Mutation and genetic variation Thymidine dimer
Point mutations
1. transitions (e.g., A → G, C → T)
2. transversions (e.g., T → A, C → G) Models of Point mutation dynamics
A α G A α G
α α α α β β β β
C α T C α T Jukes-Cantor Kimura - 2 parameter Model Model Genetic code Mutation hotspots
Cytochrome b
Molecular clock
Emile Zuckerkandl and
Linus Pauling, 1965
Synonymous substitutions v.s. Non-synonymous substitutions
Ka/Ks is an indicator of selection General classes of mutations
Point mutations
“Copy-number” mutations
Chromosomal mutations
Genome mutations Chromosomal inversions lock blocks of genes together Changes in chromosome number are common
• Robertsonian fusions and fissions are common and can have major effects on speciation
• in mammals, chromosome numbers range from N = 3 to N = 42.
• in insects, the range is from N = 1(some ants) to N = 220 (a butterfly)
• karyotypes can evolve rapidly! Muntiacus reevesi; N = 23
Muntiacus muntjac; N = 4 Genome mutations
• polyploidization events cause the entire genome to be duplicated.
• polyploidy has played a major role in the evolution of plants.
• ancient polyploidization events have also occurred in most animal lineages. Generation of a tetraploid 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
Hardy-Weinberg Equilibrium The Hardy-Weinberg Equilibrium
Godfrey Hardy Wilhelm Weinberg Gregor Mendel
Udny Yule
Reginald Punnett & William Bateson (1875-1967) (1861-1926) William Castle 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
2 A1 x A1 → A1A1 p x p = p
What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability
2 A1 x A1 → A1A1 p x p = p
A1 x A2 → A1A2 p x q = pq
What are the probabilities of matings at the gamete level?
Egg Sperm Zygote Probability
2 A1 x A1 → A1A1 p x p = p
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
2 A1 x A1 → A1A1 p x p = p
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
2 A1 x A1 → A1A1 p x p = p
A1 x A2 → A1A2 p x q = pq 2pq
A2 x A1 → A2A1 q x p = qp
2 A2 x A2 → A2A2 q x q = q
Therefore, zygotes produced in proportions:
Genotype: A1A1 A1A2 A2A2
2 2 Frequency: p 2pq q
Therefore, zygotes produced in proportions:
Genotype: A1A1 A1A2 A2A2
2 2 Frequency: p 2pq q what are the allele frequencies?
What are the allele frequencies?
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q) What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q)
= p
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q)
= p
2 Frequency of A2 = q + ½ (2pq)
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q)
= p
2 Frequency of A2 = q + ½ (2pq) 2 = q + pq
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q)
= p
2 Frequency of A2 = q + ½ (2pq) 2 = q + pq
= q(q + p)
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q)
= p
2 Frequency of A2 = q + ½ (2pq) 2 = q + pq
= q(q + p)
= q
What are the allele frequencies?
2 Frequency of A1 = p + ½ (2pq) 2 = p + pq
= p(p + q)
= p
2 Frequency of A2 = q + ½ (2pq) 2 = q + 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”.
2 2 2 • for two alleles = (p + q) = p + 2pq + q
Conclusions of the Hardy-Weinberg principle
1. Allele frequencies will not change from generation to generation.
2. Genotype proportions determined by the “square law”.
2 2 2 • for two alleles = (p + q) = p + 2pq + q
2 2 2 2 • for three alleles (p + q + r) = p + q + r + 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
2 Expected No. of A1A1 = p x N
Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
2 Expected No. of A1A1 = p 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
2 Expected No. of A1A1 = p 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
2 Expected No. of A1A1 = p 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
2 Expected No. of A1A1 = p 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
2 Expected No. of A1A1 = p 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
2 Expected No. of A2A2 = q x N
Testing for Hardy-Weinberg equilibrium
Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
2 Expected No. of A2A2 = q 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
2 Expected No. of A2A2 = q 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
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 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 n*p2 = (0.766) 2 * 1833 = 1075.52 n * 2pq = 2(0.766 * 0.234) * 1833 = 657.10 n * q2 = (0.234) 2 * 1833 = 100.36
Inbreeding: Example
Deafness in Tristan da Cunha
observed expected AA = 1228 1075.5 Aa = 352 657.10 aa = 253 100.36 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 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
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 p = 1075.5/1833 + (657.10/2)/1833 = 0.766 q = 100.36/1833 + (657.10/2)/1833 = 0.234 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:
♀ ♂ A1A1 A1
A1 What is the fitness of an allele?
Consider fitness from the gamete’s perspective:
♀ ♂ A1A1 “realized” fitness A1 w11
A1 What is the fitness of an allele?
Consider fitness from the gamete’s perspective:
♀ ♂ A1A1 “realized” fitness A1 w11
A1 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:
♀ p ♂ A1A1 “realize” fitness A1 w11
A1 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:
♀ p ♂ A1A1 “realized” fitness A1 w11
A1 ♂
A2 What is the fitness of an allele?
Consider fitness from the gamete’s perspective:
♀ p ♂ A1A1 “realized” fitness A1 w11
A1 ♂
A1A2 “realized” fitness A 2 w12
What is the fitness of an allele?
Consider fitness from the gamete’s perspective:
♀ p ♂ A1A1 “realized” fitness A1 w11
A1 q ♂
A1A2 “realized” fitness A 2 w12
This route will occur with a probability q,
since q is the frequency of the A2 allele What is the fitness of an allele?
Consider fitness from the gamete’s perspective:
♀ p ♂ A1A1 “realized” fitness A1 w11
A1 q ♂
A1A2 “realized” fitness A 2 w12
Therefore, w1 = pw11 + qw12 What is the fitness of an allele?
Consider fitness from the gamete’s perspective:
♀ p ♂ A1A1 “realized” fitness A1 w1w1
A1 q ♂
A1A2 “realized” fitness A 2 w12
Therefore, w1 = pw11 + qw12 (this is equivalent to a weighted average of the two routes) What is the fitness of an allele?
Similarly for the A2 allele:
♀ q ♂ A2A2 “realized” fitness A2 w22
A2 p ♂
A1A2 “realized” fitness A 1 w12
Therefore, w2 = qw22 + pw12 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