Multiallelic and Haplotype Transmission Tests

Why Extend Transmission Tests?

A biallelic marker offers a natural count: how many copies of one allele were transmitted to an affected child? At a marker with three or more alleles, the question is no longer a single count unless one allele or contrast has been chosen in advance. A haplotype creates a related extension: the unit being transmitted is a combination of alleles across loci on the same chromosome. Both settings require the same basic discipline: define what is counted, calculate its conditional expectation under Mendelian transmission, and compare the observed value with that expectation.

The conditional approach uses the parental genotypes to determine the possible transmissions and their probabilities under the null of no association. This tutorial develops the score for a multiallelic marker, shows how to test either all allele categories jointly or a chosen contrast, and then applies the same logic to haplotypes. When phase is uncertain, the transmission expectation must average over compatible phase configurations rather than silently treating one configuration as certain.

Definition: At a marker with \(K\) alleles, let \(X_i=(X_{i1},\ldots,X_{iK})^\mathsf{T}\) be the vector of allele counts in the genotype of offspring \(i\). Each \(X_{ik}\) is the number of copies of allele \(k\), so the counts sum to two for a diploid offspring. Given the observed parental genotypes, write \(E(X_i\mid S_i)\) for the expected count vector under Mendelian transmission, where \(S_i\) denotes the conditioning information used for that family. The centered transmission vector is \(X_i-E(X_i\mid S_i)\).

Multiallelic Scores and the Omnibus Test

For affected-offspring analyses, one can form a score by adding the centered allele-count vectors for the affected offspring. More generally, when a phenotype weight \(w_i\) is used, multiply each centered vector by that weight. For example, a quantitative-trait analysis may use a trait value centered at its null mean. In the examples below, each analyzed child is affected and has weight one; this makes the score a direct sum of centered transmitted allele counts.

U=\sum_i w_i\{X_i-E(X_i\mid S_i)\},\qquad V=\operatorname{Var}(U\mid S)

Here \(U\) is a \(K\)-component score vector and \(V\) is its conditional covariance matrix under the null. Under a correctly specified Mendelian transmission model, the conditional expectation of each centered vector is zero. Therefore \(E(U\mid S)=0\) under the null. The covariance must reflect the joint transmission possibilities: allele counts within a genotype are not independent, because each offspring has exactly two copies.

The omnibus test asks whether the allele-specific score vector departs from zero in any direction. When the null distribution is adequately approximated by a multivariate normal distribution, the quadratic statistic \(Q=U^\mathsf{T}V^{-}U\) is compared with a chi-square distribution. Here \(V^{-}\) is the inverse on the estimable score space (equivalently, a generalized inverse), and the degrees of freedom are the rank of \(V\). A singular covariance is expected: because the allele counts sum to two, their centered values sum to zero, so all \(K\) components cannot vary independently. The rank is at most \(K-1\), and can be lower if the observed parental genotypes do not permit all contrasts.

Interpretation: A significant omnibus result is evidence that the transmitted allele distribution differs from its conditional Mendelian expectation in at least one estimable direction. It does not identify the responsible allele or establish that every allele is associated. Follow-up contrasts should be specified and interpreted with attention to multiplicity and the study design.

Worked Example: A Three-Allele Marker

Suppose a hypothetical affected child has parents with genotypes A/B and A/C. Assume the two parents transmit one allele each, with each of their two alleles equally likely to be transmitted, and that the transmissions are independent. Use the allele-count order (A, B, C), and suppose the child’s observed genotype is B/C.

The possible offspring count vectors are (2, 0, 0), (1, 0, 1), (1, 1, 0), and (0, 1, 1), each with probability one quarter. Their average is

E(X\mid S)=\frac14\{(2,0,0)+(1,0,1)+(1,1,0)+(0,1,1)\} =(1,0.5,0.5).

The observed vector is \(X=(0,1,1)\). With affected-child weight one, its score contribution is

U=X-E(X\mid S)=(-1,0.5,0.5).

For these parental genotypes, the covariance of the count vector is

V= \begin{pmatrix} 0.5&-0.25&-0.25\\ -0.25&0.25&0\\ -0.25&0&0.25 \end{pmatrix}.

For example, allele A can be transmitted by either parent, each with probability one half, giving variance \(0.25+0.25=0.5\); the negative covariances with B and C arise because an allele transmitted by a parent excludes that parent’s alternative allele. The matrix has rank two, consistent with the sum-to-two constraint. The observed score is a valid conditional score contribution, but one trio is not enough to justify an asymptotic chi-square conclusion. A calibrated test would require an appropriate sample and variance calculation, or a conditional procedure based on the actual transmission distribution.

Testing a Prespecified Allele Contrast

An omnibus test is useful when no single allele is the primary target. If the scientific question names an allele or a contrast in advance, it can be more direct to project the vector score onto that contrast. Let \(c\) be a fixed vector of weights. The scalar score is \(U_c=c^\mathsf{T}U\), and its conditional variance is \(V_c=c^\mathsf{T}Vc\). For example, \(c\) can assign weight one to a candidate allele and zero to the others. A contrast comparing two alleles can assign weights one and negative one to those alleles.

Z_c=\frac{c^\mathsf{T}U}{\sqrt{c^\mathsf{T}Vc}}

When the denominator is positive and the conditional normal approximation is adequate, a two-sided test compares \(|Z_c|\) with standard normal reference values. Its sign indicates the direction of the score under the chosen coding: a positive score means that the weighted allele count was transmitted more often than its conditional expectation. The coding must be stated, because changing the sign of \(c\) reverses the sign of the score but leaves a two-sided test unchanged.

Worked Example: A Targeted Allele Score

Consider one hypothetical affected child whose parents are both heterozygous H1/H0 at a biallelic marker. Each parent transmits H1 with probability one half, independently. The child is observed to have two copies of H1. Test the prespecified H1-count score using the normal approximation only as an illustration.

The H1 count \(X\) is the sum of two independent Bernoulli variables with success probability 0.5. Thus \(E(X\mid S)=1\), \(\operatorname{Var}(X\mid S)=0.5\), and the observed score is \(U_c=2-1=1\). Substitution gives

Z_c=\frac{2-1}{\sqrt{0.5}}=1.414,\qquad Z_c^2=2.000.

A standard-normal two-sided approximation gives \(p\approx0.1573\), which is not significant at the 0.05 level. This is not reliable evidence about association: the example has only one trio, so the normal approximation is poor. In fact, under the conditional null the H1 count is 0, 1, or 2 with probabilities 0.25, 0.50, and 0.25. This small example shows why the score formula and the choice of null calibration are separate steps.

Haplotype Transmission Scores

A haplotype is the ordered combination of alleles carried together on one chromosome across specified loci. For two biallelic loci, for example, AC and GT are two possible haplotypes. In a diploid child, let \(X_{ih}\) count the copies of haplotype \(h\), and collect these counts into a vector \(X_i\). Conditional on parental haplotypes and phase, the child receives one haplotype from each parent according to the transmission probabilities. The multiallelic score construction applies directly: center the haplotype-count vector by its conditional expectation, then use either an omnibus test or a prespecified haplotype contrast.

A targeted test might use a score that counts copies of a candidate haplotype. Alternatively, a contrast can compare that haplotype with a defined group of alternatives. A haplotype score answers a question about the specified combination of alleles, not automatically about each component locus separately. The haplotype set and contrast should therefore match the scientific question and be defined without choosing them based on the observed transmission results.

Worked Example: A Candidate Haplotype

Suppose a hypothetical affected child’s parents each have known haplotypes H1/H0, and the parents transmit independently with probability one half for either haplotype. The child receives two copies of H1. Let \(X\) be the child’s H1 count and use the prespecified H1-count contrast.

The possible counts 0, 1, and 2 have probabilities 0.25, 0.50, and 0.25, so \(E(X\mid S)=1\) and \(\operatorname{Var}(X\mid S)=0.5\). The observed score and its illustrative standardized value are

U_c=2-1=1,\qquad Z_c=\frac{1}{\sqrt{0.5}}=1.414.

The two-sided normal approximation is again \(p\approx0.1573\); with one trio, it should not be treated as a dependable asymptotic result. The calculation illustrates the targeted haplotype score, not evidence that H1 is associated with the trait. The null distribution must be calibrated in a way suitable for the number and structure of the observed families.

Phase Uncertainty

At multiple loci, an individual’s unphased genotype does not always reveal which alleles lie together on each chromosome. For instance, an individual heterozygous A/G at one locus and C/T at another could have haplotypes AC/GT or AT/GC. These are different phase configurations compatible with the same unphased genotype, and they imply different possible transmitted haplotypes.

If phase is uncertain, write \(C\) for the phase configuration and \(w_c\) for its conditional probability given the genotypes and other appropriate conditioning information. The expected haplotype count is then an average across configurations. Within each configuration, first calculate the Mendelian transmission expectation; then average those expectations using the phase probabilities. The same principle applies to the conditional variance: uncertainty between phase configurations may add variation beyond the transmission variance within any one configuration.

E(X\mid S)=\sum_c w_c E(X\mid S,C=c),\qquad \sum_c w_c=1

Worked Example: Averaging Over Two Phases

Suppose a hypothetical parent is heterozygous A/G and C/T at two loci. The compatible phases are AC/GT and AT/GC. Assume that, based on the available marker information and a specified phase model, their conditional probabilities are 0.7 and 0.3, respectively. Calculate the expected number of AC haplotypes transmitted by this parent.

Under phase AC/GT, one of the two haplotypes is AC, so the parent transmits AC with probability 0.5. Under phase AT/GC, neither haplotype is AC, so the probability is zero. Averaging gives

E(X_{\mathrm{AC}}\mid S) =0.7(0.5)+0.3(0) =0.35.

Thus the expected AC transmission is 0.35 copies per transmission in this setup, not 0.5. The calculation depends on the stated phase probabilities; they are not universal constants. If the phase information or model changes, the weights may change. A test must use phase probabilities justified by the available data and the conditioning scheme, and must account for phase uncertainty in the relevant null distribution and variance.

Justification and Practical Checks

The zero-mean property follows directly from centering. For each family, under the null Mendelian model, the conditional mean of the observed transmission vector is \(E(X_i\mid S_i)\). Therefore

E\{X_i-E(X_i\mid S_i)\mid S_i\} =E(X_i\mid S_i)-E(X_i\mid S_i)=0.

Multiplying by a fixed phenotype weight and summing preserves a zero conditional mean. This is the basis for comparing the observed score with its conditional null distribution. It does not by itself guarantee a chi-square or normal approximation: that also depends on sample size, the number of informative transmissions, and the variance calculation. For uncertain phase, the conditioning and averaging must be handled consistently; plugging in a single guessed phase can omit uncertainty and alter both the expected score and its variance.

Before interpreting a result, check that the allele or haplotype categories and contrast were defined clearly, that conditional transmission probabilities match the parental information, and that the reported degrees of freedom equal the rank of the score covariance for the omnibus test. For targeted tests, report which category was counted and the direction corresponding to a positive score. An allele-count score describes transmission imbalance within the modeled families; it is not by itself a causal effect estimate.

Common Mistakes

Using all allele counts as independent. Since offspring counts sum to two, the components are constrained. Use the conditional covariance and its rank; do not assign \(K\) independent degrees of freedom automatically.
Choosing a contrast after seeing the largest score. That selection makes a nominal single-contrast p-value misleading. Prespecify the contrast or use an appropriate procedure for the selection or multiple testing.
Treating the most likely phase as certain. The most likely configuration can still have substantial uncertainty. Average over compatible phases using justified weights and account for uncertainty in the variance or null calibration.
Equating a score with a significance result. A computed \(U\) is not a p-value. The covariance and a suitable conditional or asymptotic reference distribution are also needed.
Interpreting a joint result as identifying an allele. An omnibus test detects departure in some estimable direction; it does not specify which allele or haplotype is responsible.

Key takeaway: For multiallelic markers and haplotypes, compare observed allele or haplotype counts with their Mendelian conditional expectations. Use a covariance-aware omnibus test when the direction is unspecified, a clearly defined contrast for a targeted question, and an explicit average over phase configurations when phase is uncertain.

Check Your Understanding

Try these questions without looking back.

Why does an allele-count vector for a diploid offspring have at most \(K-1\) independent components?
What does a significant multiallelic omnibus test establish, and what does it leave unresolved?
For a prespecified contrast \(c\), write the score and its conditional variance.
An unphased two-locus genotype is compatible with two phase configurations. What information is needed to average its expected haplotype transmission across them?
Why is a calculated transmission score not, on its own, enough to report a p-value?