找到DNA序列中所有重复的4聚体

你好，

我尝试编写一个程序，读取包含多个DNA序列的FASTA格式文件，识别序列中所有重复的4聚体（即，多次出现的所有4聚体），并打印出重复的4聚体以及查找它的序列的 Headers . k聚体仅仅是k个核苷酸的序列（例如，“aaca”，“gacg”和“tttt”是4聚体） .

这是我的代码：

use strict;
use warnings;

my $count = -1;
my $file = "sequences.fa";
my $seq = '';
my @header = ();
my @sequences = ();
my $line = '';
open (READ, $file) || die "Cannot open $file: $!.\n";

while ($line = <READ>){
    chomp $line;
    if ($line =~ /^>/){
        push @header, $line;
        $count++;
        unless ($seq eq ''){
            push @sequences, $seq;
            $seq = '';
        }
    } else {
        $seq .= $line;
    }
}   push @sequences, $line;

for (my $i = 0; $i <= $#sequences+1; $i++){
    if ($sequences[$i] =~ /(....)(.)*\g{1}+/g){
        print $header[$i], "\n", $&, "\n";
    }
}

我有两个请求：首先，我不知道如何设计我的正则表达式模式以获得所需的输出 . 第二，不太重要的是，我确信我的代码效率非常低，所以如果有办法缩短代码，请告诉我 .

提前致谢！

以下是FASTA文件的示例:(请注意，序列之间有一条额外的行，原始的fasta文件不是这种情况）

NC_001422.1肠杆菌噬菌体意义上的PhiX174拉托，完整基因组GAGTTTTATCGCTTCCATGACGCAGAAGTTAACACTTTttttttCGGATATTTCTGATGAGTCGAAAAAT CCCTTACTTGAGGATAtatataAATTATGTCTAATATTCAAACTGGCGCCGAGCGTATGCCGCATGACCT> NC_001501.1肠杆菌噬菌体phiX184意义上拉托，完整基因组AACGGCTGGTCAGTATTTAAGGTTAGTGCTGAGGTTGACTACATCTGTTTTTAGAGACCCAGACCTTTTA TCTCACTTCTGTTACTCCAGCTTCTTCGGCACCTGTTTTACAGACACCTAAAGCTACATCGTCAACGTTA TATTTTGATAGTTTGACGGTTAATGCTGGTAATGGTgagagagaGGTTTTCTTCATTGCATTCAGATGGA TCAACGCCGCTAATCAGGTTGTTTCTGTTGGTGCTGATATTGCTTTTGATGCCGACCCTAAATTTTTTGC CTGTTTGGTTCGCTTTGAGTCTTCTTCGGTTCCGACTACCCTCCCGACTGCCTATGATGTTTATCCTTTG> NC_001622.5肠杆菌噬菌体phiX199意义上拉托，完整基因组TTCGCTGAATCAGGTTATTAAAGAGTTGCCGAGATATTTATGTTGGTTTCATGCGGATTGGTCGTTTAAA TTGGACTTGGTGGCAAGTCTGCCGCTGATAAAGGAAAGGATAATGACCAAATCAAAGAACTCGTGATTAT CTTGCTGCTGCATTTCCTGAGCTTAATGCTTGGGAGCGTGCTGGTGCTGATGCTTCCTCTGCTGGTATGG TTGACGCCGGATTTGAGAATCAAAAATGTGAGAGAGCTTACTAAAATGCAACTGGACAATCAGAAAGAGA GATGCAA AATGAGACTCAAAAAGAGATTGCTGGCATTCAGTCGGCGACTTCACGCCAGAATACGAAAGAC CAGGTATATGCACAAAATGAGATGCTTGCTTATCAACAGAAGGAGTCTACTGCTCGCGTTGCGTCTATTA TGGAAAACACCAATCTTTCCAAGCAACAGCAGGTTTCCGAGATTATGCGCCAAATGCTTACTCAAGCTCA AACGGCTGGTCAGTATTTTACCAATGACCAAATCAAAGAAATGACTCGCAAGGTTAGTGCTGAGGTTGAC TTAGATGAGTGTTCATCAGCAAACGCAGAATCAGCGGTATGGCTCTTCTCATATTGGCGCTACTGCAAAG

1 回答

我可能更喜欢解决你的问题：

#!/usr/bin/env perl

use strict;
use warnings;

use Data::Dumper;

#set paragraph mode. Iterate on blank lines. 
local $/ = ''; 

#read from STDIN or a file specified on command line, 
#e.g. cat filename_here | myscript.pl
#or myscript.pl filename_here
while ( <> ) {
   #capture the header line, and then remove it from our data block
   my ($header) = m/\>(.*)/;
   s/>.*$//;

   #remove linefeeds and whitespace. 
   s/\s*\n\s*//g;
   #use lookahead pattern, so the data isn't 'consumed' by the regex. 
   my @sequences = m/(?=([atcg]{4}))/gi;

   #increment a count for each sequence found. 
   my %count_of;
   $count_of{$_}++ for @sequences;

   #print output. (Modify according to specific needs. 
   print $header,"\n";

   print "Found sequences:\n";
   print Dumper \@sequences;
   print "Count:\n";
   print Dumper \%count_of;

   #note - ordered, but includes duplicates. 
   #you could just use keys  %count_of, but that would be unordered. 
   foreach my $sequence ( grep { $count_of{$_} > 1 } @sequences ) {
      print $sequence, " => ", $count_of{$sequence},"\n";
   }
   print "\n";
}

我们按记录迭代记录，捕获并删除“ Headers ”行，然后将其余部分拼接在一起 . 然后捕获4的每个（重叠）序列，并对它们进行计数 .

这样，对于您的样本数据（简洁的第一节）：

NC_001422.1 Enterobacteria phage phiX174 sensu lato, complete genome 
Found sequences:
    GAGT => 2
    AGTT => 2
    TTAT => 2
    CATG => 2
    ATGA => 3
    TGAC => 2
    CGCA => 2
    AGTT => 2
    ACTT => 2
    tttt => 3
    tttt => 3
    tttt => 3
    GGAT => 2
    GATA => 2
    ATAT => 2
    TATT => 2
    ATGA => 3
    TGAG => 2
    GAGT => 2
    AAAA => 2
    AAAA => 2
    ACTT => 2
    TGAG => 2
    GGAT => 2
    GATA => 2
    tata => 2
    tata => 2
    TTAT => 2
    TATG => 2
    ATAT => 2
    TATT => 2
    GCCG => 2
    TATG => 2
    GCCG => 2
    CGCA => 2
    CATG => 2
    ATGA => 3
    TGAC => 2

注意 - 因为它基于原始序列，它基于数据中的排序，你会看到TGAC两次因为......它在那里两次 .

但是你可以改为：

foreach my $sequence ( sort { $count_of{$b} <=> $count_of{$a} }
                          grep { $count_of{$_} > 1 } 
                                 keys %count_of ) {
      print $sequence, " => ", $count_of{$sequence},"\n";
   }
   print "\n";

哪个将丢弃任何少于2个匹配，并按频率排序 .

回复于 2024-04-29T09:56:08+08:00

找到DNA序列中所有重复的4聚体 - Perl

1 回答

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