Chapter 3: Deviations from the Hardy- Weinberg equilibrium Systematic deviations Selection,...

Chapter 3: Deviations from the Hardy-

Weinberg equilibrium

• Systematic deviations

Selection, migration and mutation

• Random genetic drift

Small effective population size

Deviations from the Hardy-Weinberg law

Systematic deviations:

• Migration

• Selection

• Mutation

Deviations from theHardy-Weinberg law

Selection

Deviations from the Hardy-Weinberg law

Migration

Deviations from theHardy-Weinberg law

Mutation

Small population size, random changes

Mutation:

The selection coefficient has the symbol s

The mutation frequency has the symbol Selection mutations equilibrium occurs when:

q2 s = for the recessive genes

pq s = p s = for the dominant genes

Deviations from theHardy-Weinberg law

Genetic load

• Selection can cause the death of some individuals or make them unable to reproduce

• This cost is called a genetic load

Belgian Blue cattle

Selection against the recessive

Genotype EE Ee ee Total

Frequency p2 2pq q2 1,00Fitness 1 1 1-s Proportion p2 2pq q2 (1-s) 1-sq2

after selection

Fitness is constant (1-s). That is the opposite of selection, s

Genetic load = sq2

Selection against the recessive:Example

Example: q =0,25 and s=1:

Genotype EE Ee ee Total

Frekvens 0.5625 0.375 0.0625 1.00Fitness 1 1 0 Proportion 0.5625 0.375 0 0.9375after selection

q’ = (2pq/2 + q2 (1-s))/(1-sq2) = (0.375/2 + 0)/ 0.9375 = 0.20

Selection against the recessive:Several generations

Formula for the calculating of gene frequency in the following generation

q’ = (2pq/2 + q2 (1-s))/(1-sq2)

q

Selection against the recessive:Formula for s=1

Expansion to n generations for s=1:

qn = q0/(1+n q0)

n can be isolated

n = 1/qn - 1/q0

Example: Gene frequency changes from 0.01 to 0.005

n = 1/0.005 - 1/0.01 = 200 - 100 = 100 generations

Bedlington-terrier, example

Genotype EE Ee ee Total

Frekvens p2 2pq q2 1,00Fitness 1-s1 1 1-s2 Proportion p2 (1-s1) 2pq q2 (1-s2) 1-p2s1 - q2s2

after selection

Genetic load = p2s1 + q2s2

Selection for heterozygotes

Selection for heterozygotes: Equilibrium frequency

After selection the gene frequency is calculated by use of the gene counting method:

q' = (q2 (1-s2) + pq)/(1-p2s1 - q2s2)

And equilibrium occurs at:

q = pq(ps1- qs2)/(1-p2s1 - q2s2) = 0

for: ps1- qs2 = 0

q= s1 / (s1 + s2)

which is the equilibrium frequency

^

Selection for heterozygotes: Fitness-graph

Relative fitness by over dominance

In the population 5% is born with sickle cell anaemia.

q2 = 0.05 q = 0.22 s2 = 1

Equilibrium occurs at:

p= s2 / (s1 + s2) = 1 - qWhich solved gives:

s1 = (s2 /(1 - q)) - s2 = 0.285

^

Sickle cell anaemia in malaria areas

Selection for heterozygotes: Example

Selection against heterozygotes

The gene couting method gives:

q' = (q2 (1-s2) + pq)/(1-p2s1 - q2s2)

Equilibrium occurs at:

q = pq(ps1- qs2)/(1-p2s1 - q2s2) = 0

forq = s1 / (s1 + s2)

The equilibrium is unstable

^

Small populations

Binominal variance on p or q:

2 = (pq)/(2N),

2N is equal to the number of genes, drawn

from the population to form the new generation

Small populations: Variance on the gene frequency

Small populations, continued

The standard deviation of the gene frequency

Number of genes

Effective population size (Ne)

The number of sires and dams for the new generation has

significance for Ne

Ne the harmonic mean of the two sexes

4/Ne = 1/Nmales + 1/Nfemales

• 10 males and 10 females

4/Ne = 1/10 + 1/10 which gives Ne = 20

• 1 male and 10 females 4/Ne = 1/1 + 1/10 which gives Ne = 3.7

• 100 males and 100000 females 4/Ne = 1/100 + 0 which gives Ne = 400

Effective population size (Ne): Examples

Increase in the degree of inbreeding

(F)The increase in inbreeding per generation is dependent on the effective population size (Ne)

F = 1/(2Ne)

In a population with Ne = 20, the increase in each generation is delta F = 2.5 %.

The inbreeding coefficient F is defined in the next chapter.