AMBER Archive (2008)

Subject: AMBER: Regarding binding free energy calculation using MM_PBSA method

From: nag raj (
Date: Wed May 21 2008 - 02:01:14 CDT

Dear Amber users,
  I want to calculate the binding free energy of ligand with a, b, and g cyclodextrins(CDs). I want to compare binding free energies of ligand bound with a, and g cyclodextrins with the ligand complex of b-cyclodextrin. While, crystal structure of ligand bound with b-cyclodextrin is available confirms the ligand binding and negative free energy is expected for this particular complex. But after MD simulations I got the positive binding free energy for a and b cyclodextrins and negative in case of g cyclodextrin.
  Now my question is why the binding free energies are positive in case of a and b CDs and negative for g -CD. Am I followed any thing wrong in my procedure in calculating the binding free energy. Can any one please go through the procedure that I adopted and suggest me where the wrong step has taken and what is the correct procedure?
  (2) Also, I would like to know how the samples saved in the trajectory influences the binding free energy and also samples in MM_PBSA influence the binding energy.
  I followed the procedure given in amber tutorial (Tutorial 3)
  Here briefly giving the procedure
   Starting conformation was taken as distance between centroids of ligand and cyclodextrins close to zero. The complexes are solvated with TIP3P water BOX, with buffer radius 10. Trajectory saved for every 10ps time scale and in MM_PBSA calculation alternative structures are considered for the analysis. Input files that are considered for minimization, heating, equilibration and production dynamics are given below.
            minimise cyclodextrine
    ntr=1, restraintmask=':1-6',
    heat cyclodextrine
    cut=8.0, ntb=1,
    ntpr=10000, ntwx=10000, ntwr=10000,
    ntt=0, vlimit=16.0, tautp=1.0,
    tempi=0.0, temp0=300.0,
    ntr=1, restraintmask=':1-6',
   &wt TYPE='TEMP0', istep1=0, istep2=50000, value1=0.1, value2=300.0, / &wt TYPE='END' /
      equilibrate the cyclodextrine at constant pressure
    cut=8.0, ntb=2, ntp=1, taup=2.0,
    ntpr=10000, ntwx=10000, ntwr=10000,
    ntt=1, tautp=1.0, vlimit=16.0,
    tempi=300.0, temp0=300.0,
    Production dynamics of cyclodextrine
    cut=8.0, ntb=2, ntp=1, taup=2.0,
    ntpr=10000, ntwx=10000, ntwr=10000,
    ntt=1, tautp=1.0, vlimit=16.0,
    tempi=300.0, temp0=300.0,

  Input parameters used in MM_PBSA to calculate the binding free energy:
  PROC 2
  REFE 0
  INDI 1.0
  EXDI 80.0
  SCALE 4.0
  LINIT 1000
  PRBRAD 1.4
  ISTRNG 0.0
  SURFTEN 0.0072
  SURFOFF 0.00
  DIELC 1.0
  IGB 2
  GBSA 1
  SALTCON 0.00
  EXTDIEL 80.0
  SURFTEN 0.0072
  SURFOFF 0.00
  MAXCYC 1000
  DRMS 0.1
  PROBE 0.0

  Energy components obtained from MM_PBSA analysis for a, b, and g cyclodextrins.
            Energy components
  Delta Mean
      Free energy of binding

  Free energy of binding is calculated as:
  Free energy of binding= PBTOT―TSTOT
  Thank you in advance.
                                                         with best regards,

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