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You are working with a population of snails. During the mating season, you observe that individuals in the population will only mate with others of the same genotype. For example, Mm individuals will only mate with Mm individuals, and mm individuals will only mate with other mm individuals. There are only two alleles for this gene (M is dominant; m is recessive). You have determined that the frequency of the M allele is 0.5. After one generation, what is the expected genotype frequency for Mm individuals in this population

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marianaegarciaperedo

If matings are not random in a population and individuals mate with other individuals of similar genotype/phenotype, h0m0zyg0us frequencies increase. In this example, the genotype frequency for Mm is F(Mm) = 0.25.

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In the exposed example, one of the assumptions of Hardy-Weinberg equilibrium is not accomplished. There are non random matings.

Individuals mate with other snails of the same genotypes

MM  x  MM

Mm  x  Mm

mm  x  mm

We can assume this is an example of matings by similar phenotypes.

Eventually, this mating system leads to an increase in the h0m0zyg0us genotype frequency, at the expense of heter0zyg0us ones in loci that determine the trait.

Allelic frequencies do not change. Only genotypic frequencies do.

This mating system tends to separate the population into two subgroups, decreasing the amount of heter0zyg0us individuals.

          Matings              Progeny                              

  • MM  x  MM         4/4 MM
  • MM  x  Mm         1/4  MM + 2/4 Mm + 1/4 mm
  • mm  x  mm         4/4 mm

Zygotic population of the next generation

  • F(MM) = 4/4 MM + 1/4 Mm
  • F(Mm) = 1/2 Mm
  • F(mm) = 4/4 mm + 1/4 Mm

So, in the exposed example we know that the frequency of the dominant allele M is 0.5

f(M) = p = 0.5

knowing that p + q = 1, we can clear the equation to get the frequency of the recessive allele.

p + q = 1

0.5 + q = 1

q = 1 - 0.5

q = 0.5

f(m) = q = 0.5

Zygotic population of the next generation

  • F(MM) = 4/4 MM + 1/4 Mm

       F(MM) = p² + 1/4 (2pq)

       F (MM) = 0.5² + 1/4 (2 x 0.5 x 0.5) = 0.25 + 0.125

       F(MM) = 0.375

  • F(Mm) = 1/2 Mm

        F(Mm) = 1/2 (2pq)

        F(Mm) = 1/2 (0.5)

        F (Mm) = 0.25

  • F(mm) = 4/4 mm + 1/4 Mm

        F(mm) = q² + 1/4 (2pq)

        F(mm) = 0.5² + 1/4 (2 x 0.5 x 0.5) = 0.25 + 0.125

        F(mm) = 0.375

The expected genotype frequency for Mm individuals in this population is F(Mm) = 0.25.

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You can learn more about mating systems at

https://brainly.com/question/13007693?referrer=searchResults

https://brainly.com/question/19186330?referrer=searchResults

https://brainly.com/question/15737843?referrer=searchResults

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