Why Compare Transmitted and Untransmitted Alleles?
A family-based association test asks whether an allele is transmitted to affected offspring more often than expected under Mendelian inheritance. In a trio, the genotypes of an affected child and both biological parents allow us to identify which allele each parent transmitted. A heterozygous parent provides the key comparison: the allele passed to the child versus the allele that parent did not pass.
The transmission disequilibrium test (TDT) summarizes this comparison for a biallelic marker. Its familiar statistic is not an arbitrary difference in counts: it is the score statistic for a conditional model of transmission to affected offspring. The derivation makes clear what the counts represent, why the null value is one-half, and which assumptions connect the family data to the score test.
A homozygous parent does not contribute an informative meiosis at this marker: whichever allele is passed, the parent’s transmitted and untransmitted alleles are the same. A trio can contribute zero, one, or two informative meioses. In particular, if both parents are heterozygous, each parent contributes one transmission to the counts.
Under Mendelian segregation, a heterozygous parent transmits each allele with probability one-half before conditioning on the child’s affected status. The TDT asks whether, among affected offspring, transmission of \(A\) is disproportionately common. The null hypothesis is no transmission association, so that conditioning on the parents and affected status does not favor either allele. The derivation below gives a model under which the conditional transmission probability has a simple form.
From Transmissions to the TDT Statistic
Suppose each affected child has both parents genotyped, and condition on the parental genotypes and the fact that the child is affected. Let \(g\) denote the child’s number of copies of \(A\). Use a multiplicative penetrance model: the probability of being affected for a child with \(g\) copies is proportional to \(\theta^g\), where \(\theta>0\) is the per-copy multiplicative factor. More precisely, write the penetrance as \(\kappa\theta^g\) over the possible genotypes in the model, with values in the interval from zero to one. The null of no marker-allele effect is \(\theta=1\).
For the conditional likelihood, treat the informative parental meioses as independent, and assume the multiplicative model applies to the allele copies inherited by the child. Any copies fixed by homozygous parents contribute a constant to \(g\); they do not affect the likelihood for the informative transmissions. Under Mendelian inheritance, each possible pattern of transmissions has probability \(2^{-M}\) before conditioning on affection. Conditioning on affection weights that pattern by \(\theta^T\), where \(T\) is the number of transmitted \(A\) alleles among the informative meioses. Summing over possible transmissions gives the normalizing factor \((1+\theta)^M\). Thus the conditional likelihood, up to factors that do not depend on \(\theta\), is:
Equivalently, each informative transmission to an affected child has conditional probability \(\theta/(1+\theta)\) of being \(A\), under this model. When \(\theta=1\), this probability is one-half. The log-likelihood in terms of \(\beta=\log(\theta)\) is \(T\beta-M\log(1+e^\beta)\), up to an additive constant. Differentiating gives the score, the derivative of the log-likelihood:
At the null \(\beta=0\), the score is \(T-M/2\). The information, the negative expected second derivative of the log-likelihood, is \(M q(1-q)\), where \(q=e^\beta/(1+e^\beta)\). At the null, this is \(M/4\). Substituting \(T=b\) and \(M=b+c\) therefore gives the score statistic:
This is a score test because it evaluates the slope of the log-likelihood at the null value, using the information at that same null value. It does not first estimate \(\theta\) under the alternative. The formula requires \(b+c>0\); if there are no informative parental transmissions, the score and information are both zero and the data provide no TDT statistic. The test’s null calibration is a separate question; the calculation here derives the statistic, not a p-value or a decision threshold.
Worked Example: Counting Informative Meioses in Six Hypothetical Trios
Suppose six hypothetical affected-child trios have the following parental genotypes and transmissions. In each row, only a heterozygous parent contributes a count.
| Trio | Parent 1 | Parent 2 | Informative transmissions | Contribution to \(b\) | Contribution to \(c\) |
|---|---|---|---|---|---|
| 1 | Aa transmits A | Aa transmits a | A, a | 1 | 1 |
| 2 | Aa transmits a | aa | a | 0 | 1 |
| 3 | aa | Aa transmits A | A | 1 | 0 |
| 4 | Aa transmits A | Aa transmits A | A, A | 2 | 0 |
| 5 | Aa transmits a | AA | a | 0 | 1 |
| 6 | Aa transmits a | Aa transmits a | a, a | 0 | 2 |
The homozygous parents in trios 2, 3, and 5 contribute no informative transmission. Adding the table entries gives \(b=4\), \(c=5\), and \(M=4+5=9\). The TDT statistic is:
The signed score is \((4-5)/2=-0.5\), indicating one fewer observed transmission of \(A\) than of \(a\). The statistic is nonnegative because it squares that score. This arithmetic describes the observed imbalance; interpreting its null evidence requires a calibration method.
The Score-Test Justification
The score statistic has the standard form “score squared divided by information.” To see why, the log-likelihood derivative measures the local change in fit as \(\beta\) moves away from zero. Under the null, the expected score is zero, and its variance is the information \(I(0)\). Dividing the squared score by that variance produces a standardized squared measure of departure from the null. In this model the ingredients are obtained directly from the conditional likelihood:
Among \(M\) informative transmissions to affected offspring, \(T=b\) are \(A\). The model gives \(L(\theta)\propto\theta^b/(1+\theta)^M\).
With \(\beta=\log(\theta)\), the score is \(b-M e^\beta/(1+e^\beta)\). At \(\beta=0\), it is \(b-M/2=(b-c)/2\).
The information is \(M q(1-q)\), with \(q=1/2\) under the null, so \(I(0)=M/4\).
Squaring the score and dividing by the information gives \(((b-c)/2)^2/(M/4)=(b-c)^2/(b+c)\).
This derivation also shows why both counts matter: the null score depends on their difference, while the information depends on their total. A large difference based on very few informative transmissions does not have the same standardized score as that difference based on many transmissions.
Worked Example: Deriving the Statistic From a Conditional Likelihood
Suppose a second hypothetical set of affected trios yields \(b=15\) transmissions of \(A\) and \(c=9\) transmissions of \(a\). Then \(M=24\), and the conditional likelihood as a function of \(\beta=\log(\theta)\) is proportional to:
where \(C\) is constant in \(\beta\). Differentiating gives \(U(\beta)=15-24e^\beta/(1+e^\beta)\), so at the null \(U(0)=15-24(1/2)=3\). The null information is \(I(0)=24(1/2)(1/2)=6\). Therefore:
The positive score points toward more transmission of \(A\) in this hypothetical sample. The agreement between the likelihood calculation and the count formula verifies the score-test interpretation.
Assumptions and Interpretation
The simple count formula does not remove the need to check how the data were generated. Its interpretation as the score test above relies on a clearly specified set of conditions:
- Correctly identified alleles and transmissions. Genotypes and parent-child relationships are recorded accurately, so the transmitted allele is assigned correctly.
- Mendelian inheritance under the null. Conditional on a heterozygous parent’s genotype, each allele has probability one-half of transmission in the absence of the modeled association.
- Affected-child ascertainment and the conditional model. The likelihood conditions on parental genotypes and affected status, and uses a multiplicative model in the number of \(A\) copies. The null is \(\theta=1\).
- Appropriate independence. The derivation treats informative meioses as independent under the model and assumes the analyzed trio contributions are independent. Related offspring or other clustering can require a different variance calculation.
The TDT tests a transmission imbalance at the marker; it does not by itself establish that the tested allele causes disease. An associated marker can also be informative because it is correlated with another genetic variant. The sign of \(b-c\) identifies which allele was transmitted more often, while the squared statistic measures the strength of imbalance without retaining that direction.
Worked Example: A Common Counting Pitfall
Suppose a third hypothetical data set has \(b=18\) and \(c=10\). A student uses \((b-c)/(b+c)=8/28\approx0.2857\) as the TDT statistic, arguing that this is the observed excess transmission as a fraction of all informative meioses.
That fraction does describe a difference relative to the total count, but it is not the TDT score statistic. The score is \((b-c)/2=(18-10)/2=4\), and the null information is \((b+c)/4=28/4=7\). Thus the score statistic is:
The correction is to standardize the score using its null information, not to divide the difference by the total. Also, the counts must include only informative heterozygous-parent transmissions. Counting an allele passed by a homozygous parent would add data that cannot distinguish transmission of one allele from the untransmitted allele.
Why the Derivation Matters
The count representation makes the TDT easy to calculate, but the score derivation explains what that calculation is doing. The null compares the observed number \(b\) of transmitted \(A\) alleles with the expected number \(M/2\), and the information \(M/4\) supplies the scale for that departure. Writing the statistic as \((b-c)^2/(b+c)\) is therefore useful only when the counts are defined from the relevant informative meioses and the conditional model is appropriate.
A frequent mistake is to treat every allele observed in an affected child as a separate informative transmission, or to read the squared statistic as showing which allele is favored. The first error ignores the parental comparison; the second loses the direction when squaring. Define \(b\) and \(c\) from heterozygous parents, check that \(b+c>0\), and report the signed difference alongside the statistic when direction matters.
Check Your Understanding
Use the definitions and derivation above to answer these questions.
- Why does a homozygous parent contribute no informative meiosis to the biallelic TDT?
- If \(b=12\) and \(c=8\), calculate \(M\), the signed score, the null information, and the TDT statistic.
- Under the multiplicative penetrance model, what is the conditional probability that an informative transmission to an affected child is \(A\) when \(\theta=1\)?
- Explain why the TDT statistic is nonnegative but does not, by itself, indicate whether \(A\) or \(a\) was transmitted more often.
- What does the score-test derivation assume about the relationship between the number of \(A\) copies and penetrance?