In a population of fruit flies, viability selection on adults acts on the B locu

Question

In a population of fruit flies, viability selection on adults acts on the B locus. Eggs are produced by random mating, and there are 180 BB genotypes, 240 Bb genotypes and 80 bb genotypes in the juvenile stage Calculate genotype and allele frequencies. Freq BB = Freq Bb = Freq bb = p (at birth) = Because mating is random and selection has not yet acted, the population is at HWE. (You can confirm that if you wish, but not required here.) 80% of the BB genotypes survive to adulthood, 80% of the Bb genotypes survive, and only 4% of the bb genotypes survive. Calculate the relative fitness values for each genotype. W_BB W_Bb W_bb Now calculate the genotype frequencies at adulthood (after selection) by multiplying the allele frequencies at birth (you calculated these as observed values, but note they are also the expected values because the population is in HWE) by relative fitness and dividing by the total (mean population fitness). Freq BB after selection = Freq Bb after selection = Freq bb after selection = Calculate the frequency of B after selection (adults) and provide one sentence of interpretation.

Explanation / Answer

Answer:

From the data provided,

Number of BB genotypes = 180

Number of Bb genotypes = 240

Number of bb genotypes = 80

Total number of genotypes = 500

To Calculate

Freq BB = number of BB genotypes/total number of genotypes = 180/500 = 0.36

Freq Bb = number of Bb genotypes/total number of genotypes = 240/500 = 0.48

Freq bb = number of bb genotypes/total number of genotypes = 80/500 = 0.16

p ( at birth) =

Let p = f(B) = f(BB) + 0.5*f(Bb) = 0.36 + 0.5*(0.48) = 0.60

Let q = f(b) = f(bb) = 0.5*f(Bb) = 0.16 + 0.5*(0.48) = 0.40

80% of the BB genotypes survive to adulthood, 80% of the Bb genotypes survive and only 4% of the bb genotype survive. Calculate the relative fitness values for each genotype

WBB = 80/80 = 1

WBb = 80/80 =1

Wbb = 4/80 = 0.05

To calculate relative fitness, divide the number surviving by the highest fitness.

Mean fitness = p2*(WBB) + 2pq*(WBb) + q2*(Wbb)

= 0.602*(1) + 2*0.60*0.40*1 + 0.402*0.05 = 0.36 + 0.48 + 0.008 = 0.848

Calculate the genotype frequencies at adulthood (after selection) by multiplying the allele frequencies at birth by relative fitness and dividing by total (mean population fitness)

Freq BB after selection = (0.36*1)/0.848 = 0.4245

Freq Bb after selection = (0.48*1)/0.848 = 0.566

Freq bb after selection = (0.16*0.05)/0.848 = 0.0094

Calculate the frequency of B after selection (adults) and provide one sentence of interpretation

Let p = f(B) = f(BB) + 0.5*f(Bb) = 0.4245 + 0.5*(0.566) = 0.7075

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