[英]Generating Synthetic DNA Sequence with Substitution Rate
鑒於這些輸入:
my $init_seq = "AAAAAAAAAA" #length 10 bp
my $sub_rate = 0.003;
my $nof_tags = 1000;
my @dna = qw( A C G T );
我想生成:
一千個長度 - 10個標簽
標簽中每個位置的替代率為0.003
產量如下:
AAAAAAAAAA
AATAACAAAA
.....
AAGGAAAAGA # 1000th tags
在Perl中有一種緊湊的方式嗎?
我堅持使用這個腳本的邏輯作為核心:
#!/usr/bin/perl
my $init_seq = "AAAAAAAAAA" #length 10 bp
my $sub_rate = 0.003;
my $nof_tags = 1000;
my @dna = qw( A C G T );
$i = 0;
while ($i < length($init_seq)) {
$roll = int(rand 4) + 1; # $roll is now an integer between 1 and 4
if ($roll == 1) {$base = A;}
elsif ($roll == 2) {$base = T;}
elsif ($roll == 3) {$base = C;}
elsif ($roll == 4) {$base = G;};
print $base;
}
continue {
$i++;
}
作為一個小優化,替換:
$roll = int(rand 4) + 1; # $roll is now an integer between 1 and 4
if ($roll == 1) {$base = A;}
elsif ($roll == 2) {$base = T;}
elsif ($roll == 3) {$base = C;}
elsif ($roll == 4) {$base = G;};
同
$base = $dna[int(rand 4)];
編輯:假設替代率在0.001到1.000的范圍內:
和$roll
,生成[1..1000]范圍內的另一個(偽)隨機數,如果它小於或等於(1000 * $ sub_rate)則執行替換,否則什么都不做(即輸出'A' “)。
請注意,除非知道隨機數生成器的屬性,否則可能會引入微妙的偏差。
不完全是你想要的,但我建議你看看BioPerl的Bio :: SeqEvolution :: DNAPoint模塊。 但它並不以突變率作為參數。 相反,它詢問與您想要的原始序列同一性的下限。
use strict;
use warnings;
use Bio::Seq;
use Bio::SeqEvolution::Factory;
my $seq = Bio::Seq->new(-seq => 'AAAAAAAAAA', -alphabet => 'dna');
my $evolve = Bio::SeqEvolution::Factory->new (
-rate => 2, # transition/transversion rate
-seq => $seq
-identity => 50 # At least 50% identity with the original
);
my @mutated;
for (1..1000) { push @mutated, $evolve->next_seq }
所有1000個突變序列將存儲在@mutated數組中,它們的序列可以通過seq
方法訪問。
如果替換,您希望從可能性中排除當前基數 :
my @other_bases = grep { $_ ne substr($init_seq, $i, 1) } @dna;
$base = @other_bases[int(rand 3)];
另請參閱Mitch Wheat關於如何實施替代率的答案 。
我不知道我是否理解正確,但我會做這樣的事情(偽代碼):
digits = 'ATCG'
base = 'AAAAAAAAAA'
MAX = 1000
for i = 1 to len(base)
# check if we have to mutate
mutate = 1+rand(MAX) <= rate*MAX
if mutate then
# find current A:0 T:1 C:2 G:3
current = digits.find(base[i])
# get a new position
# but ensure that it is not current
new = (j+1+rand(3)) mod 4
base[i] = digits[new]
end if
end for
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