Correcting for a potential bias in the pedigree disequilibrium test.

To the Editor: Recently, we proposed the pedigree disequilibrium test (PDT) as a test for allelic association and linkage (linkage disequilibrium) in general pedigrees (Martin et al. 2000). We have discovered that, in certain cases in extended pedigrees, the PDT can be biased under the null hypothesis. In this letter we describe the nature of the bias and illustrate a model in which the bias arises. We offer two alternative modifications to the PDT statistic, both of which result in valid tests of linkage disequilibrium over all genetic models and family structures. In constructing the PDT statistic, we considered two types of families within a pedigree. Informative nuclear families are those in which there is at least one affected individual, with both parents genotyped at the marker and at least one parent heterozygous. Informative discordant sibships have at least one affected and one unaffected sibling with different marker genotypes. For a marker locus with two alleles, M 1 and M 2 , we defined the random variables X Tj , for the jth triad (affected individual and both parents) in the pedigree, and X Sj , for the jth discordant sib pair (DSP) in the pedigree: X Tj p (no. of M 1 transmitted) Ϫ (no. of M 1 not transmitted) and X Sj p (no. of M 1 in affected sib) Ϫ (no. of M 1 in unaffected sib), respectively. In our previous study (Mar-tin et al. 2000), we defined a measure of association (D) for a pedigree containing n T triads from informative nuclear families and n S DSPs from informative discordant sibships: n n T S 1 D p X ϩ X. (1) ͸ ͸ [ ] Tj Sj n ϩ n jp1 jp1 T S Let D i be the measure of association in the ith pedigree in a sample of N independent pedigrees. The PDT statistic is given by. The critical as-N N 2 ͱ T p S D / S D jp1 i j p1 i sumption is that the T is asymptotically normal, with mean 0 and variance 1, under the null hypothesis of no linkage disequilibrium. The difficulty that can be encountered is that, for some cases under the null hypothesis, the expected value of T may actually be different from 0, a situation that results in an inflated type I error. This is best illustrated by an example. Consider a fully …