1. 1 QTL mapping in mice Lecture 10, Statistics 246 February 24, 2004

2. 2 The mouse as a model Same genes? The genes involved in a phenotype in the mouse may also be involved in similar phenotypes in the human. Similar complexity? The complexity of the etiology underlying a mouse phenotype provides some indication of the complexity of similar human phenotypes. Transfer of statistical methods. The statistical methods developed for gene mapping in the mouse serve as a basis for similar methods applicable in direct human studies.

3. 3 Backcross experiment

4. 4 F 2 intercross experiment

5. 5 F 2 intercross: another view

6. 6 Quantitative traits (phenotypes) 133 females from our earlier (NOD  B6)  (NOD  B6) cross Trait 4 is the log count of a particular white blood cell type.

7. 7 Another representation of a trait distribution Note the equivalent of dominance in our trait distributions.

8. 8 A second example Note the approximate additivity in our trait distributions here.

9. 9 Trait distributions: a classical view In general we seek a difference in the phenotype distributions of the parental strains before we think seeking genes associated with a trait is worthwhile. But even if there is little difference, there may be many such genes. Our trait 4 is a case like this.

10. 10 Data and goals Data Phenotypes: y i = trait value for mouse i Genotype: x ij = 1/0 of mouse i is A/H at marker j (backcross); need two dummy variables for intercross Genetic map: Locations of markers Goals Identify the (or at least one) genomic region, called quantitative trait locus = QTL, that contributes to variation in the trait Form confidence intervals for the QTL location Estimate QTL effects

11. 11 Genetic map from our NOD B6 intercross

12. 12 Genotype data

13. 13 Models: Recombination We assume no chromatid or crossover interference.  points of exchange (crossovers) along chromosomes are distributed as a Poisson process, rate 1 in genetic distancce  the marker genotypes {x ij } form a Markov chain along the chromosome for a backcross; what do they form in an F 2 intercross?

14. 14 Models: Genotype  Phenotype Let y = phenotype, g = whole genome genotype Imagine a small number of QTL with genotypes g 1,…., g p (2 p or 3 p distinct genotypes for BC, IC resp). We assume E(y|g) =  (g 1,…g p ), var(y|g) =  2 (g 1,…g p )

15. 15 Models: Genotype  Phenotype, ctd Homoscedacity (constant variance)   2 (g 1,…g p )  =  2  (constant) Normality of residual variation y|g ~ N(  g,  2  ) Additivity:  (g 1,…g p ) =  + ∑  j g j (g j = 0/1 for BC) Epistasis: Any deviations from additivity.

16. 16 Additivity, or non-additivity (BC)

17. 17 Additivity or non-additivity: F 2

18. 18 The simplest method: ANOVA Split mice into groups according to genotype at a marker Do a t-test/ANOVA Repeat for each marker Adjust for multiplicity LOD score = log 10 likelihood ratio, comparing single-QTL model to the “no QTL anywhere” model.

19. 19 Exercise 1. Explain what happens when one compares trait values of individuals with the A and H genotypes in a backcross (a standard 2-sample comparison), when a QTL contributing to the trait is located at a map distance d (and recombination fraction r) away from the marker. 2. Can the location of a QTL as in 1 be estimated, along with the magnitude of the difference of the means for the two genotypes at the QTL? Explain fully.

20. 20 Interval mapping (IM) Lander & Botstein (1989) Take account of missing genotype data (uses the HMM) Interpolates between markers Maximum likelihood under a mixture model

21. 21 Interval mapping, cont Imagine that there is a single QTL, at position z between two (flanking) markers Let q i = genotype of mouse i at the QTL, and assume y i | q i ~ Normal(  qi,  2 ) We won’t know q i, but we can calculate p ig = Pr(q i = g | marker data) Then, y i, given the marker data, follows a mixture of normal distributions, with known mixing proportions (the p ig ). Use an EM algorithm to get MLEs of  = (  A,  H,  B,  ). Measure the evidence for a QTL via the LOD score, which is the log 10 likelihood ratio comparing the hypothesis of a single QTL at position z to the hypothesis of no QTL anywhere.

22. 22 Exercises 1. Suppose that two markers M l and M r are separated by map distance d, and that the locus z is a distance d l from M l and d r from M r. a) Derive the relationship between the three recombination fractions connecting M l, M r and z corresponding to d l + d r = d. b) Calculate the (conditional) probabilities p ig defined on the previous page for a BC (two g, four combinations of flanking genotypes), and an F 2 (three g, nine combinations of flanking genotype). 2. Outline the mixture model appropriate for the BC distribution of a QT governed by a single QTL at the locus z as in 1 above.

23. 23 LOD score curves

24. 24 LOD curves for Chr 9 and 11 for trait4

25. 25 LOD thresholds To account for the genome-wide search, compare the observed LOD scores to the distribution of the maximum LOD score, genome-wide, that would be obtained if there were no QTL anywhere. LOD threshold = 95th %ile of the distribution of genome-wide maxLOD,, when there are no QTL anywhere Derivations: Analytical calculations (Lander & Botstein, 1989) Simulations Permutation tests (Churchill & Doerge, 1994).

26. 26 Permutation distribution for trait4

27. 27 Epistasis for trait4

28. 28 Acknowledgement Karl Broman, Johns Hopkins