Expected Allele Sharing Versus Identity by Descent

Why Matching Alleles Do Not Always Mean Shared Ancestry

Two people may both have an allele labelled A at a genetic locus. That observation alone does not tell us whether they inherited the allele from the same ancestral copy. If A is common in the population, the two copies may have been inherited through entirely separate ancestral lines. Statistical genetics therefore distinguishes identity by state from identity by descent.

This distinction matters when interpreting genetic resemblance. Allele-state matches can be frequent simply because an allele is common. Identity by descent (IBD), by contrast, concerns the history of the allele copies: whether they descend from the same copy in a specified pedigree or ancestral reference. The probability of a match by state can be high even when the probability of sharing by descent under the pedigree model is zero.

Definition: Two allele copies are identical by state (IBS) if they have the same observed allele state, such as both being labelled A. They are identical by descent (IBD) if they are copies descended from the same ancestral allele copy. Under the usual no-mutation model, IBD copies have the same state, but IBS copies need not be IBD.

Allele-State Matches and Pedigree Sharing

IBS is defined by the observed allele values at a locus. For a diploid individual, a genotype records the two allele states, such as AA, Aa, or aa. Two individuals can be said to share at least one allele IBS when at least one allele state appears in both genotypes. Another possible IBS event is an exact genotype match; that is a stricter event. It is important to say which event is being counted.

IBD is defined by ancestry, not by whether the genotype labels happen to match. For two diploid individuals, their four allele copies can be described by IBD states: \(k_0\) is the probability that they share zero copies IBD, \(k_1\) the probability that they share one, and \(k_2\) the probability that they share both. These probabilities refer to the specified pedigree and its founder assumptions. The probability of sharing at least one copy IBD is \(k_1+k_2\).

The kinship coefficient, often denoted \(\phi\), is a related but distinct probability: it is the probability that one allele copy sampled at random from each individual is IBD. For a pair of diploid individuals, \(\phi=(k_1+2k_2)/4\). Thus it is not generally the same as the probability that the pair shares at least one copy IBD.

A pedigree calculation needs a boundary assumption. If two individuals are treated as unrelated founders, with no shared ancestry represented in the pedigree, their IBD probability under that model is zero. This does not claim that their ancestors never shared any allele in the remote past. It says that the pedigree model assigns no shared descent to their copies. IBS probabilities, in contrast, depend on allele states and their frequencies, even for such unrelated founders.

Worked Example: Two Unrelated People Both Carry a Common Allele

Suppose a hypothetical biallelic locus has alleles A and a, with population frequency \(p=0.80\) for A and \(q=0.20\) for a. Two unrelated individuals are sampled independently from a population in Hardy–Weinberg equilibrium. What is the probability that both carry at least one A? Compare this with their IBD probability under the unrelated-founder pedigree model.

Under Hardy–Weinberg equilibrium, the probability that one person has no A copies (genotype aa) is \(q^2=(0.20)^2=0.04\). Therefore the probability that one person carries at least one A is \(1-q^2=0.96\). The two sampled individuals are independent, so

P(\text{both carry A})=(1-q^2)^2=(1-0.04)^2=0.96^2=0.9216.

The probability is 0.9216, or 92.16%. This is an IBS event: both genotypes contain the state A. Under the stipulated unrelated-founder pedigree model, the probability of an IBD match is 0. The high IBS probability reflects the high frequency of A, not a shared ancestral copy identified by the pedigree.

Calculating the Chance of Any Shared Allele State

At a biallelic locus, the event that two genotypes share at least one allele state is broader than the event that both carry a particular allele. Assuming Hardy–Weinberg equilibrium and independent sampling of unrelated individuals, the only genotype pair with no allele state in common is AA paired with aa, in either order. The probability of that event is \(2p^2q^2\). Subtracting from one gives the probability of at least one IBS match.

Result: For two independently sampled, unrelated diploid individuals at a biallelic locus with allele frequencies \(p\) and \(q=1-p\), assuming Hardy–Weinberg equilibrium, the probability their genotypes share at least one allele state is \(1-2p^2q^2\). This is an IBS probability, not an IBD probability.

Worked Example: Probability of Sharing Any Allele State

Use the hypothetical allele frequencies \(p=0.80\) and \(q=0.20\). Calculate the probability that two unrelated individuals share at least one allele state, regardless of whether that state is A or a.

The probability that one individual is AA is \(p^2=0.64\), and the probability of aa is \(q^2=0.04\). For the pair to share no state, one must be AA and the other aa. There are two possible orders, so

P(\text{no shared state})=2(p^2)(q^2) =2(0.64)(0.04)=0.0512.

Taking the complement,

P(\text{at least one IBS match}) =1-0.0512=0.9488.

Thus, under the stated sampling and equilibrium assumptions, the chance of at least one allele-state match is 94.88%. The unrelated-founder pedigree model still assigns zero probability to IBD sharing. A common allele can make allele-state matches very likely without changing the pedigree relationship.

Exact Genotype Matches Are Another IBS Event

“Sharing an allele” should not be confused with having exactly the same genotype. With allele frequencies \(p\) and \(q\), the Hardy–Weinberg genotype probabilities are \(p^2\) for AA, \(2pq\) for Aa, and \(q^2\) for aa. For independently sampled individuals, an exact genotype match occurs when both are AA, both are Aa, or both are aa. Adding those mutually exclusive possibilities gives the probability of an exact IBS genotype match.

Worked Example: Exact Genotype Matching Without IBD

For two hypothetical unrelated individuals sampled from the same population with \(p=0.80\) and \(q=0.20\), calculate the probability that their genotypes match exactly.

The three genotype probabilities are \(P(AA)=p^2=0.64\), \(P(Aa)=2pq=2(0.80)(0.20)=0.32\), and \(P(aa)=q^2=0.04\). Independence means the probability that both have any particular genotype is the square of that genotype’s probability. Therefore,

\begin{aligned} P(\text{exact genotype match}) &=(0.64)^2+(0.32)^2+(0.04)^2\\ &=0.4096+0.1024+0.0016\\ &=0.5136. \end{aligned}

The exact genotype-match probability is 0.5136, or 51.36%. Even an exact match does not, by itself, establish IBD: independently inherited copies can produce the same genotype. Under the specified unrelated-founder model, the IBD probability remains zero.

A Related Pair Can Share by Descent and by State

For relatives, the pedigree supplies a route by which a particular allele copy can be transmitted. The state of the allele still matters separately: a relative may carry the same allele label through a different ancestral route. The next example isolates one transmission and compares a named-allele IBS event with inheritance of that allele by descent.

Worked Example: A Parent–Child Pair and a Common Allele

Consider a hypothetical parent with genotype Aa. The other parent is unrelated to this person, and a randomly transmitted allele from that other parent is A with probability \(p=0.80\). Assume Mendelian transmission, no mutation, and no additional shared ancestry. Find (1) the probability that the child carries A, and (2) the probability that the child receives the parent’s A copy IBD.

The Aa parent transmits its A copy with probability \(1/2\). If the child receives that copy, the child carries A. Otherwise, the child receives a from this parent and carries A only if the other parent transmits A, with probability 0.80. These outcomes are disjoint, so

P(\text{child carries A}) =\frac{1}{2}+\frac{1}{2}(0.80) =0.50+0.40=0.90.

The child therefore carries A with probability 0.90. But the probability that the child inherits the particular A copy from the Aa parent is only the transmission probability, \(1/2=0.50\). That transmission creates an A copy IBD with the parent’s A copy. In the other route to carrying A, the child receives an A from the unrelated parent; this is an A state match with the first parent’s A, but not IBD under the stated assumptions. The two probabilities answer different questions.

Why the Distinction Matters

IBS probabilities are influenced by allele frequencies. When a particular allele is common, independently sampled individuals often carry it, and unrelated people may match at many loci by state. IBD probabilities instead depend on inheritance and the pedigree or ancestral reference being used. A high frequency of A does not by itself create a shared descent path between two individuals.

The distinction also affects how evidence should be described. Observing the same allele state is evidence of a match, but it is not a record of how that state arose. To infer IBD, one needs information or assumptions about ancestry and transmission, not simply matching allele labels. Conversely, when two copies are IBD, they have a shared ancestral copy under the pedigree model; under the usual no-mutation assumption they also match by state.

The probability calculations above rely on specific assumptions. Hardy–Weinberg equilibrium gives the genotype probabilities from allele frequencies. Independent sampling of unrelated individuals permits multiplying their genotype probabilities. Pedigree-based IBD calculations rely on the stated relationships, Mendelian transmission, and founder assumptions. If individuals are sampled from a structured population or founders share ancestry not represented in the pedigree, the numerical probabilities may change. A result should therefore be reported with its event definition and assumptions.

Common Mistake

Wrong reasoning: “Both people have allele A, so they must have inherited it from a common ancestor.”
Correction: The observation establishes an A-state match, or IBS. If A is common, unrelated individuals may both carry it with high probability. IBD requires shared descent of the allele copies under a specified ancestry model; it cannot be concluded from the allele label alone.
A second source of confusion: Treating the kinship coefficient as the probability that two diploid individuals share at least one allele IBD. The kinship coefficient concerns one randomly sampled allele from each person; the probability of at least one shared IBD copy is \(k_1+k_2\).

Key takeaway: IBS describes matching allele states; IBD describes common ancestry of allele copies. Allele frequency can make IBS common, including among unrelated individuals, while IBD is determined by the pedigree and its founder assumptions.

Check Your Understanding

Use the stated allele frequencies and assumptions when calculating; distinguish each allele-state event from an ancestry event.

In your own words, what information distinguishes an IBS match from an IBD match?
At a biallelic locus with \(p=0.70\) and \(q=0.30\), what is the probability that one Hardy–Weinberg individual carries at least one A?
For two unrelated, independently sampled individuals at those frequencies, what genotype pairing produces no shared allele state?
Why does an exact genotype match between unrelated individuals not prove that their allele copies are IBD?
If a parent is Aa, what is the probability of transmitting its particular A copy to a child? How does that differ from the probability the child carries A when the other parent can also transmit A?