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Chapter 6

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 Henri-Michaelis-Menten equation:

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 normalized.
  • 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 diagram.

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 species.

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 catalyzes.

To honor the enzymologist who first identified this requirement, this is often referred to as the Haldane relationship.

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.

  1. 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.
  2. 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 formed.
  3. 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 [P].

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 inhibition.

Exercise: Formulate a rate law for Uncompetitive inhibition.

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 complex.

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 by Pt/([P]+[SP]+[ASP]):

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.

Chapter 7