Question

Sandra and John have a flock of 4500 sheep, and they started to concern about the wool color after the flock has been closed for random breeding/mating for an extended period. One day, they have counted and found that 950 sheep were white, and the remainders were black. Assume that black wool is controlled by a single locus and the allele for black color is completely dominant to white. Let B represent the black allele and b represent the while allele. Assume HW equilibrium,

A) What is the observed genotypic frequency for the bb genotype?
B) What are the expected gene frequencies for B and b alleles?
C) What are the expected genotypic frequencies and counts for each of the three genotypes?

Answer 1

Part A:

Frequency of bb genotype=0.2111

Part B:

Frequency of B alleles is denoted by x=0.5407

Frequency of b alleles is denoted by y=0.4593

Part C:

Frequency of BB genotype=0.2924

Frequency of bb genotype=0.2111

Frequency of Bb genotype=0.4966

Step-by-step explanation:

Given:

Total Sheep= 4500

White Sheep=950

B for black allele

b for white allele

Part A:

`Frequency\ of\ white\ sheep=(White\ Sheep)/(Total\ Sheep)`

`Frequency\ of\ white\ sheep=(950)/(4500)\\Frequency\ of\ white\ sheep=0.211`

Frequency of white sheep= Frequency of bb genotype=0.2111

Further:

Let frequency of b alleles is denoted by y.

`Frequency\ of\ white\ sheep=y^(2)= 0.211\\y=√(0.211)\\y=0.4593`

Frequency of b alleles is 0.4593.

Part B:

Sum of the frequencies of B and b alleles is 1

Let frequency of B alleles is denoted by x

Let frequency of b alleles is denoted by y

x+y=1

y is calculated in part A and has value=0.4593

x+0.4593=1

x=1-0.4593

x=0.5407

Frequency of B alleles is denoted by x=0.5407.

Frequency of b alleles is denoted by y=0.4593.

Part C:

Three genotypes are BB, Bb, bb

In case of three genotypes frequency is:

`F=x^2+2xy+y^2`

Now
`x^2\\` is for BB genotypes

From Part B;

x=0.5407

`x^2=(0.5407)^2\\x^2=0.2924`(Frequency of BB genotype)

Now
`y^2\\` is for bb genotypes

From Part A:

y=0.4593

`y^2=(0.4593)^2\\y^2=0.211` (Frequency of bb genotype)

From above frequency Formula:

Now
`2xy` is for Bb genotype:

2*x*y=2*0.5407*0.4593

2*x*y=2*0.2483=0.4966 (Frequency of Bb genotype)