Is this the correct phenotype frequency? Environment: Clean Forest Moths Released 810 190 1000 Typica Carbonaria Phenotype Frequency Typica Carbonaria Color Light Dark G₁ 405 72 G₂ 468 66 534 Initial Frequency 0.81 0.19 G₁ 569 64 633 G4 691 61 752 G₁ 857 56 013 Frequency G5 (Round to 2 decimal places) q 0.75

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2:09
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be4a59ef93e98e52064e89c5c...
< NATURAL SELECTION NATURAL SELECTION IN INSECTS
INTRODUCTION LABORATORY SIMULATION
Lab Data
Is this the correct phenotype frequency?
Environment: Clean Forest
Typica
Carbonaria
Total
Moths Released
Typica
Carbonaria
Allele Frequency
Phenotype Frequency
810
Moths
190
q²
Typica
2pq Carbonaria
p² Carbonaria
1000
Allele
q
P
Genotype Frequency
d
D
Color
Light
Dark
Genotype
dd
Dd
DD
Environment: Polluted Forest
Phenotype Frequency
Allele Frequency
Genotype Frequency
G₁
405
72
477
Color
G₂
Light
Dark
Dark
468
66
534
Initial Frequency
Initial Allele Frequency
0.90
0.10
0.81
180
0.19
Moths
Released
810
10
G3
569
64
633
Initial
Frequency
0.81
0.18
0.01
How to Calculate Phenotype Frequency
G4
691
61
752
o
0.75
764
Frequency Gs
...
56
- X
G5
857
Frequency G
(Round to 2 decimal places)
G5 Allele Frequency
(Round to 2 decimal places)
Number of
Moths G
Transcribed Image Text:2:09 × be4a59ef93e98e52064e89c5c... < NATURAL SELECTION NATURAL SELECTION IN INSECTS INTRODUCTION LABORATORY SIMULATION Lab Data Is this the correct phenotype frequency? Environment: Clean Forest Typica Carbonaria Total Moths Released Typica Carbonaria Allele Frequency Phenotype Frequency 810 Moths 190 q² Typica 2pq Carbonaria p² Carbonaria 1000 Allele q P Genotype Frequency d D Color Light Dark Genotype dd Dd DD Environment: Polluted Forest Phenotype Frequency Allele Frequency Genotype Frequency G₁ 405 72 477 Color G₂ Light Dark Dark 468 66 534 Initial Frequency Initial Allele Frequency 0.90 0.10 0.81 180 0.19 Moths Released 810 10 G3 569 64 633 Initial Frequency 0.81 0.18 0.01 How to Calculate Phenotype Frequency G4 691 61 752 o 0.75 764 Frequency Gs ... 56 - X G5 857 Frequency G (Round to 2 decimal places) G5 Allele Frequency (Round to 2 decimal places) Number of Moths G
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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