Question: Hardy-Weinbergy Equilibrium and SNPs
0
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 8 months ago by Biostar ♦♦ 20 • written 3.8 years ago by waynestatetutor0

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

ADD REPLYlink written 3.8 years ago by h.mon27k
0
Ibrahim Tanyalcin980 wrote:

Dear,

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 written 3.8 years ago by Ibrahim Tanyalcin980
Please log in to add an answer.

Content
Help
Access

Use of this site constitutes acceptance of our User Agreement and Privacy Policy.
Powered by Biostar version 2.3.0
Traffic: 1457 users visited in the last hour