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簡介
PAML(Phylogenetic Analysis by Maximum Likelihood)是倫敦大學的楊子恆(Yang Ziheng)教 授開發的一套基於最大似然估計來對蛋白質和核酸序列進行系統發育分析的軟件，對學術使用是 免費的。楊子恆教授維護和發布 PAML 在 UNIX/Linux/MAC OS X 平台下的 ANSI C 的源程序和MS Windows 下的可執行文件。
PAML 可實現系統發育樹的構建，祖先序列估計,進化模擬和 KaKs 計算等功能。其中分支及 位點 KaKs 的計算是本軟件包的特色功能。
由此可以下載並且安裝
1 cd paml4.5/ 2 rm bin/*.exe 3 cd src 4 make -f Makefile 5 ls -l 6 rm *.o 7 mv baseml basemlg codeml pamp evolver yn00 chi2 ../bin
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實例
1、codeml
所需文件有2
1⃣️：CDS的fasta比對文件（一定是3的倍數，且去掉終止密碼子）；
*一定是比對過后的文件，我是用clusw進行比對，進而輸出fasta格式，有說PAML可以識別phlylip格式，但是我的這個不行，不知道咋么個情況。
得到的fasta文件后，可以使用EasyCodeML 進行格式轉換，轉為paml格式（把U替換為T）。把fasta文件放入inPath運行如下命令即可。
1 java -cp EasyCodeML.jar SeqFormatConvert.seqFactory.SeqConverter -i inPath/ -iF fasta -o outPath/ -oF PAML
2⃣️：tree文件
使用phylip得到樹文件
將兩個文件放入PAML bin文件下即可
第一步：
配置文件的配置
打開codeml.ctl，其內容如下：
1 seqfile = dm_cds.pml * sequence data filename ##比對文件 2 treefile = outtree * tree structure file name##樹的文件 3 outfile = dm2.mlc * main result file name##輸出文件 4 5 noisy = 9 * 0,1,2,3,9: how much rubbish on the screen 6 verbose = 1 * 0: concise; 1: detailed, 2: too much 7 runmode = 0 * 0: user tree; 1: semi-automatic; 2: automatic 8 * 3: StepwiseAddition; (4,5):PerturbationNNI; -2: pairwise##利用自己的樹，選擇0 9 10 seqtype = 2 * 1:codons; 2:AAs; 3:codons-->AAs 11 CodonFreq = 2 * 0:1/61 each, 1:F1X4, 2:F3X4, 3:codon table 12 13 * ndata = 10 14 clock = 0 * 0:no clock, 1:clock; 2:local clock; 3:CombinedAnalysis 15 aaDist = 0 * 0:equal, +:geometric; -:linear, 1-6:G1974,Miyata,c,p,v,a 16 aaRatefile = dat/jones.dat * only used for aa seqs with model=empirical(_F) 17 * dayhoff.dat, jones.dat, wag.dat, mtmam.dat, or your own 18 19 model = 0 20 * models for codons: 21 * 0:one, 1:b, 2:2 or more dN/dS ratios for branches 22 * models for AAs or codon-translated AAs: 23 * 0:poisson, 1:proportional, 2:Empirical, 3:Empirical+F 24 * 6:FromCodon, 7:AAClasses, 8:REVaa_0, 9:REVaa(nr=189)##選擇0表示假設所有樹具有相同seqfile = dm_cds.pml * sequence data filename 25 treefile = outtree * tree structure file name 26 outfile = dm2.mlc * main result file name 27 28 noisy = 9 * 0,1,2,3,9: how much rubbish on the screen 29 verbose = 1 * 0: concise; 1: detailed, 2: too much 30 runmode = 0 * 0: user tree; 1: semi-automatic; 2: automatic 31 * 3: StepwiseAddition; (4,5):PerturbationNNI; -2: pairwise 32 33 seqtype = 2 * 1:codons; 2:AAs; 3:codons-->AAs 34 CodonFreq = 2 * 0:1/61 each, 1:F1X4, 2:F3X4, 3:codon table 35 36 * ndata = 10 37 clock = 0 * 0:no clock, 1:clock; 2:local clock; 3:CombinedAnalysis 38 aaDist = 0 * 0:equal, +:geometric; -:linear, 1-6:G1974,Miyata,c,p,v,a 39 aaRatefile = dat/jones.dat * only used for aa seqs with model=empirical(_F) 40 * dayhoff.dat, jones.dat, wag.dat, mtmam.dat, or your own 41 42 model = 0 43 * models for codons: 44 * 0:one, 1:b, 2:2 or more dN/dS ratios for branches 45 * models for AAs or codon-translated AAs: 46 * 0:poisson, 1:proportional, 2:Empirical, 3:Empirical 47 * 6:FromCodon, 7:AAClasses, 8:REVaa_0, 9:REVaa(nr=189)##選擇0表示假定樹所有的相同的dN/dS，選擇2表示一個樹具有2個或者2個以上的不同dN/dS
其中，比對文件格式如下：
5 4609 RPP4 -----ATGGCTTCTTCTTCTTCTTCTCCTAGTAGCCGGAGATACGACGTTTTCCCAAGCTTCAGTGGGGTAGATGTTCGCAAAACGTTCCTCAGCCATCTAATCGAGGCGCTCGACCGCAGATCAATCAATACA-TTCATGGATCACGGCATCG---TGAGAAGCTGCATAATCGCCGATGCGCTTATAACGGCCATTAGAGAAGCGAGGATCTCAATAGTCATCTTCTCTGAGA-ACTATGCTTCTTCAACGTGG------TGCTTGAATGAATTGGTGGAGATCCACAAG-TGTTACAAGAAAGGGGAACAAATGGTGATTCCGGTTTTCTACGGCGTTGATCCTTCTC-------ATGTTAGAAAACAGATCGGTGGCTTTGGCGATGTCTTTAAAAAGACATGCGAGGACA------------AACCAGAGGATCAGAAACAAAGA--TGGGTTAAAGCTCTCACAGATATATCAAATTTAGCCGGGGAGGATCTTCGGAACGGGCCTACTGAAGCG---TTTATGGTTAAAAAGATAGCCAATGATGTTTCGAATAAACTTT--TTCCTCTGCCAAAGGGTTTTGGTGACTTCGTCGGAATTGAAGATCATATAAAGGCAATAAAATC-AATACTTTGCTTGGAATC--CAAGGAAGCTAGAATAATGGTCGGGATTTGGGGACAGTCAGGGATTGGTAAGAGTACCATAGGAAG----AGCTCTTTTCAGTCAACTCTCTAGCCAGTTCCACCATCGCGCTTTCATAACTTATAAAAGCACCAGTGGTAGTGACGTCTCTGGCATGAAGTTGAGTTGGGAAAAAGAACTT----CTCTCGGAAATCTTAGGTCAAAAGGACATAAAGATAGATCATTTTGGTGTGG-TGGAGCAAAGGTTAAAGCACAA--GAAAGTTCTTATCCTTCTTGATGATGTGGATAATCTAGAGTTT---CTTAAGACCTTGGTGGGAAAAGCTGAATGGT---TTGGTTCTGGAA-GCAGAATAATTGTGATCACTCA----AGATAAGCAACTTCTCAAGGCTCATGAGATTGACCTTGTATATGAGGTG-GAGCTGCCATCTCAAGGTCTTGCTCTTAAGATGATATCCCAATATGCTTTTGGGAAAGACTCTCCACCTGATGATTTTAAGGAACTAGCATTTGAAGTTGCCGAGCTTGTCGGTAGTCTTCCTTTGGGTCTCAGTGT-----CT---TGGGTTCATCTTTA-AAAGGAAGG----GACAAAGATGAGTGGGTGAAGATGATGCCTAGGCTTCGAAATGATTCAGATGATAAAATTGAGGAAACACTAAGAGTCGGCTACG-ATAGGTTAAATAAAAAAAA--TAGAG--AGTTATTTAAGTGCATTGCATGTTTTTTCAATGGTTTTAAAGTC--------AGTAACGTCAAAGAATTACTTGAAGATGATGTTGGGCTTACAATGTTGGCTGATAAGTCCCTCATACGTATTACACCGGATGGAGATATAGAGATGCACAATTTGC---TAGAGAAATTGGGTAGAGAAATTGATCGTGCAAAGTCCAAGGGTAATCCTGCAAAACGTCAATTTCTGACGAATTTTGAGGATATTCAAGAAGTAGTGACCGAGAAAACTGGGACAGAAACTGTTCTTGGAATACGTGTGCCACCCACGGTATTATTTTCGACAAGG-CCGTTATTAGTAATAAACGAAGAATCGTTCAAAGGCATGCG---TAATCTCCAATATCTAGAAATTGGTCATTGGTCAGAAATTGGTCTTTGGTCAGAAATTGGTCTTTGGTCAAAAATAGATCTACCT--CAGGGCCTCGTTTATTTGCCCCTTAAACTCAAATTGCTA-AAATGGAATTATTGTCCATTGAAGTC--TTTGCCATCT--ACTTTTAAGGCGGAATATCTAG-TTAACCTCATAATGA-A-GTATAGTAAGCTTGAGAA--ACTGT--GGGAAGGAACTCTGCCCCTTGGAAG------TCTCAAGAAGATGGA--TTTGGGGTGTTCCAACAATTTGAAA-------GAAATTCCAGA---TCTTTCTTTAGCCAT-AAACCTCGAGGAATTAAATCTTTCTAAATGCGA-ATCTT---TGGTGACACT-TCCTTCCTCGAT-TCAGAATGCCATTAAACTGAGGACGTTATATTGTTCGGGGGTGCTATTAAT-AGATTTAAAATCATTAGAAGGCATGTGTAATCTCGAAT---ATCTATCAGTTGATTGGT---CAAGTATGGAAGGCACTCAAGGCCTCA--TTTACTTGCCACGTAAACTCAAAAGGCTATGGT---GGGATTATTGTC-----------CAGTGAAGCGTTTGCCTTCTAATTT------------TAAGGCTGAGTATCTAGTTGAACTCAGAATGGAGAATAGTGACCTTGAGAAGCTGTGGGATGGAACTCAGCCACTTGGAAGCCTCAAGGAGATGTA-------------TCTGCATGGTTCCAAATAT--TTGAAAGAAAT------TCCAGATCTTTCTTTAGCCATAAACCTGGAGAGACTATAT-CTTTTTGGATGCGAATCTTTGGTGACACTTCCTTCCTCGATTCAGAATGCCACTAAATTGATCAATTTAGA-TATGAGAGATTGCAAAAAGCTAGAGAGTTTTCCAACCGATCTCAACTTGGAATCTCTCGAGTACCTCAATCTCACTGGATGCCCGAATTTGAGAAATTTCCCAGCAATCAAAATG--GGATGTTCATACTTTGAAATTCTGCAAGATAGAAATGAGATCGAGGTAGAAGATTGT-TTCTGGAACAAG-AATCTCCCTGCTGGACTAGATTATCTCGACT-----GCCTTATGAGGTGTATGCCTTGTGAATTTCGCCCAGA----ATATCTCACTTTTCTCGATGTGAGCGGCTGCAA--GCATG--AGAAGCTATGGGAAGGCATCCAGTCGCTTGGAAG----------TCTCAAGAGGATGGATCTGTCAGAATCTGA--AAACCTGACAGAAATTCCAGATCTTTCGAAGGCCACCAATCTGAAGCGTTTATATCTCAACGGGTGCAAAAGTTTGGTGAC-ACTTCCTTCTACAATTGGGAATCTTCATAGATTGGTGAGGTTGGAAATGAAAGAATGCACAGGGCTGGAGCTTCTTCCAACCGATGTCAACTTGTCATCTCT------------------------------------------------------------TATCA------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------TCCTCGATCTCAGTGGTTGCTCAAGTCTGAGAACTTTTCCTCTGATTTCAACTAGAATCGAATGTCTCTATCTAGAAAACACCGCCATTGAAGAAGTTCCCTGCTGCATTGAGGATTTAACGAGGCTCAGTGTACTACTGATGTATTGTTGCCAGAGGTTGAAAAACATCTCCCCAAACATTTTCAGACTGACAAGTCTAATGGTCGCCGACTTTACAGACTGTAGAGGTGTCATCAAGGCGTTGAGTGATGCAACTGTGGTAGCGACAATGGAAGATCATGTTTCTTGTGTACCATTATCTGAAAACATTGAATATACATGTGAACGTTTCTGGGATGAGTTGTATGAAAGAAAT--TCCAGATCTA--TCTTTAGCTATAAAGATGAGGATGGCGACGTATATTGGGTAAA------TTGGGACTTA----ATGATGATGCTG-ATGTTGATA------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ RPP5 -----ATGGCGGCTTCTTCTTCTT---CTGGCAGACGGAGATACGACGTTTTTCCAAGCTTCAGTGGGGTTGATGTTCGCAAGACGTTCCTCAGCCATCTAATCGAGGCGCTCGACGGCAAATCAATCAATACA-TTCATCGATCATGGAATCG---AGAGAAGCCGCACAATCGCCCCTGAGCTTATATCGGCGATTAGAGAAGCTAGGATCTCAAT
tree的格式如下：
(RPP5:0.05837,(RPP27:0.64437,(RPP8:0.51359,RPP13:0.22395):0.18814):0.42283,RPP4:0.04920);
第二部：
進行假設檢驗
0假設：一個樹所有的分支具有相同的dN/dS
備選假設：自定義一支具有不同的dN/dS （在tree文件中進行標注，在特定位置標出$1 或者#1，具體可看b站視頻）
分別以model為0，以及model為2進行運行PAML，根據兩個文件的結果中Lnl 計算p值
譬如：model=0: lnr1
model=2: lnr2
abs(lnr1-lnr2)*2 = q 值，自由度df為np2 -np1, np 為參數的數量
利用PAML中的chi函數，即可得到p值
1 ./chi df q
p值<0.05 則拒絕原假設，接受備選假設，即你認為的支確實dN/dS 有差異
關注下方公眾號可獲得更多精彩
參考