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?

In article <[Only registered users see links. ]>,
<[Only registered users see links. ]> wrote:
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?

Does the affinity/dissociation constant of the receptor/enzyme
change with the concentration of the receptor/enzyme? That is, as you add
more or less receptor/enzyme, does the binding energy between a receptor
molecule and an inhibitor molecule change?
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Lou Hom >K'93 [Only registered users see links. ] [Only registered users see links. ]

it has been more than 10 years I passed an exam on enzyme kinetics, and
that was not my favorite topic... But common sense dictates that the
situation will very much depend on the mechanism of inhibition, relative
concentrations of the enzyme, substrate and inhibitor, and the
stoichiometry of binding. If it is a simple competetive inhibition, and
the stoichiometries of the E-S and E-I complexes are same, and you have
a classic case of E<<S, then, I think IC50 will solely depend on
relative concentrations of S and I.

On the other hand if E is not << I and you have a sort of inhibitor that
irreversibly binds your E and inactivates it (forming a dead-end
complex), for example, then its IC50 will of course directly depend on
concentration of the enzyme (E).

what kind of inhibitor is it?

otherwise, look into some good enzymology book, like Fersht's Enzyme
structure and mechanism (I remember everybody loved it).

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
concentration:

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 *
T})

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).