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



A1 x A1 → A1A1 p x p = p2

A1 x A2 → A1A2 p x q = pq



A1 x A1 → A1A1 p x p = p2

A1 x A2 → A1A2 p x q = pq

A2 x A1 → A2A1 q x p = qp



A1 x A1 → A1A1 p x p = p2

A1 x A2 → A1A2 p x q = pq 2pq




A1 x A1 → A1A1 p x p = p2

A1 x A2 → A1A2 p x q = pq 2pq


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:


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



Allele frequencies Genotype frequencies

A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04




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 = 0.80, 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


3. No migration


1. Mating is random


3. No migration

4. No mutation


1. Mating is random


3. No migration

4. No mutation

5. No selection


1. Mating is random


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



• 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


Step 1: Estimate genotype frequencies



Frequency of A1A1 = 109/364 = 0.2995



Frequency of A1A1 = 109/364 = 0.2995

Frequency of A1A2 = 182/364 = 0.5000



Frequency of A1A1 = 109/364 = 0.2995

Frequency of A1A2 = 182/364 = 0.5000

Frequency of A2A2 = 73/364 = 0.2005


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)




Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2) = 0.2005 + ½ (0.5000)




Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2) = 0.2005 + ½ (0.5000) = 0.4505




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


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.

(see Box 6.5 on pages 192-193)

Suppose we sampled a mixed population


Population A

p = A1 = 1.0 q = A2 = 0 100% A1A1


Population A Population B

p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2



p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2

Mixed population



p = A1 = 1.0 p = A1 = 0 q = A2 = 0 q = A2 = 1.0 100% A1A1 100% A2A2

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)


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 A2 allele = w2 = qw22 + pw12



One final fitness to define….




Mean population fitness




Mean population fitness = w = pw1 + qw2




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


p’ = pw1/(pw1 + qw2)


p’ = pw1/(pw1 + qw2)

p’ = p(w1/w)


p’ = pw1/(pw1 + qw2)

p’ = p(w1/w)

Let q’ = frequency of A2 allele in the next generation


p’ = pw1/(pw1 + qw2)

p’ = p(w1/w)


q’ = qw2/(pw1 + qw2)


p’ = pw1/(pw1 + qw2)

p’ = p(w1/w)


q’ = qw2/(pw1 + qw2)

q’ = q (w2/w)

An example of directional selection


Let p = q = 0.5


Let p = q = 0.5

Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90


Let p = q = 0.5


w1 = pw11 + qw22 = 0.975


Let p = q = 0.5


w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925


Let p = q = 0.5


w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950


Let p = q = 0.5



p’ = p(w1/w) = 0.513


Let p = q = 0.5



p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487


Let p = q = 0.5



p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487

In ~150 generations the A1 allele will be fixed

Lecture 8 Microevolution 1 - selection

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