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Modeling Symmetric Dimers
Cellucidate Support Dec 07, 2009
Simulations on Cellucidate are stochastic or rule-based. Here, the Stochastic Simulation Algorithm convention is used to model the binding of agents.
For the binding rate of two different agents A and B in a stochastic simulation:
A!/[(A-B)!B!]
For the binding rate of two of the same agents A in a stochastic simulation:
A!/[(A-2)!2!] = [(A)(A-1)]/2
As shown, the rate of binding of the same two agents is divided by two. Whereas in a simulation of ordinary differential equations, the binding rate for two different agents is A^B and for two of the same agents is then A^2.
Therefore, the rate of homodimerization or symmetrical binding reaction of two of the same agents must be doubled to compensate for the convention of stochastic simulations.
Comments
Hi!
What does cellucidate do in the following situation:
1. An agent R, representing a receptor, can dimerize at a (homodimerization) rate of kf.
2. R can be phosphorylated, and assume this does not affect the kinetics of the dimerization.
3. The dimerization rule is written such that the phosphorylation state is "don't care".
So the model is more or less (hope I have this right):
R(d,t),R(d,t) -> R(d!1,t)R(d!1,t) # at rate kf, independent of phosphorylation state
R(d,t~U),K->R(d,t~P),K # phosphorylation (non-MM kinetics)
R(d,t~P),P->R(d,t~U),P # de-phosphorylation (non-MM kinetics)
Logically, Kappa should "know" that for the heterodimerization, the rate constant is double.
Note that anything that breaks the symmetry (e.g. binding a ligand) leads to the same question.
Regards,
Julien
Ps. There is also a simple proof for the ODE situation.
Suppose that a receptor R homodimerizes at a rate kf and heterodimerizes with a phosphorylated version of itself, Rp, at rate kf'. Assume also that phosphorylation has a negligible effect on the kinetics. I.e. it merely "tags" the receptor.
In this case, the reaction velocity (v) of the dimerization should be the same independent of the fraction x of a total amount of receptor Rtot in the R form. By the law of mass action, the reaction velocity is
v = kf * (x * Rtot)^2 + kf * ((1-x)*Rtot)^2 + kf' * x * Rtot * (1-x) * Rtot
v = (2kf - kf') x^2 Rtot^2 - (2 kf - kf') x Rtot^2 + kf * Rtot^2
That this result is independent of the value of x implies
kf' = 2 kf
Which of course gives v = kf * Rtot ^2 as expected.
Hi Julien - I am not sure I understand your question; if the rate of dimerisation of R is insensitive to the internal state of R, then the engine will just dimerise at rate kf * R(d)^2 which indeed is independent of x = fraction of R in the ~p form. Looking at your ODE proof, I think you might be thinking of an "unfolding" of the original rule into 3 subcases:
R(d,t~u),R(d,t~u) -> R(d!1,t~u),R(d!1,t~u)
R(d,t~u),R(d,t~p) -> R(d!1,t~u),R(d!1,t~p)
R(d,t~p),R(d,t~p) -> R(d!1,t~p),R(d!1,t~p)
if you perform the above decomposition of the original rule, then the rate of the middle reaction kf' has to be twice kf if you want the system to dimerize at the same speed as the original one - is that what you had in mind?
such questions are unavoidable when you try to refine/unfold rule sets; including the important case of the ground expansion into a system of reactions (aka a Petri net); we have written two papers on how to organise this best, and what constitutes a good refinement.
Cheers
Vincent