Why Marker Genotypes Affect Linkage Evidence
A marker helps linkage analysis when its observed genotypes distinguish which parental allele was transmitted through a pedigree. A marker with many observed genotypes is not automatically informative: the value of each observation depends on the parental genotypes, allele frequencies, phase, and the other pedigree information. Missing or uncertain calls also matter. If a genotype is uncertain, the likelihood must account for each possible true genotype and its probability; choosing one as if it were known can distort the evidence for linkage.
The key principle is that observed marker data are not always the same as the underlying genetic state. A person’s true genotype, the alleles transmitted in each meiosis, and sometimes the phase are latent quantities. The pedigree likelihood averages over these possibilities, weighting each by its probability under the model.
Allele Frequencies and Marker Informativeness
For a marker with alleles \(A_1,\ldots,A_K\), let \(p_i\) be the frequency of allele \(A_i\), with \(\sum_i p_i=1\). Under random mating and Hardy–Weinberg equilibrium (HWE), a founder’s genotype probabilities are \(P(A_iA_i)=p_i^2\) for a homozygote and \(P(A_iA_j)=2p_ip_j\) for a heterozygote when \(i\ne j\). These are probabilities for founder genotypes, not independent probabilities to assign to descendants whose genotypes arise by transmission.
Expected heterozygosity, \(1-\sum_i p_i^2\), is a useful summary of the chance that a randomly sampled founder is heterozygous under HWE. Heterozygous parents can be especially useful because their two marker alleles differ, so an offspring’s genotype may reveal which allele was inherited. But even a heterozygous parent may provide little information if the child’s genotype cannot distinguish the parental contribution, or if the relevant phase is unknown.
Worked Example: Allele Frequencies and Expected Heterozygosity
Consider a hypothetical biallelic marker with alleles \(A\) and \(B\), with founder frequencies \(p_A=0.8\) and \(p_B=0.2\). Assume random mating and HWE among founders. Find the founder genotype probabilities and expected heterozygosity.
The genotype probabilities are:
They sum to \(0.64+0.32+0.04=1.00\). Expected heterozygosity is
Thus, under these assumptions, 32% of founders are expected to be heterozygous. For comparison, if \(p_A=p_B=0.5\), then \(P(AB)=2(0.5)(0.5)=0.50\). The balanced-frequency marker has a higher expected heterozygosity in this comparison. This summary alone does not determine how informative either marker will be in a particular pedigree; the observed family genotypes and inheritance structure still matter.
Allele frequencies enter a pedigree likelihood through the probability assigned to possible founder genotypes. A common allele will often make some founder configurations more probable than a rare allele, while a rare allele can be highly informative when it is observed and its transmission can be tracked. If the analysis conditions on specified founder genotypes, their frequency probabilities may not contribute to the conditional comparison. If founder genotypes are unknown or modeled probabilistically, their probabilities belong in the likelihood.
Uncertain Genotypes Belong in the Likelihood
A genotype call can be ambiguous because of low-quality data, overlapping assay signals, or an explicit genotype-likelihood output. Let \(G\) denote the true genotype and \(O\) the observed call or assay result. The observation model specifies \(P(O\mid G)\). A hard call is the special case in which the observation is treated as certain for one genotype and impossible for the others. When the call is uncertain, the likelihood uses the probabilities under the observation model rather than replacing the call by its most likely genotype.
The sum includes every possible latent state \(z\) consistent with the model. Each term is the joint probability of that state and the observed data. If the observed marker result \(O\) is uncertain given the true genotype \(G\), its contribution is \(P(O\mid G)\) inside the relevant term. A missing call that is independent of genotype and inheritance contributes no genotype-specific observation factor: it supplies no direct evidence about which marker genotype or allele was present.
Worked Example: Weighting an Uncertain Marker Call
A hypothetical mother has known marker genotype \(AB\), and the father has genotype \(AA\). The child therefore has true genotype \(AA\) if the mother transmits \(A\), and \(AB\) if she transmits \(B\). Assume ordinary Mendelian transmission, so each maternal allele is transmitted with probability \(1/2\). The child’s assay result is \(O\), with \(P(O\mid AA)=0.8\) and \(P(O\mid AB)=0.2\). Find the probability that the mother transmitted \(B\), given \(O\).
First calculate the probability of the observed assay result by summing over the two possible true genotypes:
The joint probability of observing \(O\) and maternal transmission of \(B\) is \((0.2)(0.5)=0.10\). Therefore,
The call makes transmission of \(B\) less likely than transmission of \(A\), but does not rule it out. Treating the most likely genotype \(AA\) as certain would instead assign probability zero to transmission of \(B\), discarding the assay’s stated uncertainty.
Summing Over Genotypes and Inheritance States
In parametric linkage analysis, the likelihood combines marker observations with the disease model and the probabilities of transmissions. The recombination fraction \(\theta\) specifies the probability that the disease allele and marker allele transmitted from a parent are recombinant, given their phase. Marker observations can favor one transmission path over another, and that can change the likelihood at a given \(\theta\). The LOD score compares the likelihood at a specified \(\theta\) with the likelihood at \(\theta=0.5\):
When genotypes are missing or uncertain, each possible state contributes according to its probability under the model and its compatibility with the observations. This marginalization is not an optional correction: it is what makes the likelihood account for what is known and what is not known.
Worked Example: An Informative Genotype Changes the LOD Score
Consider a hypothetical family with one informative meiosis. An affected parent has known phased haplotypes \(D\!-\!A\) and \(d\!-\!B\), where \(D\) is the disease allele and \(A,B\) are marker alleles. The other parent is known to be \(B/B\) at the marker and does not carry \(D\). Assume complete penetrance, no phenocopies, and that the child is affected and has marker genotype \(A/B\). Compare \(\theta=0.1\) with \(\theta=0.5\). All stated parental genotypes and phase are treated as known, so founder-frequency factors are conditioned out of this calculation.
For the child to be affected, the affected parent must transmit \(D\). For the child’s marker genotype to be \(A/B\), that parent must also transmit \(A\), because the other parent always transmits \(B\). With phase \(D\!-\!A/d\!-\!B\), the probability of transmitting \(D\) together with \(A\) is \((1-\theta)/2\): the nonrecombinant \(D\!-\!A\) haplotype is transmitted with probability \(1-\theta\), and either parental homologue is chosen with probability \(1/2\).
At \(\theta=0.1\), this contribution is \((1-0.1)/2=0.45\). At \(\theta=0.5\), it is \((1-0.5)/2=0.25\). The likelihood ratio and LOD score are therefore
Under the stated model and this single-meiosis calculation, the observed disease and marker data are 1.8 times as likely at \(\theta=0.1\) as at \(\theta=0.5\). The positive LOD value is evidence favoring the specified linked recombination fraction over the unlinked value for these data; it is not, by itself, a probability that linkage is true.
What a Missing Call Does—and Does Not Do
Suppose the child’s marker genotype in the preceding example is missing, and assume the missingness is independent of the true genotype, disease status, and inheritance state. The marker observation then provides no preference for maternal allele \(A\) or \(B\). To obtain the joint probability of an affected child with the marker unobserved, sum over both possible maternal marker alleles. Given transmission of \(D\), the affected parent transmits either marker allele with total probability one. Thus the contribution is \(1/2\) at every \(\theta\), the probability of transmitting \(D\), and the likelihood ratio for this one meiosis is \(0.5/0.5=1\), giving LOD \(0\). The disease status alone in this simplified example does not distinguish \(\theta=0.1\) from \(\theta=0.5\).
That result does not mean missing genotypes always leave an entire pedigree’s linkage likelihood unchanged. Other observed relatives’ genotypes may still provide marker information, and the disease phenotypes and transmissions still contribute. It means only that a genotype-independent missing call supplies no direct marker observation for that individual. If the probability of a missing call depends on the true genotype or another modeled variable, the missingness process must also be represented in the likelihood.
Why Marginalization Is Correct
The likelihood rule follows from the addition rule for mutually exclusive events. The possible latent states \(z\) partition the ways the observed data can arise. For any fixed \(\theta\), the event \(D\) can occur together with one latent state, another latent state, and so on. Since those joint events are mutually exclusive, their probabilities add:
This identity explains why one should not select a single plausible genotype or transmission and proceed as if it were certain. Doing so replaces a sum of probability-weighted contributions with just one term. It can inflate or reduce the likelihood depending on which term is selected, and it can change the comparison between \(\theta\) and \(0.5\). The same summation applies when uncertainty concerns the founder genotype, phase, or an inheritance path.
Record allele frequencies and, if used, the assumptions for founder genotype probabilities such as HWE.
Use a definite genotype for a treated-as-certain call, genotype-specific observation probabilities for an uncertain call, and no genotype-specific factor for an independent missing call.
Include the founder genotypes, transmissions, and other unobserved states needed by the pedigree model.
For each state, multiply its model probability by the probability of the observed data given that state, then add across states to obtain the likelihood.
Common Mistake: Replacing Uncertainty With a Hard Call
A tempting but incorrect shortcut is: “The assay favors genotype \(AA\), so I will enter \(AA\) as known and ignore \(AB\).” That reasoning treats a probabilistic observation as a certainty. In the worked assay example, the posterior probability of maternal transmission of \(B\) was \(0.20\), not zero. The correction is to retain both possibilities and weight them by their transmission probabilities and their genotype-specific assay probabilities. Similarly, a missing call should not be filled in with the most common genotype merely because it is common; the likelihood should sum over possible genotypes using the model’s probabilities.
A second error is to equate high expected heterozygosity with guaranteed informativeness. Heterozygosity can make parental transmissions distinguishable, but only the pedigree’s observed genotypes, phase, and inheritance structure show whether that distinction is available. Allele frequencies help determine which founder configurations are plausible; they do not substitute for evaluating the pedigree likelihood.
Check Your Understanding
Try these questions using the definitions and likelihood rules above.
- At a biallelic marker with allele frequencies \(0.7\) and \(0.3\), what is the HWE probability of a heterozygous founder genotype?
- Why can a heterozygous parent be informative about transmission, and why does heterozygosity not guarantee informativeness in every pedigree?
- An assay result has positive probability under two possible genotypes. What must be included in the likelihood calculation?
- Under what assumption does a missing genotype contribute no direct marker evidence?
- In a pedigree likelihood, what is summed over when genotypes or inheritance paths are unobserved?