KING-Robust Kinship Estimation

Why Use a Frequency-Robust Estimator?

Pairwise kinship is a measure of how likely alleles sampled from two individuals at the same locus are identical by descent (IBD), meaning they descend from the same ancestral allele. Earlier material on identity by state (IBS) and IBD distinguishes shared alleles by sequence from alleles shared through ancestry. Many genomic estimators use allele frequencies to distinguish these ideas. In a cohort with uncertain or heterogeneous ancestry, however, choosing a suitable frequency reference can be difficult.

The KING-robust estimator instead uses observed genotype patterns within each pair. In particular, it combines the count of loci where both people are heterozygous with the count where they have opposite homozygotes, and scales that contrast by their total heterozygote count. Its advantage is that it does not require an external allele-frequency estimate. “Robust” does not mean assumption-free or immune to every effect of population structure; it describes the estimator’s reduced dependence on shared frequency inputs.

Definition and Genotype Counts

Consider a pair of individuals genotyped at the same set of biallelic markers. At each marker, code genotypes by the number of copies of one designated allele: 0, 1, or 2. A genotype coded 1 is heterozygous; genotypes 0 and 2 are the two homozygotes. The choice of which allele is coded 1 or 2 does not change whether a genotype is heterozygous or whether two genotypes are opposite homozygotes.

The numerator is a contrast: opposite homozygotes subtract twice as much as jointly heterozygous genotypes add. The denominator puts that contrast on a scale determined by how often the pair’s genotypes are heterozygous. This form uses genotype counts, not a separately estimated allele frequency \(p\).

Why the Allele-Sharing Contrast Estimates Kinship

The basis for the estimator can be seen at one biallelic locus. Let the allele frequencies in the population supplying the alleles be \(p\) and \(q=1-p\). For this justification, assume random mating in that population, and that when a pair shares no allele IBD at the locus, their allele draws are independent. Let \(k_0\), \(k_1\), and \(k_2\) be the probabilities that the pair shares zero, one, or two alleles IBD at the locus, so \(k_0+k_1+k_2=1\). The kinship coefficient is $$ \phi_{ij}=\frac{k_1}{4}+\frac{k_2}{2}. $$ This definition counts the probability that one allele sampled from each person is IBD: one shared allele contributes probability \(1/4\), while two shared alleles contribute probability \(1/2\).

Conditional on sharing zero alleles IBD, the two genotypes are independent. Each is heterozygous with probability \(2pq\), so the probability both are heterozygous is \(4p^2q^2\). The probability of opposite homozygotes is \(p^2q^2+q^2p^2=2p^2q^2\). Conditional on sharing one allele IBD, the shared allele is the same for both people. If that allele has frequency \(p\), each person must independently inherit the other allele, of frequency \(q\), to be heterozygous; the probability is \(q^2\). If the shared allele has frequency \(q\), the corresponding probability is \(p^2\). Averaging over the shared allele gives \(pq^2+qp^2=pq\). Opposite homozygotes are impossible in this state because the pair has a shared allele. Conditional on sharing two alleles IBD, their genotypes match, so both are heterozygous with probability \(2pq\), and opposite homozygotes are again impossible.

This argument explains the frequency cancellation: both parts of the ratio carry the same factor \(4pq\). It also shows why the jointly heterozygous probability in the one-allele-IBD state is \(pq\), not \(p^2+q^2\). For instance, when \(p=q=1/2\), the probability that both are heterozygous in that state is \(1/4\).

The derivation is a ratio-of-expectations justification, not a claim that the expectation of every finite-sample ratio equals the true kinship exactly. Across loci with differing frequencies, the expected numerator and denominator both accumulate contributions proportional to \(4p_\ell q_\ell\); when kinship is consistent across the included loci, this common weighting supports the same target ratio. A finite marker set produces a noisy estimate.

Worked Applications

Assumptions, Interpretation, and Use

The derivation relies on a biallelic genotype at each locus, appropriate genotype calls, and the stated random-mating and allele-draw model for its frequency-based justification. The practical estimator also assumes that the markers used provide informative, comparable observations for the two people. If a locus is nearly monomorphic, it contributes little expected heterozygosity, and therefore little information to this contrast. If the total heterozygote count in the denominator is zero, the estimator is undefined; a denominator close to zero makes the estimate unstable.

KING-robust is useful when population allele frequencies are uncertain, because users do not need to plug in a single set of estimated frequencies for the cohort. However, the label should not be read as “unaffected by ancestry.” Population structure can change genotype-pattern distributions, and a pair’s ancestry composition and marker set can still influence observed counts. The estimator is not a substitute for checking data quality or interpreting pairwise estimates in the study’s population context.

For a relationship screen, the estimate is typically compared with expected kinship levels, not treated as a definitive pedigree label. A value near a familiar benchmark can prioritize a pair for review, while additional evidence is needed to resolve the relationship. An estimated value can also be negative: the numerator may be negative when twice the IBS0 count exceeds the jointly heterozygous count. This does not represent a negative probability of IBD; it is a finite-sample estimate that can arise from the contrast.

Common Mistakes

• Using \(p^2+q^2\) for both-heterozygous probability when one allele is shared IBD. Conditional on a shared allele of frequency \(p\), both nonshared alleles must be the other allele, giving probability \(q^2\). Averaging over the shared allele gives \(pq^2+qp^2=pq\), not \(p^2+q^2\).
• Subtracting IBS0 only once. The estimator’s numerator is \(N_{\mathrm{HetHet}}-2N_{\mathrm{IBS0}}\). Omitting the factor of 2 changes the estimator and its calibration.
• Calling an estimate a relationship diagnosis. A value near a pedigree benchmark is consistent with that expected level of kinship, but it does not uniquely identify a pedigree.
• Interpreting “robust” as “immune to population structure.” The estimator avoids requiring external frequency inputs, but context and data quality still matter.