#### 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?**

^{2}+2pq= 0.19

applying the Hardy-Weinberg Equation, p

^{2}+2pq+q

^{2}=1

q

^{2}=1-0.19

q

^{2}= 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?**

^{2}= 1/14,000

q

^{2}=0.000071428

q= square root of 0.000071428

q=0.0085

*we know that*p+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?**

(i) Here frequency of all dominant phenotypes, (p2+2pq) =60% =60/100 =0.6

then applying the Hardy - Weinberg Equation, p

^{2}+ 2pq +q

^{2 }=1

here p

^{2}+2pq = 0.6

then q

^{2}= 1 - (p

^{2}+2pq)

q

^{2 }= 1 - 0.6

q

^{2}= 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 ( p

^{2}+ 2pq ) = 0.6

Frequency of homozygous plant that germinate ( p

^{2}) = (0.37)

^{2 }= 0.14

Frequency of homozygous plants among plants that germinate = p

^{2}/ (p

^{2}+2pq)

= 0.14 / 0.6

__= 0.23__

Tags:
CSIR genetics questions
Hardy-Weinberg
Hardy-Weinberg equilibrium problems
Hardy-Weinberg practice problems
population genetics
population genetics problems