Skip to main content
Biology LibreTexts

Henri-Michaelis-Menten Kinetics

Rapid Equilibrium Assumption: Henri-Michaelis-Menten Kinetics

Consider the following reaction mechanism for the enzyme-catalyzed conversion of substrate \(S\) into product \(P\) (we assume that the catalyzed rate is much greater than the noncatalyzed rate.)

14 rapideqmech.gif

As we did for the derivation of the equations for the facilitated transport reactions under rapid equilibrium conditions, this derivation is based on the assumption that the relative concentrations of \(S\), \(E\), and \(ES\) can be determined by the dissociation constant, \(K_s\), for the interactions and the concentrations of each species during the early part of the reaction (i.e., under initial rate conditions). Assume also the \(S \gg E_o\). Remember that under these conditions, S does not change much with time. Is this a valid assumption? Examine the mechanism shown above. \(S\) binds to \(E\) with a second order rate constant \(k_1\).

\(ES\) has two fates: it can dissociate with a first order rate constant \(k_2\) to \(S + E\), or it can be converted to product with a first order rate constant of \(k_3\) to give \(P + E\). If we assume that \(k_2 \gg k_3\) (i.e. that the complex falls apart much more quickly than S is converted to P), then the relative ratios of \(S\), \(E\), and \(ES\) can be described by \(K_s\). Alternatively, you can think about it this way. If \(S\) binds to \(E\), most of \(S\) will dissociate, and a small aM_ount will be converted to \(P\). If it does, then \(E\) is now free, and will quickly bind \(S\) and reequilibrate since the most likely fate of bound \(S\) is to dissociate, not be converted to \(P\) (since \(k_3 \ll k_2\)). This also makes sense if you consider that the physical step characterized by \(k_2\) is likely to be quicker than the chemical step, characterized by \(k_3\). Hence the following assumptions have been used:

  • \(S \gg E_o\)
  • \(P_o = 0\)
  • \(k_3\) is rate limiting (i.e., the slowest step)

We would like to derive equations which show the velocity v as a function of the initial substrate concentration, \(S_o\) , (assuming that \(P\) is negligible over the time course of measuring the initial velocity). Also assume that the \(v_{catalyzed} \gg v_{noncatalyzed}\). In contrast to the first-order reaction of S to P in the absence of E, \(v\) is not proportional to \(S_o\), but rather to \(S_{bound}\) as we described in class with facilitated diffusion (flux proportional to AR, not free A). Therefore,

\[v \propto ES\], or \[ v = \text{const} \, [ES] = k_3 [ES] \tag{1}\]

where \(v\) is the velocity. How can we calculate \([ES]\) when we know \([S]\) (which is equal to \(S_o\)) and Etot (which is \(E_o\))? Let us assume that S is much greater than E, as is the likely biological case. We can calculate \([ES]\) using the following equations and the same procedure we used for the derivation of the binding equation, which gives the equation below:

\[ ES = \dfrac{E_oS}{(K_s + S} \]

which is analogous to

\[ML] = \dfrac{M_oL}{K_d + L}\]

Derivation

15 MMRapEqDer.gif

\[ v = \text{const} [ES] = k_3 [ES] = \dfrac{k_3 E_oS}{K_s + S} = \dfrac{V_mS}{K_s + S} \tag{5}\].

The World Famous Henri-Michaelis-Menten Equation

Just as in the case with noncatalyzed first order decay, it is easiest to measure the initial velocity of the reaction when [S] does not change much with time and the velocity is constant (i.e the slope of the \(\frac{dP}{dt}\) curve is constant). A plot of \([P]\) vs. \(t\) (called a progress curve) is made for each different substrate concentration studied. From these curves, the initial rates at each \([S]\) are determined. Alternatively, one set reaction time is determined for all the different substrate concentrations that gives a linear rise in \([P]\) with time. At the time, the reaction can be stopped (quenched) with a reagent that does not cause any change in \(S\) or \(P\). Initial rates can be easily calculated for each \([S]\) from a single data point in this case.

  • a plot of \(v\) vs. \(S\) is hyperbolic
  • \(v = 0\) when \(S = 0\)
  • \(v\) is a linear function of S when \(S \ll K_s\).
  • \(v = V_{max}\) or \(V_m\) when \(S\) is much greater than \(K_s\)
  • \(S = K_s\) when \(v = \dfrac{V_{max}}{2}\).

These are the same conditions we detailed for our understanding of the binding equation

\[ML = \dfrac{M_oL}{K_d + L} \]

As we saw in the graph of \(A\) or \(P\) vs. \(t\) for a noncatalyzed, first order reaction, the velocity of the reaction, given as the slope of those curves, is always changing. Which velocity should we use in Equation 5? The answer invariably is the initial velocity measured in the early part of the reaction when little substrate is depleted. Hence \(v\) vs. \(S\) curves for enzyme catalyzed reactions invariably are really \(v_0\) vs. \([S]\).

Analysis of the General Michaelis-Menten Equation

This equation can be simplified and studied under different conditions. First notice that \((k_2 + k_3)/k_1\) is a constant which is a function of relevant rate constants. This term is usually replaced by \(K_m\), which is called the Michaelis constant. Likewise, when \(S\) approaches infinity (i.e., \( S \gg K_m\), equation 5 becomes \(v = k_3[E_o]\) which is also a constant, called \(V_m\) for maximal velocity. Substituting \(V_m\) and \(K_m\) into equation 5 gives the simplified equation:

\[v = \dfrac{V_m[S]}{K_m+ [S]} \tag{10}\]

It is extremely impoortant to note that \(K_m\) in the general equation does not equal the \(K_s\), the dissociation constant used in the rapid equilibrium assumption! \(K_m\) and \(K_s\) have the same units of moolarity, however. A closer examination of \(K_m\) shows that under the limiting case when \(k_2 \gg k_3\) (the rapid equilibrium assumption) then,

\[K_m = \dfrac{k_2 + k_3}{k_1} \approx \dfrac{k_2}{k_1} = K_d = K_s. \tag{11}\]

If we examine Equations 9 and 10 under several different scenarios, we can better understand the equation and the kinetic parameters:

  • when \(S = 0\) and \(v = 0\).
  • when \(S \gg K_m\), \(v = V_m = k_3E_o\) (i.e. \(v\) is zero order with respect to \(S\) and first order in E). Remember, \(k_3\) has units of s-1 since it is a first order rate constant and is often called the turnover number, because it describes how many molecules of \(S\) "turn over" to \(P\) per second.
  • \(v = \dfrac{V_m}{2}\), when \(S = K_m\).
  • when \(S \ll K_m\), \(v = \frac{V_mS}{K_m} = \frac{k_3E_oS}{K_m}\) (i.e. the reaction is bimolecular, dependent on both on S and E. \(k_3/K_m\) has units of M-1s-1, the same as a second order rate constant.

Notice that equations 9 and 10 are exactly analysis to the previous equations we derived:

  • \(ML = \dfrac{M_oL}{K_d + L}\) for binding of \(L\) to \(M\)
  • \(J_o = \dfrac{J_m[A]}{K_d + [A]}\) for rapid equilibrium binding and facilitated transport of \(A\)
  • \(v_o = \dfrac{V_m[S]}{K_s + [S]}\) for rapid equilbrium binding and catalytic conversion of \(A\) to \(P\)
  • \(v_o = \dfrac{V_m[S]}{(Km+ [S]}\) for steady state binding and catalytic conversion of \(A\) to \(P\)

Note that all these equations give hyperbolic dependencies of the y-dependent variable (\(ML\), \(J_o\), and \(v)\)o) on the ligand, solute, or substrate concentration, respectively.