Why the TDT Resists Population Stratification

Why Population Stratification Can Confound an Association

Population stratification occurs when a study sample contains ancestry groups with different allele frequencies and different disease frequencies. A case–control comparison can then show an allele-frequency difference even when, within each ancestry group, the allele is not associated with disease. The difference can arise because cases and controls contain different proportions of the groups.

A family-based transmission test asks a different question. Rather than comparing allele frequencies across unrelated cases and controls, the transmission disequilibrium test (TDT) compares alleles transmitted to an affected child with alleles that could have been transmitted by that child's parents. Its protection against population stratification comes from conditioning on the parental genotypes: ancestry can affect which parental genotypes are observed, but, under the null and the Mendelian assumptions, it does not make one allele more likely than another to be transmitted by a heterozygous parent.

Definition: For a biallelic marker with alleles \(A\) and \(a\), an \(A/a\) parent is informative for the TDT because either allele could be transmitted. A transmission is counted as transmitted if \(A\) is passed to the affected child and as untransmitted if \(a\) is passed. In the conventional notation, \(b\) is the number of informative transmissions of \(A\), and \(c\) is the number of informative transmissions of \(a\) from heterozygous parents. The TDT tests whether \(b\) and \(c\) differ.

Worked Example: How Ancestry Can Create a Case–Control Difference

Suppose, hypothetically, that two ancestry groups have the same allele frequencies among cases and controls within each group: allele \(A\) has frequency 0.60 in group 1 and 0.10 in group 2. Assume disease is more common in group 2, so cases are 20% from group 1 and 80% from group 2, while controls are 80% from group 1 and 20% from group 2. Calculate the overall allele frequencies.

For cases, the weighted frequency is

$$ 0.20(0.60)+0.80(0.10)=0.12+0.08=0.20. $$

For controls, it is

$$ 0.80(0.60)+0.20(0.10)=0.48+0.02=0.50. $$

The case group therefore has an \(A\)-allele frequency of 0.20, while the control group has a frequency of 0.50. This large marginal difference does not require an association within either group: the assumed within-group case and control frequencies are identical. Here it arises because the groups have different allele frequencies and different ancestry proportions. A comparison that does not account for the group composition could mistake this pattern for an association with disease.

Conditioning on the Parents

For a heterozygous \(A/a\) parent, Mendelian segregation gives each allele probability one-half of being transmitted to a child. The parent's ancestry may affect the probability that the parent is \(A/a\), and thus may affect how often that parent contributes an informative transmission. It does not, under ordinary Mendelian inheritance, change the conditional probability of transmitting \(A\) rather than \(a\) once the parent is known to be \(A/a\).

The null relevant to the TDT is that, conditional on the parental genotypes, an allele at the marker is not preferentially transmitted to affected offspring. In the simple biallelic setting, this gives a probability of one-half for each allele from an informative heterozygous parent. If a marker is linked to a disease-related locus, transmission to affected offspring can be preferential even without a population-level association; that is a signal the TDT may detect. The stratification argument concerns the null transmission distribution, not a claim that the TDT only detects population-level association.

Key result: Under the null of no preferential transmission, and assuming Mendelian segregation, the probability of transmitting \(A\) from an \(A/a\) parent is \(1/2\), conditional on that parental genotype. The distribution of parental genotypes may differ across ancestry groups without changing this conditional probability.

This conditional statement explains the protection. Let \(Z\) denote ancestry and \(G\) the parental genotypes. Even if ancestry changes the distribution of \(G\), under the stated null and assumptions,

P(\text{transmit }A\mid G=A/a,\ Z,\ H_0)=\frac{1}{2}.

Consequently, the probability remains one-half after averaging over ancestry among parents with that genotype:

P(\text{transmit }A\mid G=A/a,\ H_0) =\sum_z P(\text{transmit }A\mid G=A/a,Z=z,H_0) P(Z=z\mid G=A/a,H_0) =\frac{1}{2}.

The ancestry weights in this sum can vary; each term being averaged is still one-half. In contrast, an unrelated case–control comparison averages allele frequencies over the ancestry composition of each group, and those compositions can differ.

The TDT Null Distribution

In the simplest setting, suppose there is one affected child per independent family and each informative heterozygous-parent transmission is counted once. Under the null, each such transmission has probability one-half of transmitting \(A\). Conditional on the number \(n=b+c\) of informative transmissions, the transmitted-\(A\) count follows a binomial distribution:

b\mid (b+c=n),H_0\sim \operatorname{Binomial}\left(n,\frac{1}{2}\right).

It follows that the expected count of transmissions of each allele is \(n/2\). For sufficiently large counts, the standard TDT statistic has an approximate chi-square distribution with one degree of freedom under the null:

X^2_{\mathrm{TDT}}=\frac{(b-c)^2}{b+c}\ \approx\ \chi^2_1.

The denominator is the total number of informative transmissions, and the numerator measures the squared difference between the two transmission counts. Exact calibration can be used instead when counts are small; the conditional binomial model supplies the null distribution. The key protection is not the particular approximation: it is that the conditional null probability remains one-half despite ancestry-related differences in parental genotypes.

Worked Example: A TDT With Different Ancestry Groups

Consider hypothetical data from independent trios, each with one affected child. In ancestry group 1 there are 40 informative transmissions: \(A\) is transmitted 24 times and \(a\) 16 times. In group 2 there are 160 informative transmissions: \(A\) is transmitted 84 times and \(a\) 76 times. The groups may have different parental genotype distributions; calculate the pooled TDT statistic.

The pooled counts are \(b=24+84=108\) transmissions of \(A\), and \(c=16+76=92\) transmissions of \(a\). Thus \(b+c=200\), and

X^2_{\mathrm{TDT}}=\frac{(108-92)^2}{108+92} =\frac{16^2}{200} =\frac{256}{200} =1.28.

Against a chi-square distribution with one degree of freedom, this gives an approximate two-sided \(p\)-value of 0.258. At a significance level of 0.05, these data do not provide statistically significant evidence of preferential transmission of \(A\). This result is about the transmission imbalance in these families; it does not establish that the allele has no effect. Most importantly for stratification, pooling transmissions from groups with different parental-genotype distributions does not by itself change the null expectation of 100 transmissions of each allele.

What the Protection Requires

The TDT is resistant to the specific confounding mechanism in which ancestry affects both allele frequencies and the proportions of cases and controls. This is not an unconditional guarantee against bias. The argument depends on a valid conditional transmission model: once parental genotypes are fixed, the null probability of transmission must be the Mendelian probability used by the test, and the affected-child ascertainment must not create a transmission imbalance under the null.

For the standard calculation above, the main assumptions include:

Correct family and genotype information. Parentage and marker genotypes are sufficiently accurate to identify heterozygous parents and their transmitted alleles.
Mendelian segregation under the null. A heterozygous parent transmits either allele with probability one-half; there is no segregation distortion that favors one allele.
No null transmission imbalance induced by ascertainment. Conditional on parental genotypes, the process that identifies affected children does not favor one marker allele under the null being tested.
Appropriate handling of dependence. The simple binomial and chi-square calculations assume independent informative transmissions. Related transmissions, such as multiple affected siblings from one family, require a method or variance calculation that accounts for their dependence.
A suitable test model. The biallelic count comparison assumes that transmissions are correctly classified and that the null being tested is the equal-transmission null for the informative parents.

These conditions clarify the scope of the protection. Conditioning removes the influence of different ancestry-specific parental genotype frequencies from the null transmission probability. It does not repair genotyping errors, non-Mendelian transmission, dependence treated as independence, or a design that induces preferential transmission under the null. Nor does it mean that ancestry is irrelevant to every aspect of a family study; the result is specifically about the conditional transmission comparison.

Worked Example: A Transmission Imbalance From Segregation Distortion

Suppose a hypothetical sample contains 120 informative transmissions from heterozygous parents, with \(A\) transmitted 72 times and \(a\) transmitted 48 times. The conventional statistic is

X^2_{\mathrm{TDT}}=\frac{(72-48)^2}{72+48} =\frac{24^2}{120} =\frac{576}{120} =4.80.

Using the usual chi-square approximation with one degree of freedom gives \(p\approx 0.0285\). If the assumptions of the test hold, this would be evidence against equal transmission. Now suppose, however, that the hypothetical marker has segregation distortion and \(A\) is transmitted more often even when it is unrelated to the disease process. The equal-transmission null is then false for reasons unrelated to disease association, so the nominal \(p\)-value does not have its intended interpretation. Conditioning on parental genotypes protects against ancestry-related allele-frequency differences; it does not protect against a genuine violation of the one-half transmission assumption.

Common Mistake and Practical Interpretation

A common mistake is to say that the TDT is protected because “families have the same ancestry.” Families in a study need not have the same ancestry, and parental genotype frequencies can differ substantially across ancestry groups. The correct reasoning is conditional: for each heterozygous parent, the null transmission probability is one-half given the parental genotype. Different frequencies of parental genotypes across groups do not alter that probability.

Another mistake is to treat a non-significant TDT as proof that population stratification or all other sources of bias are absent. A non-significant result means the observed transmission imbalance is not sufficiently inconsistent with the specified null at the chosen significance level. It is not proof of no association, and it does not validate assumptions that may have been violated. The defensible conclusion is narrower: when the conditional transmission assumptions hold, ancestry differences that would confound an unadjusted case–control allele-frequency comparison do not, by themselves, create a TDT transmission imbalance.

Key takeaway: Population ancestry can change which parental genotypes appear in a sample and can create a marginal case–control allele-frequency difference. Under the TDT null, conditioning on each heterozygous parent's genotype leaves the chance of transmitting either allele at one-half. That conditional null—not identical ancestry across families—is the basis for resistance to population stratification.

Check Your Understanding

Use the conditional transmission argument, not an assumption that all families share the same ancestry.

Why can different ancestry proportions among cases and controls produce an allele-frequency difference when there is no within-group association?
Under the TDT null, what is the probability that an \(A/a\) parent transmits \(A\), conditional on that parental genotype?
Across 150 independent informative transmissions, what are the null expected counts of transmissions of \(A\) and \(a\)?
In the usual TDT notation, what does \(b+c\) represent in the statistic \((b-c)^2/(b+c)\)?
Name one assumption whose failure could invalidate the equal-transmission calibration, even if population stratification is present.