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?

Thank you very much for your answer.


1 Answer 1


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

See for example the relevant wikipedia article.

  • $\begingroup$ @user1136 Apparent because the association/dissociation reaction is not a single step kinetics. $\endgroup$
    Jan 13, 2020 at 10:53

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