Why Ascertainment Matters
Family-based association studies often recruit families through an affected child or another relative with a particular phenotype. The resulting sample is deliberately unlike a random sample of families: affected people and their relatives are overrepresented. That fact does not automatically invalidate a transmission test. The key question is whether selection changes the distribution of the genetic transmissions used by the test under its null hypothesis.
For a standard transmission test, the reference distribution is conditional on parental genotypes. Under the relevant null, a heterozygous parent transmits either allele with probability one-half. Recruitment based only on phenotypes can preserve that conditional distribution if, under the null, those phenotypes do not depend on the transmissions being tested. Recruitment that also depends on genotype, or on a feature correlated with genotype after conditioning, need not preserve it.
The Conditional Null and Phenotype-Based Recruitment
Let \(G_p\) denote the parental marker genotypes, \(G_t\) the offspring transmissions at that marker, \(Y\) the phenotypes used in analysis, and \(A\) the event that a family is included. Let \(C\) denote any relevant family characteristics that are conditioned on, such as design strata. A common conditional null for a transmission score says that, given parental genotypes and \(C\), the tested transmissions are independent of the phenotypes:
This statement is about the null model, not a claim that phenotypes and genotypes are independent under every possible genetic model. For example, if a marker is associated with the trait, the phenotype may depend on which allele was transmitted. The test is calibrated by asking what transmission patterns would be expected if the specified null held.
Suppose inclusion is determined by observed phenotypes and conditioned family characteristics, so that \(A\) is a function of \(Y\) and \(C\). Under the conditional null, conditioning on those phenotypes does not change the transmission distribution:
The equality follows because, under the null, the distribution of \(Y\) given \(G_p,C\) does not depend on \(G_t\), and \(A\) adds no information once \(Y,C\) are fixed. Thus a test can condition on the observed phenotypes, use the Mendelian transmission distribution given parental genotypes, and remain calibrated despite phenotype-based recruitment. This is the central reason affected-family ascertainment can be compatible with a conditional family test.
Worked Example: An Affected-Child TDT
Worked Example: An Affected-Child TDT
Suppose a hypothetical study recruits families through an affected child. At a biallelic marker, there are 12 informative transmissions from heterozygous parents to affected children: the allele being tested was transmitted 7 times and not transmitted 5 times. Assume parental genotypes are correctly known, transmissions follow Mendelian segregation, and the null is no marker transmission distortion associated with the phenotype.
Under this null, conditional on the 12 informative meioses, the number \(T\) of transmissions of the tested allele has the distribution \(T\sim\operatorname{Binomial}(12,0.5)\). The TDT score is \(T-U\), where \(U=12-T\):
The usual asymptotic statistic is therefore
Compared with a chi-square distribution with 1 degree of freedom, this gives an approximate two-sided \(p\)-value of \(0.564\). With only 12 transmissions, an exact binomial calculation is also straightforward. Using the two-sided tail that counts outcomes at least as far from 6 as the observed count 7,
Neither calculation provides evidence against the stated null. Crucially, the fact that all 12 children were recruited because they were affected does not, on its own, change the one-half transmission probability under this null. The conclusion relies on the assumptions above; if the tested marker affects the phenotype, that is departure from the null rather than a reason to use this null distribution.
Conditioning Phenotype-Weighted Family Scores
Family-based association tests can use phenotype information beyond a binary affected-child indicator. A useful way to understand their conditional validity is to consider a score built from transmission deviations and phenotype weights. Let \(X_i=1\) if child \(i\) receives the allele from a heterozygous parent and \(X_i=0\) otherwise. Conditional on the parental genotypes, \(E(X_i\mid G_p)=0.5\). If \(w_i\) is a fixed weight determined by the observed phenotype, a simple score is
When the phenotype weights are held fixed and the conditional null applies, each transmission deviation has mean zero. Consequently, \(E(S\mid G_p,Y)=0\). Its variance must reflect the family transmission structure; in this simple example with independent transmissions, it is \(\sum_i w_i^2/4\). More complicated pedigrees require the appropriate dependence and variance calculation rather than treating all transmissions as independent.
Worked Example: Conditioning on an Affected Sibling Pair
Consider a hypothetical family with two children and one heterozygous parent at the marker. One child is affected and the other is unaffected; the family was recruited because it contained an affected child. For illustration, use fixed phenotype weights \(w_1=0.5\) for the affected child and \(w_2=-0.5\) for the unaffected child. Suppose the affected child received the tested allele (\(X_1=1\)), while the unaffected child did not (\(X_2=0\)).
The observed score is
Under the conditional null, and conditional on the parental genotype and the two observed phenotypes, each transmission is still Bernoulli with probability \(0.5\). With independent transmissions in this example, the conditional variance is
For this very small example, use the exact distribution rather than relying on a large-sample approximation. Since \(S=0.5(X_1-X_2)\), it takes the values \(-0.5,0,0.5\) with probabilities \(0.25,0.5,0.25\). The two-sided conditional \(p\)-value for the observed \(S=0.5\) is therefore \(P(|S|\geq 0.5)=0.25+0.25=0.500\). This calculation conditions on the phenotypes that led to recruitment; it does not treat the selected children as a random population sample.
When Selection Breaks the Reference Distribution
Phenotype-dependent recruitment is not the same as genotype-dependent selection. If investigators include families because a child is affected, that inclusion rule is based on phenotype. Under the conditional null, the phenotype does not reveal which marker allele was transmitted after parental genotypes are fixed. By contrast, if inclusion depends directly on a transmission or on a genotype-related feature, selection can distort the very transmission distribution used by the test.
Worked Example: Selection on a Transmission
Suppose, only for illustration, that under a null model three informative transmissions are independent, each with probability \(0.5\) of transmitting the tested allele. Imagine that a study includes a family only when at least one of those transmissions is of the tested allele. This is genotype-dependent selection, not ordinary affected-relative ascertainment.
Before selection, \(T\sim\operatorname{Binomial}(3,0.5)\). The inclusion probability is \(P(T\geq1)=7/8\). Among selected families, the conditional probabilities for \(T=1,2,3\) are \(3/7,3/7,1/7\), respectively. The selected-sample expectation is
The unselected null expectation is \(3(0.5)=1.5\). Thus the usual null distribution \(T\sim\operatorname{Binomial}(3,0.5)\) is not the distribution among included families: selection has shifted the mean upward. A test that ignores this rule is not calibrated by the ordinary conditional reference distribution. It would need to account for the selection event, or use a design in which inclusion does not depend on the transmissions being tested.
Practical Conditions for Valid Conditional Inference
Before applying a family-based test to an ascertained sample, make the conditioning argument explicit. The following questions help distinguish a valid conditional analysis from one that needs additional modeling or a different design.
State whether the test concerns no transmission association, no linkage under its specified model, or a conditional mean null. These are related but not interchangeable statements.
For a transmission test, this commonly includes parental genotypes and the observed phenotype information used to form the score or define the family sample.
If recruitment is a function of the conditioned phenotypes and known design variables, it can preserve the null transmission distribution. Check whether genotype, transmission, or a genotype-related technical outcome also affected inclusion.
Correct genotypes, the relevant Mendelian transmission probabilities, and a variance calculation that respects dependence are needed. Violations can invalidate calibration even when ascertainment is phenotype-based.
If the sampling probability depends on genotype after conditioning on the information used by the test, the simple conditional null may not apply. Depending on the design, valid analysis may require conditioning on the selection event, modeling the sampling mechanism, or using an appropriate sampling-weighted method. A statement that a study recruited “through cases” is not enough to decide validity: what matters is exactly how families were chosen and what the null says about phenotype and transmission.
Common Mistake and Key Takeaway
A common mistake is to reason, “The sample contains mostly affected people, so a family-based transmission test must be biased,” or, at the other extreme, “It is a family-based test, so ascertainment can never matter.” Both claims are too broad. Under a valid conditional null, phenotype-based selection can leave Mendelian transmissions balanced after conditioning on parental genotypes and observed phenotypes. Selection that depends on tested genotypes or transmissions can instead alter that conditional distribution.
Conditional validity also does not mean that a selected sample represents the population. A transmission test may support a valid null test while estimates of population disease frequency, absolute risk, or effects in a broader population remain affected by the sampling design. Keep the inferential target and the sampling scheme distinct.
Check Your Understanding
For each question, identify the conditioning argument or assumption that determines the answer.
- A TDT study recruits only affected children. What null assumption allows the one-half transmission probability to remain valid after conditioning on parental genotypes?
- Why can a transmission score use observed phenotype-based weights without automatically invalidating its conditional null distribution?
- A clinic includes families only if a tested allele is observed in at least one child. Why is the ordinary binomial transmission reference distribution no longer appropriate without adjustment?
- Does conditional validity imply that an affected-family sample estimates population disease prevalence without bias? Explain.
- Name two assumptions, other than phenotype-based ascertainment, that are needed for valid transmission-test calibration.