## Related questions with answers

The Michaelis-Menten equations describe a biochemical reaction in which an enzyme E and substrate S bind to form a complex C. This complex can then either dissociate back into its original components or undergo a reaction in which a product $P$ is produced along with the free enzyme: $\mathrm{E}+\mathrm{S} \leftrightarrow \mathrm{C} \rightarrow \mathrm{E}+\mathrm{P}$This can be expressed by the differential equations

$\begin{aligned} &\frac{d x}{d t}=-k_{f} x y M+k_{r}(1-y) M\\ &\frac{d y}{d t}=-k_{f} x y M+k_{r}(1-y) M+k_{\mathrm{cat}}(1-y) M\\ &\frac{d z}{d t}=k_{\mathrm{cat}}(1-y) M \end{aligned}$

Where $M$ is the total number of enzymes (both free and bound), $x$ and $z$ are the numbers of the substrate and product molecules, $y$ is the fraction of the enzyme pool that is free, and the $k_i's$ are positive constants. Calculate the Jacobian matrix.

Solution

VerifiedThe goal is to find the Jacobian matrix.

The Jacobian matrix of the differential system of equations has the form:

$\begin{align} J\left(x, y\right)&=\begin{bmatrix} \frac{\partial f_{1}\left(x, y\right)}{\partial x}& \frac{\partial f_{1}\left(x, y\right)}{\partial y}\\ \frac{\partial f_{2}\left(x, y\right)}{\partial x}& \frac{\partial f_{2}\left(x, y\right)}{\partial y} \end{bmatrix} \end{align}$

To find the Jacobian matrix, compute the partial derivatives of the following functions:

$\begin{aligned} f_{1}\left(x, y\right)&=-k_{f}xyM+k_{r}(1-y)M\\ f_{2}\left(x, y\right)&=-k_{f}xyM+k_{r}(1-y)M+k_{cat}(1-y)M \end{aligned}$

with respect to $x$ and $y$. We can then substitute those into Eq. $(1)$ to obtain the Jacobian matrix.

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