In the early 1900's, Leonor Michaelis and Maude Menten proposed a model for enzyme action based on Emil Fisher's conclusion that, for this action to occur, the enzyme and its substrate(s) had to come together. The model, in its simplest form, may be formulated as follows:

where E is enzyme, S is substrate, and P is product. The model shows that the first step in the reaction is the formation of an enzyme-substrate complex. This is followed by the catalytic step in which the substrate is transformed into free product and the enzyme is restored to its original form. The second step of the reaction may be subdivided into various sub-steps, as follows

but then the analysis of the model becomes more involved while not adding much to the fundamental concepts. Since the theory of reaction kinetics is sufficiently developed, a series of quantitative relations that should result in some expression for the reaction velocity may be derived from the above reaction schemes. We will do it for the simple scheme.

An inspection of this last equation and of its corresponding reaction 3, shows that it is, in effect, equal to d[P]/dt, the rate of product formation or, in other words, the velocity of the overall reaction. Thus, it may seem that by simply integrating it (which is easy) we would have the problem solved. But a quantitative model of a phenomenon is useful only if it can be validated by experimental results. It happens that the variable [ES] cannot be measured directly, so it becomes necessary to express it in terms of measurable variables and of experimentally obtainable parameters. We find that addition of the three equations1 yields the differential equation for the net rate of formation of the E-S complex.

Unfortunately, this is a nonlinear differential equation that cannot be solved. Some sort of solution of this equation must be reached if we want a quantitative description of the velocity of the reaction. If we had an expression for [E·S] to substitute in equation 3, then the problem would be solved. This expression for [E·S] is necessary because, as things stand, there is no way to determine its value. So we are back to square one. When one faces a differential equation that has no analytical solution, it is customary to limit the problem to special conditions such that the differential equation is reduced to an algebraic equation. In the age of computers, it is possible to obtain numerical solutions and graphic representations of the results of the differential equation, but this was not possible during the early part of the 20^th century. Besides, it is very satisfying to have an integrated equation, which synthesizes a lot of information about the nature of the underlying process.

In this particular case, certain experimental evidence gives us a clue as of how to proceed to get an algebraic equation: when a small concentration of enzyme is mixed with a relatively large concentration of substrate, the concentration of product formed during the early part of the reaction follows a linear curve (constant slope). This means that the velocity of the reaction (the rate of product formation) is constant, and, since this velocity is equal to k₃ · [E·S], [E·S] must remain constant during that time ( d[E·S]/dt = 0). Therefore, during the early part of the reaction [E·S] shows evidence of being in a steady state, which means that, then,

Since the velocity of the reaction is d[P]/dt = v = k₃[E·S] , the problem has now been reduced to solving the above for [E·S] and to state that expression in terms of measurable quantities. Little error is made by assuming that [S] remains constant and equal to its initial value during the early part of the reaction. On the other hand, [E] may not be measurable. The idea of mass balance comes to our rescue: [E] must be equal to the total enzyme concentration ([E_T]) minus [E·S]. When this substitution is made,

which is an expression for the initial velocity (the velocity during the early part of the reaction) in terms of the total enzyme concentration, the initial substrate concentration, and the rate constants for the separate reactions. It is called the Michaelis-Menten (M&M) equation. The initial velocity is the velocity during the early part of the reaction while little product has been formed and when the concentration of substrate has not changed much from its initial value. By definition, K_M is a constant derived from the three rate constants.

The product k₃[E_T] is the value of k₃[E·S] when the concentration of substrate is so large that the enzyme present is saturated with it, i.e., all the enzyme present in the reaction vessel is in the form E·S. This is equivalent to saying that k₃[E_T] is the maximum possible value of k₃[E·S] for a given amount of enzyme present. Then, we could write

where V_M = maximal reaction velocity possible for the concentration of enzyme present.

The last three equations show a hyperbolic shape in cartesian coordinates. Figure 1 shows a plot of this equation. The value of [S] required to reach initial velocities close to the value of V_M is very large and not achievable in practice. This curve may be obtained experimentally for an enzyme by preparing a series of tubes all with the same concentration of enzyme and each one with a different concentration of substrate. As soon as the mixture is made in each tube, the concentration of substrate begins to be measured at intervals. Then, a plot for each tube of [S] vs. time will show an early slope that is the initial velocity of the reaction at the corresponding [S]. The curve in Figure 2 is a plot of those initial velocities vs. their corresponding substrate concentrations.

As indicated in the graph, the value of [S] for which v_i = ½ V_M is equal to K_M, as shown below.

When [S] << K_M , v_i V_M[S]/K_M = (V_M/K_M ) [S] , and, since both V_M and K_M are constant, the curve should be almost linear for very low concentrations of substrate. Notice that, if the reaction ES E + S is considered to be at equilibrium, its dissociation constant K_D = k₂/k₁ would be close to K_{M =}(k₂+ k₃) / k₁ as long as k₃<<k₂. Since for many enzymes this is the case, K_M is frequently taken as a measure of the binding strength of the substrate to the enzyme: the larger its value the stronger its tendency to dissociate and the weaker the binding.

Since k₃ = V_M / [E_T] , a way of expressing the physical meaning of k₃ is to say that it is the maximum velocity of increase in product concentration per unit concentration of enzyme. It can also be said to be the maximum number of substrate molecules converted to product in unit time by an enzyme molecule. k₃ is usually referred to as the turn over number of the enzyme.

It is instructive to determine which are the units of the constants used in these equations. In Equations 1 we can see that the units of d[ES]/dt are M/t. Then, the units of k₁ in the first of these equations have to be M^-1t^-1, which when multiplied by M² will yield M/t. We similarly find that the units of k₂ and k₃ are t^-1. Then, the units of K_M = k₂+k₃/k₁ are M. This matches with the fact that the denominator of the M&M equation is the sum of K_M plus [S], an addition that is forbidden for quantities with different units.

The main value of an equation as the M&M equation, derived from a model of enzyme action is that, ideally, it should allow us to characterize an enzyme by its values of V_M and K_M. This would be accomplished by plotting experimental values of v_i and [S] and extrapolating it to infinite [S] to reach the value of V_M. Once this value is known, it is easy to get K_M = ½ V_M. The problem resides in the extrapolation to infinite [S], which could not be made with any confidence before computers and advanced statistical programs for nonlinear regression became available. It is still easier in the lab to use rearrangements of the M&M formulation that yield a linear instead of curved graph. Then, extrapolations can be made with more confidence, and reliable values of the constants can be obtained. The extrapolations can be made either graphically or by linear regression, which can be done without the use of computers if necessary.

There are several such rearrangements of the M&M equation, but we are going to limit our discussion to the one by Lineweaver and Burke, which is not, statistically speaking, the best, but can be easily used to differentiate certain types of inhibition to which the enzyme may be subjected. This transformation of the M&M equation is also called the inverse plot, because it is obtained by inverting the M&M formula. Starting with that equation

After inverting the expression, we proceed to separate the right side into two terms

The shape of this equation is that of a straight line with a y intercept (y = mx + b), where

This means that, when a set of paired values of v_i and [S] are plotted, the resultant points fall approximately on a straight line whose slope is K_M / V_M and its y-intercept is 1/V_M. Such a graph is shown in Figure 6, curve I. Simply by drawing such line and extrapolating to 1/ [S] = 0 (same as [S] = infinity), the y-intercept is obtained, which gives us the value of V_M. In order to find K_M, either calculate the slope or continue the extrapolation to find the x-intercept, which will be negative. This intercept is the point in the line where 1/V_i = 0, so, making equation 6 equal to zero we get

In this manner it is easy to obtain acceptable values for the parameters of the equation, V_M and K_M,for a given enzyme from a set of experimental values obtained with the same enzyme. Bear in mind that, since V_M = k₃ [E_T], its value depends not only on which enzyme we are talking about, but also on its concentration in the reaction vessel where it is being used.

Curves II and III represent plots for the same enzyme shown in curve I, but in the presence of two different kinds of inhibitors of the enzyme. Curve II is obtained with competitive inhibitors. It shows that the y-intercept is equal to the one obtained without any inhibitor. So, the value of V_M (and, therefore, the value of k₃) is not affected by the inhibitor. On the other hand, the value of K_M appears to be larger. This is not a real change in its value, but an apparent change that results from a reduction in the concentration of the free enzyme available to bind substrate because inhibitor molecules are bound to the active site of a fraction of the enzyme molecules present. The reactions that take place in this case are

E	+	S	º	E@S	6	E	+	P
+
I
;
E@I

where I is the inhibitor and EI the enzyme-inhibitor complex, whose dissociation constant is K_I. This scheme shows that the inhibitor binds to the enzyme in such a way that it prevents the binding of the substrate. It has been shown that this is because the inhibitor binds to the same site as the substrate, so that enzyme molecules bound to inhibitor are not available to the substrate. An analysis similar to the one performed for the M&M formulation, but realizing that, in this case, E_T = E + ES + EI, yields the equation

Notice that the only difference between this equation and the M&M equation is the addition of (1+{[I] / K_I }) as a factor in the denominator. If the concentration of the inhibitor were zero, the two equations would yield the same initial velocity. For any concentration of substrate, if [I] > 0, the initial velocity is smaller, a reflection of the fact that there is less free enzyme available. This is equivalent to increasing the value of the enzyme's K_M, which would result in a weaker binding of the substrate. So, in fact, the enzyme seems to have an apparent K_M(ap) = K_M (1+{[I] / K_I}) . As in the case of the M&M equation, the inverse of Equation 7 is a straight line

It is easy to demonstrate that, as shown by line II in Figure 6, the y-intercept is the same with a competitive inhibitor as without it. Also, the x-intercept, the point in the curve for which 1/v_i= 0, is equal to the value of the apparent K_M.

Curve III in Figure 6 represents the inverse plot for the same enzyme as before, but in the presence of a noncompetitive inhibitor. The way in which these inhibitors work is modeled in the following reaction scheme.

E	+	S	º	E@S	6	E	+	P
+				+
I				I
;				;
E@I	+	S	º	E@I@S

It shows that the enzyme-inhibitor complex can bind substrate and that the enzyme-substrate complex can also bind the inhibitor, but that these bindings do not result in the formation of product. The effect of this is, not only to reduce the concentration of free enzyme as in the case of competitive inhibitors, but also to reduce the concentration of the catalytically effective enzyme-substrate complex. By assuming that the affinity of the enzyme for the substrate is equal to the affinity of the enzyme-inhibitor complex for the substrate, and that the affinity of the enzyme for the inhibitor is equal to the affinity of the enzyme-substrate complex for the inhibitor, and applying the same analysis as before we get

which differs from the M&M equation in that V_M is being divided by (1 + {[I] / K_I}). Again, this quantity will be greater than one when [I] > 0, and then, the equation for v_i will work as if V_M had the smaller value. In the graph, while the x-intercept is the same as for the uninhibited enzyme (no change in K_M), the y-intercept is larger, which points to a smaller V_M. Although this change in V_M is usually interpreted as a reduction in k₃, the fact that K_M = (k₂ + k₃) / k₁ is not changed indicates that k₃ only appears to have a smaller value. In fact, what really happens is that the actual concentration of enzyme-substrate complex is [ES] + [ESI], while its catalytically effective concentration is [ES]. Therefore, the velocity of the reaction at any substrate concentration will be smaller in the presence of the inhibitor.

It seems obvious from this analysis that careful use of the language is necessary when talking about the effects of these kinds of inhibitors. According to the reaction schemes by which their actions are modeled, they do not alter the values of the rate constants of these reactions and, consequently leave the values of K_M and of V_M untouched. Therefore, we should not talk about changes in these values, but of apparent changes.

The M&M model considers only one substrate, while most enzymatic reactions involve two or more substrates. Frequently, reactions with two substrates obey M&M kinetics with respect to the concentration of one of the substrates when the concentration of the other is kept constant. Other reactions simply do not obey these kinetics and require more complex models. We will not enter in their discussion here.