Chemistry 112 Chemical Kinetics. Kinetics of Simple Enzymatic Reactions: The Case of Competitive Inhibition

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1 Chemistry Chemical Kinetics Kinetics of Simple Enzymatic Reactions: The Case of Competitive Inhibition Introduction: In the following, we will develop the equations describing the kinetics of a single substrate reaction in the presence of a species that competes with the substrate (S) for binding to the active site of an enzyme (E); this competing species acts slows down the reaction and, hence, is termed an inhibitor (I) of the reaction. Although the equations derived are limited in applicability to this particular case, the mathematics of the approach is similar to that employed in considerably more complex situations. The following reaction scheme describes the set of processes that we are interested in. The defined terms in the lower part of the panel (K I and k M ) will be used later in the derivation. k k K I E + S ES E + P k + - I EI [E][I] k K I k - + k [EI] M Our objective is to obtain a rate law for formation of P, both in the presence and in the absence of inhibitor I. Development of an expression for d[p]/dt for the case of competive inhibition. The rate law for formation of P can be written as d[p]/dt = k [ES]. Now ES is an intermediate; assume, therefore, that we can apply the steady state approximation to this species. Thus, we can write d[es]/dt = k [E] (k - + k )[ES], Solving for [ES], we obtain k [ES] = k [E]/(k - + k ) = [E]/K M () Therefore, we can write d[p]/dt = k [E]/K M. From the reaction scheme above, we can also write [EI] = [E][I]/K I. Now let us apply mass balance to E and examine the consequences.

2 [E] = [E] + [ES] + [EI] = [E] + [E]/K M + [E][I]/K I in which we have utilized the expressions for the concentrations of ES and EI given above. Removing [E] from each term, we obtain equation () below. [E] = [E] /{ + /K M + [I]/K I } () Now, we can rewrite the expression for d[p]/dt as below: d[p]/dt = k [ES] = k [E]/K M = k [E] /{K M ( + /K M + [I]/K I )} (3) Multiplying through by K M in the denominator, we find that d[p]/dt = k [E] /(K M + K M [I]/K I + = k [E] /{[K M ( + [I]/K I )] + } (4) In many studies, the rates of reaction are measured at the initial stage of the reaction. Then we we can make the following abbreviation: (dp/dt). Under these conditions, we can also assume that. (In the following, we will assume that and not bother to put in the subscript.) Now, let us evaluate the maximum rate of reaction, V max. This will occur when is very large. Then we can assume that >> K M ( + [I]/K I ). Under these conditions, we find that V max = k [E] (5) The quantity in equation (5) is often termed the turnover number. This corresponds to a situation where all of the enzyme active sites are saturated. Note that under these circumstances, the kinetics is zero order in substrate! Assuming that we are measuring initial rates, we can rewrite equation (4) as follows: (dp/dt) = V max /{[K M ( + [I]/K I )] + } (6) Specialization to the situation where the inhibitor concentration is zero. Equation 6 simplifies when [I] =. Then we can write: = V max /([K M + ) (7) This, of course, is the Michaelis-Menten equation that you encountered in the first problem set. Now is a good juncture to look again at the behavior of this equation at various concentrations. When >> K M, then the expected

3 maximum rate, = V max is obtained; note that the reaction is zero order in. When << K M, then = V max /K M and the reaction becomes first order in. When and K M are in the same ranges of magnitude, then the rate law is of undefined order with respect to. Note that if K M =, we can write the following important equation: = V max /. This equation implies (for those reactions described by the Michaelis-Menten equation) that determination of the concentration of S at which the rate achieves half of its maximum rate allows us to immediately obtain a value of K M. Now, let us deal with some specific data from the study of an enzymatic reaction. We shall work with the data contained in Table. We first plot the value of as a function of, the result of which is contained in Fig. below the Table.. Table, M (M/sec) Fig. 5 5 Note that when plotted in this manner, the plot is non-linear. However, by rearranging the equation, we can obtain the Lineweaver-Burk equation (8), which does give a linear graph when the data are plotted appropriately. = K M + (8) V max V max Figure below shows the same data displayed in the graph above, but plotted as a Lineweaver-Burk plot.

4 / /v o =.39/ +.86 r =.997 Lineweaver-Burk Plot Fig. 3 4 / In the Lineweaver-Burk plot, the slope is equal to K M /V max and the intercept is /V max. Thus, in the case above, we find V max = /.86 =. M/s and K M =.47 M if is in moles/liter (M). There is another interesting facet of the Lineweaver- Burk plot, namely that it predicts that the x-intercept (i.e. the value that / takes when = ) will be equal to /K M. The extended Lineweaver-Burk plot showing the x-intercept is shown below. / /v o =.39/ +.86 r =.997 Lineweaver-Burk Plot / Evaluating the x-intercept from the corresponding Lineweaver-Burk equation, we find its value to be -.; the corresponding value of K M can be calculated as.47. The Eadie-Hofstee Plot and the Hanes Plot. There are several alternative plots used by enzyme kineticists. We will discuss two of these, namely the Eadie- Hofstee plot and the Hanes plot. Manipulation of equation (7) in a different manner from that used previously leads to equation (9), which is the Eadie-Hofstee equation. (To obtain this equation, multiply equation (8) on both sides by V max and rearrange.) = K M [ ] V - max (9)

5 In graphically analyzing data using this equation, we plot on the y axis and / on the x-axis. The y-intercept is V max and the slope is K M ; the corresponding x-intercept is V max /K M. Graphing the data given in the table above in an appropriate manner yields the corresponding Eadie-Hofstee plot. Convince yourself that essentially the same values for K M and V max can be determined that were obtained from the Lineweaver-Burk plot..5 = -.487( /) +. Eadie-Hofstee Plot Fig / The Hanes equation is obtained by multiplying both sides of equation (8) by. Doing so, we obtain the following: ( ) K v = + M () V max Vmax In graphs of data according to this equation, which has the form y = mx + b, we find that the y-intercept is K M /V max, the slope is /V max and the x-intercept is K M. Plotting the appropriate data obtained from Table, we obtain the Hanes plot shown in Figure 4 below. From the linear least squares fit of the data, we see that the slope is.83, which yields a value of V max of.3 M/sec. The y-intercept of.4, in combination with the V max value, leads to K M =.55, while the x- intercept also gives a value of K M =.55. Note that the values are slightly different from those obtained from the Lineweaver-Burk plot, probably because of the different weighting of the points. Many enzyme kineticists consider the Hanes plot to be the best method for analyzing data, although much of data published by those researchers who are not enzyme kineticists is analyzed using the Lineweaver- Burk plot. For greater detail on these matters, see various monographs on the subject, such as Fundamentals of Enzyme Kinetics by A. Cornish-Bowden

6 (Portland Press, 995) and the treatise Enzyme Kinetics by I. H. Segal (Wiley, 993). 5 / = Hanes Plot Fig. 4 / The Lineweaver-Burk equation for the case of competitive inhibition when inhibitor is present. Recall from above that we had developed equation (6) describing how depends on substrate and inhibitor concentrations. (dp/dt) = V max /{[K M ( + [I]/K I )] + } (6) Rearranging this slightly, we obtain equation (): = V max /{(K M /)( + [I]/K I )] + } () We can rewrite this in the desired form, the Lineweaver-Burk equation for competitive inhibition, as equation (). K M = + V max V max [ [ I ] + ] K I [ ] () In this equation, the y-intercept is /V max and the slope is (K M /V max ){ + ([I]/K I )}. The corresponding x-intercept is K I /{K M (K I + [I])}. Note that for each set value of [I] at which the kinetics of this reaction is studied, the y intercept does not vary, but both the slope and the x-intercept do. This is illustrated in the set of graphs shown below for the set of data in Table in which various inhibitor concentrations have been added. These data inhibitor concentrations are.5, and M. Recall that is in terms of M and in terms of M/sec.

7 7.5 [I]= : / =.39(/ +.86 [I] =.5M: / =.5(/) +.86 [I] = M: / =.(/) +.86 [I] = M: / = 3.(/) +.86 K I = [I] = / 5.5 [I] = M [I] = M [I] =.5-4 / From the slopes of the lines in the absence and presence of inhibitor (see the linear least square fit equations given above the graph), we can evaluate the value of K I as. Note, that as predicted by the equation given above, the value for the slope becomes more positive as [I] increases. Also, note that the y-intercept is constant and that the x-intercept becomes less negative as [I] increases. Other cases of inhibition can also be envisioned. Two of these will be discussed in a handout distributed in class (and that will also appear on the web site as a separate supplement).