If I am given the following enzyme reaction:
where K$_{EH,1}$=K$_1$ = $10^{-4}$ and K$_{EH,2}$=K$_2$ = $10^{-8.2}$. Iwant to calculate at which pH I have the optimal kinetics by calculating $\frac{v'_{max}}{v_{max}}$ at different pH levels (this is done in Matlab with the code pH=0:0.1:14; K1=10^-4; K2=10^-8.2; v=1./(1+K2./10.^-pH+10.^-pH./K1); [m,n]=max(v); pH(n)
)
But if the rate equation is:
$v= \frac{k_2 [E]_0 [S]}{K_M(1+ [H+]/K_1 + K_2/[H+]) + [S]}$ (1)
Then how do you get v'$_{max}$?
From a similar problem, the rate equation was:
$v= \frac{k_2 [E]_0 [S]}{[S](1+ [H+]/K_1 + K_2/[H+]) + K_M}$ (2)
So then knowing that v$_{max}$ = $k_2 [E]_0$ when [S]>>K$_M$, you get from eq. (2) that v'$_{max}$ is:
$v'_{max}= \frac{v_{max}}{1+ K_2/[H+] + [H+]/K_1}$
However in my case, by doing this to eq. (1) then I will only be left with $v= k_2[E]_0$ which I can't use in the same way as is done for eq. (2).
Is there another way to calculate the pH or how should I get v'$_{max}$? All help is appreciated!