Deriving the Sib-Pair IBD Null Distribution

Why an Unlinked Locus Provides a Null Benchmark

In nonparametric linkage analysis, a key question is whether affected relatives share more genetic material identical by descent (IBD) than expected without linkage. Before assessing excess sharing, we need a baseline: the distribution of IBD sharing for an affected sibling pair at a locus that is unlinked to the disease determinants. Under standard Mendelian assumptions, this distribution is \(1:2:1\) for sharing zero, one, or two alleles IBD.

The important point is that the baseline follows from how siblings inherit alleles from their parents. It does not depend on the marker allele frequencies. It also does not require the siblings to have the same observed alleles: IBD refers to the ancestry of the inherited copies, not just whether their allele values match.

Definition: At a locus, let \(Z\) be the number of alleles that two siblings share identical by descent, so \(Z\in\{0,1,2\}\). Two alleles are IBD if they descend from the same ancestral allele copy. For a standard noninbred nuclear family, each sibling inherits one allele from each parent, and \(Z\) counts the shared paternal and maternal copies.

Deriving the 1:2:1 Distribution

Consider one parent at the locus. That parent has two homologous chromosome copies and passes one copy to each child. Under Mendelian segregation, each child has probability \(1/2\) of inheriting either copy. For two siblings, there are four equally likely combinations of transmitted parental copies. Two combinations use the same copy for both siblings, and two use different copies. Thus the siblings have probability \(1/2\) of sharing the transmitted copy IBD from that parent.

Apply the same reasoning separately to the other parent. Define \(M\) as 1 if the siblings inherit the same maternal copy and 0 otherwise, and \(F\) as 1 if they inherit the same paternal copy and 0 otherwise. Under independent Mendelian transmissions from the two parents, \(M\) and \(F\) are independent Bernoulli random variables with probability \(1/2\) of being 1. The total number shared IBD is their sum.

Z=M+F,\qquad M,F\sim\operatorname{Bernoulli}\left(\frac{1}{2}\right),\qquad M\perp F.

The four equally likely combinations of \((M,F)\) give the distribution directly. If neither parental copy is shared, \(Z=0\). If exactly one is shared, \(Z=1\). If both are shared, \(Z=2\).

\begin{aligned} P(Z=0)&=P(M=0,F=0)=\frac{1}{4},\\ P(Z=1)&=P(M=1,F=0)+P(M=0,F=1)=\frac{1}{2},\\ P(Z=2)&=P(M=1,F=1)=\frac{1}{4}. \end{aligned}

The probabilities sum to 1, as required: \(1/4+1/2+1/4=1\). The ratio \(1:2:1\) is a ratio of probabilities, not a claim that every small collection of sibling pairs must show those exact proportions.

Worked Example: Counting the Parental Transmission Outcomes

Suppose a hypothetical sibling pair is considered at one locus in a noninbred nuclear family. List the equally likely combinations of maternal and paternal copy sharing, then find the probability of each value of \(Z\).

For each parent, the siblings either inherit the same parental copy (coded 1) or different parental copies (coded 0). There are four equally likely combinations:

Maternal indicator \(M\)	Paternal indicator \(F\)	Sharing count \(Z=M+F\)
0	0	0
0	1	1
1	0	1
1	1	2

Each row has probability \(1/4\). One row gives \(Z=0\), two give \(Z=1\), and one gives \(Z=2\). Therefore \(P(Z=0)=1/4\), \(P(Z=1)=2/4=1/2\), and \(P(Z=2)=1/4\).

Why Affection Does Not Change This Null Distribution

The distribution above describes transmissions at a locus. For an affected sibling pair, it is a null distribution only when the locus is unlinked to the genetic determinants relevant to the disease, and when the assumptions connecting transmission to ascertainment are appropriate. At unlinked loci, inheritance at the marker locus is independent of inheritance at the disease locus under the standard model. Conditioning on both siblings being affected can select particular disease-locus inheritance patterns, but it does not thereby select marker-locus transmissions.

For this argument, assume Mendelian segregation, no relevant distortion of transmission, and no selection based on the marker genotype or its inheritance. Assume also that the marker is unlinked to the disease determinants under consideration, and that the standard noninbred nuclear-family model applies. Under these conditions, the marker’s \(Z\) distribution remains the same after conditioning on both siblings being affected.

P(Z=z\mid \text{both siblings affected}) = P(Z=z) = \begin{cases} \frac{1}{4},&z=0,\\ \frac{1}{2},&z=1,\\ \frac{1}{4},&z=2. \end{cases}

This is a statement about a conditional distribution under specified assumptions, not a guarantee that observed sharing in a finite set of affected pairs will appear in exact proportions. Marker linkage to a disease locus, population structure, nonrandom selection related to marker inheritance, or departures from the assumed transmission model can undermine the argument. The unlinked-locus distribution is a baseline, not a universal law for every family sample.

Worked Example: Conditioning on Both Siblings Being Affected

Suppose a hypothetical disease locus is on a different chromosome from a marker under study. Assume standard Mendelian transmission, no marker-based ascertainment, and a noninbred nuclear family. What is the null probability that an ascertained affected sibling pair shares exactly one marker allele IBD?

At the marker, the two parental indicators are independent, each with probability \(1/2\) of indicating shared transmission. The event \(Z=1\) occurs when exactly one indicator equals 1:

P(Z=1)=P(M=1,F=0)+P(M=0,F=1) =\left(\frac{1}{2}\cdot\frac{1}{2}\right) +\left(\frac{1}{2}\cdot\frac{1}{2}\right) =\frac{1}{2}.

Because the marker is unlinked to the disease locus and the stated ascertainment assumptions hold, conditioning on both siblings being affected leaves this probability unchanged. The null probability of sharing exactly one marker allele IBD is therefore \(0.5\), or 50%.

Null Mean and Variance

The sharing count \(Z\) is a discrete random variable with three possible values. Its null mean is the probability-weighted average of those values. Its variance is the expected squared deviation from the mean; equivalently, it is \(E(Z^2)-[E(Z)]^2\).

\begin{aligned} E(Z) &=0\left(\frac{1}{4}\right)+1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right) =1,\\ E(Z^2) &=0^2\left(\frac{1}{4}\right)+1^2\left(\frac{1}{2}\right)+2^2\left(\frac{1}{4}\right) =\frac{3}{2},\\ \operatorname{Var}(Z) &=E(Z^2)-[E(Z)]^2 =\frac{3}{2}-1^2 =\frac{1}{2}. \end{aligned}

Thus, under the unlinked-locus null, an affected sibling pair shares an average of one allele IBD, with variance \(1/2\) in units of squared allele counts. The standard deviation is \(\sqrt{1/2}\), approximately \(0.7071\) shared alleles. These are properties of the distribution for one pair. They do not by themselves specify the uncertainty of an average across a collection of pairs, which also depends on how those pairs are sampled and whether they are independent.

Worked Example: Checking the Mean and Variance Directly

For the null probabilities \(1/4, 1/2, 1/4\), calculate the mean and variance of the IBD sharing count using the definitions.

The mean is

E(Z)=0\left(\frac{1}{4}\right)+1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)=0+\frac{1}{2}+\frac{1}{2}=1.

Using this mean, the squared deviations are \(1\), \(0\), and \(1\) for \(Z=0,1,2\), respectively. Therefore,

\operatorname{Var}(Z) =(0-1)^2\left(\frac{1}{4}\right) +(1-1)^2\left(\frac{1}{2}\right) +(2-1)^2\left(\frac{1}{4}\right) =\frac{1}{4}+0+\frac{1}{4} =\frac{1}{2}.

The direct calculation agrees with \(E(Z^2)-[E(Z)]^2\). Both methods give a null mean of 1 shared allele and a null variance of 0.5 squared shared alleles.

From Pair Probabilities to Expected Counts

The pair-level distribution can also describe expected category counts across a collection of pairs. If \(n\) independent affected sibling pairs each follow the same null distribution, the expected numbers with \(Z=0,1,2\) are \(n/4,n/2,n/4\), respectively. These are expectations, not fixed counts. Independence matters for calculations about the variability of the total counts; pairs from the same family may not be independent.

Worked Example: Expected Sharing Categories in Independent Families

Imagine a hypothetical set of 24 affected sibling pairs, with one pair sampled from each of 24 unrelated families. Assume each pair follows the unlinked-locus null distribution. Find the expected number of pairs in each sharing category.

For \(Z=0\), multiply the number of pairs by its null probability: \(24(1/4)=6\). For \(Z=1\), the expected count is \(24(1/2)=12\). For \(Z=2\), it is \(24(1/4)=6\). The three expected counts sum to \(6+12+6=24\), the total number of pairs.

These results do not predict that the observed counts must be exactly 6, 12, and 6. For example, counts of 5, 13, and 6 still sum to 24 and can occur under the null. The calculated values describe long-run expectations across repetitions of the sampling process.

Assumptions and Common Pitfalls

The simple derivation is useful because it separates a basic inheritance calculation from more elaborate linkage procedures. Its simplicity also makes it easy to apply too broadly. Keep the biological and sampling assumptions visible when interpreting the \(1:2:1\) baseline.

Confusing IBD with allele matching. Two siblings can carry alleles with the same observed value without those copies descending from the same ancestor. Conversely, IBD is about descent, not whether a genotype label happens to match.
Assuming exact sample proportions. The ratio \(1:2:1\) gives probabilities. A finite set of pairs will generally not contain exactly one-quarter, one-half, and one-quarter in the three categories.
Ignoring ascertainment. The unchanged distribution for affected pairs relies on independence between marker inheritance and disease-related ascertainment under the unlinked-locus null. If selection depends on marker inheritance, the conditional distribution need not be the same.
Treating all pairs as independent. Pairs within one extended family can share relatives and inheritance paths. The single-pair distribution does not imply that multiple pair observations from such a family are independent.
Confusing the mean with a guarantee for a pair. A mean of one does not mean that each pair shares exactly one allele. Individual pairs have \(Z=0\), \(Z=1\), or \(Z=2\), with probabilities \(1/4\), \(1/2\), and \(1/4\).

Key takeaway: In a standard noninbred nuclear family, each parent independently contributes a \(1/2\) chance that two siblings share the transmitted parental copy. Summing the two indicators gives the unlinked-locus IBD probabilities \(1/4,1/2,1/4\), with null mean 1 and variance \(1/2\). For affected pairs, this is the null distribution when disease ascertainment does not select marker inheritance and the marker is unlinked to the disease determinants.

Check Your Understanding

Try these questions using the definitions and assumptions above.

For a sibling pair, what do the maternal and paternal transmission indicators represent, and how do they determine \(Z\)?
Show how the two parental transmission indicators produce the probability \(P(Z=1)=1/2\).
Calculate \(E(Z)\) from the three possible sharing counts and their null probabilities.
Why does conditioning on both siblings being affected leave the marker sharing distribution unchanged under the unlinked-locus null?
For 40 independent affected sibling pairs from unrelated families, what are the expected numbers in the \(Z=0\), \(Z=1\), and \(Z=2\) categories? Why need not the observed counts equal those expectations?