pH for optimal kinetic for enzyme reaction

If I am given the following enzyme reaction:



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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!




I don't think the rate equation is



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> $v= \frac{k\_2 [E]\_0 [S]}{K\_M(1+ [H+]/K\_1 + K\_2/[H+]) + [S]}$
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I suspect it is rather:


$$v= \frac{k\_2 [E]\_0 [S]}{(K\_M + [S]) (1+ [\ce{H+}]/K\_1 + K\_2/[\ce{H+}])}$$


The term $(1+ [\ce{H+}]/K\_1 + K\_2/[\ce{H+}])^{-1}$ describes the fraction of the enzyme in the productive protonation state.


You don't have to plot this to figure out the optimal pH. The optimal pH is always at the midpoint between the two $\mathrm{p}K\_\mathrm{a}$ values. That is because that is the pH value with the highest fraction of the once-protonated enzyme.


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