population genetics i. basic principles a. definitions: b. basic computations:

Population Genetics

I. Basic Principles

A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions

Population Genetics

I. Basic Principles


1. 2 alleles in diploids: (p + q)^2 = p^2 + 2pq + q^2

Population Genetics

I. Basic Principles



2. More than 2 alleles (p + q + r)^2 = p^2 + 2pq + q^2 + 2pr + 2qr + r^2

Population Genetics

I. Basic Principles



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


- Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes.

Population Genetics

I. Basic Principles



- 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




- 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




- 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



- In f1: p(m) = 1.0, p(f) = 0.5

Population Genetics

I. Basic Principles

II. X-linked Genes

A. Issue B. Example



- 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



- 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



- 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



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



p = 0.4, q = 0.6 AA Aa aa


prob. of survival (fitness) 0.8 0.8 0.2

Population Genetics

I. Basic Principles

II. X-linked Genes



p = 0.4, q = 0.6 AA Aa aa



Relative Fitness 1 1 0.25

Population Genetics

I. Basic Principles

II. X-linked Genes



p = 0.4, q = 0.6 AA Aa aa




Survival to Reproduction 0.16 0.48 0.09

Population Genetics

I. Basic Principles

II. X-linked Genes



p = 0.4, q = 0.6 AA Aa aa




Survival to Reproduction 0.16 0.48 0.09 = 0.73

Population Genetics

I. Basic Principles

II. X-linked Genes



p = 0.4, q = 0.6 AA Aa aa





Geno. Freq., breeders 0.22 0.66 0.12 = 1.00

Population Genetics

I. Basic Principles

II. X-linked Genes



p = 0.4, q = 0.6 AA Aa aa






Gene Freq's, gene pool p = 0.55 q = 0.45

Population Genetics

I. Basic Principles

II. X-linked Genes



p = 0.4, q = 0.6 AA Aa aa







Genotypes, F1 0.3025 0.495 0.2025 = 100



Δp = spq2/1-sq2



Δp = spq2/1-sq2

- in our previous example, s = .75, p = 0.4, q = 0.6



Δp = spq2/1-sq2


- Δp = (.75)(.4)(.36)/1-[(.75)(.36)] = . 108/.73 = 0.15



Δp = spq2/1-sq2


- Δp = (.75)(.4)(.36)/1-[(.75)(.36)] = . 108/.73 = 0.15

p0 = 0.4, so p1 = 0.55 (check)



Δp = spq2/1-sq2



Δp = spq2/1-sq2

- next generation: (.75)(.55)(.2025)/1 - (.75)(.2025)

- = 0.084/0.85 = 0.1



Δp = spq2/1-sq2

- next generation: (.75)(.55)(.2025)/1 - (.75)(.2025)

- = 0.084/0.85 = 0.1

- so:



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



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...




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....



B. Selection for an Incompletely Dominant Allele


p = 0.4, q = 0.6 AA Aa aa



Relative Fitness 1 0.5 0.25


Geno. Freq., breeders 0.33 0..50 0.17 = 1.00


Genotypes, F1 0.34 0..48 0.18 = 100


- 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.




C. Selection that Maintains Variation


1. Heterosis - selection for the heterozygote

p = 0.4, q = 0.6 AA Aa aa



Relative Fitness 0.5 (1-s) 1 0.25 (1-t)




Genotypes, F1 0.24 0.50 0.26 = 100



- 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







- 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.




- substituting (1-p) for q, ps = (1-p)tps = t - ptps +pt = tp(s + t) = tpeq = t/(s + t)




- 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



- so, peq = .75/1.25 = 0.6

p = 0.6, q = 0.4 AA Aa aa






Gene Freq's, gene pool p = 0.6 q = 0.4 CHECK

Genotypes, F1 0.36 0.48 0.16 = 100



- 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






Gene Freq's, gene pool p = 0.65 q = 0.35 CHECK

Genotypes, F1 0.42 0.46 0.12 = 100



- so, peq = .75/1.25 = 0.6

- so, if p > 0.6, it should decline to this peq

0.6



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??





- 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.



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...






- 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





4. Selection Against the Heterozygote

p = 0.4, q = 0.6 AA Aa aa




Corrected Fitness 1 + 0.5 1.0 1 + 0.25

formulae 1 + s 1 + t


p = 0.4, q = 0.6 AA Aa aa







- peq = t/(s + t)

p = 0.4, q = 0.6 AA Aa aa







- peq = t/(s + t)

- here = .25/(.50 + .25) = .33

p = 0.4, q = 0.6 AA Aa aa







- 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







- 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!!!!

population genetics i. basic principles a. definitions: b. basic computations:

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