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 = p2 + 2pq + q2
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 = p2 + 2pq + q2 + 2pr + 2qr + r2
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 = p4 + 4p3q + 6p2q2 + 4pq3 + q4
(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 = ½[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
- p(eq) = 2/3p(f) + 1/3p(m)
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
prob. of survival (fitness)
0.8
0.8
0.2
= 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
prob. of survival (fitness)
0.8
0.8
0.2
Relative Fitness
1
1
0.25
= 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
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
= 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
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
= 1.00
= 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
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
= 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
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
= 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
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
Genotypes, F1
0.3025
= 1.00
q = 0.45
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
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
Genotypes, F1
0.34
= 1.00
q = 0.42
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
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
Genotypes, F1
0.24
= 1.00
q = 0.51
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)t
ps = t - pt
ps +pt = t
p(s + t) = t
peq = 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)t
ps = t - pt
ps +pt = t
p(s + t) = t
peq = 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
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
0.16
= 100
0.48
= 1.00
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
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
0.12
= 100
0.46
= 1.00
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