Question

A particular population of chickens has a frequency for the dominant allele as 0.70 and a frequency for the recessive allele as 0.30. Which expression is the correct way to calculate the frequency of individuals that are heterozygous?

A.
(0.70) × (0.30)
B.
(0.70) – (0.30)
C.
2 × (0.70) × (0.30)
D.
2 × (0.70 – 0.30)

Answer 1

Answer:

C is the correct answer

As the formula

P^2 + q^2 + 2pq = 0

Answer 2

2 × (0.70) × (0.30).

Explanation:

• The Hardy Weinberg equilibrium is used to study the population genetics and can also be applied on the human population but for the limited traits only.
• This equilibrium states that the allele and genotype frequency of the population remains constant and stable if no external force like selection, mutation and recombination occurs in the population.
• The p represents the dominant allele frequency and q represent the recessive allele frequency.
• Then, the heterozygous frequency is 2pq. The correct option is option (c).

# Learn more with brainly about Hardy Weinberg equilibrium:

https://brainly.in/question/6746691.

https://brainly.in/question/13446998.