3.14 a Mathematical Model for Enzymatic Reactions

§3.14 A Mathematical Model for Enzymatic Reactions

The goal of this subsection is threefold. First, we will learn how to model biochemical reactions; second, we will see how mathematical models can be used to understand empirical observation; third, we will introduce an important method, namely by making

Figure 14.1 A schematic description of an enzymatic reaction.

Appropriate assumptions, the number of variables in a model can sometimes be reduced, which typically facilitates the analysis of the model.

The class of biochemical reactions we will study are enzymatic reactions, which are ubiquitous on the living world. Enzymes are proteins that act as catalysts in chemical reactions by reducing the required activation energy for the reaction. Enzymes are not altered by the reaction; they aid in the initial steps of the reaction and control the rate of the reaction by binding the reactants (called substrates) to the active site of the enzyme-substrate complex that then allows the substrates to react and to form the product. This is illustrated in Figure 14.1.

If we denote the substrate by , the enzyme by , and the product of the reaction by , then an enzymatic reaction can be described by

where , , and are the reaction rates in the corresponding reaction steps.

We can translate the schematic description of the reaction into a system of differential equations. We use the following notation.

enzyme concentration at time

substrate concentration at time

enzyme substrate complex concentration at time

product concentration at time

Using the mass action law,

Michaelis and Menten (1913) were instrumental in the description of enzyme kinetics, through both experimental and theoretical work. On the experimental side, they developed techniques that allowed them to measure reaction rates under controlled conditions. Their experiments showed a hyperbolic relationship between the rate of the enzymatic reaction and the substrate concentration , which they described as

where is the saturation constant and is the half-saturation constant (that is, if , then .)

On the theoretical side, they developed a mathematical model for enzyme kinetics that predicted the observed hyperbolic relationship between the substrate concentration and the initial rate at which the product is formed.

In the following, we will analyze , which is a system of four equations. It is not easy to analyze; we will simplify it, and in the end, we will obtain . We will also use to illustrate that it is sometimes possible to reduce the number of equations in a system.

That is,

which implies that

where is a constant. Since is constant, we say that the sum is a conserved quantity. The advantage of this is that if we know that initial concentrations and and one of the two quantities or , we immediately know the other, since . This reduces the number of equations from four to three.

To reduce the number of equations even further, we make another assumption. Whereas the existence of a conserved quantity followed from the system of equations, and we could have obtained it without knowing the meaning of these equations, the next assumption requires a thorough understanding of the enzymatic reaction itself, and cannot be deduced from the set of equations . Briggs and Haldane (1925) who had this understanding, proposed that the rate of formation balances the rate of breakdown of the complex, that is, they assumed

This assumption yields the equation

which can be rewritten as

We denote this ratio by , that is

and, therefore,

Solving for , that is, , and plugging into , we find

which we can solve for . This yields

which allows us to rewrite the equation for the rate t which the product is formed. Since , it follows from that

We can interpret the factor . Namely, if all of the enzyme is complexed with the substrate, then and therefore (since is a conserved quantity). This implies that the rate at which the product is formed, , is fastest when , in which case . We can therefore interpret as the maximum rate at which this reaction can proceed. We introduce the notation

and rewrite as

Equation is known as the Michaelis-Menten law and describes the velocity of an enzymatic reaction. We see from equation that the reaction rate is limited by the availability of the substrate .

Equations and are the same. Whereas was derived from fitting a curve to data points that related the measured substrate concentration to the velocity of the reaction , is derived from a mathematical model. The mathematical model allows us to interpret the constants and in terms of the enzymatic reaction, and the experiments allow us to measure and .

Microbial Growth in a Chemostat — An Application of Substrate-Limited Growth A rather simplistic view of microbial growth is that microbes convert substrate through enzymatic reactions into products that are then converted into microbial biomass.

Figure 14.2 A chemostat.

In the following, we will investigate a mathematical model for microbial growth in a chemostat. The growth of microbes will be limited by the availability of the substrate.

A chemostat is a growth chamber in which sterile medium with concentration of the substrate enters the chamber at a constant rate . Air is pumped into the growth chamber to mix and aerate the culture. To keep the volume in the growth chamber constant, the content of the growth chamber is removed at the same rate as new medium enters. A sketch of a chemostat can be seen in Figure 14.2.

We denote the microbial biomass at time by , and the substrate concentration at time by . Jacques Lucien Monod was very influential in the development of quantitative microbiology; in (1950) he derived the following system of differential equations to describe the growth microbes in a chemostat:

where is the substrate concentration of entering medium, is the rate at which medium enters or leaves the chemostat, and is the yield constant. The function is the rate at which microbes consume the substrate; this rate depends on the substrate concentration. The yield factor can thus be interpreted as a conversion factor of substrate into biomass.

Monod (1942) showed empirically that the uptake rate fits the hyperbolic relationship

where is the substrate level and is the half-saturation constant (that is, ). A graph of is shown in Figure 14.3. It later occurred to Monod that is identical to the Michaelis-Menten law , which might suggest that microbial growth is governed by enzymatic reactions.

In the following, we will determine possible equilibria of and analyze their stability. There is always, of course, the trivial equilibria, which is obtained when substrate enters a growth chamber void of microbes (that is, when ).

Figure 14.3 A graph of .

In that case, there will be no microbes at later times and, hence, for all times . The substrate equilibrium is then found by setting with :

which has solution . Hence, one equilibrium is

to obtain a nontrivial equilibrium , we will look for an equilibrium with . To find this, we solve the simultaneous equations

and

It follows from that

We see immediately from Figure 14.3 that has a solution if . Using in , we can compute , namely

which yields

which is indeed positive, provide that . Using in , we find

or

from which we can see that , provided that . Under the assumption that and , we therefore have a nontrivial equilibrium

There are no other equibria.

To analyze the stability if the two equilibria and , we find the Jacobi matrix associated with the system ,

We analyze the stability of he trivial equilibrium first:

Since the Jacobi matrix is in upper triangular form, the eigenvalues are the diagonal elements, and we find

provided that

Therefore, the equilibrium

For the nontrivial equilibrium , we find

Using , this simplifies to

Now,

and

for , since is an increasing function. Therefore, if the nontrivial equilibrium exists, that is, if both and , then it is locally stable.

We saw that the nontrivial equilibrium exists, provided that and . These two conditions can be summarized as

and

If the first inequality holds, then the denominator in the second inequality is positive. Solving the second inequality for , we then find

We can now summarize our results. The chemostat has two equilibria, a trivial one in which microbes are absent, and a nontrivial one that allows stable microbial growth. If , then the trivial equilibrium is the only biologically reasonable equilibrium and it is locally stable. If , both equilibria are biologically reasonable; the trivial one is now unstable and the nontrivial is the locally stable one. Stable microbial growth is therefore possible, provided that the rate at which medium enters and leaves the growth chamber is between 0 and .