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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?

Respuesta :

zainsubhani

Answer:

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

Explanation:

Given:

Total Sheep= 4500

White Sheep=950

B for black allele

b for white allele

Part A:

[tex]Frequency\ of\ white\ sheep=\frac{White\ Sheep}{Total\ Sheep}[/tex]

[tex]Frequency\ of\ white\ sheep=\frac{950}{4500}\\Frequency\ of\ white\ sheep=0.211[/tex]

Frequency of white sheep= Frequency of bb genotype=0.2111

Further:

Let frequency of b alleles is denoted by y.

[tex]Frequency\ of\ white\ sheep=y^{2}= 0.211\\y=\sqrt{0.211}\\y=0.4593[/tex]

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:

[tex]F=x^2+2xy+y^2[/tex]

Now [tex]x^2\\[/tex] is for BB genotypes

From Part B;

x=0.5407

[tex]x^2=(0.5407)^2\\x^2=0.2924[/tex](Frequency of BB genotype)

Now [tex]y^2\\[/tex] is for bb genotypes

From Part A:

y=0.4593

[tex]y^2=(0.4593)^2\\y^2=0.211[/tex]  (Frequency of bb genotype)

From above frequency Formula:

Now [tex]2xy[/tex] is for Bb genotype:

2*x*y=2*0.5407*0.4593

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

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