From: Peter Freddolino (petefred_at_ks.uiuc.edu)
Date: Tue Sep 27 2005 - 17:27:44 CDT
Dear Georgios,
to answer your questions sequentially:
> -Where did you derive the above formula from?
Consider the full hamiltonian of the bound and unbound systems (ignoring
entropy throughout these discussions because the sampling would be
difficult to do properly).
E(bound) = K(bound) + U(bound)
where the first term is kinetic energy and the second term is potential
energy. Likewise, the unbound system has energy
E(unbound) = K(unbound) + U(unbound)
Assuming that both systems have the same number of particles and are
coupled to a heat bath the kinetic energies will be equal. This means
that the difference in energy betwen the bound and unbound states is
ÄE=U(bound)-U(unbound)
You can then split up the system into subsystems for conceptual purposes
-- this is completely arbitrary and won't change the total as long as
you still account for all interactions within the system. In our case we
choose to consider the proteins and water separately, so the terms become
ÄE=(U(complex) + U(water) + U(complex/water)) - (U(P1) + U(P2) +
U(water) + U(P1/P2) + U(P1/water) + U(P2/water))
Here again, U(x) is the internal energy of x and U(x/y) is the potential
energy of the interaction of x and y; also, U(complex) is equivalent to
(U(P1) + U(P2) + U(P1/P2)). To get the expression I gave before you can
make a couple of assumptions for simplification: Assume that U(water) is
roughly the same in each case (this is the internal energy of the
solvent), and assume that in the unbound state P1 and P2 are far enough
separated to be noninteracting, so U(P1/P2) in the unbound state is
zero. Removing these terms gives the expression
ÄE=(U(complex) + U(complex/water)) - (U(P1) + U(P2) + U(P1/water) +
U(P2/water))
Getting this far is fairly simple; we're just breaking up the total
potential energy of the system in its bound and unbound states to find
the terms that are likely to change. The tricky part is actually getting
these energies, which I'll address in the third point. Let me know if
this still doesn't make sense.
> -How many (which) independent equilibrations do you need?
This is a very important question, but there's no easy answer. The
methods I outlined above are designed to give you the enthalpy of
binding for one bound conformation, and thus should be done for the
"right" binding conformation compared with a single equilibrated
structure in solution. You can also compare several binding
conformations by comparing their binding energies (see, for example,
Vaidehi /et al.* */ PNAS 2002 99: 12622-12627, which focuses on small
ligands binding to protein, but uses a conceptually similar approach).
I'm not familiar with the sampling techniques that would be needed to
calculate the entropy involved in binding, but for this sort of
calculation one bound and one unbound conformation should be enough --
you just need to be sure you have the bound conformation you're
interested in, or else do several and use the binding energy to score
them.**
> -Can you please explain how do you use the results of the simulations to
> calculate the above energies?
All of these energies can be calculated by using the pairInteraction
lkeyword in NAMD
(http://www.ks.uiuc.edu/Research/namd/current/ug/node37.html); you can
do internal energies by specifying one group (and self energies), and
interactions by specifying two.
So, the typical picture (at least from my experience) is generating one
or more bound conformations, one good unbound conformation, and then
calculating all the required energies using pairInteractions, and
calculating the enthalpy of binding from that. Please let me know if
this is still unclear.
Best, Peter
This archive was generated by hypermail 2.1.6 : Wed Feb 29 2012 - 15:41:10 CST