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)
http://amber.scripps.edu/tutorials/advanced/tutorial3/index.htm
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 10Å. 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
&cntrl
imin=1,maxcyc=1000,ncyc=300,
cut=8.0,ntb=1,
ntc=1,ntf=1,
ntpr=100,
ntr=1, restraintmask=':1-6',
restraint_wt=2.0
/
heat cyclodextrine
&cntrl
imin=0,irest=0,ntx=1,
nstlim=50000,dt=0.001,
ntc=2,ntf=2,
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',
restraint_wt=2.0,
nmropt=1
/
&wt TYPE='TEMP0', istep1=0, istep2=50000, value1=0.1, value2=300.0, / &wt TYPE='END' /
&ewald
&end
END
equilibrate the cyclodextrine at constant pressure
&cntrl
imin=0,irest=1,ntx=5,
nstlim=450000,dt=0.001,
ntc=2,ntf=2,
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,
/
&ewald
&end
END
Production dynamics of cyclodextrine
&cntrl
imin=0,irest=1,ntx=5,
nstlim=1000000,dt=0.001,
ntc=2,ntf=2,
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,
/
&ewald
&end
END
Input parameters used in MM_PBSA to calculate the binding free energy:
#####
.PB
PROC 2
REFE 0
INDI 1.0
EXDI 80.0
SCALE 4.0
LINIT 1000
PRBRAD 1.4
ISTRNG 0.0
RADIOPT 0
NPOPT 1
CAVITY_SURFTEN 0.0072
CAVITY_OFFSET 0.00
#
SURFTEN 0.0072
SURFOFF 0.00
#
####
.MM
DIELC 1.0
#####
.GB
IGB 2
GBSA 1
SALTCON 0.00
EXTDIEL 80.0
INTDIEL 1.0
#
SURFTEN 0.0072
SURFOFF 0.00
#####
.NM
DIELC 4
MAXCYC 1000
DRMS 0.1
#####
.MS
#
PROBE 0.0
Energy components obtained from MM_PBSA analysis for a, b, and g cyclodextrins.
Energy components
Delta Mean
a-CD
b-CD
g-CD
ELE
-2.05
-2.62
-3.57
VDW
-27.76
-24.82
-25.46
INT
-0.00
0.00
-0.00
GAS
-29.81
-27.44
-29.03
PBSUR
-3.59
-3.70
-3.65
PBCAL
15.11
16.15
16.08
PBSOL
11.52
12.46
12.43
PBELE
13.07
13.53
12.51
PBTOT
-18.28
-14.98
-16.60
GBSUR
-3.59
-3.70
-3.65
GB
14.93
15.04
16.20
GBSOL
11.34
11.34
12.55
GBELE
12.88
12.42
12.63
GBTOT
-18.47
-16.10
-16.48
TSTRA
-12.42
-12.45
-12.47
TSROT
-9.97
-9.97
-9.83
TSVIB
2.48
6.57
6.32
TSTOT
-19.91
-15.84
-15.97
Free energy of binding
1.63
0.86
-0.63
Free energy of binding is calculated as:
Free energy of binding= PBTOT―TSTOT
Thank you in advance.
with best regards,
Nagaraju
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Received on Sun May 25 2008 - 06:07:03 PDT
