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Pedigrees and Genetic Relatedness · Tutorial 65 of 989

Recursive Kinship Calculation From Parents

Use parental kinship values and Mendelian transmission probabilities to calculate kinship recursively through a pedigree.

Beginner 9 min read

What You'll Learn

  • Define the kinship coefficient as an identity-by-descent probability for two sampled alleles.
  • Derive the parental recursion using the law of total probability.
  • Distinguish the recursion for distinct individuals from self-kinship values.
  • Set founder values when founders are unrelated and non-inbred.
  • Calculate kinship through a pedigree in a valid order.
  • Recognize how incorrect founder assumptions can change a result.

Why Calculate Kinship Recursively?

In a pedigree, two individuals may be related through several generations of parents and offspring. A direct calculation can quickly become difficult if it requires listing every possible path by which alleles were transmitted. A simpler strategy is to calculate kinship in pedigree order: first establish kinship among founders, then use the known kinship values of a person's parents to calculate that person's kinship with other individuals.

The key is to condition on the parental origin of a randomly sampled allele. For a diploid individual, the sampled allele came from one parent or the other, each with probability one half under Mendelian transmission. This gives a recursion that works for any pair of distinct individuals when the parents of one member of the pair are known.

Definition: The kinship coefficient \(\phi(i,j)\) is the probability that one allele sampled at random from individual \(i\) and one allele sampled at random from individual \(j\), at the same autosomal locus, are identical by descent (IBD). The alleles are sampled independently, one from each individual.

Kinship is a probability, so its value lies between 0 and 1. It concerns IBD: the alleles are copies of the same ancestral allele. Merely having the same allele state does not establish that two alleles are IBD. The recursion below calculates the probability of IBD using the pedigree's parentage and transmission structure.

Deriving the Parental Recursion

Let \(j\) be an individual with known parents \(s\) and \(d\), and let \(i\) be a distinct individual. Sample one allele from \(i\) and one from \(j\). The allele sampled from \(j\) was transmitted by \(s\) with probability \(1/2\), or by \(d\) with probability \(1/2\). Conditional on its parental origin, the transmitted allele is a randomly selected allele from that parent. Therefore, the probability that it is IBD with the sampled allele from \(i\) is \(\phi(i,s)\) or \(\phi(i,d)\), respectively.

The law of total probability combines those two mutually exclusive transmission events:

$$ \begin{aligned} \phi(i,j) &=P(\text{IBD})\\ &=P(\text{IBD}\mid\text{transmitted by }s)P(\text{transmitted by }s) +P(\text{IBD}\mid\text{transmitted by }d)P(\text{transmitted by }d)\\ &=\frac{1}{2}\phi(i,s)+\frac{1}{2}\phi(i,d). \end{aligned} $$
Recursion: If \(j\) has parents \(s\) and \(d\), and \(i\ne j\), then $$ \phi(i,j)=\frac{\phi(i,s)+\phi(i,d)}{2}. $$ The kinship of \(i\) with \(j\) is the average of its kinship with \(j\)'s two parents. The same rule applies whether or not the parents are related or the child is inbred, provided the pedigree and transmission assumptions are appropriate.

The argument uses only the parental origin of the sampled allele and equal Mendelian transmission from the two parents. It does not require the parents to be unrelated. If a pedigree includes related founders, their kinship values must be used rather than set to zero. If parentage is unknown, the recursion cannot be applied to that individual until the missing parental information is supplied or modeled.

There is an important boundary case. The displayed recursion is for distinct individuals; it must not be applied by setting \(i=j\). Self-kinship is a diagonal value defined by sampling two alleles within the same individual. For a non-inbred founder, the two homologous alleles are not IBD, so its self-kinship is \(1/2\). More generally, a diagonal value must be supplied using the appropriate self-kinship rule. In the examples below, all founders are assumed non-inbred.

Setting the Starting Values and Choosing an Order

A recursion needs starting values. A common pedigree convention is to treat the listed founders as unrelated and non-inbred. For distinct founders \(u\) and \(v\), this gives \(\phi(u,v)=0\); for each such founder \(u\), \(\phi(u,u)=1/2\). These are assumptions about the pedigree boundary, not consequences of the recursion. If the founders are known to be related, their nonzero kinship values replace the unrelated-founder values.

Calculate values in an order that ensures the parents' kinship with the individual of interest are already known. For each non-founder, apply the recursion to the pair of interest and that non-founder, using its two parents. If one parent is the queried individual, the calculation may require that individual's diagonal self-kinship. Symmetry, \(\phi(i,j)=\phi(j,i)\), lets you choose which member of a pair to expand when both have known parents.

1
Set the pedigree boundary.
Record founder kinships and self-kinships from the stated assumptions or other supplied information.
2
Choose the individual to expand.
For a pair of distinct individuals, use the recursion on the member whose parents are known.
3
Substitute the parental values.
Average the queried individual's kinship with the two parents; recursively calculate any values not yet available.
4
Check the result.
Confirm the parental values, arithmetic, boundary assumptions, and that the result is between 0 and 1.

Worked Calculations

Worked Example: A Kinship Value Through Two Generations

Suppose founders \(P,Q,R,S\) are unrelated and non-inbred. Individual \(A\) is the child of \(P\) and \(Q\); \(B\) is the child of \(R\) and \(S\); and \(C\) is the child of \(A\) and \(B\). Calculate \(\phi(P,C)\).

First calculate \(\phi(P,A)\). Because \(A\)'s parents are \(P\) and \(Q\), use the recursion for the distinct pair \(P,A\):

$$ \phi(P,A)=\frac{\phi(P,P)+\phi(P,Q)}{2} =\frac{1/2+0}{2} =\frac{1}{4}. $$

The values are \(1/2\) for the non-inbred founder's self-kinship and 0 for the kinship between unrelated founders. Next, calculate \(\phi(P,B)\) from \(B\)'s parents \(R,S\):

$$ \phi(P,B)=\frac{\phi(P,R)+\phi(P,S)}{2} =\frac{0+0}{2} =0. $$

Finally, \(C\)'s parents are \(A\) and \(B\), so

$$ \phi(P,C)=\frac{\phi(P,A)+\phi(P,B)}{2} =\frac{1/4+0}{2} =\frac{1}{8}=0.125. $$

Under the stated founder assumptions, the probability that a randomly sampled allele from \(P\) and one from \(C\) are IBD is \(0.125\). This calculation depends on the founder assumptions and the specified parentage.

Worked Example: Expanding a Kinship Calculation Recursively

Let \(A,B,C,D\) be unrelated, non-inbred founders. Let \(E\) be the child of \(A,B\), \(F\) the child of \(C,D\), \(G\) the child of \(E,F\), \(H\) the child of \(A,C\), and \(I\) the child of \(G,H\). Calculate \(\phi(B,I)\).

Since \(I\)'s parents are \(G,H\), start with \(\phi(B,I)=\{\phi(B,G)+\phi(B,H)\}/2\). Calculate the needed terms from their parents. First, \(E\) is the child of \(A,B\):

$$ \phi(B,E)=\frac{\phi(B,A)+\phi(B,B)}{2} =\frac{0+1/2}{2} =\frac{1}{4}. $$

Since \(F\) is the child of the unrelated founders \(C,D\), both of whom are unrelated to \(B\),

$$ \phi(B,F)=\frac{\phi(B,C)+\phi(B,D)}{2} =\frac{0+0}{2} =0. $$

Now expand the kinship with \(G\), whose parents are \(E,F\), and with \(H\), whose parents are \(A,C\):

$$ \begin{aligned} \phi(B,G)&=\frac{\phi(B,E)+\phi(B,F)}{2} =\frac{1/4+0}{2} =\frac{1}{8},\\ \phi(B,H)&=\frac{\phi(B,A)+\phi(B,C)}{2} =\frac{0+0}{2} =0. \end{aligned} $$

Substitution for \(I\) gives

$$ \phi(B,I)=\frac{\phi(B,G)+\phi(B,H)}{2} =\frac{1/8+0}{2} =\frac{1}{16}=0.0625. $$

Thus, under the stated pedigree and founder assumptions, the probability of IBD for the two sampled alleles is \(0.0625\). The intermediate values make clear why the calculation should proceed generation by generation rather than treating \(I\)'s kinship with \(B\) as a direct founder comparison.

Worked Example: Using Related-Founder Kinship Values

Suppose founders \(A\) and \(B\) are known to have \(\phi(A,B)=1/8\), while founder \(C\) is unrelated to both. Assume all three founders are non-inbred. Let \(D\) be the child of \(A,C\), and let \(E\) be the child of \(B,D\). Calculate \(\phi(A,E)\).

The founders are not all unrelated, so it would be incorrect to replace \(\phi(A,B)\) with zero. First calculate the kinship between \(A\) and \(D\), using \(D\)'s parents \(A,C\):

$$ \phi(A,D)=\frac{\phi(A,A)+\phi(A,C)}{2} =\frac{1/2+0}{2} =\frac{1}{4}. $$

Then use \(E\)'s parents, \(B,D\):

$$ \phi(A,E)=\frac{\phi(A,B)+\phi(A,D)}{2} =\frac{1/8+1/4}{2} =\frac{3/8}{2} =\frac{3}{16} =0.1875. $$

The result is \(0.1875\), not the value obtained by treating \(A\) and \(B\) as unrelated. Correct founder values are essential because every later recursion carries them forward.

Assumptions, Interpretation, and a Common Pitfall

The recursion is a standard consequence of Mendelian transmission, but its use still depends on the pedigree model. The individuals must be diploid at the autosomal locus under consideration; the listed parents must be the genetic parents; and, conditional on a parent transmitting an allele, each of that parent's two alleles must have probability \(1/2\) of transmission. The calculation also treats transmission as having no mutation that changes the allele's genealogical identity. These assumptions define the transmission model behind the IBD probability.

The kinship coefficient is an expected pedigree-based probability, not a statement that the two observed genotypes must match. Two alleles can have the same state without being IBD, and IBD alleles need not be distinguishable from genotype data alone. The recursion calculates kinship from parentage and starting kinship values; it does not itself estimate kinship from observed genotypes.

Common mistake: A learner sees that \(j\) has parents \(s,d\) and writes \(\phi(i,j)=\phi(i,s)+\phi(i,d)\), reasoning that either parent might transmit an IBD allele. This adds the two conditional probabilities without weighting them by the probabilities of the two transmission events, and can even produce a value greater than 1. The correction is to use the law of total probability: each parent transmits the sampled allele with probability \(1/2\), so the two parental kinships are averaged. Another related error is to use this recursion for \(\phi(i,i)\); diagonal self-kinship is a separate boundary value, not a distinct-individual recursion.

Recursive calculation is useful because each result can be checked locally: identify the two parents, retrieve the two relevant kinship values, and average. This structure also makes assumptions visible. A changed founder relationship or a corrected parentage can affect values down the pedigree, while the recursion itself remains the same.

Key takeaway: For a distinct individual \(i\) and a person \(j\) with known parents \(s,d\), \(\phi(i,j)\) is the average of \(\phi(i,s)\) and \(\phi(i,d)\). Start from appropriate founder and diagonal values, then calculate in pedigree order.

Check Your Understanding

Answer each question using the recursion and its assumptions.

  1. Individual \(J\) has parents \(K,L\). Write the recursion for \(\phi(M,J)\), where \(M\ne J\).
  2. Two unrelated, non-inbred founders have kinship zero. A non-inbred founder's self-kinship is \(1/2\). Use these values to calculate the kinship between a founder and that founder's child, given the other parent is an unrelated founder.
  3. Why does the recursion average the parental kinship values rather than add them?
  4. When is it incorrect to use zero as the kinship between two founders?
  5. Explain why the parental recursion cannot be used directly to calculate \(\phi(i,i)\).
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