Populationgenetics

POPULATION GENETICS
Predicting inheritance in a population
© 2008 Paul Billiet ODWS

Predictable patterns of inheritance in a
population so long as…
 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

SUMMARY
 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

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)

Iceland
Population
 313 337 (2007 est)
Area
 103 000 km2
Distance from mainland
Europe
 970 km
Google Earth

Example Icelandic population: The
MN blood group
2 Mn
alleles
per
person
1 Mn
allele
per
person
1 Mm
allele
per
person
2 Mm
alleles
per
person
Contribution
to gene pool
129385233Numbers747
Mn
Mn
Mm
Mn
Mm
Mm
Genotypes
Type NType MNType MPhenotypesSample
Population

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%

In general for a diallellic gene A and a
(or Ax
and Ay
)
If the frequency of the A allele = p
and the frequency of the a allele = q
Then p+q = 1

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

Mn
Mn
0.18
Mm
Mn
0.25
Mm
Mn
0.25
Mm
Mm
0.32
Mn
0.43
Mm
0.57
Mn
0.43Mm
0.57
Assuming all the individuals mate
randomly
SPERMS
EGGS
NOTE the gene frequencies are the gamete frequencies too

Close enough for us to assume genetic
equilibrium
Genotypes Expected
frequencies
Observed
frequencies
Mm
Mm
0.32 233 ÷ 747 = 0.31
Mm
Mn
0.50 385 ÷ 747 = 0.52
Mn
Mn
0.18 129 ÷ 747 = 0.17

SPERMS
A p a q
EGGS
A p AA p2
Aa pq
a q Aa pq aa q2
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

THE HARDY WEINBERG EQUATION
So the genotype frequencies are:
AA = p2
Aa = 2pq
aa = q2
or p2
+ 2pq + q2
= 1

DEMONSTRATING GENETIC
EQUILIBRIUM
Using the Hardy Weinberg Equation to
determine the genotype frequencies from the
gene frequencies may seem a circular
argument

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
100 36 48 16
100 50 20 30
100 60 0 40

0.40.6
0.40.6
0.40.6
40060100
302050100
164836100
0.40.608020100
aAaaAaAA
Gene frequenciesGenotypesPopulation sample

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

SICKLE CELL ANAEMIA IN WEST AFRICA,
A BALANCED POLYMORPHISM
β haemoglobin gene
 Normal allele HbN
 Sickle allele HbS
Phenotypes Normal Sickle
Cell Trait
Sickle Cell
Anaemia
Alleles
Genotypes HbN
HbN
HbN
HbS
HbS
HbS
HbN
HbS
Observed
frequencies
0.56 0.4 0.04
Expected
frequencies

SICKLE CELL ANAEMIA IN WEST AFRICA,
A BALANCED POLYMORPHISM
β haemoglobin gene
 Normal allele HbN
 Sickle allele HbS
0.060.360.58
0.240.76
Expected
frequencies
0.040.40.56Observed
frequencies
HbS
HbN
HbS
HbS
HbN
HbS
HbN
HbN
Genotypes
AllelesSickle Cell
Anaemia
Sickle
Cell Trait
NormalPhenotypes

SICKLE CELL ANAEMIA IN THE
AMERICAN BLACK POPULATION
Phenotypes Normal Sickle
Cell Trait
Sickle Cell
Anaemia
Alleles
Genotypes HbN
HbN
HbN
HbS
HbS
HbS
HbN
HbS
Observed
frequencies
0.9075 0.09 0.0025
Expected
frequencies

SICKLE CELL ANAEMIA IN THE
AMERICAN BLACK POPULATION
0.00810.160.8281
0.090.91
Expected
frequencies
0.00250.090.9075Observed
frequencies
HbS
HbN
HbS
HbS
HbN
HbS
HbN
HbN
Genotypes
AllelesSickle Cell
Anaemia
Sickle
Cell Trait
NormalPhenotypes

RECESSIVE ALLELES
EXAMPLE ALBINISM IN THE BRITISH
POPULATION
Frequency of the albino phenotype
= 1 in 20 000 or 0.00005

Phenotypes Genotypes Hardy
Weinberg
frequencies
Observed
frequencies
Normal AA p2
Normal Aa 2pq
Albino aa q2
A = Normal skin pigmentation allele
Frequency = p
a = Albino (no pigment) allele
Frequency = q

Weinberg
frequencies
Observed
frequencies
Normal AA p2
0.99995
Normal Aa 2pq
Albino aa q2
0.00005
A = Normal skin pigmentation allele
Frequency = p
a = Albino (no pigment) allele
Frequency = q

Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q = ?

Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q = √ (0.00005) = 0.007 or 7%

HOW MANY PEOPLE IN BRITAIN ARE
CARRIERS FOR THE ALBINO ALLELE
(Aa)?
a allele = 0.007 = q
A allele = p
But p + q = 1
Therefore p = 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%

Heterozygotes for rare recessive
alleles can be quite common
 Genetic inbreeding leads to rare recessive
mutant alleles coming together more
frequently
 Therefore outbreeding is better
 Outbreeding leads to hybrid vigour

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)?

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

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

Rhesus blood group
 Frequency of the rhesus positive phenotype
= RhRh + Rhrh
 = p2
+ 2pq
 = (0.6)2
+ (2 x 0.6 x 0.4)
 = 0.84 or 84%

Rhesus blood group
 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
Weinberg
frequencies
Observed
frequencies
Rhesus positive RhRh p2
0.84
Rhesus positive Rhrh 2pq
Rhesus negative Rhrh q2
0.16

Populationgenetics

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