L16 Population Genetics-2
Transcript of L16 Population Genetics-2
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Population genetics
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Types of polymorphism
Restriction fragment length polymorphism (RFLP)
Single nucleotide polymorphism (SNP) 10 millions
Minisatellites/variable number tandem repeats (VNTR)
Microsatellites or short tandem repeats (STR)
Frequencies of polymorphic alleles may varysignificantly in different populations
Currently a new type of the genomic variation named CNV(copy number variation) is extensively studied.
CNV range from Kb to MbMost CNV are rare variants
Some CNV can be associated with the diseases.
Just to remind
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Your class of 100 people has been typed for a two-allele polymorphism (alleles labeled A and a), and the
following 3 genotype counts were obtained:Genotype Number of individuals
A,A 30A,a 50
a,a 20Total number of allele = 100 x 2 = 200
Number of allele A = (30 x 2)+ 50 = 110
Number of allele a= 50 + (20 x 2) = 90
Frequency of allele A = 110/200 = 0.55
Frequency of allele a = 90/200 = 0.45
Allele frequency ?
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Neutral polymorphism : SNP rs213934 A/G(HapMap project)
Japanese 0.434Caucasians 0.292African ancestry in SW USA 0.035Nigerian 0
Frequencies of polymorphic alleles mayvary significantly in different populations
Polymorphism associated with disease:
sickle cell mutation
African Americans 0.05Caucasians 0
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Hardy-Weinberg equilibrium/law:I. Genotypes distribution
Frequency of allele A = pFrequency of allele a = q
p+q = 1
1= p2
+ 2pq + q2
p2= frequency of homozygote AA
2pq = frequency of heterozygote Aa
q2 = frequency of homozygote aa
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Hardy-Weinberg Equilibrium:Single Nucleotide Polymorphism
SNP A/C
1 = p2
+ 2pq + q2
p2= frequency of homozygote AA= 0.72= 0.492pq = frequency of heterozygote AC = 2x0.7x0.3= 0.42
q2 = frequency of homozygote CC = 0.32= 0.09
p = 0.7 (frequency of allele A)q = 0.3 (frequency of allele C)
Expected distribution of genotypes in 100 individualsAA= 0.49 x 100 = 49 individualsAC = 0.42 x 100 = 42 individualsCC = 0.09 x 100 = 9 individuals
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Single nucleotide polymorphism SNP17361 G/T:
the frequency of a minor allele T is 0.4. What is
the frequency of heterozygotes G/T?
A. 18%
B. 28%
C. 38%
D. 48%
E. 58%
Q
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Single nucleotide polymorphism SNP17361 G/T:
the frequency of a minor allele T is 0.4. What isthe frequency of heterozygotes G/T?
A. 18%
B. 28%C. 38%
D. 48%
E. 58%
q = 0.4p=1 - 0.4 = 0.6
2pq = 2 x 0.6 x 0.4 = 0.48 (48%)
A
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Hardy-Weinberg equilibrium inMendelian diseases
1 = p2
+ 2pq + q2
p2= frequency of normal homozygote NN
2pq = frequency of heterozygote NM
q
2
= frequency of mutant homozygote MM
Frequency of normal allele N = pFrequency of mutant allele M = q
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Hardy-Weinberg equilibriumAutosomal dominant disease
1 = p2+ 2pq + q2p2= frequency ofhealthy (normal homozygote NN)
2pq = frequency of affected(heterozygote NM)
q2 = frequency of affected(mutant homozygote MM)
Frequency of normal allele N = pFrequency of mutant allele M = q
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Hardy-Weinberg equilibriumAutosomal recessive disease
Frequency of normal allele N = pFrequency of mutant allele M = q
1 = p2+ 2pq + q2
p2= frequency ofhealthy (normal homozygote NN)
2pq = frequency of healthy(heterozygote NM)
q2 = frequency of affected(mutant homozygote MM)
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Hardy-Weinberg equilibriumX-linked recessive disease
Frequency of normal allele N = pFrequency of mutant allele M = q
Males:p = frequency of healthy(normal hemizygote N)
q = frequency of affected (mutant hemizygote M)
Females:1 = p2+ 2pq + q2
p2
= frequency ofhealthy (normal homozygote NN)
2pq = frequency of healthy(heterozygote NM)
q2 = frequency of affected(mutant homozygote MM)
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Most typical exam questions
q is given
Find carrier frequency
Find disease frequency
Disease frequency (incidence) is given
Find disease allele frequency
Find carrier frequency
Carrier frequency is givenFind disease allele frequency
Find disease frequency
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Estimation of carrier frequency forautosomal recessive disease
It is difficult and often impossible to identifycarrier frequency directly
Hardy-Weinberg equilibrium allows for estimationof carrier frequency from disease incidence
Incidence of phenylketonuria (PKU) in Caucasians :
1 in 10,000Mutant allele frequency?
Carrier frequency ?
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Frequency of MM = q2 = 1/10,000q= square root of 1/10000= 1/10000 = 1/100p= 1 1/100 = 0. 99 1 (approximation can beused because the disease/mutation is rare)
Carrier frequency = 2pq = 2 x1 x 1/100 = 1/50
p2+ 2pq + q2p2= frequency of homozygote NN
2pq = frequency of heterozygote NM
q2 = frequency of homozygote MM
Incidence of phenylketonuria (PKU) in Caucasians :1 in 10,000
Mutant allele frequency
Carrier frequency ?p= frequency of normal alleleq = frequency of mutant allele
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PKU incidence in China : 1 in 100,000
Carrier frequency ?
optional
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PKU incidence in China : 1 in 100,000
Frequency of MM = q2 = 1/100,000q=1/100000 = 1/316p= 1 1/316 = 0. 999 1Carrier frequency = 2pq = 2 x1 x 1/316 = 1/158
p2+ 2pq + q2
p2= frequency of homozygote NN2pq = frequency of heterozygote NM
q2 = frequency of homozygote MM
p= frequency of normal alleleq = frequency of mutant allele
Mutant allele frequency?Carrier frequency ?
optional
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Frequencies of mutant PKU allele andPKU carriers are different in Caucasians
and Chinese
PKU incidence in Caucasians : 1 in 10,000q= 1/100
Carrier frequency = 1/50
PKU incidence in China : 1 in 100,000q= 1/316
Carrier frequency = 1/158
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From carrier frequency to
disease and allelefrequency
A biochemical or enzymatic carriertesting is available just for a limited
number of genetic disorders, particularlyfor Tay-Sachs disease
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Tay-Sachs disease is an autosomal recessive
disorder for which the carrier frequency inindividuals of Ashkenazi Jewish descent is
about 1 in 30.
How frequent is Tay-Sachs disease in this
population?
1 = p2+ 2pq + q2
p2= frequency ofhealthy (normal homozygote NN)
2pq = frequency of healthy(heterozygote NM)
q2
= frequency of affected(mutant homozygote MM)
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Tay-Sachs disease is an autosomal recessivedisorder for which the carrier frequency inindividuals of Ashkenazi Jewish descent is about
1 in 30.How frequent is Tay-Sachs disease in this
population?
2pq = 1/30In case of rare genetic disease the frequency ofmutant allele (q) is small, thus p 12pq 2q = 1/30
q = 1/60The frequency of disease = q2 =( 1/60)2 = 1/3600
Answer: 1 in 3600
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Hardy-Weinberg equilibriumX-linked diseases
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Because males have only a single X chromosome, each
affected male has one copy of the disease-causing
recessive mutation. Thus, the incidence of an X-linked
disease in males is a direct estimate of the gene
frequency in the population.
X-linked diseases
Disease frequency in males = q
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X-linked recessive:
Males:q affectedp - unaffected
Females:q2affected2pq - carrierp2- unaffected
X-linked dominant:
Males:q affectedp - unaffected
Females:
q2affected2pq - affectedp2 - unaffected
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Color blindness
8% Caucasian males have color blindnessq= ?
p = ?Females
Normal non-carrier = ?
Normal carrier = ?
Color-blindness = ?
X-linked recessive
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X-linked recessive
Color blindness
Males
q= 0.08p = 0.92
8% Caucasian males have color blindness
Females
Normal non-carrier = p2= 0.922=0.846
Normal carrier = 2pq = 2 x 0.92 x 0.08 = 0.1472
Color-blindness = q
2
= 0.08
2
=0.0064
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G6PD deficiency in 70 of 1080 males (70/1080= 0.065)q =p=
FemalesNormal non-carrier =Carrier =
G6PD deficiency
Blood Cells Mol Dis, 2003, 31 (2) , 201-205
optional
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G6PD deficiency in 70 of 1080 males (70/1080= 0.065)q = 0.065
p= 1- 0.065= 0.935
FemalesNormal non-carrier = p2=0.9352 =0.874 (87.4%)Carrier = 2pq = 2 x 0.935 x 0.065 = 0.121 (12,1%)G6PD deficiency = q2= 0.0652= 0.0042 (0.4%)
Blood Cells Mol Dis, 2003, 31 (2) , 201-205
optional
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Hardy-Weinberg equilibrium
Autosomal dominant disease
1 = p2+ 2pq + q2
p2= frequency ofhealthy (normal homozygote NN)
2pq = frequency of affected(heterozygote NM)
q2 = frequency of affected(mutant homozygote MM)
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Autosomal dominant diseases
1= p2+ 2pq + q2
Most dominant diseases are rare, it means q is smalland an approximation p 1 can be usedMost patients with AD disease are heterozygotesDisease frequency = 2pq + q2 2pq 2q
q= disease frequency/2
Familial hypercholesterolemia in USA : 1/500 or 0.002
Therefore q= 0.001Frequency of homozygotes = q2= (0.001)2= 0.000 0011 in a million
optional
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Hardy-Weinberg Law:II. The population genotype frequencies from
generation to generation will remain constant, if
allele frequency remain constantIf you want to know why it is true see the
following table
optional
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Large population
Random mating
No migration
Assumptions underlying the
Hardy-Weinberg law
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Factors that disturb Hardy-Weinberg equilibrium
Non-random matingStratificationAssortative mating
Consanguinity/inbreedingGenetic drift in small population
Founder effect
Mutation selection
Migration and gene flow
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Non-random mating
Assortative mating
Consanguinity/inbreeding
Stratification
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Non-random mating: Assortative marriage
The choice of the mate who possesses
some particular trait
Congenital deafness
Congenital blindness
Achondroplasia
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Non-random mating: Consanguinity
Marriage between relatives
Non-random mating: Inbreeding
In genetic isolates the chance of mating withanother carrier may be as high as in consanguineous
marriages
Geographically isolated populations
Religious groups (Old Order Amish, Mennonites etc)
Slide 1
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Non-random mating: Stratification
US population: many groups(strata)WhitesAfrican AmericansNative AmericansAsian AmericansHispanics
African Americas 10% of US populationFrequency of SC mutation in African Americans qAA=0.05Frequency of SC mutation in other group is practically 0Frequency of SC mutation in US population qtotal= 0.05/10 =0.005
(optional)
Sickle cell (SC) mutation in the USA
Stratification Slid 2
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Stratification
Frequency of the SC disease expected from Hardy-Weinberg equilibrium in US population:
qtotal2= 0.0052= 0.000025
However the mating is not random - most marriages areinside the group (stratum) and the real estimate of SCdisease in African Americans is
qAA2= 0.052 = 0.0025
Because African Americans are 10% of the US population,
then the real frequency of the disease in the USpopulation is0.0025/10= 0.00025
This is 10 times more than expected from Hardy-Weinberg equilibrium
African Americas(AA) 10% of US populationFrequency of sickle cell (SC) mutation in AA = qAA=0.05Frequency of SC mutation in US population qtotal= 0.05/10 = 0.005
Slide 2(optional)
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Factors that disturb Hardy-Weinberg equilibrium
Non-random matingStratificationAssortative mating
Consanguinity/inbreedingGenetic drift in small population
Founder effect
Mutation selection
Migration and gene flow
Just to remindyou where weare
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Increased fertility or survival of mutationcarriers for reason
unrelated to carrying the mutation
leads to a
change in allele frequency
Genetic drift in small populations -chance event
In small populations allele frequencies canfluctuate form generation to generation by
chance (genetic drift)
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The founder effect is the loss of genetic
variation that occurs when a new population is
established by a very small number of individuals
from a larger population
Founder effect
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Mutation selection
Positive selection ( increased fertilityor survival because of mutation)
Negative selection ( decreased fertilityor survival because of mutation)
Eff t f t ti l ti i diff t i
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PKU incidence in Caucasians : q2= 1/10,000
Carrier frequency = 2pq = 2 x1 x 1/100 = 1/50
Dominant disease:all mutant alleles are under negative selective
pressure
Autosomal recessive diseases:selection has less effect
because most mutant alleles are in heterozygotesand therefore escape negative selection
Number of mutant PKU alleles in carriers 1/50----------------------------= ---------- = 100Number of mutant PKU alleles in affected 2/10,000
Effect of mutation selection is different indominant and recessive diseases
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Positive selection for heterozygotes(Heterozygote advantage)
Resistance to malariaSickle cell anemia/Thalassemia/G6PD deficiency
Resistance to effect of chloride-secreting diarrhea
Cystic fibrosis
Resistance to some infectionsTay-Sachs disease
fl
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Migration and gene flow
Migration can change allele frequency by gene flow slow diffusion of genes across a barriers
Map of Ancient human migration
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Ethnic differences in the frequency of geneticdiseases may be due to
Genetic drift
Founder effect
Positive selection for heterozygotes(Heterozygote advantage)
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Tay-Sachs disease is common in
Ashkenazi Jews because of:
Founder effect
Inbreeding
Heterozygote advantage
Ashkenazi Jews (those originating from theWestern and Eastern Europe diaspora), who make upmore than 80 percent of world Jews and are
believed to be descended from about 1,500 Jewishfamilies dating back to the 14th century
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The end