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From: Julien Michel <j.michel.soton.ac.uk>

Date: Fri, 01 Jun 2007 17:48:55 +0100

Hi Sean,

The correct way to write Keq is

Keq = ( |AB|/|C0| ) / [ ( |A|/|C0| ) * ( |B|/|C0| ) ]

which you can also rewrite as

Keq = ( |AB| / (|A|*|B|) ) * ( ( |C0|*|C0| ) / |C0| )

and so Keq in 2 and 3 are unitless once the standard state is considered.

In an absolute binding free energy calculations using thermodynamic

integration or related methods you have to correct the computed free

energy of "decoupling" your ligand from a binding site by a term that is

proportional to the ratio of the volume your decoupled ligand could

occupy in the simulation to the standard state volume.

See for instance Boresch JPCB, 2003, 107, 9535-9551 or Gilson Biophys J

1997, 72, 1047-1069.

It is not clear to me at which stage a standard state correction would

occur in a MMPBSA calculation. That should be less of an issue if you

are interested in ranking ligands or differences in equilibrium

constants rather than their absolute values.

Julien

Sean Rathlef wrote:

*> Dear Amber,
*

*>
*

*> A quick question regarding Keq computations via MMPBSA. I have an
*

*> association reaction (A + B --> AB). I am modeling the free energies
*

*> for AB, A, and B, and generating my deltaG from:
*

*>
*

*> 1 delG(AB) - [delG(A) + delgG(B)].
*

*>
*

*> Keq is then computed via:
*

*>
*

*> 2 Keq = e^(-delG/RT).
*

*>
*

*> In this instance, Keq will have units of M-1, because of the
*

*> relationship to the concetrations of the species, i.e.,
*

*>
*

*> 3 Keq = [AB] / [A]*[B]
*

*>
*

*> The Keq leveraged from equation 2 however is unitless, so would it be
*

*> appropriate to simply divide this value by standard state concentraion,
*

*> C_0, of 1M?
*

*>
*

*> Thanks for any responses,
*

*> Sean
*

*>
*

Date: Fri, 01 Jun 2007 17:48:55 +0100

Hi Sean,

The correct way to write Keq is

Keq = ( |AB|/|C0| ) / [ ( |A|/|C0| ) * ( |B|/|C0| ) ]

which you can also rewrite as

Keq = ( |AB| / (|A|*|B|) ) * ( ( |C0|*|C0| ) / |C0| )

and so Keq in 2 and 3 are unitless once the standard state is considered.

In an absolute binding free energy calculations using thermodynamic

integration or related methods you have to correct the computed free

energy of "decoupling" your ligand from a binding site by a term that is

proportional to the ratio of the volume your decoupled ligand could

occupy in the simulation to the standard state volume.

See for instance Boresch JPCB, 2003, 107, 9535-9551 or Gilson Biophys J

1997, 72, 1047-1069.

It is not clear to me at which stage a standard state correction would

occur in a MMPBSA calculation. That should be less of an issue if you

are interested in ranking ligands or differences in equilibrium

constants rather than their absolute values.

Julien

Sean Rathlef wrote:

-- Julien Michel http://www.julienmichel.net Dr J. W. Essex research group University of Southampton United Kingdom ----------------------------------------------------------------------- The AMBER Mail Reflector To post, send mail to amber.scripps.edu To unsubscribe, send "unsubscribe amber" to majordomo.scripps.eduReceived on Sun Jun 03 2007 - 06:07:38 PDT

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