correlation of IC50 to receptors/enzyme molecules

Dr Engelbert Buxbaum engelbert_buxbaum at hotmail.com
Mon Feb 9 10:27:56 EST 2004

interpreneur_org at yahoo.com wrote:

> In a binding or kinetic assay, how does the IC50 of a compound change
> with respect to the number of receptors/enzyme molecules available? Is
> it linear (e.g: 10 times more protein leads to a 10 fold increase in the
> IC50)? Or not? And what is the mathematical equation explaining this?

Maximal binding (or enzymatic activity) depends on the substrate
concentration, Kd (or Km) does not, at least if the concentration of
protein molecules is much smaller than that of the free ligand F:

B = Bmax * F / (Kd + F)

However, if we use the usual approximation of F = T and increase the
protein concentration to more than 0.1*F, a change in _apparent_  Kd
will be observed, because binding of the ligand to the protein will
significantly reduce the concentration of free ligand. If you replace
the free ligand concentration F by (T-B) (total minus bound ligand) and
solve the resulting quadratic equation, the resulting "Langmuir
isotherm" will correctly describe binding under these conditions, and
you will see that the _true_ Kd is still independent of protein

   B  = Bmax * F / (K_d + F) = Bmax * (T-B) / (K_d + T - B) 

Separation of variables yields:

   0  = -B^2 + B*(K_d + T + Bmax) - Bmax * T

which is a quadratic equation in standard form. The solution is:
   B  = -1/2 * (-(K_d + T + Bmax) + sqrt{(K_d + T + Bmax)^2 - 4* Bmax *

Note that of the two solutions of the quadratic equation only the one
given here is physically meaningfull, as there is no such thing as a
negative concentration.

This consideration is of course important only in binding studies, in an
enzymatic assay the protein concentration is virtually always much lower
than that of the substrate (otherwise measuring initial velocities would
become technically very demanding). 

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