Question

1. A population of cats can be either black or white; the black allele (B) has complete dominance over the white allele (b). Given a population of 2,000 cats, 1540 black and 460 white, Determine the allele frequencies of the two characteristics.

2. A population of cats can be either black or white; the black allele (B) has complete dominance over the white allele (b). Given a population of 2,000 cats, 1540 black and 460 white, determine the frequency of individuals with the dominant genotype and the heterozygous genotype.

3. A population of cats can be either black or white; the black allele (B) has complete dominance over the white allele (b). Given a population of 2,000 cats, 1540 black and 460 white, Determine frequency of individuals with the recessive genotype

Answer 1

Question 1

Answer: Given, Black allele (B) is dominant and white allele (b) is recessive. In a population of 2000 cats, 1540 are black that include both the genotype BB and Bb and 460 white cats that include genotype bb.

According to Hardy Weinberg equilibrium

p + q =1 and p² + 2pq +q² = 1

p = frequency of dominant allele (B)

q = frequency of recessive allele (b)

p² = frequency of homozygous dominant genotype individuals (BB)

2pq = frequency of heterozygous dominant genotype individuals (Bb)

q² = frequency of homozygous recessive genotype individuals (bb).

To solve this question, first we need to calculate frequency of white cats because it has only one genotype bb. It is difficult to calculate frequency of black cats based on the information given in the question because black cats have two genotypes BB and Bb.

Frequency of individuals (white cats; q²) = Number of white cats/ total number of individuals = 460/2000 = 0.23

Frequency of recessive allele q can be calculated as below

q² = 0.23; q = √ 0.23 = 0.48

According to Hardy Weinberg equilibrium

p + q =1

Frequency of dominant allele p can be calculated as below

p = q -1; p = 0.48 -1 = 0.52

So, Allele frequency of dominant character is 0.52 and allele frequency of recessive character is 0.48.

Question 2

Answer: According to hardy weinberg equilibrium

p² + 2pq +q² = 1

p² = frequency of homozygous dominant genotype individuals (BB)

2pq = frequency of heterozygous dominant genotype individuals (Bb)

q² = frequency of homozygous recessive genotype individuals (bb).

From the first question we know that allele frequency of dominant (B) character, p= 0.52

Allele frequency of recessive (b) character, q is 0.48.

Frequency of homozygous dominant genotype individuals (BB) = p² = (0.52)² = 0.27

Frequency of heterozygous dominant genotype individuals (Bb) = 2pq = 2 X 0.52 X 0.48 = 0.5

Question 3

Answer: From the question 1, we know thatallele frequency of recessive (b) character, q is 0.48.

Frequency of homozygous recessive genotype individuals (bb) = q² = (0.48)² = 0.23