Answer:
C. 0.29
Explanation:
Hello!
The Hardy-Weinberg equilibrium establishes that the frequencies of the different genotypes that are formed during mating depend solely on the allelic frequencies of the population. Considering an autosomal locus, when the population is infinitely large and is not subject to any evolutionary force, both allelic and genotypic frequencies will remain constant from generation to generation, being able to predict their values for future generations.
When the population is in Hardy-Weinberg equilibrium, genotypic frequencies are given following the binomial model:
p²= f(A₁A₁) 2pq= f(A₁A₂) q²= f(A₂A₂)
and
p²= P 2pq= H q²= Q
Considering the autosomal locus with two alleles A₁ A₂, where A₁ shows complete dominance over A₂(resesive)
P: f(A₁A₁), H: f(A₁A₂) and Q: f(A₂A₂) represent the genotipic frequencies and P+H+Q=1
and p: f(A₁)= P and q: f(A₂) represent the allelic frequencies and p + q= 1
The given information is:
Phenotipe: "Ability to taste PTC" is dominant to the Phenotipe: "Incapable of tasting PTC"
This means that both homozygous-dominant and heterozygous individuals present the phenotipe "Ability to taste PTC" and only homozygous-resesive individuals present the phenotipe "Incapable of tasting PTC"
Sample size: 780
708 individuals were able to taste PTC ⇒ These represent the homozygous-dominant and heterozygous individuals of the population.
70 individuals were incapable of tasting PTC ⇒ These represent the homozygous-resesive individuals of the population.
The frequency of the non taster allele is:
The correct option is C
I hope it helps!