2
$\begingroup$

I'm working my way through the chapter on enzymes in "Biochemistry" with Stryer et. al, 9th edition. In the current paragraph it's claimed that the fraction of active sites bound to substrate in an enzyme can be approximated by: $$f_{ES}=\frac{V}{V_\mathrm{max}}$$

And fair enough, that sounds reasonable. It also claims, however that:

$$f_{ES}=\frac{[S]}{[S]+K_M}$$

This i can't quite grasp. Why would this be the case?

I tried to arrive at this expression by rewriting $V_0=V_\mathrm{max}\frac{[S]}{[S]+K_M}$ in terms of $V_\mathrm{max}$ and then dividing the expression for $V_0$ with the expression for $V_\mathrm{max}$. So far, no such luck.

$\endgroup$
1
  • 1
    $\begingroup$ Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists. $\endgroup$
    – Poutnik
    Oct 5 '21 at 10:38
3
$\begingroup$

If you already know that

$$V_0=V_\mathrm{max}\frac{[S]}{[S]+K_M}$$

you divide by $V_\mathrm{max}$ to get

$$\frac{V_0}{V_\mathrm{max}} = \frac{[S]}{[S]+K_M}$$

Both sides of the equation tell you what fraction of active sites is occupied by substrate.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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