chapter 3: deviations from the hardy- weinberg equilibrium systematic deviations selection,...
TRANSCRIPT
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.