POPULATION GENETICS
Predicting inheritance in a population
© 2008 Paul Billiet ODWS
Predictable patterns of inheritance in a
population so long as…
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the population is large enough not to show the
effects of a random loss of genes by chance events
i.e. there is no genetic drift
the mutation rate at the locus of the gene being
studied is not significantly high
mating between individuals is random
(a panmictic population)
new individuals are not gained by immigration or
lost be emigration
i.e. there is no gene flow between neighbouring
populations
the gene’s allele has no selective advantage or
disadvantage
© 2008 Paul Billiet ODWS
SUMMARY
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Genetic drift
Mutation
Mating choice
Migration
Natural selection
All can affect the
transmission of genes
from generation to
generation
Genetic Equilibrium
If none of these factors is operating then the
relative proportions of the alleles (the GENE
FREQUENCIES) will be constant
© 2008 Paul Billiet ODWS
THE HARDY WEINBERG
PRINCIPLE
Step 1
 Calculating the gene frequencies from the
genotype frequencies
 Easily done for codominant alleles (each
genotype has a different phenotype)
© 2008 Paul Billiet ODWS
Iceland
Population
 313 337 (2007 est)
Area
 103 000 km2
Distance from mainland
Europe
 970 km
Google Earth
© 2008 Paul Billiet ODWS
Example Icelandic population: The
MN blood group
Sample
Phenotypes
Population
Genotypes
747
Numbers
Contribution
to gene pool
© 2008 Paul Billiet ODWS
Type M
Type MN
Type N
MmMm
MmMn
MnMn
233
385
129
2 Mm
alleles
per
person
1 Mm
allele
per
person
1 Mn
allele
per
person
2 Mn
alleles
per
person
MN blood group in Iceland
Total Mm alleles = (2 x 233) + (1 x 385) = 851
Total Mn alleles = (2 x 129) + (1 x 385) = 643
Total of both alleles
=1494
= 2 x 747
(humans are diploid organisms)
Frequency of the Mm allele = 851/1494 = 0.57
or 57%
Frequency of the Mn allele = 643/1494 = 0.43
or 43%
© 2008 Paul Billiet ODWS
In general for a diallellic gene A and a
(or Ax and Ay)
If the frequency of the A allele =
and the frequency of the a allele =
Then p+q = 1
© 2008 Paul Billiet ODWS
p
q
Step 2
Using the calculated gene frequency to
predict the EXPECTED genotypic
frequencies in the NEXT generation
OR
 to verify that the PRESENT population is
in genetic equilibrium

© 2008 Paul Billiet ODWS
Assuming all the individuals mate
randomly
NOTE the gene frequencies are the gamete frequencies too
SPERMS
Mm 0.57
Mn 0.43
Mm
0.57
MmMm
0.32
MmMn
0.25
Mn
0.43
MmMn
0.25
MnMn
0.18
EGGS
© 2008 Paul Billiet ODWS
Close enough for us to assume genetic
equilibrium
Genotypes
Expected
frequencies
Observed
frequencies
MmMm
0.32
233  747 = 0.31
MmMn
0.50
385  747 = 0.52
MnMn
0.18
129  747 = 0.17
© 2008 Paul Billiet ODWS
In general for a diallellic gene A and a
(or Ax and Ay)
Where the gene frequencies are p and q
Then p + q = 1
and
SPERMS
A
p
a
q
A
p
AA
p2
Aa
pq
a
q
Aa
pq
aa
q2
EGGS
© 2008 Paul Billiet ODWS
THE HARDY WEINBERG EQUATION
So the genotype frequencies are:
AA
Aa
aa
=
=
=
p2
2pq
q2
or p2 + 2pq + q2 = 1
© 2008 Paul Billiet ODWS
DEMONSTRATING GENETIC
EQUILIBRIUM
Using the Hardy Weinberg Equation to
determine the genotype frequencies from the
gene frequencies may seem a circular
argument
© 2008 Paul Billiet ODWS
Only one of the populations below is in genetic
equilibrium. Which one?
Population sample
Genotypes
Gene frequencies
AA
Aa
aa
100
20
80
0
100
36
48
16
100
50
20
30
100
60
0
40
© 2008 Paul Billiet ODWS
A
a
Only one of the populations below is in genetic
equilibrium. Which one?
Population sample
Genotypes
Gene frequencies
AA
Aa
aa
A
a
100
20
80
0
0.6
0.4
100
36
48
16
0.6
0.4
100
50
20
30
0.6
0.4
100
60
0
40
0.6
0.4
© 2008 Paul Billiet ODWS
Only one of the populations below is in genetic
equilibrium. Which one?
Population sample
Genotypes
Gene frequencies
AA
Aa
aa
A
a
100
20
80
0
0.6
0.4
100
36
48
16
0.6
0.4
100
50
20
30
0.6
0.4
100
60
0
40
0.6
0.4
© 2008 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN WEST AFRICA,
A BALANCED POLYMORPHISM
 haemoglobin gene
 Normal allele HbN
 Sickle allele HbS
Phenotypes
Normal
Sickle
Cell Trait
Sickle Cell
Anaemia
Genotypes
HbNHbN
HbN HbS
HbS HbS
Observed
frequencies
0.56
0.4
0.04
Expected
frequencies
© 2008 Paul Billiet ODWS
Alleles
HbN
HbS
SICKLE CELL ANAEMIA IN WEST AFRICA,
A BALANCED POLYMORPHISM
 haemoglobin gene
 Normal allele HbN
 Sickle allele HbS
Phenotypes
Alleles
Normal
Sickle
Cell Trait
Sickle Cell
Anaemia
Genotypes
HbNHbN
HbN HbS
HbS HbS
HbN
HbS
Observed
frequencies
0.56
0.4
0.04
0.76
0.24
Expected
frequencies
0.58
0.36
0.06
© 2008 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN THE
AMERICAN BLACK POPULATION
Normal
Sickle
Cell Trait
Sickle Cell
Anaemia
Alleles
Genotypes
HbNHbN
HbN HbS
HbS HbS
HbN HbS
Observed
frequencies
0.9075
0.09
0.0025
Phenotypes
Expected
frequencies
© 2008 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN THE
AMERICAN BLACK POPULATION
Normal
Sickle
Cell Trait
Sickle Cell
Anaemia
Alleles
Genotypes
HbNHbN
HbN HbS
HbS HbS
HbN HbS
Observed
frequencies
0.9075
0.09
0.0025
0.91 0.09
Expected
frequencies
0.8281
0.16
0.0081
Phenotypes
© 2008 Paul Billiet ODWS
RECESSIVE ALLELES
EXAMPLE ALBINISM IN THE BRITISH
POPULATION
Frequency of the albino phenotype
= 1 in 20 000 or 0.00005
© 2008 Paul Billiet ODWS
A = Normal skin pigmentation allele
Frequency = p
a = Albino (no pigment) allele
Frequency = q
Phenotypes
Genotypes
Hardy
Weinberg
frequencies
Normal
AA
p2
Normal
Aa
2pq
Albino
aa
q2
© 2008 Paul Billiet ODWS
Observed
frequencies
A = Normal skin pigmentation allele
Frequency = p
a = Albino (no pigment) allele
Frequency = q
Phenotypes
Genotypes
Hardy
Weinberg
frequencies
Normal
AA
p2
Normal
Aa
2pq
Albino
aa
q2
© 2008 Paul Billiet ODWS
Observed
frequencies
0.99995
0.00005
Albinism gene frequencies
Normal allele
Albino allele
© 2008 Paul Billiet ODWS
=A=p=?
=q=?
Albinism gene frequencies
Normal allele
Albino allele
© 2008 Paul Billiet ODWS
=A=p=?
= q =  (0.00005) = 0.007 or 7%
HOW MANY PEOPLE IN BRITAIN ARE
CARRIERS FOR THE ALBINO ALLELE
(Aa)?
a allele = 0.007
A allele
But
p+q
Therefore p
=q
=p
=1
= 1- q
= 1 – 0.007
= 0.993 or 99.3%
The frequency of heterozygotes (Aa) = 2pq
= 2 x 0.993 x 0.007
= 0.014 or 1.4%
© 2008 Paul Billiet ODWS
Heterozygotes for rare recessive
alleles can be quite common
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Genetic inbreeding leads to rare recessive
mutant alleles coming together more
frequently
Therefore outbreeding is better
Outbreeding leads to hybrid vigour
© 2008 Paul Billiet ODWS
Example: Rhesus blood group in
Europe
What is the probability of a woman who
knows she is rhesus negative (rhrh)
marrying a man who may put her child at
risk (rhesus incompatibility Rh– mother and
a Rh+ fetus)?
© 2008 Paul Billiet ODWS
Rhesus blood group
A rhesus positive foetus is possible if the father
is rhesus positive
RhRh x rhrh
 100% chance
Rhrh x rhrh
 50% chance
© 2008 Paul Billiet ODWS
Rhesus blood group
Rhesus positive allele is dominant Rh
Frequency = p
Rhesus negative allele is recessive rh
Frequency = q
Frequency of rh allele = 0.4 = q
If
p+q=1
Therefore Rh allele = p = 1 – q
= 1 – 0.4
= 0.6
© 2008 Paul Billiet ODWS
Rhesus blood group
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Frequency of the rhesus positive phenotype
= RhRh + Rhrh
= p2 + 2pq
= (0.6)2 + (2 x 0.6 x 0.4)
= 0.84 or 84%
© 2008 Paul Billiet ODWS
Rhesus blood group
Phenotypes
Genotypes
Hardy
Weinberg
frequencies
Rhesus positive
RhRh
p2
Observed
frequencies
0.84
Rhesus positive
Rhrh
2pq
Rhesus negative
Rhrh
q2
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0.16
Therefore, a rhesus negative, European woman in Europe has
an 84% chance of having husband who is rhesus positive…
of which 36% will only produce rhesus positive children and 48%
will produce rhesus positive child one birth in two
© 2008 Paul Billiet ODWS