FBAT With Extended Pedigrees

Why Extended Pedigrees Change the Calculation

A nuclear family often makes the relevant transmission easy to describe: condition on the parents’ genotypes, then consider which alleles could have been transmitted to an offspring. In an extended pedigree, an untyped parent, a shared ancestor, or several related branches can change the possible transmission paths. The basic FBAT idea remains the same, but the conditional expectation used to center each genotype score must reflect the pedigree.

The key is to distinguish the observed genotype score from its expectation under the conditional null. The observed score records the alleles carried by the tested relatives. Its expectation is calculated from the possible Mendelian transmissions given the genotypes and other information being conditioned on. In an extended pedigree, those possibilities may involve several meioses jointly.

Definition: Let $T_i$ be a genotype score for relative $i$, such as the number of copies of a specified allele (the allele dosage). Let $S_i$ be a phenotype weight fixed under the conditional null. Given conditioning information $\mathcal{C}$, an FBAT score can be written as $$ U=\sum_i S_i\left\{T_i-E(T_i\mid\mathcal{C})\right\}. $$ Here $\mathcal{C}$ contains the pedigree information conditioned on for the test, such as available parental or founder genotypes and the relevant phenotype information. The conditional expectation is taken under the null transmission model. For an affected-only analysis, for example, affected individuals may receive weight $S_i=1$; other traits use the phenotype score specified for the analysis.

This expression is a framework, not a claim that every relative contributes independently. The score’s centering is determined by the conditional genotype distribution, while the distribution of the whole score also depends on how transmissions are related. A positive $U$ means that the phenotype-weighted observed allele scores exceed their conditional expectations; it is not by itself a p-value or an effect-size estimate.

Informative Meioses and Conditional Expectations

At a single biallelic locus, a heterozygous parent $A/a$ transmits either allele with probability $1/2$, assuming Mendelian transmission and conditioning on that parent’s genotype. A homozygous parent $A/A$ transmits $A$ with probability one, so that meiosis supplies no variation in transmission of the allele being tested. A meiosis is informative for a transmission score when the available pedigree information leaves more than one possible transmitted genotype with positive probability.

An informative meiosis need not be directly observed as a parent-to-child transmission. If an intermediate relative is untyped, the allele reaching a descendant may depend on several meioses. The conditional expectation must average over the possible paths and genotypes of the untyped relative. When multiple descendants share that relative, their genotype scores can also be dependent because they share part of the same transmission history.

For a specified conditioning set, the general calculation is an average over compatible transmission configurations. If $M$ denotes the configuration of meioses and $m$ ranges over configurations compatible with the conditioning information, then

E(T_i\mid\mathcal{C})=\sum_m T_i(m)P(M=m\mid\mathcal{C}).

At a single locus with known heterozygous parental genotypes and no additional constraints, each transmitted allele has probability $1/2$. With missing genotypes, shared ancestors, or marker information, the probabilities in this sum must reflect all relevant compatible paths. It is important not to condition on the tested offspring transmission itself when calculating its null expectation: doing so would make that transmission known rather than random.

Worked Example: Two Informative Transmissions From a Typed Parent

Suppose a hypothetical pedigree has a typed parent with genotype $A/a$, whose mate is $a/a$. Their two children are both affected. At this locus, each child’s allele dosage $T_i$ is either 0 or 1. Use affected-only weights $S_1=S_2=1$. Assume the genotypes and relationships are correct, inheritance is Mendelian, and the parent genotypes are the conditioning information; the children’s tested transmissions are not conditioned on.

For either child, the $A/a$ parent transmits $A$ with probability $1/2$, while the $a/a$ parent always transmits $a$. Therefore $P(T_i=1\mid\mathcal{C})=1/2$, and $E(T_i\mid\mathcal{C})=1/2$. Both meioses from the heterozygous parent are informative.

Suppose the observed child genotypes are $A/a$ and $a/a$, so $T_1=1$ and $T_2=0$. The score is $$ U=1(1-0.5)+1(0-0.5)=0.5-0.5=0. $$ The two affected children’s observed dosages, taken together, equal their conditional expected total dosage of 1. This score gives no directional excess of the allele in this example. It is not a calculation of a p-value.

When an Intermediate Relative Is Untyped

A missing genotype can make a descendant’s conditional distribution different from the one obtained by treating the descendant as the child of a known heterozygote. The missing person’s genotype must be averaged over, rather than silently replaced by a known genotype. This distinction can change both the expected dosage for each descendant and the joint distribution of their dosages.

Worked Example: Marginalizing Over an Untyped Parent

Consider a hypothetical pedigree in which a typed grandparent has genotype $A/a$ and the other grandparent is $a/a$. Their child, who is the parent of two affected grandchildren, is untyped. That untyped parent has a typed $a/a$ mate. Use $S_1=S_2=1$, and condition on the grandparent genotypes, not on the tested grandchildren’s transmissions. Assume Mendelian inheritance at a single locus.

The untyped parent receives $A$ from the $A/a$ grandparent with probability $1/2$. If that parent has $A/a$, each grandchild receives $A$ from them with probability $1/2$; if they have $a/a$, neither can receive $A$ from them. Thus, for either grandchild, $$ P(T_i=1\mid\mathcal{C})=(1/2)(1/2)+(1/2)(0)=1/4, \qquad E(T_i\mid\mathcal{C})=0.25. $$

Suppose both observed grandchildren carry one $A$ allele, so $T_1=T_2=1$. Their combined score is $$ U=(1-0.25)+(1-0.25)=0.75+0.75=1.50. $$ The positive score says that the observed affected-grandchild dosages exceed their conditional expectations by 1.50 allele copies in total. It does not establish statistical significance.

The shared untyped parent also means these two dosages are not independent under this conditioning scheme. Both grandchildren can receive $A$ only if the untyped parent first inherited $A$, and that shared event has probability $1/2$. In fact, the probability both carry $A$ is $(1/2)(1/2)(1/2)=1/8$: the first factor is the grandparent-to-parent transmission, and the other two are the transmissions to the grandchildren. Multiplying the marginal probabilities would give $(1/4)(1/4)=1/16$, which is different. The marginal expectations used in $U$ do not remove this dependence.

How Pedigree Information Determines the Distribution

The same descendant can have different conditional genotype distributions under different available pedigree information. For example, if the intermediate parent in the preceding pedigree were typed $A/a$ and that genotype were conditioned on, each grandchild’s probability of receiving $A$ from that parent would be $1/2$, rather than $1/4$. If the parent is untyped and only the grandparents’ genotypes are conditioned on, the probability is $1/4$, because the parent must first inherit $A$.

This is not a contradiction. The conditioning sets differ: a known heterozygous parent guarantees that the $A$ allele is available for transmission, whereas an untyped parent may not have inherited it. In a real analysis, the conditioning scheme must be stated and applied consistently. Genotypes used to infer possible latent parental states can inform the conditional distribution, but the tested offspring transmission must remain random under the null.

For a biallelic allele-dosage score, a child of two typed parents can have dosage 0, 1, or 2 depending on the parental genotypes. A parent with two copies of $A$ contributes $A$ with certainty; a heterozygous parent contributes $A$ with probability $1/2$. In an extended pedigree, one can enumerate the possible transmissions through each relevant branch, calculate the conditional probability of each compatible configuration, and then obtain $E(T_i\mid\mathcal{C})$ for each scored individual. If phase or marker data affect which paths are compatible, those constraints belong in the conditional distribution as well.

Key result: Under the conditional null, if $\mathcal{C}$ is the stated conditioning information and the phenotype weights $S_i$ are fixed given that information, then $$ E(U\mid\mathcal{C})= \sum_i S_i\left\{E(T_i\mid\mathcal{C})-E(T_i\mid\mathcal{C})\right\}=0. $$ This centering follows from the conditional expectation used in the score. It does not imply that the individual transmissions are independent, nor does it determine the variance of $U$.

A Short Justification of the Centering

The score is centered to remove the expected contribution of each genotype under the conditional null. For fixed phenotype weights, linearity of conditional expectation gives

\begin{aligned} E(U\mid\mathcal{C}) &=E\left[\sum_i S_i\{T_i-E(T_i\mid\mathcal{C})\}\mid\mathcal{C}\right]\\ &=\sum_i S_i\left\{E(T_i\mid\mathcal{C})-E(T_i\mid\mathcal{C})\right\}\\ &=0. \end{aligned}

The argument uses the conditional expectation appropriate to each relative and a phenotype weight that is fixed under the conditioning. It does not require the $T_i$ values to be independent. Consequently, the zero-mean property is not a justification for adding up separate nuclear-family test statistics as if every branch were unrelated.

Practical Workflow and Assumptions

Before calculating an extended-pedigree score, specify the allele or genotype score, the phenotype weights, the conditioning information, and which transmissions remain random under the null. Then identify informative meioses and calculate the relevant conditional expectations. This makes the role of each typed and untyped relative explicit.

Specify the score.
Define $T_i$ (for example, allele dosage) and the phenotype weight $S_i$ for each individual contributing to the test.

State what is conditioned on.
List the genotypes and phenotype information treated as fixed, and ensure the tested transmissions are not made known by the conditioning.

Trace compatible inheritance paths.
Mark meioses with more than one possible transmitted genotype and account for shared ancestors or untyped intermediate relatives.

Center and sum.
Calculate $E(T_i\mid\mathcal{C})$ for each scored relative and form $U=\sum_i S_i\{T_i-E(T_i\mid\mathcal{C})\}$.

These calculations assume that pedigree relationships and the genotypes treated as observed are accurate, and that the conditional transmission probabilities follow the stated Mendelian model. If founders are untyped, their possible genotypes require an appropriate model rather than being treated as known. Mutation, genotyping error, or unaccounted marker or phase information can also make a simple transmission calculation inappropriate. For an extended pedigree, the score’s uncertainty must reflect shared transmission paths; a zero conditional mean alone is not enough to justify an independence-based variance calculation.

Common Mistakes

Using the known-heterozygous-parent probability when that parent is untyped. The wrong reasoning is that every child has a $1/2$ chance of inheriting $A$ because the pedigree includes an $A/a$ grandparent. The correction is to include the chance that the untyped parent first inherited $A$; in the example, each grandchild’s probability is $1/2\times1/2=1/4$.
Treating related transmissions as independent. The wrong reasoning is to multiply marginal probabilities or add independent family contributions just because there are multiple children. Shared ancestors can create dependence. The conditional distribution must preserve the common transmission history.
Calling every meiosis informative. A homozygous parent transmits the same allele at the locus with probability one. That meiosis does not distinguish alternative transmissions for this allele score.
Conditioning on the tested transmission itself. If the offspring’s observed allele is used as though it were fixed when computing its own null expectation, there is no remaining transmission variation to test. Use conditioning information that leaves the tested transmission random.
Reading a positive score as evidence of significance. A positive score indicates an excess in the scored direction, but statistical significance requires an appropriate calibration that accounts for the conditional distribution and dependence.

Check Your Understanding

Use the conditional-transmission reasoning in each question.

A typed $A/a$ parent has an $a/a$ mate. What is the conditional probability that a child receives $A$, and is the heterozygous parent’s meiosis informative?
An untyped parent is the child of typed $A/a$ and $a/a$ grandparents and mates with an $a/a$ individual. Under the single-locus Mendelian model, what is the probability that their child receives $A$ through this path?
Why can two grandchildren’s allele dosages be dependent when they share an untyped parent, even if their own meioses from that parent are separate?
In the score $U=\sum_i S_i\{T_i-E(T_i\mid\mathcal{C})\}$, what does a positive value describe, and what does it not establish by itself?
Why should the tested offspring transmission not be included as fixed information when calculating its conditional null expectation?

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