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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%

Respuesta :

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%.
  1. 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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