population genetics2.pdf
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Human and PopulationGenetics
M. K. Tadjudin
Fakultas Kedokteran dan Ilmu Kesehatan
UIN Syarif Hidayatullah
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Gene A discrete particle of inheritancecontrolling a trait (pea shape or peaappearance)
The segment of DNA involved inproducing polypetide chain (cistron). Itincludes regions preceding andfollowing the coding region (leader andtrailer) as well as intervening sequences(introns) between individual codingsegments (exons).
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Allele = An alternative form of agene (R or r)
Phenotype = a trait exhibited byan allele that distinguishes oneindividual from another (round vs.
wrinkled) Haplotype = particular
combination of alleles in a defined
region of a chromosome
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P= G+ E
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Polymorphism
Simultaneous occurrence in thepopulation of genomes showing
variations at a given population Polymorphism is present if the
frequency of an allele is > 1 %
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Multiple alleles Multiple alleles are different forms
of the same gene/locus
The sequence of the bases isslightly different in the geneslocated on the same place of the
chromosome
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Polygenic traits
The result of the interaction
of several genes
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Gametes ABC ABc AbC Abc aBC aBc abC abc
ABC 6 5 5 4 5 4 4 3
ABc 5 4 4 3 4 3 3 2
AbC 5 4 4 3 4 3 3 2
Abc 4 3 3 2 3 2 2 1
aBC 5 4 4 3 4 3 3 2
aBc 4 3 3 2 3 2 2 1
abC
4
3
3
2
3
2
2
1
abc 3 2 2 1 2 1 1 0
Polygenic inheritance
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Pleiotropy The effect of a single gene
on more than onecharacteristic/effects
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Disease risk
Prevention ofinherited traits:Decision not toreproduce
Fetal destructionAbortion
Environmentalexposure
Genetic mutationsand polymorphisms
Reduction in
mutagens
Mutagens
Geneticrisks
Gene
therapy
To preventsecond hit
Environmentalrisk
Radiation andenvironmentalexposure:BiologicalPhysicalSocioeconomic
Behavioral changes:SmokingDietExerciseDrugs & alcohol
Other measures:Chemoprevention
Early detectionPopulationscreeningRemoval of targetorgans
Disease
Inheritance
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P= G+ E
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Definitions (1)Population = all individuals of a
speciesMendelian Population = all
individuals of a species which can
interbreed within a definedgeographical boundary
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Definitions (2)Gene pool/gamete pool =
hypothetical mixture of geneticunits from which the nextgeneration will arise
Gene frequency= proportion of anallele in a population
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PopulationInterbreeding groups oforganisms that are usuallysubdivided into partiallyisolated breeding groups
called demesor localpopulation
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Study of human inheritance
Inborn characteristics of humanbeings
Differences between human and non-human beings
Characteristics of certain human
populations Man is not an ideal species for the
study of genetics
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22Your ancestors
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Eugenics Eu =well; gen = grow;
eugenes = well born Primary aim = to produce a
race of physically perfecthuman beings
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Positive eugenics
Attempts to increaseconsistently bettergermplasm and thuspreserve the best
germplasm of the society
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Early marriage for those having desirabletraits
Subsidizing the fit Gamete banks
Avoiding germinal wastage
Improvement of environmental condition Genetic counseling
Promotion of genetic research
Positive eugenicsmeasures
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Sexual separation of the defective
Sterilization Control of immigration
Regulation of marriage
Negative eugenicsmeasures
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A Nazi-era high-schoolbiology book warns
that "a hereditarily illperson costs 50,000reichsmarks onaverage up to the age
of sixty." From theU.S. HolocaustMemorial Museum,Washington, D.C.
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The Nazi regime awardedbronze medals to "fit"Germanic women who hadfour or five children, silvermedals to those who had
six or seven, and goldmedals to those with eightor more. From the U.S.Holocaust Memorial
Museum, Washington, D.C.
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EuphenicsSymptomatic treatment ofgenetic diseases:
Prevention
Replacement therapy Research in human genetics
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Mendelian population
Attributes:
Gene frequenciesGene pool
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Phenotype frequency
Genotype frequency
Gene frequency
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Gene flowNew organisms entering apopulation by migration andmating in the newpopulation, thus bringing
new alleles to the localgene pool
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Genetic polymorphism
In a polymorphic locus thefrequency of a second allele >
1 %
39 % of loci are polymorphic
12 % of loci are heterozygous Polymorphism is maintained by
heterozygous advantage
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Hardy-Weinberg equilibrium
1. The allelic frequencies at an autosomal locusin a population will not change from onegeneration to the next (allelic frequencyequilibrium)
2. The genotypic frequencies of the populationare determined in a predictive way by theallelic frequencies (genotypic frequencyequilibrium)
3. The equilibrium if perturbed will bereestablished within one generation ofrandom mating at the new allelic frequencies
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Assumptions of the Hardy-
Weinberg equilibrium1. Random mating
2. Large and constant population size
3. No difference in fitness between thedifferent genotypes
4. No mutation5. No migration
6. No natural selection/selection
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Genotype TTFreq.pxp =p2
Gamete TFreq.p
Gamete TFreq.p
Gamete tFreq. q
Gamete tFreq. q
Genotype TtFreq.px q =pq
Genotype TtFreq.px q =pq
Genotype ttFreq. qx q = q2
Genotype frequencies of TT, Tt, and tt
TT =p2Tt = pq+ pq= 2pqtt = q2
M ti f i
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Mating frequencies
GenotypeFrequencies
TTp2
ttq2
Tt2pq
Fema
les
TTp2
p4
2p3q
p2q2
Tt2pq
2p3q
4p2q2
2pq3
ttq2
p2q2
2pq3
q4
M a l e s
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TT X TT p2
X p2
=p4 p4 - -
TT X Tt(2)
2(p2X 2pq) =4p3q
2p3q 2p3q -
Matingtypes
Matingfrequencies TT Tt tt
TT X tt
(2)
2(p2X q2) =
2p
2
q
2
- 2p2q2 -
Tt X Tt2pq X 2pq =
4p2q2p2q2 2p2q2 p2q2
Tt X tt(2)
2(2pq X q2) =4pq3
- 2pq3 2pq3
tt X tt q2X q2=q4
- - q4
Frequency of offspring of different mating types
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p4 + 4p3q + 6p2q2 + 4pq3+ q4=
p2(p2 + 2pq + q2)+ 2pq(p2 + 2pq + q2)
+ q2(p2 + 2pq + q2) =
p2 + 2pq + q2= (p + q)2= 1
Total number of matings
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TT:p4 + 2p3q +p2q2=p2(p2 + 2pq + q2) =p2 (1) =p2
Total frequency of offspring
Tt: 2p3q + 4p2q2+ 2pq3= 2pq(p2 + 2pq + q2) =2pq(1) = 2pq
tt:p2q2+ 2pq3+ q4= q2(p2 + 2pq + q2) = q2 (1) = q2
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Contributions of genes to
the next generation (1)Parent population: [AA = 36] - [Aa = 39] - [aa = 25]Total = 100
Number of genes in parent population:A = (2 X 36) + (1 X 39) = 111a= (1 X 39) + (2 X 25) = 89 = 0.45 = qTotal number of genes = 200
Gene frequencies in parent population:A = 111/200 =0.55 = pa = 89/200 = 0.45 = q
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Contributions of genes to
the next generation (2)Genotype frequencies in the nextgeneration:
AA = (0.55)2 = 0.3025
Aa = 2 X 0.55 X 0.45 = 0.495
aa = (0.45)2= 0.2025
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H2=4DR
(2pq)2=4p2q2
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D = 10 = AA
H = 80 = Aa
R = 10 = aa
H2= 0.64
4 DR = 4 X 0.1 X 0.1 = 0.0449
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D = AA = 10[A] = (2 X 10) + (1 X 80) = 100
H = Aa = 80[a] = (2 X 10) + 1 X 80 = 100
R = aa = 10
Frekwensi gen [A] = 100/200 = 0.5 = p
Frekwensi gen [a] = 100/200 = 0.5= qFrekwensi genotip [AA] = 0.52= 0.25 = p2Frekwensi genotip [Aa] = 2 X 0.5X 0.5 = 0.50 = 2pqFrekwensi genotip [aa] = 0.52= 0.25 = q2
50
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H- W Equilibrium in
X-linked genes (1) In equilibrium gene frequencies in males and
females are the same
In a population: Males have 1/3 of all X-chromosomes while females have 2/3
Gene frequency in males is the same as thegene frequency in the mothers generation
Gene frequency in females =
2
mofa pp
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H- W Equilibrium in
X-linked genes (2)Male genotypes:
XHY = p
Xh
Y = q
Female genotypes:
XHXH = p2
XHXh = 2pq
XhXh = q2
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H- W Equilibrium in
X-linked genes (3)If the number of male = number of females,
then the gene frequency is:
p= 1/3 (pmales) + 2/3 (pfemales) =
3)2( femalesmales pp
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Males 0.2 0.5 0.35 0.425 0.3875 .40625 0.39675
Females 0.5 0.35 0.425 0.3875 0.40625 0.39675 0.401562
0 1 2 3 4 5 6
Generations
Zigzag change in gene frequenciesbetween males and females
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XHY p p3 2p2q pq2
XhY q p2q 2pq3 q3
XHXH XHXh XhXh
p2 2pq q2
Mothers genotype
Fathers
genotype
Freq
Mating frequencies of X-linked genes
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Estimation of equilibriumin dominance
Heterozygotes cannot bedistinguished from dominants
When recessive phenotypes arerare carrier heterozygotes arepresent in relatively highfrequency
Snyders ratio to check for H-Wequilibrium
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p2AA p4 2p3q p2q2
2pqAa 2p3q 4p2q2 2pq3
q2aa p2q2 2pq3 q4
p2AA 2pqAa q2aa
Dominant Recessive
Dominant
Recessive
Male
Parent
Female Parent
Snyders dominant ratio (1)
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Snyders dominant ratio (2)
Proportion of recessive offspring indominant X dominant matings:
p2q2
p4+ 4p3q + 4p2q2
=p2q2
p2(p2+ 4pq + 4q2)
=q2
p2+ 4pq + 4q2
=q2
[(p + q) + q]2
q2
(1 + q)2
=
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p2AA p4 2p3q p2q2
2pqAa 2p3q 4p2q2 2pq3
q2aa p2q2 2pq3 q4
p2AA 2pqAa q2aa
Dominant Recessive
Dominant
Recessive
Male
Parent
Female Parent
Snyders recessive ratio (1)
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Snyders recessive ratio (2)
Proportion of recessive offspring indominant X recessive matings:
2pq3
2p2q2+ 4pq3
=2pq3
2pq2(p + 2pq3)
= =
=
q
p + 2q
q
p + q + q
q
1 + q
Snyders experiment
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Freq. of Non-tasters among
offspring
ParentsOffspring
MatingNo.
CouplesTasters
NonTasters
Total Obs. Exp.
TasterX
Taster
425 929 130 1069 0.123 0.122
TasterX Non-Taster
289 483 278 761 0.365 0.349
Non-
TasterX Non-Taster
86 (5) 218 223
Total 800 1417 626 2043
SDR
SRR
PTC = PTU
Snyder s experiment
Comparison of observed and expected
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Exp. SDR = = 0.5372/1.5372= 0.122
The frequency of the non-tasting allele is derived fromthe frequency of homozyogous recessives among theparents or [2 X number of recessive homozygotes(2R)]+ [(number of heterozygotes(1H)] among (2 X number
of parents) = (2R + 1H)/2N = q2
q =
Comparison of observed and expectedfrequencies of recessive phenotypes
2R + 1H
2Nq = 461/1600 = 0.288 = 0.537
Exp. SRR = = 0.537/1.537 = 0.349q
1 + q
q2
(1 + q)2
TUGAS
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Freq. of Non-tasters among
offspring
ParentsOffspring
MatingNo.
CouplesTasters
NonTasters
Total Obs. Exp.
TasterX
Taster
525 950 130 1080 0.12 0.12
TasterX Non-Taster
389 450 278 728 0.382 0.346
Non-
TasterX Non-Taster
86 0 218 218
Total 1000 1400 626 2026
SDR
SRR
PTC = PTU
TUGAS
H W ilib i i
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H-W equilibrium in
multiple alleles (1) The equilibrium condition is
described by the multinomial
expansion (p + q + r + )2 If all the alleles are co-dominant
each genotype has its own distinct
phenotype and genotypicfrequencies can be easily scored
A = p A = q A = r
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A1 = p A2= q A3= rA1A1 A1A2 A1A3
A2A2 A2A3A3A3
p = (2 A1A1+ A1A2 + A1A3)/2N
q = (2 A2A2+ A1A2 + A2A3)/2N
r = (2 A3A3+ A1A3 + A2A3)/2N
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Human blood groups
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Genotype AA AO AB BB BO OO
Frequency p2
2pr 2pq q2
2qr r2
Phenotypes
Human blood groups
In equilibrium frequency of B + O phenotypes =
q2+ 2qr + r2 = (q + r)2
As (p + q + r) = 1 1 (q + r) = p
So: 1 - (q + r)2= p - or: 1 - (B+O) phenotypes/N= p
Similarly 1 - (A+O) phenotypes/N= q
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Genotype AA AO AB BB BO OO
Frequency p2 2pr 2pq q2 2qr r2
280 13 19 288
In equilibrium frequency of B + O phenotypes =
q2+ 2qr + r2 = (q + r)2
As (p + q + r) = 1 1 (q + r) = p
So: 1 - (q + r)2= p - or: 1 - (B+O) phenotypes/N= p
Similarly 1 - (A+O) phenotypes/N= q
Comparison of observed and expected frequencies
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Observednumber 63 31 6 92 192
Phenotype
frequencies 0.3281 0.1615 0.0312 0.4792 1.000
Equilibriumfrequencies
(p2+2pr)N (q2+2qr)N (2pq)N (r2)N N
Expectednumber 61 29 8 94 192
2
(with Yatescorrection)
0.037 0.078 0.282 0.024 0.421
A B AB O Total
Phenotypes
Comparison of observed and expected frequencies
Calculation of gene frequencies with Bernsteins correction
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Phenoty-pes
Numberobserved
Frequency SQRT offrequency
Estimate Correctionformula
Correctedvalues
B + O 123 0.6407 0.8004p =
0.1996p(1+1/2d)
p =0.2003
A + O 155 0.8703 0.8985 q =0.1015
q(1+1/2d) q =0.1018
O 92 0.4792 0.6923r =
0.6923(r+1/2d) X(1+1/2d)
r =0.6979
370 0.9934 d =0.0066
1.000
Bernsteins correction factor: d = 1 0.9934
Calculation of gene frequencies with Bernstein s correction
TUGAS
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Observednumber 263 231 106 400 1000
Phenotypefrequencies 0.263 0.231
0.106=2pq
0.4 = r2
r=0.632 1
Equilibriumfrequencies 0.632
Expectednumber 530 146 64 5652
(with Yatescorrection)
A B AB O Total
Phenotypes
TUGAS
TUGAS
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Phenoty-pes
Numberobserved
FrequencySQRT of
frequencyEstimate
Correctionformula
Correctedvalues
B + O231 +
400=631
0.631=
(q+r)2
0.7944=
(q+r)
p= 1-(q+r)=0.2056
p(1+1/2d)=
0.2056-0.0837p = 0.1219
A + O263 +
400=663
0.663=
(p+r)20.8142
q=1-(p+r)=0.1858
q(1+1/2d)=0.1858-0.0837
q = 0.1021
O 400 0.4 0.6325 r = 0.6923
(r+1/2d) X(1+1/2d)=
(0.6923-0.0837)x(1-0.0837)
r = 0.5577
Yatescorrecti
on
1000 1.0837d=
-0.0837
0.7816
TUGAS
A ti f th H d
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Assumptions of the Hardy-Weinberg equilibrium
1. Random mating
2. Large and constant population size
3. No difference in fitness between thedifferent genotypes
4. No mutation
5. No migration
6. No natural selection/selection
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Changes in gene frequencies
Mutation
Selection:
Gametic selection Zygotic selection
Artificial selection
Natural selection
Migration
M t tio
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Mutation The mere appearance of new genes
is no guarantee they will persist
A newly mutated gene has a verysmall chance of survival
Size of offspring and mutationrates influence chances for survival
of a newly mutated gene
Probability of elimination of mutant genes (1)
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# ofoffsping/fam 0 1 2 3 4 5
Freq of fam e-2 2e-2 (22/2! )e-2 (23/3! )e-2 (24/4! )e-2 (25/5! )e-2
Prob of elim 1 1/2 ()2 ()3 ()4 ()5
Freq of familyX prob of elim
e-2 e-2 (1/2!)e-2 (1/3!)e-2 (1/4!)e-2 (1/5!)e-2
Tot prob ofelim in 1stgen
e-2
(1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ) =e-2(e) = e-1= 1/2.718 = 0.3679
y g ( )
e = natural number = 2.303
Probability of elimination of mutant genes (2)
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y g ( )
Probability of elimination in the 2ndgeneration =
e-(1-0.3679)= e-0.6321= 0.5315
Probability of elimination in the 3rd generation =
e-(1-0.5315)= e-0.4685= 0.6159
According to Fisher chances of a mutant genesurviving after n generations = 2/n
The persistence and increase of many mutantsis due to recurrent mutations or the frequencyof mutations.
Mutation rate (1)
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Mutation of A a - Forward mutationForward mutation rate = u
Mutation of a A - Backward mutation/Reverse mutation
Reverse mutation rate = v
Mutation rate (1)
Mutation rate (2)
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In equilibrium:Forward mutation = Reverse mutationup = vq
Since p = 1q up = u(1q)So: u(1q) = vq
uuq = vq
u = uq + vq = q(u + v)
q =
Mutation rate (2)
u
u + v
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Selection (1) Mechanisms for modifying thereproductive success of a genotype
Artificial selection
Natural selection
Gametic selection Zygotic selection
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Selection (2) FITNESSRelative reproductivesuccessAdaptive value -Selective value
SELECTION COEFFICIENT (s):The force acting on eachgenotype to reduce its adaptivevalue
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Gametic selection (1)
Selection at the gamete levelAs gametes are haploids the
gene frequency at the gametelevel = the gamete genotypefrequency
The rules for gametic selectionare also valid for haploidorganisms
G ti l ti (2)
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Gametic selection (2)
Initial freq po q
o 1
Fitness 1 1s
After
selection p q(1 s)
p + q sq
= 1 - sq
Gene freqafter sel.
p/(1 sq) q(1 s)
(1 sq)
A a Total
Gamete genotypes
G ti l ti (3)
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Change in gene frequency = q = q1 qo
= - q = - q
=
= =
Gametic selection (3)
q(1 s)
1 sq
(q sq)
1 sq
q sq q + sq 2
1 sq
sq + sq2
1 sq
sq(1 q)q1 sq
G ti l ti (4)
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Gametic selection (4)If s is small sq in the denominator can be ignoredsq= 0 and the denominator becomes = 1
The relationship for n generations calculated by
calculus is then:
sn = loge = 2.303 log10
2.303 log10
n =
q0(1 qn)
qn(1 q0)
q0(1 qn)
qn(1 q0)
q0(1 qn)
qn(1 q0)
s
G ti l ti (5)
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Gametic selection (5)If q0= 0.25 and s = 0.1, then the number ofgenerations (n) needed to reduce q0to qn= 0.10 is:
0.1n = 2.303 log10
0.1n = 2.303 log10
0.1n = 2.303 log103 = 2.303(0.47712) = 1.09861
n = 1.09861/0.1 = 110 generations
0.25(1 0.10)
0.10(1 0.25)
0.25(0.90)
0.10(0.75)
Z ti l ti (1)
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Zygotic selection (1)
Initialfreq.
p2 2pq q2 1 q
Adaptivevalue 1 1 1s
Freq afterselection p2 2pq q2(1s)
(p2+2pq)=p[p+(p+q)]=
p(1+q)
{ }=
=Rel. freq. = =
0
AA Aa aa Total [a]Genotypes
p2
p(1 + q)
p
(1 + q)
2pq
p(1 + q)
2q
(1 + q)
2pq
p(1 + q)
pqp(1 + q)
q
(1 + q)
s = 1
Z ti l ti (2)
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Zygotic selection (2)
q= - q = -q
(1 + q)
q
(1 + q)
q(1 + q)
(1 + q)
= - = -q
(1 + q)
q + q 2
(1 + q)
q 2
(1 + q)
Gene (a) is lethal in homozygous condition s = 1
Z gotic selection (3)
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Zygotic selection (3)
Initialfreq.
p2 2pq q2 1 q
Adaptivevalue 1 1 1s
Freq afterselection p2 2pq q
2(1s)(p2+2pq+q2)-sq2=1-sq2 =
=
=
Rel. freq.
AA Aa aa Total [a]Genotypes
p2
(1 sq2)
2pq
(1 sq2)
pq+q2(1-s)
1 sq2)
pq+q2-sq
1 sq2
q(p+q-sq)
(1 sq2)
s < 1
q(1-sq)
(1 sq2)
q2( 1 s)
(1 sq2)
(1 sq2)
Z gotic selection (4)
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Zygotic selection (4)
q= - q = -q(1-sq)
(1 sq2)
q(1-sq)
(1 sq2)
q(1-sq2)
(1 sq2)
= = =q sq2q + sq3
(1 sq2)
-sq2+ sq3
(1 sq2)
Gene (a) is not lethal in homozygous condition s < 1
-sq2(1 q)
(1 sq2)
Generations necessary to reach a given change in[q] of a deleterious recessive gene under different
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q0 qn q02 qn
2
s=1 s=.80 s=.50 s=.10 s=.01 s=.001
.99 .75 .980 .562 1 5 8 38 382 3,820
.75 .50 .562 .250 1 2 3 18 176 1,765
.50 .25 .250 .062 2 4 6 31 310 3,099
.25 .10 .062 .010 6 9 14 71 710 7,099
.10 .01 .010 .0001 90 115 185 924 9,240 92,398
.01 .001 .0001.0000
01900 1,128 1,805 9,023 90,231 902,314
.001 .0001.0000
01.00000001
9,000 11,515 18,005 90,023 900,230 9,002,304
Change ingene freq
Change infreq of R
Number of generations necessary forreaching new gene freq. at differentselection coefficients (s)
[q] of a deleterious recessive gene under differentselection coefficients
Selection against dominants(1)
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Selection against dominants(1)
Initialfreq.
p2 2pq q2 1 p
Adaptivevalue
1s 1s 1
p2
(1-s)+2pq(1-s)+q2=p2-p2s+2pq-2pqs+q2=(p2+2pq+q2)-p2s-2pqs=1-p2s-2p(1-p)s=1-p2s-2ps+2p2s=
1-sp(2-p)
Freq afterselection p
2(1-s) 2pq(1-s) q2Numerator = P2(1-s)+ pq(1-s)=p2p2s+pq-pqs=p2-p2s+p(1-p)-sp(1-p)=p2-p2s+p-p2-sp+sp2 =
pspRel. freq.
AA Aa aa Total [A]Genotypes
p2(1-s)
1-sp(2-p)
2pq(1-s)
1-sp(2-p)
q2
1-sp(2-p)
psp1sp(2p)
S l ti i t d i t (2)
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Selection against dominants (2)
p= -p=ps p
1 sp(2-p)
= =
= =
p sp p + sp2(2 p)
1 sp(2 p)
Gene (A) is not lethal in homozygous condition s < 1
ps p p{1 sp(2 p)}
1 sp(2-p)
sp + 2sp2sp3)
1 sp(2 p)
sp (1 2p2+ p2)
1 sp(2 p)
sp (1 p)2
1 sp(2 p)
Selection in no dominance (1)
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Selection in no dominance (1)
Initialfreq.
p2 2pq q2 1 p
Adaptivevalue
1 1s 12s
p2-p2s+ 2pq-2pqs+q2-2sq2=(p2+2pq+q2)-2pqs-2sq2=1-2sq(p+q)=1-2sq
Freq afterselection
p2 2pq(1-s) q2(12s) + =
=Rel. freq.
AA Aa aa Total [A]Genotypes
p2
1-2sq
2pq(1-s)
1-2sq
q2(1-2s)
1-2sq
pq(1-s)
1-2sq
q2(1-2s)
1-2sq
pq(1-s)+q2-2q2s
1-2sqq-sq(1+q)
1-2sq
Selection in no dominance (2)
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Selection in no dominance (2)
q= - q= -qsq(1+q)
1 2sq
= = =
Fitness of heterozygotes is exactly intermediatebetween the two homozygotes
qsq(1+q)
1 2sq
q(1-2sq)
1 2sq
qsq- sq2-q+2sq2
1 2sq
-sq + sq2
1 2sq
-sq(1 q)
1 2sq
Generations necessary to reach a given change in[q] under different selection coefficients in the
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q0 qn q0 qn s=1 s=.80 s=.50 s=.10 s=.01 s=.001
.99 .75 .01 .25 3 4 7 35 350 1,496
.75 .50 .25 .50 1 1 2 11 110 1,099
.50 .25 .50 .75 1 1 2 11 110 1,099
.25 .10 .75 .90 1 1 2 11 110 1,099
.10 .01 .90 .99 2 3 5 24 240 2,398
.01 .001 .990 .999 2 3 5 23 231 2,314
.001 .0001 .999 .9999 2 3 5 23 230 2,304
Change ingene freq
(del. allele)
Change infreq of [a]
(fav. allele)
Number of generations necessary forreaching new gene freq. at differentselection coefficients (s)
[q] under different selection coefficients in theabsence of dominance (co-dominance)
Heterozygous advantage (1)
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Heterozygous advantage (1)
Heterozygote has superiorreproductive fitness to bothhomozygotes
Overdominance Permit equilibrium with both
alleles remaining in the population,provided the selection coefficientsremain constant balancedpolymorphism
Heterozygous advantage (2)
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Heterozygous advantage (2)
Equilibrium is achieved when ps = qt.
If so then: ps + qs= qt + qss(p + q) = q(s + t)
Since p + q = 1 q = ss + t
Also ps + pt= qt + pt p(s + t) = t (p + q)
Since p + q = 1 p =t
s + t
Heterozygous advantage (3)
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Initialfreq.
p2 2pq q2 1 p
Adaptivevalue
1s 1 1t
p2-p2s+ 2pq+q2-q2t=(p2+2pq+q2)-p2s-q2t=
1-p2s-q2t
Freq afterselection
p2(1s) 2pq q2(1t) =
= =Rel. freq.
AA Aa aa Total [A]Genotypes
p2(1-s)
1-p2s-q2t
2pq
1-p2s-q2t
pq+q2(1-t)
1-p2s-q2t
Heterozygous advantage (3)
q2(1-t)
1-p2s-q2t
pq+q2-qt
1-p2s-q2t
q{(p+q)-qt}
1-p2s-q2t
q(1-qt)
1-p2s-q2t
Heterozygous advantage (4)
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q= - q=
= = =
Heterozygous advantage (4)
q(1-qt)
1-p2s-q2t
q(1-qt) q(1-p2s-q2t)
1-p2s-q2t
q+q2tpq2t-q2t)
1-p2s-q2t
qpq2t
1-p2s-q2t
pq(psqt)
1-p2s-q2t
Equilibrium between
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Equilibrium betweenmutation and selection (1)
-sq2(1 q)
(1 sq2)
For a deleterious recessive gene the loss per generation =
In equilibrium the loss of (a) genes through selection isexactly balanced by the gain of (a) genes through
mutation. The frequency of newly mutated (a) genes =u X [A] or up or u(1q).
If s is small the denominator can be considered = 1
Equilibrium between
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Equilibrium betweenmutation and selection (2)
Thus: sq2(1q) = u(1q)
sq2 = u
q2= u/sq = u/s
If s = 1 q = u or q2= u The frequency of
homozygous lethals at equilibrium = the frequency ofnew genes introduced by mutation.
Equilibrium between
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Equilibrium betweenmutation and selection (3)
For a deleterious dominant gene the reduction p can besimplified to sp(1p)2. Since q = (1p) the mutationrate of the recessive allele to dominant = u(1p).
Equilibrium occurs when:sp(1p)2 = u(1p)
p(1p) = u/s
Since small values of p can usually be expected whenthe dominant gene is selected against (1p) = 1At equilibrium p = u/s
Estimation of mutation rates
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Estimation of mutation ratesand equilibrium frequencies
In dominance: p = u/s u = ps
When dominance is lacking the reduction of genefrequencies per generation for low values of q is veryclose to sq(1q). As the mutation rate = u(1q)The selection mutation equilibrium is:
sq(1q) = u(1q)
q = u/s
Migration (1)
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Migration (1)
Recipient population
q0
Qm
m = proportion of newlyintroduced gene
Genes in population aftermigration = q0(1m) + mQ
Difference in gene frequencyafter migration = (1m)(q
0Q)
Q
Migration (2)
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Migration (2)Gen Hybrid Mig
Gene freq diff betweenhybrids and migrants
0 q0 Q q0Q
1 q0(1-m)+mQ=q0-mq+mQ Q q1-Q=q0-mq+mQ-Q=(1-m)(q0-Q)
2 (q0-mq+mQ)(1-m)+mQ Q q2-Q=q0-2mq+m2q0-Q=(1-m)2(q0-Q)
n qn-1(1-m)+mQ Q qn-Q=(1-m)n(q0-Q)
Migration (3)
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Migration (3)When after n generations the gene frequency in ahybrid population becomes qn, then:
qnQ = (1m)n
(q0Q)
(1m)n =q0- Q
qn- Q
If:qn= 0.446Q = 0.028q0 = 0.630n = 10
m = 0.036
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Mahram
((22
((23
[:2223]
Dan janganlah kamu menikahi perempuan-perempuan yang telah
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Dan janganlah kamu menikahi perempuan perempuan yang telahdinikahi oleh ayahmu, kecuali (kejadian pada masa) yang telahlampau. Sungguh, perbuatan itu sangat keji dan dibenci (oleh Allah)
dan seburuk-buruk jalan (yang ditempuh). Diharamkan atas kamu(menikahi) ibu-ibumu, anak-anakmu yang perempuan, saudara-saudaramu yang perempuan, saudara-saudara ayahmu yangperempuan, saudara-saudara ibumu yang perempuan, anak-anakperempuan dari saudara-saudaramu yang laki-laki, anak-anak
perempuan dari saudara-saudaramu yang perempuan, ibu-ibumuyang menyusui kamu, saudara-saudara perempuanmu sesusuan, ibu-ibu istrimu (mertua), anak-anak perempuan dari istrimu (anak tiri)
yang dalam pemeliharaanmu dari istri yang telah kamu campuri,tetapi jika kamu belum campur dengan istrimu itu (dan sudah kamuceraikan), maka tidak berdosa kamu (menikahinya), (dan diharamkan
bagimu) istri-istri anak kandungmu (menantu), dan (diharamkan)mengumpulkan (dalam pernikahan) dua perempuan yang bersaudara,kecuali yang telah terjadi pada masa lampau. Sungguh, Allah MahaPengampun, Maha Penyayang. (An-nisa 2223)
Ptolomeus V Cleopatra I
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Ptolomeus VI Cleopatra II
Cleopatra II
Ptolomeus VIII
Cleopatra III
Ptolomeus X Cleopatra IV Ptolomeus IX Cleopatra V
Berenice III Ptolomeus XII
Cleopatra VI Cleopatra VII
Inbreeding
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Inbreeding Occurs when two genes in a zygote
are identical
The probability that two genes in ahybrid are identical is given by theinbreeding coefficient
Inbreeding coefficient: The
probability that two genes in azygote are identical
A1A1 aa
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A A aa
aa A1a A1a A2a
A1a A1A2
A1A1 A1A2 A1a
identical similar different
Combinations of alleles from first-cousin mating
1/2 1/2
1/21/2
1/24
[c] = 0 01 [C] = 0 99
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[c] = 0.01 [C] = 0.99
Cc = 2pq = 2 x 0.99 x 0.01 = 0.0198
Cc X Cc2pq x 2pq = 0.01982 =
0.00039204
cc = 0.25 x 0.00039204 = 0.00009801113
Inbreeding coefficient (1)
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Inbreeding coefficient (1)
Gives the extent of mating between relatives
Chances in a diploid population of twoidentical gametes coming together is for any
generation = 1/2N probability of newlyarisen identical homozygotes = 1/2N
The probability of the remaining zygotes,1(1/2N), will have identical genes is theinbreeding coefficient of the previousgeneration
Inbreeding coefficient (2)
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Inbreeding coefficient (2)
F0= 0
F1= 1/2N
F2= 1/2N + (11/2N)F1
F3= 1/2N + (11/2N)F2Fn= 1/2N + (11/2N)Fn-1
Panmictic index (1)
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Panmictic index (1)
Panmictic index or outbred state
If F is the inbreeding or fixation
index, then 1F is the panmicticindex = P P is a measure of the relative
amount of random-mating
heterozygosity that is diminishedby inbreeding
Panmictic index (2)
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Panmictic index (2)Fn= 1/2N + (11/2N)Fn-1
1Pn= 1/2N + (11/2N)(1Pn-1)
Pn= -1 + 1/2N + 11/2NPn-1) + (1/2N)Pn-1
Pn=Pn-1+ (1/2N)Pn-1
Pn=Pn-1{1 + (1/2N)Pn-1}
Pn= Pn-1{1 - (1/2N)Pn-1}
Pn= P0(1 - 1/2N)n!
Inbreeding pedigrees (1)
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Inbreeding pedigrees (1)
V V
X Y
Z
Brother-sisterfull-sib mating
1/2 1/2
1/22
Inbreeding pedigrees (2)
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Inbreeding pedigrees (2)
R S
U V
Z
First-cousinmating
X Y
T W1
2
3
4
5
6
F = ()4
() + ()4
()
Inbreeding pedigrees (3)
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b eed g ped g ees (3)
U
V
Z
X Y
W
1
2
3
5
4 6
U = ()4() = ()5V = ()4() = ()5
W =()2() = ()3
Z = U + V + W
Effects of non-
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Effects of non-
random mating
Inbreeding ----->Increased homozygosity
Effects of changes in
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population size
Sampling error:The allelicfrequencies of a population
fluctuates from generation togeneration -----> GENETICDRIFT
In very small population ----->FIXATION
Gene frequencies
Probabilities of mating combinations
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AA X AA X = 1/16 1 0
(2) AA X Aa 2 X X = 0.75 0.25
(2) AA X aa 2 X X = 1/8 0.50 0.50
Aa X Aa X = 0.50 0.50
(2) Aa X aa 2 X X = 1/4 0.25 0.75
aa X aa X = 1/16 0 1
A aProbabilityMating
qin mating parents
Genetic drift
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Genetic driftMeasured mathematically by the
standard deviation of a proportion
=
pq/2N
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