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zhaoqingshan金虫 (初入文坛)
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【求助】autodock结果分析
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请教：小弟将autodock运行完毕以后，出来以下结果，但是不明白什么意思，请教各位前辈帮忙分析一下： utputting structurally similar clusters, ranked in order of increasing energy. ________________________________________________________________________________ Number of distinct conformational clusters found = 1, out of 10 runs, Using an rmsd-tolerance of 2.0 A CLUSTERING HISTOGRAM ____________________ ________________________________________________________________________________ | | | | | Clus | Lowest | Run | Mean | Num | Histogram -ter | Binding | | Binding | in | Rank | Energy | | Energy | Clus| 5 10 15 20 25 30 35 _____|___________|_____|___________|_____|____:____|____:____|____:____|____:___ 1 | -4.38 | 3 | -4.33 | 10 |########## _____|___________|_____|___________|_____|______________________________________ Number of multi-member conformational clusters found = 1, out of 10 runs. RMSD TABLE __________ _____________________________________________________________________ | | | | | | Rank | Sub- | Run | Binding | Cluster | Reference | Grep | Rank | | Energy | RMSD | RMSD | Pattern _____|______|______|___________|_________|_________________|___________ 1 1 3 -4.38 0.00 135.62 RANKING 1 2 9 -4.37 0.05 135.63 RANKING 1 3 1 -4.35 0.17 135.60 RANKING 1 4 7 -4.34 0.48 135.63 RANKING 1 5 10 -4.33 0.39 135.80 RANKING 1 6 8 -4.32 0.19 135.62 RANKING 1 7 6 -4.32 0.49 135.77 RANKING 1 8 5 -4.32 0.36 135.80 RANKING 1 9 4 -4.29 0.32 135.66 RANKING 1 10 2 -4.29 0.77 135.63 RANKING _______________________________________________________________________ INFORMATION ENTROPY ANALYSIS FOR THIS CLUSTERING ________________________________________________ Information entropy for this clustering = 0.00 (rmstol = 2.00 Angstrom) _______________________________________________________________________ STATISTICAL MECHANICAL ANALYSIS _______________________________ Partition function, Q = 10.07 at Temperature, T = 298.15 K Free energy, A ~ -1368.57 kcal/mol at Temperature, T = 298.15 K Internal energy, U = -4.33 kcal/mol at Temperature, T = 298.15 K Entropy, S = 4.58 kcal/mol/K at Temperature, T = 298.15 K _______________________________________________________________________ LOWEST ENERGY DOCKED CONFORMATION from EACH CLUSTER ___________________________________________________ Keeping original residue number (specified in the input PDBQ file) for outputting. MODEL 3 USER Run = 3 USER Cluster Rank = 1 USER Number of conformations in this cluster = 10 USER USER RMSD from reference structure = 135.620 A USER USER Estimated Free Energy of Binding = -4.38 kcal/mol [=(1)+(2)+(3)-(4)] USER Estimated Inhibition Constant, Ki = 619.83 uM (micromolar) [Temperature = 298.15 K] USER USER (1) Final Intermolecular Energy = -5.20 kcal/mol USER vdW + Hbond + desolv Energy = -5.06 kcal/mol USER Electrostatic Energy = -0.14 kcal/mol USER (2) Final Total Internal Energy = -0.54 kcal/mol USER (3) Torsional Free Energy = +0.82 kcal/mol USER (4) Unbound System's Energy = -0.54 kcal/mol USER USER USER USER DPF = D:\end\4\4v4.dpf USER NEWDPF move 4v4.pdbqt USER NEWDPF about -0.341000 0.193700 -0.053100 USER NEWDPF tran0 20.820406 128.994963 42.464737 USER NEWDPF axisangle0 -0.783365 0.619413 0.051643 84.293037 USER NEWDPF quaternion0 -0.525660 0.415644 0.034654 0.741431 USER NEWDPF ndihe 3 USER NEWDPF dihe0 23.14 80.58 17.92 USER USER x y z vdW Elec q RMS ATOM 1 C1 RES d 1 22.396 127.384 43.103 -0.18 +0.02 +0.270 135.620 ATOM 2 l3 RES d 1 22.250 125.997 44.241 -0.38 -0.00 -0.098 135.620 ATOM 3 l4 RES d 1 24.023 127.460 42.350 -0.18 -0.01 -0.098 135.620 ATOM 4 C2 RES d 1 22.111 128.694 43.865 -0.22 +0.09 +0.255 135.620 ATOM 5 N5 RES d 1 20.847 129.206 43.819 -0.18 -0.07 -0.316 135.620 ATOM 6 C7 RES d 1 20.578 130.402 44.631 -0.33 +0.07 +0.155 135.620 ATOM 7 C32 RES d 1 19.614 128.813 43.045 -0.43 -0.01 +0.135 135.620 ATOM 8 C8 RES d 1 19.058 130.385 44.685 -0.32 +0.02 +0.191 135.620 ATOM 9 O6 RES d 1 18.683 129.836 43.420 -0.20 +0.01 -0.359 135.620 ATOM 10 C9 RES d 1 19.063 127.456 43.507 -0.61 -0.02 +0.071 135.620 ATOM 11 O12 RES d 1 23.005 129.224 44.513 -0.52 -0.24 -0.270 135.620 ATOM 12 C10 RES d 1 19.845 128.940 41.523 -0.40 -0.01 +0.065 135.620 ATOM 13 C11 RES d 1 19.250 130.165 40.793 -0.44 +0.00 -0.018 135.620 ATOM 14 C23 RES d 1 20.174 131.393 40.783 -0.27 -0.00 +0.009 135.620 ATOM 15 C25 RES d 1 18.888 129.758 39.353 -0.41 -0.00 +0.009 135.620 TER ENDMDL AVSFLD: # AVS field file AVSFLD: # AVSFLD: # Created by AutoDock AVSFLD: # AVSFLD: ndim=2 # number of dimensions in the field AVSFLD: nspace=1 # number of physical coordinates AVSFLD: veclen=7 # vector size AVSFLD: dim1=15 # atoms AVSFLD: dim2=1 # conformations AVSFLD: data=Real # data type (byte,integer,Real,double) AVSFLD: field=uniform # field coordinate layout AVSFLD: label= x y z vdW Elec q RMS AVSFLD: variable 1 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 5 stride = 12 AVSFLD: variable 2 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 6 stride = 12 AVSFLD: variable 3 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 7 stride = 12 AVSFLD: variable 4 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 8 stride = 12 AVSFLD: variable 5 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 9 stride = 12 AVSFLD: variable 6 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 10 stride = 12 AVSFLD: variable 7 file = D:\end\4\4v4.dlg.pdb filetype = ascii offset = 11 stride = 12 AVSFLD: # end of file >>> Closing the docking parameter file (DPF)... |
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zhaoqingshan
金虫 (初入文坛)
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2楼2010-03-29 09:55:04