin.
Equation (11) means that, for an enzyme acting according to the simple model
(1), a plot of V0 versus [S] will be a rectangular hyperbola. When enzymes
exhibit this kinetic behavior, unless we find other evidence to the contrary, we assume
that they act according to model (1), and call them Michaelis-Menten enzymes.
QUIZ for USM Students
Quiz at first class on enzyme kinetics: Derive equation (11) from model
(1).
Ten minutes.
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