Transmission Tests for Quantitative Traits

Why Use a Transmission Score for a Quantitative Trait?

A family-based association test can ask whether an allele is transmitted to offspring in a way that tracks a measured trait, such as blood pressure or height. For a quantitative trait, the evidence is not simply whether an offspring inherited a particular allele. The score combines how the child’s inheritance differs from what the parents could have transmitted with how far the child’s trait value lies from a reference level.

The comparison is made within families. Parental genotypes determine the possible transmissions, and Mendelian inheritance supplies their expected frequencies. This makes the score different from a comparison of allele frequencies between unrelated people. It does not, by itself, establish that the tested allele causes the trait: an association can reflect a causal effect or correlation with a nearby causal variant.

The centering term \(G_i-m_i\) measures the offspring’s excess or deficit of transmitted \(A\) alleles relative to the parental-genotype expectation. The factor \(Y_i-c\) gives that transmission deviation more weight when the child’s phenotype is farther from the reference value. The value \(c\) must be chosen and reported as part of the analysis; it is expressed in the same units as the phenotype. A common choice is a prespecified trait mean or a mean estimated under the null model.

Expected Allele Counts and Informative Parents

At a biallelic marker, write parental genotypes as counts of \(A\): \(AA\) has count 2, \(Aa\) has count 1, and \(aa\) has count 0. Under Mendelian segregation, a heterozygous parent transmits \(A\) with probability \(1/2\); a homozygous parent transmits the same allele in every meiosis. Thus, if the mother and father have \(a_m\) and \(a_f\) copies of \(A\), respectively, the expected transmitted count is

This formula follows because each parent contributes one allele. A homozygous parent’s contribution is fixed, while a heterozygous parent’s expected contribution is \(1/2\). For example, \(AA\times Aa\) gives \(m_i=2/2+1/2=1.5\) only if using parental allele counts as written would be incorrect: each parent’s transmitted-allele expectation is its allele count divided by two, so the correct value is \((a_m+a_f)/2\). For \(AA\times Aa\), this is \((2+1)/2=1.5\). The offspring can have one or two copies of \(A\), with mean 1.5.

If \(h_i\) is the number of heterozygous parents, each contributes a Bernoulli transmission with probability \(1/2\). Assuming the parents’ meioses are independent,

A trio with two homozygous parents has no transmission variation: \(G_i=m_i\) for every possible child, so its score contribution is zero. A trio with one or two heterozygous parents can be informative. Informative does not mean that every such trio contributes a nonzero observed score; the realized transmitted count might equal its expectation.

Why the Conditional Null Expectation Is Zero

The relevant null hypothesis is that, conditional on the parental genotypes, transmission of the tested allele is not associated with the offspring’s phenotype. Under this conditional null, and assuming Mendelian segregation, the observed transmission count has expectation \(m_i\). For a trio whose phenotype is held fixed in the conditioning, \(Y_i-c\) is a constant. Therefore,

The last equality follows because the conditional null gives \(E(G_i\mid P_i,Y_i)=E(G_i\mid P_i)=m_i\). Summing over trios gives \(E(U\mid \text{parental genotypes and phenotypes})=0\). A nonzero observed score is possible under the null because transmissions vary randomly; the zero is the expected score over repetitions of those transmissions, not a claim that every sample score must equal zero.

If transmissions are independent across trios, the conditional variance follows from the same setup. Each phenotype deviation is fixed, so multiplying a transmission deviation by \(Y_i-c\) multiplies its variance by \((Y_i-c)^2\). The variances add across independent trios:

This variance is conditional on the parental genotypes and the observed phenotypes. It is not the variance of the trait itself. A standardized score can be formed as \(Z=U/\sqrt{\operatorname{Var}(U\mid P,Y)}\), provided the denominator is positive. A standard normal approximation for \(Z\) is most credible when information is spread across many independent transmissions; a few trios do not generally justify relying on that approximation for a p-value.

Direction, Scale, and Interpretation

The sign of \(U\) summarizes how transmission deviations align with trait deviations from \(c\). A positive contribution results when an offspring above \(c\) inherits more \(A\) alleles than expected, or when an offspring below \(c\) inherits fewer than expected. A negative contribution results from the opposite alignments. The total score combines these contributions across families; it is not an estimated change in the trait per allele and should not be reported in trait units alone. Its units are allele-count deviations multiplied by trait units.

Assumptions and a Common Pitfall

Interpreting the score requires that the parental genotypes and offspring phenotype be measured correctly, that the relevant transmissions follow Mendelian probabilities under the null, and that the conditional null of no transmission–phenotype association is appropriate. The variance formula above additionally assumes independent meioses across the trios being summed. Related offspring or other sources of dependence require a variance calculation that accounts for that dependence; simply adding individual variances would then be unjustified.

A positive score is evidence in a particular direction only relative to the chosen allele \(A\), the trait reference \(c\), and the stated score definition. Changing which allele is tested reverses the sign of each transmission deviation. Changing \(c\) can also change the score, so the reference must be chosen transparently rather than selected after inspecting which value gives the desired result.

• Mistake: “The parents are heterozygous, so the child’s observed allele count is the expected count.” Correction: Heterozygous parents define a probability distribution over transmissions; the child’s realized count can differ from its expectation.
• Mistake: “A zero contribution means the trio provided no transmission information.” Correction: The observed deviation \(G_i-m_i\) may be zero even when the parental genotypes allow more than one transmission outcome. In that case the trio can still contribute to the score variance.
• Mistake: “A positive score proves that allele \(A\) raises the trait.” Correction: The score summarizes an association pattern under a specified coding and reference. It is not a causal effect estimate, and linkage to another variant can also produce a pattern.
• Mistake: “The standardized score is automatically standard normal.” Correction: Standardization alone does not ensure a normal null distribution. The approximation needs enough independent transmission information and should not be trusted uncritically in a very small sample.