Worked FBAT Analysis With Missing Parents

Why Analyze Siblings When Parents Are Missing?

Family-based association tests use genetic variation within families to test whether allele transmission is associated with a trait. When parental genotypes are missing, a transmission cannot be directly labelled as transmitted or untransmitted. For full siblings, however, a conditional analysis can compare each sibling’s genotype with the genotypes and trait values of their co-siblings. The comparison uses variation within a family rather than differences between unrelated families.

This tutorial works through an offspring-only, sibling-based conditional score for a quantitative trait. It conditions on the observed multiset of sibling genotypes in each family. This is a particular way to obtain a family-based test when parents are unavailable; the conditioning and assumptions must be stated, rather than treating missing parental genotypes as if they had been observed.

Definition: Let $f$ index families and $j$ index full siblings in family $f$. Let $G_{fj}$ be the child’s allele dosage at a biallelic locus, coded 0, 1, or 2, and let $Y_{fj}$ be the child’s quantitative trait value. Define $\bar G_f$ and $\bar Y_f$ as the mean genotype dosage and mean trait value within family $f$. The sibling-based conditional score is $$ U=\sum_f U_f,\qquad U_f=\sum_{j=1}^{m_f}(G_{fj}-\bar G_f)(Y_{fj}-\bar Y_f), $$ where $m_f$ is the number of genotyped siblings in family $f$. A family with no variation in $G_{fj}$ has $U_f=0$ and contributes no information about a within-family genotype effect.

The centering has two useful consequences. First, family-specific differences in average genotype and average trait do not create a score: only the pairing of genotype differences and trait differences among siblings matters. Second, centering the trait by $\bar Y_f$ is equivalent here to centering it by any common value $\mu$, since $\sum_j(G_{fj}-\bar G_f)=0$. Thus, the score is not driven by an arbitrary choice of trait origin.

Conditional Null and Assumptions

The null hypothesis is that, within each family, the assignment of the observed sibling genotypes to the siblings is unrelated to their quantitative trait values. Conditional on the family’s observed genotype multiset and its observed trait values, the null treats the genotype assignments as exchangeable among siblings. In particular, for a family with $m_f$ siblings, each permutation of the genotype assignments has the same probability under this conditional null.

That exchangeability is justified when siblings are full siblings with the same two biological parents, Mendelian transmissions to the siblings are independent given those parents, and the null of no genotype–trait association holds. It does not require knowing the parental genotypes or allele frequency: conditional on the parental genotypes, siblings’ genotype draws have the same distribution. Conditioning further on the observed genotype multiset makes all assignments equally likely under the null. The inference also requires reliable genotyping and a sampling or ascertainment process that does not make sibling inclusion depend differently on genotype and trait. If these conditions are implausible, this particular conditional analysis may not be appropriate.

Conditional variance: Under the exchangeability null, the conditional mean of $U_f$ is zero and its variance is $$ V_f=\frac{\left[\sum_j(G_{fj}-\bar G_f)^2\right] \left[\sum_j(Y_{fj}-\bar Y_f)^2\right]}{m_f-1}. $$ For independent families, $V=\sum_f V_f$ is the conditional variance of $U$. A large positive score indicates that higher allele dosage tends to pair with higher trait values within families; a large negative score indicates the reverse.

Why the Conditional Variance Has This Form

For one family, write $g_j=G_{fj}-\bar G_f$ and $y_j=Y_{fj}-\bar Y_f$. Under a uniformly random permutation $\pi$ of genotype assignments, the score can be written as $U_f=\sum_j g_j y_{\pi(j)}$, with the $y_j$ values held fixed. Each assigned trait deviation has mean zero, so $E(U_f)=0$.

Let $A=\sum_j y_j^2$. A randomly assigned deviation has second moment $A/m_f$. For two distinct positions, the expected product is $-A/[m_f(m_f-1)]$: the sum of all deviations is zero, so the sum of their pairwise products is $-A/2$. Expanding the squared score and using $\sum_j g_j=0$ gives

\begin{aligned} \operatorname{Var}(U_f) &=\frac{A}{m_f}\sum_j g_j^2 -\frac{A}{m_f(m_f-1)}\sum_{j\ne k}g_jg_k\\ &=\frac{A}{m_f}\sum_jg_j^2 +\frac{A}{m_f(m_f-1)}\sum_jg_j^2\\ &=\frac{\left(\sum_jg_j^2\right)\left(\sum_jy_j^2\right)}{m_f-1}. \end{aligned}

The second line uses $\sum_{j\ne k}g_jg_k=-\sum_jg_j^2$, which follows by squaring $\sum_jg_j=0$. This derivation explains why the conditional variance depends on both within-family genotype variation and within-family trait variation.

Worked Example: A Two-Sibling Family

Suppose a hypothetical family has two genotyped siblings, although neither parent has a genotype available. Their allele dosages are 0 and 2, and their trait values are 4 and 8 units.

The family means are $\bar G_f=(0+2)/2=1$ and $\bar Y_f=(4+8)/2=6$. The score is $$ U_f=(0-1)(4-6)+(2-1)(8-6)=2+2=4. $$ The within-family genotype sum of squares is $1^2+1^2=2$; the trait sum of squares is $(-2)^2+2^2=8$. Therefore, $$ V_f=\frac{2(8)}{2-1}=16,\qquad \sqrt{V_f}=4. $$

The score divided by its null standard deviation is $4/4=1$. But with two siblings, the only two genotype assignments simply reverse the signs of the score: the scores are $4$ and $-4$. Both have the same absolute value, so a two-sided exact permutation test has $p=1$. This family has a nonzero within-family slope, but by itself it provides very little evidence against the conditional null.

Worked Analysis Across Three Families

Consider the following hypothetical data. All parents’ genotypes are missing; all nine siblings are genotyped and have a quantitative trait measurement in the stated units.

Family	Sibling	Allele dosage $G$	Trait $Y$ (units)
A	1	0	2
A	2	1	4
A	3	2	8
B	1	0	1
B	2	0	3
B	3	1	4
C	1	1	3
C	2	1	5
C	3	2	6

Worked Example: Computing the Conditional Score and Its Uncertainty

For Family A, $\bar G_A=1$ and $\bar Y_A=14/3$. Its score and genotype sum of squares are $$ U_A=(-1)(-8/3)+0(-2/3)+(1)(10/3)=6,\qquad \sum_j(G_{Aj}-\bar G_A)^2=2. $$ The trait sum of squares is $64/9+4/9+100/9=56/3$, so $V_A=2(56/3)/(3-1)=56/3$.

For Family B, $\bar G_B=1/3$ and $\bar Y_B=8/3$. The centered genotypes are $-1/3,-1/3,2/3$, and the centered traits are $-5/3,1/3,4/3$. Thus $$ U_B=\frac{5}{9}-\frac{1}{9}+\frac{8}{9}=\frac{4}{3}. $$ The genotype and trait sums of squares are $2/3$ and $14/3$, respectively, giving $V_B=(2/3)(14/3)/2=14/9$. Family C has the same centered values up to the pairing shown in the table; direct calculation gives $U_C=4/3$ and $V_C=14/9$.

All three families are informative because each has variation in allele dosage. Summing their contributions, $$ U=6+\frac{4}{3}+\frac{4}{3}=\frac{26}{3}=8.667,\qquad V=\frac{56}{3}+\frac{14}{9}+\frac{14}{9}=\frac{196}{9}=21.78. $$ The conditional standard deviation of the score under the null is $\sqrt V=14/3=4.667$, and the standardized score is $$ Z=\frac{U}{\sqrt V}=\frac{26/3}{14/3}=1.857. $$ Using the standard normal approximation, the two-sided $p$-value is approximately $0.0633$. At the 0.05 significance level, this approximation does not provide evidence against the conditional null. With only three families, the normal approximation may be rough; an exact test can instead enumerate genotype permutations within each family.

A descriptive within-family slope estimate is the total score divided by total within-family genotype sum of squares: $$ \widehat\beta=\frac{U}{\sum_f\sum_j(G_{fj}-\bar G_f)^2} =\frac{26/3}{2+2/3+2/3}=2.600\text{ trait units per allele-dosage unit}. $$ The null score’s standard deviation translated to this slope scale is $\sqrt V/(10/3)=1.400$ trait units per dosage unit. This is a null-based measure of score uncertainty, not a general-purpose confidence interval for $\beta$. The estimated slope describes the observed within-family trend; the score test evaluates how unusual that trend is under the specified conditional null.

Recognizing Informative and Uninformative Families

A family contributes only if sibling genotypes differ. If every sibling has dosage 1, then $\bar G_f=1$, every centered genotype is zero, and both $U_f$ and $V_f$ are zero regardless of the trait values. The family can still inform other questions, but it cannot inform this within-family allele-dosage score.

Worked Example: A Family With No Genotype Information

Suppose three siblings have dosages $1,1,1$ and trait values $2,5,8$ units. Their means are $\bar G_f=1$ and $\bar Y_f=5$. The score is $$ U_f=(1-1)(2-5)+(1-1)(5-5)+(1-1)(8-5)=0, $$ and the genotype sum of squares is $0^2+0^2+0^2=0$, so $V_f=0$. No within-family slope can be estimated from this family: there is no genotype contrast to pair with the trait differences. The trait variation does not make the family informative for this test by itself.

Interpreting the Estimate and Test

The dosage coding gives the estimated slope a direct scale: it is the estimated change in trait units associated with one additional copy of the allele, based on sibling differences within families. It is not a causal effect merely because the comparison is within-family. Interpretation also depends on the additive dosage model used for the slope; the score test itself measures a direction of within-family association under its conditional null.

A $p$-value is the probability, under the conditional null and the stated sampling assumptions, of a score statistic at least as extreme as the observed one. It is not the probability that the null is true. A nonsignificant result does not establish no association, particularly when few families are informative. Conversely, a small $p$-value does not by itself measure the size or practical importance of an association. Report the estimated slope and its units alongside the test result, and do not present the null-based score standard deviation as though it were a confidence interval.

Common Pitfalls

Using between-family differences as evidence. Families can differ in both average genotype and average trait for reasons unrelated to the tested association. The conditional score uses deviations from each family’s own means.
Counting every family as informative. A family with identical sibling dosages has zero within-family genotype variation and contributes no score information.
Ignoring the missing-parent assumption. Offspring-only conditioning is not a license to use any family structure or sampling scheme. Check full-sibling status, exchangeability, and ascertainment before interpreting the test.
Calling the slope a causal effect. A within-family association can reduce some sources of population-level confounding, but the score and its estimate do not establish causation.
Overinterpreting a normal approximation. With few families or a small number of possible genotype assignments, the conditional null distribution may be discrete and far from normal. Exact within-family permutation can be preferable.

Key takeaway: With missing parental genotypes, a sibling-based conditional score can test a quantitative-trait association by pairing within-family genotype differences with within-family trait differences. Its validity depends on exchangeability under the conditional null, and its uncertainty must be interpreted on the scale and under the assumptions actually used.

Check Your Understanding

Work through these questions using the definitions and assumptions above.

Why does a family in which all siblings have the same allele dosage contribute no information to this within-family score?
For one family, explain why the conditional null mean of $U_f$ is zero when genotype assignments are exchangeable.
What assumptions about siblings and ascertainment are needed for the offspring-only conditional test?
In the three-family example, what does the estimated slope of 2.600 trait units per dosage unit describe?
Why is the score-scale standard deviation under the null not automatically a confidence interval for the slope?