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Hardy Weinberg equilibrium Problems and Solutions

Population Genetics: Hardy-Weinberg equilibrium 

Hardy Weinberg Equation
Hardy Weinberg Equation


What is the frequency of heterozygotes Aa in a randomly mating population in which the frequency of all dominant phenotypes is 0.19?
Here (dominant phenotypes) p2+2pq= 0.19
applying the Hardy-Weinberg Equation, p2+2pq+q2=1
q2= 0.81
we know that p+q=1, then p=1-q
here p=1-0.9
p =0.1
Frequency of heterozygyotes Aa (2pq) = 2 x 0.1 x 0.9

Problem: 2

αβγ is an autosomal recessive disorder of man. The frequency of affected newborn infants is about 1 in 14,000. Assuming random mating, what is the frequency of heterozygotes?
Here q2= 1/14,000
q= square root of 0.000071428
we know thatp+q=1 , then p=1-q
Frequency of heterozygotes i.e, 2pq= 2 x 0.9915 x0.0085

Problem: 3

In a plant species the ability to grow in soil contaminated with nickel is determined by a dominant allele.
(i) If 60% of the seeds in a randomly mating population are able to germinate in contaminated soil, what is the frequency of the resistance allele?
(ii) Among plants that germinate, that proportion are homozygous?
(i) Here frequency of all dominant phenotypes, (p2+2pq) =60% =60/100 =0.6
then applying the Hardy - Weinberg Equation, p2 + 2pq +q2 =1
here p2+2pq = 0.6
then q2= 1 - (p2+2pq)
q2 = 1 - 0.6
q2 = 0.4
q = square root of 0.4
q = 0.63
Frequency of resistance allele p= 1-q
p = 1- 0.63
p = 0.37
(ii) Frequency of plants that germinate ( p2 + 2pq ) = 0.6
Frequency of homozygous plant that germinate ( p2 ) = (0.37)2 = 0.14
Frequency of homozygous plants among plants that germinate = p2 / (p2+2pq)
= 0.14 / 0.6
= 0.23
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