Lecture 7 Mutation and genetic variation
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)
Indels are insertions and deletions
STOP making sense: effective frameshifts
“Copy-number” mutations
“Copy-number” mutations
• these mutations change the numbers of genetic elements.
“Copy-number” mutations
• these mutations change the numbers of genetic elements.
• gene duplication events create new copies of genes.
“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.
Unequal crossing-over can generate gene duplications
Unequal crossing-over can generate gene duplications
“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.
Whole-genome data yields data on gene families
Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Retrogenes may also be created
• retrogenes have identical exon structures to their “progenitors” but lack introns!
Example: jingwei in Drosophila yakuba
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
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
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
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”
Where do new genes come from?
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
Lecture 8 Microevolution 1 - selection
The Hardy-Weinberg Equilibrium
Godfrey Hardy Wilhelm Weinberg
William Castle
Gregor Mendel
Udny Yule
Reginald Punnett & William Bateson (1875-1967) (1861-1926)
Reginald Punnett Godfrey Hardy
The Hardy-Weinberg-Castle Equilibrium
Godfrey Hardy
Wilhelm Weinberg
William Castle
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?
What are the probabilities of matings at the gamete level?
What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2
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
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
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
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
Therefore, zygotes produced in proportions:
Genotype: A1A1 A1A2 A2A2
Frequency: p2 2pq q2
Therefore, zygotes produced in proportions:
Genotype: A1A1 A1A2 A2A2
Frequency: p2 2pq q2
what are the allele frequencies?
What are the allele frequencies?
What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
What are the allele frequencies?
Frequency of A1 = p2 + ½ (2pq)
= p2 + pq
= p(p + q)
= p
Frequency of A2 = q2 + ½ (2pq)
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
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)
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
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!!
Conclusions of the Hardy-Weinberg principle
Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
2. Genotype proportions determined by the “square law”.
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
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
Conclusions of the Hardy-Weinberg principle
3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies
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
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
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
Assumptions of Hardy-Weinberg equilibrium
Assumptions of Hardy-Weinberg equilibrium
1. Mating is random
Assumptions of Hardy-Weinberg equilibrium
1. Mating is random… but some traits experience positive assortative mating
Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift)
Assumptions of Hardy-Weinberg equilibrium
1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration
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
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
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!
Does Hardy-Weinberg equilibrium ever exist in nature?
Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
as a juvenile…
Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
… and as an adult
Does Hardy-Weinberg equilibrium ever exist in nature?
Example: Atlantic cod (Gadus morhua) in Nova Scotia
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)
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
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?
Testing for Hardy-Weinberg equilibrium
Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies
Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995
Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000
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
Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2)
Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000)
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
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)
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)
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
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
Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
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
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
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
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
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
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
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
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
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
Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes
Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
A1A1 109 109.9
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
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
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.
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.
Suppose we sampled a mixed population
Suppose we sampled a mixed population Population A p = A1 = 1.0 q = A2 = 0 100% A1A1
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
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
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)
Suppose we sampled a mixed population
50% from population A (all A1A1) 50% from population B (all A2A2)
↓
Sample 1000 individuals
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
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
Sampling a mixed population generates a deficiency of
heterozygotes This is called a Wahlund effect
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
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
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
A simple model of directional selection
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
A simple model of directional selection
• consider a single locus with two alleles A1 and A2
• let p = frequency of A1 allele
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
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
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:
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
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
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
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
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
What is the fitness of an allele? Consider fitness from the gamete’s perspective:
♀
A1
A1
♂ A1A1 “realized” fitness w11
♂
A2
p
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
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
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
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)
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
The fitness of the A1 allele = w1 = pw11 + qw12
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12
The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define….
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
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
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.)
Let p’ = frequency of A1 allele in the next generation
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2)
Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w)
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
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)
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)
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)
An example of directional selection
An example of directional selection Let p = q = 0.5
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2
An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90
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
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
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
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
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
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