When a ligand and receptor interact with a given kd, one can easily determine % of ligand bound to the receptor when the kd, total concentration of ligand, and total concentration of receptor are known. When there is a mixture of multiple ligands and receptors where each ligand and each receptor can interact with eachother, this is not so simple. For example, lets say you have the following set of information:

$$K_{11} = \frac{[R_1][L_1]}{[R_1L_1]}$$

$$K_{12} = \frac{[R_1][L_2]}{[R_1L_2]}$$

$$K_{21} = \frac{[R_2][L_1]}{[R_2L_1]}$$

$$K_{22} = \frac{[R_2][L_2]}{[R_2L_2]}$$

along with the $K_d$'s ($K_{ij}$) and the total concentrations of $R_1$, $R_2$, $L_1$, and $L_2$. I have not managed to find any publications where this has been demonstrated but I would be surprised if it hasn't been figured out by someone.

  • $\begingroup$ Are you familiar with differential equations? $\endgroup$ Mar 29 '15 at 5:04

The answer is here, but depending on your level of comfort with the math I'm not sure how enlightening it will be. I think that the reason people tend to stick to one ligand/one receptor models is that they capture all the intuition without the tedious algebra.

It's interesting that you are asking for the fraction of ligand bound to receptors ($[RL]/[L]_{tot}$). Typically one is interested in the fraction of receptors that have ligand bound ($[RL]/[R]_{tot}$). This is for two reasons:

  1. Typically the fraction of bound receptor is the physiologically relevant variable for a cell, determining the response of the cells to the ligand stimulus. Since the cell can only sense the concentration of ligand when it binds to the receptor, the fraction of bound receptor is the natural variable to study.
  2. In most contexts, the number of receptors is much smaller than the number of ligands. This motivates the approximation found in pretty much all treatments of receptor-ligand equilibrium that the concentration of receptor-ligand complex is negligible compared to the total concentration of ligand. For cells, this is a fairly reasonable assumption. In this limit, the fraction of ligands bound to receptors ($[RL]/[L]_{tot}$) is always negligible. It's the fraction $[RL]/[R]_{tot}$ that is interesting. Another reason this assumption might be adopted is that in practice, ligands tend to be soluble while receptors are insoluble. In this case, it is easier to control the concentration of ligand, adding it in excess to a pellet of receptors, and then measure the amount of bound receptors (maybe by radioactivity): in this experiment the total concentration of receptors is measured, not controlled, so you want your equation to be a function of your controlled variable, the concentration of ligand. This requires adopting the approximation above.

The situation in which there are two ligands binding to one receptor has been solved (in the limit that $[RL]\ll[L]_{tot}$): this case is known as competitive inhibition. In your notation, the fraction of receptor 1 bound to ligand 1 (ligand 2 is the "competitive inhibitor," i.e. rival ligand):

$$\frac{[RL]}{[R]_{tot}} = \frac{[L_1]}{[L_1]+K_{11}(1+\frac{[L_2]}{K_{12}})}$$

So I'm going to generalize to the case that there are $N$ ligands and I am interested in the fraction of a given receptor $R_i$ bound to ligand $L_j$. The equilibrium constant for this reaction is $K_{ij}\equiv [R_i][L_j]/[R_iL_j]$.

The receptor is also binding to all the other ligands, so that the free concentration of receptor $[R_i]$ is equal to the difference between the total receptor concentration $[R_i]_{tot}$ and the concentration of each bound complex $[R_iL_k]$ for $k$ from 1 to $N$. And the ligand is also binding to all the other receptors. In order to get a reasonable equation, we have to invoke an approximation very similar in spirit to the one mentioned above, namely, that the concentration of all ligand-receptor complexes is small compared to the total concentration of each ligand individually. Thus I can ignore the difference between $[L_j]_{free}$ and $[L_j]_{tot}$.

In this approximation, I expand out the expression for the equilibrium constant (using $[L_j]$ as shorthand to stand for $[L_j]_{tot}$).

$$K_{ij} = \frac{([R_i]_{tot}-[R_iL_j]-\sum_{k\neq j}^{N}[R_iL_k])[L_j]}{[R_iL_j]}=\frac{[R_i]_{tot}[L_j]}{[R_iL_j]}-[L_j]-\sum_{k\neq j}^N\frac{L_kK_{ij}}{K_{ik}}$$

Note that the sum is over $k$ from 1 to $N$ except $k=j$, which I pulled out of the sum early. I also used the definition of each dissociation constant to write the sum in terms of dissociation constants instead of $[R_iL_k]$, which makes the algebra a lot simpler.

This can then be simplified to solve for the fraction of bound receptors:

$$\frac{[R_iL_j]}{[R_i]_{tot}} = \frac{[L_j]}{[L_j]+K_{ij}(1+\sum_{k\neq j}^N\frac{[L_k]}{K_{ik}})}$$

Suppose instead of being interested in the fraction of receptor $i$ bound to a particular ligand $j$, one is interested in the fraction of receptor $i$ bound to any ligand. I'll call this quantity $[R_i\mathbf{L}]/[R_i]_{tot}$. We can figure this out by noting that

$$[R_i\mathbf{L}]\equiv \sum_{k=1}^N[R_iL_k] = [R_i]_{tot} - [R_i]_{free}$$

We obtain $[R_i]_{free}$ from the formula for each $K_d$:

$$[R_i]_{free}\equiv[R_i] = \frac{K_{ij}[R_iL_j]}{[L]_j}$$

And so using our equation for $[R_iL_j]$, we obtain (I'll spare you some of the gory algebra details)

$$\frac{[R_i\mathbf{L}]}{[R_i]_{tot}} = \frac{\sum_{k=1}^N\frac{[L]_k}{K_{ik}}}{1+\sum_{k=1}^N\frac{[L]_k}{K_{ik}}}$$

(these sums are again unrestricted, from $k=1$ to $N$ including $k=j$).

Happily, if there is only one ligand, these formula reduce to the regular result, namely $[L_1]/([L_1]+K_{i1})$. As another check, if all the $K_d$s are equal, then the fraction of receptors bound to any type of ligand is the same as the bound fraction for the same amount of a single type of ligand with the same concentration as the whole pool of different ligands ($\sum_{k=1}^N[L_k]$), and the fraction of receptor bound to a particular ligand $L_j$ is smaller by the fraction of the total ligand pool that is species $j$, $[L_j]/\sum_{k=1}^N[L_k]$.

If you are really interested in the fraction of ligands bound to receptors---that is, you are working in the limit in which there are tons of receptors compared to ligands, so that the number of receptor-ligand complexes is tiny compared to the number of receptors---then you can use the above equations just by switching $L$'s and $R$'s. In this case you would obtain

$$\frac{[R_iL_j]}{[L_j]_{tot}} = \frac{[R_i]}{[R_i]+K_{ij}(1+\sum_{k\neq i}^{N}\frac{[R_k]}{K_{kj}})}\qquad [R_i]\equiv[R_i]_{free}\approx[R_i]_{tot}$$

A caveat

To derive these equations, I made a pretty strong assumption that the total concentration of all receptor-ligand complexes (for different receptors as well as different ligands) is small compared to the total concentration of each ligand individually. In this limit, we can assume that the pools of free ligands of each species are so large that each receptor equilibrates with this pool independently of each other receptor (there's plenty of ligands to go around). You might be interested in the limit in which ligand binding and unbinding to one receptor affects the binding and unbinding to another receptors, so that receptors are not independent. I am not sure how to approach this problem. I'm skeptical that there will be any sort of closed form solution; in my preliminary thinking about this limit, I think I can imagine situations in which stationary equilibria are not possible: depending on rates, receptor A might partially equilibrate, depleting the pool of ligands, affecting the equilibration of receptor B, which would cause receptor B to "pull" more ligands away from A, in some sort of oscillation. However, that is just speculation, and I'd be very curious if someone could provide more insight to this case, in which receptors are able to "interact" through the pool of free ligand.

One of the take-aways of this whole analysis is that it is very helpful to be aware of the assumptions made in standard derivations that lead to the nice results we learn in biochemistry class, and how those assumptions might have to change if we generalize the system somewhat.

  • $\begingroup$ dzjeezz, my math is shaky, and I'm not really capable of upvoting your answer in terms of its content, but heck +1. Nice. Very nice. I really hadn't expected this one to be dealt with any time soon. kudos. $\endgroup$
    – AliceD
    Mar 29 '15 at 10:43

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