What the TDT Does—and What It Assumes
The transmission disequilibrium test (TDT) compares, among affected children, how often a heterozygous parent transmits one allele with how often that parent transmits the other. Its within-family comparison avoids a particular problem: differences in allele frequencies between population groups can create an apparent association in an ordinary case-control comparison. That protection is valuable, but it is not a guarantee against every source of bias.
For a biallelic marker, call the allele being tested \(A\). Among informative parental transmissions, let \(b\) be the number in which \(A\) was transmitted to an affected child and \(c\) the number in which \(A\) was not transmitted. A parent is informative when their genotype makes it possible to distinguish whether \(A\) or the other allele was passed on. The usual TDT statistic is
Under the standard null model, and with enough informative transmissions for the asymptotic approximation to be suitable, this statistic is compared with a chi-square distribution with one degree of freedom. A small p-value indicates that the observed transmission imbalance would be unusual under that null model. It does not establish that the marker itself causes disease.
These conditions clarify what a TDT result means. The test evaluates transmission within the observed families, conditional on the parental genotypes. It does not automatically correct errors in those genotypes, departures from Mendelian transmission, biased selection into the analysis, or an analysis that treats dependent observations as independent.
Why the Standard Calibration Works
The conditional null gives a simple reference distribution. For each informative transmission, define an indicator that equals 1 if \(A\) is transmitted and 0 otherwise. If there are \(m=b+c\) independent informative transmissions and each has probability \(1/2\) of transmitting \(A\), then \(b\) has a binomial distribution with parameters \(m\) and \(1/2\). Consequently, \(E(b)=m/2\) and \(\operatorname{Var}(b)=m/4\). Since \(c=m-b\), the signed difference is \(b-c=2b-m\).
Thus, when the normal approximation is adequate, \((b-c)/\sqrt{m}\) is approximately standard normal. Squaring it gives the usual one-degree-of-freedom chi-square statistic:
The calculation depends on the stated model. If the actual probability of transmitting \(A\) is not \(1/2\), or if the recorded transmissions do not behave like the observations used to obtain this variance, the reference distribution may no longer calibrate the p-value correctly.
Worked Example: A Mendelian Transmission Imbalance
Suppose a hypothetical study records \(b=120\) transmissions of \(A\) and \(c=80\) non-transmissions of \(A\) from informative heterozygous parents. Assume that genotypes and phenotypes are accurate, families were included without regard to the transmission outcome, and the transmissions can be treated as independent for this calculation.
There are \(m=120+80=200\) informative transmissions. The statistic is
Against a chi-square distribution with one degree of freedom, this gives an approximate two-sided p-value of \(0.00468\). Under the stated assumptions, the data provide evidence against equal transmission of the two alleles to affected children. The result alone does not show that \(A\) causes disease; the imbalance could reflect association with a causal variant, or a failure of an assumption used to interpret the test.
Errors and Biological Departures Can Mimic Association
A transmission count is only as reliable as the genotypes used to construct it. A genotyping error can turn a heterozygous parent into an apparent homozygote, misidentify which allele was transmitted, or create an apparent transmission that is inconsistent with the family genotypes. If errors affect the two alleles or transmission categories differently, they can create an imbalance even when the true transmissions are balanced. Other error patterns may instead blur a genuine imbalance and reduce power.
Worked Example: An Allele-Specific Genotyping Error
Imagine 400 true informative transmissions in a hypothetical sample: 200 transmissions of \(A\) and 200 non-transmissions. Suppose 30 of the non-transmissions are mistakenly recorded as transmissions of \(A\), while the remaining calls are correct. The observed counts are therefore \(b=230\) and \(c=170\).
The observed statistic is
The one-degree-of-freedom chi-square approximation gives \(p\approx 0.00270\). That small value is calculated from the recorded counts, but it is not evidence that the true transmissions were imbalanced: by construction, the true counts were equal. The error mechanism shifted the observed counts. A TDT p-value cannot diagnose or remove that problem by itself.
The \(1/2\) transmission probability is also a biological assumption, not merely a convenient arithmetic choice. For example, a marker allele could be transmitted more often because of meiotic drive or another form of transmission distortion. If that distortion occurs independently of disease risk, it may generate an apparent TDT imbalance without disease association. A significant TDT therefore needs interpretation in light of possible departures from ordinary Mendelian transmission.
Phenotype Classification and Selection
The TDT is commonly applied to affected children, so the definition of “affected” matters. If phenotype misclassification is unrelated to the marker and transmission, it can dilute a real association by mixing affected and unaffected children into the case group. It does not necessarily create a systematic imbalance under a genuine null. But if the chance of being labelled affected depends on genotype, allele knowledge, or a factor associated with transmission, the selected affected group can have an imbalance that is not due to a true disease association.
Worked Example: Genotype-Dependent Phenotype Misclassification
Consider a hypothetical set of 200 truly affected children, with 100 having inherited \(A\) and 100 having not inherited \(A\). Suppose the marker is not associated with the true disease status, but the diagnostic process is more likely to label children who inherited \(A\) as affected: the probability of receiving that label is \(0.80\) for those children and \(0.20\) for those who did not inherit \(A\).
The expected numbers included as affected are \(100(0.80)=80\) after transmission of \(A\) and \(100(0.20)=20\) after non-transmission. If the resulting sample has those counts, the TDT statistic is
The approximate p-value is about \(2.0\times10^{-9}\), but the example’s imbalance arises from genotype-dependent classification, not from an association with true disease status. This is why accurate, consistently applied phenotype definitions are part of the inferential argument.
Ascertainment of families because they contain an affected child is not, by itself, the same as selecting on the transmitted allele. The concern is whether the probability of inclusion or successful analysis depends on the transmission outcome, or on variables that induce such a dependence. For example, retaining only families with a particular observed genotype pattern, or dropping families after seeing their transmission counts, changes the set of observations in a way the ordinary TDT calibration does not account for.
Dependence and the Unit of Analysis
The ordinary calculation treats the informative transmissions contributing to \(b\) and \(c\) as independent Bernoulli observations, or otherwise requires a variance calculation appropriate to their dependence. Family data can violate that simplification. Multiple affected siblings share parents and family circumstances; in some designs, their outcomes or selection into the sample are correlated. Counting each child as an independent observation when the analysis has not accounted for that dependence can make the reported standard error too small and the p-value too optimistic.
Worked Example: Counting the Same Trio Twice
Suppose a hypothetical data file contains 20 unique trios, with \(b=15\) and \(c=5\). A data-processing error duplicates every trio, so the analysis treats the file as 40 independent records, with \(b=30\) and \(c=10\).
Using the unique trios, the statistic is
Using the duplicated records as if they were independent gives
The duplicated analysis appears to provide stronger evidence because it has counted each observation twice without adding independent information. The example is deliberately simple: it shows why identifying unique families and choosing an analysis that reflects the actual dependence structure are essential to valid uncertainty estimates.
Interpreting a TDT Result Carefully
A TDT is useful because it compares transmissions within families rather than comparing unrelated people across potentially different ancestry backgrounds. But that strength addresses population stratification; it does not protect against the other problems discussed here. In particular, the test does not establish that the tested marker is causal. A marker can be associated with disease because it is correlated with a causal variant, and biased measurement or selection can also produce an apparent association.
When reviewing a trio-based result, ask whether the parental and child genotypes were measured reliably, whether the transmission mechanism plausibly follows the Mendelian null, how affected status was defined, and how families entered the analysis. Then check whether repeated or related observations were handled in a way that gives a valid variance and p-value. A statistically significant result answers a conditional question under a model; it cannot repair a mismatch between that model and the data.
- Genotyping error: can alter the observed transmission counts or which transmissions are considered informative.
- Transmission distortion: can violate the null probability of one-half even without a disease association.
- Phenotype misclassification: may dilute an association, or create one if classification depends on genotype or transmission.
- Selection: can bias the comparison if inclusion depends on the transmission outcome or related information.
- Nonindependence: can make a standard variance too small when repeated or correlated observations are treated as independent.
Check Your Understanding
Use the assumptions and examples above to assess each claim.
- In the standard TDT null, what probability does an informative heterozygous parent have of transmitting either allele?
- Why can allele-specific genotyping error create an apparent transmission imbalance even if the true transmission counts are equal?
- How can phenotype misclassification create a false TDT signal, and why does nondifferential misclassification have a different interpretation?
- Why might selecting families based on observed transmission outcomes invalidate the ordinary TDT p-value?
- What is wrong with treating duplicated trio records as independent observations?
- Does a significant TDT establish that the tested marker causes the disease? Explain.