I have beads in solution at concentration $B$ which have $N$ independent binding sites for a single target molecule at concentration $T$. $T$ is small compared to $B$, so most beads will probably not have anything bound at all. The on and off rates for a single target binding a single site on the bead are $k_{on}$ and $k_{off}$.
I want to determine the equilibrium distribution of the number of bound targets per bead. I expect it to be a Poisson distribution (right?) with an average binding count that depends on the parameters above.
Define $B_n$ to be the concentration of beads with $n$ bound targets. First order kinetics suggest we should have the coupled system of equations:
$\frac{dB_n}{dt}=k_{on}T(B_{n-1}-B_n)+k_{off}(B_{n+1}-B_n)$,
$\frac{dT}{dt}=\sum_{n=0}^{N-1}\left(-k_{on}TB_n+k_{off}B_{n+1}\right)$
with $B_{-1}\equiv 0\equiv B_{N+1}$
First question: is there an analytical solution for this system of equations? I am perfectly happy to simulate it, but if there is a way to avoid that, it would be nice.
Second question: the wikipedia page for Dissociation Constant has some ambiguous wording suggesting that even if the receptors are all identical, the rate constants will be dependent on $n$ as well, via
$K_d(n) =K_d \frac{n}{N-n+1}$
where $K_d=\frac{k_{off}}{k_{on}}$
How does that play here? Are my equations in need of modification, to something like:
$\frac{1}{k_{on}}\frac{dB_n}{dt}=T(B_{n-1}-B_n)+K_d \frac{n}{N-n+1}(B_{n+1}-B_n)$.
If so, could someone explain where that modification comes from?
EDIT: it seems that this is probably due to the fact that there is a higher change of dissociation if a bead has many targets bound since there is an equal probability for each target to unbind. However, it seems that the fraction presented is 1 for $n=N$, which suggests that the dissociation constant is defined in terms of a fully bound bead and not on a per-site basis. Could someone show me the derivation?