Mapping multiple QTL in experimental crosses

Transcription

1 Mapping multiple QTL in experimental crosses Karl W Broman Department of Biostatistics and Medical Informatics University of Wisconsin Madison [ Teaching Miscellaneous lectures]

2 Example Sugiyama et al. Genomics 71:70-77, male mice from the backcross (A B) B Blood pressure after two weeks drinking water with 1% NaCl Blood pressure 2

3 Genetic map Location (cm) X Chromosome 3

4 Genotype data X 200 Individuals Markers 4

5 Goals Identify quantitative trait loci (QTL) (and interactions among QTL) Interval estimates of QTL location Estimated QTL effects 5

6 LOD curves 8 6 LOD score 4 5% Chromosome 6

7 Estimated effects Chr 48 cm Chr 30 cm Chr 24 cm Chr 20 cm blood pressure BB BA BB BA BB BA BB BA Genotype Genotype Genotype Genotype 7

8 Modeling multiple QTL Reduce residual variation increased power Separate linked QTL Identify interactions among QTL (epistasis) 8

9 Epistasis in BC Additive Epistatic H H Ave. phenotype A QTL 2 Ave. phenotype A QTL A H A H QTL 1 QTL 1 9

10 Epistasis in F 2 Additive Epistatic B B 80 H QTL 2 80 QTL 2 Ave. phenotype A Ave. phenotype H A A H B QTL 1 A H B QTL 1 10

11 2-dim, 2-QTL scan For all pairs of positions, fit the following models: H f : y = µ + β 1 q 1 + β 2 q 2 + γq 1 q 2 + ɛ H a : y = µ + β 1 q 1 + β 2 q 2 + +ɛ H 1 : y = µ + β 1 q 1 + ɛ H 0 : y = µ + ɛ log 10 likelihoods: l f (s, t) l a (s, t) l 1 (s) l 0 11

12 2-dim, 2-QTL scan LOD scores: LOD f (s, t) = l f (s, t) l 0 LOD a (s, t) = l a (s, t) l 0 LOD i (s, t) = l f (s, t) l a (s, t) LOD 1 (s) = l 1 (s) l 0 12

13 Results: LOD i and LOD f Chromosome Chromosome 13

14 Results: LOD i and LOD f Chromosome Chromosome 14

15 Summaries Consider each pair of chromosomes, (j, k), and let c(s) denote the chromosome for position s. M f (j, k) = M a (j, k) = M 1 (j, k) = max LOD f(s, t) c(s)=j,c(t)=k max LOD a(s, t) c(s)=j,c(t)=k max LOD 1(s) c(s)=j or k M i (j, k) = M f (j, k) M a (j, k) M fv1 (j, k) = M f (j, k) M 1 (j, k) M av1 (j, k) = M a (j, k) M 1 (j, k) 15

16 Results: LOD i and LOD fv Chromosome Chromosome 16

17 Thresholds A pair of chromosomes (j, k) is considered interesting if: M f (j, k) > T f and { M fv1 (j, k) > T fv1 or M i (j, k) > T i } or M a (j, k) > T a and M av1 (j, k) > T av1 where the thresholds (T f, T fv1, T i, T a, T av1 ) are determined by a permutation test with a 2d scan 17

18 2d scan summary pos1f pos2f lod.full lod.fv1 lod.int c1:c c6:c c1:c pos1a pos2a lod.add lod.av1 c1:c c6:c c1:c

19 Estimated effects 1 x 4 6 x Chr 1 genotype 110 Chr 6 genotype BB BA BB BA blood pressure BB BA BB BA Chr 4 genotype Chr 15 Genotype 19

20 Chr 1: LOD i and LOD av Position (cm) Position (cm)

21 Hypothesis testing? In the past, QTL mapping has been regarded as a task of hypothesis testing. Is this a QTL? Much of the focus has been on adjusting for test multiplicity. It is better to view the problem as one of model selection. What set of QTL are well supported? Is there evidence for QTL-QTL interactions? Model = a defined set of QTL and QTL-QTL interactions (and possibly covariates and QTL-covariate interactions). 21

22 Model selection Class of models Additive models + pairwise interactions + higher-order interactions Model comparison Estimated prediction error AIC, BIC, penalized likelihood Bayes Model fit Maximum likelihood Haley-Knott regression extended Haley-Knott Multiple imputation Markov chain Monte Carlo Model search Forward selection Backward elimination Stepwise selection Randomized algorithms 22

23 Target Selection of a model includes two types of errors: Miss important terms (QTLs or interactions) Include extraneous terms Unlike in hypothesis testing, we can make both errors at the same time. Identify as many correct terms as possible, while controlling the rate of inclusion of extraneous terms. 23

24 What is special here? Goal: identify the major players A continuum of ordinal-valued covariates (the genetic loci) Association among the covariates Loci on different chromosomes are independent Along chromosome, a very simple (and known) correlation structure 24

25 Exploratory methods Condition on a large-effect QTL Reduce residual variation Conditional LOD score: { } Pr(data q1, q 2 ) LOD(q 2 q 1 ) = log 10 Pr(data q 1 ) Piece together the putative QTL from the 1d and 2d scans Omit loci that no longer look interesting (drop-one-at-a-time analysis) Study potential interactions among the identified loci Scan for additional loci (perhaps allowing interactions), conditional on these 25

26 Controlling for chr 4 8 Interval mapping Control for chr 4 6 LOD score Chromosome 26

27 Drop-one-QTL table df LOD %var :

28 Automation Assist inexperienced analysts Understand performance Many phenotypes 28

29 Additive QTL Simple situation: Dense markers Complete genotype data No epistasis y = µ + β j q j + ɛ which β j 0? plod(γ) = LOD(γ) T γ 29

30 Additive QTL Simple situation: Dense markers Complete genotype data No epistasis y = µ + β j q j + ɛ which β j 0? plod(γ) = LOD(γ) T γ 0 vs 1 QTL: plod( ) = 0 plod({λ}) = LOD(λ) T 29

31 Additive QTL Simple situation: Dense markers Complete genotype data No epistasis y = µ + β j q j + ɛ which β j 0? plod(γ) = LOD(γ) T γ For the mouse genome: T = 2.69 (BC) or 3.52 (F 2 ) 29

32 Experience Controls rate of inclusion of extraneous terms Forward selection over-selects Forward selection followed by backward elimination works as well as MCMC Need to define performance criteria Need large-scale simulations Broman & Speed, JRSS B 64: ,

33 Epistasis y = µ + β j q j + γ jk q j q k + ɛ plod(γ) = LOD(γ) T m γ m T i γ i T m = as chosen previously T i =? 31

34 Idea 1 Imagine there are two additive QTL and consider a 2d, 2-QTL scan. T i = 95th percentile of the distribution of max LOD f (s, t) max LOD a (s, t) 32

35 Idea 1 Imagine there are two additive QTL and consider a 2d, 2-QTL scan. T i = 95th percentile of the distribution of max LOD f (s, t) max LOD a (s, t) For the mouse genome: T m = 2.69 (BC) or 3.52 (F 2 ) T H i = 2.62 (BC) or 4.28 (F 2 ) 32

36 Idea 2 Imagine there is one QTL and consider a 2d, 2-QTL scan. T m + T i = 95th percentile of the distribution of max LOD f (s, t) max LOD 1 (s) 33

37 Idea 2 Imagine there is one QTL and consider a 2d, 2-QTL scan. T m + T i = 95th percentile of the distribution of max LOD f (s, t) max LOD 1 (s) For the mouse genome: T m = 2.69 (BC) or 3.52 (F 2 ) T H i = 2.62 (BC) or 4.28 (F 2 ) T L i = 1.19 (BC) or 2.69 (F 2 ) 33

38 Models as graphs A C B D 34

39 Results 1 4 LOD =

40 Results LOD = T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

41 Results T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

42 Profile LOD curves Profile LOD score 8 6 1@68.3 6@ @ Chromosome 36

43 Drop-one-QTL table df LOD %var :

44 Add an interaction? T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

45 Add an interaction? T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

46 Add an interaction? T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

47 Add an interaction? T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

48 Add an interaction? T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

49 Add another QTL? b T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

50 Add another QTL? T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

51 Add another QTL? b 6 15 T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

52 Add a pair of QTL? 1 4 3a b 6 15 T m = 2.69 T i H = 2.62 T i L = 1.19 T m + T i H = 5.31 T m + T i L = T m =

53 Summary QTL mapping is a model selection problem The criterion for comparing models is most important We re focusing on a penalized likelihood method, with penalties derived from permutation tests with 1d and 2d scans Broman and Speed, J Roy Stat Soc B 64: , 2002 Manichaikul et al., Genetics 181: ,