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 reduced fertility 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:

What is the fitness (w) of each genotype?

Genotypes SS Ss ss Number of Adults 40 28 12 Number of offspring after infestation 120 104 64

Explanation / Answer

a) The population is in H-W equilibrium.

The allele frequency for the allele S = 0.8

The allele frequency for the allele s = 0.2

Therefore, the genotype frequency of genotype SS = S2 = (0.8)2 = 0.64,

the genotype frequency of genotype Ss = 2Ss = 2 X (0.8) X (0.2) = (0.32)

and the genotype frequency of genotype ss = s2 = (0.2)2 = 0.04

The genotype with the greatest fitness is given a value of 1. The fittest genotype is the SS genotype with 120 offspring. Therefore fitness of SS = 120/120 = 1; fitness of Ss = 104/120 =0.867 and fitness of ss = 64/120 = 0.534.

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