Monozygotic Twins, Dizygotic Twins, and Ordinary Siblings

Why Compare Twins and Full Siblings?

Monozygotic (MZ) twins, dizygotic (DZ) twins, and ordinary full siblings have different developmental histories, but their genetic relatedness can be compared precisely under a standard autosomal inheritance model. The key is to separate two quantities: the IBD state—how many alleles the pair shares identity by descent at a locus—and kinship—the probability that one allele sampled from each person is IBD. These quantities are related, but they are not interchangeable.

The comparison is useful because the same IBD-state distribution applies to DZ twins and ordinary full siblings under the usual assumptions. MZ twins, by contrast, share both alleles IBD at the locus in the standard model. Converting those states to kinship requires care: even when two people share both alleles IBD, independently selected alleles from them need not be the same ancestral copy.

Definition: At an autosomal locus, let \(K\) be the number of allele copies one diploid person shares IBD with another, under the usual noninbred, unrelated-founder model. The possible states are \(K=0,1,2\), with probabilities \(k_0=P(K=0)\), \(k_1=P(K=1)\), and \(k_2=P(K=2)\). The probabilities satisfy \(k_0+k_1+k_2=1\). Kinship, denoted \(\phi_{ij}\) for individuals \(i\) and \(j\), is the probability that one allele chosen at random from each person is IBD.

From IBD State to Kinship

For the kinship definition above, choose one of the two allele copies from each person independently, each with probability \(1/2\). There are four equally likely cross-person allele pairs. If \(K=0\), none of those pairs are IBD. If \(K=1\), exactly one of the four pairs is IBD, so the conditional probability is \(1/4\). If \(K=2\), both homologous pairings are IBD, so two of the four pairs are IBD and the conditional probability is \(1/2\).

Taking the probability-weighted average over the three states gives the kinship formula. In particular, \(K=2\) does not make kinship equal to 1 under this definition: the independently chosen alleles must also select corresponding homologues, which occurs with probability \(1/2\).

\phi_{ij} = k_0(0)+k_1\left(\frac{1}{4}\right)+k_2\left(\frac{1}{2}\right) = \frac{k_1}{4}+\frac{k_2}{2}.

Key result: When kinship is defined as the probability that one independently selected allele from each person is IBD, \(\phi_{ij}=k_1/4+k_2/2\). The conditional probabilities for \(K=0,1,2\) are \(0,1/4,1/2\), respectively.

Why Full Siblings Have Three Possible IBD States

Assume the two parents are unrelated and noninbred, their founder alleles are not IBD with one another, and transmissions to the siblings follow ordinary Mendelian segregation. At one autosomal locus, each sibling receives one allele from the mother and one from the father. For the maternal contributions, the two siblings receive the same one of their mother's two homologues with probability \(1/2\). The corresponding probability for their paternal contributions is also \(1/2\). These two matching events are independent under the model.

Let \(M=1\) when the siblings inherited the same maternal homologue and \(M=0\) otherwise; define \(F\) in the same way for paternal inheritance. Then \(K=M+F\). If neither parental contribution matches, the siblings share zero alleles IBD. If exactly one matches, they share one. If both match, they share two. Multiplying the probabilities of the independent parental events yields the distribution.

\begin{aligned} k_0 &= P(M=0,F=0)=\left(\frac12\right)\left(\frac12\right)=\frac14,\\ k_1 &= P(M=1,F=0)+P(M=0,F=1) =\left(\frac12\right)\left(\frac12\right)+\left(\frac12\right)\left(\frac12\right)=\frac12,\\ k_2 &= P(M=1,F=1)=\left(\frac12\right)\left(\frac12\right)=\frac14. \end{aligned}

The probabilities sum to \(1/4+1/2+1/4=1\), as required. This is a single-locus result. Across a chromosome, recombination can change which parental segments are shared, but that does not change the standard expected state probabilities at a locus.

Worked Example: Deriving Full-Sibling Kinship

Consider a hypothetical pair of ordinary full siblings whose parents are unrelated and noninbred. Their standard IBD-state probabilities are \(k_0=1/4\), \(k_1=1/2\), and \(k_2=1/4\). Calculate the kinship of the pair using the random-allele-pair definition.

Substitute the state probabilities into the kinship formula:

\begin{aligned} \phi_{ij} &= \frac{k_1}{4}+\frac{k_2}{2}\\ &= \frac{1/2}{4}+\frac{1/4}{2}\\ &= \frac18+\frac18\\ &= \frac14=0.250. \end{aligned}

Thus, at this locus under the stated model, the probability that one randomly selected allele from each sibling is IBD is \(0.250\). The result is not \(0.50\): \(k_1=0.50\) is the probability that the pair shares exactly one allele IBD, not the probability that a randomly selected cross-person allele pair is IBD.

MZ Twins, DZ Twins, and Ordinary Full Siblings

In the standard autosomal model, MZ twins arise from the same fertilized egg. They therefore have the same two inherited ancestral copies at a locus, so \(K=2\) with probability 1. Their IBD-state probabilities are \(k_0=0\), \(k_1=0\), and \(k_2=1\). The kinship calculation then gives \(\phi_{ij}=0/4+1/2=1/2\).

DZ twins arise from two separately fertilized eggs. At an autosomal locus, their inheritance is like that of any two children of the same parents: each receives one allele from each parent through separate transmissions. Consequently, DZ twins have the same standard IBD-state probabilities as ordinary full siblings: \(k_0=1/4\), \(k_1=1/2\), and \(k_2=1/4\). Their kinship is \(1/4\), just as it is for ordinary full siblings.

Pair type	\(k_0\)	\(k_1\)	\(k_2\)	Kinship \(\phi_{ij}\)
MZ twins	0	0	1	\(1/2\)
DZ twins	\(1/4\)	\(1/2\)	\(1/4\)	\(1/4\)
Ordinary full siblings	\(1/4\)	\(1/2\)	\(1/4\)	\(1/4\)

The equality between DZ twins and ordinary full siblings concerns their expected autosomal IBD states under this model; it does not claim that their environments or observed traits are identical. The table also uses kinship, not the additive relationship coefficient. When that coefficient is defined as \(2\phi_{ij}\), it is \(1\) for MZ twins and \(1/2\) for DZ twins and ordinary full siblings.

Worked Example: Comparing Kinship for MZ and DZ Twins

Suppose a hypothetical pedigree analysis asks for kinship at an autosomal locus for one MZ twin pair and one DZ twin pair. Use the standard state probabilities and report the probability for a randomly selected allele from each twin to be IBD.

For the MZ pair, \(k_2=1\) and \(k_1=0\). Therefore,

\phi_{\mathrm{MZ}} = \frac{0}{4}+\frac{1}{2}(1) =0.500.

For the DZ pair, \(k_1=1/2\) and \(k_2=1/4\). Therefore,

\phi_{\mathrm{DZ}} = \frac{1/2}{4}+\frac{1/4}{2} = \frac18+\frac18 =0.250.

Under these assumptions, the random cross-person allele pair is twice as likely to be IBD for the MZ pair as for the DZ pair. The MZ result remains \(0.500\), not \(1\), because each twin has two homologues and the independently chosen alleles must select corresponding copies.

Worked Calculation and a Common Confusion

A reliable way to avoid coefficient errors is to keep the state probabilities and the kinship calculation on separate lines. First identify \(k_0,k_1,k_2\). Then apply the conditional probabilities \(0,1/4,1/2\). Do not substitute the chance of sharing exactly one or two alleles for the chance that a randomly chosen pair of alleles is IBD.

Worked Example: Checking a Full-Sibling Kinship Report

A hypothetical report gives kinship \(0.25\) for a pair of ordinary full siblings under the unrelated, noninbred-parent model. Check the report by averaging the random-allele-pair probabilities over the three IBD states.

The state probabilities are \(1/4,1/2,1/4\). Conditional on those states, the random allele pair is IBD with probabilities \(0,1/4,1/2\), respectively. Thus,

\begin{aligned} \phi_{ij} &= \left(\frac14\right)(0) +\left(\frac12\right)\left(\frac14\right) +\left(\frac14\right)\left(\frac12\right)\\ &=0+\frac18+\frac18\\ &=\frac14=0.250. \end{aligned}

The reported value is correct. The full siblings have probability \(1/2\) of being in the \(K=1\) state, but that state contributes only \(1/4\) to the chance of an independently selected allele pair being IBD. The \(K=2\) state contributes \(1/2\), and occurs with probability \(1/4\).

Assumptions, Scope, and Why the Distinction Matters

The sibling and DZ-twin state probabilities above assume unrelated, noninbred parents and the standard autosomal transmission model. If the parents are related or inbred, their alleles may themselves be IBD, and the simple full-sibling distribution need not apply. The MZ calculation assumes both twins carry the same two ancestral copies at the locus; post-zygotic mutation or other biological complications can make observed sequence or genotype calls differ without changing the standard pedigree calculation.

The kinship formula follows from its stated sampling definition and the three IBD states. It does not depend on whether the pair are twins or ordinary siblings once their state probabilities are specified. This separation is important in pedigree calculations: state probabilities describe inheritance, while kinship summarizes the chance of IBD for a randomly selected allele pair. The additive relationship coefficient, when used, is a different scale equal to twice kinship.

Do not equate \(k_1\) with kinship. For full siblings, \(k_1=1/2\), whereas kinship is \(1/4\).
Do not set kinship to 1 when \(k_2=1\). With one allele chosen independently from each person, only half of cross-person choices match corresponding homologues.
Do not distinguish DZ twins from full siblings on autosomal IBD grounds under this model. Their standard state probabilities and kinship are the same.
Keep coefficient scales explicit. Kinship is \(\phi_{ij}\); the additive relationship coefficient defined as \(2\phi_{ij}\) is twice that value.

Key takeaway: MZ twins have \((k_0,k_1,k_2)=(0,0,1)\) and kinship \(1/2\). DZ twins and ordinary full siblings have \((k_0,k_1,k_2)=(1/4,1/2,1/4)\) and kinship \(1/4\), under the standard unrelated, noninbred-parent autosomal model. The conversion is \(\phi_{ij}=k_1/4+k_2/2\).

Check Your Understanding

Use the random-allele-pair definition of kinship when answering.

For a pair with \(k_0=0\), \(k_1=1\), and \(k_2=0\), calculate kinship. Explain why the answer is not \(1/2\).
What assumptions lead to the standard IBD-state probabilities for ordinary full siblings?
Why do DZ twins and ordinary full siblings have the same expected autosomal IBD states under the standard model?
Calculate kinship for a pair with \(k_0=0\), \(k_1=0\), and \(k_2=1\). What relationship type in this tutorial has those state probabilities?
For ordinary full siblings, distinguish the probability of the \(K=1\) state from the kinship coefficient.