Question

In a species of bird, the allele for the striped trait is recessive to the allele for the speckled trait. Suppose that a coastal bird population has a striped phenotype frequency of and an inland population has a striped phenotype frequency of The two populations are isolated and in Hardy–Weinberg equilibrium. Let represent the allele frequency of the coastal population and represent the allele frequency of the inland population. Calculate the allele frequency for the striped allele for each population. Round your answers to the nearest hundredth.

Answer 1

Answer:

The answer is "0.42"

Explanation:

Please find the complete question in the attached file.

Coastal community striped allele intensity $= q = 0.62$

Landlocked community, strip intensity allele$=q'=0.40$

Its eliminated allele frequency for movement $= q"=0.42 \ in \ indoors$community And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts$=q_2=0.39$

Heritable (striped) allele frecency:

$\to q=\sqrt{q_2}$

      $=\sqrt{0.39} \\\\ =0.624 \\\\ =0.62$

(round up to the closest cent) is therefore the only frequency of coastal people.

The genome of pulling function in host population$= q'_2=0.16$ Heterozygous feature regularly.

Therefore the inland community rate of recessed (striped), allele rate:

$\to q'= \sqrt{q'_{2}}= \sqrt{0.16}=0.40$

Following migration;

Its percentage of coastal migrants:

$m=10\%$

   $= \frac{10}{100} \\\\= 0.1$

Coastal population Non-immigrant percentage of coastal residents:

$=1- m\\\\=1-0.1\\\\ =0.9$

Stripping coastal community of allele rate after immigration:

$(q")=\text{(stripping of migrant only freq)}\times (m)+\text{(stripping allele baud rate for non-immigrants)} \times ( 1-m)$

       $= [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42$