Question 2 (4 pts). A population of squirrels in H-W equilibrium has allele freq

Question

Question 2 (4 pts). A population of squirrels in H-W equilibrium has allele frequencies of 0.8 for allele S and 0.2 for allele s. A parasite infests the population, causing mortality and reducedftilty so that there is a selection pressure on the population. a. What is the genotype frequency of EACH GENOTYPE BEFORE the parasite infests the population? (1pt) b. (3 pts) If the number of adults and the number of offspring is as follows: (3 pts)If Genotypes Ss 28 104 12 Number of adults 40 Number of offspring120 after infestation 64 What is the fitness (w) of each genotype?

Explanation / Answer

Part a-

Frequency of allele S =0.8

Frequency of allele s =0.2

There are three gentoypes for two alleles; SS, Ss, and ss.

Frequency of genotype SS =0.8 x0.8 =0.64

Frequency of genotype Ss =0.8 x 0.2 =0.16

Frequency of genotype ss =0.2 x0.2 =0.04

Part b-

Now, after infestation in the offspring we calculate the frequency of all genotypes.

Frequency of genotype SS = Offspring of the genotype ÷Total offspring

= 120 ÷288

= 0.42

We here conclude that SS is top survived with maximum frequency, on the reference of this we will calculate the relative fitness w.

Frequency of genotype Ss = Offspring of the genotype ÷Total offspring

= 104 ÷288

= 0.36

Frequency of genotype ss = Offspring of the genotype ÷Total offspring

= 64 ÷288

= 0.22

Now relative fitness of genotype SS after infestation = Frequency after infestation÷ Frequency of top survived

= 0.42 ÷ 0.42

=1

Relative fitness of genotype Ss after infestation = Frequency after infestation÷ Frequency of top survived

= 0.36 ÷ 0.42

=0.857

Relative fitness of genotype ss after infestation = Frequency after infestation÷ Frequency of top survived

= 0.22 ÷ 0.04

=0.523

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