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14 September, 09:46

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%

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Answers (2)
  1. 14 September, 10:00
    0
    Its 4%.

    process of elimination.

    if 20% of the population have a, then only 4% have both recessive alleles.
  2. 14 September, 10:04
    0
    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%.
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