Answer
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Hint: With the aid of the Hardy Weinberg equilibrium, the question can be solved. In a non-evolving, stable population, according to the Hardy Weinberg equilibrium, the square of the number of allele frequencies gives the individual genotype frequencies.
Complete answer:
The genotype frequencies in a population at equilibrium for a locus of two alleles A and a having frequencies p and q, respectively, are:
$AA={ p }^{ 2 }$,
$Aa=2pq$,
$aa={ q }^{ 2 }$
and
$p+q=1$
According to Hardy-Weinberg equilibrium,
${ p }^{ 2 }+2pq+{ q }^{ 2 }\rightarrow 1$
Where ${ p }^{ 2 }$ is equal to the frequency of homozygous genotype (AA)
pq is the frequency of heterozygous genotype (Aa)
${ q }^{ 2 }$ is the frequency of homozygous genotype (aa)
The total no of individuals is 1000
${ p }^{ 2 }=\dfrac { 360 }{ 1000 } =0.36$
${ p }q=\dfrac { 240 }{ 1000 } =0.24$
${ q }^{ 2 }=\dfrac { 160 }{ 1000 } =0.16$
We know,
${ p }^{ 2 }=0.36$
$\Rightarrow { p }=\sqrt { 0.36 } =0.6$
Thus, the frequency of allele A in the population is 0.6.
Note: The Hardy-Weinberg equilibrium is a theory which states that in the absence of disturbing factors, genetic variation in a population will remain constant from one generation to the next. In a large population with no destructive conditions, when mating is random, the law assumes that both genotype and allele frequencies will stay constant since they are in balance.
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. These factors include genetic drift, choice of mate, assortative mating, sexual selection, mutation, gene flow, natural selection, genetic hitchhiking, meiotic drive, the bottleneck of the population, inbreeding, and founder effect.
Complete answer:
The genotype frequencies in a population at equilibrium for a locus of two alleles A and a having frequencies p and q, respectively, are:
$AA={ p }^{ 2 }$,
$Aa=2pq$,
$aa={ q }^{ 2 }$
and
$p+q=1$
According to Hardy-Weinberg equilibrium,
${ p }^{ 2 }+2pq+{ q }^{ 2 }\rightarrow 1$
Where ${ p }^{ 2 }$ is equal to the frequency of homozygous genotype (AA)
pq is the frequency of heterozygous genotype (Aa)
${ q }^{ 2 }$ is the frequency of homozygous genotype (aa)
The total no of individuals is 1000
${ p }^{ 2 }=\dfrac { 360 }{ 1000 } =0.36$
${ p }q=\dfrac { 240 }{ 1000 } =0.24$
${ q }^{ 2 }=\dfrac { 160 }{ 1000 } =0.16$
We know,
${ p }^{ 2 }=0.36$
$\Rightarrow { p }=\sqrt { 0.36 } =0.6$
Thus, the frequency of allele A in the population is 0.6.
Note: The Hardy-Weinberg equilibrium is a theory which states that in the absence of disturbing factors, genetic variation in a population will remain constant from one generation to the next. In a large population with no destructive conditions, when mating is random, the law assumes that both genotype and allele frequencies will stay constant since they are in balance.
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. These factors include genetic drift, choice of mate, assortative mating, sexual selection, mutation, gene flow, natural selection, genetic hitchhiking, meiotic drive, the bottleneck of the population, inbreeding, and founder effect.
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