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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 Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Any help? Pls with the boxes in red and this is dealing with Hardy -Weinberg equilibrium
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
Transcribed Image Text: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
How to Calculate Phenotype Frequency 1) How to Calculate Genotypic Ratios By using phenotypic ratios of a characteristic like math color in a parent population, we can predict 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 Hardy Weinberg aquation. Heterozygous (Carbonaria, Da) is represented by the 200 value in the Hardy-Weinberg equation. 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% Carbonarie Allele Frequencies Hardy-Weinberg Equation P 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% Carbonarie Allele Frequencies Hardy-Weinberg Equation: P Generation: Allele Frequencies Hardy-Weinberg Equation Phenotypic Ratio: 20% Carbonarie D-011, 4) Now that we know a la 0.89, pla 1.0-0.89. Therefore, p=0.11. P Generation: p²+210)q²-1 Allele Frequencies: F, Generation Genotypes Genotype Frequencies: Hardy-Weinberg Equation: ifa=0.8, then q p2|pq) q-1 Phenotypic Ratio: 20% Carbonaria D 0.11 P Generation: 5) Now that we knowp-0.11 and 0-0.89 in the parental generation, we can plug these numbers into the Hardy-Weinberg equation to predict the genotypic frequencies in the next generation. F Generation Genotypes: Genotype Frequencies: Hardy-Weinberg Equation: HOA TY CH DD p+c-10 p+ 0.89-10 or p-1.0-0.89-0.m p²2jpg) q-1 Phenotypic Retic 20% Carbonarie Allele Frequencies D=0.11 ĐƠN TY DỊCH d-0.89 0.89 DD B0% TY DICH d-0.89 0.01 6) Here we see that p² (homozygous dominant) la 0.01 and of (homozygous recessive) is 0.79. Lastly, 200 (heterozygous) is 0.20 as shown below. BON. Typica d=0.89 0.01 0.20 0.79 10.11 + 2(0.11 x 0.89) (0.899-10 0.01 +0.20 +0.79-1.0 p+21pq) q² =1 Dd ĐỘNG TY DỊCH d-0.80 dd Dd 0.20 p²+20pql+q-1 X dd 0.79
Transcribed Image Text:How to Calculate Phenotype Frequency 1) How to Calculate Genotypic Ratios By using phenotypic ratios of a characteristic like math color in a parent population, we can predict 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 Hardy Weinberg aquation. Heterozygous (Carbonaria, Da) is represented by the 200 value in the Hardy-Weinberg equation. 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% Carbonarie Allele Frequencies Hardy-Weinberg Equation P 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% Carbonarie Allele Frequencies Hardy-Weinberg Equation: P Generation: Allele Frequencies Hardy-Weinberg Equation Phenotypic Ratio: 20% Carbonarie D-011, 4) Now that we know a la 0.89, pla 1.0-0.89. Therefore, p=0.11. P Generation: p²+210)q²-1 Allele Frequencies: F, Generation Genotypes Genotype Frequencies: Hardy-Weinberg Equation: ifa=0.8, then q p2|pq) q-1 Phenotypic Ratio: 20% Carbonaria D 0.11 P Generation: 5) Now that we knowp-0.11 and 0-0.89 in the parental generation, we can plug these numbers into the Hardy-Weinberg equation to predict the genotypic frequencies in the next generation. F Generation Genotypes: Genotype Frequencies: Hardy-Weinberg Equation: HOA TY CH DD p+c-10 p+ 0.89-10 or p-1.0-0.89-0.m p²2jpg) q-1 Phenotypic Retic 20% Carbonarie Allele Frequencies D=0.11 ĐƠN TY DỊCH d-0.89 0.89 DD B0% TY DICH d-0.89 0.01 6) Here we see that p² (homozygous dominant) la 0.01 and of (homozygous recessive) is 0.79. Lastly, 200 (heterozygous) is 0.20 as shown below. BON. Typica d=0.89 0.01 0.20 0.79 10.11 + 2(0.11 x 0.89) (0.899-10 0.01 +0.20 +0.79-1.0 p+21pq) q² =1 Dd ĐỘNG TY DỊCH d-0.80 dd Dd 0.20 p²+20pql+q-1 X dd 0.79
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