In a sample of 200 individuals from a population which is expected to be at Hard

Question

In a sample of 200 individuals from a population which is expected to be at Hardy-Weinberg equilibrium for a locus with 3 alleles, the observed numbers Oi for the 6 possible genotypes are given in the table below:

What do you expect these six genotypes to be under HWE?
If each genotype is a Poisson variable with expected value Mi from b) and standard deviation si, compute the 95% confidence intervals for the expected number of samples in each genotype. (Hint: Think about how si is related to Mi for the Poisson distribution).
Compute the quantity: C = Si=16 (Oi – Mi)2/si2 which is distributed as Chi-squared with 5 degrees of freedom. Given the genotype frequencies computed in b), test the null hypothesis that the observed data is 95% likely to be in HWE.
Genotype Oi = Observed Number of Individuals
A1-A1 76
A1-A2 54
A1-A3 33
A2-A2 18
A2-A3 16
A3-A3 3

Explanation / Answer

Genotype Oi = Observed Number of Individuals A1-A1 76 A1-A2 54 A1-A3 33 A2-A2 18 A2-A3 16 A3-A3 3 (A1+ A2+ A3)^2=(p+ q +r)^2 P^2= A1-A1 = 76/200=0.38 q ^2= A2-A2 =18/200=0.09 r^2 =A3-A3= 3/200=0.015 2pq =A1-A2= 54/200=0.27 2pr= A1-A3= 33/200=0.165 2qr= A2-A3 =16/200=0.08 Frequency of A1 allele= =P^2 +2pq/2+2pr/2= 0.38 +0.27/2+0.165/2= 0.38 +0.135+0.0825 = 0.5975 Frequency of A2 allele= = q ^2+2pq/2+2qr/2=0.09+0.135+0.04=0.265 Frequency of A3 allele= = r^2 +2pr/2+2qr/2=0.015 +0.0825 +0.04=0.1375 p+ q +r=1=0.5975+0.265+0.1375=1

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.