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
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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 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 • 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
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
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
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
Population Genetics I. Basic Principles II. X-linked Genes III. Modeling Selection A. Selection for a Dominant Allele
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 - 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 - 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 - 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)t ps = t - pt ps +pt = t p(s + t) = t peq = t/(s + t)
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