I just want to see if I filled out everything correctly. Thanks. Does my explana
ID: 94674 • Letter: I
Question
I just want to see if I filled out everything correctly. Thanks. Does my explanation make sense?
1. In an electrophoretic study of enzyme variation in a species of pelican, you find 77 A1A1, 45 A1A2, and 18 A2A2 individuals at a particular locus in a sample of 140. What are the allele frequencies for the A1 and A2 alleles? Calculate the genotype frequencies for this locus.
Genotypes
A1A1
A1A2
A2A2
Number of individuals
77
45
18
Genotype frequencies
0.71 x 0.71 x 140 = 70.57
2 x 0.71 x 0.29 =
57.65
0.29 x 0.29 x 140 =
11.77
Frequency of A1 allele
199/280 = 0.71
Frequency of A2 allele
81/280 = 0.29
Show whether these genotype frequencies are in Hardy-Weinberg equilibrium.
The genotype frequencies calculated are what is to be expected of each genotype, however the number of individuals reflected shows that while it is close to theoretical values, it is not exact and so the population is not in expected Hardy – Weinberg equilibrium.
Genotypes
A1A1
A1A2
A2A2
Number of individuals
77
45
18
Genotype frequencies
0.71 x 0.71 x 140 = 70.57
2 x 0.71 x 0.29 =
57.65
0.29 x 0.29 x 140 =
11.77
Explanation / Answer
Frequency of A1 allele = F(A1A1) + F(A1A2)/2
= 0.77 + (0.45)/2
= 0.77 + 0.225
= 0.995
Frequency of A2 allele = F(A2A2) + F(A1A2)
= 0.18 + 0.225
= 0.405
When the population is in Hardy-Weinberg's equilibrium,
p + q = 1 and p2 + 2pq + q2 =1
Here, p + q = 0.995 + 0.405 = 1.4
p2 + 2pq + q2
= (0.995)2 + 2(0.995)(0.405) + (0.405)2
= 0.99 + 0.80 + 0.16
= 1.95
Both values are not equal to 1. So, the population is not in Hardy-Weinberg's equilibrium.