Subject: Ionizable residues affect the free energy calc.?

From: Masakatsu Ito <m-ito_at_jp.fujitsu.com>
Date: Tue, 04 Jan 2005 23:48:19 +0900 (JST)

Dear CCLers,

I am wondering if an ionizable residue affects the accuracy of free
energy calculation. 

  We are trying to calculate the binding free energies of peptides to
their receptor molecule using the 'double coupling' method (*) which
requires two simulations. The first yields the free energy change, G1,
for a transfer of a protonated peptide molecule, PH(+), from water to
gas.

    PH(+)(aq) + Cl(-)(aq) -> PH(+):Cl(-)(gas)

One chloride ion is added to neutralize the system.
  The second yields the free energy change, G2, for decoupling the
peptide from the binding site of the solvated receptor, R(-)(aq) .

    R(-):PH(+)(aq) + Na(+)(aq) + Cl(-)(aq) -> 
             R(-)(aq) + Na(+)(aq) +  PH(+):Cl(-)(gas)

Here a sodium ion is added to neutralize the system.
  The binding free energy is then G2 - G1 .

  But the problem is that the calculation would be more complicated if
we consider all the charge configurations of the solvated peptide,
PH(+)(aq) and P(aq). Because the protonated residue is arginine so
far, I suppose the deprotonated configuration does not have to be
considered. But this approximation is bad for lysine or histidine, and
how can I estimate the error? Or how can I resolve the problem?

  I would be very grateful if you give me some advice or point me to
articles dealing with similar problems. ( Please tell me if there is
any paper estimating the binding affinities of ioniable ligands into a
receptor. ) 

Thanks in advance for any help.  
Kind Regards,
Masakatsu

(*) M.K. Gilson et al. Biophys. J. 72, 1047 (1997) ; J. Herman and
S, Shankler, Isr. J. Chem. 27,225 (1986) ; L. Zhang and J. Hermans,
Proteins: Struct. Func. Gen. 24, 433 (1996); Roux et
al. Biophys. J. 71, 670 (1996)

--------------------------------------------------------------------
Masakatsu Ito 

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