From pedigree data to a linkage conclusion
Parametric linkage analysis is a chain of model-based calculations. A pedigree’s observed phenotypes and marker genotypes contribute evidence about unobserved transmissions; those transmissions have probabilities determined by a genetic map and inheritance model. The final likelihood compares how well a proposed disease-locus position explains the observations with how well an unlinked model explains them.
A numerical result is meaningful only relative to its assumptions. Before calculating, specify the disease-allele frequency and penetrance, marker-allele frequencies, phase information, recombination model, and treatment of missing or uncertain genotypes. Also state whether the pedigree was ascertained through affected relatives and, if so, how that ascertainment is handled. The examples below use deliberately small hypothetical data sets so the likelihood calculations can be checked by hand.
Specify the model before evaluating the pedigree
A parametric disease model assigns probabilities to phenotypes conditional on genotype. For example, a penetrance function gives \(P(\text{phenotype}\mid\text{disease genotype})\), including probabilities for unaffected carriers or affected noncarriers when incomplete penetrance or phenocopies are allowed. The marker model specifies allele frequencies and observed genotypes. Phase specifies which alleles lie together on each parental chromosome. Recombination fractions specify the chance that a transmitted chromosome switches from one parental homologue to the other across an interval.
For a full pedigree, let \(Y\) denote the observed phenotypes and \(G\) the observed marker genotypes. The inheritance paths, unobserved genotypes, and sometimes phase are latent. The likelihood sums over their possible values:
Here \(\mathbf{z}\) denotes inheritance paths, \(\mathbf{g}_{\mathrm{mis}}\) denotes missing genotypes, \(\mathbf{h}\) denotes phase configurations, and \(\boldsymbol{\theta}\) contains the relevant recombination fractions. The joint model probability weights phase configurations according to the model. In a simple fully informative example, observed phase and genotypes identify each path, so the sum reduces to the probability of the observed transmission pattern. In realistic pedigrees, it generally does not: compatible paths must be weighted and summed, not selected as if they were known.
Worked Example: Two-point likelihood with known phase
Worked Example: Two-point likelihood with known phase
Suppose a hypothetical pedigree supplies four independent, informative transmissions. Under a specified disease model, each transmitted disease-associated chromosome can be classified as recombinant or nonrecombinant relative to a fully informative marker. Phase is known, and one of the four transmissions is recombinant. Assume no genotyping error or mutation and that the phenotype information has been used only to assign the disease-associated chromosome; the calculation below is conditional on those assignments.
For recombination fraction \(\theta\), with \(0\leq\theta\leq0.5\), the likelihood of the ordered observations is
At the unlinked value \(\theta=0.5\), \(L(0.5)=(0.5)^4=0.0625\). The maximum likelihood estimate must respect the allowed interval. For these data, \(R/n=1/4=0.25\), which is admissible, so \(\hat{\theta}=0.25\). The likelihood there is \(0.25(0.75)^3=0.10546875\), and
Under this simplified conditional model, the observations are about \(1.69\) times as likely at the best-fitting recombination fraction as at \(\theta=0.5\). This is a small likelihood ratio, not strong evidence of linkage. It does not establish that the disease locus lies at the marker.
Why the two-point estimate is constrained
In the known-phase model, ignoring the binomial coefficient that is constant in \(\theta\), the likelihood for \(R\) recombinants among \(n\) informative meioses is \(L(\theta)=\theta^R(1-\theta)^{n-R}\). For an interior value \(0<\theta<0.5\), differentiating the log likelihood gives
The derivative changes from positive to negative at \(\theta=R/n\), when that value lies in the allowed interval. If \(R/n>0.5\), the derivative is positive throughout the interior up to the boundary, so the maximum occurs at \(0.5\). If \(R/n\leq0.5\), the maximum on the closed interval is at \(\theta=R/n\), including the boundary cases \(R=0\) and \(R/n=0.5\). Thus the constrained estimate is
The constraint matters. With three recombinants among four meioses, the unconstrained ratio \(R/n=0.75\) is impossible as a recombination fraction. The likelihood increases up to the boundary, so the constrained estimate is \(0.5\), not \(0.75\).
Worked Example: A three-marker multipoint likelihood
Worked Example: A three-marker multipoint likelihood
Consider a hypothetical, fully informative pedigree with a disease-associated parental chromosome and two markers, \(M_1\) and \(M_2\). For four independent meioses, assume phase is known and the observed inheritance paths identify one switch between the disease-associated transmission state and \(M_1\), and one switch between \(M_1\) and \(M_2\). The four observed switch patterns, in interval order, are \(00,00,10,01\). Assume the candidate disease position gives recombination fraction \(\theta_1=0.10\) to \(M_1\); the marker interval has fixed recombination fraction \(\theta_2=0.20\). This conditional calculation assumes no genotyping error, mutation, or unresolved transmissions.
The probability of a path with no switch in an interval is \(1-\theta\), and with a switch is \(\theta\). By independence of recombination events in the two specified intervals, the four path probabilities under linkage are \((1-\theta_1)(1-\theta_2)\), \((1-\theta_1)(1-\theta_2)\), \(\theta_1(1-\theta_2)\), and \((1-\theta_1)\theta_2\). Therefore,
For the unlinked comparison, transmission at the disease locus is independent of marker inheritance. The switch between disease transmission and \(M_1\) is then a fair event with probability \(0.5\), while the \(M_1\)-to-\(M_2\) marker switch remains governed by the map value \(0.20\). The four pattern probabilities are \(0.5(0.8),0.5(0.8),0.5(0.8),0.5(0.2)\), giving
This candidate position makes the observed paths only about \(1.17\) times as likely as the unlinked model does. The marker-to-marker transition must remain in the null likelihood: treating both intervals as independent fair switches under the unlinked model would incorrectly discard the linkage between the markers themselves.
Multipoint profiles and candidate positions
In multipoint analysis, a candidate disease position determines transition probabilities along a map. The likelihood at that position accounts jointly for the marker genotypes and the disease phenotypes, summing over compatible inheritance paths. Repeating this calculation over positions yields a LOD profile. The profile’s maximum is the best-fitting position among those evaluated, conditional on the model; it is not automatically a precise estimate of physical location.
The next example illustrates how to read such a profile. Its likelihoods are hypothetical outputs from a fully specified model after summing over latent paths; they are supplied to focus on the comparison and interpretation. Suppose the unlinked likelihood is \(0.00250\) at all three candidate positions.
Worked Example: Reading a multipoint LOD profile
A hypothetical pedigree analysis gives the following likelihoods at positions A, B, and C. Each LOD is the base-10 logarithm of the candidate-position likelihood divided by the same unlinked likelihood.
| Position | Linkage likelihood | Likelihood ratio | LOD |
|---|---|---|---|
| A | 0.00300 | 0.00300 / 0.00250 = 1.20 | log10(1.20) = 0.0792 |
| B | 0.0100 | 0.0100 / 0.00250 = 4.00 | log10(4.00) = 0.6021 |
| C | 0.00375 | 0.00375 / 0.00250 = 1.50 | log10(1.50) = 0.1761 |
Position B is the profile maximum among these candidates, but its likelihood is only four times the unlinked likelihood. The profile favors B relative to A and C in this calculation; it does not show that B is the true location or quantify uncertainty in map distance. A complete analysis would inspect the profile between candidates and check whether the result is robust to defensible changes in model assumptions.
Assumption checks and calibrated interpretation
A linkage result is conditional on the model and the data being analyzed. Before reporting it, check that the pedigree structure, sex and affection status, and genotype calls have been verified. Confirm that the assumed phase is supported by the data or that phase uncertainty was included in the likelihood. Check that marker frequencies and disease penetrance are plausible for the population and that missing genotypes or genotyping errors are not being treated as certain observations. Review whether the genetic map and recombination model suit the region and the relevant meioses. If recruitment depended on affected family members, an analysis that ignores ascertainment can misstate the likelihood.
A LOD score is a log likelihood ratio, not a calibrated measure of statistical significance by itself. A LOD of 3 corresponds to a likelihood ratio of \(10^3=1000\) in favor of the specified linkage model over its specified unlinked comparison. That ratio alone does not account for searching across many positions, testing several models, or uncertainty in the disease and marker assumptions. A significance statement requires calibration under an appropriate null procedure that reflects the analysis and its search, such as a suitable simulation-based calibration. Do not translate a LOD directly into a p-value or a probability that linkage is true.
A useful report names the model, the comparison, and the result together: for example, “Under the stated penetrance, allele-frequency, phase, and map assumptions, the maximum LOD was 0.60 at position B; the candidate likelihood there was four times the unlinked likelihood. This is a model-conditional likelihood comparison, not a calibrated significance conclusion.” If assumptions are uncertain, report how the profile changes under reasonable alternatives rather than presenting one value as definitive.
Common mistakes
- Using an unconstrained estimate. Reporting \(R/n=0.75\) as a recombination fraction violates the bound \(0\leq\theta\leq0.5\). Use the constrained estimate in the simple known-phase model.
- Multiplying only the most likely inheritance paths. If several latent paths are compatible with the data, the likelihood sums their probabilities, weighted by the observed-data model.
- Breaking the marker map under the unlinked comparison. A disease locus unlinked to the markers does not make linked markers independent of each other. Keep marker-to-marker transitions in the null likelihood.
- Calling a profile peak proof of location. The maximum identifies the best fit among the positions evaluated under the assumptions; it does not establish a causal locus or guarantee a sharply localized position.
- Calling a LOD a p-value. A LOD is a base-10 log likelihood ratio. Statistical calibration must account for the null model and, where relevant, the position search.
Check Your Understanding
Use the stated assumptions in each question, and distinguish a likelihood calculation from a significance conclusion.
- In the simple known-phase model, 3 of 4 informative meioses are recombinant. What is the constrained maximum likelihood estimate of \(\theta\), and why is \(R/n\) inadmissible?
- Why must a multipoint likelihood sum over compatible inheritance paths when phase or genotypes are uncertain?
- Under an unlinked disease model, why can marker-to-marker recombination probabilities still be less than \(0.5\)?
- A candidate has a LOD of \(0.60\). What likelihood ratio does this represent, approximately, and what does it not establish?
- Name two model or data checks that could change the interpretation of a pedigree’s linkage evidence.