Question

In a population of 100 rabbits 40 are short-eared. Short ears are recessive to long ears. There are only two alleles for this gene. Long- eared is 60 which is in a 1:1 ratio of homozygous and heterozygous. Find out the dominant allelic frequency?
(a) 0.55
(b) 0.45
(c) 0.09
(d) 0.37

Answer 1

Hint: The Hardy-Weinberg principle, also known as the Hardy- Weinberg equilibrium, model, theorem, or law, states in population genetics that allele and genotype frequencies in a population will remain constant in the absence of other evolutionary factors from generation to generation.

Complete answer:
Let’s assume that the given population is in Hardy-Weinberg equilibrium.
According to the Hardy- Weinberg principle,
${ p }+{ q }={ 1 }$
${ p }^{ 2 }+{ q }^{ 2 }+{ 2pq }={ 1 }$
Where frequencies of individuals (f) under random mating are
p is the dominant allele frequency.
Q is the recessive allele frequency.
f(AA) = ${ p }^{ 2 }$ for the AA homozygotes (dominant),
f(aa) = ${ q }^{ 2 }$ for the aa homozygotes(recessive),
and f(Aa) = 2pq for the heterozygotes.
Given,
${ q }^{ 2 }={ \cfrac { 40 }{ 100 } }$
So,
${ q }=\sqrt { 40/100 }$
${ q }={0.63 }$
Using equation ${ p }+{ q }={ 1 }$,
${ p }={ 1 }- { q }$
${ p }={ 0.37 }$
So, the correct answer is ‘${ 0.37 }$’.

Note:
- Here, the ratio of the frequency of dominant homozygotes and heterozygotes is given to create confusion.
- Without solving for the ratio of dominant homozygous and heterozygous, one can solve just using the data given for q ( homozygous recessive).
- The heterozygous individuals have one dominant allele and recessive allele but the phenotype is of the dominant allele.