Question

In: Biology

1.Imagine a locus with two alleles, A and a, in a wild population of touch-me-not (Impatiens...

1.Imagine a locus with two alleles, A and a, in a wild population of touch-me-not (Impatiens capensis) plants,which has both out crossing and selfing flowers.40 AA, 2 Aa, and 18 aa genotypes are found among 60 individuals.

a.Calculate the allele frequencies p= freq(A) and q=freq(a)

.b.How many individuals of each genotype would be expected if the HW assumption were met?

c.Is this population in HWE? Explain your answer.

d.Uh oh,bad year for pollinators!What would be the genotype frequencies in the next generation if the plant only produced selfed progeny? (Hint: the coefficient of inbreeding for selfing in a diploid is 0.5.This is because the chance of the two alleles in the selfed offspring being identical-by-descent one generation back is 1/2).

e.Oh no, it looks like the pollinators are gone forever!What would be the frequency of heterozygotes at equilibrium (i.e. after many,many generations of selfing)

Solutions

Expert Solution

a. Allele frequency can be calculated for a population by adding twice of number of individuals homozygous for the allele to the number of heterozygous individuals and then dividing it by twice of total population. This twice of total population is the total number of alleles in the population.

Here , p = [(2×40) + 2]÷ 60×2

= 82÷120 =0.68

q = [(2×18)+2]÷60×2

=38÷120 = 0.32

b. According to HWE, p2 = frequency of homozygous dominant individuals

= (0.68)2

= 0.46

Number = .46× 60=28

2pq = heterozygous individuals frequency

= 2 ×0.68×0.32

= 0.44

Number = 0.44×60 = 26

q2 = homozygous recessive individuals frequency

= (0.32)2

= 0.10

Number = .10 ×60 = 6

c. The population is not in HWE because the expected number of individuals according to rule and the actual number of individuals for each genotype are different.

d. Genotype frequency during inbreeding can be calculated by using the formula , where F is inbreeding coefficient

AA = p2 + Fpq = (0.68)2 + 0.5 ×0.68×0.32

= 0.57

Aa = 2pq (1-F) = 2 × 0.68 × 0.32 (1-0.5)

= 0.22

aa = q2 + Fpq = (0.32)2+ 0.5 ×0.68×0.32

= .21

Sum of all the frequencies should be 1.

gladiator answered 3 months ago
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