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let's suppose the following reaction:
enter image description here
At equilibrium we can calculate the Kd (Dissociation constant) using the following formula:
Kd = [P][L] / [PL] or kd / ka
We can then use the Hill equation to calculate the fraction of binding sites which are occupied:
occupied binding sites = [L] / [L] + Kd

When [L] euals Kd, half of the ligand binding sites are occupied (Lehninger principles of biochemistry)

Why is this true? This would mean that at equilibrium half of the binding sites are occupied?

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As the other answer says, it is true for a simplified model. The model you describe has only one ligand binding site per protein, which makes it the most simple model there is.

It doesn't mean that this is the case at equilibrium as you asked. This is because Kd is not necessarily equal to [L] at equilibrium. Hopefully this simple derivation will help:

As you said, we know the formula for the dissociation constant: Kd = [P][L]/[PL]

Suppose [L] = Kd. If we plug back into the equation above we get: [L] = [P][L]/[PL]

Doing some algebra we can derive: [P] = [PL]

Therefore, at equilibrium, when [L] = Kd, the amount of free protein ([P]) and the amount of ligand-bound protein ([PL]) are equal. In other words, half of the protein is bound to ligand, which is exactly what you were asking about.

But now let's suppose that [L] =/= Kd. The reaction will still come to equilibrium, finding concentrations of [P] and [PL] that satisfy the equation Kd = [P][L]/[PL]. However, these values will not be equal. Either more or less than half of the protein will be bound to ligand.

Only in the specific case of [L] = Kd will half of the protein be bound to ligand when equilibrium is reached.

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The Hill equation in that link uses the simplifying assumption that all ligands bind simultaneously and that the binding sites are identical. The reaction is then P + nL = PL, and Kd = ( [P][L]^n )/[PL] Then by rearranging you get the expression in the link.

The equation "fraction occupied binding sites = [L] / [L] + Kd" Is not the Hill equation. It is called the binding isotherm. You can see that if Kd = [L] then the fraction of occupied binding sites (let's call it theta) = 1/2. Thus half of the binding sites are occupied.

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  • $\begingroup$ It is the Hill equation where the Hill coefficient is 1, which kind of defeats the purpose of using the Hill equation. $\endgroup$ – stords Dec 28 '16 at 4:14

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