Question

Allele frequencies at a wing length locus are measured for a natural population of a migratory cricket species as P(A₁) = 0.4 and Q (A₂) = 0.6. Observed genotype frequencies in the field population are shown below. Compute the expected frequency of the three genotypes if the population is in Hardy-Weinberg Equilibrium for this locus?

                                     Observed frequency                  Expected frequency      Show your work

             A₁A₁ =          0.28                

             A₁A₂ =          0.22                

             A₂A₂ =             0.50      

          

The expected frequency of the genotypes are:          

a) A₁A₁ =0.25              A₁A₂ = 0.50                 A₂A₂ = 0.25

b) A₁A₁ =0.28             A₁A₂ = 0.22                 A₂A₂ = 0.50

c) A₁A₁ =0.16              A₁A₂ = 0.48                 A₂A₂ = 0.36

d) A₁A₁ =0.36             A₁A₂ = 0.48                 A₂A₂ = 0.16

e) None of the above

Compute the value of the inbreeding coefficient (F) for this population. The inbreeding coefficient is

A) 0

B) 0.043

C) 0.43

D) 0.54           

E) 1

Please show work, Thanks!

Based on your calculation is there evidence for assortative mating in this population?

a) Yes                          b) No                c) Not enough information

Please explain what led to your answer, Thanks!

Answer 1

In population genetics, the Hardy–Weinberg principle, also known as the Hardy–Weinberg equilibrium, which states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

In the case of a single locus with two alleles denoted A for dominant and a for recessive with frequencies f(A) = p and f(a) = q, respectively, the expected genotype frequencies under random mating are f(AA) = p2 for the AA homozygotes, f(aa) = q2 for the aa homozygotes, and f(Aa) = 2pq for the heterozygotes.

As, here in this question the allele frequencies are given P(A1) = 0.4 and Q (A2) = 0.6

So, the genotypic frequency will be A1A1=0.4*0.4=0.16, A1A2= 2*0.4*0.6=0.48, and A2A2=0.6*0.6=0.36

The expected frequency of the genotypes are: (C) A1A1 =0.16              A1A2 = 0.48                 A2A2 = 0.36

The inbreeding coefficient (F) is one minus the observed frequency of heterozygotes over that expected from Hardy–Weinberg equilibrium.

Here, E(f(A1A2)) is the expected genotypic frequency of A1A2= 0.48 and O(f(A1A2) is the observed genotypic frequency of A1A2=0.22 (given in question)

So, F= (0.48-0.22)/0.48=0.541

So, the correct option will be (D).

Assortative mating ( homogamy) is a mating pattern and form of sexual selection in which individuals with similar phenotypes mate with one another more frequently than would be expected under a random mating pattern.Disassortative mating (also known as negative assortative mating or heterogamy) means that individuals with dissimilar genotypes or phenotypes mate with one another more frequently than would be expected under random mating.

As, the genotypic frequency from this Hardy weinberg equilibrium is not similar, observed and expected frequencies are different. So, there is no evidence of assortive mating.

So, option (b) will be correct.