In a populaiton of Southwestern Jackalopes, you obseve the following numbers of
ID: 198419 • Letter: I
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 (WAA = 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)
Explanation / Answer
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
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
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