Editing BG Lab 2 (section)

=== Reading a vcf file ===
To walk you through how to read a vcf I ran the following command, which will show the contents of the file minus the header, and minus the annotation column. This just cleans things up by removing parts of the file that we don't need.

<syntaxhighlight lang="bash">
zgrep -v '##' hu916767_20170324191934.1kgALTallele.withHeader.snpEff.vcf.gz | cut -f-5,9- | head
</syntaxhighlight>

The output is here:
 #CHROM	POS	ID	REF	ALT	FORMAT	hu916767_20170324191934
 1	82154	rs4477212	A	.	GT	0/0
 1	752566	rs3094315	G	A	GT	1/0
 1	752721	rs3131972	A	G	GT	0/1
 1	768448	rs12562034	G	A	GT	0/0
 1	776546	rs12124819	A	G	GT	0/1
 1	798959	rs11240777	G	A	GT	1/0
 1	800007	rs6681049	T	C	GT	1/1
 1	838555	rs4970383	C	A	GT	0/0
 1	846808	rs4475691	C	T	GT	0/1

The the columns are 
# chromosome
# position
# unique identifier (rs ID)
# the reference allele (the allele found in the human reference genome)
# the alternate allele (an allele discovered in other individuals)
# the FORMAT the individuals genotypes are in (in this case they are coded in the "GT" format, which is 0/0, 0/1, 1/0, or 1/1; believe it or not there are other useful formats). <span style="color:#ff0000"> THIS COLUMN DOES NOT PROVIDE THE INDIVIDUAL'S GENOTYPES AND CAN BE SAFELY IGNORED! </span>
# the genotype of this 23andme individual

To decode the genotype, you must combine the last column with the REF and ALT allele information. Take the last line, rs4475691. The REF allele is "C", and the ALT allele is "T". The genotype is 0/1, which tells you that one chromosome of this individual carries 0 ALT alleles (i.e., 1 reference allele) and the other chromosome carries 1 ALT allele. So the genotype is C/T.

For another example, take rs6681049. The REF allele is T and the ALT is C. The genotype is 1/1. That means that one chromosome of this individual carries 1 ALT allele (i.e., a C) and the other chromosome also carries 1 ALT allele (i.e., a C). So the genotype for this individual at that site is C/C.