Chapter 3: Deviations from the
HardyWeinberg 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 the
Hardy-Weinberg law
Selection
Deviations from the HardyWeinberg law
Migration
Deviations from the
Hardy-Weinberg law
Mutation
Small population size, random changes
Deviations from the
Hardy-Weinberg law
Mutation:
The selection coefficient has the symbol s
The mutation frequency has the symbol m
Selection mutations equilibrium occurs when:
q2  s = m
for the recessive genes
pq  s = p  s = m
for the dominant genes
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
Frequency
p2
Fitness
1
Proportion
p2
after selection
Ee
2pq
1
2pq
ee
q2
1-s
q2 (1-s)
Total
1,00
1-sq2
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
Frekvens
0.5625
Fitness
1
Proportion
0.5625
after selection
Ee
0.375
1
0.375
q’ = (2pq/2 + q2 (1-s))/(1-sq2)
= (0.375/2 + 0)/ 0.9375 = 0.20
ee
0.0625
0
0
Total
1.00
0.9375
Selection against the recessive:
Several generations
Formula for the calculating of gene frequency in the
following generation
q
q’ = (2pq/2 + q2 (1-s))/(1-sq2)
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
Selection for heterozygotes
Genotype
EE
Frekvens
p2
Fitness
1-s1
Proportion
p2 (1-s1)
after selection
Ee
2pq
1
2pq
Genetic load = p2s1 + q2s2
ee
Total
q2
1-s2
q2 (1-s2)
1,00
1-p2s1 - q2s2
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)
Selection for heterozygotes:
Fitness-graph
Relative fitness by over dominance
Selection for heterozygotes:
Example
Sickle cell anaemia in malaria areas
In the population 5% is born with sickle cell anaemia
q2 = 0.05 
q = 0.22
s2 = 1
Equilibrium occurs at:
^ s / (s + s ) = 1 - q
p=
2
1
2
Which solved gives:
s1 = (s2 /(1 - q)) - s2 = 0.285
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
for
q^ = s1 / (s1 + s2)
The equilibrium is unstable
Small populations
Small populations:
Variance on the gene frequency
Binominal variance on p or q:
s2 = (pq)/(2N),
2N is equal to the number of genes, drawn
from the population to form the new
generation
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
Effective population size (Ne):
Examples
• 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
= 1/100 + 0 which gives Ne = 400
4/Ne
Increase in the degree of inbreeding
(F)
The increase in inbreeding per generation is
dependent on the effective population size (Ne)
 F = 1/(2Ne)

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.