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Hardy Weinberg equilibrium Problems and Solutions

Population Genetics: Hardy-Weinberg equilibrium 

Hardy Weinberg Equation

Problem:1

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=1-0.19
q2= 0.81
q=0.9
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
2pq=0.18
 

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
q2=0.000071428
q= square root of 0.000071428
q=0.0085
we know thatp+q=1 , then p=1-q
p=1-0.0085
p=0.9915
Frequency of heterozygotes i.e, 2pq= 2 x 0.9915 x0.0085
2pq=0.017
 

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?
Answer:
(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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