Inferring Marker IBD From Relative Genotypes

From Genotypes to Descent

At a marker, a relative’s genotype records the alleles observed, but not necessarily which ancestral allele copies were inherited. This distinction matters in linkage analysis: the evidence of interest is whether relatives inherited the same allele copy from a common ancestor, not merely whether their genotypes contain alleles with the same label.

For a pair of full siblings, let \(K\) be the number of allele copies they share identical by descent (IBD) at a locus. The possible states are \(K=0,1,2\). Under Mendelian inheritance, with unrelated, noninbred parents and no selection on the marker, their prior probabilities at one locus are \(P(K=0)=1/4\), \(P(K=1)=1/2\), and \(P(K=2)=1/4\). These are prior probabilities before using the marker genotypes. Genotype evidence updates them; it does not automatically reveal the state.

Definition: Two allele copies are identical by descent (IBD) if they are copies of the same ancestral allele inherited through the pedigree. They are identical by state (IBS) if they have the same observed allele value, whether or not they descend from the same copy. For a relative pair, the IBD sharing state \(K\) is the number of allele copies shared IBD at the locus: 0, 1, or 2.

A diploid genotype is usually unordered: \(A/B\) records one copy of allele \(A\) and one of \(B\), but does not say which parent supplied either copy. Thus, matching genotypes can be IBS without establishing IBD. Conversely, a known transmission path can reveal IBD even when the relatives’ complete genotypes are not identical.

What Genotype Configurations Tell Us

With error-free genotyping and no mutation, a pair with no allele values in common cannot share an allele IBD: an allele inherited from a common ancestor must have the same marker value in both relatives. The converse does not hold. Sharing a value may result from inheritance of the same copy or from two distinct copies that happen to carry the same allele value.

If two relatives have identical genotypes, \(K=2\) is possible, but so are \(K=1\) and \(K=0\). If their genotypes differ, \(K=2\) is ruled out, but the pair may still share one allele IBD. These deductions assume accurate genotypes and no mutation; genotyping error or mutation can break the direct correspondence between observed values and inherited copies.

A formal way to use genotype evidence is through a likelihood. Let \(G_1\) and \(G_2\) denote the two observed genotypes, and let \(L_k=P(G_1,G_2\mid K=k)\). The conditional probability of an IBD state after observing the genotypes is

P(K=k\mid G_1,G_2) = \frac{L_k P(K=k)} {\sum_{j=0}^{2} L_j P(K=j)}.

The likelihood depends on allele frequencies and on the assumed inheritance model. One way to compute it is to generate the alleles under a specified IBD state, then sum the probabilities of all allele assignments that produce the observed unordered genotypes. For \(K=0\), the four copies are independent draws from the population allele frequencies under the usual founder model. For \(K=1\), the pair shares one ancestral copy and has one separate copy each. For \(K=2\), both ancestral copies are shared. Parental genotypes can provide additional evidence by constraining which alleles could have been transmitted.

Worked Example: Parental Genotypes Identify One Shared Copy

Suppose a father has genotype \(A/B\), a mother has genotype \(C/D\), and their two children have genotypes \(A/C\) and \(A/D\). Assume the allele labels are distinct, genotyping is accurate, and the parents are the biological parents.

For the first child, the paternal contribution must be \(A\) and the maternal contribution must be \(C\). For the second child, the paternal contribution must also be \(A\), while the maternal contribution must be \(D\). Because the father has only one \(A\) copy, both children inherited that same paternal copy. Their maternal contributions differ, so they do not share the same maternal copy.

Therefore, at this marker, the pair shares exactly one allele copy IBD: \(K=1\). Their genotypes share the allele value \(A\), but the inference of IBD comes from the parental genotypes and the unique transmission paths, not from the matching value alone.

Ambiguous Phase and the Value of Parents

Phase describes how alleles at a locus, or across loci, are assigned to the two homologous chromosome copies. At a single marker, the practical ambiguity is often parental origin: an unordered genotype does not reveal which allele came from which parent. If several transmissions are consistent with the observed genotypes, the IBD state may remain uncertain even when parental genotypes are available.

Heterozygous parents can be especially informative when their alleles distinguish the possible transmissions. If both parents are \(A/B\) and a child is \(A/B\), that child may have received paternal \(A\) and maternal \(B\), or paternal \(B\) and maternal \(A\). The genotype alone does not distinguish these phases. For two such children, the possible phase assignments can imply different IBD states.

Worked Example: Identical Genotypes, Different Possible IBD States

Suppose both parents are \(A/B\), and both siblings are also \(A/B\). Assume the parents are unrelated, noninbred founders; transmissions from each parent are Mendelian; and there is no genotyping error. For each child, the possible parental origins are paternal \(A\) with maternal \(B\), or paternal \(B\) with maternal \(A\).

Conditional on the child being \(A/B\), each of those two phase assignments has probability \(1/2\), since the four equally likely paternal–maternal transmission combinations include two that produce \(A/B\). The two children’s transmissions are independent conditional on the parental genotypes. There are therefore four equally likely joint phase assignments:

If both children receive paternal \(A\) and maternal \(B\), they share both parental copies, so \(K=2\).
If both receive paternal \(B\) and maternal \(A\), they also share both parental copies, so \(K=2\).
If one receives paternal \(A\) and maternal \(B\), and the other receives paternal \(B\) and maternal \(A\), they share neither transmitted copy, so \(K=0\).

Two of the four assignments give \(K=2\), and two give \(K=0\). Thus, conditional on these parental and sibling genotypes, \(P(K=2)=1/2\), \(P(K=0)=1/2\), and \(P(K=1)=0\) under the stated model. The siblings’ identical \(A/B\) genotypes do not identify whether they share zero or two copies.

This example also illustrates why allele labels alone are insufficient: an \(A\) inherited from the father and an \(A\) inherited from the mother are not necessarily the same ancestral copy. Parental information helps only to the extent that it distinguishes allele copies and their transmissions. Homozygous parents, for example, may provide little information about which of their two homologous copies was transmitted.

Using Allele Frequencies When Phase Is Unknown

When parental genotypes or phase are unavailable, allele frequencies help quantify how plausible a genotype pair is under each IBD state. Consider full siblings under the prior probabilities above, with allele \(A\) having population frequency \(q\). Assume unrelated, noninbred founders, random mating, independent allele draws for copies that are not IBD, and no mutation or genotyping error.

For two siblings both observed as \(A/A\), the likelihood under \(K=0\) is \(q^4\): all four independent copies must be \(A\). Under \(K=1\), the shared copy and the two separate copies must all be \(A\), giving likelihood \(q^3\). Under \(K=2\), the two shared ancestral copies must be \(A\), giving likelihood \(q^2\). Bayes’ rule then combines these likelihoods with the prior probabilities.

Worked Example: Updating IBD Probabilities for Two Homozygotes

Suppose the siblings are both \(A/A\), and the frequency of \(A\) is \(q=0.10\). The likelihoods are

L_0=(0.10)^4=0.0001,\qquad L_1=(0.10)^3=0.001,\qquad L_2=(0.10)^2=0.01.

Multiplying each likelihood by its prior probability gives unnormalized weights:

\begin{aligned} w_0&=(1/4)(0.0001)=0.000025,\\ w_1&=(1/2)(0.001)=0.0005,\\ w_2&=(1/4)(0.01)=0.0025. \end{aligned}

The total weight is \(0.000025+0.0005+0.0025=0.003025\). Dividing each weight by this total gives

P(K=0\mid A/A,A/A)=0.008264,\quad P(K=1\mid A/A,A/A)=0.1653,\quad P(K=2\mid A/A,A/A)=0.8264.

Under this model and allele frequency, the matching rare homozygous genotypes provide substantial evidence for sharing both copies, but do not prove it: the posterior probability of \(K=2\) is about \(82.64\%\), not \(100\%\). The values depend on the allele-frequency and inheritance assumptions.

A Practical Evaluation Sequence

A disciplined sequence keeps genotype matching separate from inference about descent. First record the genotypes as unordered allele pairs. Then identify what the observations rule out, inspect parental data for possible transmissions, and account for any remaining ambiguity. If no unique state follows, use an explicit likelihood model rather than treating IBS as certainty about IBD.

Record the observed genotypes.
Keep allele values and parental origins distinct; an unordered genotype does not specify phase.

Check for exclusions.
With accurate genotypes and no mutation, disjoint allele sets rule out sharing an allele IBD; different genotypes rule out \(K=2\).

Use parental genotypes where available.
List the compatible transmitted alleles and determine whether they identify particular ancestral copies.

Represent unresolved possibilities.
If more than one transmission pattern fits, retain the possible IBD states and their probabilities rather than selecting one without justification.

Quantify genotype evidence under stated assumptions.
Use allele frequencies and state-specific likelihoods, then combine them with the appropriate prior probabilities.

Why This Distinction Matters

IBD inference is an input to family-based linkage analysis: conclusions about marker sharing depend on how much the observed genotypes actually inform the inherited copies. A marker with many distinguishable alleles may often be informative, while common alleles or unresolved phase can leave multiple IBD states plausible. Missing or uninformative parental genotypes can widen that uncertainty. A genotype-based IBD probability is therefore model-based evidence, not a direct observation of descent.

A common mistake is to reason, “The siblings have the same genotype, so they share both alleles IBD.” The correction is to separate the observed allele values from their origins: identical genotypes can arise from different transmitted copies, and can even occur with no IBD sharing. Conversely, parental genotypes can sometimes identify a shared transmitted copy even when the siblings’ genotypes differ. State exactly which assumptions and transmissions support each inference.

Key takeaway: Genotypes show allele values; IBD concerns ancestral copies. Use parental transmissions when they resolve copy origin, and use a stated likelihood model when phase or descent remains ambiguous.

Check Your Understanding

For each question, distinguish what the genotype values show from what can be concluded about descent.

Two siblings have genotypes \(A/B\) and \(C/D\), with all four allele labels distinct. Under error-free genotyping and no mutation, which IBD states are excluded?
Why can two siblings with identical \(A/A\) genotypes still have \(K=0\)?
A father is \(A/B\), a mother is \(C/D\), and two children are \(A/C\) and \(A/D\). What makes the shared paternal allele informative about IBD?
In the \(A/B\) parents and \(A/B\) siblings example, why is the genotype evidence insufficient to distinguish \(K=0\) from \(K=2\)?
In the homozygous example, what role does the allele frequency play in the posterior probability of \(K=2\)?