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Enzyme-Catalyzed Reaction Kinetics

Enzyme-Catalyzed Reaction Kinetics

We have previously derived equations for the reversible binding of a ligand to a macromolecule. Next we derived equations for the receptor-mediated facilitated transport of a molecule through a semipermeable membrane. This latter case extended the former case by the addition of a physical transport step. Now, in what hopefully will seem like déjà vu, we will derive almost identical equations for the chemical transformation of a ligand, commonly referred to as a substrate, by an enzyme. Two scenarios will be studied.

  1. Rapid Equilibrium Assumption: enzyme \(E\) (macromolecule) and substrate \(S\) (ligand) concentrations can be determined using the dissociation constant since \(E\), \(S\), and \(ES\) are in rapid equilibrium, as we previously used in our derivation of the equations for facilitated transport. Sorry about the switch from \(A\) to \(S\) in designation of substrate. Biochemists use \(A\), \(B\), and \(S\) to represent \(S\) and usually \(P\) and \(Q\) to represent products.
  2. Steady State Assumption (more general): enzyme and substrate concentrations are not those determined using the dissociation constant.

Enzyme kinetics experiments, as we will see in the next several chapters, must be used to determine the detailed mechanism of the catalyzed reaction. Using kinetics analysis you can determine the order of binding/dissociation of substrates and products, the rate constants for individual steps, and clues as the to methods used by the enzyme in catalysis.

Experimental Determination of \(V_m\) and \(K_M\)

How can \(V_m\) and \(K_m\) be determined from experimental data? 

From initial rate data: The most common way is to determine initial rates, \(v_0\), from experimental values of \(P\) or \(S\) as a function of time. Hyperbolic graphs of \(v_0\) vs. \([S]\) can be fit or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine Kd. These included:

  • nonlinear hyperbolic fit
  • double reciprocal plot
  • Scatchard plot

The double-reciprocal plot is commonly used to analyze initial velocity vs. substrate concentration data. When used for such purposes, the graphs are referred to as Lineweaver-Burk plots, where plots of 1/v vs. 1/S are straight lines with slope \(m = \frac{KM}{V_{max}}\), and y-intercept \(b = \frac{1}{V_{max}}\). These plots can not be analyzed using linear regression, however, since that method assumes constant error in the y axis (in this case \(\frac{1}{v}\)) data. A weighted linear regression or even better, a nonlinear fit to a hyperbolic equation should be used. The Mathcad template below shows such a nonlinear fit. In the laboratory, we will use a series of programs developed by W. W. Cleland specifically designed to analyze initial rate data of enzyme catalyzed reactions. A rearrangement of the corresponding Scatchard equations in the Eadie-Hofstee plot is also commonly used.

Common Error in Biochemistry Textbooks: The Shape of the Hyperbola

From integrated rate equations: \(K_M\) and \(V_m\) can be extracted from progress curves of \(A\) or \(P\) as a function of \(t\) at one single \(A_0\) concentration by deriving an integrated rate equation for \(A\) or \(P\) as a function of \(t\), as we did in equation 2 which shows the integrated rate equation for the conversion of \(A \rightarrow P\) in the absence of enzyme. In principle this method would be better than the initial rates methods. Why? One reason is that is not easy to be certain you are measuring the initial rate for each and every \([S]\) which should vary over a wide range. It's also time intensive. In addition, think how much of the data is discarded if you take an entire progress curve at each substrate concentration, especially if you quench the reaction at a given time point, which effectively limits the data to one time point per substrate. 

In practice, the mathematics are complicated as it is not possible to get a simple explicit function of \([P]\) or \([S]\) as a function of time. Nevertheless, progress has been made in progress curve analysis. Let us consider the simple case of a single substrate \(S\) (or \(A\)) being converted to product \(P\) in an enzyme catalyzed reaction. The analogous equations for first order, non-catalyzed rates were

\[A=A_oe^{-k_1t} \]

or

\[P = A_0(1-e^{-k_1t}).\]

Now let's derive the equations for the enzyme-catalyzed reaction. 

21.1 steadystatemech.gif

The derivation of the relevant equations are shown below. 

\[ v = -\dfrac{dS}{dt} = - \dfrac{dP}{dt} = \dfrac{V_MS}{K_M+S} \tag{15}\]

\[ \int_{S_o}^S \dfrac{K_M+S}{V_MS}dS = \int_0^t t = -t=\dfrac{S+K_M\ln S-S_o-K_M\ln S_o}{V_M}\]

which on rearrangment gives

\[S_0- S+K_M\ln\dfrac{S_0}{S}-V_Mt \tag{16}\]

This equation is an implicit equation, not explicit, as it does not give S(t) explicitly as s function of \(t\). Equations 16 can be written with respect to the protudct \(P\) using

\( P=S_o-S\) or \( S=S_o-P\)

\[ S_o-(S_o-P)+K_M\ln \dfrac{S_o}{S_o-P}=V_Mt\]

\[ P -K_M \ln \left( \dfrac{S_o-P}{S_o} \right) = V_Mt \tag{17}\]

Rearranging equation 17 gives

\[ \dfrac{P}{t}=\dfrac{K_M \ln \left( \dfrac{S_o-P}{S_o}\right)}{t} + V_M\]

or

\[ \dfrac{P}{t}=\dfrac{K_M \ln \left( 1-\dfrac{P}{S_o}\right)}{t} + V_M \tag{18}\]

A graph of \(P/t\) vs. \((\ln (1-P/S_0)]/t\) (shown below) from equation 18 gives a straight line with a slope of \(K_M\) and a \(y\)-intercept of \(V_m\). Note that the calculated values of \(V_m\) and \(K_M\) are derived from only one substrate concentration, and the values may be affected by product inhibition

Figure: Enzyme Kinetics Progress Curve

23 EnzKinProgCurve.gif

Can a simple explicit equation of \(P\) vs. \(t\) be derived? The answer is no. However, it can be represented by an explicit Lambert function as shown in the derivation below.

Rearrange equations 16 to get equation 19

\[ S+K_M\ln S = s_o + K_M \ln S_o - V_Mt \tag{19}\]

Let \(\phi = \frac{S}{K_M}\) noting that \(S\) changes with time. Substitute into equation 19.

\[ \phi K_M + K_M \ln (\phi \, K_M) = S_o + K_M \ln(S_o) - V_m \, t\]

Divide by \(K_m\) and rearrange to get

\[ \phi + \ln(\phi) = \dfrac{S_o}{K_m} + \ln \left( \dfrac{S_o}{K_M} \right) - \dfrac{V_M\, t}{K_M} \tag{20}\]

Now consider the righthand side of equation 20 and note the following equality:

\[x= \ln ( e^{x}) = x\]

Apply this identity to equations 20

\[  \dfrac{S_0}{K_M} + \ln \left( \dfrac{S_o}{K_M} \right) - \dfrac{V_M\, t}{K_M} = \ln \left[  e^{\dfrac{S_0}{K_M} + \ln \left( \dfrac{S_o}{K_M} \right) - \dfrac{V_M\, t}{K_M} } \right] = \ln \left[ \dfrac{S_o}{K_M} e^{\frac{S_o-V_Mt}{K_M}} \right]= \ln (x)  \tag{21}\]

where

\[x=\dfrac{S_o}{K_M}e^{\frac{S_o-V_M\, t}{K_M}}\]

Now substitute equations 21 into equation 22

\[ \phi + \ln \left[ \dfrac{S_o}{K_M} e^{\frac{S_o-V_Mt}{K_M}}\right] = \ln (x) \tag{22}\]

Equation 22 is analogous to the Labert \(W\) function.

\[ W(x) + \ln\{W(x)\}=\ln(x) \tag{23}\]

Equating the two left most terms of equation 22 and 23 gives

\[ \phi = W(x) = W \left( \dfrac{S_o}{K_M} e^{\frac{S_o-V_Mt}{K_M}}\right) \tag{24}\]

and since \(\phi = \dfrac{S}{K_M}\)

\[ S = K_M W \left( \dfrac{S_o}{K_M} e^{\frac{S_o-V_Mt}{K_M}}\right) \tag{25}\]

which give \(S(t)\) as a function fo \(t\) and the constancts \(V_M\) and \(K_M\).