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Select initial allele frequencies

2Click Next generation to wait a year for first generation of moths

3Click Capture moths to monitor population numbers

4Calculate phenotype frequencies in 5th generation. Record in Lab Data

5Calculate allele frequencies in 5th generation. Record in Lab Data

6Calculate genotype frequencies and number of moths in 5th generation. Record in Lab Data

Select initial allele frequencies 2Click Next generation to wait a year for first generation of moths 3Click Capture moths to monitor population numbers 4Calculate phenotype

Answer :

Final answer:

The question outlines an exercise in population genetics, focusing on tracking changes in allele and phenotype frequencies over generations with a hypothetical moth population. It allows the visualization of concepts like natural selection and genetic drift.

Explanation:

The steps you've provided outline a process for studying population genetics, specifically tracking the frequency of certain alleles and phenotypes within a population. In this exercise, you'd start with initial allele frequencies, then simulate the passing of multiple generations by clicking 'Next generation'. After a few generations (in this case, five), you capture moths to take note of the population numbers. At this point, you'll calculate the phenotype frequencies, allele frequencies, and the genotype frequencies, along with the total number of moths in the population. This process can help you understand how certain alleles and phenotypes change over time. It seems like an effective way to visualize concepts related to natural selection and genetic drift.

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Rewritten by : Barada

p₀=0.5, q₀ = 0.5.// p₅=0.86, q₅ = 0.14.// p₅² = 0.74, 2pq₅ = 0.24, q₅² = 0.02. // 1062 Black individuals DD, 344 Black, 29 White individuals dd.

Note: Due to technical problems, in the attached files you will find the complete explanation.

1) Phenotype frequencies

⇒ F(white) = 0.02

⇒ F(black) ≅ 0.98

2) Allele Frequencies

⇒ f₀(D) = p = 0.5

⇒ f₀(d) = q = 0.5

⇒ f₅(D) = p = 0.86

⇒ f₅(d) = q = 0.14

3) Genotypic Frequencies

⇒ F₅(DD) = p² ≅ 0.74

⇒ F₅(Dd) = 2pq = 0.24

⇒ F₅(dd) = q² ≅ 0.02

4) Number of individuals

⇒ DD Black individuals → ≅ 1062

⇒ Dd Black individuals → ≅ 344

⇒ dd White individuals → ≅ 29

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