Question: Hardy-Weinbergy Equilibrium and SNPs
gravatar for waynestatetutor
4.6 years ago by
United States
waynestatetutor0 wrote:

" Assuming Hardy-Weinberg equilibrium, SNPs with minor allelic frequency between 0.4 and 0.5 would be found homozygous in two people 23% to 25% of the time for both alleles. The probability of both donor and recipient having a different allele is therefore 11.5% to 12.5% using accepted estimation models for calculation of exclusion probabilities" This is from an article I'm reading. I do not follow how they got their number. Can someone explain? Thanks!

ADD COMMENTlink modified 19 months ago by Biostar ♦♦ 20 • written 4.6 years ago by waynestatetutor0

It would probably be helpful if you link the paper in question.

ADD REPLYlink modified 7 months ago by RamRS27k • written 4.6 years ago by h.mon30k
gravatar for Ibrahim Tanyalcin
4.6 years ago by
Ibrahim Tanyalcin1.1k wrote:


Assuming you are familiar with Hardy-Weinberg (which to me is simply binomial opening of (a+b)^n), the first part calculates the probabilty of being homozygous for a allele of let's say frequency 0.5. Then, the change of having both alleles the same would be 0.5*0.5 = 0.25. If the frequency would be 0.4 than it would be 0.4*0.4 = 0.16.

Next it calculates what are the chances that 2 people are having different alleles. Here I assume they calculate the probablity of both being homozygous for different alleles. Than, if you take the number above, 0.25 which is probability of being homozygous for 1 allele, than consider the other person begin homozygous for the other allele, which is again 0.25. So this situation occurs if:

person a is homozygous for alele 1 and the other person is homoozygous for allele 2: 0.25*0.25 = 0.0625

OR vice versa : 0.0625

add them up 0.125.

This is all I could understand based on this passage,

I hope it helps.

ADD COMMENTlink modified 7 months ago by RamRS27k • written 4.6 years ago by Ibrahim Tanyalcin1.1k
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