Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,

Lecture 7 Mutation and genetic variation

Point mutations

There are four categories of point mutations:

1.  transitions (e.g., A → G, C → T)

2. transversions (e.g., T → A, C → G)

3. insertions (e.g., TTTGAC → TTTCCGAC)

• in coding regions, point mutations can involve silent (synonymous) or replacement (nonsynonymous) changes.

• in coding regions, insertions/deletions can also cause frameshift mutations.

4. deletions (e.g., TTTGAC → TTTC)

Indels are insertions and deletions

STOP making sense: effective frameshifts

“Copy-number” mutations


• these mutations change the numbers of genetic elements.



• gene duplication events create new copies of genes.




• one important mechanism generating duplications is unequal crossing over.

Unequal crossing-over can generate gene duplications

Unequal crossing-over can generate gene duplications




• one mechanism believed responsible is unequal crossing over.

• over time, this process may lead to the development of multi-gene families.

Whole-genome data yields data on gene families

Retrogenes may also be created

• retrogenes have identical exon structures to their “progenitors” but lack introns!



Example: jingwei in Drosophila yakuba




Alcohol dehydrogenase (Adh)

→

Chromosome 2 Chromosome 3





→


→ mRNA





→


→ mRNA

↓ cDNA





→


→ mRNA

↓ cDNA → “jingwei”

Where do new genes come from?

Where do new genes come from? An example: the antifreeze glycoprotein (AFGP) gene in the Antarctic fish, Dissostichus mawsoni

Convergent evolution of an AFGP gene in the arctic cod, Boreogadus saida

Lecture 8 Microevolution 1 - selection

The Hardy-Weinberg Equilibrium

Godfrey Hardy Wilhelm Weinberg

William Castle

Gregor Mendel

Udny Yule

Reginald Punnett & William Bateson (1875-1967) (1861-1926)

Reginald Punnett Godfrey Hardy

The Hardy-Weinberg-Castle Equilibrium

Godfrey Hardy

Wilhelm Weinberg

William Castle

The Hardy-Weinberg Equilibrium

consider a single locus with two alleles A1 and A2

• three genotypes exist: A1A1, A1A2, A2A2

• let p = frequency of A1 allele

• let q = frequency of A2 allele

• since only two alleles present, p + q = 1

Question: If mating occurs at random in the population, what will the frequencies of A1 and A2 be in the next generation?

What are the probabilities of matings at the gamete level?


Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2


Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2 A1 x A2 → A1A2 p x q = pq


Egg Sperm Zygote Probability A1 x A1 → A1A1 p x p = p2 A1 x A2 → A1A2 p x q = pq A2 x A1 → A2A1 q x p = qp



2pq A2 x A1 → A2A1 q x p = qp



2pq A2 x A1 → A2A1 q x p = qp A2 x A2 → A2A2 q x q = q2

Therefore, zygotes produced in proportions:

Genotype: A1A1 A1A2 A2A2

Frequency: p2 2pq q2

Therefore, zygotes produced in proportions:

Genotype: A1A1 A1A2 A2A2

Frequency: p2 2pq q2

what are the allele frequencies?

What are the allele frequencies?


Frequency of A1 = p2 + ½ (2pq)



= p2 + pq



= p2 + pq

= p(p + q)



= p2 + pq

= p(p + q)

= p



= p2 + pq

= p(p + q)

= p

Frequency of A2 = q2 + ½ (2pq)



= p2 + pq

= p(p + q)

= p


= q2 + pq



= p2 + pq

= p(p + q)

= p


= q2 + pq

= q(q + p)



= p2 + pq

= p(p + q)

= p


= q2 + pq

= q(q + p)

= q



= p2 + pq

= p(p + q)

= p


= q2 + pq

= q(q + p)

= q

ALLELE FREQUENCIES DID NOT CHANGE!!

Conclusions of the Hardy-Weinberg principle


1. Allele frequencies will not change from generation to generation.



2. Genotype proportions determined by the “square law”.




• for two alleles = (p + q)2 = p2 + 2pq + q2




• for two alleles = (p + q)2 = p2 + 2pq + q2

• for three alleles (p + q + r)2 = p2 + q2 + r2 + 2pq + 2pr +2qr


3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies


3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04


3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 A1 = 0.50, A2 = 0.50 A1A1 = 0.25, A1A2 = 0.50, A2A2 = 0.25


3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 A1 = 0.50, A2 = 0.50 A1A1 = 0.25, A1A2 = 0.50, A2A2 = 0.25 A1 = 0.10, A2 = 0.90 A1A1 = 0.01, A1A2 = 0.18, A2A2 = 0.81

Assumptions of Hardy-Weinberg equilibrium


1. Mating is random


1. Mating is random… but some traits experience positive assortative mating


1. Mating is random 2. Population size is infinite (i.e., no genetic drift)


1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration


1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation


1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation 5. No selection


1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation 5. No selection

The Hardy-Weinberg equilibrium principle thus predicts that no evolution will occur unless one (or more) of these assumptions are violated!

Does Hardy-Weinberg equilibrium ever exist in nature?


Example: Atlantic cod (Gadus morhua) in Nova Scotia



as a juvenile…



… and as an adult




Example: Atlantic cod (Gadus morhua) in Nova Scotia • a sample of 364 fish were scored for a single nucleotide polymorphism (SNP)



A1A1 = 109 A1A2 = 182 A2A2 = 73



A1A1 = 109 A1A2 = 182 A2A2 = 73

Question: Is this population in Hardy-Weinberg equilibrium?

Testing for Hardy-Weinberg equilibrium

Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies

Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995

Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000

Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000 Frequency of A2A2 = 73/364 = 0.2005

Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies


Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2)


Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000)


Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495



Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2)




= 0.2005 + ½ (0.5000)




= 0.2005 + ½ (0.5000) = 0.4505




= 0.2005 + ½ (0.5000) = 0.4505

Check that p + q = 0.5495 + 0.4505 = 1


Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium



Expected No. of A1A1 = p2 x N



Expected No. of A1A1 = p2 x N = (0.5495)2 x 364



Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9



Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9 Expected No. of A1A2 = 2pq x N



Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9 Expected No. of A1A2 = 2pq x N = 2(0.5495)(0.4505) x 364



Expected No. of A1A1 = p2 x N = (0.5495)2 x 364 = 109.9 Expected No. of A1A2 = 2pq x N = 2(0.5495)(0.4505) x 364 = 180.2



Expected No. of A2A2 = q2 x N



Expected No. of A2A2 = q2 x N = (0.4595)2 x 364



Expected No. of A2A2 = q2 x N = (0.4595)2 x 364 = 73.9

Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes

Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected


A1A1 109 109.9


A1A1 109 109.9 A1A2 182 180.2


A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9


A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9

χ2 = Σ (Obs. – Exp.)2

Exp.


A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9

χ2 = Σ (Obs. – Exp.)2 = 0.036

Exp.

Suppose we sampled a mixed population

Suppose we sampled a mixed population Population A p = A1 = 1.0 q = A2 = 0 100% A1A1

Suppose we sampled a mixed population Population A Population B p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2


æ å Mixed population


æ å Mixed population

50% from population A (all A1A1) 50% from population B (all A2A2)



↓

Sample 1000 individuals



↓


Observed

A1A1 = 500 A1A2 = 0 A2A2 = 500



↓


Observed Expected

A1A1 = 500 A1A1 = 250 A1A2 = 0 A1A2 = 500 A2A2 = 500 A2A2 = 250

Sampling a mixed population generates a deficiency of

heterozygotes This is called a Wahlund effect

Deafness in Tristan da Cunha

observed AA = 1228 Aa = 352 aa = 253 Total: 1833 p = 1228/1833 + 352/2*1833 = 0.670 + 0.096 = 0.766 q = 253/1833 + 352/2*1833 = 0.138 + 0.096 = 0.234

Inbreeding: Example


observed AA = 1228 Aa = 352 aa = 253 Total: 1833 p = 1228/1833 + 352/2*1833 = 0.670 + 0.096 = 0.766 q = 253/1833 + 352/2*1833 = 0.138 + 0.096 = 0.234

Inbreeding: Example

expected 1075.5 657.10 100.36


observed expected AA = 1228 1075.5 Aa = 352 657.10 aa = 253 (13.8%) 100.36 (5.4%) Total: 1833 p = 1228/1833 + 352/2*1833 = 0.670 + 0.096 = 0.766 q = 253/1833 + 352/2*1833 = 0.138 + 0.096 = 0.234

Inbreeding: Example

A simple model of directional selection


• consider a single locus with two alleles A1 and A2












• relative fitnesses are:

A1A1 A1A2 A2A2

w11 w12 w22






A1A1 A1A2 A2A2

w11 w12 w22

• it is also possible to determine relative fitnesses of the A1 and A2 alleles:






A1A1 A1A2 A2A2

w11 w12 w22


let w1 = fitness of the A1 allele






A1A1 A1A2 A2A2

w11 w12 w22




What is the fitness of an allele? Consider fitness from the gamete’s perspective:

♀

A1


♀

A1

A1

♂


♀

A1

A1

♂ A1A1


♀

A1

A1

♂ A1A1 “realized” fitness w11


♀

A1

A1


This route will occur with a probability p, since p is the frequency of the A1 allele


♀

A1

A1

♂ A1A1 “realize” fitness w11

This route will occur with a probability p, since p is the frequency of the A1 allele

p


♀

A1

A1


♂

A2

p


♀

A1

A1


♂

A2

p

A1A2 “realized” fitness w12


♀

A1

A1


♂

A2

p


This route will occur with a probability q, since q is the frequency of the A2 allele

q


♀

A1

A1


♂

A2

p


Therefore, w1 = pw11 + qw12

q


♀

A1

A1

♂ A1A1 “realized” fitness w1w1

♂

A2

p


Therefore, w1 = pw11 + qw12

q

(this is equivalent to a weighted average of the two routes)

What is the fitness of an allele? Similarly for the A2 allele:

♀

A2

A2


♂

A1

q


Therefore, w2 = qw22 + pw12

p

The fitness of the A1 allele = w1 = pw11 + qw12

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define….

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness = w = pw1 + qw2

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness = w = pw1 + qw2 (This too is a weighted average of the two allelic fitnesses.)

Let p’ = frequency of A1 allele in the next generation

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2) q’ = q (w2/w)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2) q’ = q (w2/w)

An example of directional selection

An example of directional selection Let p = q = 0.5

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw12 = 0.975 w2 = qw22 + pw12 = 0.925 w = pw1 + qw2 = 0.950

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487 In ~150 generations the A1 allele will be fixed

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Lecture 7 Mutation and genetic variation · 10/8/2015 · Point mutations There are four categories of point mutations: 1. transitions (e.g., A → G, C → T) 2. transversions (e.g.,