Reversible Michaelis Menten Kinetics

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A more realistic description of a enzymatic reactions than pure Michaelis Menten Kinetics is given by considering the product forming reaction step as reversible. \begin{equation} E+S \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} ES \underset{k_{-2}}{\overset{k_2}{\rightleftarrows}}E+P \end{equation} The Enzyme(E) forms a Complex (ES) with the Substrate (S) and modifies it to the Product (P). In the reversible Michaelis-Menten reaction the Product can bind the enzyme again and react back to the substrate. This process is described by the ODE system \begin{equation} \frac{dS}{dt}=-k_1E\cdot S + k_{-1}ES \end{equation} \begin{equation} \frac{dES}{dt}=k_1 E\cdot S +k_{-2} E \cdot P -(k_{-1}+k_{2})ES \tag{1} \end{equation} \begin{equation} \frac{dE}{dt}=-k_1 E\cdot S -k_{-2} E \cdot P + (k_{-1}+k_{2})ES \tag{2} \end{equation} \begin{equation} v=\frac{dP}{dt}=k_2 ES -k_{-2} E \cdot P=v_f - v_b \tag{3} \end{equation} To get an equation for $v$, we need expressions for $ES$ and $E$. We get them by assuming a (quasi) steady state for (1) and (2), which is justified iff (if and only if) \begin{equation} k_1,k_{-1}\gg k_2,k_{-2} \end{equation} and iff the initial substrate concentration is much larger then the enzyme concentration. This assumptions allow to solve for the rate of product formation $v$ \begin{equation} v=-\frac{dS}{dt}=\frac{dP}{dt} \end{equation} to this end, we add (1) and (2) \begin{equation} \frac{dES}{dt}+\frac{dE}{dt}=0 \hspace{1cm}or \hspace{1cm} E_{total}=E+ES=constant. \end{equation} inserting $E=E_{total}-ES$ into (1) yields \begin{align} \frac{dES}{dt}=0&=k_1 (E_{total}-ES) \cdot S +k_{-2} (E_{total}-ES) \cdot P -(k_{-1}+k_{2})ES\\ &= (k_1 S +k_{-2} P) \cdot E_{total} - (k_{-1}+k_{2}+k_1 S +k_{-2} P)ES\\ ES&=\frac{(k_1 S +k_{-2} P) \cdot E_{total} }{(k_{-1}+k_{2}+k_1 S +k_{-2} P)} \tag{4} \end{align} Similarly inserting $ES=E_{total}-E$ into (2) \begin{align} \frac{dE}{dt}=0&=-k_1 E \cdot S -k_{-2} E \cdot P + (k_{-1}+k_{2})(E_{total}-E)\\ &=-E\cdot (k_1 S + k_{-2} P + (k_{-1}+k_{2}))+(k_{-1}+k_{2})E_{total} \end{align} which results in \begin{align} E&=\frac{(k_{-1}+k_{2})E_{total}}{(k_1 S + k_{-2} P + k_{-1}+k_{2})} \tag{5} \end{align} inserting (5) and (4) together into (3) gives the product formation rate

\begin{align} v=\frac{dP}{dt}&=k_2 ES -k_{-2} E \cdot P\\ &= \frac{k_2(k_1 S +k_{-2} P) \cdot E_{total} }{(k_{-1}+k_{2}+k_1 S +k_{-2} P)} - \frac{k_{-2}(k_{-1}+k_{2})E_{total}\cdot P}{(k_1 S + k_{-2} P + k_{-1}+k_{2})} \\ \end{align} the forward rate $v_f$ and the backwared rate $v_b$ just happen to have a common denominator, so the equation is readily simplified. \begin{align} v&= \frac{k_2(k_1 S +k_{-2} P) \cdot E_{total}-k_{-2}(k_{-1}+k_{2})E_{total}\cdot P}{(k_{-1}+k_{2}+k_1 S +k_{-2} P)} \\ &= \frac{k_2 k_1 S \cdot E_{total} -k_{-2}k_{-1}E_{total}\cdot P}{(k_{-1}+k_{2}+k_1 S +k_{-2} P)} \\ \end{align} Divide by $(k_{-1}+k_{2})$ and use $K_{m,1}=(k_{-1}+k_{2})/k_1$, $K_{m,2}=(k_{-1}+k_{2})/k_{-2}$ \begin{align} v&= \frac{k_2(k_1 S +k_{-2} P) \cdot E_{total} -k_{-2}(k_{-1}+k_{2})E_{total}\cdot P}{(k_{-1}+k_{2}+k_1 S +k_{-2} P)} \\ &= \frac{k_2\cdot E_{total} S/K_{m,1}-k_{-1}E_{total}\cdot P/K_{m,2}}{(1+S/K_{m,1} +P/K_{m,1})} \\ \end{align} Finally we use $v_f^{max}=k_2\cdot E_{total}$ and $v_b^{max}=k_{-1}\cdot E_{total}$ to get the common form for the reversible Michaelis Menten equation

\begin{align} v&= \frac{v_f^{max} \, S/K_{m,1}-v_b^{max} \, P/K_{m,2}}{(1+S/K_{m,1} +P/K_{m,1})} \\ \end{align} Notice, at the very beginning $P=0$ and thus $v$ reduces to simple Michaelis Menten Kinetics. For increasing pruduct concentrations the right term becomes larger and reduces the velocity compared to simple Michaelis Menten Kinetics.



Further reading:

  • Philip W. Kuchel $\&$ Peter J. Mulquiney - Modelling Metabolism with Mathematica: Analysis of Human Erythrocyte[1]
  • Michael A. Savageau - Biochemical Systems Analysis: A study of Function and Design in Molecular Biology
  • Edda Klipp et al. - Systems Biology: A Textbook