Dear Sergei,
Thank you so much for your quick reply.
I think I finally get some idea about HyPhy ^-^. Really appreciate you!
But I need your help again.
Because I am not sure how to read the results.
I got the following results, after running HyPhy using the GA partition gotten from [Raw] link.
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HYPHY 1.0020080508beta(MP) for Windows (Win32) console saved on Tue Dec 2 23:05:11 2008
Loaded 22 model templates from E:\Software\HYPHY_Win32(1)\HYPHY_Win32\SubstitutionModels
Loaded 12 genetic code tables from E:\Software\HYPHY_Win32(1)\HYPHY_Win32\GeneticCodes
2 Intel x86 architecture processors detected.
You can change the number of processors utilized by HyPhy via the HyPhy settings dialog.
Please cite S.L. Kosakovsky Pond, S. D. W. Frost and S.V. Muse. (2005) HyPhy: hypothesis testing using phylogenies. Bioinformatics 21: 676-679 if you use HyPhy in a publication
If you are a new HyPhy user, the tutorial located at Multimedia File Viewing and Clickable Links are available for Registered Members only!! You need to
may be a good starting point.
Loaded 2 partitions
Sequences :8
Sites :6462
Variable :818
f(A) = 0.289326
f(C) = 0.194309
f(G) = 0.206747
f(T) = 0.309618
c-AIC = 25997
Log Likelihood = -12977.4383434101;
Shared Parameters:
GT=0.160283=0.160283
CT=2.43=2.43
CG=0.0611013=0.0611013
AT=0.402894=0.402894
AC=0.158056=0.158056
alpha=0.010005=0.010005
betaQ=1.59101=1.59101
betaP=0.681693=0.681693
Tree givenTree=((((124_Bat_SARS_273:0.0502577,125_Bat_SARS_279:0.0502527)Node3:0.0610
553,130_PC03_SZ3:0.000585082)Node2:0.000196157,3_HP03E_GZ02:0.000312407)Node1:0,
(15_HP03M_BJ02:0.000620001,31_HP03L_Tor2:0.000150991)Node8:0.000150991,(106_HP04
_GZ0402:0.000620001,PC04_PC4_136_AY613949:0.000312407)Node11:0.00108391);
Mean divergence : 5.25787%
Fitting a single-tree, multiple partition model
Log Likelihood = -10325.9738761955;
Shared Parameters:
S_1=1.40178
betaP=0.681693=0.681693
betaQ=1.59101=1.59101
alpha=0.010005=0.010005
AC=0.158056=0.158056
AT=0.402894=0.402894
CG=0.0611013=0.0611013
CT=2.43=2.43
GT=0.160283=0.160283
Tree tree_0=((((124_Bat_SARS_273:0.0448527,125_Bat_SARS_279:0.0353642)Node3:0.0525925
,130_PC03_SZ3:0.000655097)Node2:0.00020522,3_HP03E_GZ02:0.000171432)Node1:0,(15_
HP03M_BJ02:0.000343505,31_HP03L_Tor2:0.000171517)Node8:0,(106_HP04_GZ0402:0.0005
16532,PC04_PC4_136_AY613949:0.000343924)Node11:0.000688349);
Tree tree_1=((((124_Bat_SARS_273:0.0628736,125_Bat_SARS_279:0.0495728)Node3:0.073723,
130_PC03_SZ3:0.000918301)Node2:0.000287673,3_HP03E_GZ02:0.00024031)Node1:0,(15_H
P03M_BJ02:0.000481519,31_HP03L_Tor2:0.000240429)Node8:0,(106_HP04_GZ0402:0.00072
4064,PC04_PC4_136_AY613949:0.000482105)Node11:0.000964913);
Fitting a mutilple tree, multiple partition model
Log Likelihood = -10315.2583699075;
Shared Parameters:
betaP=0.681693=0.681693
betaQ=1.59101=1.59101
alpha=0.010005=0.010005
AC=0.158056=0.158056
AT=0.402894=0.402894
CG=0.0611013=0.0611013
CT=2.43=2.43
GT=0.160283=0.160283
Tree tree_0=(((((124_Bat_SARS_273:0.0502045,125_Bat_SARS_279:0.0345148)Node4:0.048335
7,130_PC03_SZ3:0.000949729)Node3:9.48274e-05,(106_HP04_GZ0402:0.000783531,PC04_P
C4_136_AY613949:0)Node8:0.000782496)Node2:0,3_HP03E_GZ02:0)Node1:0,15_HP03M_BJ02
:0.000260537,31_HP03L_Tor2:0);
Tree tree_1=((((124_Bat_SARS_273:0.0457419,125_Bat_SARS_279:0.0533774)Node3:0.0870497
,130_PC03_SZ3:0)Node2:0.000706954,31_HP03L_Tor2:0.000707831)Node1:0,(3_HP03E_GZ0
2:0.000706903,15_HP03M_BJ02:0.000707295)Node8:0,(106_HP04_GZ0402:0,PC04_PC4_136_
AY613949:0.00142015)Node11:0.000707243);
Versus the single partition model: c-AIC = 20698.9
Delta AIC = 5298.13
Versus the single tree/multiple partition model: Delta AIC = -2.78213
Partition 1 : 3837 sites
Partition 2 : 1422 sites
Fitting tree 1 to partition 1
Log Likelihood = -7368.17401533097;
Shared Parameters:
betaP=0.681693=0.681693
betaQ=1.59101=1.59101
alpha=0.010005=0.010005
AC=0.158056=0.158056
AT=0.402894=0.402894
CG=0.0611013=0.0611013
CT=2.43=2.43
GT=0.160283=0.160283
Tree aTree=(((((124_Bat_SARS_273:0.0502044,125_Bat_SARS_279:0.0345152)Node4:0.0483353
,130_PC03_SZ3:0.00094992)Node3:9.48549e-05,(106_HP04_GZ0402:0.000783767,PC04_PC4
_136_AY613949:0)Node8:0.000782828)Node2:0,3_HP03E_GZ02:0)Node1:0,15_HP03M_BJ02:0
.000260704,31_HP03L_Tor2:0);
Fitting tree 2 to partition 1
Log Likelihood = -7368.1740159421;
Shared Parameters:
betaP=0.681693=0.681693
betaQ=1.59101=1.59101
alpha=0.010005=0.010005
AC=0.158056=0.158056
AT=0.402894=0.402894
CG=0.0611013=0.0611013
CT=2.43=2.43
GT=0.160283=0.160283
Tree aTree=((((124_Bat_SARS_273:0.0502049,125_Bat_SARS_279:0.034515)Node3:0.0483355,1
30_PC03_SZ3:0.000949661)Node2:9.47293e-05,31_HP03L_Tor2:0)Node1:0,(3_HP03E_GZ02:
0,15_HP03M_BJ02:0.000260946)Node8:0,(106_HP04_GZ0402:0.000783819,PC04_PC4_136_AY
613949:0)Node11:0.000782895);
KH Testing partition 1
Tree 2 base LRT = 1.22224e-06. p-value = 0.4817
Fitting tree 1 to partition 2
Log Likelihood = -2947.08435430648;
Shared Parameters:
betaP=0.681693=0.681693
betaQ=1.59101=1.59101
alpha=0.010005=0.010005
AC=0.158056=0.158056
AT=0.402894=0.402894
CG=0.0611013=0.0611013
CT=2.43=2.43
GT=0.160283=0.160283
Tree aTree=(((((124_Bat_SARS_273:0.0457418,125_Bat_SARS_279:0.0533773)Node4:0.0870492
,130_PC03_SZ3:0)Node3:0.000707349,(106_HP04_GZ0402:0,PC04_PC4_136_AY613949:0.001
41987)Node8:0.000707103)Node2:0,3_HP03E_GZ02:0.000706771)Node1:0,15_HP03M_BJ02:0
.000707307,31_HP03L_Tor2:0.000707425);
Fitting tree 2 to partition 2
Log Likelihood = -2947.08435419031;
Shared Parameters:
betaP=0.681693=0.681693
betaQ=1.59101=1.59101
alpha=0.010005=0.010005
AC=0.158056=0.158056
AT=0.402894=0.402894
CG=0.0611013=0.0611013
CT=2.43=2.43
GT=0.160283=0.160283
Tree aTree=((((124_Bat_SARS_273:0.0457417,125_Bat_SARS_279:0.0533769)Node3:0.0870492,
130_PC03_SZ3:0)Node2:0.000706985,31_HP03L_Tor2:0.000707755)Node1:0,(3_HP03E_GZ02
:0.00070661,15_HP03M_BJ02:0.00070704)Node8:0,(106_HP04_GZ0402:0,PC04_PC4_136_AY6
13949:0.00141997)Node11:0.000707026);
KH Testing partition 2
Tree 1 base LRT = 2.32327e-07. p-value = 0.4839
A total of 0/1 significant couplings
Mean splits identify: 1
+++++++++++++++++
Do we just see these parts?
-------------------
KH Testing partition 1
Tree 2 base LRT = 1.22224e-06.
p-value = 0.4817KH Testing partition 2
Tree 1 base LRT = 2.32327e-07.
p-value = 0.4839A total of 0/1 significant couplings
Mean splits identify: 1
----------------------
Does these mean that the KH test indicates there is no significant evidence (P>0.05) of recombination? The breakpoints are not real recombination sites?
Thank you very much!!
Best regards
Sundy