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:
a. (3 points) What are the frequencies of the three alleles A1, A2, A3?
b. (3 points) What do you expect these six genotypes to be under HWE?
c. (7 points) 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).
d. (7 points) 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

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