# Calculate the flow rate in the steady state of a dynamic equilibrium

I have the following situation:

I am interested of the reactions rate of an enzymatic reaction, depending of the substrate concentration. The thing is, that I don't add the substrate directly to the enzymes; the substrate has first to pass (I think diffuse) through a membrane.

I assume an infinite amount of substrate (outside of the membrane), so it won't run out. Further I assume that the product of the enzymatic reaction is eliminated immediately.

$Substrate_{outside} \leftrightarrow Substrate_{inside} \rightarrow Product$

The first reaction ($\leftrightarrow$) is k1 and the second ($\rightarrow$) is k2 (I didn't figure out how to name the arrows, sorry)

k1 is the diffusion trough the membrane which I think should be proportional (fick's law) to the difference of $Substrate_{outside} - Substrate_{inside}$, so it is:

$k1 = D * Substrate_{outside} - Substrate_{inside}$, where $D$ is some constant.

k2 is Michaelis-Menten:

$k2 = \frac{V_{max} * Substrate_{inside}}{K_m +Substrate_{inside}}$

when I solve eq. 1 to $Substrate_{inside}$ and insert it to eq. 2, I got :

$k2 = \frac{V_{max} * (D *Substrate_{outside} -- k1)}{K_m + D *Substrate_{outside} -- k1}$

in the steady state is k1 = k2 so:

$k = \frac{V_{max} * (D *Substrate_{outside} -- k)}{K_m + D *Substrate_{outside} -- k}$

but if I solve it, I get some strange function, which does not looks like I think it should look.

In short words: I am looking for $k(Substrate_{outside},D,K_m,V_{max})$, whereby k is the speed of the reaction.

I am not really into this topic, so sorry if I was unclear or not 100 % correct. Thanks for your help.