Solutions for Population Genetics Practice Problems

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since 26 March 2008

1. Suppose the frequencies for alleles 11 and 12 (at Marker 1) are 0.2 and 0.3, and the frequency of allele 6 at Marker 2 is 0.1.
   • Under Hardy-Weinberg and the product rule, what is the probability of an individual being 11,12 and 6,6
     (2*0.2*0.3)*(0.1²) = 0.0012 or 1/833.

   • Now use NRC recommendation 4.1, and assume theta = 0.05. What is this match probability?
     (2*0.2*0.3)*(0.1² + 0.05*0.1*0.9) = 0.0017 or 1/574

   • Now use NRC recommendation 4.2 also with theta = 0.05. What is this match probability (use Eq. 4.10 in the notes)
     for 11,12, 2*(0.05 + (1-0.05)*0.2)*(0.05 + (1-0.05)*0.3)/(1+0.05)*(1+2*0.05) = 0.1392 (as opposed to 0.12 under HW)

     for 6,6, (2*0.05 + (1-0.05)*0.1)*(3*0.05 + (1-0.05)*0.1)/(1+0.05)*(1+2*0.05) = 0.0414 (as opposed to 0.01 under HW)

     Hence, the joint probablity if 0.1392*0.0414 = 0.00575 or 1/174