Answer/calculate the missing parts in the image attached Part 4: Peppered Moth Allele and Gene FrequenciesThe dark allele in peppered moths is dominant (B= p). Light moths arehomozygous recessive (bb= q²). Use the data from Table 1 in Lab 1 to calculatethe allele and genotype frequencies for peppered moths in year 1 and year 10.Year 1 data: 55664q² = frequency of bb (light) =_;q=frequency of the dominant allele B is p = 1 -q; p = 1Calculate genotypic makeup of the population.[0.054 (p) + 0.946_(q)]²= 1; 0.232 _(p²) +Genotypic frequencies are:_light moths,__0.896dark moths0.9460.9460.0540.102 (2pq) + 0.896_(q²) = 1]p² = frequency of BB (dark)2pq = frequency of Aa (dark)_q² = frequency of aa(light)

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Description: Answer/calculate the missing parts in the image attached Part 4: Peppered Moth Allele and Gene FrequenciesThe dark allele in peppered moths is dominant (B= p). Light moths arehomozygous recessive (bb= q²). Use the data from Table 1 in Lab 1 to calcul

According to Hardy Weinberg equation p2 + q2 + 2pq =1 and p + q = 1 p = frequency of the dominant…

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Part 4: Peppered Moth Allele and Gene Frequencies The dark allele in peppered moths is dominant (B= p). Light moths are homozygous recessive (bb= q²). Use the data from Table 1 in Lab 1 to calculate the allele and genotype frequencies for peppered moths in year 1 and year 10. 64 Year 1 data: 556 dark moths q² = frequency of bb (light) = ;q= 0.946 frequency of the dominant allele B is p = 1 - q; p = 1 - 0.946 Calculate genotypic makeup of the population. [0.054 _(p) + 0.946_(q)]²= 1; 0.232 _(p²) +_ Genotypic frequencies are: _light moths, 0.896 0.102 0.054 (2pq) + 0.896_(q²) = 1] p² = frequency of BB (dark) 2pq = frequency of Aa (dark)_ q² = frequency of aa(light)

Part 4: Peppered Moth Allele and Gene Frequencies The dark allele in peppered moths is dominant (B= p). Light moths are homozygous recessive (bb= q²). Use the data from Table 1 in Lab 1 to calculate the allele and genotype frequencies for peppered moths in year 1 and year 10. 64 Year 1 data: 556 dark moths q² = frequency of bb (light) = ;q= 0.946 frequency of the dominant allele B is p = 1 - q; p = 1 - 0.946 Calculate genotypic makeup of the population. [0.054 _(p) + 0.946_(q)]²= 1; 0.232 _(p²) +_ Genotypic frequencies are: _light moths, 0.896 0.102 0.054 (2pq) + 0.896_(q²) = 1] p² = frequency of BB (dark) 2pq = frequency of Aa (dark)_ q² = frequency of aa(light)

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter12: Mendel's Experiments And Heredity

Section: Chapter Questions

Problem 16RQ: A farmer raises black and white chickens. To his surprise, when the first generation of eggs hatch...

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