Excess-Sharing Tests For Affected Sib Pairs

Why Test Excess IBD Sharing?

A marker can help assess linkage to a disease locus because affected siblings may inherit marker alleles together with the disease-associated region. When the marker is unlinked to the disease locus, a pair of siblings has the usual Mendelian IBD distribution. If the marker is linked, affected sibling pairs may share more alleles IBD than expected under that distribution. The basic question is whether the observed sharing is unusually high, not whether the marker genotypes merely look similar.

An affected-sib-pair test does not require specifying a disease-allele frequency or penetrance model. It does require a valid IBD state for each pair and a clearly stated sampling unit. The tests below treat pairs from independent families as independent observations; repeated pairs from one family cannot automatically be counted as independent.

Null Distribution and Test Setup

Definition: For an affected sibling pair, let \(Z_i\) be the number of alleles the siblings share identical by descent (IBD) at the marker, where \(Z_i\in\{0,1,2\}\). Under the no-linkage null hypothesis, the IBD probabilities are \(P(Z_i=0)=1/4\), \(P(Z_i=1)=1/2\), and \(P(Z_i=2)=1/4\). For \(m\) independent pairs, let \(N_0,N_1,N_2\) be the numbers in these three classes, so \(N_0+N_1+N_2=m\).

The null distribution applies when the marker is unlinked to the disease locus and Mendelian transmission is correctly represented. The pairs must be independent for the simple calculations below; in particular, the observations should not include multiple correlated pairs from the same family. The IBD states are treated as known without error. If IBD information is uncertain, replacing it with an unexamined hard classification can distort the test.

The no-linkage null is \(H_0:E(Z_i)=1\). The directional linkage alternative for an excess-sharing test is \(H_a:E(Z_i)>1\). The direction matters: these tests ask whether sharing is greater than expected, not whether the distribution differs in either direction.

Test Based on Mean IBD Sharing

For \(m\) independent affected sibling pairs, the observed mean sharing is \(\bar{Z}=(N_1+2N_2)/m\). Under the null, a pair’s expected sharing is \(0(1/4)+1(1/2)+2(1/4)=1\). Its variance is \(0^2(1/4)+1^2(1/2)+2^2(1/4)-1^2=1/2\). Therefore, the null standard error of the mean is \(\sqrt{(1/2)/m}\). For a sufficiently large number of independent pairs, a one-sided standardized statistic is compared with the standard normal distribution.

Z_{\text{mean}}=\frac{\bar{Z}-1}{\sqrt{1/(2m)}}=\frac{N_2-N_0}{\sqrt{m/2}}.

The second expression follows because \(N_1+2N_2-m=N_2-N_0\). Thus, the mean-sharing test is a standardized contrast between the two-IBD and zero-IBD classes. A positive statistic indicates that the observed count in the two-IBD class exceeds the count in the zero-IBD class. A small one-sided p-value is evidence against the no-linkage null in the direction of excess mean sharing.

Worked Example: Mean Sharing in 40 Independent Pairs

Suppose a hypothetical analysis includes \(m=40\) affected sibling pairs from 40 independent families. The observed counts are \(N_0=7\), \(N_1=18\), and \(N_2=15\). These sum to 40, as required. Under the null, the expected class counts are 10, 20, and 10, respectively, so each class has a reasonable expected count for a normal approximation.

The observed mean number of alleles shared IBD is

\bar{Z}=\frac{N_1+2N_2}{m} =\frac{18+2(15)}{40} =\frac{48}{40} =1.200.

Under the null, \(SE=\sqrt{1/(2m)}=\sqrt{1/80}=0.1118\). The standardized statistic is

Z_{\text{mean}}=\frac{1.200-1}{0.1118}=1.789.

The one-sided standard normal p-value is approximately \(P(Z\geq1.789)=0.0368\), where \(Z\) here denotes a standard normal variable. At the 0.05 significance level, this is evidence of excess mean IBD sharing under the stated model and assumptions. It is not a 3.68% probability that the null hypothesis is true, nor does it establish that the marker causes disease.

Test Based on Sharing Two Alleles

A second approach focuses only on the proportion of pairs with \(Z_i=2\). Under the null, each independent pair has probability \(1/4\) of sharing two alleles IBD. Consequently, \(N_2\) has a binomial distribution with \(m\) trials and success probability \(1/4\). For the one-sided alternative of excess two-IBD sharing, the exact p-value is the probability, under this null distribution, of observing at least the observed number of two-IBD pairs.

N_2\sim\operatorname{Binomial}(m,1/4), \qquad p\text{-value}=P(N_2\geq N_{2,\mathrm{obs}}\mid H_0) =\sum_{t=N_{2,\mathrm{obs}}}^{m}\binom{m}{t}(1/4)^t(3/4)^{m-t}.

Because this is a discrete distribution, the exact p-value need not equal a chosen significance level. The test uses the full binomial tail rather than a normal approximation, which is particularly useful when the number of pairs is small or the expected number of two-IBD pairs is low.

Worked Example: Exact Two-IBD Test for 12 Pairs

In a hypothetical set of \(m=12\) independent affected sibling pairs, suppose \(N_2=7\) pairs share two alleles IBD. Under the null, \(N_2\sim\operatorname{Binomial}(12,0.25)\). The one-sided exact p-value is

\begin{aligned} P(N_2\geq7\mid H_0) &=\sum_{t=7}^{12}\binom{12}{t}(0.25)^t(0.75)^{12-t}\\ &\approx 0.01147127+0.00238985+0.000354052\\ &\quad+0.000035405+0.000002146+0.0000000596\\ &=0.01425278. \end{aligned}

Rounded to four significant figures, the p-value is 0.01425. At \(\alpha=0.05\), this is evidence against the no-linkage null in the direction of excess two-IBD sharing. The calculation assumes the 12 pairs are independent and that each pair’s IBD class is correctly determined. The proportion observed is \(7/12=0.5833\), while the null expectation is 0.25; the p-value assesses how unusual a count at least this large would be under the null, not the probability that linkage is present.

What the Three IBD Classes Reveal

The three class counts contain more information than either a mean or a two-IBD proportion alone. Under the null, their expected counts are \(m/4,m/2,m/4\). The mean test compresses the data to \(N_2-N_0\), so it does not respond to every possible departure from those expected frequencies. The two-IBD test instead targets an increase in \(N_2\) without directly using \(N_0\) or \(N_1\).

A general comparison of the observed class counts with their null expectations can be made with a Pearson goodness-of-fit statistic:

X^2=\frac{(N_0-m/4)^2}{m/4} +\frac{(N_1-m/2)^2}{m/2} +\frac{(N_2-m/4)^2}{m/4}.

When the expected counts are sufficiently large, this statistic is approximately chi-square distributed with \(df=2\): there are three categories, and their probabilities sum to one, leaving two independent deviations. This is an omnibus, non-directional test. It can detect departures in patterns that do not represent excess mean sharing, so a small p-value alone does not say that sharing is increased. The direction and the class counts must be examined.

Worked Example: A Class-Distribution Departure Without Excess Mean

Consider hypothetical counts from \(m=24\) independent pairs: \(N_0=4\), \(N_1=16\), and \(N_2=4\). The counts sum to 24. Under the null, the expected counts are 6, 12, and 6, all large enough for the usual chi-square approximation.

The observed mean is

\bar{Z}=\frac{16+2(4)}{24}=1, \qquad N_2-N_0=4-4=0.

Thus the mean-sharing statistic is zero: this pattern gives no excess mean sharing. The observed proportion sharing two alleles is \(4/24=0.1667\), below the null value 0.25. For the omnibus comparison, the Pearson statistic is

X^2=\frac{(4-6)^2}{6}+\frac{(16-12)^2}{12}+\frac{(4-6)^2}{6} =\frac{4}{6}+\frac{16}{12}+\frac{4}{6} =2.667.

With \(df=2\), the approximate upper-tail p-value is 0.2636. This example shows why the choice of statistic must match the question: a general distribution test and a directional excess-sharing test answer different questions. These data do not show excess mean sharing, and the omnibus result is not statistically significant at \(\alpha=0.05\).

Choosing and Interpreting a Test

The mean test uses both the zero- and two-IBD classes through their contrast; the binomial test asks specifically whether the two-IBD class is overrepresented; and the goodness-of-fit test checks the full distribution against the null. The methods can yield different results because they emphasize different features of the data. A test should be chosen to reflect the prespecified scientific question, rather than selected after looking for whichever p-value is smallest.

The calculations also rely on the independent-pair setup. If several affected pairs come from the same family, their sharing states may be correlated. Treating each as an independent binomial trial or using the simple mean-test standard error can then give an incorrect reference distribution. The unit of independence and the way families enter the analysis must be accounted for before interpreting a p-value.

Key takeaway: Under no linkage, sibling-pair IBD sharing has probabilities \(1/4,1/2,1/4\) for zero, one, and two alleles. Affected-sib-pair excess-sharing tests ask whether observed sharing departs in a specified direction or in the overall distribution; the statistic, assumptions, and interpretation must remain aligned.

Common Mistake

A common mistake is to see many two-IBD pairs and conclude that the null probability is the observed proportion, or to treat an omnibus departure from the three-class distribution as proof of excess sharing. The correction is to calculate the reference distribution under \(H_0\), use the statistic that matches the alternative, and interpret the p-value as a tail probability conditional on that null model. For a directed excess test, the direction and the independence assumptions must also be justified.

Check Your Understanding

Use the null distribution and assumptions stated above to answer these questions.

For one affected sibling pair under no linkage, what are the probabilities of sharing zero, one, and two alleles IBD?
For \(m\) independent pairs, explain why the null standard error of mean IBD sharing is \(\sqrt{1/(2m)}\).
What count has a binomial distribution in the test for excess sharing of two alleles, and what is its null success probability?
Show algebraically why the mean-sharing numerator is proportional to \(N_2-N_0\), not to \(N_2\) alone.
Why can an omnibus test of the three IBD classes be significant without demonstrating excess mean sharing?
What problem arises if several sibling pairs from one family are treated as independent observations?