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Can someone please help me? I have been denied/rejected two times on this problem. I need someone to help me calculate the “G5 Allele Frequency” I have included all the information in the picture and I put a red box around the section I need help with. The instructions only show me how to calculate the “Frequency G5” but NOT the “G5 Allele Frequency”. any help? Pls stop rejecting me and just try to help me? Pls

Can someone please help me? I have been denied/rejected two times on this problem. I need someone to help me calculate the “G5 Allele Frequency” I have included all the information in the picture and I put a red box around the section I need help with. The instructions only show me how to calculate the “Frequency G5” but NOT the “G5 Allele Frequency”. any help? Pls stop rejecting me and just try to help me? Pls

Curren'S Math For Meds: Dosages & Sol
Curren'S Math For Meds: Dosages & Sol
11th Edition
ISBN:9781305143531
Author:CURREN
Publisher:CURREN
Chapter2: Multiplication And Division Of Decimals
Section: Chapter Questions
Problem 6.2P
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Can someone please help me? I have been denied/rejected two times on this problem. I need someone to help me calculate the “G5 Allele Frequency” I have included all the information in the picture and I put a red box around the section I need help with. The instructions only show me how to calculate the “Frequency G5” but NOT the “G5 Allele Frequency”. any help? Pls stop rejecting me and just try to help me? Pls
3:49 × image.png Environment: Clean Forest Typica Carbonaria Total Phenotype Frequency Typica Carbonaria q Р Moths Released Allele Frequency Allele d D Color Light Dark G₁ 166 308 474 G₂ 259 254 513 Initial Frequency Initial Allele Frequency G3 372 234 606 G4 521 210 731 73 ... G5 851 199 1050 Frequency G5 (Round to 2 decimal places) 0.81 0.19 Gs Allele Frequency (Round to 2 decimal places) 0
Transcribed Image Text:3:49 × image.png Environment: Clean Forest Typica Carbonaria Total Phenotype Frequency Typica Carbonaria q Р Moths Released Allele Frequency Allele d D Color Light Dark G₁ 166 308 474 G₂ 259 254 513 Initial Frequency Initial Allele Frequency G3 372 234 606 G4 521 210 731 73 ... G5 851 199 1050 Frequency G5 (Round to 2 decimal places) 0.81 0.19 Gs Allele Frequency (Round to 2 decimal places) 0
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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