找到DNA序列中所有重復的4聚體 - Perl

Question

你好，

我嘗試編寫一個程序，讀取包含多個DNA序列的FASTA格式文件，識別序列中所有重復的4聚體（即，多次出現的所有4聚體），並打印出重復的4聚體以及查找它的序列的標題。 k聚體僅僅是k個核苷酸的序列（例如，“aaca”，“gacg”和“tttt”是4聚體）。

這是我的代碼：

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";
    }
}

我有兩個請求：首先，我不知道如何設計我的正則表達式模式以獲得所需的輸出。 第二，不太重要的是，我確信我的代碼效率非常低，所以如果有辦法縮短代碼，請告訴我。

提前致謝！

以下是FASTA文件的示例:(請注意，序列之間有一條額外的行，原始的fasta文件不是這種情況）

> NC_001422.1腸桿菌噬菌體phiX174 sensu lato，完整基因組GAGTTTTATCGCTTCCATGACGCAGAAGTTAACACTTTttttttCGGATATTTCTGATGAGTCGAAAAAT CCCTTACTTGAGGATAtatataAATTATGTCTAATATTCAAACTGGCGCCGAGCGTATGCCGCATGACCT

> NC_001501.1腸桿菌噬菌體phiX184意義上拉托，完整基因組AACGGCTGGTCAGTATTTAAGGTTAGTGCTGAGGTTGACTACATCTGTTTTTAGAGACCCAGACCTTTTA TCTCACTTCTGTTACTCCAGCTTCTTCGGCACCTGTTTTACAGACACCTAAAGCTACATCGTCAACGTTA TATTTTGATAGTTTGACGGTTAATGCTGGTAATGGTgagagagaGGTTTTCTTCATTGCATTCAGATGGA TCAACGCCGCTAATCAGGTTGTTTCTGTTGGTGCTGATATTGCTTTTGATGCCGACCCTAAATTTTTTGC CTGTTTGGTTCGCTTTGAGTCTTCTTCGGTTCCGACTACCCTCCCGACTGCCTATGATGTTTATCCTTTG

> NC_001622.5腸桿菌噬菌體phiX199意義上拉托，完整基因組TTCGCTGAATCAGGTTATTAAAGAGTTGCCGAGATATTTATGTTGGTTTCATGCGGATTGGTCGTTTAAA TTGGACTTGGTGGCAAGTCTGCCGCTGATAAAGGAAAGGATAATGACCAAATCAAAGAACTCGTGATTAT CTTGCTGCTGCATTTCCTGAGCTTAATGCTTGGGAGCGTGCTGGTGCTGATGCTTCCTCTGCTGGTATGG TTGACGCCGGATTTGAGAATCAAAAATGTGAGAGAGCTTACTAAAATGCAACTGGACAATCAGAAAGAGA GATGCAAAATGAGACTCAAAAAGAGATTGCTGGCATTCAGTCGGCGACTTCACGCCAGAATACGAAAGAC CAGGTATATGCACAAAATGAGATGCTTGCTTATCAACAGAAGGAGTCTACTGCTCGCGTTGCGTCTATTA TGGAAAACACCAATCTTTCCAAGCAACAGCAGGTTTCCGAGATTATGCGCCAAATGCTTACTCAAGCTCA AACGGCTGGTCAGTATTTTACCAATGACCAAATCAAAGAAATGACTCGCAAGGTTAGTGCTGAGGTTGAC TTAGATGAGTGTTCATCAGCAAACGCAGAATCAGCGGTATGGCTCTTCTCATATTGGCGCTACTGCAAAG

Answer 1

我可能更喜歡解決你的問題：

#!/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";
}

我們按記錄迭代記錄，捕獲並刪除“標題”行，然后將其余部分拼接在一起。 然后捕獲4的每個（重疊）序列，並對它們進行計數。

這樣，對於您的樣本數據（簡潔的第一節）：

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 - 因為它基於原始序列，它基於數據中的排序，你會看到TGAC兩次因為......它在那里兩次。

但是你可以改為：

   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";

哪個將丟棄任何少於2個匹配，並按頻率排序。