**From:** Peter Freddolino (*petefred_at_ks.uiuc.edu*)

**Date:** Tue Sep 27 2005 - 17:27:44 CDT

**Next message:**ioana_at_pegasus.arc.nasa.gov: "random seed"**Previous message:**Jason Lee: "NAMD running problem"**In reply to:**Georgios Papadopoulos: "calculate enthalpy of binding"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

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

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