find a DNA barcode with mismatches in sequence

Question

I have 36-nt reads like this: atcttgttcaatggccgatcXXXXgtcgacaatcaa in the fastq file with XXXX being the different barcodes. I want to search for a barcode in the file at exact position(21 to 24) and print the sequences with up to 3 mismatches in sequence not barcode.

For example: I have barcode: aacg search that barcode between position 21 to 24 in fastq file with allowing 3 mismatches in the sequence like:

atcttgttcaatggccgatcaacggtcgacaatcac # it has 1 mismatch
ttcttgttcaatggccgatcaacggtcgacaatcac # it has 2 mismatch
tccttgttcaatggccgatcaacggtcgacaatcac # it has 3 mismatch

I was trying to find unique lines first using awk and look for mismatches but it is very tedious for me to look and find them.

awk 'NR%4==2' 1.fq |sort|uniq -c|awk '{print $1"\t"$2}' > out1.txt

Is there any quick way i can find?

Thank you.

Answer 1

Using Python:

strs = "atcttgttcaatggccgatcaacggtcgacaatcaa"

with open("1.fq") as f:
    for line in f:
        if line[20:24] == "aacg":
            line = line.strip()
            mismatches = sum(x!=y for x, y  in zip(strs, line))
            if mismatches <= 3:
                print line, mismatches

atcttgttcaatggccgatcaacggtcgacaatcac 1
ttcttgttcaatggccgatcaacggtcgacaatcac 2
tccttgttcaatggccgatcaacggtcgacaatcac 3

Answer 2

Using Python:

import re
seq="atcttgttcaatggccgatcaacggtcgacaatcaa"
D = [ c for c in seq ]
with open("input") as f:
    for line in f:
        line=line.rstrip('\n')
        if re.match(".{20}aacg", line):
            cnt = sum([ 1 for c,d in zip(line,D) if c != d])
            if cnt < 4:
                print cnt, line

Answer 3

Using python regex module allows you to specify the number of mismatches

import regex #intended as a replacement for re
from Bio import SeqIO
import collections

d = collections.defaultdict(list)
motif = r'((atcttgttcaatggccgatc)(....)(gtcgacaatcaa)){e<4}' #e<4 = less than 4 errors
records = list(SeqIO.parse(open(infile), "fastq"))
for record in records:
    seq = str(record.seq)
    match = regex.search(motif, seq, regex.BESTMATCH)
    barcode = match.group(3)
    sequence = match.group(0)
    d[barcode].append(sequence) # store as a dictionary key = barcode, value = list of sequences
for k, v in d.items():
    print("barcode = %s" % (k))
    for i in v:
        print("sequence = %s" % (i))

using capture groups, the fourth group (3), will be the barcode