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060310391 0560565

6

2015/11/6 13:00-14:45

@1 - 4

1

(associa4on analysis)

(associa4on study)

……T……C……A……G……T………A…………A………

• 

DNA

……T……C……A……A……T………A…………A………

……T……A……G……G……T………A…………A………

……G……C……A……A……T………C…………T………

……T……A……A……G……T………A…………A………

……T……C……A……A……T………A…………A………

……T……C……A……G……T………A…………T………

……T……A……G……G……T………C…………A………

……G……C……A……G……C………A…………T………

……T……A……A……G……C………A…………A………

……T……C……A……G……C………C…………T………

……T……A……A……A……C………C…………A………

……T……C……G……G……C………A…………T………

……G……A……A……G……C………A…………A………

……T……C……A……A……C………C…………A………

A

B

C

:

: :

DNA

200

70/80

23/120

200

2

AD2 T( Ab N 1 73 7 0

0

0

0

T

0

C

• 

•  QTL

• 

•  GWAS

•  χ

2

•  Fisher

• 

• 

•  1 2

• 

3

•  3

AA, Aa, aa 3

30,000

• 

• 

2005

4

(2)

QTL

DNA

QTL

• 

• 

• 

• 

×

6

DNA

SNP

• 

• 

• 

ATCGAG TAGACT TATACG

ATCGAG TAGACT TATACG

ATCGAG TAGACA TATACG

ATCGAG TAGACT TATACG

ATCGAG TAGACT TATACG

ATCGAG TAGACT TATACG

ATCGAG TAGACA TATACG

ATCGAG TAGACA TATACG

ATCGAG TAGACT TATACG

ATCGAG TAGACA TATACG

7

(single nucleo4de polymorphisms: SNPs)

GeneChip Rice 44K SNP Genotyping Array

•  44,100 SNPs (10kb 1 SNP)

Tung et al. (2010) Rice 3:205-217 8

QTL

DNA

GWAS

DNA

Morrell et al. (2012) Nature Review Gene4cs 13:85

QTL vs.

(3)

10

QTL

Yes No

linkage disequilibrium: LD

11

QTL

(Modified from Balding 2006)

SNP

( )

SNP

( )

SNP

13

(4)

(Rafalski 2002

B b

A pAB pAb pA

a paB pab pa

pB pb

r2= D

2

pApapBpb

(●● 1,●● 0

)

D = p

AB

− p

A

p

B

= p

AB

p

ab

− p

Ab

p

aB

r2= 0.25

2

0.5× 0.5 × 0.5 × 0.5

= 1

AB

AB

AB

r2= 0.1024 r2= 0

A B

14

• 

• 

(Rafalski 2002 )

15

LD

Gupta et al. (2005)

• 

• 

16

LD

(Zhu et al. 2007)

LD

→ LD

17

(5)

AA

aa

AA aa

χ2 Fisher Fisher’s exact test

18

χ

2

(R) (S)

AA f11 (11) f12 (4) f1. (15) aa f21 (3) f22 (7) f2. (10) f.1 (14) f.2 (11) n (25)

p(R) = f.1 / n = 14 / 25 = 0.56 p(S) = f.2 / n = 11 / 25 = 0.44 AA p(AA) = f1. / n = 15 / 25 = 0.60 aa p(aa) = f2. / n = 10 / 25 = 0.40

R AA n × p(R) × p(AA) = 25 × 0.56 × 0.60 = 8.4 R aa n × p(R) × p(aa) = 25 × 0.56 × 0.40 = 5.6 S AA n × p(S) × p(AA) = 25 × 0.44 × 0.60 = 6.6 S aa n × p(S) × p(aa) = 25 × 0.44 × 0.40 = 4.4

χ2= (obs− exp)

2

exp =(11− 8.4)2

8.4 +

(3− 5.6)2

5.6 +

(4− 6.6)2

6.6 +

(7− 4.4)2 4.4 = 4.57

(r-1)(c-1) χ2 r c

5%

χ0.012 (1) = 6.63 >χ2= 4.57 >χ0.052 (1) = 3.84

5

19

-4 -2 0 2 4 6 8

mQQ mqq

QTL 㑇ఏᏊᆺ

QQ qq

AA

aa

40 10

10 40

-4 -2 0 2 4 6 8

₯ᅾⓗ࡞ΰྜศᕸ

-4 -2 0 2 4 6 8

࣐࣮࣮࢝

⾲⌧ᆺศᕸ

mAA

-4 -2 0 2 4 6 8

maa

y i = u + β j x ij + e i

x y

-0.2 0.2 0.6 1.0

-20246

Marker genotype

Phenotype

i AA xi=2

aa xi=0

xi yi

N0 0

( ( CA

0 2 20

-3 -2 -1 0 1 2 3

455055

x

y

y

i

= α + β x

i

+ ε

i

= y ˆ

i

+ ε

i

y

dependent (response) variable

independent (explanatory) variable

SNP

β

regression coefficient

ε

residuals yi

ˆ

y i=α+βxi

xi εi

α

intercept, constant term y

21

(6)

The method of least squares

ε

i

= y

i

− (α + βx

i

)

( 2

SSE = εi 2

i n

= (yiα − βxi)2

i n

:

∂SSE

∂β =−2 i (yiα − βxi)xi

n

= 0

∂SSE

∂α =−2 i (yiα − βxi)

n

= 0

a = yi

i

n n− b ixi

n n = y − bx b =

xiyi

i

n ixi

n iyi

n n

xi2xi

i

n

( )

2 n

i

n =

(xi− x )(yi− y )

i

n

(xi− x )2

i

n

n xi

i

n

xi

i

n xi 2 i

n

#

$

%

%

&

' ( ( a b

#

$ %

& ' ( = iyi

n

xiyi

i

n

#

$

%

%

&

' (

-3 -2 -1 0 1 2 3 (

455055

x

y

yi ˆ

y i=α+βxi

xi εi

22

(a b α β

y i = u + β j x ij + e i

1311

SNPs

All materials can be downloaded from hkp://ricediversity.org/

SNPs …

0e+00 1e+08 2e+08 3e+08

05102030

position (bp)

−log10(p)

LD

LD

Rafalski and Morgante 2004 Oraguzie et al. 2007) 25

(7)

p 

… suppose that a would-be gene7cist set out to

study the “trait” of ability to eat with chops7cks in the

San Francisco popula7on by performing an

associa7on study with the HLA complex. The allele

HLA-A1 would turn out to be posi7vely associated

with ability to use chops7cks … because the allele

HLA-A1 is more common among Asians than

Caucasians.”

Lander and Schork (1994)

HLA Human Leukocyte Antigen

26

?

1

2

(Modified from Balding 2006)27

•  Indica

Japonica

y i = u + β j x ij + v k q ik

k=1

K

+ e i

1.0

0.0 0.5

Bayesian

Structure Pritchard et al. (2000) Gene4cs 155:945– 959

4

6 K=6

2

qi1 = 0.78, qi3 = 0.32, 0

(8)

0e+00 1e+08 2e+08 3e+08

0510152025

position (bp)

−log10(p)

31

A

•  Yu et al. (2006) Nat. Genet. 38: 203

y i = u + β j x ij + v k q ik

k=1

K

+ α i + e i

a ~ N (0, Aσ α 2 )

var(α

i

) = a(i, i) σ

α2

cov( α

i

, α

i'

) = a(i, i ') σ

α2

0e+00 1e+08 2e+08 3e+08

01234

position (bp)

−log10(p)

(9)

1SNP

SNP

SNP

A A

B B

B A

34

B A

SNPs

SNP

A

A B

B A

35

B

Bayesian

QTL

y i = u + β j γ j x ij

j=1

J

+ v k q ik

k=1

K

+ α i + e i

•  Iwata et al. (2007) Theor Appl Genet 114:1437–1449

γ

j

= 1

0

! "

#

j SNP

j SNP

γ

j

~ B(1, p

j

)

0e+00 1e+08 2e+08 3e+08

0.00.20.40.60.81.0

position (bp)

Posterior prob. of QTL

Bayesian

γ j

(10)

Chr. 3

GS3

63 kb

id3008333

546 kb

id3008127

Chr. 5

qSW5

id5002699

0.95 1.0

66 kb 0.88

0e+00 1e+08 2e+08 3e+08

0.00.20.40.60.81.0

position (bp)

Posterior prob. of QTL

SNPs

GWAS & GS

GWAS

GS

GS

• 

• 

•  NA

• 

•  NA

(11)

•  AC 0

Q, K

•  Q

–  Structure (Pritchard et al. 2000)

–  Price et al. 2006)

–  Zhu and Yu 2009)

•  K:

–  Loiselle et al. (1995),

Ritland et al. (1996))

–  Zhao et al. 2007

– 

43

GWAS

Atwell et al. (2010) Nature 465: 627

→ a

45

(12)

A B (

C ( D

false posi4ve rate: FPR = B / (A + B)

false nega4ve rate: FNR = C / (C + D)

false discovery rate: FDR = B / (B + D)

46

•  FDR: false discovery rate

1.  K P

2.  P

3.  FDR q* 5%

i

4.  1 i

Benjamini and Hochberg (1995)



) ( )

2 ( ) 1

(

P P

K

P ≤ ≤ !

K

P

i

iq

* )

(

•  p

•  K P

p K × P

• 

P

(i)

iq

*

K ⇔ q

*

KP

(i)

i

•  q* i

K × P

•  q* 5%

GWAS

49

•  GWAS Hd6

Yano et al. (2016) Nature Gene4cs 48: 927

(13)

Yano et al. (2016) Nature Gene4cs 48: 927 51 53

Yang et al. (2003) Am. J. Hum. Genet 73: 627

•  ACTN3 α-c4nin-3

•  X α-c4nin-3

•  a-ac4nin-3 a-ac4nin-2 ACTN3

(14)

126,559 GWAS

•  Fig. 1 3 SNPs

Rietveld et al. 2014, PNAS 13790, Ward et al. 2014 PLoS ONE e100248)

•  R2 0.02%

•  Fig. 2

54

Rietveld et al. (2013) Science 340: 1467

• 

• 

• 

• 

• 

•  DNA

QTL

55

• 

• 

QTL

• 

•  :

(2000/01)

•  ISBN-10: 4130602063

•  ISBN-13: 978-4130602068

56

hkp://www.amazon.co.jp/

6

1.  100 2

A, B AABB,

aaBB, AAbb, aabb 40, 10, 10, 40

AB D r2

2.  AB

AB 1

57

χ

0.012

(1) = 6.63, χ

0.05

2

(1) = 3.84

(15)

PDF MS-Word PDF

e

[email protected]

2016 11 14

58

Fisher

Fisher’s exact test

• 

“ ”

• 

• 

59

R S

AA a c g

aa b d h

e f n

e, f, g, h P(e, f ,g,h) = n

e

"

# $ %

& ' peqf× n

g

"

# $ %

&

' rgsh= (n!)

2

e! f !g!h!p

eqfrgsh

p = e/n, q = f/n, r = g/n, s = h/n

a, b, c, d e, f, g, h

P(a,b,c, d e, f ,g,h) = p(a,b,c,d,e, f ,g,h) / p(e, f ,g,h) =e! f !g!h! n!

1 a!b!c!d!

e, f, g, h

a, b, c, d P(a,b,c, d,e, f ,g,h) = n

e

"

# $ %

& ' peqf × e

a

"

# $ %

& ' rasb× f

c

"

# $ %

& ' rcsd = n!

e! f ! e! a!b!

f ! c!d!p

eqfrgsh

60

P(B)

P(A ∩ B)

P(A | B) = P(A ∩ B) /P(B)

a,b,c,d

11 4

3 7

R S

AA 11 4 15

aa 3 7 10

14 11 25

px

6

px=14!11!15!10! 25!

1 11!3!4!7!

"

# $ %

& ' = 0.0367

Fisher

χ2

5%

12 3

2 8

13 2

1 9

14 1

0 10

10 5

4 6

9 6

5 5

8 7

6 4

7 8

7 3

6 9

8 2

5 10

9 1

4 11

10 0

p =14!11!15!10! 25!

1 11!3!4!7!+

1 12!2!3!8!+

1 13!1!2!9!+

1 14!0!1!10!+

1 5!9!10!1!+

1 4!10!11!0

"

# $ %

& '

= 0.0486

px

61

(16)

-3 -2 -1 0 1 2 3

455055

-3 -2 -1 0 1 2 3

455055

-3 -2 -1 0 1 2 3

455055

y

i

ˆ

y

i

y

(yi− y )2

i

n = ( ˆ y i− y ) 2 i

n + (yi− ˆ y i) 2 i

n

(a)

SSY

(a)

(b)

(c)

(b)

SSR

(c)

SSE F

(b)

SSR 1 MSR =

SSR

MSR /MSE

(c)

SSE n - 2 MSE = SSE /(n-2)

(a)

SSY n - 1

F

y

i

y

ˆ

y

i

(

62

J = (yiα − βxi)2

i=1

N

y

i

= α + β x

i

+ ε

i

ε

i

~ N(0, σ

2

)

0 σ2

y

i

~ N( α + β x

i

, σ

2

)

y α+βx σ2

L = p(y1, y2,..., yN) = exp{−(yi−α − βxi)

2/2σ2

} 2πσ2

i=1 N

=exp{− (yi−α − βxi)

2/2σ2 i=1

N }

(2πσ2)N

0.00.10.20.30.4

α + βxi yi

p(yi) = 1 2πσ2exp

(yi−α − βxi)2 2σ2 ' ( )

* + ,

ln L = − 1

2 (yi−α − βxi)

2 i=1

N N2log2πσ2

ln L

∂β = 1

σ2 i(yiαβxi)xi

n

= 0

∂ ln L

∂α =− 1

σ2 i(yi−α − βxi)

n

= 0

0 σ2

63

8a

参照

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