The 7th Workshop on QTL Mapping and Breeding Simulation, 20-21 Jan., Mexico

Frequently Asked Questions in QTL Mapping

Jiankang WangJiankang WangCIMMYT China and CAAS

E-mail: wangjk@caas.net.cn or jkwang@cgiar.org

1

E mail: wangjk@caas.net.cn or jkwang@cgiar.org

Q1: What is LOD?Q1: What is LOD?

2

H th i t t i QTL iHypothesis test in QTL mapping

LOD > LOD0, T itiF l iti

LOD LOD

0i.e., accept Ha

True positives, no errors

False positives, Type I errors

False negati esLOD < LOD0, i.e., accept H0

True negatives, no errors

False negatives, Type II errors

H0: there is no QTL at a genomic

iti th

Ha: there is one QTL at the

i3

position on the trait in interest

genomic position

Likelihood ratio test (LRT)

Definition of LRT )ln(2 0

ALLLRT −=

Definition of LOD (likelihood of odd)A

L )log()log()log( 00

LLLLLOD A

A −==

Relationship between LOD and LRT

or61.4)10ln(2

LRTLRTLOD ≈= LODLRT 61.4≈

4

Q2: How to choose a Qthreshold value of LOD?

5

T i h th i t tTwo errors in hypothesis test

Type I error rate = P {Reject H0 |Type I error rate P {Reject H0 | True H0}

T II t P {A t H |Type II error rate = P {Accept H0 |False H0}False H0}

6

Significance level (α) in hypothesisSignificance level (α) in hypothesis test – The control of Type I error

Significance level for N times of gindependent tests: 1-(1- α)N

B f i dj t t / NBonferroni adjustment: ≈ α / N

Problem: Multiple and dependent tests exist in QTL mapping! Permutation test in QTL mappingPermutation test in QTL mapping

7

Choice of the threshold of LODChoice of the threshold of LOD

For one test：α (e.g., 0.1, 0.05, 0.01)N times of independent tests: 1-(1-α)N

Empirical LOD threshold for an overallEmpirical LOD threshold for an overall significance level of 0.05: 2.0 – 3.0

8

Q3: Which method to use?Q3: Which method to use?

9

P f t ti ti l t tPower of a statistical test

The power of a statistical test is the pprobability that the test will reject a false null hypothesis (i e it will notfalse null hypothesis (i.e., it will not make a Type II error).P 1 0 T IIPower= 1.0 – Type II error

10

QTL mapping from IM and ICIMQTL mapping from IM and ICIM

36

2427303336

e

IMICIM (PIN=0.01)

Under LOD threshold 3 0

1215182124

OD score Under LOD threshold 3.0,

IM identified 3 QTL, and ICIM identified 9 QTL

369

12LO

Q

0

1H

1H

1H

1H

1H

1H

2H

2H

2H

2H

2H

2H

2H

3H

3H

3H

3H

3H

4H

4H

4H

4H

4H

4H

5H

5H

5H

5H

5H

5H

5H

6H

6H

6H

6H

6H

6H

7H

7H

7H

7H

7H

7H

Testing position on the barley genome, step = 1 cM

11

g p y g , p

QTL mapping in a simulated population20

10

15

20

D s

core

完备区间作图 ICIM

0

1

2

ted

effe

ct

完备区间作图 ICIM

0

5LOD

‐2

‐1

Estim

a

10

15

20

scor

e

复合区间作图 CIM

0

1

2

ed e

ffect 复合区间作图 CIM

0

5

10

LOD

s

‐2

‐1

0

Estim

ate

15

20

core

区间作图 IM

1

2

d ef

fect

区间作图 IM

0

5

10

LOD

sc

‐2

‐1

0Es

timat

ed

12Testing every 1 cM on six chromosomes Testing every 1 cM on six chromosomes

0

11111112222223333334444445555556666662

1111111222222333333444444555555666666

Power from 3 simulated l i h f 200 RILpopulations, each of 200 RILs

Pop Chr. Pos. Effect LOD PVE (%) CI=10 cM( )1 2 25.1 0.19 2.56 3.48 False QTL

5 51.1 0.29 6.05 8.14 IQ56 60.0 0.30 6.72 8.86 IQ67 40.0 0.20 2.94 3.71 False QTL7 70 0 0 42 11 87 16 64 IQ77 70.0 0.42 11.87 16.64 IQ7

2 2 30.5 0.27 5.35 7.78 IQ25 45 0 0 27 5 25 7 94 False QTL5 45.0 0.27 5.25 7.94 False QTL6 59.1 0.26 4.94 7.50 IQ67 59.4 0.38 9.84 15.61 False QTL

3 2 30.0 0.21 2.50 3.96 IQ26 55.4 0.29 4.47 7.81 IQ6

13

7 70.0 0.28 4.42 7.14 IQ77 90.0 0.25 3.39 5.41 False QTL

Power comparison: ICIM vs. IMPower comparison: ICIM vs. IM

90100 20 40

60 80100 120

4050607080

ower (%

)

100 120140 160180 200220 240260 280300 320

010203040

Po

300 320340 360380 400420 440460 480500 520

100 20 40

1 2 3 4 5 10 20 30 FDRPercentage of phenotypic variation  explained (PVE)

500 520540 560580 600

5060708090

er (%

)

60 80100 120140 160180 200220 240

1020304050

Powe 260 280

300 320340 360380 400420 440460 480

14

0

1 2 3 4 5 10 20 30 FDRPercentage of phenotypic variation  explained (PVE)

460 480500 520540 560580 600

Pop lation si e eq i edPopulation size required

ProbabilityPVE (%) 0 9 0 8 0 7 0 6PVE (%) 0.9 0.8 0.7 0.6

1 >600 5402 600 420 340 2802 600 420 340 2803 430 280 230 2004 340 250 190 1605 280 200 160 1305 280 200 160 130

10 160 120 100 8020 100 80 60 50

15

20 100 80 60 5030 100 60 50 40

FDR: false discovery ratey(False QTL out of all positives)

90100 20 40

60 80100 120

607080

%)

100 120140 160180 200

405060

ower (% 220 240

260 280300 320

203040P 300 320

340 360380 400420 440

010 420 440

460 480500 520

16

1 2 3 4 5 10 20 30 FDR

Percentage of phenotypic variation  explained (PVE)

540 560580 600

Q4: What are the ways that can improve mapping efficiency?improve mapping efficiency?

17

Possible ways

Large populationg p pPrecision phenotyping and

t igenotyping Efficient methodEfficient methodHigh marker density?!Two-stage mapping strategy?!

18

The length of empirical 95%The length of empirical 95% confidence intervals of QTLconfidence intervals of QTL

PVE (%) PS 200 PS 400PVE (%) PS=200 PS=400MD=5 cM

MD=10 cM

MD=20 cM

MD=40 cM

MD=5 cM

MD=10 cM

MD=20 cM

MD=40 cM

1 93.30 104.55 103.10 120.03 47.00 61.94 76.83 88.242 54.14 62.37 73.66 86.08 37.55 32.38 38.34 47.593 52 65 47 12 50 02 48 18 25 64 21 95 18 78 33 363 52.65 47.12 50.02 48.18 25.64 21.95 18.78 33.364 38.89 46.14 41.94 56.72 25.01 18.46 22.74 36.575 25.99 37.83 44.73 59.74 16.35 16.39 22.54 36.30

10 10.31 8.35 11.29 46.45 3.72 4.78 7.92 22.7420 8.55 10.19 14.78 26.97 4.90 6.04 8.70 15.5630 5 33 8 23 11 56 18 62 3 18 4 78 6 62 12 62

19

30 5.33 8.23 11.56 18.62 3.18 4.78 6.62 12.62

Q5: How to calculate theQ5: How to calculate the contribution of individual QTL?Q

20

Contribution of a QTL onContribution of a QTL on phenotypic variationp yp

PVE = Phenotypic variation explained (%)PVE Phenotypic variation explained (%)PVEg =Vg/Vp * 100%

BC, DH and RIL，Vg=a2 (a is the additive effect)F2, Vg=a2/2+d2/4 (d is the dominance effect), g / / ( )

21

D hi h ff t hi h PVE?Does high effect mean high PVE?

In DH or RIL, when there is segregation di idistortion,

Vg=(1-q)*a2+q*a2-[(1-2q)*a]2=4q(1-q)a2Vg (1 q) a +q a [(1 2q) a] 4q(1 q)a

Vg depends on effect and allele frequency

When p=q=0.5, Vg is maximized; otherwise, smaller than that of non-distortion

It is possible that one higher-effect QTL has lower PVEhas lower PVE

22

Non-additive PVE

For two random variables X and YE(X+Y) = E(X) + E(Y)V(X+Y) = V(X) + V(Y) +2Cov(X, Y)( ) ( ) ( ) ( , )

When QTL are unlinked, PVE of multiple QTL i th f i di id l PVEQTL is the sum of individual PVEWhen QTL are linked PVE of multipleWhen QTL are linked, PVE of multiple QTL is not equal to the sum of

d d lindividual PVE23

PVE can be more than 100%PVE can be more than 100%基因型 频率 基因型值

AABB )1(21 r− a1+a2

AAbb r1 a1-a2AAbb r2a1 a2

aaBB r21 -a1+a2

In the RIL population

aabb )1(21 r− -a1-a2

In the RIL populationGenetic variance of A-a: a1

2

Genetic variance of B-b: a22

Total genetic variance : a12 +a2

2+2(1-2r)a1a2ota ge et c a a ce a1 a2 ( )a1a2

24

A simulated populationA simulated population of 200 DH lines

30405060

 scoreTwo QTL are located

t 25 M d 36 M01020

LOD

1 2

at 25 cM and 36 cMon a chromsome of 120 M Th i dditi

‐0 40

0.40.81.2

Effect

120 cM. Their additive effects are 1.0 and 1 0 R d

‐1.2‐0.80.4E

100120

−1.0. Random error variance is 0.4. M k i t l i 2 M

20406080100

PVE (%

)Marker interval is 2 cM.

25

0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

Scanning every 1 cM

Linkage in couplingA: a1=1, a2=1, r=0.5; V1=1, V2=1, Vg=2, Ve=0.4; H2=0.83 B: a1=1, a2=1, r=0.1; V1=1, V2=1, Vg=3.6, Ve=0.4; H2=0.90Estimated R2 from regression = 0.78 Estimated R2 from regression = 0.82

60 60

1020304050

LOD score

1020304050

LOD score

0

00.40.81.2

fect

0

00.40.81.2

fect

‐1.2‐0.8‐0.4

0

Eff

100120

‐1.2‐0.8‐0.4

0

Eff

100120

020406080100

PVE (%)

020406080100

PVE (%)

26

0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

Scanning every 1 cM

0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

Scanning every 1 cM

Linkage in repulsionC: a1=1, a2=-1, r=0.5; V1=1, V2=1, Vg=2, Ve=0.4; H2=0.83 D: a1=1, a2=-1, r=0.1; V1=1, V2=1, Vg=0.4, Ve=0.4; H2=0.50Estimated R2 from regression = 0.89 Estimated R2 from regression = 0.51

60 60

102030405060

LOD score

102030405060

LOD score

0

00.40.81.2

ect

0

00.40.81.2

ect

‐1.2‐0.8‐0.4

0

Effe

100120

‐1.2‐0.8‐0.4

0

Effe

100120

020406080100

PVE (%)

020406080100

PVE (%)

27

0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

Scanning every 1 cM

0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

Scanning every 1 cM

Q6: How to determine the Q6 o o dsource of favorable alleles?

28

Source of favorable allelesSource of favorable alleles

Definition of additive and dominanceDefinition of additive and dominance genetic effects

Coding in QTL mapping: 2 (P1) 0 (P2) 1 (F1)Coding in QTL mapping: 2 (P1), 0 (P2), 1 (F1)P1: m+a; F1: m+d; P2: m-a

基因型 A2A2 A1A2 A1A1

中亲值m基因型值 G22 G12 G11

加性效应a 加性效应a

29显性效应d

Q7: Is selective genotypingQ7: Is selective genotyping still useful?still useful?

30

Selective genotypingSelective genotyping

Low tail High tail

LLHH

LH

ppppppt

)1()1( −+

−−

=

31L

LL

H

HH

Npp

Npp

2)(

2)(+

Comparison of SGM with IM and ICIMp(PVE=5%, MD=5cM and both tails have the

selected proportion of 10%)p p )

SGM has higher detection power than theSGM has higher detection power than the conventional IM but lower detection power

than ICIMthan ICIMSGM may still be useful!

Q8: Can composite traits be Q8 o pos sused in QTL mapping?

33

Genetic effects of composite traitsGenetic effects of composite traitsEffect Trait I Trait II Addition Subtraction Multiplication DivisionEffect Trait I Trait II Addition Subtraction Multiplication DivisionMean 25 20 45 5 500 1.2563A1 1 0 1 1 20 0.0503A 1 0 1 1 20 0 0503A2 1 0 1 1 20 0.0503A3 0 1 1 -1 25 -0.0631A4 0 1 1 -1 25 -0.0631A 0 0 0 0 0 0A12 0 0 0 0 0 0A13 0 0 0 0 1 -0.0025A14 0 0 0 0 1 -0.0025A23 0 0 0 0 1 -0.0025A24 0 0 0 0 1 -0.0025A34 0 0 0 0 0 0.006334A123 0 0 0 0 0 0A124 0 0 0 0 0 0A134 0 0 0 0 0 0.0003

34

A134 0 0 0 0 0 0.0003A234 0 0 0 0 0 0.0003A1234 0 0 0 0 0 0

Composite traits reduced powerComposite traits reduced power and increased FDRQTL Trait I Trait II Addition Subtraction Multiplication Division

Model I Power (%) Q1 95.10 69.60 69.30 55.20 50.50Q2 94 80 69 80 70 40 54 10 50 90Q2 94.80 69.80 70.40 54.10 50.90Q3 92.50 67.20 65.30 76.90 75.20Q4 94.50 68.40 65.40 77.80 75.20

(%)FDR (%) 21.63 22.98 27.42 28.05 28.07 29.68Model II Power (%) Q1 95.40 67.40 65.60 54.80 49.90

Q2 92.90 62.40 66.00 50.00 49.90 Q3 93.70 69.90 67.00 79.20 74.90 Q4 91.90 62.40 64.90 73.50 72.90

FDR (%) 21.35 22.18 28.76 28.59 28.07 28.89( )Model III Power (%) Q1 95.20 66.60 52.40 53.60 37.70

Q2 95.00 69.20 51.60 54.70 36.40 Q3 92 90 63 40 47 80 69 70 56 20

35

Q3 92.90 63.40 47.80 69.70 56.20 Q4 92.60 61.50 49.90 72.60 58.00

FDR (%) 19.78 23.44 28.83 27.71 29.74 30.18

Q9: Does the phenotype haveQ9: Does the phenotype have to be normally distributed?

36

Quantitative traits are normally distributed under the polygene hypothesis F2

h 2 0 90 h 2 0 10B1

h 2 0 00B2

h 2 0 95 h 2 0 05F2:3

2 2RIL

h 2 0 88 h 2 0 12

h mg2=0.90, h pg

2=0.10 h mg2=0.00

h pg2=1.00

h mg2=0.95, h pg

2=0.05 h mg2=0.89, h pg

2=0.11 h mg2=0.88, h pg

2=0.12

F2h mg

2=0.80, h pg2=0.10

B1h mg

2=0.002

B2h mg

2=0.87, h pg2=0.05

F2:3h mg

2=0.822

RILh mg

2=0.84, h pg2=0.13

h pg2=0.38 h pg

2=0.12

F2h mg

2=0.70, h pg2=0.10

B1h mg

2=0.00h pg

2=0.23

B2h mg

2=0.78, h pg2=0.05

F2:3h mg

2=0.74h pg

2=0.12

RILh mg

2=0.78, h pg2=0.13

Phenotypic di t ib ti

F2h mg

2=0.60, h pg2=0.10

B1h mg

2=0.00h pg

2=0.17

B2h mg

2=0.69, h pg2=0.05

F2:3h mg

2=0.66h pg

2=0.12

RILh mg

2=0.72, h pg2=0.14

distribution under one major

F2h mg

2=0.50, h pg2=0.10

B1h mg

2=0.00h pg

2=0.13

B2h mg

2=0.59, h pg2=0.05

F2:3h mg

2=0.57h pg

2=0.13

RILh mg

2=0.65, h pg2=0.16

gene model

F2h mg

2=0.40, h pg2=0.10

B1h mg

2=0.00h pg

2=0.11

B2h mg

2=0.49, h pg2=0.05

F2:3h mg

2=0.47h pg

2=0.13

RILh mg

2=0.57, h pg2=0.17Random errors

have to beF2

h mg2=0.30, h pg

2=0.10B1

h mg2=0.00

h pg2=0.09

B2h mg

2=0.38h pg

2=0.06

F2:3h mg

2=0.37h pg

2=0.14

RILh mg

2=0.47, h pg2=0.19

have to be normally di t ib t d d

37

F2h mg

2=0.20h pg

2=0.10

B1h mg

2=0.00h pg

2=0.08

B2h mg

2=0.26h pg

2=0.06

F2:3h mg

2=0.26h pg

2=0.15

RILh mg

2=0.34h pg

2=0.21

distributed and independent!

One QTL with PVE = 80%One QTL with PVE = 80%Located at 25 cM and a=1.0

40

50

cy

0

10

20

30Freq

uen

0

8.5 9 9.5 10 10.5 11 11.5 12Phenotypic value

1.2100

0.40.60.81

mated

 effect

ICIMIM

20

40

60

80

LOD score

ICIMIM

‐0.20

0.2

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

120

125

130

135

140

145

150

155

160

Estim

1D‐scanning on one chromosome, step=1cM

0

20

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

105

110

115

120

125

130

135

140

145

150

155

160

1D‐scanning on one chromosome, step=1cM

38

Q10: Can the precision be Q pimproved by adding more

k ?markers?

39

Empirical marker density

In linkage mappingg pp g10-20 cM, covering the whole genomeM k d it l l tiMarker density + large population

In linkage mappingIn linkage mappingThe more, the better to exploit the

i i LD i th i l tiremaining LD in the mapping population

40

Power and FDR for two marker d i i 10 M ( ) d 20 M (d )densities: 10 cM (up), and 20 cM (down)

(Confidence interval is the whole chromosome)

60708090

100

er

20406080100120 8

10

12

14

overy rate

1020304050

Powe 140

160180200220240260 0

2

4

6

False disco

0

1 2 3 4 5 10 20 30Phenotypic variation explained (PVE) (%)

260280300320340

0

20 80 140

200

260

320

380

440

500

560

Population size

60708090

100

r

20406080100120 8

10

12

14

very rate

102030405060

Powe 140

160180200220240260 0

2

4

6

False discov

41

010

1 2 3 4 5 10 20 30Phenotypic variation explained (PVE) (%)

260280300320340

020 80 140

200

260

320

380

440

500

560

Population size

Power and FDR for two marker densities: 10 cM (up), and 20 cM (down)

(10 cM confidence interval, true QTL at the center of CI)

708090

100 20406080100 60

708090

ery rate

2030405060

Power

120140160180200220240 10

20304050

False discove

010

1 2 3 4 5 10 20 30Phenotypic variation explained (PVE) (%)

240260280300320340

0

20 80 140

200

260

320

380

440

500

560

Population size

708090

100 20406080100 60

708090

ry rate

203040506070

Power

100120140160180200220240 10

2030405060

False discover

4201020

1 2 3 4 5 10 20 30Phenotypic variation explained (PVE) (%)

240260280300320340

010

20 80 140

200

260

320

380

440

500

560

Population size

Q11: What is effect of missing markers?

43

Mi i d t i QTL iMissing data in QTL mapping

Missing markersss g a e sImputation using the linkage map

Missing phenotypelMean replacement

DeletionDeletion

44

Effect of missing markersec o ss g a e s(First simulated F2 population from QTL distribution

model I and population size 500)

No missing markers 5% of missing

10% of missing 15% of missing

45

Power analysis of various ylevels of missing markers

A

80

100

%)

0 5% 10% 15% 20% 25% 30% C

80

100

)

20

40

60

Po

wer

(%

20

40

60

Po

wer

(%)

B

0qPH1-1qPH1-2qPH3-1qPH3-2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

100 D

0qHD1 qHD3 qHD4 qHD5 qHD8 qHD11 FDR

100

40

60

80

Po

wer

(%

)

40

60

80

100

ow

er

(%)

C

0

20

qPH1-1qPH1-2qPH3-1qPH3-2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

P

0

20

40

qHD1 qHD3 qHD4 qHD5 qHD8 qHD11 FDR

Po

A,QTL for plant height, population size=180B,QTL for plant height, population size=500

C, QTL for heading days, population size=180D, QTL for heading days, population size=500

Effect of missing markers is similar gto the reduction in population size

Population size 180, with various missing marker levels 100100

40

60

80 180 171 162 153 144 135 126

40

60

80

ower (%

)

0% 5% 10% 15% 20% 25% 30%

0

20

qPH1‐1qPH1‐2qPH3‐1qPH3‐2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

0

20

qPH1‐1 qPH1‐2 qPH3‐1 qPH3‐2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

Po

Population size 500, with various missing marker levels

80

100 500 475 450 425 400 375 350

80

100

%)

0% 5% 10% 15% 20% 25% 30%

20

40

60

20

40

60

Power (%

47

0

qPH1‐1qPH1‐2qPH3‐1qPH3‐2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

0

qPH1‐1 qPH1‐2 qPH3‐1 qPH3‐2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

Q12: What is the effect of Q s osegregation distortion?

48

Segregation distortion

P1 (AA) X P2 (aa)无选择

P1BC1 AA A 1 1P1BC1: AA:Aa=1:1P2BC1: Aa:aa=1:1F2: AA:Aa:aa=1:2:1DH RIL: AA:aa=1:1DH, RIL: AA:aa=1:1

Reasons for distortion Random driftSelection in gametes and zygotes g yg

49

Ratio of AA:aa in a barleyRatio of AA:aa in a barley DH population

1 0

1.2

1.4

1.6

1.8

0.0

0.2

0.4

0.6

0.8

1.0

0.0

Act8

A

aHor

2

ABG

464

iP

gd2

A

tpbA

A

BC26

1

Aga

7

MW

G84

4

WG

516

M

WG

520A

B

CD

351B

BC

D11

1

MW

G86

5

MW

G88

2

ABG

317

AB

G61

3

ABC

171

AB

G47

1

Ugp

1

MW

G84

7

MW

G20

40

ABC

166

M

WG

838

A

BC17

2

WG

622

C

DO

669

M

WG

716

AB

G71

5

MW

G65

5C

ABG

601

AB

A306

B

MW

G50

2

MW

G92

0A

Dor

5

BC

D41

0B

ksuA

1B

mSr

h

MW

G91

4

RZ

404

C

DO

504

Tu

bA3

BC

D29

8

ABG

463

A

BC30

9

PSR

167

AB

G37

8

PSR

106

M

WG

916

BC

D10

2

MW

G82

0

Am

y1

ABG

711

M

WG

658

dR

pg1

BC

D12

9

Prx1

A

ABG

380

Br

z

MW

G51

1

VAt

p57A

M

WG

889B

P

SR12

9

ABG

608

M

WG

635B

50

Segregation distortion in anSegregation distortion in an actual rice F2 population

RP12920

RP178

RM159RM147

91

16

20

(-lo

g 10P

)

RM143

4 RM

304 R

M49

8

12

nce

leve

l

RM

44 R RM552

4

8

sign

ifica

n

0

1 1 1 1 2 2 2 2 3 3 3 3 4 4 5 5 5 5 6 6 6 6 7 7 8 8 9 910 10 11 11 11 12 12

Markers on the 12 rice chromosomesMarkers on the 12 rice chromosomes

When segregation distortion markers are not linked with QTL

AA

80

100

%)

No distortion SDM1 SDM2 SDM3SDM4 SDM5 SDM6 SDM7SDM8 SDM9A,QTL for

l h i h

20

40

60Po

wer

(%plant height, population i 180

B

0

20

qPH1-1 qPH1-2 qPH3-1 qPH3-2 qPH4 qPH5 qPH6 qPH7 qPH12 FDR

size=180

B QTL f B

80

100B,QTL for heading d

40

60

Pow

er (%

)days, population i 180

0

20

qHD1 qHD3 qHD4 qHD5 qHD8 qHD11 FDR

Psize=180

When segregation distortion g gmarkers are linked with QTL

A

80

100 No distortion SDM1 SDM2 SDM3SDM4 SDM5 SDM6 SDM7SDM8 SDM9

C

80

100

)

20

40

60

Pow

er (%

)

20

40

60

Pow

er (%

)

B

0qPH1-1 qPH1-2 qPH3-1 qPH3-2 qPH4 qPH5 qPH6 qPH7 qPH12

80

100 D

0qHD1 qHD3 qHD4 qHD5 qHD8 qHD11

100

120

40

60

80

Po

wer

(%)

40

60

80

Pow

er (%

)

0

20

qPH1-1 qPH1-2 qPH3-1 qPH3-2 qPH4 qPH5 qPH6 qPH7 qPH12

0

20

qHD1 qHD3 qHD4 qHD5 qHD8 qHD11

A,QTL for plant height, population size=180B,QTL for plant height, population size=500

C, QTL for heading days, population size=180D, QTL for heading days, population size=500

Effect of segregation distortionEffect of segregation distortion markers (SDM) on QTL mapping

If the SDM is not closely linked with any l t h i ht h di d t QTLplant height or heading date QTL, no

significant effects were observed on the detection power.Otherwise SDM may increase or decreaseOtherwise, SDM may increase or decrease the QTL detection power.

l l fIn large-size populations, say size of 500, the effect of SDM was minor even the SDM was closely linked with QTL.

奇异分离对作图功效的影响奇异分离对作图功效的影响（F2群体）

2211021

220202 )()(2])([ dffadfffaffffVG −+−−−−+= 110210202 )()(2])([ dffadfffaffffVG ++

a b dc

k=1k 1

r=0Prob(k>1) ≈ 47%

r=0.5Prob(k>1) ≈ 51%

r=0.75 Prob(k>1) ≈ 52%

r=1 Prob(k>1) ≈ 50%

k=1k=1k=1f qq

k=1 k=1

k=1

k=1

k=1( ) ( ) ( ) ( )

fe g hfQQ fQQfQQ fQQ

r=1.5 r=2 r=5 r=10

f qq

k=1k=1 k=1

k=1

r 1.5 Prob(k>1) ≈ 37%

r 2 Prob(k>1) ≈ 29%

r 5Prob(k>1) ≈ 19%

r 10 Prob(k>1) ≈ 14%

fQQ fQQfQQ fQQ

k=1

In F1 and BC derived DHIn F1 and BC derived DH populations

1 2

1.4 

1.0 

1.2 

ce (k)

0.6 

0.8 

of varianc

0.4 

Ratio o

BC1‐derived DH population

0.0 

0.2 F1‐derived DH population

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

0.65

0.70

0.75

0.80

0.85

0.90

0.95

1.00

Frequency of QQ