# How to derive Hill equation (one specific part)

There is just one specific step in the derivation of the Hill equation for haemoglobin which I can't understand.

Step from: $$Y = \frac{(p\ce{O2})^n}{K_d + (p\ce{O2})^n}$$
To: $$Y = \frac{(p\ce{O2})^n}{(K_d)^n + (p\ce{O2})^n}$$

I don't understand where $$(K_d)^n$$ comes from. (I can go from this second equation to the Y/(1-Y) equation afterwards).

How I derived my equation is the following.

$$\ce{Hb.(O2)_n <=> Hb + nO2}$$

Hence the dissociation constant $$K_d$$ should be:

$$K_d = \frac{[\ce{Hb}][\ce{O2}]^n}{[\ce{Hb.(O2)_n}]}$$

And Y, which is the fractional $$\ce{O2}$$ saturation would be:

$$Y = \frac{[\ce{Hb.(O2)_n}]}{[\ce{Hb.(O2)_n}] + [\ce{Hb}]}$$

If you multiply the top and bottom by $$K_d$$, the following is obtained:

$$Y = \frac{(p\ce{O2})^n}{K_d + (p\ce{O2})^n}$$

This is where I became stuck. What am I doing wrong?

In your second step $$(K_d)^n$$ should be $$(K_A)^n$$.
• $$K_d$$ is the apparent dissociation constant and
• $$K_{A}$$ is the ligand concentration that results in half occupation