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Lecture 1 Introduction to Quantitative

GeneticsPopulation Genetics Foundation

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Quantitative Genetics

• Quantitative Traits– Continuous variation– Varies by amount rather than kind– Height, weight, IQ, etc

• What is the Basis for Quantitative Variation?

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Mendelian bases for Quantitative Genetics

Early experiments by Nilsson-Ehle (1908)Wheat color

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0

0.1

0.2

0.3

0.4

0.5

0.6

Dark red med DR mediumred

pale red white0

0.2

0.4

0.6

0.8

1

1.2

Dark red med DR mediumred

pale red white

What Mode of Inheritance Would Explain This?

0.0625

0.25

0.365

0.25

0.0625

00.050.1

0.150.2

0.250.3

0.350.4

Dark red med DR mediumred

pale red white

Parents F1

F2 Blending?

Would Blending Explain this?

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5

Hypothesis: 2 loci acting independently and cumulatively on one trait?

Dark RedAABB

Whiteaabb

Medium RedAaBb

X

Medium RedAaBb

X

Medium RedAaBbAAbbaaBB

Dark RedAABB

Whiteaabb

MediumDark RedAABbAaBB

Pale RedAabbaaBb

1/16 4/16 6/16 4/16 1/16

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Gene Effects

Gene 1 Trait 1

Gene 2 Trait 2

Usual Mendelian Concept

Gene 1 Trait 1

Trait 2

Pleiotropy

Genetic CorrelationBetween Traits

Gene 1Trait 1

Gene 2Polygenic Trait

Simple Traits

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What happens to the distribution as the number of loci increases?

A continuous distribution emerges

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Stability of Distribution

F2

From Previous Example with Wheat

What is the expected Distribution of the F3? F4?

Need to know concepts of probability

0.0625

0.25

0.365

0.25

0.0625

00.050.1

0.150.2

0.250.3

0.350.4

Dark red med DR mediumred

pale red white

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Probability

Important ConceptsIndependence

Mutual Exclusivity

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Compound Events

Pr(A and B)=Pr(A)xPr(B)

Two Events are Independent if Knowledge of one event tells us nothing about the probability of occurrence of

the other event

Pr(This AND That) Pr(This given that)

Pr(A and B)=Pr(A|B)xPr(B)

Pr(A|B)=Pr(A)

Pr(A|B)=Pr(A and B)/Pr(B)

Pr(A and B)=Pr(A)xPr(B)

Pr(A|B)=Pr(A)

Only for the case of Independence

With IndependenceWith Independence

ThusThus

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Independence

• By Design– Mutli-factor experiment

• Want to estimate effects INDEPENDENT of the other

• Factorial experiment• Each level of each factor

occurs equally in each other factor

– The Effect of A can be estimated independent of the effect of B

xxxB2

xxxB1B

A3A2A1

AFactor

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Independence• By Nature

– Factors are naturally independent

– Loci inherited on DIFFERENT chromosomes

– The probability of an allele being inherited at the first locus is independent of what is inherited at the other locus

– What is the Pr(A1B1) gamete A2B2A1B2B2

A2B1A1B1B1

Locus B

A2A1

Locus AGametes produced by double heterozygote A1B1/A2B2 Unlinked Loci

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Independence

A2B2

¼A1B2

¼B2½

A2B1

¼A1B1B1

½Locus B

A2½

A1½

Locus AGametes produced by double heterozygote A1B1/A2B2Unlinked Loci

Pr(This AND That)

Pr(A and B)=Pr(A|B)xPr(B)

Pr(A1|B1)=Pr(A1)=1/2

Pr(A1 and B1)= Pr(A1|B1)xPr(B1)

Pr(A1 and B1)= Pr(A1)xPr(B1)

Pr(A1 and B1)= ½ x ½ =1/4

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Non-Independence• By Poor Design or missing data

– The Effect of Factor A for some combinations of B is dependent of the level of B

– A3 is confounded with the effect of B1– A1 is confounded with the effect of B2– If B were an unknown, unaccounted

for factor, could lead to spurious correlation

– Factor A may not have any effect, all results could be due to unseen Factor B effects

– Example: Factor A genotypic frequencies at A locus, Factor B proportion of 2 different subpopulations B1 and B2

xxB2

xxB1B

A3A2A1

AFactor

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Non-Independence• By Nature

– Factors are naturally dependent

– Loci inherited on SAME chromosomes perfectly linked

– If the allele inherited at the A locus was A1 then the allele at the second locus must be B1 and vice versa

– What is the Pr(A1B1) gamete? A2B2A1B2B2

½

A2B1A1B1B1½

Locus B

A2½

A1½

Locus AGametes produced by double heterozygote A1B1/A2B2Linked Loci

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Non-Independence

A2B2

½A1B2

0 B2 ½

A2B1

0A1B1B1

½

Locus B

A2½

A1½

Locus AGametes produced by double heterozygote A1B1/A2B2Linked Loci

Pr(This AND That)

Pr(A and B)=Pr(A|B)xPr(B)

Pr(A1|B1)=1

Pr(A1 and B1)= Pr(A1|B1)xPr(B1)

Pr(A1 and B1)= Pr(A1)x1

Pr(A1 and B1)= ½ x 1 =1/2

Pr(A1|B2)=0

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Multiple EventsThis OR That

Pr(A or B)=Pr(A)+Pr(B)

Mutually Exclusive Events The occurrence of one event excludes

the other

P(A and B)=0

Pr(A or B)=Pr(A)+Pr(B)-P(A and B) General Formula

Requires knowledge of Exclusivity and Independence

Not mutually exclusive but independent Pr(A or B)=Pr(A)+Pr(B)-P(A)P(B)

Not mutually exclusive nor independent Pr(A or B)=Pr(A)+Pr(B)-P(A)P(B|A)

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Single Rose Walnut Pea

12 12 12 12

1212 12 12

There were 24 birds of each of 4 comb types in a pen of 96 birds, ½ of each comb type are black, the other white. Chose 1 bird. What is the probability of choosing either a Single or Rose comb bird?

P(Single OR Rose)=Pr(Single)+Pr(Rose)-Pr(Single and Rose)

P(Single or Rose)=(24/96)+(24/96)-(0/96)=48/96

Rose 24Single 24

Mutually exclusive

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Single Rose Walnut Pea

12 12 12 12

1212 12 12

There were 24 birds of each comb type in a pen of 96 birds, ½ of each comb type are black, the other white. What is the probability of choosing either a Black bird or one with a Single comb?

P(Black OR Single)=Pr(black)+Pr(single)-Pr(black and single)

P(Black OR Single)=(48/96)+(24/96)-(48/96)(24/96)=60/96

Black48

Single 24 Independence by Design

Example of not mutually exclusive but independent events

Total96

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Single Rose Walnut Pea

12 12 12 12

1212 12 12

There were 24 birds of each of 4 comb types in a pen of 96 birds, ½ of each comb type are black, the other white. Chose 2 birds with replacement. What is the probability of choosing at least one rose comb bird?

P(B1=Rose or B2=Rose)Easy way and hard way

1/4 +1/4-(1/4)(1/4)= 7/16

1-P(B1=not Rose and B2=not Rose)

1-¾ x ¾ = 1-9/16 = 7/16

Example of not mutually exclusive but independence events

P(Rose1)+P(Rose2)-P(Rose1 and Rose2)

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Single Rose Walnut Pea

12 12 12 12

012 12 12

There were 24 birds of each comb type in a pen of 96 birds, ½ of each comb type are black, the other white. All black birds with pea combs died. What is the probability of choosing either a Black bird or one with a Single comb?

P(Black OR Single)=Pr(black)+Pr(single)-Pr(black and single)

P(Black OR Single)=(36/84)+(24/84)-(36/84)(12/36)=30/60=48/84

Black36

Single 24 Independence by Design

Example of not mutually exclusive nor independent events

Total84

White48

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Application of Probabilities to Populations: Mating

• Assume 3 genotypes at the A locus– AA, Aa, aa

• Put a female of genotype AA in a large population of males with frequencies of the genotypes P(AA)=xP(Aa)=yP(aa)=zx+y+z=1

• What is the probability she will mate with a given genotype of male

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Mating PreferencesIndependence (no preference)

• P(AA male| AA female)=x• P(Aa male| AA female)=y• P(aa male| AA female)=z

– Extreme negative assortative• P(AA male| AA female)=0• P(Aa male| AA female)=0• P(aa male| AA female)=1

– Extreme positive assortative• P(AA male| AA female)=1• P(Aa male| AA female)=0• P(aa male| AA female)=0

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What is the Consequence of Random Mating on Genotypic Frequencies

• Assume a Perfect World– No Forces Changing Gene Frequency– Equal Gene Frequencies in the Sexes– Autosomal Inheritance– Random Mating

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GENERATION 0

• Allow The Genotypic Frequencies To Be Any Arbitrary Values

genotypic frequencyP( AA ) = XP( Aa ) = YP( aa ) = Z

such that X + Y + Z = 1

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Allele Frequencies

Y21 + X = )A P(

p + q = 1

p=

Y21 + Z= ) a P(

q=

P(Aa) P(AA) = )A P( 21+

P(Aa) P(aa) = ) a P( 21+

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Frequency of Matingmale genotype

female genotype AA Aa aafrequency ( X ) ( Y ) ( Z )

AA ( X ) X2 XY XZ

Aa ( Y ) XY Y2 YZ

aa ( Z ) XZ YZ Z2

Random Mating=independence

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Expected genotypic frequencies that result from matings (Gen 1).

PossibleMatings

AA x AA

AA x Aa

AA x aa

Aa x Aa

Aa x aa

aa x aa

Frequency ofMating

X2

2XY

2XZ

Y2

2YZ

Z2

Expected Frequency of Offspring

AA Aa aa

1

1/2

0

1/4

0

0

0

0

0

0

1/2 0

1

1/2 1/4

1/2 1/2

1

Conditional Probabilities given genotypes of parents

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GENERATION 1

( ) + X =

Y + XY + X =

) Y ( + ) 2XY ( + ) X 1( = ) AA P(

221

2412

241

212

offsping

Y

[ ]

2offspring

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2parents

) AA P( 1 generationfor Therefore

= Y + X = )A P( BECAUSE

)A P( =

p

p

=

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( )( )pq2 =

Y + Z Y + X 2 = YZ + Y + 2XZ + XY =

) 2YZ ( + ) Y ( + ) 2XZ 1( + ) 2XY ( = ) Aa P(

21

21

221

212

21

21

offsping

( )2

221

2241

2212

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offsping

=

Y+ Z =

Z+ YZ + Y =

) Z1( + ) 2YZ ( + ) Y ( = ) aa P(

q

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Generation 2Frequency of Matings

male genotypefemale genotype AA Aa aa

frequency ( p 2 ) ( 2pq ) ( q 2 )

AA ( p 2 ) p 4 2p 3q p 2q 2

Aa ( 2pq ) 2p 3q 4p 2q 2 2pq 3

aa ( q 2 ) p 2q 2 2pq 3 q 4

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Expected genotypic frequencies that result from matings (Gen 1).

PossibleMatings

AA x AA

AA x Aa

AA x aa

Aa x Aa

Aa x aa

aa x aa

Frequency ofMating

Expected Frequency of Offspring

AA Aa aa

1

1/2

0

1/4

0

0

0

0

0

0

1/2 0

1

1/2 1/4

1/2 1/2

1

p4

4p3q

2p2q2

4p2q2

4pq3

q4

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Overall Genotypic Frequencies

( )( )

2

22

222

2234

2234

ppqpp

qpqppqpqpp

qpqpp

= ) 1 ( = + =

+ 2 + = + 2 + =

) 4 (41 + ) 4 ( 2

1 + ) 1( = ) AA P(

2

offsping

P( Aa ) = offsping 2 pq

P( aa ) = offsping2q

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Summary of genotypic frequencies by Generation

genotype gen 0 gen 1 gen 2

P( AA ) X p 2 p 2

P( Aa ) Y 2pq 2pqP( aa ) Z q 2 q 2

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( ) + = + ( 2 + A ap q p pq qAA Aa aa2 2 2)

after one generation of random mating and will remain in that distribution until acted upon by other forces

Hardy-Weinberg Equilibriumor the Squared Law

If a population starts with any arbitrary distribution of genotypes, provided they are equally frequent in the two sexes, the proportions of genotypes (AA, Aa, aa) with initial gene frequencies p and q will be in the proportion

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Limitations

• If genotypic frequencies conform to the squared law, does that mean that all the assumptions hold?

• NO– Selection can be occurring after the time the

population was counted

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What if Organisms Do not Mate But Release Gametes to Search Out

Each Other?

Do the Same Properties Hold?

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Fertilization

If not IndependenceP(A male | a female)= 0 to 1

What is the probability that a male gamete (sperm) carrying the ‘A’ alleleWill fertilize an egg given that the egg is carrying the ‘a’ allele

If IndependenceP(A male | a female)=P(A male)

Incompatibility

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What is the Consequence of Random Union of Gametes

• Assume a Perfect World– No Forces Changing Gene Frequency– Equal Genotypic Frequencies in the Sexes– Autosomal Inheritance

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GENERATION 0

lets allow the genotypic frequencies to be any arbitrary value and the gene frequencies to be the appropriate function of those values.

genotypic frequencyP( AA ) = XP( Aa ) = YP( aa ) = Z

such that X + Y + Z = 1

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Allele Frequency

Y21 + X = )A P(

p + q = 1

p=

Y21 + Z= ) a P(

q=

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With Independence

male gamete/frequencyA a

( p ) ( q )female gamete/ A AA Aafrequency ( p ) ( p 2 ) ( pq )

a Aa aa( q ) ( pq ) ( q 2 )

Random Union of Gametes Produces the Same Outcome as Random Mating

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Assumptions

• Random Mating– Mates Chose Partners Independent of

Genotype• Random Union of Gametes

– Gametes Pair Independent of the Alleles Which they Carry

• Then– Random Mating= Random Union of Gametes

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What Happens If The Allele Frequencies Are Not Equal Between The Sexes?

Generation 0Gametic Frequencies

Male GameteFrequency

Female Gametefrequency

A

A

a

a

0mp 0

mq

0fq

0fp 00

fm pp

00fmqq00

fmqp

00mf qp

AA Aa

Aaaa

Genotypic Frequencies in Generation 1

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Gametic Frequencies Produced by Adults of First Generation

( )

( )0000001

0000001

2121

fmfmfmf

fmfmfmm

pqqpppp

pqqpppp

++=

++=

Frequency of Homozygous Class + ½ frequency of heterozygous class

Note, gametic frequencies are now equal in sexes

111 ppp fm ==

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Generation 2

Generation 1Gametic Frequencies

Male GameteFrequency

Female Gametefrequency

A

A

a

a

1p 1q

1q

1p ( )2111 ppp =

( )2111 qqq =

11qpAA Aa

Aa aa

Genotypic Frequencies in Generation 211qp

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Example Cross Between Populations

Population 1Males

¾ AA ¼ aa

Population 2Females

¼ AA ¾ aa

G1

G2

Random Mate

G0

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Generation 1

Population 1Males

¾ AA ¼ aa

Population 2Females

¼ AA ¾ aa

¾ A

¼ a

¼ A ¾ a

3/16 AA 9/16 Aa

1/16 Aa 3/16 aa

p1=3/16 + ½(9/16 + 1/16)=8/16= ½

G1 Genotypic Freq

G1 Gene Freq

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Generation 2

G1Males 3/16 AA 10/16

Aa 3/16 aa

½ A ½ a

¼ AA ¼ Aa

¼ Aa ¼ aa

p2= ¼ + ½( ¼ + ¼ )= ½

G2 Genotypic Freq

G2 Gene Freq

G1Females 3/16 AA 10/16 Aa

3/16 aa

½ A

½ a

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SummaryPopulation 1

Males ¾ AA ¼ aa

Population 2Females

¼ AA ¾ aa

3/16 AA 10/16 Aa 3/16aap1= ½

4/16 AA 8/16 Aa 4/16aa

p2 = ½

G3 ?

G0

G1

G2 Gametic Frequencies Same

Genotypic Frequencies Different

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Summary• Random Mating

– Autosomal Loci– Equal Allele Frequency in Sexes

• Equilibrium Established After 1 Generation– Unequal Allele Frequencies in Sexes

• One Generation Required to Establish Equal Allele Frequencies in Sexes

• Second Generation Required to Establish Equilibrium• Excess of Heterozygotes Produced in First Generation

– Indicative of Crossing between populations– Useful in Plant and Animal Breeding– May result in Heterosis

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Lecture 1 Problems

1. Two separate populations of equal size are in equilibrium for the same pair of alleles because of random mating within each. In population I, pA = 0.6,

while in population II, pA = 0.2, with q = 1 - p in each population.• (a) If a random sample of females from one population is crossed to a

random sample of males from the other population, what would be the expected genotypic frequencies among the progeny? If these progeny are then allowed to mate at random, what would be the expected gene and genotypic frequencies in the next-generation? What happens to heterozygote frequencies between the F1 and F2 generations?

• (b) If equal numbers of both sexes from each population are combinedand allowed to mate at random, what would be the expected gene and genotypic frequencies in the next-generation?

• (c) Compare results in part a and b, what conclusions can you draw from this.