Conditional Inference From Within-Family Transmissions

Why Condition on Families?

A population association compares genotypes across people who may differ in ancestry, environment, or other characteristics. A family-based comparison instead asks a narrower question: given the alleles a child’s parents could transmit, which allele was passed to a child with the phenotype of interest? Fixing the parental genotypes defines the available alleles within that family. The child’s phenotype then identifies which transmissions are being examined.

For a simple example, suppose a parent has alleles \(A\) and \(a\), and a child is affected. The parent passed one of these alleles to that child; the other was not transmitted. Under the null of no association between the marker allele and the phenotype, the affected child should not systematically receive \(A\) more often than \(a\) from this heterozygous parent. The comparison is within the family, not between the child and a separately sampled control.

The purpose here is to define that conditional comparison and its null probabilities. It is not yet to construct a test statistic or calibrate a test.

Conditional Transmission and the Null

In a single meiosis, Mendelian segregation gives an \(A/a\) parent probability \(1/2\) of transmitting \(A\) and probability \(1/2\) of transmitting \(a\). An \(A/A\) parent transmits \(A\) with probability 1, while an \(a/a\) parent transmits \(a\) with probability 1. Thus, only a heterozygous parent offers a choice between two different marker alleles.

For a family-based association comparison, the key null assumption is that, after fixing the parental genotypes, the transmission at the marker is independent of the offspring’s phenotype. In particular, let \(Y=1\) denote an affected offspring, and let \(G_m\) and \(G_f\) denote the maternal and paternal genotypes. If the mother is \(A/a\), the null specifies:

The same probabilities apply to a heterozygous father. A homozygous parent has no allele contrast: for example, \(P(\text{father transmits }A\mid G_f=A/A)=1\), whether the child is affected or not. The null does not say every child has a 50% probability of each possible genotype. It says that each heterozygous parent’s two alleles have equal transmission probabilities, and that conditioning on the affected phenotype does not favor one of those alleles.

How the Within-Family Comparison Arises

For a heterozygous parent, the allele not transmitted to a particular child is known from the parental genotype once the transmitted allele is identified. If an \(A/a\) mother transmits \(A\), then \(a\) is her nontransmitted allele for that child; if she transmits \(a\), then \(A\) is the nontransmitted allele. This creates a paired comparison between the allele received by the affected child and the alternative allele available from the same parent.

The nontransmitted allele is not an observed genotype for the child. It is the within-parent alternative that could have been passed in that meiosis. That distinction matters: the family comparison concerns what the same parent could transmit, not a second child who is assumed to have the same phenotype or circumstances.

Conditioning on parental genotypes also determines which transmissions carry information. A heterozygous parent can transmit either of two marker alleles, so the data can reveal whether one is overrepresented among affected offspring. A homozygous parent always transmits the same allele and therefore contributes no such contrast. This is why a trio can be only partly informative: one parent may be heterozygous while the other is homozygous.

Why the Null Has Equal Probabilities

The null transmission probabilities follow directly from Mendelian segregation and the assumption that the marker transmission does not depend on affection once the parental genotypes are fixed. Let \(T\) denote the allele transmitted by a heterozygous parent. Under Mendelian segregation, \(P(T=A\mid G_m=A/a,G_f)=1/2\). Under the null, affection status \(Y\) does not alter this probability. The conditional-probability calculation makes the implication explicit:

The second equality uses the null independence condition: the probability of affection is the same for either transmitted marker allele, given the parental genotypes. The denominator is assumed positive, so conditioning on an affected offspring is defined. Replacing \(A\) by \(a\) gives the same result, and the two probabilities sum to 1.

This argument is specific to the conditional null. Mendelian segregation alone states the 1/2 probability before conditioning on phenotype. It does not, by itself, guarantee that the probability remains 1/2 among affected offspring. If the marker allele is associated with the phenotype, or is linked to a causal variant in a way that makes transmission relevant to affection, the phenotype-conditioned transmission probability can depart from 1/2. That departure is the signal a family-based association analysis is designed to examine.

Assumptions and Interpretation

The equal-probability null has clear assumptions. The marker alleles must be correctly observed and parental genotypes correctly specified; transmission must follow Mendelian segregation without distortion; and, under the null, marker transmission must be independent of the offspring phenotype conditional on parental genotypes. The simple product calculations for two parents also assume their meioses are independent conditional on those genotypes. These conditions define the probability model; they should not be silently inferred from an observed family table.

A family-based comparison asks whether allele transmission to affected offspring departs from the conditional null. It does not establish that an associated marker allele causes the phenotype. A marker may track another genetic variant, and a transmission imbalance is evidence about association under the model rather than proof of a biological mechanism.

Conditioning is useful because it makes the relevant comparison explicit: among the alleles that these parents could transmit, does one allele appear preferentially among affected offspring? This is distinct from comparing allele frequencies between unrelated affected and unaffected groups. The family provides its own set of possible transmitted alleles, and the null states the probabilities for those choices.

Common Mistakes

• Wrong: “Every child genotype has probability \(1/2\).” Correction: The \(1/2\) null applies to each allele transmitted by a heterozygous parent. A child-genotype probability depends on both parents; for two heterozygous parents, the probabilities are \(1/4\), \(1/2\), and \(1/4\).
• Wrong: “The affected phenotype can be ignored because meiosis is Mendelian.” Correction: Mendelian segregation gives probabilities before conditioning on phenotype. The no-association null is what allows those probabilities to remain equal among affected offspring.
• Wrong: “A homozygous parent contributes the same transmission contrast as a heterozygous parent.” Correction: A homozygous parent transmits only one marker allele, so there is no alternative allele from that parent to compare.
• Wrong: “The nontransmitted allele is a control child’s genotype.” Correction: It is the alternative allele available from that parent, not an observed genotype or a separate person.