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Pedigrees and Genetic Relatedness · Tutorial 66 of 989

Deriving Kinship for Parent–Offspring and Full Siblings

Use allele-copy transmission to derive standard kinship coefficients and identify how related parents change them.

Beginner 10 min read

What You'll Learn

  • Define kinship as the probability that randomly sampled allele copies from two individuals are identical by descent.
  • Derive parent–offspring kinship by conditioning on which parent transmitted the sampled allele.
  • Derive the full-sibling IBD states from the parental transmissions.
  • Explain why full siblings from unrelated, noninbred parents have kinship one quarter.
  • Adjust parent–offspring and full-sibling calculations when the parents are related.
  • Avoid assuming the second parent contributes no IBD route without checking relatedness.

Why the Standard Coefficients Need Assumptions

Parent–offspring pairs and full siblings are familiar pedigree relationships, but a relationship label alone does not always determine a kinship coefficient. The usual values—one quarter for a parent and child, and one quarter for full siblings—apply under particular assumptions about parental relatedness and inbreeding. Following the transmitted allele copies makes both the calculation and its assumptions explicit.

Here, identity by descent (IBD) means that two allele copies descend from the same ancestral allele copy. It is a statement about ancestry, not merely about having the same allele value. Two copies can have the same value without being IBD. Unless stated otherwise, assume ordinary diploid inheritance, random transmission of one of each parent's two allele copies, and independent transmissions at the locus under consideration.

Definition: The kinship coefficient \(\phi(i,j)\) is the probability that one allele copy sampled at random from individual \(i\) and one sampled at random from individual \(j\) are IBD. For a noninbred individual \(i\), \(\phi(i,i)=1/2\): two allele copies sampled from that individual are the same copy with probability one half, and distinct noninbred founder copies are not IBD.

The word sampled matters: kinship concerns a random pair of allele copies, one from each individual. It is not the probability that the individuals share at least one IBD allele, and it is not the proportion of their observed allele values that happen to match. For the standard coefficients below, founders are assumed unrelated and noninbred unless a different pedigree condition is specified.

Parent–Offspring Kinship by Conditioning on the Transmitting Parent

Let \(P\) be a parent of child \(C\), and let \(Q\) be the child's other parent. A random allele copy from \(C\) has probability one half of having come from \(P\), and probability one half of having come from \(Q\). Conditional on its source, its probability of being IBD with a random allele copy from \(P\) is the kinship of \(P\) with that source parent. Therefore,

$$ \phi(P,C)=\frac{1}{2}\phi(P,P)+\frac{1}{2}\phi(P,Q). $$

This decomposition does not presume that \(P\) and \(Q\) are unrelated. If \(P\) is noninbred, then \(\phi(P,P)=1/2\), so the formula becomes \(\phi(P,C)=1/4+\phi(P,Q)/2\). The usual value of one quarter follows only if the other parent \(Q\) is also unrelated to \(P\), so that \(\phi(P,Q)=0\).

Worked Example: A Parent and Child With Unrelated Parents

Suppose \(P\) and \(Q\) are unrelated, noninbred parents of child \(C\). Find the kinship coefficient between \(P\) and \(C\).

The assumptions give \(\phi(P,P)=1/2\) and \(\phi(P,Q)=0\). Substituting into the parent–offspring formula,

$$ \phi(P,C)=\frac{1}{2}\left(\frac{1}{2}\right)+\frac{1}{2}(0) =\frac{1}{4}=0.2500. $$

Thus, a random allele copy from \(P\) and a random allele copy from \(C\) have probability \(0.2500\) of being IBD under these pedigree assumptions. The contribution is through the allele that \(C\) may have inherited from \(P\); the unrelated other parent contributes no additional IBD route.

Full Siblings: Enumerating Parental Transmissions

Let \(S_1\) and \(S_2\) be distinct full siblings with the same parents, father \(A\) and mother \(B\). At the locus, each sibling receives one of the father's two allele copies and one of the mother's two allele copies. For each parent, the two siblings' transmissions match in ancestral copy with probability one half: each transmission selects one of two copies, and the two selections agree in one of the two equally likely cases. The father's and mother's transmission events are independent under Mendelian inheritance.

Consequently, the number of parental allele copies shared IBD by the siblings can be zero, one, or two. The table shows the probability of each case. “Father matches” and “mother matches” mean that the siblings inherited the same allele copy from that parent; different copies from a parent are not IBD merely because they are both from that parent.

Father's transmitted copiesMother's transmitted copiesIBD copies sharedProbability
DifferentDifferent01/4
SameDifferent11/4
DifferentSame11/4
SameSame21/4

To obtain kinship from these cases, also account for the random allele-copy sampling within each sibling. If they share exactly one transmitted copy, both random samples must select that copy, with probability \(1/2 \times 1/2=1/4\). If they share both their paternal and maternal transmitted copies, the random samples must select the same parental origin, with probability \(1/2\); selecting opposite origins gives copies from different founder alleles under the unrelated-founder assumption. If they share no transmitted copy, the sampled copies are not IBD under that assumption. Weighting these conditional probabilities by the case probabilities gives the standard result:

$$ \phi(S_1,S_2) =\frac{1}{4}(0)+\frac{1}{4}\left(\frac{1}{4}\right) +\frac{1}{4}\left(\frac{1}{4}\right)+\frac{1}{4}\left(\frac{1}{2}\right) =\frac{1}{4}=0.2500. $$

Equivalently, the probabilities of sharing zero, one, or two allele copies IBD are \(1/4\), \(1/2\), and \(1/4\). Those are probabilities about the siblings' overall IBD state. They are not themselves the kinship coefficient: the kinship coefficient additionally accounts for the chance that the two randomly sampled copies select a shared IBD pair.

Worked Example: Full Siblings From Unrelated, Noninbred Parents

A hypothetical pedigree has two unrelated, noninbred parents \(A\) and \(B\) and two children \(S_1\) and \(S_2\). Calculate their kinship by conditioning on the transmissions from each parent.

For each parent, the probability the siblings inherit the same ancestral allele copy is \(1/2\); the probability they inherit different copies is also \(1/2\). Independence gives probability \(1/4\) for each of the four combinations: both match, father only matches, mother only matches, or neither matches. In those cases, the probabilities that a random pair of sibling allele copies is IBD are respectively \(1/2, 1/4, 1/4,\) and \(0\).

The weighted probability is

$$ \phi(S_1,S_2) =\left(\frac{1}{4}\right)\left(\frac{1}{2}\right) +\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) +\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) +\left(\frac{1}{4}\right)(0) =\frac{1}{4}=0.2500. $$

The siblings' kinship is therefore \(0.2500\), conditional on the stated unrelated, noninbred-parent assumptions. The calculation is an expectation over transmission patterns, not a claim that every realized sibling pair shares exactly one quarter of its allele copies IBD.

How Parental Relatedness Changes the Calculation

The same reasoning works when the parents are related or inbred. For parent and child, the co-parent's relationship to the named parent contributes through \(\phi(P,Q)\), as the conditional formula shows. For two full siblings, conditioning first on which parent supplied a random allele to \(S_2\) gives \(\phi(A,S_2)=\{\phi(A,A)+\phi(A,B)\}/2\) and \(\phi(B,S_2)=\{\phi(B,A)+\phi(B,B)\}/2\). A random allele from \(S_1\) comes from \(A\) or \(B\) with probability one half, so averaging these quantities yields the general expression:

$$ \phi(S_1,S_2) =\frac{1}{4}\left\{\phi(A,A)+2\phi(A,B)+\phi(B,B)\right\}. $$

For noninbred parents, \(\phi(A,A)=\phi(B,B)=1/2\), and this reduces to \(\phi(S_1,S_2)=1/4+\phi(A,B)/2\). Thus the standard one-quarter value requires unrelated parents. If the parents are related, their shared ancestry can create additional IBD paths between the siblings.

Worked Example: A Parent and Child Whose Parents Are Related

Consider a hypothetical child \(C\) whose parents are \(P\) and \(Q\). Suppose \(P\) and \(Q\) are full siblings from unrelated, noninbred founders, and \(P\) is noninbred. The full-sibling result gives \(\phi(P,Q)=1/4\), while \(\phi(P,P)=1/2\). Find the kinship between \(P\) and \(C\).

Both possible sources of the child's random allele must be included:

$$ \phi(P,C) =\frac{1}{2}\phi(P,P)+\frac{1}{2}\phi(P,Q) =\frac{1}{2}\left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{4}\right) =\frac{1}{4}+\frac{1}{8} =\frac{3}{8}=0.3750. $$

The parent–child kinship is \(0.3750\), not \(0.2500\). The additional \(0.1250\) contribution arises because the allele \(C\) inherits from \(Q\) can be IBD with a random allele from \(P\). This example shows why the relationship between a parent and the co-parent must be checked before using the familiar one-quarter coefficient.

A Short Derivation of the Recursions

The conditioning formulas follow directly from the definition of kinship. For any child \(J\) of parents \(A\) and \(B\), a random allele copy from \(J\) came from \(A\) with probability \(1/2\) and from \(B\) with probability \(1/2\). For a second individual \(i\), conditional kinship is therefore \(\phi(i,A)\) or \(\phi(i,B)\), respectively. Averaging the two exhaustive possibilities gives the recursion used in the calculations:

$$ \phi(i,J)=\frac{1}{2}\phi(i,A)+\frac{1}{2}\phi(i,B). $$

For parent and child, set \(i=P\), with \(J=C\) and parents \(P,Q\), to recover \(\phi(P,C)=\{\phi(P,P)+\phi(P,Q)\}/2\). For siblings, apply the same recursion first to \(S_2\) relative to \(A\) and \(B\), then average over the parent of origin of a random allele from \(S_1\). This gives the full-sibling formula above. Each recursion averages over all possible parental sources; omitting one source is justified only when its kinship contribution is known to be zero.

Common Mistake and Practical Use

A common mistake is to reason that a parent and child share one transmitted allele, so their kinship must always be one quarter. That argument silently discards the child's other parental allele. If the other parent is related to the named parent, that second route can also produce IBD. The correction is to condition on the child's allele source and calculate both terms, \(\phi(P,P)/2\) and \(\phi(P,Q)/2\). Likewise, for full siblings, the usual one-quarter coefficient presumes unrelated, noninbred parents; related parents can increase the kinship.

These coefficients summarize expected IBD sharing implied by a pedigree. They are useful in pedigree calculations because they distinguish a relationship's transmission structure from a particular observed genotype. Before applying a standard value, state the pedigree assumptions: who the parents are, whether they are related, and whether any individual is inbred. If those assumptions differ from the unrelated, noninbred-founder setting, use the conditional formulas rather than memorizing the standard coefficient.

Key takeaway: Under unrelated, noninbred-parent assumptions, parent–offspring kinship and full-sibling kinship are both \(1/4\). In general, parent–offspring kinship is \(\{\phi(P,P)+\phi(P,Q)\}/2\), and full-sibling kinship is \(\{\phi(A,A)+2\phi(A,B)+\phi(B,B)\}/4\). Include every parental transmission route.

Check Your Understanding

Use the transmission assumptions in each question, and distinguish a random-pair kinship probability from a whole-pair IBD state.

  1. If \(P\) is a noninbred parent and is unrelated to the child's other parent \(Q\), what are \(\phi(P,P)\) and \(\phi(P,Q)\), and what parent–offspring kinship follows?
  2. For full siblings, what are the probabilities of sharing zero, one, or two parental allele copies IBD when their parents are unrelated and noninbred?
  3. Why does the full-sibling state “both transmitted copies match” contribute a conditional random-pair IBD probability of one half rather than one?
  4. If \(P\) is noninbred and \(\phi(P,Q)=1/4\), calculate \(\phi(P,C)\) for their child \(C\). Explain why the unrelated-co-parent result does not apply.
  5. Write the general full-sibling formula in terms of \(\phi(A,A)\), \(\phi(A,B)\), and \(\phi(B,B)\). Which assumptions reduce it to one quarter?
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