Why Standardize Allele Sharing?
When a chromosomal region is linked to a disorder, affected relatives may inherit the same ancestral chromosome segment more often than expected by chance. Nonparametric linkage (NPL) analysis asks whether this excess sharing is present without specifying a disease-allele frequency or penetrance model. The central task is to summarize sharing among affected relatives, then compare the summary with what that particular pedigree would be expected to show under no linkage.
A raw count is not directly comparable across pedigrees of different sizes or structures. A family with many affected relatives has more opportunities to share alleles than a small family. Whittemore–Halpern scores address this by centering a sharing statistic at its no-linkage expectation and scaling it by its no-linkage standard deviation. Two useful summaries emphasize different features: sharing common to all affected relatives, and the sum of sharing across affected-relative pairs.
For example, a shared allele lineage carried by every affected member contributes to \(S_{\mathrm{all}}\). In \(S_{\mathrm{pairs}}\), it contributes once for each pair of affected members who carry it. Consequently, \(S_{\mathrm{pairs}}\) can be large because several pairs share, even when no single allele lineage is common to everyone. Neither statistic is simply a count of matching marker alleles: IBD concerns descent from the same ancestral allele, not merely alleles that happen to have the same observed state.
Standardizing Against the Pedigree’s No-Linkage Distribution
For either sharing statistic \(S\), the basic Whittemore–Halpern standardized score has the form
Here \(E_0(S)\) and \(\operatorname{Var}_0(S)\) are the expectation and variance under the no-linkage reference distribution for the pedigree. Under that reference, disease status does not make the relatives’ inheritance pattern at the locus more similar; the inheritance pattern follows the Mendelian transmission probabilities appropriate to the pedigree. Pedigree structure matters because it determines both which sharing configurations are possible and how likely they are.
When the inheritance state is fully known, \(S\) can be calculated directly. In practice, marker data often leave uncertainty about IBD. The sharing statistic can then be calculated for each possible inheritance state and averaged using the state probabilities conditional on the marker data. This produces an expected sharing statistic rather than a claim that one uncertain state is known. The score’s interpretation remains an excess relative to the no-linkage reference, with the uncertainty and calibration handled according to the method being used.
Worked Example: Deriving the Null Distribution for Three Siblings
Worked Example: Deriving the Null Distribution for Three Siblings
Suppose three affected full siblings are considered at a locus. Under no linkage, each parent independently transmits either of their two alleles to each child with probability \(1/2\). Assume ordinary Mendelian segregation and no segregation distortion. Derive the no-linkage means and variances for \(S_{\mathrm{all}}\) and \(S_{\mathrm{pairs}}\).
Total sharing. For one parental origin, all three siblings receive the same one of the parent's two alleles with probability \(2(1/2)^3=1/4\): there are two all-same transmission patterns among eight equally likely patterns. Let \(A\) be 1 if that parental origin supplies an allele shared by all three, and 0 otherwise. Then \(A\) is Bernoulli with probability \(1/4\), so \(E_0(A)=1/4\) and \(\operatorname{Var}_0(A)=(1/4)(3/4)=3/16\). The paternal and maternal contributions are independent under the stated model. Thus \(S_{\mathrm{all}}\) is the sum of two such indicators:
Pairwise sharing. For one parental origin there are three sibling pairs. If all three transmissions match, all three pairs share that parent's allele; this happens with probability \(1/4\), giving a contribution of 3. Otherwise, two siblings receive one parental allele and the third receives the other; exactly one pair shares, giving a contribution of 1. Therefore, for one parental origin, the sharing contribution \(B\) has \(P(B=3)=1/4\) and \(P(B=1)=3/4\). Its mean and variance are
The two parental origins are independent, so \(S_{\mathrm{pairs}}\) is the sum of two independent contributions with this distribution. Hence
The derivation also shows why the three pairwise counts must not be treated as independent observations: their values are linked by the same three transmissions. The variances above account for that joint structure.
Worked Example: A Realizable Sharing Pattern
Worked Example: A Realizable Sharing Pattern
Continue with three affected full siblings. Suppose the paternal transmission gives all three siblings the same paternal allele lineage. The first and second siblings also receive the same maternal allele lineage, while the third receives the other maternal allele lineage. Assume this inheritance pattern is known for the illustration.
The pairwise IBD counts are \(K_{12}=2\), \(K_{13}=1\), and \(K_{23}=1\). Siblings 1 and 2 share both parental-origin lineages; each shares the paternal lineage, but not the maternal lineage, with sibling 3. This is a realizable pattern because the two pairs involving sibling 3 have the same count, as required when siblings 1 and 2 share both lineages.
One paternal allele lineage is shared by all three, and no maternal lineage is shared by all three. Thus \(S_{\mathrm{all}}=1\). The pairwise total is
Using the no-linkage means and variances derived above, the standardized scores are
Both scores are positive because the pattern has more sharing than the corresponding no-linkage mean. Their equality here is a feature of this particular pattern and null distribution, not a general identity between the two statistics. Neither value alone supplies a calibrated significance level.
Worked Example: Sharing Inferred With Uncertainty
Worked Example: Averaging Over Possible Inheritance States
Suppose marker information in the same three-sibling pedigree leaves two inheritance states plausible. For illustration, assume the probabilities conditional on the marker data are 0.75 for a state with \(S_{\mathrm{all}}=1\) and \(S_{\mathrm{pairs}}=4\), and 0.25 for a state with \(S_{\mathrm{all}}=0\) and \(S_{\mathrm{pairs}}=2\). These are hypothetical probabilities and states, chosen to illustrate the calculation.
The marker-inferred expected total sharing is
The corresponding expected pairwise sharing is
Using the same pedigree-specific no-linkage reference means and standard deviations as before gives the illustrative standardized excesses
The fractional sharing values are expectations over uncertain inheritance states, not literal fractions of an allele. This example uses the basic standardized construction to show how probabilistic IBD information enters the statistic. It does not imply that all analysis software uses the same uncertainty treatment or that these standardized values have a standard normal distribution in every pedigree.
What the Two Statistics Emphasize
| Feature | \(S_{\mathrm{all}}\) | \(S_{\mathrm{pairs}}\) |
|---|---|---|
| Summary | Allele lineages shared by every affected relative | Total IBD sharing summed over affected-relative pairs |
| Emphasizes | Sharing common to the whole affected group | Sharing distributed across the pairs |
| Dependence issue | Possible shared lineages constrain who can share | Pairwise counts are dependent because pairs overlap in relatives |
| Reference for standardization | Pedigree-specific no-linkage mean and variance | Pedigree-specific no-linkage mean and variance |
These statistics can respond differently to the same pedigree. A pattern where one subset shares strongly but another affected relative does not may increase the pairwise sum without producing an allele lineage common to everyone. Conversely, a lineage shared by all affected relatives directly contributes to the total-sharing statistic. Their different emphases are why the choice of statistic affects the evidence being summarized.
Common Pitfalls
- Assuming every combination of pairwise IBD counts is possible. Pairwise sharing values arise from a single compatible inheritance pattern. For example, if two siblings share both parental-origin lineages, a third sibling’s IBD count with each of them must be the same. Specify a realizable inheritance pattern before using pairwise counts.
- Using a universal null mean and variance. The reference distribution depends on the pedigree and the statistic. Derive or calculate it from the relevant inheritance possibilities rather than applying the three-sibling values to a different family.
- Treating overlapping pairs as independent. Pairs such as (1,2) and (1,3) share a relative and depend on that person’s transmissions. The variance of a pairwise sum must reflect the joint inheritance distribution.
- Calling the NPL score a p-value. A positive standardized score describes excess sharing relative to a reference mean in standard-deviation units. Its null distribution may be discrete or otherwise differ from a standard normal distribution, so significance requires appropriate calibration.
- Confusing IBD with matching marker alleles. Two relatives can carry marker alleles with the same observed state without inheriting them from the same ancestor. NPL sharing concerns IBD, inferred from pedigree and marker information.
Check Your Understanding
Work through these questions using the definitions and calculations above.
- In your own words, distinguish \(S_{\mathrm{all}}\) from \(S_{\mathrm{pairs}}\).
- For three affected full siblings under the stated no-linkage model, why is the probability that one parental origin is shared by all three equal to \(1/4\)?
- For a realizable three-sibling pattern with pairwise counts 2, 1, and 1, what are \(S_{\mathrm{pairs}}\) and \(S_{\mathrm{all}}\) if the two-sharing pair shares both parental-origin lineages?
- Why can the pairwise counts within a pedigree not generally be treated as independent?
- What does a positive standardized NPL score say, and what does it not say?