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I have a ligand $\ce{L}$ and a receptor $\ce{R}$. As can easily be shown, with dissociation constant $K_\mathrm{D} = \frac{[\ce{L}][\ce{R}]}{[\ce{LR}]}$ the fraction of receptors occupied by ligand is

$$\theta = \frac{[\ce{LR}]}{[\ce{LR}] + [\ce{R}]} = \frac{[\ce{L}]}{[\ce{L}] + K_\mathrm{D}}$$

It's easy to see that the fractional saturation of receptors is independent of receptor concentration. Now here's where I get confused. Let's say we have a constant volume of $\pu{1 L}$ with $\pu{5 mol}$ ligand and a $K_\mathrm{D}$ of 5:

$$L = 5, \quad K_\mathrm{D} = 5 \quad \to \quad \theta = 0.5$$

If we have $\pu{10 mol}$ of receptor total ($[\ce{LR}] + [\ce{R}] = 10$), then $\pu{5 mol}$ will be bound to ligand. Now let's say we increase the number of receptors by $10\times$. Our formula for $\theta$ is independent of $[\ce{R}]$ or $[\ce{LR}]$. We now have $\pu{50 mol}$ receptor bound to ligand.

That being said, we only had $\pu{5 mol}$ ligand to start with, which is itself a paradox. Can someone explain to me where I went wrong?

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    $\begingroup$ What do you mean by Kd=5; 5M? $\endgroup$ – WYSIWYG Aug 24 '15 at 5:33
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I see where I made my mistake. I set $[\ce{L}] = 5$, but this "$L$" is referring to the equilibrium concentration of free ligand -- not the total ligand concentration. In order to do that I should have constrained the system to $[\ce{L}] + [\ce{LR}] = 5$. Working through what this would do currently.


So in the first case, my assumption actually dictated that there be $\pu{10 mol}$ ligand total instead of the $\pu{5 mol}$ I erroneously stated. I incorrectly equated total ligand and free ligand the first time I did this.


So I see where I messed up, but now I'm curious if the values of $[\ce{L}]$, $[\ce{R}]$, and $[\ce{LR}]$ can be solved as functions of $[\ce{L_{total}}]$, $[\ce{R_{total}}]$ and $K_\mathrm{D}$.

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    $\begingroup$ If you still have queries then you should add this as an edit to your question. $\endgroup$ – WYSIWYG Aug 24 '15 at 5:39
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    $\begingroup$ Three unknowns (L, R, and LR) per three equations (total L, total R, and the dissociation constant) - should be solvable, don't you think? $\endgroup$ – Ivan Neretin Sep 23 '15 at 7:28

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