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 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

= p(p + q)

= p2 + pq

= p(p + q)

= p2 + pq

= p(p + q)

Frequency of A2 = q2 + ½ (2pq)

= p2 + pq

= p(p + q)

= q2 + pq

= p2 + pq

= p(p + q)

= q2 + pq

= q(q + p)

= p2 + pq

= p(p + q)

= q2 + pq

= q(q + p)

= p2 + pq

= p(p + q)

= q2 + pq

= q(q + p)

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 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 = 0.50, 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

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 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

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

χ2 = Σ (Obs. – Exp.)2

A1A1 109 109.9 A1A2 182 180.2 A2A2 73 73.9

χ2 = Σ (Obs. – Exp.)2 = 0.036

(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

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:

♂ A1A1

♂ A1A1 “realized” fitness w11

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

♂ A1A1 “realize” fitness w11

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

A1A2 “realized” fitness w12

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

Therefore, w1 = pw11 + qw12

♂ A1A1 “realized” fitness w1w1

Therefore, w1 = pw11 + qw12

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

Similarly for the A2 allele:

Therefore, 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….

Mean population fitness

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’ = 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)

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

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

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