Suppose a population is carrying a condition controlled by two alleles: A (dominant) and a (recessive). Only homozygous individu
als 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%
Answer: The correct answer is - B. 4% Explanation: As per the given information in the question, the genetic condition is homozygous recessive (aa).
The frequency of dominant (A) and recessive allele (a) are given, which are 20% (that is 0.20) and 80% (0.80) respectively. According to Hardy-Weinberg Principle -
p+q=1 , p{2} +2pq+q^{2} =1
Where, p and q denotes frequency of dominant and recessive alleles respectively.
p2 = frequency of homozygous dominant genotype (AA)
q2= frequency of homozygous recessive gentotype (aa)
2pq= frequency of heterozygous genotypes (Aa).
We have given value of -
p= 0.80 and q= 0.20
To calculate the frequency of homozygous recessive condition (aa) that is represented by q2-
q2 = 0.20 × 0.20 = 0.0400 In percentage - [0.0400/10000] ×100
= 4%
Thus, option B) is the right answer.
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