You have a population of 1000 people, and you are looking at the population genetics of a blood marker that has two alleles. The dominant allele (Y) makes the protein, and the recessive allele (y) does not make the protein. When you test the blood of the population, you find:
490 people have the genotype YY
420 people have the genotype Yy
90 people have the genotype yy
What are the allele frequencies (Y and y)?
What are the expected genotype frequencies (YY, Yy, and yy)?
Is the population in equilibrium? Yes or no?
If the population is not in equilibrium, which of the assumptions of Hardy-Weinberg do you think may have been violated?

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Adepeju Adepeju · Answer 1 · 2020-02-04T14:04:20+01:00

Answer:

Genotypic frequency (How often the allele combination shows YY, Yy or yy)

YY - 490/1000 = 0.49

Yy - 420/1000 = 0.42

yy - 90/1000 = 0.09

Allelic frequency (How often the allele shows Y or y)

P = Frequency of Y = (490+420)/1000 = 0.91

q = Frequency of Y =(420+90)/1000 = 0.51

The population isn't in equilibrium according to Hardy-Weinberg because p + q is more than one.

Hardy-Weinberg equation is p+q =1

P^2 +2pq + q^2 = 1

The Hardy-Weinberg assumption the population violated is that there is gene flow as seen in the Allelic frequency that is more than 1

Explanation:

When a population is in Hardy-Weinberg equilibrium for a gene, it is not evolving, and allele frequencies will stay the same across generations.

There are five basic Hardy-Weinberg assumptions: no mutation, random mating, no gene flow, infinite population size, and no selection.

If the assumptions are not met for a gene, the population may evolve for that gene (the gene's allele frequencies may change).