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I am provided a one-dimensional reaction system showing zero-order ultra sensitivity.

$$\ce{X_1 ->[f_1] X_2}, \quad \quad \quad \ce{X_2 ->[f_2] X_1}$$

The reaction rate function is the reduced enzyme reaction according to Michaelis-Menten's law.

$$f_1(x) = E\left(\frac{x_1}{K_1+x_1}\right)$$

$$f_2(x) = \left(\frac{x_2}{K_2+x_2}\right)$$

$$K_1=K_2=K$$

$$x_1(t) + x_2(t) = x_0$$

where $x_0$ is a constant. Let the steady state of $x_2$ be denoted by $x_2^{\text{st}}$. I need to find the value of $E$ that satisfies

$$x_2^{\text{st}}(E) = \frac{x_0}{2}$$

and the logarithmic sensitivity. I tried to look into some published papers/textbooks but could not find similar problems. Can anyone give some suggestions on directions to look into?

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  • $\begingroup$ Why not write $\ce{X_1 <=>[f_1][f_2] X_2}$ instead? $\endgroup$ – Rodrigo de Azevedo Jun 28 at 16:36
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I found the answer.

First, we need to find balanced equation for reaction. Then express the concentration of $xˆst$ in the steady state through E.

Then we can arrive to this kind of solution for the value of E.

$\displaystyle{E\frac{x_0 - \frac{x_o}{2}}{K + x_0 - \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} = 0.$

$\displaystyle{E\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}}= 0. $

E = 1

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