Introduction to Transmission Disequilibrium Test Yixuan Chen4/25/08

Outlines Transmission Disequilibrium Test Fundamentals Description of the test Pros and Cons Haplotype-based TDT 2

Transmission Disequilibrium Test (TDT) Spielman RS, McGinnis RE, Ewens WJ. (1993) Transmission test for linkage disequilibrium: the insulin gene region and insulin-dependent diabetes mel itus (IDDM). Am J Hum Genet. 52(3):506-16. A family-based association test to test for the presence of genetic linkage between a genetic marker and a trait. 3

Background An association between disease and marker may be due to disequilibrium between linked loci. Association can occur in the absence of linkage For example, as a result of population stratification Thus it is not valid to use the presence of association as a test for linkage if population stratification is a possibility. 4

TDT Overview The TDT considers parents who are heterozygous for an al ele associated with disease and evaluates the frequency with which that al ele or its alternate is transmitted to affected offspring. The advantage It does not require data either on multiple affected family members or on unaffected sibs. The disadvantage It can detect linkage between the marker and the disease locus only if association (due to linkage disequilibrium) is present. 5

One Affected Child per Family Assume that a disease locus D, with disease al ele D , and a 1 normal al ele D2 a marker locus with codominant al eles M and 1 M . 2 Suppose that a sample of n such single-child families at the marker M, a total of 4n parental al eles 2n are transmitted and the other 2n are not transmitted 6

One Affected Child per Family (c1) 2n 2n 2n 1 2 Total (1,1) (1,2) Transmitted 1 1 2 Nontransmitted 2 0 2 (1,2) Total 3 1 4 7

Chi-square Test n 2 (O E ) 2 i i χ − = ￥ 1 E i= i Chi-square Test O = an observed frequency i E = an expected (theoretical) frequency, asserted by i the nul hypothesis n = the number of possible outcomes of each event. 4n(w-y)2/[(w+y)(4n-w-y)] 8

Validation of Statistic Ott J (1989) Statistical properties of the haplotype relative risk. Genet Epidemiol 6:127-130 (1,1) (1,2) (1,2) 1 2 Total 1 1 0 1 2 1 0 1 Total 2 0 2 9

Validation of Statistic (c1) δ = freq(M D − mp) 1 1 Only the data from heterozygous M M parents should be used in the test 1 2 10

Valid Test Statistic 11

Two Affected Children per Family 12

More than Two Affected Children in a Family 13

TDT Advantages The TDT may be much more sensitive than haplotype sharing tests, with the same data. The TDT uses simpler family data. The TDT is not affected by population stratification. 14

TDT Disadvantages The TDT will not detect linkage between disease and marker unless there is also population association (linkage disequilibrium). The TDT does not make use of all family information, especially of founders. 15

Extension to Haplotypes The simplest way is to treat haplotypes as multi-allelic markers. 16

FBAT FBAT implements a broad class of Family Based Association Tests, adjusted for population admixture. Horvath S, Xu X, Laird NM. 2001. The family based association test method: strategies for studying general genotype-phenotype associations. Eur J Hum Genet 9:301–306. Horvath S, Xu X, Lake SL, Silverman EK, Weiss ST, Laird NM. 2004. Family-based tests for associating haplotypes with general phenotype data: application to asthma genetics. Genet Epidemiol 26:61–69. 17

Haplotype-based FBAT Find the compatible haplotype configurations (mating types) for the families Compute the conditional probabilities of the families The statistic S = Y ￥ X( ij G ) ￥ i ij V = Var(S ) i i j T − U = { ￥ 1 S − E(S )} U V U i i i 18

Haplotype-based FBAT (c1) It is natural to estimate the weights w by the Gjk conditional probability of observing Gjk given that it is compatible with gj. FBAT uses an expectation-maximization (EM) algorithm to estimate haplotype frequencies from which the weights are calculated. 19

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