population genetics2.pdf

8/10/2019 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

10


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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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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

=

= =


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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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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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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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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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.


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



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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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Heterozygote has superiorreproductive fitness to bothhomozygotes

Overdominance Permit equilibrium with both

alleles remaining in the population,provided the selection coefficientsremain constant balancedpolymorphism



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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



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99

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.


p2(1-s)

1-p2s-q2t

2pq

1-p2s-q2t

pq+q2(1-t)

1-p2s-q2t


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



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q= - q=

= = =


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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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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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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109

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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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



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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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V V

X Y

Z

Brother-sisterfull-sib mating

1/2 1/2

1/22



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R S

U V

Z

First-cousinmating

X Y

T W1

2

3

4

5

6

F = ()4

() + ()4

()



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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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population genetics2.pdf

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