The EGLN1 gene helps regulate body oxygen levels. One allele H is dominant over h, and helps increase fitness at high altitude (where oxygen is low). In a study of families in Tibet, researchers found 422 HH individuals, 377 Hh, and 201 hh individuals. The study showed that infant mortality was lower in Hh individuals than hh, and lowest in HH individuals. What is the genotype frequency for heterozygotes?

Respuesta :

Answer:

Genotype frequency for heterozygotes is equal to [tex]37.7[/tex] %

Explanation:

Given

Number of individuals with genotype HH - [tex]422[/tex]

Number of individuals with genotype hh - [tex]201[/tex]

Number of individuals with genotype Hh - [tex]377[/tex]

Sum of all individuals with varied genotypes

[tex]422 + 377 + 201\\= 1000[/tex]

Genotype frequency of any  specific genotype with in a population is equal to  total number of individuals with that genotype divided by total population.

Hh is a heterozygous genotype.

Thus, genotype frequency for heterozygotes

[tex]\frac{377}{1000} * 100\\= 37.7[/tex]

Genotype frequency for heterozygotes is equal to [tex]37.7[/tex] %

To get the frequency of heter0zyg0tes, we just need to divide the total number of heter0zyg0us individuals by the population size. In this example, the genotype frequency for heterozygotes is 0.377 = 37.7%.

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Available data:

  • allele H is dominant over h
  • 422 HH individuals
  • 377 Hh individuals
  • 201 hh individuals
  • Total number of individuals = N = 422 + 377 + 201 = 1000

We need to get the genotypic frequency of heter0zyg0us individuals in this Tibet population.

We already know the number of heter0zyg0us individuals and the total number of individuals in the population. These data are enough to calculate genotypic frequency -F(Hh)-.

All we need to do is to divide the number of heter0zyg0us individuals by the population size.

  • Hh = 377 individuals
  • N = 1000 individuals

F(Hh) = Hh/N = 377 / 1000 = 0.377 = 37.7%

So, the genotype frequency for heter0zyg0tes is 0.377 or 37.7%.

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