Answer and Explanation:
a) Frequencies of A and G lalleles are as follows:
f(A) =( 84+24)/174 = 0.62
f(B) = (42+24)/174 =0.38
b) Expected genotype frequencies:
f(AA) = (0.62) (0.62) = 0.384
f (AG)= 2(0.62) (0.38) =0.471
f(GG) = (0.38) (0.38) = 0.144
c) Genotype Observed Expected O-E (O-E)2 (O-E)2/E
AA 42 33 9 81 2.45
AG 24 41 17 289 7.05
GG 21 13 8 64 4.92
Chi squared = 14.42
The number of degrees of freedom is the number of genotypes minusthe number of alleles= 3-2 =1
The p value is much less than 0.05, therefore we reject the hypothesis that these genotype frequencies may be expected from HArdy Weinberg equilibrum