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POPULATION GENETICS Predicting inheritance in a population © 2016 Paul Billiet ODWS

Powerpoint Presentation: Population Genetics

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Page 1: Powerpoint Presentation: Population Genetics

POPULATION GENETICS

Predicting inheritance in a population

© 2016 Paul Billiet ODWS

Page 2: Powerpoint Presentation: Population Genetics

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

Page 3: Powerpoint Presentation: Population Genetics

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

Page 4: Powerpoint Presentation: Population Genetics

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

Page 5: Powerpoint Presentation: Population Genetics

Iceland

Population 313 337 (2007 est)Area 103 000 km2

Distance from mainland Europe

970 km

© 2016 Paul Billiet ODWS

Page 6: Powerpoint Presentation: Population Genetics

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

Page 7: Powerpoint Presentation: Population Genetics

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

Page 8: Powerpoint Presentation: Population Genetics

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

Page 9: Powerpoint Presentation: Population Genetics

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

Page 10: Powerpoint Presentation: Population Genetics

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

Page 11: Powerpoint Presentation: Population Genetics

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

Page 12: Powerpoint Presentation: Population Genetics

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

Page 13: Powerpoint Presentation: Population Genetics

THE HARDY WEINBERG EQUATION

So the genotype frequencies are:

AA = p2

Aa = 2pqaa = q2

or p2 + 2pq + q2 = 1

© 2016 Paul Billiet ODWS

Page 14: Powerpoint Presentation: Population Genetics

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

Page 15: Powerpoint Presentation: Population Genetics

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

Page 16: Powerpoint Presentation: Population Genetics

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

Page 17: Powerpoint Presentation: Population Genetics

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

Page 18: Powerpoint Presentation: Population Genetics

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

Page 19: Powerpoint Presentation: Population Genetics

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

Page 20: Powerpoint Presentation: Population Genetics

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

Page 21: Powerpoint Presentation: Population Genetics

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

Page 22: Powerpoint Presentation: Population Genetics

RECESSIVE ALLELES

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

© 2016 Paul Billiet ODWS

Page 23: Powerpoint Presentation: Population Genetics

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

Page 24: Powerpoint Presentation: Population Genetics

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

Page 25: Powerpoint Presentation: Population Genetics

Albinism gene frequencies

Normal allele = A = p = ?Albino allele = q = ?

© 2016 Paul Billiet ODWS

Page 26: Powerpoint Presentation: Population Genetics

Albinism gene frequencies

Normal allele = A = p = ?Albino allele = q = (0.00005) = 0.007

or 0.7%

© 2016 Paul Billiet ODWS

Page 27: Powerpoint Presentation: Population Genetics

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

Page 28: Powerpoint Presentation: Population Genetics

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

Page 29: Powerpoint Presentation: Population Genetics

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

Page 30: Powerpoint Presentation: Population Genetics

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

Page 31: Powerpoint Presentation: Population Genetics

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

Page 32: Powerpoint Presentation: Population Genetics

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

Page 33: Powerpoint Presentation: Population Genetics

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