Suppose a population is carrying a condition controlled by two alleles: A (dominant) and a (recessive). Only homozygous individuals that have two copies of the recessive allele have the condition. If the a allele has a frequency of 20%, and the A allele had a frequency of 80%, what percentage of the population will have the condition?

A. 20%

B. 4%

C. 80%

D. 64%

Question

Biology

Guest

17 April, 17:23

Suppose a population is carrying a condition controlled by two alleles: A (dominant) and a (recessive). Only homozygous individuals that have two copies of the recessive allele have the condition. If the a allele has a frequency of 20%, and the A allele had a frequency of 80%, what percentage of the population will have the condition?

A. 20%

B. 4%

C. 80%

D. 64%

+3

Answers (2)

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Emiliano Martin · Answer 1

Let a be Q, all which is homozygous recessive = Q∧2

A = p, AA which is homozygous recessive = Q∧2

2pq = heterozygous

It is derived from p + q = 1

All those in a population which is Q = 20%

All A in the population (p) = 80%

Now that the disease is homozygous recessive therefore,

aa = qq or q * q = 0.20 * 2 = 4%

Then the answer is 4%.

Tough Guy · Answer 2

Its 4%.

process of elimination.

if 20% of the population have a, then only 4% have both recessive alleles.