# solve steady state of simple product reaction

consider these biochemical reactions (simpler than usual michaelis-menten setup):

$E + S \rightarrow^{k_{f}} ES$

$ES \rightarrow^{k_{r}} E + S$

$E$ and $S$ reversibly make $ES$. the forward/reverse rate constants are equal: $k_{f} = k_{r} = k$. how can you solve for steady state concentration of $[ES]$ given initial amounts $[E]_{0}, [S]_{0}$ and assuming $[ES]_0 = 0$?

$d[ES]/dt = k[E][S] - k[ES] = 0$

rate constant cancels:

$[ES] = [E][S]$

then $[E] = E_{t} - [ES]$ where $E_{t}$ is total enzyme concentration. by substitution:

$[ES] = [S]([E]_{t} - [ES])$

$[ES] = [S][E]_{t} - [ES][S]$

$[ES] + [ES][S] = [S][E]_{t}$

$[ES] = \frac{[S][E]_{t}}{1 + [S]}$

but if $E_{0} = 10, S_{0} = 5$ then $[ES] = (5 * 10) / (5 + 1) \approx 8.3$ but the right answer is closer to 4. what's wrong with this?

• You considered $[E]+[ES]=E_0$. I think you also need $[S]+[ES]=S_0$? Another weird thing is the enzyme doesn't convert the substrate to any product. – Zhuoran He Jan 14 '18 at 19:00
• it's not weird not to have a product. there is no product in this system. can you explain how substituting [S]_0 = [S] + [ES] helps? – user8379674 Jan 14 '18 at 20:14

There are $3$ equations for $3$ unknowns $[E],[S],[ES]$ given by
$$\left\{\begin{array}{l} k_f[E][S]=k_r[ES],\\ [E]+[ES]=E_0,\\ [S]+[ES]=S_0. \end{array}\right.$$
We can use the last two equations to cancel $[E]=E_0-[ES]$ and $[S]=S_0-[ES]$ and plug them into the first equation to get a quadratic equation for $[ES]$ given by
$$k_f(E_0-[ES])(S_0-[ES])=k_r[ES].$$
Then we plug in numbers $k_r/k_f=1$, $E_0=10$, $S_0=5$ to obtain $[ES]=8\pm\sqrt{14}$. Since we have $[ES]\leq\min\{E_0,S_0\}=5$, the solution is $[ES]=8-\sqrt{14}\approx 4.26$.