HARDY-WEINBERG 113 TABLE 10.3. Collared lizards in the Ozark region. Generation
ID: 188755 • Letter: H
Question
HARDY-WEINBERG 113 TABLE 10.3. Collared lizards in the Ozark region. Generation 1 Phenotype Orange Yellow Blue AzA 27 Total Number of individuals Observed genotype frequencies Allele frequencies Expected genotype frequencies Are these lizards in Hardy-Weinberg equilibrium (yes/no)? Generation 2 Phenotype Genotype Number of individuals Observed genotype frequencies Allele frequencies Expected genotype frequencies Are these lizards in Hardy-Weinberg equilibrium (yes/no)? Generation 3 Phenotype Genotype Number of individuals 34 29 =90 1.0 =1.0 = 1.0 Yellow A1A2 18 Total Orange AIAI 23 Blue AzA2 21 = 1.0 =1.0 =1.0 Blue Total Orange A Ai 24 Yellow A A2 16 = 74 34 Observed genotype frequencies Allele frequencies = 1.0 Expected genotype frequencies Are these lizards in Hardy-Weinberg equilibrium (yes/no)PExplanation / Answer
1. Generation 1
Generation 1
Phenotype
Orange
Yellow
Blue
Total
Genotype
A1A1
A1A2
A2A2
Number of individuals
34
29
27
=90
Observed genotype frequencies
34/90=0.377
29/90=0.322
0.3
=1.0
Allele frequencies
p=0.539
q= 0.461
=1.0
Expected genotype frequencies
0.29
0.5
0.21
=1.0
Are these lizards in Hardy-Weinberg equilibrium= Yes
For Allele frequencies and observed genotype frequencies
There are two A1 alleles in A1A1 = 34 X 2= 68
There are one A1 alleles in A1A2= 29
Ozark lizards are diploid.
Frequency of allele A1 in generation 1= (p= 68+29)/90+90= 0.539
Frequency of allele A2 in generation 1= q= 1-p= 1-0.539=0.461
For Expected Frequencies:
The Hardy Weinberg equation is given as:
p2+2pq+q2=1
p2 = frequency of the homozygous genotype A1A1, q2 = the frequency of the homozygous genotype A2A2, and 2pq = frequency of the heterozygous genotype A1A2.
Expected frequency of A1A1= p2= 0.539 X 0.539= 0.291
Expected frequency of A1A2= 2pq= 2X 0.539 X 0.461= 0.5
Expected frequency of A1A2= q2= 0.461 X 0.461= 0.213
Chi-square Test:
Oberved
Expected
(O-E)
(O-E)2
(O-E)2/E
A1A1
0.3777
0.29
0.0877
0.007691
0.026522
A1A2
0.322
0.5
-0.178
0.031684
0.063368
A2A2
0.3
0.21
0.09
0.0081
0.038571
Total
0.9997
1
-0.0003
0.047475
0.128461
Degree of freedom= Number of genotypes-1= 3-1=2
Probability of chi square at d.f 2= 0.93200= 93.2%
A Chi Square value between 1%-5% should allow you to reject the null hypothesis. However, 93.2% is a large P Value. Hence, the null hypothesis is accepted. The frequencies are in Hardy-Weinberg equilibrium.
2. Generation 2
Generation 1
Phenotype
Orange
Yellow
Blue
Total
Genotype
A1A1
A1A2
A2A2
Number of individuals
23
18
21
=62
Observed genotype frequencies
23/62=0.37
18/62=0.29
21/62=0.34
=1.0
Allele frequencies
p=0.52
q= 0.48
=1.0
Expected genotype frequencies
0.27
0.5
0.23
=1.0
Are these lizards in Hardy-Weinberg equilibrium= Yes
For Allele frequencies and observed genotype frequencies
There are two A1 alleles in A1A1 = 23 X 2= 46
There are one A1 alleles in A1A2= 18
Ozark lizards are diploid.
Frequency of allele A1 in generation 1= (p= 46+18/62+62= 0.52
Frequency of allele A2 in generation 1= q= 1-p= 1-0.52=0.48
For Expected Frequencies:
The Hardy Weinberg equation is given as:
p2+2pq+q2=1
p2 = frequency of the homozygous genotype A1A1, q2 = the frequency of the homozygous genotype A2A2, and 2pq = frequency of the heterozygous genotype A1A2.
Expected frequency of A1A1= p2= 0.52 X 0.52= 0.27
Expected frequency of A1A2= 2pq= 2X 0.52 X 0.48= 0.5
Expected frequency of A1A2= q2= 0.461 X 0.461= 0.23
Chi-Square Test:
Oberved
Expected
(O-E)
(O-E)2
(O-E)2/E
A1A1
0.37
0.27
0.1
0.01
0.037037
A1A2
0.29
0.5
-0.21
0.0441
0.0882
A2A2
0.34
0.23
0.11
0.0121
0.052609
Total
1
1
-2.8E-17
0.0662
0.177846
Degree of freedom= Number of genotypes-1= 3-1=2
Probability of chi square at d.f 2= 0.9150= 91.5%
A Chi Square value between 1%-5% should allow you to reject the null hypothesis. However, 91.5% is a large P Value. Hence, the null hypothesis is accepted. The frequencies are in Hardy-Weinberg equilibrium.
3. Generation 3
Generation 1
Phenotype
Orange
Yellow
Blue
Total
Genotype
A1A1
A1A2
A2A2
Number of individuals
24
16
34
=74
Observed genotype frequencies
24/74=0.32
16/34=0.22
34/74=0.46
=1.0
Allele frequencies
p=0.43
q= 0.57
=1.0
Expected genotype frequencies
0.19
0.49
0.33
=1.0
Are these lizards in Hardy-Weinberg equilibrium= Yes
For Allele frequencies and observed genotype frequencies
There are two A1 alleles in A1A1 = 24 X 2= 48
There are one A1 alleles in A1A2= 16
Ozark lizards are diploid.
Frequency of allele A1 in generation 1= p= 48+16/74+74= 0.43
Frequency of allele A2 in generation 1= q= 1-p= 1-0.52=0.57
For Expected Frequencies:
The Hardy Weinberg equation is given as:
p2+2pq+q2=1
p2 = frequency of the homozygous genotype A1A1, q2 = the frequency of the homozygous genotype A2A2, and 2pq = frequency of the heterozygous genotype A1A2.
Expected frequency of A1A1= p2= 0.43 X 0.43= 0.19
Expected frequency of A1A2= 2pq= 2X 0.57 X 0.43= 0.49
Expected frequency of A1A2= q2= 0.461 X 0.461= 0.33
Chi-Square Test:
Oberved
Expected
(O-E)
(O-E)2
(O-E)2/E
A1A1
0.37
0.19
0.18
0.0324
0.170526
A1A2
0.29
0.49
-0.2
0.04
0.081633
A2A2
0.34
0.33
0.01
0.0001
0.000303
Total
1
1.01
-0.01
0.0725
0.252462
Degree of freedom= Number of genotypes-1= 3-1=2
Probability of chi square at d.f 2= 0.881= 88.1%
A Chi Square value between 1%-5% should allow you to reject the null hypothesis. However, 88.1% is a large P Value. Hence, the null hypothesis is accepted. The frequencies are in Hardy-Weinberg equilibrium.
Generation 1
Phenotype
Orange
Yellow
Blue
Total
Genotype
A1A1
A1A2
A2A2
Number of individuals
34
29
27
=90
Observed genotype frequencies
34/90=0.377
29/90=0.322
0.3
=1.0
Allele frequencies
p=0.539
q= 0.461
=1.0
Expected genotype frequencies
0.29
0.5
0.21
=1.0
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