Question

In a populaiton of Southwestern Jackalopes, you obseve the following numbers of genotypes for the A locus:

AA = 10

Aa = 15

aa = 5

Given these amounts, what are the genotype and allele frequencies in this population? (please round to 2 significant figures)

p² =

2pq =

q²=

p =

q =

Assuming the AA genotype has a fitness of 0.5 (W_AA = 0.5), what are the expected allele and genotype frequencies in the next generation?

p =

q =

p²=

2pq=

q²=

Did out population of Southwestern Jackalopes evolve? (yes or no)

Answer 1

Information provided:

AA = 10

Aa = 15

aa = 5

Total no of individuals = 30

Frequency of AA genotype = 10/30 = 0.33

Frequency of Aa genotype = 15/30 = 0.5

Frequency of aa genotype = 5/30 = 0.17

The observed allele frequency is calculated as follows:

Total no. of alleles = (10 x 2) + (15 x 2) + (5 x 2) = 20 + 30 + 10 = 60

No. of A alleles = (AA genotype x 2) + (Aa genotype) = (10 x 2) + 15 = 35

A allele frequency = 35/60 = 0.58

Frequency of a allele = (15+10) /60 = 0.42

Therefore, p² = 0.33, 2pq = 0.5, q² = 0.17, p = 0.58, q = 0.42

Genotypes	AA	Aa	aa	alleles	Frequency
Initial freq	0.33	0.5	0.17		p = 0.58 q = 0.42
Fitness(w)	0.5	1	1
Product	0.165	0.5	0.17	0.165+0.5+0.17 = 0.835
Frequency after selection	0.197	0.6	0.21		p’ = 0.197=(0.6/2) = 0.497 q’ = 0.20+0.3 = 0.51
Frequency among offspring	(p’)2= 0.25	2pq = 0.506	(q’)2 = 0.26