Multipoint IBD Probabilities Along A Chromosome

Why Multipoint IBD Probabilities Use the Whole Chromosome

At a single marker, relatives’ genotypes can leave several possible inheritance patterns, especially when parental genotypes or haplotypes are not fully informative. Nearby markers add evidence: a chromosome segment tends to be inherited continuously, with changes in the transmitted parental chromosome occurring through recombination. Multipoint inference uses both facts together. It estimates the probability of each inheritance pattern at a position from the marker genotypes across the region, not just from the genotype at the nearest marker.

The target is a posterior distribution over inheritance states. For affected relatives, that distribution describes how likely each IBD-sharing state is given the pedigree marker data and the model of transmission along the chromosome. It is not itself a test statistic, and it does not say that sharing caused the relatives’ affection status.

Definition: An inheritance vector at a marker records, for each nonfounder in a pedigree, which of the two parental homologues was transmitted by each parent. A particular inheritance vector is one possible joint pattern of transmissions. Multipoint IBD probabilities are posterior probabilities of IBD-sharing states at a chromosome position, conditional on the pedigree’s marker data across multiple positions and the assumed transmission model.

Inheritance States and Their IBD Meaning

For two siblings, represent the inheritance vector at marker \(j\) by \(V_j=(b_{1p},b_{1m},b_{2p},b_{2m})\). Each bit indicates which parental homologue a sibling inherited: \(p\) denotes paternal and \(m\) maternal transmission, while siblings are indexed by 1 and 2. The convention for coding the two homologues as 0 or 1 is arbitrary, provided it is used consistently.

The siblings share the paternal allele IBD if their paternal bits agree, and share the maternal allele IBD if their maternal bits agree. Thus their number of alleles shared IBD is \(K(V_j)=I(b_{1p}=b_{2p})+I(b_{1m}=b_{2m})\), where \(I(\cdot)\) equals 1 when its condition is true and 0 otherwise. The possible values are 0, 1, and 2. This is identity by descent, not merely identity by state: alleles that happen to have the same marker value are not necessarily inherited from the same ancestral copy.

For a larger pedigree, the inheritance vector has one bit for every parental transmission. The IBD state for a selected pair or group of relatives is a function of the relevant bits. In either case, the calculation proceeds by estimating probabilities of inheritance vectors and then summing the probabilities of all vectors that imply the IBD state of interest.

How Recombination Connects Neighboring Positions

Consider adjacent marker positions \(j\) and \(j+1\), with recombination fraction \(\theta_j\) between them. Under the standard transmission model, each transmitted homologue remains the same with probability \(1-\theta_j\) and switches with probability \(\theta_j\). Switches in distinct meioses are modeled as independent. Conditional on the inheritance vector at position \(j\), these switches determine the inheritance vector at \(j+1\).

Transition probability: If \(V_j\) and \(V_{j+1}\) contain \(M\) transmission bits and each has recombination fraction \(\theta_j\) over the interval, then \(P(V_{j+1}=w\mid V_j=v)=\prod_{r=1}^{M}\theta_j^{I(v_r\ne w_r)}(1-\theta_j)^{I(v_r=w_r)}\). If recombination fractions differ by meiosis, use the corresponding fraction for each bit. This transition model assumes independent transmissions and uses the recombination fraction for the interval, not the physical distance alone.

The formula is a product because each bit either switches or stays the same. For example, with four bits and \(\theta_j=0.10\), the probability that a particular vector remains unchanged is \(0.90^4=0.6561\). The probability of a specified vector differing in exactly one named bit is \(0.10(0.90)^3=0.0729\). These are probabilities of particular destination vectors; the probability of any one-bit change is the sum over the four possible bits.

Worked Example: Probability That Siblings’ Maternal Sharing Changes

Suppose two siblings’ maternal transmissions are observed across one interval with recombination fraction \(\theta=0.10\). Let \(S\) mean that the siblings inherited the same maternal homologue at a position, and \(D\) mean that they inherited different maternal homologues. Find the probability that the state changes from \(S\) to \(D\) across the interval.

For the siblings to change from same to different, exactly one of their two maternal transmissions must recombine. Either the first recombines and the second does not, or the second recombines and the first does not. By independence,

P(D_{j+1}\mid S_j) =\theta(1-\theta)+(1-\theta)\theta =2\theta(1-\theta) =2(0.10)(0.90) =0.18.

The probability of remaining in the same-sharing state is therefore \(1-0.18=0.82\). By symmetry, the probability of changing from different to same is also 0.18. The resulting two-state transition matrix, with rows and columns ordered \(S,D\), is

T= \begin{pmatrix} 0.82 & 0.18\\ 0.18 & 0.82 \end{pmatrix}.

This two-state transition describes only the parity of the two maternal transmissions. It is not the transition probability for a single transmission bit, which is 0.10. Confusing those probabilities would understate the chance that the siblings’ sharing state changes.

Marker Genotypes Update the Inheritance States

A transition model alone describes how states tend to persist; marker data indicate which states are compatible with the observed genotypes. Let \(G_j\) denote the genotypes observed in the pedigree at marker \(j\). For a candidate state \(v\), define the marker likelihood \(L_j(v)=P(G_j\mid V_j=v)\), conditional on the available pedigree information and the marker model. With known parental genotypes and phase, this likelihood follows from Mendelian transmission. If phase, founder alleles, or genotypes are uncertain, those uncertainties must be included or consistently marginalized when calculating the likelihood.

A likelihood need not sum to one across inheritance states. It measures how well the observed genotypes fit each state. If genotypes are uninformative about the state, the likelihoods are equal and the marker does not change their relative probabilities. If some states are incompatible with the observed genotypes, they have likelihood zero.

For a single interval, the update has two stages. First propagate the state probabilities through the recombination transition. Then multiply each predicted probability by its marker likelihood and normalize. The next example works through both stages.

Worked Example: Updating Maternal IBD With Two Markers

Suppose the state is \(S\) or \(D\) for whether two affected siblings share the same maternal homologue. At marker 1, use prior probabilities \(P(S)=P(D)=0.50\). Hypothetical marker genotypes have likelihoods \(L_1(S)=0.80\) and \(L_1(D)=0.20\). Between markers, let the recombination fraction for each maternal transmission be 0.10, so the state-switch probability from the previous example is 0.18.

At marker 1, the unnormalized posterior weights are \(0.50(0.80)=0.40\) for \(S\) and \(0.50(0.20)=0.10\) for \(D\). Their sum is 0.50, giving \(P(S\mid G_1)=0.40/0.50=0.80\) and \(P(D\mid G_1)=0.20\).

Propagate these probabilities to marker 2 before using its genotype data:

P(S_2\mid G_1)=0.80(0.82)+0.20(0.18)=0.692, \qquad P(D_2\mid G_1)=0.80(0.18)+0.20(0.82)=0.308.

Now suppose the marker 2 genotype likelihoods are \(L_2(S)=0.30\) and \(L_2(D)=0.90\). The unnormalized weights are \(0.692(0.30)=0.2076\) for \(S\) and \(0.308(0.90)=0.2772\) for \(D\), with total 0.4848. Therefore

P(S_2\mid G_1,G_2)=\frac{0.2076}{0.4848}=0.4282, \qquad P(D_2\mid G_1,G_2)=\frac{0.2772}{0.4848}=0.5718.

For this simplified example, suppose the siblings’ paternal transmissions are known to match at marker 2. They then share two alleles IBD if the maternal state is \(S\), and one allele IBD if it is \(D\). The multipoint probabilities are consequently \(P(K=2\mid G_1,G_2)=0.4282\) and \(P(K=1\mid G_1,G_2)=0.5718\), with \(P(K=0\mid G_1,G_2)=0\). These are posterior probabilities conditional on the stated data and assumptions, not probabilities that the siblings’ affection status was caused by the marker.

Using Evidence From Both Sides of a Position

At an interior chromosome position, marker genotypes to the right can be informative as well as those to the left. A forward calculation summarizes evidence from markers up through the current position; a backward calculation summarizes evidence from markers after it. Combining the two gives the posterior state distribution at that position. This is the central multipoint step.

Forward-backward result: For a sequence of markers, let \(\alpha_j(v)=P(G_1,\ldots,G_j,V_j=v)\) and \(\beta_j(v)=P(G_{j+1},\ldots,G_J\mid V_j=v)\). Then \(P(V_j=v\mid G_1,\ldots,G_J)\) is proportional to \(\alpha_j(v)\beta_j(v)\); normalize these products over all states. The recursions are \(\alpha_j(v)=L_j(v)\sum_u\alpha_{j-1}(u)P(v\mid u)\) and \(\beta_j(u)=\sum_v P(v\mid u)L_{j+1}(v)\beta_{j+1}(v)\).

The forward recursion starts from the distribution of inheritance states before the first marker, usually the Mendelian transmission distribution for unselected transmissions. Each step sums over all possible previous states, applies the transition probability, and incorporates the current marker likelihood. The backward recursion starts with \(\beta_J(v)=1\) at the last marker and sums over possible next states. If the number of inheritance states is large, probabilities can be rescaled at each step to avoid numerical underflow; the same scaling must be accounted for when normalizing the final posterior.

Worked Example: Flanking Markers Inform IBD at an Interior Marker

Consider the two-state maternal-sharing model \(S,D\) at three markers, with state-switch probability 0.18 between each adjacent pair. The initial probabilities are 0.50 for each state. Suppose the hypothetical marker likelihoods are \(L_1(S)=0.80,L_1(D)=0.20\); \(L_2(S)=L_2(D)=1\); and \(L_3(S)=0.40,L_3(D)=0.80\). Marker 2 is uninformative, but both flanking markers can still affect the estimate there.

After marker 1, the unnormalized forward weights are \(\alpha_1(S)=0.50(0.80)=0.40\) and \(\alpha_1(D)=0.50(0.20)=0.10\). Propagating to marker 2 gives

\alpha_2(S)=0.40(0.82)+0.10(0.18)=0.346, \qquad \alpha_2(D)=0.40(0.18)+0.10(0.82)=0.154.

For the evidence to the right of marker 2, the backward weights are

\beta_2(S)=0.82(0.40)+0.18(0.80)=0.472, \qquad \beta_2(D)=0.18(0.40)+0.82(0.80)=0.728.

Multiply the forward and backward weights. The unnormalized posterior weights at marker 2 are \(0.346(0.472)=0.163312\) for \(S\), and \(0.154(0.728)=0.112112\) for \(D\). Their sum is 0.275424, so

P(S_2\mid G_1,G_2,G_3) =\frac{0.163312}{0.275424}=0.5929, \qquad P(D_2\mid G_1,G_2,G_3)=0.4071.

If paternal sharing is known to match, these are also the probabilities of sharing two versus one alleles IBD at marker 2. The right-hand marker shifts the estimate from the forward-only probability \(0.346/(0.346+0.154)=0.692\) to 0.5929. Thus a marker’s contribution depends not only on its own genotype but also on how inheritance states can persist between positions.

From Inheritance-Vector Probabilities to IBD Estimates

In the full pedigree calculation, each state can imply a different IBD configuration for the affected relatives. Let \(\mathcal{V}_k\) be the set of inheritance vectors that imply \(k\) shared alleles IBD for a selected sibling pair. Once the posterior over vectors is calculated, the IBD probabilities are

P(K=k\mid G_1,\ldots,G_J) =\sum_{v\in\mathcal{V}_k}P(V_j=v\mid G_1,\ldots,G_J), \qquad k=0,1,2.

For a group of affected relatives, the IBD configuration can be more complex than a single count, but the same principle applies: add posterior probabilities for all inheritance vectors that produce the configuration being reported. These estimates describe uncertainty about descent given the observed markers. They are inputs to later nonparametric linkage summaries, not substitutes for those summaries.

In a nonparametric analysis, affection status identifies the relatives or pedigrees being examined for excess sharing; the calculation above uses marker data and transmission probabilities to estimate their IBD sharing. It does not require a specified disease-allele frequency or penetrance model. Adding a phenotype likelihood would change the conditioning and would be a different model-based calculation.

Assumptions and Common Pitfalls

Treating adjacent marker states as independent. Nearby inheritance states are connected by recombination. The transition probabilities must reflect the genetic interval; assuming a fresh independent inheritance vector at every marker discards the linkage information.
Using the single-meiosis recombination fraction for a pair’s sharing state. For two independent transmissions from the same parent, the pair’s same-versus-different state changes if exactly one transmission recombines. Its switch probability is \(2\theta(1-\theta)\), not \(\theta\).
Using only the closest marker. A multipoint posterior at an interior position incorporates marker evidence on both sides, propagated through the intervening transitions. A single-marker estimate is not generally the multipoint estimate.
Calling IBS evidence IBD certainty. Genotypes that match can arise from distinct ancestral copies. Marker likelihoods inform the inheritance vector, but the posterior can remain uncertain, especially at uninformative markers or with missing genotypes.
Interpreting an IBD posterior as causal evidence. A high posterior probability of sharing concerns descent conditional on the model and marker data. It does not establish that the shared region contains a causal variant or that sharing caused affection.

The transition model should match the map and transmission structure, and marker likelihoods should reflect the pedigree genotypes and any uncertainty in founder phase or alleles. These conditions matter because an incorrect transition or likelihood can move probability among inheritance states and distort the resulting IBD estimates.

Check Your Understanding

Work through these questions by tracing how state transitions and marker evidence contribute to the posterior.

For two siblings’ transmissions from one parent, why is the probability of changing from same to different sharing \(2\theta(1-\theta)\), rather than \(\theta\)?
If two inheritance states have equal marker likelihoods, what effect does that marker have on their relative posterior probabilities?
In a forward-backward calculation, what information is represented by \(\alpha_j(v)\), and what additional information is represented by \(\beta_j(v)\)?
For two siblings with known matching paternal transmissions, how do maternal states \(S\) and \(D\) map to IBD counts?
Why does a high posterior probability of IBD sharing not establish that a shared region caused the relatives’ disease?