Any help? Pls with the boxes in red and this is dealing with Hardy -Weinberg equilibrium Allele FrequencyqРAlleleMothsdGenotype Frequencyq²Typica2pq Carbonariap² CarbonariaDGenotypeddDdDDInitial Allele FrequencyColorLightDarkDark0.700.30MothsReleased49042090InitialFrequency0.490.420.09G5 Allele Frequency(Round to 2 decimal places)0.91Frequency G50.011.80.90.1Number ofMoths G5848838 How to Calculate Phenotype Frequency1) How to Calculate Genotypic RatiosBy using phenotypic ratios of a characteristic like math color in a parent population, we canpredict the genotypic ratios in the next generation. There are 3 genotypes present.Homozygous dominant (Carbonaria, DD) represented by the p value in the Hardy-Weinberg equation.▪ Homozygous recessive (Typica, do) represented by the of value in the HardyWeinberg aquation.Heterozygous (Carbonaria, Da) is represented by the 200 value in the Hardy-Weinbergequation.2)This shows a population where 20% of the moths have the dominant dark color(Carbonaria) and 20% have a light color (Typica).P Generation:Phenotypic Ratio 20% CarbonarieAllele FrequenciesHardy-Weinberg EquationP Generation:3)Using the phenotypic ratio, we can determine allele frequencies in the parental generation.If the homozygous tralt (a) is 0.8 than a la 0.89.Phenotypic Retic: 20% CarbonarieAllele FrequenciesHardy-Weinberg Equation:P Generation:Allele FrequenciesHardy-Weinberg EquationPhenotypic Ratio: 20% CarbonarieD-011,4)Now that we know a la 0.89, pla 1.0-0.89. Therefore, p=0.11.P Generation:p²+210)q²-1Allele Frequencies:F, Generation GenotypesGenotype Frequencies:Hardy-Weinberg Equation:ifa=0.8, then qp2|pq) q-1Phenotypic Ratio: 20% CarbonariaD 0.11P Generation:5)Now that we knowp-0.11 and 0-0.89 in the parental generation, we can plug thesenumbers into the Hardy-Weinberg equation to predict the genotypic frequencies in the nextgeneration.F Generation Genotypes:Genotype Frequencies:Hardy-Weinberg Equation:HOA TY CHDDp+c-10p+ 0.89-10 or p-1.0-0.89-0.mp²2jpg) q-1Phenotypic Retic 20% CarbonarieAllele FrequenciesD=0.11ĐƠN TY DỊCHd-0.890.89DDB0% TY DICHd-0.890.016)Here we see that p² (homozygous dominant) la 0.01 and of (homozygous recessive) is0.79. Lastly, 200 (heterozygous) is 0.20 as shown below.BON. Typicad=0.890.010.200.7910.11 + 2(0.11 x 0.89) (0.899-100.01 +0.20 +0.79-1.0p+21pq) q² =1DdĐỘNG TY DỊCHd-0.80ddDd0.20p²+20pql+q-1Xdd0.79

Allele Frequency q Р Allele Moths d Genotype Frequency q² Typica 2pq Carbonaria p² Carbonaria D Genotype dd Dd DD Initial Allele Frequency Color Light Dark Dark 0.70 0.30 Moths Released 490 420 90 Initial Frequency 0.49 0.42 0.09 G5 Allele Frequency (Round to 2 decimal places) 0.91 Frequency G5 0.01 1.8 0.9 0.1 Number of Moths G5 848 838

Allele Frequency q Р Allele Moths d Genotype Frequency q² Typica 2pq Carbonaria p² Carbonaria D Genotype dd Dd DD Initial Allele Frequency Color Light Dark Dark 0.70 0.30 Moths Released 490 420 90 Initial Frequency 0.49 0.42 0.09 G5 Allele Frequency (Round to 2 decimal places) 0.91 Frequency G5 0.01 1.8 0.9 0.1 Number of Moths G5 848 838

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 10QP

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