Worked Two-Point LOD Calculations for Small Pedigrees

From Pedigree Evidence to a LOD Profile

A two-point linkage calculation asks whether a disease locus and a marker locus are inherited together more often than expected if they assort independently. The calculation is made over possible values of the recombination fraction, $\theta$, between the two loci. A small $\theta$ corresponds to few recombinations between them; $\theta=0.5$ is the unlinked reference in the two-point model.

In a small pedigree, the evidence comes from the transmissions that can be inferred from relatives’ disease and marker information. Some transmissions are informative: given the transmitting parent’s haplotypes and phase, the observed child’s inheritance can be classified as recombinant or nonrecombinant. Other transmissions may not distinguish those possibilities. The worked calculations below first use the informative, clearly classified case, then show how likelihoods from separate families combine.

Definition: The two-point likelihood $L(\theta)$ is the probability of the observed pedigree data under a specified model when the recombination fraction between the disease locus and marker locus is $\theta$. The two-point LOD score at $\theta$ is $$ \operatorname{LOD}(\theta)=\log_{10}\left(\frac{L(\theta)}{L(0.5)}\right). $$ Here, $L(0.5)$ is the likelihood under the unlinked reference. A positive LOD score means the observed data are more likely at that $\theta$ than under the reference; it is not itself a probability.

A Classified-Meioses Likelihood

For the simple calculations in this tutorial, assume that phase is known, the relevant parental transmissions are informative, and each can be classified as recombinant or nonrecombinant. Also assume the recombination fraction is the same for the meioses considered and that their recombination outcomes are independent under the specified two-point model. Other disease-model parameters, such as allele frequency and penetrance, are treated as fixed. These assumptions make each classified informative meiosis contribute a factor of $\theta$ if recombinant and $1-\theta$ if nonrecombinant.

Likelihood for classified informative meioses: If $R$ informative meioses are recombinant and $N$ are nonrecombinant, then $m=R+N$ and, up to factors that do not depend on $\theta$, $$ L(\theta)\propto \theta^R(1-\theta)^N,\qquad 0\leq\theta\leq 0.5. $$ The proportionality is enough for a LOD ratio because any common factor cancels between $L(\theta)$ and $L(0.5)$. If $m=0$, then $R=N=0$, and this likelihood is constant in $\theta$; every $\theta$ in the allowed range is a maximizer.

The range stops at $0.5$ because values above one-half do not represent a greater degree of two-point linkage in this model. The constrained likelihood maximum therefore needs attention to the boundary. When $m>0$, the log-likelihood, omitting constants, is $\ell(\theta)=R\log(\theta)+N\log(1-\theta)$. For an interior value $0<\theta<0.5$, its derivative is $\ell'(\theta)=R/\theta-N/(1-\theta)$. Setting this equal to zero gives $\theta=R/(R+N)=R/m$. If $R=0$, then $N=m>0$ and $L(\theta)\propto(1-\theta)^N$, which decreases on $0\leq\theta\leq0.5$, so its maximum is at $\theta=0$. The log-likelihood is concave because $\ell''(\theta)=-R/\theta^2-N/(1-\theta)^2\leq 0$, so this stationary point, when it lies in the allowed interval, is a maximum. If $R/m>0.5$, the likelihood is still increasing up to the allowed boundary, so the maximum is at $0.5$. In short, for $m>0$, the maximizing value is $R/m$ when $R/m\leq0.5$, and $0.5$ otherwise. This rule does not apply when $m=0$, since $R/m$ is then undefined.

Worked Calculations

Worked Example: Three Nonrecombinant and One Recombinant Transmission

Suppose a hypothetical nuclear family provides four informative transmissions from a parent whose marker and disease-locus phase is known. Three are classified as nonrecombinant and one as recombinant. Use the classified-meioses likelihood to calculate the LOD score at its maximum.

Here $N=3$, $R=1$, and $m=4$. The likelihood, up to a factor common to all values of $\theta$, is $$ L(\theta)\propto \theta(1-\theta)^3. $$

The estimated recombination fraction is within the allowed range: $$ \hat{\theta}=\frac{R}{m}=\frac{1}{4}=0.25. $$

At this value, $L(0.25)\propto(0.25)(0.75)^3=0.10546875$. At the unlinked reference, $L(0.5)\propto(0.5)(0.5)^3=0.0625$. Thus $$ \operatorname{LOD}(0.25) =\log_{10}\left(\frac{0.10546875}{0.0625}\right) =\log_{10}(1.6875) \approx 0.2272. $$

Under this simplified model, the data are about $1.69$ times as likely at $\theta=0.25$ as at $\theta=0.5$. The maximum LOD is modest; it is not a probability that the loci are linked.

Worked Example: More Recombinants Than Nonrecombinants

In another hypothetical family, five informative transmissions are classified: four recombinant and one nonrecombinant. Calculate the maximum LOD over the permitted range.

Here $R=4$, $N=1$, and $m=5$, so the unconstrained proportion of recombinant transmissions is $R/m=0.8$. That value is outside the model’s allowed range $0\leq\theta\leq0.5$. The likelihood is proportional to $\theta^4(1-\theta)$, and the constrained maximum is therefore at $\theta=0.5$.

At the maximum and at the unlinked reference, the likelihood is the same: $$ L(0.5)\propto (0.5)^4(0.5)=(0.5)^5=0.03125. $$ Consequently, $$ \operatorname{LOD}(0.5)=\log_{10}\left(\frac{L(0.5)}{L(0.5)}\right)=\log_{10}(1)=0. $$

This example does not support a preferred value below one-half. It does not establish that the loci are unlinked; it says that, for these classified transmissions and this model, the maximum within the allowed range is the reference value.

Worked Example: Combining Evidence From Two Small Pedigrees

Consider two hypothetical, independent nuclear families analyzed under the same fixed two-point model and compatible phase assumptions. Family A contributes three nonrecombinant informative transmissions. Family B contributes two nonrecombinant and one recombinant informative transmission. Find the combined maximum LOD.

Across the families, $N=3+2=5$, $R=0+1=1$, and $m=6$. Independence means the joint likelihood is the product of the family likelihoods. In terms of the classified transmissions, this gives $$ L_{\mathrm{joint}}(\theta)\propto(1-\theta)^3(1-\theta)^2\theta =\theta(1-\theta)^5. $$

The maximizing value is $\hat{\theta}=R/m=1/6\approx0.1667$, which is in the allowed range. At that value, $L_{\mathrm{joint}}(1/6)\propto(1/6)(5/6)^5\approx0.06698$. At the reference, $L_{\mathrm{joint}}(0.5)\propto(0.5)^6=0.015625$. Therefore $$ \operatorname{LOD}(1/6) =\log_{10}\left(\frac{0.06698}{0.015625}\right) \approx \log_{10}(4.287) \approx 0.632. $$

The combined data are about $4.29$ times as likely at the fitted recombination fraction as at the unlinked reference, under the stated assumptions. Multiplying the family likelihoods is essential: the shared $\theta$ is evaluated against all the transmissions, not estimated separately and then averaged.

What Changes When a Transmission Is Not Clearly Classified?

The product $\theta^R(1-\theta)^N$ is appropriate when the relevant informative meioses can be classified as recombinant or nonrecombinant under the stated assumptions. Pedigree data do not always permit that classification. A genotype may be missing, phase may be uncertain, or an observed phenotype may be compatible with more than one inherited disease allele because penetrance is incomplete or phenocopies are possible.

In such cases, one should not label an ambiguous transmission as recombinant merely to use the simple count formula. Instead, the pedigree likelihood sums over the possible unobserved inheritance configurations. In general terms, if $I$ denotes a possible configuration of transmissions and genotypes, then $$ L(\theta)=\sum_I P(\text{observed data}\mid I)\,P(I\mid\theta). $$ The terms $P(\text{observed data}\mid I)$ account for the specified observation and disease model, while $P(I\mid\theta)$ accounts for the inheritance probabilities. The LOD is still the base-10 log likelihood ratio against $L(0.5)$, but it must be calculated from this full likelihood. The examples here use the simpler classified case so that every likelihood factor and maximization can be seen directly.

Why the Maximum and the Reference Both Matter

A LOD profile is the set of LOD scores across candidate values of $\theta$. Its maximum summarizes the strongest fit within the permitted range, while the value at $\theta=0.5$ is the unlinked reference. Because $\theta=0.5$ is included in the maximization range, the largest LOD in that range cannot be negative: the score at the reference itself is zero. A score at a particular non-reference value can be negative if that value gives a lower likelihood than the reference.

The maximum-likelihood estimate and its LOD score answer related but different questions. The estimate identifies the recombination fraction that gives the greatest likelihood under the model. The LOD compares that likelihood with the unlinked likelihood. Neither quantity removes dependence on the pedigree model, phase assumptions, genotype information, or disease penetrance assumptions.

A particularly important edge case is a pedigree with no informative meioses. Then $m=0$, and the classified-meioses likelihood is proportional to $\theta^0(1-\theta)^0=1$ throughout the allowed interval. Its LOD is zero for every $\theta$, and every $\theta$ is a maximizer. There is no basis for reporting $R/m$, and the data provide no recombination-fraction information through this likelihood.

Common Mistakes

Using $R/m$ without checking the boundary. If $m>0$ but $R/m>0.5$, the constrained maximum is $0.5$, not a value above one-half.
Dividing when there are no informative meioses. If $m=0$, the likelihood is constant in $\theta$; $R/m$ is undefined and every allowed value maximizes the likelihood.
Calling a LOD score a probability. A LOD of $0.632$ corresponds to a likelihood ratio of about $10^{0.632}=4.29$, not a 63.2% probability of linkage.
Counting every child as informative. A transmission belongs in $R$ or $N$ only if the available data and phase assumptions actually distinguish its recombinant status.
Ignoring the model behind the calculation. The simple count likelihood depends on its stated phase, classification, and independence assumptions. Ambiguous data require a likelihood that accounts for possible inheritance configurations.

Key takeaway: For $m>0$ clearly classified informative meioses, calculate $L(\theta)\propto\theta^R(1-\theta)^N$, maximize it over $0\leq\theta\leq0.5$, and compare that likelihood with $L(0.5)$ using a base-10 logarithm. For $m=0$, the likelihood is constant and every allowed value is a maximizer.

Check Your Understanding

For each question, state any assumption needed to use the classified-meioses likelihood.

A hypothetical pedigree has two recombinant and six nonrecombinant informative meioses. What is the maximizing $\theta$, and is it within the allowed range?
If $R=3$ and $N=2$, where does the constrained likelihood attain its maximum? Explain why.
A pedigree has no informative meioses. What does its classified-meioses likelihood imply about the maximizing value of $\theta$?
In one sentence, explain what a LOD score of $1$ says about the likelihood at that value of $\theta$ relative to the unlinked reference.
Why can an ambiguous transmission not automatically be counted as recombinant or nonrecombinant?