📂Dynamical Systems

Michaelis–Menten Model

Model¹

The following governing equation, which describes the reaction between an enzyme and a substrate $x$, is called the Michaelis–Menten model. $$ \dot{x} = j_{x} - {\frac{V_{\max} x}{K_{m} + x}} $$

Parameters

• $j_{x}$: the influx rate of the substrate $x$.
• $V_{\max}$: the maximum reaction rate of the enzyme.
• $K_{m}$: a constant called the Michaelis–Menten constant, representing the affinity of the enzyme.

Explanation

The Michaelis–Menten model describes the most basic dynamical system of enzyme kinetics.

• I have never studied its biological meaning, but just from the formula one can easily guess that it means the reaction proceeds quickly when $x$ is small, and saturates and slows down as $x$ becomes large.
• In dynamics research, despite being $1$-dimensional, it has strong nonlinearity due to its fractional term, so it is used in various benchmarks and shows up surprisingly often².

Derivation

Rather than its intuitive meaning, I was curious how the right-hand side ends up in fractional form, so I looked into it. $$ E + S \overset{k_{1}}{\underset{k_{-1}}{\rightleftharpoons}} ES \overset{k_{2}}{\longrightarrow} E + P $$ Suppose, as above, a reaction in which an enzyme $[E]$ and a substrate $[S]$ bind to form an enzyme–substrate complex $[ES]$, which then decomposes back into the enzyme and a product $[P]$. By the law of mass action, the concentrations can be expressed by the simplest differential equation as follows. $$ {\frac{ d[ES] }{ dt }} = k_1 [E][S] - k_{-1}[ES] - k_2 [ES] $$ Solving for $[ES]$ at the fixed point $d[ES]/dt = 0$ of this system, it can be written simply as follows using the constant $K_{M} := (k_{-1} + k_2) / k_1$. $$ [ES] = \frac{[E][S]}{K_{M}} $$ The initial enzyme concentration $[E]_{0}$ does not change even when the enzyme binds with the substrate to form the complex, so we can set $[E]_{0} = [E] + [ES]$. Since we are interested not in $[E]$ but in the substrate $[S]$, we will manipulate the equation in a way that eliminates $[E]$. $$ \begin{align*} K_{m} [ES] = & [E] [S] \\ \implies K_{m} [ES] =& ([E]_{0} - [ES]) [S] \\ \implies \left( K_{m} + [S] \right) [ES] =& [E]_{0} [S] \\ \implies [ES] =& \frac{[E]_{0} [S]}{K_{m} + [S]} \end{align*} $$ Since the reaction $ES \to P$, that is, $k_{2} [ES]$, is the reaction rate, multiplying both sides by $k_{2}$ gives the following for $V_{\max} := k_{2} [E]_{0}$. $$ k_{2} [ES] = \frac{k_{2} [E]_{0} [S]}{K_{m} + [S]} = \frac{V_{\max} [S]}{K_{m} + [S]} $$ Finally, replace the left-hand side with $\dot{x}$ and $[S]$ on the right-hand side with $x$.

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