[GNOMAD v2]How to count the number of heterozygous individuals
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18 months ago
leeray00151 ▴ 10

Hi:

I want count the number of heterozygous individuals.

for autosomes:

I use (AN/2 - nhomalt) as the heterozygous individuals. Is it correct ?

for X :

I use (AN_female/2 - nhomalt_female) as the female heterozygous individuals, as for male , I think AN = nhomalt + AC. so AC = male heterozygous individuals. but the data tell me it's wrong.

I don't know how to count the number of heterozygous individuals....maybe there's something wrong with my thinking....pls help me...

Thanks in advance!

SNP genome gnomad heterozygous individuals • 881 views
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16 months ago
scrap • 0

For non-sex chromosome (and female X chrom) variants I think the correct calculation for number of heterozygous individuals would be AC - (2 x nhomalt).

Each homozygous individual contributes 2 alternate alleles to the allele count (AC), so all remaining AC must come from heterozygous individuals who contribute 1 to the AC.

I'm not certain about the Y chromosome variants for males at the moment, but I hope this helps.

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Thanks for your answer.

But i wander that what's the diffence between the (AN/2 - nhomalt) and AC - (2 x nhomalt).

I think they are the same answer.

But obviously, the results are different

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I think it is because nhomalt is the number of individuals homozygous for the alternate allele.

AN/2 = total number of individuals, but that includes individuals homozygous for the reference allele, heterozygotes, and individuals homozygous for the alternate allele (nhomalt). So, AN/2 - nhomalt = heterozygotes + reference homozygotes , not only the number of heterozygotes.

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