Writing Rate Laws for Enzymatically Catalyzed Processes
The essential principles for writing enzyme rate laws can be
seen in the derivation for a simple 1-reactant, 1-product
enzyme-catalyzed process where S is the substrate, and P is the
product. Binding is immediately seen as essential because an
enzyme must bind to its substrate in order to catalyze its
conversion to product. The enzyme-substrate complex is designated
C; it is formed by second order reactions from the free enzyme,
E, and S as well as from E and P. The complex may break down, via
first order kinetics, into E plus S or E plus P.
This is the enzymatically catalyzed reaction that was first
analyzed by Henri and by Michaelis and Menten. Based on the
principles of mass action and conservation of mass, we first
write the differential equation for ES:
The "pseudo steady state" assumption is invoked
next. It asserts that very shortly after the reaction is
initiated, dES/dt becomes zero and remains zero. This has turned
out to be experimentally verifiable in many cases. So even though
S is decreasing and P is increasing, ES is found to be constant.
This means only that the conversion of S to ES is balanced by the
conversion of ES to P.
A further simplification, somewhat less defensible, is
introduced next. Michaelis and Menten argued that the term
involving k-2 is negligible. There are several common
rationalizations for this assumption. First, there may be
negligible accumulation of product during a short experiment, so
[P] may be approximately zero. Second, it is sometimes argued
that k-2 itself is negligibly small. In living cells,
these simplifications are rarely justified, but they are
essential to the Michaelis-Menten derivation. In later sections
we will explore rate laws that do not require this assumption.
If the "pseudo steady state" assumption and the
"short experiment" assumption are made, they lead to
the following algebraic equation
Next, Km is defined as
and conservation of mass is used to write [E] as a function of
the total amount of enzyme present, Etot, and ES.
Namely, [Etot]-[ES]=[E]. This is substituted for [E]
in the equation defining Km
Solving this for [ES] yields
and since dP/dt = k2[ES] (still assuming that the
process governed by k-2 is negligible)
Finally, recognizing that k2[Etot] is
the maximum rate at which P can be formed, we define Vmax=
k2[Etot] and arrive at the famous
The derivation is thus complete, but what about significance?
First, notice that [S]/(Km+[S]) is a unitless fraction
whose value always lies between zero and one. It defines the
fraction of maximal velocity that is attained at a given [S]. The
Michaelis-Menten equation is the simplest algebraic form that
displays saturation. Second, the units of Vmax define
the units of dP/dt. Third, notice that this one simple equation
behaves as a first order process or as a zeroth order process
depending on the value of [S]. When [S] is small (say, less than
10%) compared to Km, the process appears linear with a
rate constant equal to Vmax/Km. When [S] is
large (say, 10 times) compared to Km, the process
appears zeroth order, producing P at a constant rate equal to Vmax.
Finally, you should compare this equation to the equation for
binding developed earlier. The algebraic form is the same, but
there are important differences:
- The Michaelis-Menten equation is a differential equation
while the binding isotherm is algebraic.
- Vmax has units of flux (mass/time) usually
normalized in some way. Btot has units of mass, usually
- Km and KD both have units of
concentration, but only KD represents the
equilibrium constant of a binding reaction. Km
should not be confused with the equilibrium constant
enzyme-substrate binding reaction; it is different
because of the presence of k2 in the
expression for Km.
The methods of steady state enzyme kinetics allow experimental
determination of kcat and Km, where kcat
is that factor in the maximal velocity that is independent of the
concentration of enzyme. In other words, Vmax = kcat[E]tot;
it is the rate at which product can be produced if all of the
available enzyme is in the form of ES complex. The value of kcat
(which is equal to k2 in the above example) is thus a
quantitative measure of the efficiency of the enzyme; it answers
the question: How many molecules of product can be produced in a
given time by one molecule of substrate-saturated enzyme?
Writing Rate Laws for Reversible Reactions
The study of enzyme kinetics advanced, at first, through the
study of initial rates of reaction that are well-characterized by
the Michaelis-Menten rate law because the assumption that [P] is
zero was satisfied. In living systems, however, the assumption
that the concentration of product is zero is frequently violated.
In these situations, the correct rate law must account for both
forward and reverse reactions.
We will make the rapid equilibrium assumption, namely that the
binding of S or P to E is rapid compared to the catalytic rates
converting ES to EP or EP to ES. Under this assumption, any rate
law may be written down by inspection of the symbol and arrow
First, write down the velocity equation as:
Now, divide both sides of the equation by the total enzyme
concentration, Et, but on the right hand side, express
the total concentration as the sum of all enzyme containing
Next, express each E-containing species in terms of [E], by
using the corresponding equilibrium expressions. The
concentration of each E-containing species can be written as the
product of the individual concentrations of its constituents
divided by the product of the equilibrium dissociation constants
between that E-containing species and free [E].
In the present case
Then cancelling [E], and recognizing that k+Et
is equal to the maximal forward velocity and k-Et
is equal to the maximal velocity of the reverse reaction, yields
Notice that at equilibrium, since there are no chemical
potential gradients, the net production of P must be zero.
Consequently, the numerator of this rate expression must be zero.
This leads to a famous relationship among the enzyme's kinetic
constants and the equilibrium constant of the reaction it
To honor the enzymologist who first identified this
requirement, this is often referred to as the Haldane
Exercise: It is often assumed that when the catalyzed
reaction has an equilibrium constant that strongly favors the
formation of P (that is, Keq is large),
the maximal velocity of the reverse reaction can be neglected
compared to the maximal velocity of the forward reaction. What
else must be true for this assumption to be valid?
The rapid equilibrium assumption can also be used to construct
useful rate laws for processes that are modulated by inhibitors
or activators. These are explored in the next two sections.
Writing Rate Laws That Include Inhibition
To inhibit a process that is mediated by a protein, P, the
inhibitor molecule must bind to the protein. Three main classes
of inhibition may be defined by analogy with enzyme kinetics.
- Competitive inhibition: occurs when PI complexes form,
and prevent the binding of the substrate molecule, S. In
other words, I and S bind to the same site on P.
- Noncompetitive inhibition: occurs when I binds to a
separate site on P and may do so whether or not S is
bound. In other words, both PI and SPI complexes are
- Uncompetitive inhibition: occurs when I can bind only to
the SP complex, so only SPI complexes are formed.
We derive here the rate law for the common case of competitive
inhibition. You should be able to apply this method to write rate
laws for the other cases.
Always start with the symbol and arrow diagram:
Write the velocity equation:
Multiply the right hand side by Pt/([P]+[SP]+[RP]+[IP]).
As before, this is multiplication by a quantity equal to one, a
time-honored mathematical trick.
Express each bracketed term in terms of [P].
where the maximal velocities are defined as before and the Kk
are the dissociation constants of the Pk complexes. That is, KS
is the dissociation constant for the SP complex. Finally, cancel
Use the classical engineering approach to explore this rate
law. Consider what happens to the reaction rate if [I] becomes
much greater than KI, then leaving [I] at its elevated value,
what happens if [S] is increased to an enormous multiple of KS?
Think about another situation in which [S] is small, but [R] is
initially large. Is the inhibitor capable of inhibiting the
reverse reaction? Go back to the symbol and arrow diagram for
this system and explain why or why not. Suggest an improvement to
the symbol and arrow diagram that would make it easier to explain
the action of the inhibitor on the reverse process.
Exercise: Formulate a rate law for Noncompetitive
Exercise: Formulate a rate law for Uncompetitive
Writing Rate Laws for Processes With Activators
An activator provides an alternative pathway from reactants to
products. Two common cases are 1) an activator that binds to the
substrate creating a "better" substrate, and 2) an
activator that binds to a protein-substrate complex creating a
complex that yields product more readily than the unmodified
A symbol and arrow diagram for the first case is shown here.
To simplify the algebra, we consider the case for which chemical
equilibrium strongly favors formation of the product, R. The
Haldane relation then shows that Vmax- can be
neglected compared to Vmax+. S is the substrate, P is
the protein mediating the process, and A is the activator.
There are two terms, both positive, in the velocity equation:
where b is the factor by which the activator increases the
catalytic efficiency of the process. Multiply the right hand side
Express bracketed terms as functions of [P] and cancel [P]:
This can be written in terms of free activator concentration
by substituting for [AS] using the equilibrium constant for the
binding of A to S:
It is worth remembering that both inhibitors and activators
will add terms to the denominator of the rapid-equilibrium rate
law since both contribute new species of P, but only activators
will add terms to the numerator because they provide new pathways
for formation of R.
Exercise: Write a rapid-equilibrium rate law for the
process whose symbol and arrow diagram is shown below.
Uppercase Ki are dissociation
constants, kr is a rate constant, and f
is the factor by which the dissociation constant for B is altered
when A is also bound. Consider whether it is possible for f to be
different when A dissociates first. I is an uncompetitive
inhibitor of the process.