1
$\begingroup$

If I am given the following enzyme reaction:

enter image description here

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!

$\endgroup$

1 Answer 1

1
$\begingroup$

I don't think the rate equation is

$v= \frac{k_2 [E]_0 [S]}{K_M(1+ [H+]/K_1 + K_2/[H+]) + [S]}$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.