Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four...
Transcript of Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four...
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Lecture 7 Mutation and genetic variation
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Point mutations
There are four categories of point mutations:
1. transitions (e.g., A → G, C → T)
2. transversions (e.g., T → A, C → G)
3. insertions (e.g., TTTGAC → TTTCCGAC)
• in coding regions, point mutations can involve silent (synonymous) or replacement (nonsynonymous) changes.
• in coding regions, insertions/deletions can also cause frameshift mutations.
4. deletions (e.g., TTTGAC → TTTC)
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Indels are insertions and deletions
STOP making sense: effective frameshifts
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“Copy-number” mutations
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“Copy-number” mutations
• these mutations change the numbers of genetic elements.
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“Copy-number” mutations
• these mutations change the numbers of genetic elements.
• gene duplication events create new copies of genes.
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“Copy-number” mutations
• these mutations change the numbers of genetic elements.
• gene duplication events create new copies of genes.
• one important mechanism generating duplications is unequal crossing over.
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Unequal crossing-over can generate gene duplications
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Unequal crossing-over can generate gene duplications
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“Copy-number” mutations
• these mutations change the numbers of genetic elements.
• gene duplication events create new copies of genes.
• one mechanism believed responsible is unequal crossing over.
• over time, this process may lead to the development of multi-gene families.
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Whole-genome data yields data on gene families
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Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
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Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Example: jingwei in Drosophila yakuba
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Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Example: jingwei in Drosophila yakuba
Alcohol dehydrogenase (Adh)
→
Chromosome 2 Chromosome 3
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Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Example: jingwei in Drosophila yakuba
Alcohol dehydrogenase (Adh)
→
Chromosome 2 Chromosome 3
→ mRNA
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Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Example: jingwei in Drosophila yakuba
Alcohol dehydrogenase (Adh)
→
Chromosome 2 Chromosome 3
→ mRNA
↓ cDNA
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Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Example: jingwei in Drosophila yakuba
Alcohol dehydrogenase (Adh)
→
Chromosome 2 Chromosome 3
→ mRNA
↓ cDNA → “jingwei”
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Where do new genes come from?
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Where do new genes come from? An example: the antifreeze glycoprotein (AFGP) gene in the Antarctic fish, Dissostichus mawsoni
Convergent evolution of an AFGP gene in the arctic cod, Boreogadus saida
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Lecture 8 Microevolution 1 - selection
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The Hardy-Weinberg Equilibrium
Godfrey Hardy Wilhelm Weinberg
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William Castle
Gregor Mendel
Udny Yule
Reginald Punnett & William Bateson (1875-1967) (1861-1926)
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Reginald Punnett Godfrey Hardy
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The Hardy-Weinberg-Castle Equilibrium
Godfrey Hardy
Wilhelm Weinberg
William Castle
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The Hardy-Weinberg Equilibrium
consider a single locus with two alleles A1 and A2
• three genotypes exist: A1A1, A1A2, A2A2
• let p = frequency of A1 allele
• let q = frequency of A2 allele
• since only two alleles present, p + q = 1
Question: If mating occurs at random in the population, what will the frequencies of A1 and A2 be in the next generation?
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What are the probabilities of matings at the gamete level?
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What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2
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What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2 A1 x A2 → A1A2 p x q = pq
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What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2 A1 x A2 → A1A2 p x q = pq A2 x A1 → A2A1 q x p = qp
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What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2 A1 x A2 → A1A2 p x q = pq
2pq A2 x A1 → A2A1 q x p = qp
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What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2 A1 x A2 → A1A2 p x q = pq
2pq A2 x A1 → A2A1 q x p = qp A2 x A2 → A2A2 q x q = q2
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Therefore, zygotes produced in proportions:
Genotype: A1A1 A1A2 A2A2
Frequency: p2 2pq q2
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Therefore, zygotes produced in proportions:
Genotype: A1A1 A1A2 A2A2
Frequency: p2 2pq q2
what are the allele frequencies?
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What are the allele frequencies?
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
Frequency of A2 = q2 + ½ (2pq)
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
Frequency of A2 = q2 + ½ (2pq)
= q2 + pq
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
Frequency of A2 = q2 + ½ (2pq)
= q2 + pq
= q(q + p)
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
Frequency of A2 = q2 + ½ (2pq)
= q2 + pq
= q(q + p)
= q
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What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
Frequency of A2 = q2 + ½ (2pq)
= q2 + pq
= q(q + p)
= q
ALLELE FREQUENCIES DID NOT CHANGE!!
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Conclusions of the Hardy-Weinberg principle
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Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
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Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
2. Genotype proportions determined by the “square law”.
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Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
2. Genotype proportions determined by the “square law”.
• for two alleles = (p + q)2 = p2 + 2pq + q2
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Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
2. Genotype proportions determined by the “square law”.
• for two alleles = (p + q)2 = p2 + 2pq + q2
• for three alleles (p + q + r)2 = p2 + q2 + r2 + 2pq + 2pr +2qr
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Conclusions of the Hardy-Weinberg principle
3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies
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Conclusions of the Hardy-Weinberg principle
3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04
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Conclusions of the Hardy-Weinberg principle
3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 A1 = 0.50, A2 = 0.50 A1A1 = 0.25, A1A2 = 0.50, A2A2 = 0.25
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Conclusions of the Hardy-Weinberg principle
3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 A1 = 0.50, A2 = 0.50 A1A1 = 0.25, A1A2 = 0.50, A2A2 = 0.25 A1 = 0.10, A2 = 0.90 A1A1 = 0.01, A1A2 = 0.18, A2A2 = 0.81
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Assumptions of Hardy-Weinberg equilibrium
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random… but some traits experience positive assortative mating
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift)
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation 5. No selection
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Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation 5. No selection
The Hardy-Weinberg equilibrium principle thus predicts that no evolution will occur unless one (or more) of these assumptions are violated!
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Does Hardy-Weinberg equilibrium ever exist in nature?
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Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
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Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
as a juvenile…
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Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
… and as an adult
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Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
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Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia • a sample of 364 fish were scored for a single nucleotide polymorphism (SNP)
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Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia • a sample of 364 fish were scored for a single nucleotide polymorphism (SNP)
A1A1 = 109 A1A2 = 182 A2A2 = 73
![Page 69: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/69.jpg)
Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia • a sample of 364 fish were scored for a single nucleotide polymorphism (SNP)
A1A1 = 109 A1A2 = 182 A2A2 = 73
Question: Is this population in Hardy-Weinberg equilibrium?
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Testing for Hardy-Weinberg equilibrium
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Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies
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Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995
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Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000
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Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000 Frequency of A2A2 = 73/364 = 0.2005
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2)
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000)
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495
Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2)
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495
Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2)
= 0.2005 + ½ (0.5000)
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495
Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2)
= 0.2005 + ½ (0.5000) = 0.4505
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Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495
Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2)
= 0.2005 + ½ (0.5000) = 0.4505
Check that p + q = 0.5495 + 0.4505 = 1
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A1A1 = p2 x N
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A1A1 = p2 x N = (0.5495)2 x 364
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9 Expected No. of A1A2 = 2pq x N
![Page 88: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/88.jpg)
Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9 Expected No. of A1A2 = 2pq x N = 2(0.5495)(0.4505) x 364
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9 Expected No. of A1A2 = 2pq x N = 2(0.5495)(0.4505) x 364 = 180.2
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A2A2 = q2 x N
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Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A2A2 = q2 x N = (0.4595)2 x 364
![Page 92: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/92.jpg)
Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Expected No. of A2A2 = q2 x N = (0.4595)2 x 364 = 73.9
![Page 93: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/93.jpg)
Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes
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Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
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Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
A1A1 109 109.9
![Page 96: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/96.jpg)
Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
A1A1 109 109.9 A1A2 182 180.2
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Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9
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Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9
χ2 = Σ (Obs. – Exp.)2
Exp.
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Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9
χ2 = Σ (Obs. – Exp.)2 = 0.036
Exp.
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Suppose we sampled a mixed population
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Suppose we sampled a mixed population Population A p = A1 = 1.0 q = A2 = 0 100% A1A1
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Suppose we sampled a mixed population Population A Population B p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2
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Suppose we sampled a mixed population Population A Population B p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2
æ å Mixed population
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Suppose we sampled a mixed population Population A Population B p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2
æ å Mixed population
50% from population A (all A1A1) 50% from population B (all A2A2)
![Page 105: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/105.jpg)
Suppose we sampled a mixed population
50% from population A (all A1A1) 50% from population B (all A2A2)
↓
Sample 1000 individuals
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Suppose we sampled a mixed population
50% from population A (all A1A1) 50% from population B (all A2A2)
↓
Sample 1000 individuals
Observed
A1A1 = 500 A1A2 = 0 A2A2 = 500
![Page 107: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/107.jpg)
Suppose we sampled a mixed population
50% from population A (all A1A1) 50% from population B (all A2A2)
↓
Sample 1000 individuals
Observed Expected
A1A1 = 500 A1A1 = 250 A1A2 = 0 A1A2 = 500 A2A2 = 500 A2A2 = 250
![Page 108: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/108.jpg)
Sampling a mixed population generates a deficiency of
heterozygotes This is called a Wahlund effect
![Page 109: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/109.jpg)
Deafness in Tristan da Cunha
observed AA = 1228 Aa = 352 aa = 253 Total: 1833 p = 1228/1833 + 352/2*1833 = 0.670 + 0.096 = 0.766 q = 253/1833 + 352/2*1833 = 0.138 + 0.096 = 0.234
Inbreeding: Example
![Page 110: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/110.jpg)
Deafness in Tristan da Cunha
observed AA = 1228 Aa = 352 aa = 253 Total: 1833 p = 1228/1833 + 352/2*1833 = 0.670 + 0.096 = 0.766 q = 253/1833 + 352/2*1833 = 0.138 + 0.096 = 0.234
Inbreeding: Example
expected 1075.5 657.10 100.36
![Page 111: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/111.jpg)
Deafness in Tristan da Cunha
observed expected AA = 1228 1075.5 Aa = 352 657.10 aa = 253 (13.8%) 100.36 (5.4%) Total: 1833 p = 1228/1833 + 352/2*1833 = 0.670 + 0.096 = 0.766 q = 253/1833 + 352/2*1833 = 0.138 + 0.096 = 0.234
Inbreeding: Example
![Page 112: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/112.jpg)
A simple model of directional selection
![Page 113: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/113.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
![Page 114: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/114.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
![Page 115: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/115.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
• let q = frequency of A2 allele
![Page 116: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/116.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
• let q = frequency of A2 allele
• relative fitnesses are:
A1A1 A1A2 A2A2
w11 w12 w22
![Page 117: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/117.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
• let q = frequency of A2 allele
• relative fitnesses are:
A1A1 A1A2 A2A2
w11 w12 w22
• it is also possible to determine relative fitnesses of the A1 and A2 alleles:
![Page 118: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/118.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
• let q = frequency of A2 allele
• relative fitnesses are:
A1A1 A1A2 A2A2
w11 w12 w22
• it is also possible to determine relative fitnesses of the A1 and A2 alleles:
let w1 = fitness of the A1 allele
![Page 119: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/119.jpg)
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
• let q = frequency of A2 allele
• relative fitnesses are:
A1A1 A1A2 A2A2
w11 w12 w22
• it is also possible to determine relative fitnesses of the A1 and A2 alleles:
let w1 = fitness of the A1 allele
let w2 = fitness of the A2 allele
![Page 120: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/120.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
![Page 121: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/121.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂
![Page 122: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/122.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1
![Page 123: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/123.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
![Page 124: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/124.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
This route will occur with a probability p, since p is the frequency of the A1 allele
![Page 125: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/125.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realize” fitness w11
This route will occur with a probability p, since p is the frequency of the A1 allele
p
![Page 126: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/126.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
♂
A2
p
![Page 127: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/127.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
♂
A2
p
A1A2 “realized” fitness w12
![Page 128: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/128.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
♂
A2
p
A1A2 “realized” fitness w12
This route will occur with a probability q, since q is the frequency of the A2 allele
q
![Page 129: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/129.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
♂
A2
p
A1A2 “realized” fitness w12
Therefore, w1 = pw11 + qw12
q
![Page 130: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/130.jpg)
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w1w1
♂
A2
p
A1A2 “realized” fitness w12
Therefore, w1 = pw11 + qw12
q
(this is equivalent to a weighted average of the two routes)
![Page 131: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/131.jpg)
What is the fitness of an allele? Similarly for the A2 allele:
♀
A2
A2
♂ A2A2 “realized” fitness w22
♂
A1
q
A1A2 “realized” fitness w12
Therefore, w2 = qw22 + pw12
p
![Page 132: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/132.jpg)
The fitness of the A1 allele = w1 = pw11 + qw12
![Page 133: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/133.jpg)
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12
![Page 134: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/134.jpg)
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define….
![Page 135: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/135.jpg)
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness
![Page 136: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/136.jpg)
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness = w = pw1 + qw2
![Page 137: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/137.jpg)
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness = w = pw1 + qw2 (This too is a weighted average of the two allelic fitnesses.)
![Page 138: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/138.jpg)
Let p’ = frequency of A1 allele in the next generation
![Page 139: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/139.jpg)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2)
![Page 140: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/140.jpg)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w)
![Page 141: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/141.jpg)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation
![Page 142: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/142.jpg)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2)
![Page 143: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/143.jpg)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2) q’ = q (w2/w)
![Page 144: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/144.jpg)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2) q’ = q (w2/w)
![Page 145: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/145.jpg)
An example of directional selection
![Page 146: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/146.jpg)
An example of directional selection Let p = q = 0.5
![Page 147: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/147.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2
![Page 148: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/148.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90
![Page 149: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/149.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975
![Page 150: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/150.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925
![Page 151: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/151.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw12 = 0.975 w2 = qw22 + pw12 = 0.925 w = pw1 + qw2 = 0.950
![Page 152: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/152.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513
![Page 153: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/153.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487
![Page 154: Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,](https://reader034.fdocuments.net/reader034/viewer/2022050203/5f8df9bcdd48da7c04415eea/html5/thumbnails/154.jpg)
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487 In ~150 generations the A1 allele will be fixed