STAT100, Module 3: Statistics in geneticsbouchard/pub/module3-lecture3.pdfSTAT100, Module 3: Statistics in genetics Dr. Alexandre Bouchard-Côté Feb 8, 2011 Next topics 1.Review:

STAT100, Module 3:Statistics in genetics

Dr. Alexandre Bouchard-CôtéFeb 8, 2011

Next topics

1. Review: mendelian inheritance & parentage testing

2. Hardy-Weinberg principle

Review

Preview of the clicker questionSuppose A and a are two alleles on chromosome 1; and B and b are two alleles on chromosome 2.

Given the following parental profiles, what it the probability that their child has the AaBB profile?

Father: AaBbMother: AaBB a) 1/3

b) 1/4

c) 1/2

d) 0

Review: locus and genotype

Alleles (version) at locus 1(for example, a SNP)

A a

Locus: address in the genome where there is a variation hot spot between individuals(Plural: loci)

Example of a locus: 6p21.3

Review: locus and genotype

Alleles (version) at locus 1(for example, a SNP)

A a

Locus: address in the genome where there is a variation hot spot between individuals(Plural: loci)

Example of a locus: 6p21.3

Genotype: the unordered combination of the allele from mother and father

Notation:Aa

Genotype that we can distinguish (one locus)

or oraa AaAA

Review: profileProfile: the genotype at each of the locus

Alleles at locus 1

Alleles at locus 2

A a

B b

Notation:AaBB

or: AaBB

AaBB

AaBb

AaBB

F M

C

Given the following parental profiles, what it the probability that their child has the AaBB profile?

Father: AaBbMother: AaBB P(C | F, M)

AaBB

AaBb

AaBB

F M

C

P(C | F, M)

3 steps: 1) do the computation for locus 1

AaBB

AaBb

AaBB

F M

C

P(C | F, M)

3 steps: 1) do the computation for locus 12) do the computation for locus 2

AaBB

AaBb

AaBB

F M

C

P(C | F, M)


3) multiply the two numbers (using indep. assumption)

AaBB

AaBb

AaBB

F M

C

P(C | F, M)




Aa

F

C ProbabilityAA 1/4aa 1/4Aa 1/2

Aa

M

Possibilities:

Both parents have both alleles

aa

F

C ProbabilityAA 0aa 1Aa 0

aa

M

Both parents have only one alleles

Aa

F

C ProbabilityAA 0aa 1/2Aa 1/2

aa

M

On parent has both, one has only one


C ProbabilityAA 1/4aa 1/4

Aa 1/2



Aa F AaM


aa F aaM


Aa F aaM


Father: AaBbMother: AaBBChild: AaBB

=> 1/2





Aa F AaM


aa F aaM


Aa F aaM



=> 1/2 x 1/2

2) do the computation for locus 2





Aa F AaM


aa F aaM


Aa F aaM



=> 1/2 x 1/2


Questions?

Clicker question for creditsSuppose A and a; B and b; and C and c are pairs of alleles located on different chromosomes.

Given the following parental profiles, what it the probability that their child has the AABBCc profile?

Father: AaBbCcMother: AaBBCc a) 0

b) 1/4

c) 1/16

d) 1/3

Clicker question for creditsSuppose A and a; B and b; and C and c are pairs of alleles located on different chromosomes.

Given the following parental profiles, what it the probability that their child has the AABBCc profile?

Father: AaBbCcMother: AaBBCc a) 0

b) 1/4

c) 1/16

d) 1/3

Preview of the clicker question

Aa

BB

aa

Bb

aa

BB

Aa

Bb

or

F1 F2M

C

H1 H2

Likelihood ratio

Find the likelihood

ratio

a) 1/2

b) 1

c) 2

d) 0NB: the SNPs are on different chromosomes

Pr( Data | H1)

Pr( Data | H2)

Difference between H and FAa

BB

aa

Bb

aa

BB

Aa

Bb

or

F1 F2M

C

H1 H2

F1 Event that the first man has genotype aaBbF2 Event that second man has genotype AaBbH1 Event that the first man is the biological fatherH2 Event that the second man is the biological father

Comparing probabilities

Pr( Data | H1)

Pr( Data | H2)Can use a ratio:

Interpretation

- Greater than 1: evidence for H1: (the first man being the biological father)- Less than 1: evidence for H2: (the second man being the biological father)

What is Pr(Data | H1) ?

AaBB

aaBb

aaBB

AaBb

F1 F2M

C

Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)

H1

Probability of all the observed profiles given that F1 is the father’s genotype (hypothesis 1, i.e. H1)

= P(adults, C | H1)


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)

= P(adults, C | H1)What we know:

P(C|adults,H1) = P(C|F1,M)P(C|adults,H2) = P(C|F2,M)(How?)

What we want:


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)



What we want:Allele contributed by mother

Allele contributed by mother

A (1/2) a (1/2)

Allele contributed

by father

A (1/2) AA (1/4) Aa (1/4)Allele contributed

by father a (1/2) aA (1/4) aa (1/4)


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)



What we want:

P(adults) = P(F1)P(F2)P(M)(How?)


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)



What we want:

P(adults) = P(F1)P(F2)P(M)(How?)

AAbb

aabb

Estimated value for Pr( AA ) = # of times AA is observed

# genotypes for SNP1

AaBB

AABb

AABb

= 3

5


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)

Chain rule:

= P(adults, C | H1)

= P(adults| H1) x

P(C|adults,H1)

What we know:

P(C|adults,H1) = P(C|F1,M)P(C|adults,H2) = P(C|F2,M)

P(adults) = P(F1)P(F2)P(M)


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)

Chain rule:

= P(adults, C | H1)

= P(adults| H1) x

P(C|F1,M)

What we know:


P(adults) = P(F1)P(F2)P(M)Known


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)

Chain rule:

= P(adults, C | H1)

= P(adults| H1) x

P(C|F1,M)

What we know:


P(adults) = P(F1)P(F2)P(M)Not quite the same

Extra assumption

Under what condition P(adults|H1) = P(adults)?

How can we interpret this assumption?

> Independence

> Consider the extreme case: genotype AA implies infertility

‘neutral alleles’


Pr( Data | H1)

Pr( Data | H2)

= P(M, F1, F2, C | H1)

Pr( Data | H1)

Chain rule:

= P(adults, C | H1)

= P(adults| H1) x

P(C|F1,M)

What we know:


P(adults) = P(F1)P(F2)P(M) = P(adults) x

P(C|F1,M)

Neutrality

What is the ratio ?

Pr( Data | H1)

Pr( Data | H2)=

P(adults) x P(C|F1,M)

P(adults) x P(C|F2,M)

=P(C|F1,M)

P(C|F2,M)

What is the ratio ?

=P(C|F1,M)

P(C|F2,M)

Aa

BB

aa

Bb

aa

BB

Aa

Bb

F1 F2M

C




=(1/2) x (1/2)

What is the ratio ?

=P(C|F1,M)

P(C|F2,M)

Aa

BB

aa

Bb

aa

BB

Aa

Bb

F1 F2M

C




(1/2) x (1/2)

(1/4) x (1/2)=

2=

What is the ratio ?

=P(C|F1,M)

P(C|F2,M)

Aa

BB

aa

Bb

aa

BB

Aa

Bb

F1 F2M

C




(1/2) x (1/2)

(1/4) x (1/2)=

2=

Questions?

Clicker question for credits

AaBB

AABb

aaBB

AaBB

F1 F2M

C H1

Likelihood ratio

Find the likelihood

ratio

a) 1/2

b) 1

c) 2


Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

P(C|F2,M)H2


AaBB

AABb

aaBB

AaBB

F1 F2M

C H1

Likelihood ratio

Find the likelihood

ratio

a) 1/2

b) 1

c) 2


Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

P(C|F2,M)H2


Aaaa

aa

? ?

or

F1 UnkM

CH1 H2

Find the likelihood

ratio

a) 1/13

b) 10/3

c) 5/2

d) 1


Aaaa

aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Find the likelihood

ratio

a) 1/13

b) 10/3

c) 5/2

d) 1


Aaaa

aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

E( P(C | U, M) )

Find the likelihood

ratio

a) 1/13

b) 10/3

c) 5/2

d) 1


Aaaa

aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

E( P(C | U, M) )




Aaaa

aa

? ?

or

F1 UnkM

C

H1 H2


Aaaa

aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

E( P(C | U, M) )

Different computation needed here...

Aaaa

aa

? ?

or

F1 UnkM

C

H1 H2

What is Pr(Data | H2) ?Pr( Data | H1)

Pr( Data | H2)

Pr( Data | H2) = P( M, F1, FAA, C | H2 )

Probability of all the observed profiles given H2, summing over the possible values of the unknown genome

+ P( M, F1, FAa, C | H2 )

+ P( M, F1, Faa, C | H2 )


Pr( Data | H2)



+ P( M, F1, FAa, C | H2 )

+ P( M, F1, Faa, C | H2 )

Event of the observed mother

genome (Aa)

Event that the unknown genome is aa (pretending

we know it’s aa)


Pr( Data | H2)



+ P( M, F1, FAa, C | H2 )

+ P( M, F1, Faa, C | H2 )

Event of the observed mother

genome (Aa)

Event that the unknown genome is aa (pretending

we know it’s aa)

Very useful principle: sum over the uncertainty

(marginalize)

What is Pr(Data | H2) ?Pr( Data | H2) = P( M, F1, FAA, C | H2 )

+ P( M, F1, FAa, C | H2 )

+ P( M, F1, Faa, C | H2 )

P( M, F1, F, C | H2 ) = P(M) P(F1) P(F) P(C | F, M)

Earlier result: for any genotype F...

What is Pr(Data | H2) ?Pr( Data | H2) = P(M) P(F1) P(FAA) P(C | FAA, M)

+ P( M, F1, FAa, C | H2 )

+ P( M, F1, Faa, C | H2 )




+ P(M) P(F1) P(FAa) P(C | FAa, M)

+ P( M, F1, Faa, C | H2 )





+ P(M) P(F1) P(Faa) P(C | Faa, M)






= P(M) x P(F1) x

( P(FAA) P(C|FAA, M) + P(FAa) P(C|FAa, M)

+ P(Faa) P(C|Faa, M) )




= P(M) x P(F1) x



Gets cancelled by factors in the

numerator, P(Data | H1)

Important part!

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############################################2444



Slide from Corinne Riddell

Possible valuesPossible values

http://www.stat.ubc.ca/~jenny/


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I J K L M N


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############################################2444




Possible valuesProbabilities



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I J K L M N


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############################################2444

E( P(C|Unknown genome, M) ) = ( P(FAA) P(C|FAA, M) + P(FAa) P(C|FAa, M)



Possible valuesExpectation notation






AA

aa

Aa

AA

AA

P(FAA) = 3/5, ...

E( P(C|Unknown genome, M) ) =

E( P(C|Unknown genome, M) ) =




Aaaa

aa

? ?

or

F1 UnkM

C

H1 H2

1) pretend ?? = AA

2) use the usual mendelian inheritance probabilities

Example

Aaaa

aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C | F1, M)

E( P(C | U, M) )

=1/2

Example

Aaaa

aa

AA

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C | F1, M)

E( P(C | U, M) )

=1/2

P(AA) x P(C | M, pretending Unk=AA) + ...

Example

Aaaa

aa

AA

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C | F1, M)

E( P(C | U, M) )

=1/2

P(AA) x 0 + ...

Example

Aaaa

aa

??

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C | F1, M)

E( P(C | U, M) )

=1/2

P(AA) x 0 + P(Aa) x 1/4 + P(aa) x 1/2=

1/2

1/5 x 1/4 + 1/5 x 1/2

= 10/3

Example

Aaaa

aa

??

or

F1 UnkM

CH1 H2

AA

aa

Aa

AA

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C | F1, M)

E( P(C | U, M) )

=1/2

P(AA) x 0 + P(Aa) x 1/4 + P(aa) x 1/2=

1/2

1/5 x 1/4 + 1/5 x 1/2

= 10/3Questions?


Aaaa

Aa

? ?

or

F1 UnkM

CH1 H2

Find the likelihood

ratio

a) 1/3

b) 7/2

c) 5/3

d) 1


Aaaa

Aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

Aa

AA

SNP survey data

Find the likelihood

ratio

a) 1/3

b) 7/2

c) 5/3

d) 1


Aaaa

Aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

Aa

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

E( P(C | U, M) )

Find the likelihood

ratio

a) 1/3

b) 7/2

c) 5/3

d) 1


Aaaa

Aa

? ?

or

F1 UnkM

CH1 H2

AA

aa

Aa

Aa

AA

SNP survey data

Pr( Data | H1)

Pr( Data | H2)=

P(C|F1,M)

E( P(C | U, M) )

Find the likelihood

ratio

a) 1/3

b) 7/2

c) 5/3

d) 1

Hardy-Weinberg principle: the genotype structure of simplified

populations

Motivation

Suppose A vs. a allele corresponds to red hair

phenotype

AA or Aa : No red hair

aa : Red hair

a: ‘Recessive allele’A: ‘Dominant allele’

Assignment question: A couple want to estimate the probability that their first child has red hair. The mother has red hair, but not the father. You only know that red hair is controlled by a recessive allele, and that 1/5 people in the population have red hair.

Motivation

What we needed: P(AA)P(Aa)P(aa)

1- Direct estimation

2- Indirect estimation

How we got these numbers

aa? ?

aa

F1 M

CAA

aa

Aa

AA

SNP survey

P(C) = E P(C|M, father genotype)

Recall: direct estimation

AA

aa

Estimated value for Pr( AA ) = # of times AA is observed

# genotypes for SNP1

Aa

AAAA

Motivation

What we needed: P(AA)P(Aa)P(aa)

1- Direct estimation

2- Indirect estimation

How we got these numbers

aa? ?

aa

F1 M

C AA

aa

Aa

AA

SNP survey

P(C) = E P(C|M, father genotype)

Indirect estimation

No red hair

No red hair

No red hair

Red hair No red hair

Known: red hair gene is recessive

Hair color statistics

AA or Aa : No red hair

aa : Red hair

This gives us Pr(aa). How can we get the other genotype probabilities, Pr(AA), Pr(Aa) from this information?

In the first generation, not enough information

Generation 1

Known:P(aa)

Fraction of Aa versus AA???


Generation 1

Known:P(aa)

Could be like this...

P(Aa)

P(AA)


Generation 1

Known:P(aa)

or like this:

P(Aa)

P(AA)

In the second generation, can be inferred exactly

Generation 1 Generation 2

Known:P(aa)

Fraction of Aa versus AA???

Hardy-WeinbergIn the next

generation, the fraction of Aa, aa and AA can be

determined under idealized conditions

What is missing: a large population has been mixing


Random mating: Assume each individual in the next generation

has a father taken uniformly at random from the previous generation, and a

mother taken independently at

random

AA

aaAa

Individuals with at least one copy of allele A

Individuals with at least one copy of allele a

AA

Individuals with red hairPreliminaries

Allele frequencies 5/8

3/8

aaAa

AA

P( a )

P( A )

Genotype frequencies

2/4

1/4P( Aa )

P( AA )

AA

1/4P( aa )

Clicker Question



What is the allele frequency of A,

Pr(A) ?

a) 1/6

b) 1/2

c) 7/12

b) 2/3

Clicker Question



What is the allele frequency of A,

Pr(A) ?

a) 1/6

b) 1/2

c) 7/12

b) 2/3

Clicker Question



What is the genotype frequency of AA,

Pr(AA) ?

a) 1/6

b) 1/2

c) 7/12

b) 2/3

Clicker Question



What is the genotype frequency of AA,

Pr(AA) ?

a) 1/6

b) 1/2

c) 7/12

b) 2/3

Allele frequencies 5/8

3/8P( a )

P( A )

Genotype frequencies

2/4

1/4P( Aa )

P( AA )

1/4P( aa )

Useful formula: P( A ) = P( AA ) + P( Aa ) / 2

5/8 1/2 1/8

OverviewWhat we need:

P(red hair) = P(aa)P(Aa)

P(AA)

How we will do it:

p = P(A) [ and hence q = P(a) ]

why? P(Aa)

P(aa)

P(AA)

Hardy-WeinbergWhat we need:


P(AA)

How we will do it:

P(A) = pP(a) = q = 1-p

P(Aa) = 2 x p x q

P(aa) = q x q

P(AA) = p x p

When a large population has been mixing

under random mating for at

least one generation

Hardy-Weinberg: application

How we will do it:

No red hair

No red hair

No red hair



P(Aa) = 2 x p x q

P(aa) = q x q

P(AA) = p x p

P(A) = pP(a) = q = 1-p


How we will do it:

No red hair

No red hair

No red hair



= 1/5

P(Aa) = 2 x p x q

P(aa) = q x q

P(AA) = p x p

P(A) = pP(a) = q = 1-p


How we will do it:

No red hair

No red hair

No red hair



= 1/5=> p = 1 - 1/ √5

q = 1 / √5 P(Aa) = 2 x p x q

P(aa) = q x q

P(AA) = p x p

P(A) = pP(a) = q = 1-p


How we will do it:

No red hair

No red hair

No red hair



P(aa) = 1/5

P(AA) = 2(3-√5)/5

P(Aa) = 1-1/5-P(AA)=> p = 1 - 1/ √5

q = 1 / √5

P(A) = pP(a) = q = 1-p

Where do the formulas come from?Part 1: stability of allele frequencies


50 individuals:Suppose there are:

70 copies of the A allele, 30 copies of the a allele

What is the probability that the allele inherited from

the father is A?

A: 3/7B: 3/5C: 3/10D: 7/10

F

M

Hardy-WeinbergPart 1: stability of allele frequencies


What is the probability that the allele inherited from

the father is A?

D: 7/10 =

70 / ( 30 + 70 )

F

M

Suppose there are: 70 copies of the A allele, 30 copies of the a allele



Suppose there are still 50 peoples (100 allele copies) at generation 2. What is your best guess for the number

of copies of the A allele in generation 2?A: 70B: About 70C: Not enough informationD: Less than 70

F

M




Suppose there are still 50 peoples (100 allele copies) at generation 2. What is your best guess for the number

of copies of the A allele in generation 2?A: 70B: About 70C: Not enough informationD: Less than 70

F

M


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In a large population the expectation is close to the observed fractionConsequence: in a large population, the allele frequency stays relatively constant across generations




Hardy-WeinbergPart 2: genotype frequencies


F

M


Let’s find P’(AA)

Finding P’(aa) uses the same argument, and P’(Aa) = 1-P’(AA)-P’(aa)

The prime means:‘in generation 2’



F

M



? ? ? ?FM

CAA



? ?FM

CAA

P’(AA) ≈ P(CAA)

Large population approximation (as in

part 1)

? ?



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)

Same argument as in parentage testing (marginalize over

unknown variables)

? ?



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)? ?= P(FAA) P(MAA) P(CAA| FAA,MAA)

+ P(FAA) P(MAa) P(CAA| FAA,MAa)

+ P(FAa) P(MAA) P(CAA| FAa,MAA)

+ P(FAa) P(MAa) P(CAA| FAa,MAa)

+ 0 + 0 + ...Some parental genotype

cannot lead to AA (e.g. aa & Aa)



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)? ?= P(FAA) P(MAA) x 1

+ P(FAA) P(MAa) x (1/2)

+ P(FAa) P(MAA) x (1/2)

+ P(FAa) P(MAa) x (1/4)



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)? ?= P(AA) P(AA) x 1

+ P(AA) P(Aa) x (1/2)

+ P(Aa) P(AA) x (1/2)

+ P(Aa) P(Aa) x (1/4)



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)? ?= P(AA) P(AA) x 1

+ P(AA) P(Aa)

+ P(Aa) P(Aa) x (1/4)



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)? ?= P(AA) P(AA)

+ P(AA) P(Aa)

+ P(Aa) P(Aa) x (1/4)

= (P(AA) + P(Aa) / 2)2

x2 + xy + y2/4 = (x + y/2)2



? ?FM

CAA

P’(AA) ≈ P(CAA)

= E P(CAA| parents)? ?= P(AA) P(AA)

+ P(AA) P(Aa)

+ P(Aa) P(Aa) x (1/4)

= (P(AA) + P(Aa) / 2)2Useful formula:

P( A ) = P( AA ) + P( Aa ) / 2

= (P(A))2


Let’s find P’(AA): Got: (P(A))2 = p 2



For P’(aa): Do exactly the same argument, writing ‘a’ instead of ‘A’, and ‘A’ instead of ‘a’

Get: (P(a))2 = q 2



For P’(aa): Do exactly the same argument, writing ‘a’ instead of ‘A’, and ‘A’ instead of ‘a’

Get: (P(a))2 = q 2

For P’(Aa): P’(Aa) = 1 - P’(AA) - P’(aa)

= 1 - p2 - q2

= 2pq Using p + q = 1

Hardy-WeinbergWhat we need:


P(AA)

How we will do it:

P(A) = pP(a) = q = 1-p

P(Aa) = 2 x p x q

P(aa) = q x q

P(AA) = p x p

When a large population has been mixing

under random mating for at

least one generation

STAT100, Module 3: Statistics in geneticsbouchard/pub/module3-lecture3.pdfSTAT100, Module 3: Statistics in genetics Dr. Alexandre Bouchard-Côté Feb 8, 2011 Next topics 1.Review:

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