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Population Genetics
I. Basic Principles
A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions
Population Genetics
I. Basic Principles
A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions
1. 2 alleles in diploids: (p + q)^2 = p^2 + 2pq + q^2
Population Genetics
I. Basic Principles
A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions
1. 2 alleles in diploids: (p + q)^2 = p^2 + 2pq + q^2
2. More than 2 alleles (p + q + r)^2 = p^2 + 2pq + q^2 + 2pr + 2qr + r^2
Population Genetics
I. Basic Principles
A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions
1. 2 alleles in diploids: (p + q)^2 = p^2 + 2pq + q^2
2. More than 2 alleles (p + q + r)^2 = p^2 + 2pq + q^2 + 2pr + 2qr + r^2
3. Tetraploidy: (p + q)^4 = p^4 + 3p^3q + 6p^2q^2 + 3pq^3 + q^4(Pascal's triangle for constants...)
Population Genetics
I. Basic Principles
II. X-linked Genes
Population Genetics
I. Basic Principles
II. X-linked Genes A. Issue
Population Genetics
I. Basic Principles
II. X-linked Genes A. Issue
- Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes.
Population Genetics
I. Basic Principles
II. X-linked Genes A. Issue
- Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes.
- As a consequence, Females will carry 2/3 of these genes in a population, and males will only carry 1/3.
Population Genetics
I. Basic Principles
II. X-linked Genes A. Issue
- Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes.
- As a consequence, Females will carry 2/3 of these genes in a population, and males will only carry 1/3.
- So, the equilibrium value will NOT be when the frequency of these alleles are the same in males and females... rather, the equilibrium will occur when: p(eq) = 2/3p(f) + 1/3p(m)
Population Genetics
I. Basic Principles
II. X-linked Genes A. Issue
- Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes.
- As a consequence, Females will carry 2/3 of these genes in a population, and males will only carry 1/3.
- So, the equilibrium value will NOT be when the frequency of these alleles are the same in males and females... rather, the equilibrium will occur when: p(eq) = 2/3p(f) + 1/3p(m)
- Equilibrium will not occur with only one generation of random mating because of this imbalance... approach to equilibrium will only occur over time.
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
1. Calculating Gene Frequencies in next generation:
p(f)1 = 1/2(p(f)+p(m)) Think about it. Daughters are formed by an X from the mother and an X from the father. So, the frequency in daughters will be AVERAGE of the frequencies in the previous generation of mothers and fathers.
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
1. Calculating Gene Frequencies in next generation:
p(f)1 = 1/2(p(f)+p(m)) Think about it. Daughters are formed by an X from the mother and an X from the father. So, the frequency in daughters will be AVERAGE of the frequencies in the previous generation of mothers and fathers.
p(m)1 = p(f) Males get all their X chromosomes from their mother, so the frequency in males will equal the frequency in females in the preceeding generation.
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
2. Change over time:
- Consider this population: f(A)m = 0, and f(A)f = 1.0.
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
2. Change over time:
- Consider this population: f(A)m = 0, and f(A)f = 1.0.
- In f1: p(m) = 1.0, p(f) = 0.5
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
2. Change over time:
- Consider this population: f(A)m = 0, and f(A)f = 1.0.
- In f1: p(m) = 1.0, p(f) = 0.5
- In f2: p(m) = 0.5, p(f) = 0.75
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
2. Change over time:
- Consider this population: f(A)m = 0, and f(A)f = 1.0.
- In f1: p(m) = 1.0, p(f) = 0.5
- In f2: p(m) = 0.5, p(f) = 0.75
- In f3: p(m) = 0.75, p(f) = 0.625
Population Genetics
I. Basic Principles
II. X-linked Genes
A. Issue B. Example
2. Change over time:
- Consider this population: f(A)m = 0, and f(A)f = 1.0.
- In f1: p(m) = 1.0, p(f) = 0.5
- In f2: p(m) = 0.5, p(f) = 0.75
- In f3: p(m) = 0.75, p(f) = 0.625
- There is convergence on an equilibrium = p = 0.66
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Relative Fitness 1 1 0.25
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Relative Fitness 1 1 0.25
Survival to Reproduction 0.16 0.48 0.09
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Relative Fitness 1 1 0.25
Survival to Reproduction 0.16 0.48 0.09 = 0.73
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Relative Fitness 1 1 0.25
Survival to Reproduction 0.16 0.48 0.09 = 0.73
Geno. Freq., breeders 0.22 0.66 0.12 = 1.00
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Relative Fitness 1 1 0.25
Survival to Reproduction 0.16 0.48 0.09 = 0.73
Geno. Freq., breeders 0.22 0.66 0.12 = 1.00
Gene Freq's, gene pool p = 0.55 q = 0.45
Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.8 0.2
Relative Fitness 1 1 0.25
Survival to Reproduction 0.16 0.48 0.09 = 0.73
Geno. Freq., breeders 0.22 0.66 0.12 = 1.00
Gene Freq's, gene pool p = 0.55 q = 0.45
Genotypes, F1 0.3025 0.495 0.2025 = 100
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
- in our previous example, s = .75, p = 0.4, q = 0.6
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
- in our previous example, s = .75, p = 0.4, q = 0.6
- Δp = (.75)(.4)(.36)/1-[(.75)(.36)] = . 108/.73 = 0.15
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
- in our previous example, s = .75, p = 0.4, q = 0.6
- Δp = (.75)(.4)(.36)/1-[(.75)(.36)] = . 108/.73 = 0.15
p0 = 0.4, so p1 = 0.55 (check)
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
- next generation: (.75)(.55)(.2025)/1 - (.75)(.2025)
- = 0.084/0.85 = 0.1
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
- next generation: (.75)(.55)(.2025)/1 - (.75)(.2025)
- = 0.084/0.85 = 0.1
- so:
III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
- next generation: (.75)(.55)(.2025)/1 - (.75)(.2025)
- = 0.084/0.85 = 0.1
- so:
p0 to p1 = 0.15
p1 to p2 = 0.1
III. Modeling Selection
A. Selection for a Dominant Allele
so, Δp declines with each generation.
III. Modeling Selection
A. Selection for a Dominant Allele
so, Δp declines with each generation.
BECAUSE: as q declines, a greater proportion of q alleles are present in heterozygotes (and invisible to selection). As q declines, q2 declines more rapidly...
III. Modeling Selection
A. Selection for a Dominant Allele
so, Δp declines with each generation.
BECAUSE: as q declines, a greater proportion of q alleles are present in heterozygotes (and invisible to selection). As q declines, q2 declines more rapidly...
So, in large populations, it is hard for selection to completely eliminate a deleterious allele....
III. Modeling Selection
A. Selection for a Dominant Allele
B. Selection for an Incompletely Dominant Allele
B. Selection for an Incompletely Dominant Allele
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.4 0.2
Relative Fitness 1 0.5 0.25
Survival to Reproduction 0.16 0.24 0.09 = 0.49
Geno. Freq., breeders 0.33 0..50 0.17 = 1.00
Gene Freq's, gene pool p = 0.58 q = 0.42
Genotypes, F1 0.34 0..48 0.18 = 100
B. Selection for an Incompletely Dominant Allele
- deleterious alleles can no longer hide in the heterozygote; its presence always causes a reduction in fitness, and so it can be eliminated from a population.
III. Modeling Selection
A. Selection for a Dominant Allele
B. Selection for an Incompletely Dominant Allele
C. Selection that Maintains Variation
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.4 0.8 0.2
Relative Fitness 0.5 (1-s) 1 0.25 (1-t)
Survival to Reproduction 0.08 0.48 0.09 = 0.65
Geno. Freq., breeders 0.12 0.74 0.14 = 1.00
Gene Freq's, gene pool p = 0.49 q = 0.51
Genotypes, F1 0.24 0.50 0.26 = 100
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- Consider an 'A" allele. It's probability of being lost from the population is a function of:
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- Consider an 'A" allele. It's probability of being lost from the population is a function of:
1) probability it meets another 'A' (p)
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- Consider an 'A" allele. It's probability of being lost from the population is a function of:
1) probability it meets another 'A' (p)
2) rate at which these AA are lost (s).
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- Consider an 'A" allele. It's probability of being lost from the population is a function of:
1) probability it meets another 'A' (p)
2) rate at which these AA are lost (s).
- So, prob of losing an 'A' allele = ps
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- Consider an 'A" allele. It's probability of being lost from the population is a function of:
1) probability it meets another 'A' (p)
2) rate at which these AA are lost (s).
- So, prob of losing an 'A' allele = ps
- Likewise the probability of losing an 'a' = qt
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- Consider an 'A" allele. It's probability of being lost from the population is a function of:
1) probability it meets another 'A' (p)
2) rate at which these AA are lost (s).
- So, prob of losing an 'A' allele = ps
- Likewise the probability of losing an 'a' = qt
- An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt.
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt.
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt.
- substituting (1-p) for q, ps = (1-p)tps = t - ptps +pt = tp(s + t) = tpeq = t/(s + t)
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- An equilibrium will occur, when the probability of losing A an a are equal; when ps = qt.
- substituting (1-p) for q, ps = (1-p)tps = t - ptps +pt = tp(s + t) = tpeq = t/(s + t)
- So, for our example, t = 0.75, s = 0.5
- so, peq = .75/1.25 = 0.6
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- so, peq = .75/1.25 = 0.6
p = 0.6, q = 0.4 AA Aa aa
Parental "zygotes" 0.36 0.48 0.16 = 1.00
prob. of survival (fitness) 0.4 0.8 0.2
Relative Fitness 0.5 (1-s) 1 0.25 (1-t)
Survival to Reproduction 0.18 0.48 0.04 = 0.70
Geno. Freq., breeders 0.26 0.68 0.06 = 1.00
Gene Freq's, gene pool p = 0.6 q = 0.4 CHECK
Genotypes, F1 0.36 0.48 0.16 = 100
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- so, peq = .75/1.25 = 0.6
- so, if p > 0.6, it should decline to this peq
p = 0.7, q = 0.3 AA Aa aa
Parental "zygotes" 0.49 0.42 0.09 = 1.00
prob. of survival (fitness) 0.4 0.8 0.2
Relative Fitness 0.5 (1-s) 1 0.25 (1-t)
Survival to Reproduction 0.25 0.48 0.02 = 0.75
Geno. Freq., breeders 0.33 0.64 0.03 = 1.00
Gene Freq's, gene pool p = 0.65 q = 0.35 CHECK
Genotypes, F1 0.42 0.46 0.12 = 100
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
- so, peq = .75/1.25 = 0.6
- so, if p > 0.6, it should decline to this peq
0.6
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism -
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism -
- equilibrium can occur if AA and aa are each fit in a given niche, within the population. The equilibrium will depend on the relative frequencies of the niches and the selection differentials...
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism -
- equilibrium can occur if AA and aa are each fit in a given niche, within the population. The equilibrium will depend on the relative frequencies of the niches and the selection differentials...
- can you think of an example??
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism -
- equilibrium can occur if AA and aa are each fit in a given niche, within the population. The equilibrium will depend on the relative frequencies of the niches and the selection differentials...
- can you think of an example??
Papilio butterflies... females mimic different models and an equilibrium is maintained; in fact, an equilibrium at each locus, which are also maintained in linkage disequilibrium.
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism
3. Frequency Dependent Selection
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism
3. Frequency Dependent Selection
- the fitness depends on the frequency...
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism
3. Frequency Dependent Selection
- the fitness depends on the frequency...
- as a gene becomes rare, it becomes advantageous and is maintained in the population...
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism
3. Frequency Dependent Selection
- the fitness depends on the frequency...
- as a gene becomes rare, it becomes advantageous and is maintained in the population...
- "Rare mate" phenomenon...
- Morphs of Heliconius melpomene and H. erato
Mullerian complex between two distasteful species... positive frequency dependence in both populations to look like the most abundant morph
C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism
3. Frequency Dependent Selection
4. Selection Against the Heterozygote
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.4 0.6
Relative Fitness 1 0.5 0.75
Corrected Fitness 1 + 0.5 1.0 1 + 0.25
formulae 1 + s 1 + t
4. Selection Against the Heterozygote
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.4 0.6
Relative Fitness 1 0.5 0.75
Corrected Fitness 1 + 0.5 1.0 1 + 0.25
formulae 1 + s 1 + t
4. Selection Against the Heterozygote
- peq = t/(s + t)
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.4 0.6
Relative Fitness 1 0.5 0.75
Corrected Fitness 1 + 0.5 1.0 1 + 0.25
formulae 1 + s 1 + t
4. Selection Against the Heterozygote
- peq = t/(s + t)
- here = .25/(.50 + .25) = .33
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.4 0.6
Relative Fitness 1 0.5 0.75
Corrected Fitness 1 + 0.5 1.0 1 + 0.25
formulae 1 + s 1 + t
4. Selection Against the Heterozygote
- peq = t/(s + t)
- here = .25/(.50 + .25) = .33
- if p > 0.33, then it will keep increasing to fixation.
p = 0.4, q = 0.6 AA Aa aa
Parental "zygotes" 0.16 0.48 0.36 = 1.00
prob. of survival (fitness) 0.8 0.4 0.6
Relative Fitness 1 0.5 0.75
Corrected Fitness 1 + 0.5 1.0 1 + 0.25
formulae 1 + s 1 + t
4. Selection Against the Heterozygote
- peq = t/(s + t)
- here = .25/(.50 + .25) = .33
- if p > 0.33, then it will keep increasing to fixation.
- However, if p < 0.33, then p will decline to zero... AND THERE WILL BE FIXATION FOR A SUBOPTIMAL ALLELE....'a'... !! UNSTABLE EQUILIBRIUM!!!!