Michaelis Menten Kinetics

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Derivation of the Michaelis Menten Kinetics

Michaelis Menten Kinetics describes quantitative the enzymatic mechanisms for one substrate reactions. Here we consider kinetics without backward reaction and effectors \begin{equation} E+S \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} ES \overset{k_2}{\rightarrow}E+P \end{equation} The Enzyme(E) forms a Complex (ES) with the Substrate (S) and modifies it to the Product (P). The Product is then released and the Enzyme is free to perform another modification. 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_1E\cdot S-(k_{-1}+k_{2})ES \tag{1} \end{equation} \begin{equation} \frac{dE}{dt}=-k_1E\cdot S+(k_{-1}+k_{2})ES \tag{2} \end{equation} \begin{equation} \frac{dP}{dt}=k_2 ES \tag{3} \end{equation} The system cannot be solved analytically. Michaelis and Menten considered a quasi equilibrium between E and the ES-Complex to simplify the system. They assumed that the ES formation is much faster than the E+P formation, for the rate constants this means \begin{equation} k_1,k_{-1}\gg k_2 \end{equation} Briggs and Haldane assumed the so called quasi steady state were the ES concentration is constant. \begin{equation} \frac{dES}{dt}=0 \end{equation} This assumption is reasonable iff (if and only if) the initial substrate concentration is much larger then the enzyme concentration. This assumptions allow to solve for the rate of product formation or the negative decay rate $v$ \begin{equation} v=-\frac{dS}{dt}=\frac{dP}{dt} \end{equation} for this purpose 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_{-1}+k_{2})ES\\ k_1E_{total} \cdot S=ES(S + k_{-1}+k_2)\\ ES=\frac{k_1E_{total} \cdot S}{k_1\cdot S + k_{-1}+k_2}=\frac{E_{total} \cdot S}{ S + (k_{-1}+k_2)/k_1} \tag{4} \end{align} So we get for the reaction rate through inserting (4) into (3) \begin{equation} v=\frac{v_{max} \cdot S}{S + K_m} \hspace{0.5cm} with \hspace{0.5cm} v_{max}=k_2 E_{total} \hspace{0.5cm} and \hspace{0.5cm} K_m=\frac{(k_{-1}+k_2)}{k_1} \text{ ...Michaelis constant} \end{equation} Notice, that the $v$ has the form of a Hill function with Hill coefficient $n=1$

Time Course

The Michaelis Menten Equation for $v$ \begin{equation} S'(t)=-\frac{v_{max} \cdot S}{S + K_m} \end{equation} can be solved analytically since it is a first order autonomous ODE, which are always seperable. \begin{align} \int_{S_0}^{S_t} \left( 1 + \frac{K_m}{S} \right) dS &= \int_0^t -v_{max} dt\\ S_t-S_0 + K_m (\ln S_t - \ln S_0) &= - v_{max} \cdot t\\ \end{align} The implicit solution for the equation reads \begin{equation} K_m \cdot \ln \left(\frac{S_0}{S_t} \right) + S_0 - S_t - v \cdot t =0 \end{equation} The inverse function $W(x)$ is called Lambert function defined as $z=W(z)\cdot e^{W(z)}$ and allows the explicit formulation of $S(t)$ \begin{equation} S(t)=K_m \cdot W\left[ \frac{S(0)}{K_m} \cdot exp\left( \frac{S(0) - v_{max} \cdot t}{K_m} \right) \right] \end{equation}


The following graphic shows a time course of the substrate concentration $S(t)$

Degradation of Substrate $S$ with increasing time. Parameters: $v_{max}=10^{-4}$, $K_m=5 \cdot 10^{-3}$, $S(t=0)=10^{-2}$

Parameter estimation

The Parameters $v_{max}$ and $K_m$ are usually found by measuring the initial reaction rate $v_0$ for various substrate concentrations $S_0$ and linear fitting of the data after transforming. A common method is the Lineweaver-Burk plot, where $1/v_0$ is plottet against $1/[S_0]$, fitting a line with slope $\frac{K_m}{v_{max}[S_0]}$ which is shifted from the origin by $-\frac{1}{K_m}$ to the left and crosses the ordinate at $\frac{1}{v_{max}}$. \begin{equation} \frac{1}{v_0}=\frac{K_m}{v_{max}[S_0]}+\frac{1}{v_{max}} \end{equation}

Lineweaver-Burke.png

Once the parameters are found the reaction rate $v$ can be determined for every Substrateconcentration $[S]$ \begin{equation} v=-\frac{dS}{dt}=\frac{dP}{dt}=\frac{v_{max} \cdot [S]}{[S] + K_m} \end{equation} with Parameters $v_{max}=10^{-6}\mu mol/s$ and $K_m=0,00044mM$ we get the following Michaelis Menten curve

Michaelis Menten Plot with Parameters $v_{max}=10^{-6}\mu mol/s$ and $K_m=0,00044mM$

Notice, for high substrate concentrations $[S]$ the reaction rate is maximal $v=v_{max}$ for $[S]=K_m$ the reactionvelocity is $v=v_{max}/2$

Database

The constants $v_{max}$ and $K_m$ can be looked up at the public data base BRENDA. The rate konstant $k_2$ describes how many substrate molecules are transformed into product molecules per second and active site, it is thus called turnover number.

\begin{equation} v_{max}=k_{2} \cdot E_{total}=k_{cat} \cdot E_{total}\\ \end{equation} However in general an enzyme could be oligomeric with several active sites. If each subunit of the enzyme has $n$ active sites and the holoenzyme consists of $m$ subunits, then $E_{total}$ consists of \begin{equation} v_{max}=k_{2} \cdot m \cdot n \cdot E_{total}\\ \end{equation} where $E_{total}$ is the total number of holoenzymes. Then the $k_{cat}$ is \begin{equation} k_{cat}=k_{2} \cdot m \cdot n\\ \end{equation} $m \dots$ number of subunits
$n \dots$ number of active sites per subunit





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