Hi Tom,
----- Original Message -----
From: "Thomas Cheatham" <tec3.utah.edu>
> > In this regard, the term "standard state" refers to the concentration of
> > binding ligand. Of course I am looking at larger sets of reactions (as
in
>
> That is not the definition of standard state.
>
The term "standard state" in this context is referring to the concentration
of binding ligand as defined in accordance with enzyme kinetic theory. In a
multistep kinetic mechanism, as shown below.
E + A --> EA --> EB --> E + B
.the difference in energy between A and B translates in to Keq for the
enzyme reaction in question. However, in simulating the kinetic constants
of this reaction using first principles & thermodynamics, we need to treat
each step sequentially. Hence, our first step would be to model E+ A -->
EA, and so forth.
Were I to model the E+A --> EA step using 1M concentrations, I know for a
fact that these "standard state" concentrations will not reflect the system
in equilibrium, which is what is needed in order to estaimte the kinetic
constants for the system using rate coefficients and the methods of King &
Altman. Hence, as described by Cleland, the problem is the choice of
standard states.
We have developed a protocol to allow convergence to equilibrium through
adjustments in standard state concentration of reactant and product. It
involves an advanced treatment of solvent using mm-pbsa to elucidate the dG
of the "transition states" (here, binding intermediates). But to implement
it, we need to be able to *choose* the concentration of binding ligand.
David Case went over this previously. My big question for the day is, can
we arbitrarily assign a concentration for A in the reaction E+A --> EA?
(i.e., something other than 1M)? Can this be done in Amber? If it can,
then we have an entire method that will compute enzyme kinetic constants
using mm-pbsa with modifications of the solvent dielectric (there is more
talk of this on the new bioblade site that I am hosting on yahoo).
Can we set the "standard state" from 1M to something else? That is all I
need to know for now.
Thanks in advance,
Sean
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Received on Wed Oct 25 2006 - 06:07:17 PDT
