POPULATION GENETICS Predicting inheritance in a population © 2008 Paul Billiet ODWSODWS.

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 emigrationi.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 EquilibriumIf 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

MnMnMmMnMmMmGenotypes

Type NType MNType MPhenotypesSamplePopulation


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%


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


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 gene frequencies are the gamete frequencies too


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


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 gene frequencies are p and qThen p + q = 1and


THE HARDY WEINBERG EQUATION

So the genotype frequencies are:

AA = p2

Aa = 2pq

aa = q2

or p2 + 2pq + q2 = 1


DEMONSTRATING GENETIC EQUILIBRIUMUsing 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 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


0.56 0.4 0.04

Expected frequencies


SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED

POLYMORPHISM haemoglobin gene Normal allele HbN

Sickle allele HbS

0.060.360.58

0.240.76


0.040.40.56Observed frequencies

HbSHbNHbS HbSHbN HbSHbNHbNGenotypes

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 HbNHbN HbN HbS HbS HbS HbN HbS


0.9075 0.09 0.0025



SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION

0.00810.160.8281

0.090.91


0.00250.090.9075Observed frequencies

HbSHbNHbS HbSHbN HbSHbNHbNGenotypes

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


Normal AA p2

Normal Aa 2pq

Albino aa q2

A = Normal skin pigmentation alleleFrequency = p

a = Albino (no pigment) alleleFrequency = q



frequencies


Normal AA p2

0.99995Normal Aa 2pq

Albino aa q2 0.00005

A = Normal skin pigmentation alleleFrequency = p

a = Albino (no pigment) alleleFrequency = 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 = 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%


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 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


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


frequencies


Rhesus positive RhRh p2

0.84Rhesus positive Rhrh 2pq

Rhesus negative Rhrh q2 0.16


POPULATION GENETICS Predicting inheritance in a population © 2008 Paul Billiet ODWSODWS.

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