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 640 BB genotypes, 320 Bb genotypes and 40 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.)

60% of the BB genotypes survive to adulthood, 60% of the Bb genotypes survive, and only 15% of the bb genotypes survive. Calculate the relative fitness values for each genotype.

WBB

WBb

Wbb

Now calculate the genotype frequencies at adulthood (after selection) by multiplying the genotype 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 p, the frequency of B after selection (adults) and provide one sentence of interpretation.

Show that the population of adults (post selection) has deviated somewhat from HWE.   Explain.

Explanation / Answer

Calculate genotype frequencies:

Freq BB = BB genotype/total = 640/1000 = 0.64 or 64%

Freq Bb = Bb genotype/total = 320/1000 = 0.32 or 32%

Freq bb = bb genotype/total = 40/1000 = 0.04 or 4%

Allele frequencies:

Frequency of allele B= (640*2)+(320*1)/Total alleles = 1600/2000 = 0.8 or 80%

Frequency of allele b= (320*1)+(40*2)/Total alleles = 400/2000 = 0.2 or 20%

p (at birth) = 80% (Frequency of allele B)

Relative Fitness (w) is the survival rate of a genotype relative to the maximum survival rate of other genotypes in the population.

WBB = 60/60 =1 (maximum survival rate)

WBb= 60/60 =1

Wbb= 15/60 =0.25

Genotype frequencies at adulthood:

Freq BB after selection = 60% survives= 60/100 *640=384

Freq Bb after selection = 60% survives= 60/100 *320=192

Freq bb after selection =15% survives= 15/100 *40=6

(Frequency of allele B) =p (adult) = (384*2)+(192*1)/1164(Total allele)=960/1164=0.82 or 82%

Frequency of allele B has increased by 2% approx.

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