Consider a deleterious recessive allele that is present in t

Consider a deleterious recessive allele that is present in two different populations. In population 1, the recessive allele is present at a frequency of 0.1, whereas in population 2, it is present at a frequency of 0.9. me that the genotype frequencies in each population are initially in Hardy-Weinberg equilibrium. a. of all the copies of the deleterious recessive alleles in population 1, what percentage of those copies are found in homozygous individuals? Show your calculations, but you do not need to write out any explanation. b. of all the copies of the deleterious recessive alleles in population 2, what percentage of those copies are found in homozygous individuals? Show your calculations, but you do not need to write out any explanation. c. If individuals that are homozygous for the recessive allele have a relative fitness of 0.7 (in other words, W 1: W 1; and W 0.7), calculate the expected change in allele frequency caused by selection after one generation in each of the two populations. Show your work and state which of the populations is expected to experience a higher absolute rate of change in allele frequency [no further explanation requiredj d. Briefly describe how your answers in parts (a) and (b) can explain your conclusion for part (c). In other words, how does the current frequency of a recessive allele affect how fast the allele frequency can change in response to selection?

Solution

c) Population 1, p = 0.9, q = 0.1

Genotype AA Aa aa
Genotype Frequency p² = 0.81 2pq = 0.18 q² = 0.01

Relative Fitness W_AA =1    W_Aa = 1 W_aa = 0.7

W = average fitness = (p²W_AA)+ (2pqW_Aa)+ (q²W_Aa)

= (0.81 x 1) + (0.18 x 1) + (0.01 x 0.7)

W = 0.997

Frequency of A after one generation of selection:
p\' = p² W_AA/W + pqW_Aa/W

= 0.812 + 0.18 = 0.992

Frequency of a after one generation of selection:

(1-p’) or q\'= q² Waa/W + pqWAa/W

q\' = 1- p = 1 - 0.992 = 0.008

Population 2, p = 0.1, q = 0.9

Genotype AA Aa aa
Genotype Frequency p² = 0.01 2pq = 0.18 q² = 0.81

Relative Fitness W_AA =1    W_Aa = 1 W_aa = 0.7

W = average fitness = (p²W_AA)+ (2pqW_Aa)+ (q²W_Aa)

= (0.01 x 1) + (0.18 x 1) + (0.81 x 0.7)

W = 0.757

Frequency of A after one generation of selection:
p\' = p² W_AA/W + pqW_Aa/W

= 0.013 + 0.237 = 0.25

Frequency of a after one generation of selection:

(1-p’) or q\'= q² Waa/W + pqWAa/W

q\' = 1- p = 1 - 0.25 = 0.75

In population 1, p = 0.9, p\' = 0.992, q = 0.1, q\' = 0.008

In population 2, p = 0.1, p\' = 0.25, q = 0.9, q\' = 0.75

Population is expected to experience higher absolute rate of change in allele frequency because p has changed from 0.1 to 0.25.

d) Only 1 % is homozygous recessive so there is least chance for change in allele frequency after selection.

About 81 % is homozygous recessive so there is higher chance for change in allele frequency after selection.