We have a population of 400 animals that are being selected and bred for their g

Question

We have a population of 400 animals that are being selected and bred for their genotype at a specific locus. Animals are only bred to other animals with the same genotype at that locus. So animals with HH are only bred to other HH animals, Hh animals are only bred to Hh animals and hh animals are only bred to hh

a.If the parental generation has 100 HH, 200 Hh and 100 hh animals, what are the genotypic frequencies?

b. If the parental generation has 100 HH, 200 Hh and 100 hh animals, what are the allele frequencies?

c.After 2 generations of mating, what are the allele frequencies?

d.After 2 generations of mating, what are the genotypic frequencies?

e.Are the F2s exhibiting Hardy-Weinberg equilibrium? Explain why or why not.

3. What is the allele and genotypic frequency of the F3s if they are randomly mated to one another?

4. What is the type of mating described in 3 called? What is the type of mating in #2 called?

Explanation / Answer

In the given case, animals are mated on non-random or selective mating. So, they do not follow Hardy's-Weinberg equilibrium.

a). The parental genotypic frequencies of HH, Hh and hh are given as 1:2:1 ratio. The genotype HH are only mated with HH type. Thus, the resulting offspring are 100% HH type.

If two parents with genotype Hh are crossed together, the resulting offspring will have the genotypes HH, Hh and hh in 1:2:1 ratio. Means, the population of heterozygotes decline by 50%, the other 50% will be homozygotes.

Similarly, the offspring resulting from hh cross will be 100% hh.

2). In non-random mating, the genotypic frequencies do change but the allele frequencies do not change, i.e. the sum of alleles is always equal to 1. Given that the frequency of HH genotype (P1) is 100 (or 0.25 since 100/400 = 0.25), Hh genotype (P2) is 200 (or 0.5), and hh genotype (P3) is 100 (or 0.25). The allele frequencies can be calculated as below.

The frequency of H allele = P1 + 1/2 (P2) = 0.25 + 1/2 (0.5) = 0.5.

The frequency of h allele = P3 + 1/2 (P2) = 0.25 + 1/2 (0.5) = 0.5.

c) The allele frequencies do not change after two or twenty generations of mating, they remain same, i.e. p = q = 0.5

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