population genetics i. basic principles a. definitions: b. basic computations:
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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 - PowerPoint PPT PresentationTRANSCRIPT
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Population Genetics
I. Basic Principles
A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions
![Page 2: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/2.jpg)
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
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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
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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...)
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Population Genetics
I. Basic Principles
II. X-linked Genes
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Population Genetics
I. Basic Principles
II. X-linked Genes A. Issue
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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.
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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.
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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)
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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.
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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.
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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.
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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.
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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
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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
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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
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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
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Population Genetics
I. Basic Principles
II. X-linked Genes
III. Modeling Selection
A. Selection for a Dominant Allele
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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
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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
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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
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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
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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
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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
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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
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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
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III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
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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
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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
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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)
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III. Modeling Selection
A. Selection for a Dominant Allele
Δp = spq2/1-sq2
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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
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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:
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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
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III. Modeling Selection
A. Selection for a Dominant Allele
so, Δp declines with each generation.
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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...
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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....
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III. Modeling Selection
A. Selection for a Dominant Allele
B. Selection for an Incompletely Dominant Allele
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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
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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.
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III. Modeling Selection
A. Selection for a Dominant Allele
B. Selection for an Incompletely Dominant Allele
C. Selection that Maintains Variation
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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
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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:
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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)
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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).
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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
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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
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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.
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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.
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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)
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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
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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
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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
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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
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C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism -
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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...
![Page 57: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/57.jpg)
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??
![Page 58: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/58.jpg)
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.
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C. Selection that Maintains Variation
1. Heterosis - selection for the heterozygote
2. Multiple Niche Polymorphism
3. Frequency Dependent Selection
![Page 60: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/60.jpg)
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...
![Page 61: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/61.jpg)
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...
![Page 62: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/62.jpg)
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...
![Page 63: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/63.jpg)
- 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
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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
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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
![Page 66: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/66.jpg)
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
![Page 67: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/67.jpg)
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
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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
![Page 69: Population Genetics I. Basic Principles A. Definitions: B. Basic computations:](https://reader035.vdocuments.site/reader035/viewer/2022081603/56815414550346895dc21144/html5/thumbnails/69.jpg)
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!!!!