Family-based association analysis is a converging point of both. It can be viewed from both linkage analysis's perspective and linkage disequilibrium anlysis' perspective.

From linkage's perspective: it enhances the linkage signal by knowing which phase is more likely.

[COMMENT: there is a well known result that if a pedigree has only one (affected) offspring, this pedigree could not provide any linkage signal. Why, because of the problem of unknown phase (is the disease allele on the same haplotype with marker allale "a" or marker allele "A"?), both phases are assigned a 50% probability in the "LOD" calculation.

On the other hand, the most welll-known design for family-based association contains only one affected offspring. This pedigree potentially could contribute to the linkage signal because two phases are assigned different probabilities. ]

From association's perspective: it deals the problem of population heterogeneity/stratification.

If a family-based association result is significant, both ancestral and current recombination contribute to the signal.
Without transmission bias in the current recombination (i.e., no linkage), there would be no significant result in family-based association analysis
Similarly, without a linkage disequilibrium between the marker allele and the disease allele (hitchhike), there would be no significant result in family-based association analysis
Both linkage and linkage disequilibrium ("AND") are required!

So in some sense, if a family-based association analysis on the trio data and a population-based association analysis lead to different conclusions, do not think there is something wrong!

• Considering both parents: the transmitted genotype and the untransmitted genotype
• Focusing on a particular allele. A genotype is 1 if it contains this allele, and 0 if it does not (dominant, because two copies of this allele is equivalent to 1 copy of the allele)
• Two parents contribute four allele samples (haplotype-based HRR)
• Explicitly admitting the pairing between one transmitted and one untransmitted allele. Two parents contribute two pairs. If we admit the pairing, then 0-0 pair and 1-1 pair do not contribute. In other words, parents with homozygous genotype (e.g. aa, AA) do not contribute to the signal .

Example: suppose we have (total 50 parents, or 100 alleles)
10 parents whose genotype is homozygous aa
20 parents whose genotype is homozygous AA
15 parents transmit A, left out a
5 parents transmit a, left out A

If we do not admit the pairing, there are 35 "A" alleles and 15 "a" alleles in the transmitted ("case") group, and 25 "A" and 25 "a" alleles in the untransmitted ("control") group. Using chi-square test, p-value=0.0412, odd-ratio= 2.333 (95%CI is (1.03, 5.30)).

If we admit the pairing, all homozygous parents are thrown away. We only look at these two numbers:

McNemar test calculates: TDT = (5-15)²/(5+15)= 100/20=5. TDT should follow the chi-square distribution with one degree of freedom. The p-value for TDT=5 is p-value=0.025347. (1- pchisq(5, df=1) in "R" )

The name TDT (transmission disequilibrium test, something like "linkage and linkage disequilibrium test") is a re-invention of the McNemar test. The two are exactly identical.

In general, pairing/matching reduces the sample size, thus reduce the power of the test

Three alleles

The simplicity and easy-to-use feature of TDT in parent(s)-affected-offspring trios/pairs is destroyed if there are other pedigree members are available. There are two schools of opinions in dealing with this situation.

1. Correcting the simplest TDT

see: TG Schulze, FJ McMahon (2002), "Genetic association mapping at the crossroads: which test and why? overview and practical guidelines", American Journal of Medical Genetics, 114:1-11. PDF

2. Using linkage program but adding linkage disequilibrium as an extra parameter

If the simplicity of TDT is lost when other pedigree members are included, why do we still use TDT? Why not use the linkage analysis program (since TDT is, after all, to detect BOTH linkage and linkage disequilibrium)?

see: HHH Goring, JD Terwilliger (2000), "Linkage analysis in the presence of errors IV: joint pseudomarker analysis of linkage and/or linkage disequilibrium on a mixture of pedigree and singletons when the mode of inheritance cannot be accurately specified", American Journal of Human Genetics, 66:1310-1327. PDF

Interactions

From Hwang et al (1995), "Association study of transforming growth factor alpha (TGF alpha) TaqI polymorphism and oral clefts: indication of gene-environment interaction in a population-based sample of infants with birth defects", American Journal of Epidemiology, 141:629-636.

ODd-ratio due to having the risk allele (not exposed to environmental risk factor, i.e. non-smoking): OR_g = (7*167)/(36*34) = 0.955.
Odd-ratio due to the exposure to environmental risk factor (but do not carry the risk allele): OR_e = (13*167)/(36*69)=0.874.
Odd-ratio due to both risk allele and risk environmental factor: OR_ge = (13*167)/(36*11)=5.48. with 95% CI (2.27, 13.21)

2-by-2-by-2 table is reduced to 2-by-2 table:

Odd-ratio= 5.14, 95% CI (1.68, 15.71).

Since TDT for one environment has only two independent counts: the number of parents that transmits A but not a, and the number of parents that transmits a but not A, for two environmental conditions, it is a 2-by-2 table:

Further reading: Q Yang, MJ Khoury (1997), "Evolving methods in genetic epidemiology. III. Gene-environment interaction in epidemiologic research", Epidemiology Review, 19:33-43.

Other Topics Not Covered

• Admixture mapping
• Tagging SNP and its selection
• DNA pooling
• Quantitative traits: no way to distinguish "cases" and "controls", individuals (for case-control) or trios (for TDT) are grouped according to genotypes (for case-control) or transmitted/untransmitted genotypes (for TDT). Then whether the quantitative trait is different in these groups can be tested by ANOVA (analysis of variance).