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A large population of laboratory animals has been allowed to breed randomly for a number of generations. After several generations, 25 % of the animals display a recessive trait (a a) , the same percentage as at the beginning of the breeding program. The rest of the animals show the dominant phenotype, with heterozygotes indistinguishable from the homozygous dominants.
What proportion of the population is probably heterozygous (A a) for this trait? (A) 0.05 (B) 0.25 (C) 0.50 (D) 0.75



Answer :

The correct option is (C) 0.50

0.50 proportion of the population is heterozygous  (Aa) for this trait.

Indistinguishable from homozygous dominants are heterozygotes.

As a result, children having the genotypes Aa and AA would have the same phenotype.

We are aware that 0.25 percent of the population has recessive individuals.

Taking into account the population's Hardy-Weignbergs equilibrium,

q² = 0.25

q = 0.5

We know that based on the first equilibrium equation of Hardy-Weinberg.

p+q = 1

0.5+p =1

p = 1 - 0.5

p = 0.5

Here, p stands for the population's frequency of the A allele.

Therefore, the proportion of heterozygous is 0.50

Learn more about the Hardy-weinberg's equilibrium with the help of the given link:

https://brainly.com/question/16823644

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