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D4: The Hardy-Weinberg D4: The Hardy-Weinberg PrinciplePrinciple2 hours2 hours
D.4.1 Explain how the Hardy–D.4.1 Explain how the Hardy–Weinberg equation is derived.Weinberg equation is derived.
• TT = pTT = p22 = freq of homoz dom = freq of homoz dom• Tt = pq = freq of heterozygoteTt = pq = freq of heterozygote• tt = qtt = q22 = freq of homoz rec = freq of homoz rec
• Punnett Square... Tt x TtPunnett Square... Tt x Tt• pp22 + 2pq + q + 2pq + q22 = 100% (1) = 100% (1)
• p = freq of dominant allelep = freq of dominant allele• q = freq of recessive alleleq = freq of recessive allele
D.4.2 Calculate allele, genotype and D.4.2 Calculate allele, genotype and phenotype frequencies for two alleles of a phenotype frequencies for two alleles of a gene, using the Hardy–Weinberg equation.gene, using the Hardy–Weinberg equation.• Examples from class!Examples from class!
D.4.3 D.4.3 State the assumptions made when State the assumptions made when the Hardy–Weinberg equation is used. the Hardy–Weinberg equation is used. It must be assumed that:It must be assumed that:--a population is large --a population is large
--with random mating --with random mating
--constant allele frequency over --constant allele frequency over time time
This implies This implies
--no allele-specific mortality --no allele-specific mortality
--no mutation --no mutation
--no emigration --no emigration
--no immigration--no immigration
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
© 2008 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 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
© 2008 Paul Billiet ODWS
Google Earth
Example Icelandic population: The AB 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
BBABAAGenotypes
Type BType ABType APhenotypesSamplePopulation
© 2008 Paul Billiet ODWS
AB blood group in Iceland
Total A alleles = (2 x 233) + (1 x 385) = 851Total B alleles = (2 x 129) + (1 x 385) = 643Total of both alleles =1494
= 2 x 747 (humans are diploid organisms)p=Frequency of the A allele = 851/1494 = 0.57
or 57%q=Frequency of the B allele = 643/1494 = 0.43
or 43%
© 2008 Paul Billiet ODWS
In general for a diallellic gene A and aIf the frequency of the A allele = p
and the frequency of the a allele = q
Then p+q = 1
© 2008 Paul Billiet ODWS
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
BB 0.18AB 0.25
AB 0.25AA 0.32
B 0.43
A 0.57
B 0.43A 0.57
Assuming all the individuals mate randomly
SPERMS
EGGS
NOTE the ALLELE frequencies are the gamete frequencies too
© 2008 Paul Billiet ODWS
p*p= p2 p*q
p*q q*q= q2
Close enough for us to assume genetic equilibrium
Genotypes Expected frequencies
Observed frequencies
AA p2 = 0.32 233 747 = 0.31
AB 2pq =0.50 385 747 = 0.52
BB q2 =0.18 129 747 = 0.17
© 2008 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 aWhere the gene frequencies are p and qThen p + q = 1and
© 2008 Paul Billiet ODWS
THE HARDY WEINBERG EQUATIONSo the genotype frequencies are:
AA = p2
Aa = 2pq
aa = q2
or p2 + 2pq + q2 = 1
© 2008 Paul Billiet ODWS
DEMONSTRATING GENETIC EQUILIBRIUMUsing 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 Allele frequencies
AA Aa aa A a
100 20 80 0
100 36 48 16
100 50 20 30
100 60 0 40
© 2008 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
Gene frequenciesGenotypesPopulation sample
© 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
Alleles
Genotypes HbNHbN HbN HbS HbS HbS HbN HbS
Observed frequencies
0.56 0.4 0.04
Expected frequencies
© 2008 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
© 2008 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
© 2008 Paul Billiet ODWS
SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION
0.00810.160.8281
0.090.91
Expected frequencies
0.00250.090.9075Observed frequencies
HbSHbNHbS HbSHbN HbSHbNHbNGenotypes
AllelesSickle Cell Anaemia
Sickle Cell Trait
NormalPhenotypes
© 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
Phenotypes Genotypes Hardy Weinberg
frequencies
Observed frequencies
Normal AA p2
Normal Aa 2pq
Albino aa q2
A = Normal skin pigmentation alleleFrequency = p
a = Albino (no pigment) alleleFrequency = q
© 2008 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 alleleFrequency = p
a = Albino (no pigment) alleleFrequency = q
© 2008 Paul Billiet ODWS
Use p+q=1 to determine
Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q = ?
© 2008 Paul Billiet ODWS
Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q = ?
If q2 = 0.00005…then q = …
(0.00005) = 0.007 or 0.7%
© 2008 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%
© 2008 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
© 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 groupR = Rh+ r = Rh-
A rhesus positive fetus is possible if the father is rhesus positive
RR x rr 100% chance
Rr x rr 50% chance
© 2008 Paul Billiet ODWS
Rhesus blood group
Rhesus positive allele is dominant RFrequency = p
Rhesus negative allele is recessive rFrequency = q
Frequency of r allele = 0.4 = qIf p + q = 1Therefore R allele = p = 1 – q
= 1 – 0.4 = 0.6
© 2008 Paul Billiet ODWS
Rhesus blood group
Frequency of the rhesus positive phenotype = RR + Rr
= p2 + 2pq = (0.6)2 + (2 x 0.6 x 0.4) = 0.36 + 0.48 = 0.84 or 84%
© 2008 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 RR p2 0.36 (0.84 total)
0.48Rhesus positive Rr 2pq
Rhesus negative Rr q2 0.16
© 2008 Paul Billiet ODWS
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