Powerpoint Presentation: Population Genetics
Transcript of Powerpoint Presentation: Population Genetics
POPULATION GENETICS
Predicting inheritance in a population
© 2016 Paul Billiet ODWS
Predictable patterns of inheritance in a population so long as… there is no genetic drift
The population is large enough not to show the effects of a random loss of alleles by chance events
the mutation rate at the locus of the gene being studied is not significantly high
mating between individuals is random no gene flow between neighbouring populations
New individuals are not gained by immigration or lost by emigration
the gene’s allele has no selective advantage or disadvantage (no natural selection).
© 2016 Paul Billiet ODWS
SUMMARY Genetic drift Mutation Mating choice Migration Natural selection
All can affect the transmission of genes from generation to generation
Genetic EquilibriumIf none of these factors is operating then the relative proportions of the alleles (the ALLELE FREQUENCIES) will be constant.
© 2016 Paul Billiet ODWS
THE HARDY WEINBERG PRINCIPLE Step 1 Calculating the allele frequencies from the
genotype frequencies Easily done for codominant alleles (each
genotype has a different phenotype).
© 2016 Paul Billiet ODWS
Iceland
Population 313 337 (2007 est)Area 103 000 km2
Distance from mainland Europe
970 km
© 2016 Paul Billiet ODWS
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
MnMnMmMnMmMmGenotypes
Type NType MNType MPhenotypesSamplePopulation
© 2016 Paul Billiet ODWS
MN blood group in IcelandTotal Mm alleles = (2 x 233) + (1 x 385) = 851Total Mn alleles = (2 x 129) + (1 x 385) = 643Total 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%
© 2016 Paul Billiet ODWS
In general for a diallellic gene A and a (or Ax and Ay) If the frequency of the A allele = pand the frequency of the a allele = q
Then p+q = 1
© 2016 Paul Billiet ODWS
Step 2
Using the calculated allele frequency to predict the EXPECTED genotypic frequencies in the NEXT generation
OR to verify that the PRESENT population is
in genetic equilibrium.
© 2016 Paul Billiet ODWS
MnMn 0.18
MmMn 0.25
MmMn 0.25
MmMm 0.32
Mn 0.43
Mm 0.57
Mn 0.43Mm 0.57
Assuming all the individuals mate randomly
SPERMS
EGGS
NOTE the allele frequencies are the gamete frequencies too
© 2016 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
© 2016 Paul Billiet ODWS
SPERMS
A p a q
EGGSA 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 allele frequencies are p and qThen p + q = 1and
© 2016 Paul Billiet ODWS
THE HARDY WEINBERG EQUATION
So the genotype frequencies are:
AA = p2
Aa = 2pqaa = q2
or p2 + 2pq + q2 = 1
© 2016 Paul Billiet ODWS
DEMONSTRATING GENETIC EQUILIBRIUMUsing the Hardy Weinberg Equation to determine the genotype frequencies from the allele frequencies may seem a circular argument.
© 2016 Paul Billiet ODWS
Only one of the populations below is in genetic equilibrium. Which one?
Population sample Genotypes Allele frequencies
AA Aa aa A a
100 20 80 0
100 36 48 16
100 50 20 30
100 60 0 40
© 2016 Paul Billiet ODWS
Only one of the populations below is in genetic equilibrium. Which one?
0.40.6
0.40.6
0.40.6
40060100
302050100
164836100
0.40.608020100
aAaaAaAA
Allele frequenciesGenotypesPopulation sample
© 2016 Paul Billiet ODWS
Only one of the populations below is in genetic equilibrium. Which one?
Population sample Genotypes Allele 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
© 2016 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
Alleles
Genotypes HbNHbN HbN HbS HbS HbS HbN HbS
Observed frequencies
0.56 0.4 0.04
Expected frequencies
© 2016 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM haemoglobin gene Normal allele HbN
Sickle allele HbS
0.060.360.58
0.240.76
Expected frequencies
0.040.40.56Observed frequencies
HbSHbNHbS HbSHbN HbSHbNHbNGenotypes
AllelesSickle Cell Anaemia
Sickle Cell Trait
NormalPhenotypes
© 2016 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION
Phenotypes Normal Sickle Cell Trait
Sickle Cell Anaemia
Alleles
Genotypes HbNHbN HbN HbS HbS HbS HbN HbS
Observed frequencies
0.9075 0.09 0.0025
Expected frequencies
© 2016 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION
0.00250.0950.9025
0.050.95
Expected frequencies
0.00250.090.9075Observed frequencies
HbSHbNHbS HbSHbN HbSHbNHbNGenotypes
AllelesSickle Cell Anaemia
Sickle Cell Trait
NormalPhenotypes
© 2016 Paul Billiet ODWS
RECESSIVE ALLELES
EXAMPLE ALBINISM IN THE BRITISH POPULATION Frequency of the albino phenotype = 1 in 20 000 or 0.00005
© 2016 Paul Billiet ODWS
Phenotypes Genotypes Hardy Weinberg
frequencies
Observed frequencies
Normal AA p2
Normal Aa 2pq
Albino aa q2
A = Normal skin pigmentation allele Frequency = pa = Albino (no pigment) allele Frequency = q
© 2016 Paul Billiet ODWS
Phenotypes Genotypes Hardy Weinberg
frequencies
Observed frequencies
Normal AA p2
0.99995Normal Aa 2pq
Albino aa q2 0.00005
A = Normal skin pigmentation allele Frequency = pa = Albino (no pigment) allele Frequency = q
© 2016 Paul Billiet ODWS
Albinism gene frequencies
Normal allele = A = p = ?Albino allele = q = ?
© 2016 Paul Billiet ODWS
Albinism gene frequencies
Normal allele = A = p = ?Albino allele = q = (0.00005) = 0.007
or 0.7%
© 2016 Paul Billiet ODWS
HOW MANY PEOPLE IN BRITAIN ARE CARRIERS FOR THE ALBINO ALLELE (Aa)?a allele = 0.007 = qA allele = pBut p + q = 1Therefore 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%
© 2016 Paul Billiet ODWS
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.
© 2016 Paul Billiet ODWS
Example: Rhesus blood group in EuropeWhat 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)?
© 2016 Paul Billiet ODWS
Rhesus blood group
A rhesus positive foetus is possible if the father is rhesus positive RhRh x rhrh 100% chanceRhrh x rhrh 50% chance
© 2016 Paul Billiet ODWS
Rhesus blood group
Rhesus positive allele is dominant RhFrequency = p
Rhesus negative allele is recessive rhFrequency = q
Frequency of rh allele = 0.4 = qIf p + q = 1Therefore Rh allele = p = 1 – q
= 1 – 0.4 = 0.6
© 2016 Paul Billiet ODWS
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%
© 2016 Paul Billiet ODWS
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.
Phenotypes Genotypes Hardy Weinberg
frequencies
Observed frequencies
Rhesus positive RhRh p2
0.84Rhesus positive Rhrh 2pq
Rhesus negative rhrh q2 0.16
© 2016 Paul Billiet ODWS