bioe 109 summer 2009 lecture 5- part i hardy- weinberg equilibrium
TRANSCRIPT
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BIOE 109Summer 2009
Lecture 5- Part IHardy- Weinberg Equilibrium
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The Hardy-Weinberg-Castle Equilibrium
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The Hardy-Weinberg-Castle Equilibrium
Godfrey Hardy
Wilhelm Weinberg
William Castle
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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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Hardy-Weinberg principle: A null model
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 specifies conditions under which the population will NOT evolve.
In other words, H-W principle identifies the set of events that can cause evolution in real world.
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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
• 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 = 109A1A2 = 182A2A2 = 73
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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 = 109A1A2 = 182A2A2 = 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
Step 2: Estimate allele frequencies
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Testing for Hardy-Weinberg equilibrium
Step 1: Estimate genotype frequencies
Step 2: Estimate allele frequencies
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
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Testing for Hardy-Weinberg equilibrium
Step 1: Estimate genotype frequencies
Step 2: Estimate allele frequencies
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
Step 4: Compare observed and expected numbers of genotypes
2 = (Obs. – Exp.)2 Exp.
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A simple model of directional selection
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Persistent selection changes allele frequencies over generations
(Obvious) Conclusion:Natural selection can cause rapid evolutionary change!
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A simple model of directional selection
• consider a single locus with two alleles A and a
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A simple model of directional selection
• consider a single locus with two alleles A and a
• let p = frequency of A allele
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A simple model of directional selection
• consider a single locus with two alleles A and a
• let p = frequency of A allele• let q = frequency of a allele
![Page 38: BIOE 109 Summer 2009 Lecture 5- Part I Hardy- Weinberg Equilibrium](https://reader036.vdocument.in/reader036/viewer/2022062407/56649cff5503460f949cfeca/html5/thumbnails/38.jpg)
A simple model of directional selection
• consider a single locus with two alleles A and a • let p = frequency of A allele• let q = frequency of a allele • relative fitnesses are:
AA Aa aa w11 w12 w22
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A simple model of directional selection
• consider a single locus with two alleles A and a
• let p = frequency of A allele• let q = frequency of a allele
• relative fitnesses are:
AA Aa aa w11 w12 w22
• it is also possible to determine relative fitness of the A and a alleles:
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A simple model of directional selection
• consider a single locus with two alleles A and a • let p = frequency of A allele• let q = frequency of a allele • relative fitnesses are:
AA Aa aa w11 w12 w22
• it is also possible to determine relative fitness of the A and a
alleles:
let w1 = fitness of the A allele
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A simple model of directional selection
• consider a single locus with two alleles A and a • let p = frequency of A allele• let q = frequency of a allele • relative fitnesses are:
AA Aa aa w11 w12 w22
• it is also possible to determine relative fitnesses of the A and a
alleles:
let w1 = fitness of the A allele
let w2 = fitness of the a allele
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The fitness of the A allele = w1 = pw11 + qw12
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The fitness of the A allele = w1 = pw11 + qw12
The fitness of the a allele = w2 = qw22 + pw12
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Directional selection
• let p = frequency of A allele• let q = frequency of a allele
• relative fitness of different genotypes are:
AA Aa aa w11 w12 w22
• it is also possible to determine relative fitness of the A and a alleles:
The fitness of the A allele = w1 = pw11 + qw12The fitness of the a allele = w2 = qw22 + pw12
• Mean population fitness = w = pw1 + qw2