The Michaelis-Menten kinetics reaction is given by $$ E+S \frac{k_{3}}{k_{1}}+E S \underset{k_{2}}{ } E+P $$ The resulting system of equations for the chemical concentrations is $$ \begin{aligned} \frac{d[S]}{d t} &=-k_{1}[E][S]+k_{3}[E S] \\ \frac{d[E]}{d t} &=-k_{1}[E][S]+\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[E S]}{d t} &=k_{1}[E][S]-\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[P]}{d t} &=k_{3}[E S] \end{aligned} $$ In chemical kinetics, one seeks to determine the rate of product formation \(\left(v=d[P] / d t=k_{3}[E S]\right)\). Assuming that \([E S]\) is a constant, find \(v\) as a function of \([S]\) and the total enzyme concentration \(\left[E_{T}\right]=[E]+[E S] .\) As a nonlinear dynamical system, what are the equilibrium points?

Step by step solution

01

Determine v as a function of [S] and [E_T]

Assuming that [ES] is constant, we can replace [E] in the expression for \(v = k_{3}[ES]\) with \(E = E_{T} - [ES]\). If we recall that \(\frac{d[ES]}{dt} = k_1[E][S] - \left(k_2 + k_3\right)[ES]\), and solve for [ES] to get \([ES] = \frac{k_1[E][S]}{k_2 + k_3}\), we can substitute \([E] = E_{T} - [ES]\) into this equation and rearrange to give \([ES] = \frac{k_1E_{T}[S]}{k_2 + k_3 + k_1[S]}\). Substituting this back into the expression for v then gives \(v = \frac{k_1k_3E_{T}[S]}{k_2 + k_3 + k_1[S]}\).

02

Finding the Equilibrium Points

At equilibrium, the rates of change of [S], [E], [ES] and [P] are all zero. Setting the given differential equations to zero will provide a system of equations that can be solved for equilibrium concentrations for [S], [E], [ES] and [P]. These concentrations form the equilibrium points of the system.

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