Sibling-Based Tests When Parents Are Missing

Why Compare Siblings?

A family-based association test can use genotype differences between affected and unaffected siblings even when neither parent has been genotyped. The comparison is made within a family: it asks whether the sibling with the phenotype of interest tends to carry more copies of a specified allele than a sibling who does not have that phenotype. Because siblings share parents and much of their family background, this contrast can reduce confounding from factors shared within the family.

This is not the same as observing which allele a parent transmitted. Without parental genotypes, the test generally cannot identify the transmitting parent or reconstruct every transmission. It instead tests whether genotype variation among siblings is associated with phenotype under stated assumptions.

Definition: For a biallelic variant, let \(G_{fi}\) be the allele dosage for sibling \(i\) in family \(f\), coded as 0, 1, or 2 copies of the allele being tested. Let \(Y_{fi}=1\) indicate an affected sibling and \(Y_{fi}=0\) an unaffected sibling. A within-family association contrast compares the genotype values of affected and unaffected siblings in the same family. With this coding, a positive contrast means that affected siblings carry more copies, on average, of the tested allele.

Model and Conditions for a Valid Comparison

A simple design consists of families with at least one affected and at least one unaffected sibling, with the relevant siblings' phenotypes and genotypes observed. The test conditions on each family's observed genotypes and, in its basic form, on the number \(m_f\) of affected siblings among \(n_f\) siblings. It asks whether the affected labels are unusually aligned with the genotypes within families.

The central null assumption is exchangeability: under no within-family association, conditional on the family genotype vector, the affected count, and any design variables needed for the comparison, each assignment of the affected labels to the siblings is equally likely. For a pair containing one affected and one unaffected sibling, this means that under the null the two possible label assignments are equally likely. This is a conditional null; it does not require treating siblings as genetically independent.

This assumption needs scrutiny. Age, sex, birth order, or another sibling-specific factor may affect disease risk and also be related to genotype or to who is included in the analysis. Such factors can make the phenotype labels nonexchangeable unless the design or analysis accounts for them. Ascertainment must also be compatible with the conditioning: selection that favors one particular sibling's phenotype can invalidate a simple label permutation. Genotype missingness and family inclusion should not select siblings according to the unobserved genotype–phenotype contrast.

Families are ordinarily treated as independent sampling units when scores from separate families are combined. The contrast controls shared family characteristics, such as much of the ancestry background, but it does not remove every possible source of bias. It does not control factors that differ between siblings, correct genotyping errors, or establish that a genetic association is causal.

Discordant Sibling Pairs

When each family contributes one affected and one unaffected sibling, define \(D_f\) as the affected sibling's allele dosage minus the unaffected sibling's dosage. Under exchangeability, swapping the two phenotype labels reverses the sign of \(D_f\) but leaves its magnitude unchanged. A family with \(D_f=0\) provides no directional genotype contrast.

Definition: For \(F\) independent affected–unaffected sibling pairs, the paired score is \(U=\sum_{f=1}^{F}D_f\). Conditional on the observed genotypes, its null variance is \(V=\sum_{f=1}^{F}D_f^2\), because each pair's contrast is equally likely to have either sign. When \(V>0\), the standardized score is \(Z=U/\sqrt{V}\); the corresponding one-degree-of-freedom statistic is \(U^2/V\).

For small numbers of informative pairs, use the exact conditional distribution rather than relying on a normal approximation. Under the null, enumerate the possible independent sign reversals of the nonzero \(D_f\)'s. For a two-sided test, the exact p-value is the proportion of those sign assignments whose absolute score is at least as large as the observed absolute score. For a large number of independent informative pairs, \(Z\) is approximately standard normal under the null, and \(U^2/V\) is approximately chi-square with one degree of freedom.

Worked Example: An Exact Test With Discordant Pairs

Suppose five hypothetical sibling pairs have affected-minus-unaffected allele dosage contrasts \(1, 1, 1, -1,\) and \(0\). The zero contrast comes from a pair whose siblings have the same dosage, so it cannot distinguish which sibling is affected.

The observed score and conditional variance are \(U=1+1+1-1+0=2\) and \(V=1^2+1^2+1^2+(-1)^2+0^2=4\). Thus \(Z=2/\sqrt{4}=1\), and the asymptotic statistic is \(U^2/V=4/4=1\).

For an exact two-sided test, only the four nonzero contrasts matter. Under the null their signs are independently reversible, giving \(2^4=16\) equally likely sign assignments. A score with absolute value at least 2 occurs when all four signs agree (2 assignments) or when exactly one sign differs from the other three (8 assignments). The exact p-value is therefore \((2+8)/16=0.625\). These data do not provide strong evidence against the conditional null. The asymptotic chi-square approximation would give a p-value of about 0.317, illustrating why exact calibration can matter with few informative pairs.

Families With More Than Two Siblings

A family may contribute several siblings, including more than one affected sibling. Let \(n_f\) be the number of genotyped siblings in family \(f\), and let \(m_f=\sum_iY_{fi}\) be its number of affected siblings. Write \(\bar{G}_f\) for the mean allele dosage among these siblings. A convenient family score is \(U_f=\sum_i(Y_{fi}-m_f/n_f)G_{fi}\). It equals the total dosage among affected siblings minus \(m_f\bar{G}_f\). Consequently, it is positive when affected siblings carry more copies than expected from that family's genotype distribution under exchangeability.

Result: Conditional on the observed genotypes and the number \(m_f\) affected in family \(f\), the score has null mean zero and variance \(V_f=\{m_f(n_f-m_f)/[n_f(n_f-1)]\}\sum_i(G_{fi}-\bar{G}_f)^2\), for \(n_f>1\). Across independent families, set \(U=\sum_f U_f\) and \(V=\sum_f V_f\). If \(V>0\), \(U^2/V\) is an asymptotic one-degree-of-freedom test statistic.

Families with no affected siblings or no unaffected siblings have \(m_f=0\) or \(m_f=n_f\), and contribute zero to this conditional score. A family in which all siblings have identical dosage also contributes zero, even if it has both affected and unaffected siblings. Both situations lack the within-family genotype variation needed for this test.

Why the Conditional Variance Has This Form

Under the null, exactly \(m_f\) of the \(n_f\) siblings are labelled affected, with all assignments equally likely. Thus \(E(Y_{fi})=m_f/n_f\), so \(E(U_f)=0\). To obtain the variance, let \(q=m_f/n_f\). For any sibling, \(\operatorname{Var}(Y_{fi})=q(1-q)\). For two different siblings, the fixed affected count makes their indicators negatively associated:

\operatorname{Cov}(Y_{fi},Y_{fj}) = -\frac{q(1-q)}{n_f-1} \qquad (i\ne j).

Since \(U_f=\sum_i(Y_{fi}-q)G_{fi}\), its variance is the sum of the genotype-weighted variances and covariances. Substituting the expressions above and collecting terms gives

\begin{aligned} \operatorname{Var}(U_f) &=q(1-q)\left[ \sum_iG_{fi}^2-\frac{1}{n_f-1}\sum_{i\ne j}G_{fi}G_{fj} \right]\\ &=\frac{m_f(n_f-m_f)}{n_f(n_f-1)} \sum_i(G_{fi}-\bar{G}_f)^2. \end{aligned}

The final expression is nonnegative and is zero if the family has no affected–unaffected contrast or no genotype variation. Adding the family variances is justified when families are independent. The resulting standardized score is an asymptotic procedure; if only a few families or a few possible label assignments are available, a conditional randomization calculation is preferable.

Worked Example: Combining Two Multiple-Sibling Scores

Consider two hypothetical families, each with three genotyped siblings and one affected sibling. In Family 1 the dosages are \(0,1,2\), and the sibling with dosage 2 is affected. The family mean is \(\bar{G}_1=(0+1+2)/3=1\), so \(U_1=2-1=1\). The sum of squared deviations is \((0-1)^2+(1-1)^2+(2-1)^2=2\), and the variance is \(V_1=[1(3-1)/(3(3-1))]\times2=2/3\).

In Family 2 the dosages are \(1,1,2\), and the sibling with dosage 2 is affected. Here \(\bar{G}_2=(1+1+2)/3=4/3\), so \(U_2=2-4/3=2/3\). The squared deviations sum to \((1-4/3)^2+(1-4/3)^2+(2-4/3)^2=2/3\), giving \(V_2=[1(3-1)/(3(3-1))]\times(2/3)=2/9\).

Combining the families gives \(U=1+2/3=5/3\) and \(V=2/3+2/9=8/9\). Therefore \(U^2/V=(25/9)/(8/9)=25/8=3.125\). A chi-square reference distribution with one degree of freedom would give a p-value of about 0.077. Because this illustration has only two families, that asymptotic p-value should not be treated as reliable evidence; an exact conditional calculation would be more appropriate. The calculation shows how the family contrasts and their conditional variances combine, not a well-calibrated large-sample conclusion.

What Missing Parental Genotypes Limit

Sibling-based contrasts require genotype variation among siblings, not observed parental genotypes. They can therefore provide evidence when parental samples are unavailable. Under the exchangeability assumptions, a systematic excess of the tested allele among affected siblings is evidence of within-family association. Shared ancestry is less likely to explain such a contrast than an association based only on unrelated cases and controls, because siblings come from the same family background.

However, these tests do not generally identify which parent supplied a copy, determine phase, or reconstruct a complete transmission history. A dosage of 1 records one copy of the tested allele, but without parental genotypes it does not reveal its parental origin. The test also loses information when siblings share the same genotype, when a family has no phenotype discordance, or when few independent families contribute informative contrasts. A null result in such a design can reflect limited within-family information; it does not establish that the allele has no effect.

The dosage score tests an additive trend in the number of allele copies: its directional alternative is that affected siblings tend to have higher dosage. If the scientific question concerns a different genetic model, the genotype coding and corresponding score must be chosen to represent that model. Whatever the coding, the result is an association test under its conditional assumptions, not a proof of causation.

Common Mistake

Wrong reasoning: “The parents were not genotyped, so the observed sibling contrast tells us which parent transmitted the risk allele.” A sibling dosage comparison does not identify parental origin.
Correction: Interpret the result as evidence about within-family association between the chosen genotype coding and affection, conditional on the design and exchangeability assumptions. Parental origin and transmission reconstruction require information this contrast does not supply.
Wrong reasoning: “A family with several siblings always contributes more evidence than a pair.” A larger family helps only if it has both phenotype groups and genotype variation that can contribute to the conditional contrast.

Key takeaway: When parents are missing, affected–unaffected sibling contrasts can test within-family genotype–phenotype association. Their validity depends on appropriate conditional exchangeability, and their information is limited to genotype variation observed among siblings.

Check Your Understanding

Try these questions without looking back.

For a pair with affected-sibling dosage 2 and unaffected-sibling dosage 1, what is the signed contrast, and what does its sign mean?
Why does a pair whose siblings have identical dosage contribute no directional information to the paired test?
State the exchangeability condition that supports a conditional test among siblings.
In a family with three siblings, dosages \(0,1,2\), and one affected sibling with dosage 0, calculate the family score \(U_f\).
Name one conclusion a sibling contrast can support and one conclusion it cannot support when parental genotypes are unavailable.