blast -m1
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当database是10个物种(A、B、C、E、F、G、H、J、I、K)时,进行all vs all 比对结果是:
此时reference是物种A的第一个基因:即用10个物种的genome中的所有基因比对物种A的基因1
eg:如下
$ cat A_R00001_A.fa.70.blastp.m1|grep ‘1_0‘ -A 2 > r.sh Query= A_R00001_A (423 letters) Database: all.10 165,229 sequences; 72,505,122 total letters Searching..................................................done Score E Sequences producing significant alignments: (bits) Value A_R00001_A 855 0.0 B_R00001_B 766 0.0
1_0 1 PITLWLFYTHWLPGSPTTHHRVQGTTAMKDMVKCRVLKAHPCQECGKSFSKKGNLKRHQR 60 A_R00001_A 1 ............................................................ 60 B_R00001_B 4 ..l.......w..dAA....L..rv......................... 53 -- ... 1_0 361 GKCFPHKRHLIKHQLLHSRGGAYKCGVCGKRYRLKKYLRRHQKIHMREGTAPCSKPGYTT 420 A_R00001_A 361 ............................................................ 420 B_R00001_B 357 ..........................d..................t.........wee.. 416 -- 1_0 421 QTA 423 A_R00001_A 421 ... 423 B_R00001_B 417 R.G 419
1.Sequences producing significant alignments:大致情况,并计算score值,E值
关于score值:
he bit score gives an indication of how good the alignment is; the higher the score, the better the alignment. In general terms, this score is calculated from a formula that takes into account the alignment of similar or identical residues, as well as any gaps introduced to align the sequences.
2.具体情况,
1_0:表示reference的的第一个基因,即A物种的基因1
B_R00001_B:表示物种B的基因1与该reference(A物种的基因1) 的比对情况,从第4个AA开始到第53个AA结束
..l.......w..dAA....L..rv........................长度为53-4+1=50个;.是比对成功;空白是B物种基因1没有序列;60个为一组
如果物种A有10个基因,则将会把10个基因全部比对一遍:
$cat A_R00001_A.fa.70.blastp.m1|grep ‘Query‘ >> r.sh Query= A_R00001_A Query= A_R00002_A Query= A_R00003_A Query= A_R00004_A Query= A_R00005_A Query= A_R00006_A Query= A_R00007_A Query= A_R00008_A Query= A_R00009_A Query= A_R00010_A
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