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

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.

• Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists. Oct 5 '21 at 10:38

$$V_0=V_\mathrm{max}\frac{[S]}{[S]+K_M}$$
you divide by $$V_\mathrm{max}$$ to get
$$\frac{V_0}{V_\mathrm{max}} = \frac{[S]}{[S]+K_M}$$