1. Thoughts about the TDT

2. Contribution of TDT: Finding Genes for 3 Complex Diseases PPAR-gamma in Type 2 diabetes Altshuler et al. Nat Genet 26:76-80, 2000 NOD2 in Crohn’s Disease Hugot et al., Nature 411: 599-603, 2001 ADAM33 in asthma Van Eerdewegh et al., Nature 418: 426-430, 2002

3. The common PPAR-gamma Pro12Ala polymorphism is associated with decreased risk of type 2 diabetes Altshuler et al. Nat Genet 26:76-80, 2000 *

4. NOD2 Variants and Susceptibility to Crohn’s Disease Hugot et al., Nature 411: 599-603, 2001 SNP13: p=6x10 -6 Chrom 16q

5. Van Eerdewegh et al., Nature 418: 426-430, 2002 ADAM33 Gene: Asthma and Bronchial Hyperresponsiveness P= 3x10 -6 to 0.04 Chrom 20p

7. Population distributions of (a) disease given genotype, and (b) genotype given disease. Affected GenotypeYesNoGenotypeYesNo M1M1M1M1 a1 – a M1M1M1M1 dg M1M2M1M2 b1 – b M1M2M1M2 eh M2M2M2M2 c1 – c M2M2M2M2 fi (a)(b)

8. Clayton Odds Ratio He calls this the relative risk. Confusing! Ott

9. DM θ D1M1D2M2D1M1D2M2

10. Null hypothesis: θ = ½ (Disease and marker loci unlinked) Alternative hypothesis: θ < ½ (Disease and marker loci linked)

11. freq (D 1 M 1 ) ≠ freq (D 1 ) × freq (M 1 ) δ = freq (D 1 M 1 ) – freq (D 1 ) × freq (M 1 )

12. We assume that we observe the marker locus genotypes, either M 1 M 1, M 1 M 2, or M 2 M 2, of both parents and the affected sibs in all families in the data.

13. Probabilities for transmitted and non-transmitted marker alleles M 1 and M 2 from any parent of an affected child. Non-transmitted allele Transmitted Allele M 1 M 2 Total M 1 P(11)P(12) P(1.) M 2 P(21)P(22) P(2.) TotalP(.1)P(.2) 1

14. P(11) = q 2 + q δ / p P(12) = q (1 – q) + (1 – θ – q) δ / p P(21) = q (1 – q) + (θ – q) δ / p P(22) = (1 – q) 2 – (1 – q) δ / p

15. Numbers of transmitted and non-transmitted marker alleles M 1 and M 2 among the parents of the affected sibs Non-transmitted allele Transmitted AlleleM 1 M 2 M 1 n 11 n 12 M 2 n 21 n 22 Put n 12 + n 21 = n

16. Only P(12) and P(21) depend on θ. Also, when θ = ½, P(12)=P(21) So the “natural” (TDT) test statistic is This (McNemar statistic) has an asymptotic 1 df χ 2 distribution when the null hypothesis is true.

17. Note that this statistic depends only on n 12 and n 21 only, and ignores n 11 and n 22. This makes sense: the statistic uses data only from M 1 M 2 parents, and only these are informative for linkage. We call these ‘informative” parents. So at the end of the day we consider only transmissions from informative parents.

18. We will focus entirely on the denominator, n, of the TDT statistic. It is remarkable how many questions one can ask about this. But before we ask these, we first ask, where does this denominator come from?

19. Assuming the null hypothesis is true, n 12 has a binomial (n, ½) distribution. Note: this is true even if the data contain several affected children from the same family. Thus the variance of n 12 - n 21 (= 2n 12 – n) is 4n/4 = n.

20. We will examine three situations, all focusing on the question: “Is n the correct (variance) denominator for the situation at hand?”.

21. Situation 1. Testing for association. Here the null hypothesis is “no association”, or The problem here is that transmissions to different affected sibs in the same family are not independent under this null hypothesis. Thus when there are several families in the data with more than one affected sib, n 12 does not have a binomial distribution.

22. If H 0, δ =0, is true, the cell probabilities for the simple random-mating case are P(11) = q 2,P(12) = q(1 – q), P(21) = q(1 – q), P(22) = (1 – q) 2 (Thus should we not be testing this H 0 by using both n 11 n 22 – n 12 n 21 andn 12 – n 21 and a 2 degrees of freedom test ?) Let’s ignore this point for now.

23. P(11) = ( Σ i α i ( p i 2 q i 2 + δ i p i q i ) ) / (Σ i α i p i 2 ) P(12) = ( Σ i α i ( p i 2 q i (1 – q i ) + δ i p i (1 – θ – q i ) ) ) / (Σ i α i p i 2 ) P(21) = ( Σ i α i ( p i 2 q i (1 – q i ) + δ i p i (θ – q i ) ) ) / (Σ i α i p i 2 ) P(22) = ( Σ i α i ( p i 2 (1 – q i ) 2 – δ i p i (1 – q i ) ) ) / (Σ i α i p i 2 ) α i = relative size of subpopulation i δ i = linkage disequilibrium in subpopulation i p i = frequency of D 1 in subpopulation i q i = frequency of M 1 in subpopulation i

24. Suppose that in family j, M 1 is transmitted n 12j times, M 2 is transmitted n 21j times, from M 1 M 2 parents. Define D j as n 12j – n 21j The test statistic is n 12 – n 21

25. Suppose that there is only one affected child in each family. Then D j = ±1 (for all j) T T 2 = TDT χ 2

26. Situation 2. Suppose we have families in the data where both parents are dead, (so we do not know their marker locus genotypes), but where there are two affected sibs, one being M 1 M 1, the other M 2 M 2. We therefore can infer that both parents were informative. Should we use the data from these families in the analysis, using the standard TDT statistic?

27. The answer is “no”. Why is this so? Because the very fact that we can infer the parental genotypes unambiguously means that one sib MUST be M 1 M 1 and the other MUST be M 1 M 1. In such families there is zero variance, rather than some binomial variance, for the number of M 1 genes in the two sibs.

28. Philosophical question: is there any difference between the actions you take in directly observing an event and having unambiguous evidence that the event occurred? In this case, “yes there is”.

29. Situation 3. Suppose that we have two affected sibs, one informative (i.e. M1M2) parent, in each family in the data. Numbers of transmission from the informative parents 2M12M1 1M 1, 1M 2 2M22M2 Total # familiesijkn H 0 means n/4n/4 n/2n/2 n/4n/4 n

33. Suppose that a sharing Χ 2 has been carried out, correctly, as a one-sided test. Given i + k = s, what is the distribution of Χ 2 TDT ?

34. ,

35. One affected sib, two informative (M 1 M 2 ) parents Genotype of affected child M 1 M 1 M 2 M2M2M2M2 Total # families pqrn Expected when H 0 (θ=½) n/4n/4 n/2n/2 n/4n/4 n

37. Gametic DisequilibriumΔ 2 = Δ 1 (1–θ) Gametic DisequilibriumΔ 3 = Δ 2 (1–θ) Gametic DisequilibriumΔ 1 Parents of generation 1 mate only within their subpopulation Parents of generation 2 mate at random throughout population Parents of generation 3 mate at random throughout population Subpopulation 12……i……k Relative Sizeα 1 α 2 ……α i ……α k Coefficient of gametic Disequilibriumδ 1 δ 2 ……δ i ……δ k Generation 0 Generation 1 Generation 2 Generation 3

38. Generation 1 Gametic Disequilibrium Δ 1 Generation 2 Gametic Disequilibrium Δ 2 Generation 3 Gametic Disequilibrium Δ 3 Generation 4, etc Generation 0

39. The value of the TDT statistic in two models 1. Immediate admixture Generation 11.48 Generation 22.07 Generation 315.34 Generation 412.43 2. Gradual admixture Generation 11.48 Generation 22.07 Generation 38.53 Generation 46.99