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.
In this model, the substrate S reversibly associates with the enzyme E in a first step, and some of the resulting complex ES is allowed to break down and yield the product P and the free enzyme back. 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. The latter stands for the appearance of the product P in solution (+ d[P]/dt) whose phenomenological rate equation (first-order) is given by
v0 = kcat[ES] (2),
containing an experimentally measurable (dependent) variable - v0, a kinetic parameter - kcat, and another variable unknown to us - [ES].
Before proceeding, one should state (and remember) some implicite assumptions:
assumptions, which hold in most kinetic experiments performed in test tubes
at low enzyme concentration, are convenient when considering the mass
conservation equations for the reactants
We want to express v0 in terms of measurable (experimentally defined, independent) variables, like [S] and [E]total , so we can see how to test the mechanism by experiments in kinetics. So we must replace the unknown [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] (one second
order kinetic step) equals its rate of consumption (two first order
kinetic steps). 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] (3)
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, and the concentrations (variables) in (3):
(k-1 + kcat) [ES] = k1 [E][S],
(k-1 + kcat)/k1 = [E][S]/[ES]
To simplify (4), first group the kinetic constants by defining them as Km:
Km = (k-1 + kcat)/k1 (5)
and then express [E] in terms of [ES] and [E]total, to limit the number of unknowns:
[E] = [E]total - [ES] (6)
Substitute (5) and (6) into (4):
Km = ([E]total - [ES]) [S]/[ES] (7)
Solve (7) for [ES]:
First multiply both sides by [ES]
(no Black Magic involvement here...):
[ES] Km = [E]total[S]
Then collect terms containing [ES]
on the left:
[ES] Km + [ES][S] = [E]total[S]
Factor [ES] from the left-hand
[ES](Km + [S]) = [E]total[S]
and finally, divide both sides by
(Km + [S]):
[ES] = [E]total [S]/(Km + [S]) (8)
Substitute (8) into (2):
v0 = kcat[E]total [S]/(Km + [S]) (9)
The maximum velocity Vmax occurs when the enzyme is saturated -- that is, when all enzyme molecules are tied up with S, or [ES] = [E]total. Thus,
Vmax = kcat [E]total (10)
Substitute Vmax into (9) 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, Vmax and 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
This is the equation of a rectangular hyperbola, just like the saturation equation for the binding of oxygen to myoglobin.
Mathematically, the function v0 presents two asymptotes:
* one parallel to the [S] axis at v0
= Vmax , represents the velocity at infinite [S] (saturation),
* the second parallel to the v0 axis at [S] = - Km, has no physical meaning (no negative concentrations).
Further analysis reveals the physical meaning of Km: the concentration of substrate at which the velocity is half Vmax. Indeed, substituting Km for [S] in (11) yields
v0 = 1/2 Vmax.
Thus, a low value for Km may indicate a high affinity of the enzyme for its substrate.
Another physically meaningful limit of this function is found at vanishingly small values of [S] (--> 0),
where v0 --> Vmax
In this case, the velocity becomes proportional to the (low, relative to Km) substrate concentration, displaying pseudo-first order kinetics in [S].
The parameter Vmax /Km (or rather its constant part kcat /Km), often referred to as the catalytic ability of the enzyme, is a direct measure of the efficiency of the enzyme in transforming the substrate S.
kcat /Km recombines the two traditionally-separated aspects of enzyme catalysis:
* the effectiveness of
transformation of bound product (catalysis per se, kcat )
* the effectiveness of productive substrate binding (affinity, 1/Km = k1 /(k-1 + kcat))
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 at first class on enzyme kinetics: Derive equation (11) from model (1).
Last update: Dec 1999- Claude Aflalo
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