Question

A population of dogs can be either brown or black; the black allele (B) has complete dominance over the brown allele (b). Given a population of 2,000 dogs, 1660 black and 340 brown, determine The frequency of individuals with the dominant phenotype.

Answer 1

Answer:

The frequency of individuals with the dominant phenotype is 0.83.

Explanation:

We are provided with:

Black allele has completed dominance over brown allele

T. no of dogs (dominant)= 2000

No. of black dogs (dominant) = 1660

From Hardy-weinberg equilibrium

Frequency of individual - (individual/ Total population)

Frequency of black dogs = 1660/2000 =

0.83

So, The frequency of individuals with the dominant phenotype is 0.83

Answer 2

Answer: The frequency is p=0.911

Explanation: Black allele is dominant over brown. According to the data, there are 1660 black in a population of 2000 dogs. The Hardy-Weinberg Law states that $p^{2} + 2pq + q^{2} = 1$, where:

p is the frequency of the dominant allele;

q is the frequency of the recessive allele;

So, for the dominant allele:

p² = $\frac{1660}{2000}$

$p^{2} = 0.83$

$p = \sqrt{0.83}$

p = 0.911

Thus, individuals with the dominant phenotype has a frequency of p = 0.911 of happening.