Compare explanations of inheritance with a likelihood ratio
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Full chapter index · Independent practice · Genetics subject guide
Sixteen parental transmissions and four recombinant transmissions suggest that two markers may be inherited together more often than expected under no linkage. A LOD score makes that comparison explicit. It does not turn the observation into a percentage probability that a biological explanation is true.
This guide builds a simple linkage likelihood from counted transmissions, explains logarithms through evidence multiplication, and shows why separately maximised family scores cannot always be added. All families, counts and marker labels are fictional teaching constructions. The model concerns marker transmission, not diagnosis or individual health prediction.
Continue from The Gold Standard of Genetics within the Gold Standard of Education and How X Works learning library.
Full chapter index
Read the observations · Chapters 1–2
Build the model · Chapters 3–4
Calculate and compare · Chapters 5–6
Test the assumptions · Chapters 7–8
Suppose two markers have alleles A/a and B/b. A parent carries haplotypes AB and ab, and the phase is known. A suitable cross or family structure lets us identify whether the transmitted chromosome retains AB or ab, or carries a recombinant combination Ab or aB.
An informative transmission is one whose marker pattern permits the relevant parental or recombinant classification. Merely counting every family member does not establish the number of informative observations. Homozygosity, missing genotypes and uncertain parental phase can leave some observations uninformative for this particular comparison.
In our opening dataset there are twenty independent informative transmissions: sixteen non-recombinants and four recombinants. Write N = 16, R = 4 and n = 20. The observed recombinant fraction is R/n = 0.20. We assume correct classification and known phase throughout the first calculation.
These assumptions matter because a family pedigree is not automatically a collection of independent coin tosses. Real linkage analyses calculate probabilities through the pedigree while accounting for the information available. Our deliberately simple example isolates one likelihood comparison without pretending to reproduce every pedigree analysis.
The first checkpoint is to identify what was observed. We counted transmission categories, not a physical distance, a population haplotype correlation or a functional effect of an allele. The existing linkage-disequilibrium guide addresses a different observation structure: association between alleles in sampled chromosomes.
A learner should explain why an offspring is informative before using it in n. If that explanation is missing, the denominator remains uncertain and a precise logarithm will only conceal the problem.
Let θ denote the recombination probability for the two endpoints. The no-linkage reference model uses θ = 0.5. Our specified linked comparison uses θ = 0.2. Under that comparison, a transmission is recombinant with probability 0.2 and non-recombinant with probability 0.8.
A likelihood answers how well a specified parameter value supports the observed data under the model. It is treated as a function of θ after the data are fixed. It is not the probability distribution of θ, and it does not become a probability of linkage simply because it has been normalised against another likelihood.
For a particular ordered sequence with N non-recombinants and R recombinants, the likelihood is L(θ) = (1 − θ)^N θ^R. Under the reference model it is L(0.5) = 0.5^(N + R). The ratio compares those two probabilities for the same observed sequence.
If we instead describe the data as an unordered count of R recombinants, the binomial coefficient multiplies both likelihoods. Because both refer to the same n and R, that coefficient cancels in their ratio. Omitting it in this ratio calculation is therefore justified; omitting data-dependent factors without checking cancellation would not be.
The LOD score at a specified θ is log10[L(θ)/L(0.5)]. A positive score favours that linked model relative to the reference. A negative score favours the reference over that particular linked value. A zero score means the two likelihoods are equal for the data and comparison being made.
Name the comparison before calculating. “The LOD is positive” is incomplete if the tested θ and model have disappeared from the explanation. The evidence is always relative to explicitly defined alternatives.
For N = 16, R = 4 and θ = 0.2, the likelihood ratio is (0.8^16 × 0.2^4)/0.5^20. Equivalently, it is (0.8/0.5)^16 × (0.2/0.5)^4 = 1.6^16 × 0.4^4.
Taking base-10 logarithms converts multiplication into addition. The score is 16 log10(1.6) + 4 log10(0.4). These terms are approximately 16 × 0.204120 and 4 × (−0.397940), giving 1.674160.
The likelihood ratio is 10^1.674160, approximately 47.2. Under the stated independent-transmission model, the observed categories are about 47 times as likely at θ = 0.2 as at θ = 0.5. That is the interpretation supported by the calculation.
Do not report a 98% probability that the loci are linked by converting 47.2 into 47.2/(1 + 47.2). Such a conversion would introduce a prior-odds assumption and a particular comparison between models. Neither has been supplied by the simple LOD calculation.
The positive and negative terms are instructive. Each non-recombinant supports θ = 0.2 relative to 0.5 because 0.8 exceeds 0.5. Each recombinant opposes that θ because 0.2 is smaller than 0.5. The final score balances both categories rather than treating non-recombinants as the only evidence.
For an arithmetic check, calculate the ratio directly and then take its logarithm. It should agree with the sum of logarithms, subject to rounding. The log form becomes particularly useful when many observations would make the direct probabilities extremely small.
| Non-recombinants N | Recombinants R | Tested θ | LOD |
|---|---|---|---|
| 8 | 2 | 0.2 | 0.837080 |
| 16 | 4 | 0.2 | 1.674160 |
| 32 | 8 | 0.2 | 3.348319 |
| 5 | 5 | 0.1 | −2.218487 |
The same dataset can be evaluated at several θ values. At θ = 0.5, the ratio is exactly one and the LOD is zero by construction. At θ = 0.2, our dataset yields about 1.674. At θ = 0.1, the score is about 1.289, lower than at 0.2.
Why does a smaller recombination probability not automatically produce stronger evidence? Four recombinants were observed. Making θ extremely small makes those observations increasingly surprising. The best-fitting value must balance the observed numbers of both categories.
For this simple model with known phase and independent classified transmissions, the unconstrained maximum occurs at θhat = R/n. Restricting θ to the usual interval from zero to 0.5 gives the corresponding constrained estimate. In our example, R/n = 0.2 already lies inside that interval.
If all observations were non-recombinant, the simple likelihood would peak at the boundary θ = 0. That does not prove that recombination is biologically impossible. A finite sample can contain no observed recombinant even when the true probability is positive. Boundary estimates need a careful interpretation of sample information.
A LOD curve preserves more information than a single peak number. It shows which θ values receive comparable support and which fit poorly relative to the reference. A narrow peak and a broad peak may have similar maximum heights while supporting different levels of parameter precision.
Connect this distinction with Haldane mapping. Estimating θ from transmissions is one task. Converting θ into a model-dependent genetic distance is another. A maximum likelihood fraction does not remove the assumptions needed for that conversion.
Suppose two independent families each provide eight non-recombinants and two recombinants, with known phase and the same parameter θ = 0.2. Each family has a LOD of approximately 0.837080. Adding the scores gives 1.674160, matching the pooled sixteen-and-four example.
This works because independent likelihoods multiply when evaluated under the same shared parameter and compatible models. Taking log10 turns that product into a sum. The additivity is a property of the model and its factorisation, not an instruction to add any numbers labelled LOD.
The reference likelihood also multiplies across those independent datasets. The joint likelihood ratio is [L1(θ)L2(θ)]/[L1(0.5)L2(0.5)]. Its logarithm is the sum of the two individual log likelihood ratios at that same θ.
Dataset size influences evidence strength. Repeating the original proportions in an independent dataset of forty transmissions, with thirty-two non-recombinants and eight recombinants, doubles the LOD to about 3.348319. The estimated θ remains 0.2. More observations can increase support without changing the estimated fraction.
Duplicating the same records in a spreadsheet is different. It does not create independent evidence and should not double the score in a valid analysis. Family identifiers, shared individuals and pedigree structure help distinguish genuinely new information from repeated representations of the same observations.
The checkpoint is to state why the pieces may be multiplied. If they share observations or depend on an unmodelled common process, the simple product may overstate evidence. Combining data requires a justified relationship between the datasets, not merely matching column headings.
Consider two invented families. Family A has nine non-recombinants and one recombinant. Its simple best-fitting θ is 0.1. Family B has five non-recombinants and five recombinants, so its best-fitting constrained θ is 0.5.
If a joint model requires one shared θ, those two family-specific optima cannot both be used simultaneously. Adding the separately maximised scores would combine two different parameter choices while labelling the result as evidence for one shared recombination fraction.
Pool the counts for the shared-θ model: fourteen non-recombinants and six recombinants. The common optimum is θhat = 6/20 = 0.3. The joint score is 14 log10(0.7/0.5) + 6 log10(0.3/0.5), approximately 0.7147.
Family A’s separate maximum is approximately 1.5985, while Family B’s maximum is zero. Their sum exceeds the shared-model maximum. The difference arises because separate maximisation allows more flexibility, not because an arithmetic error has occurred in either individual calculation.
This illustrates a general rule: sum likelihood contributions at each common parameter value, then maximise the combined curve. Do not maximise each curve first and assume the resulting sum represents the same model. In symbols, the maximum of a sum is generally no greater than the sum of separate maxima.
A heterogeneous model can legitimately allow different parameters or other family-specific features, but that is a changed scientific model. Its interpretation and evidence assessment need to reflect the extra flexibility. The simple example does not diagnose the cause of family differences; it demonstrates why model compatibility matters when evidence is combined.
A likelihood ratio describes how the observed data compare under specified alternatives. Posterior odds also depend on prior odds: posterior odds = likelihood ratio × prior odds in a suitable two-model Bayesian comparison. The prior information is not supplied by taking a logarithm.
For a purely mathematical illustration, a likelihood ratio of 1,000 multiplies prior odds of 1:1 into posterior odds of 1,000:1. The same ratio multiplies prior odds of 1:10,000 into posterior odds of 1:10. The data comparison is identical while the posterior comparison differs.
These numbers demonstrate the distinction, not a proposed prior for a genetics study. Real model comparisons can involve composite alternatives and integration over uncertain parameters. A maximised likelihood ratio should not automatically be treated as a Bayes factor with the same numerical interpretation.
A LOD of 3 corresponds arithmetically to a likelihood ratio of 10^3, or 1,000, for the stated comparison. That does not establish a universal 0.1% error rate or a 99.9% probability that the biological hypothesis is correct. Different statistical questions require different calculations.
Similarly, a p-value describes the probability of results at least as extreme under a specified null procedure. It is not the same object as a likelihood ratio. Converting between them requires a sampling theory and procedure; it is not achieved by simply reversing an exponent.
The practical checkpoint is to write “relative support under these models” beside a LOD calculation. Then identify whether the question asks for parameter estimation, evidence against a reference, a calibrated testing rule or a posterior belief. The same dataset can support several analyses, but the resulting numbers must retain their different meanings.
LOD thresholds appear in genetics literature, including traditional references around 3 and genome-scan discussions using approximately 3.3 under particular assumptions. These should not be presented as universal pass marks for every linkage study, organism, marker set or analysis.
Testing many positions creates more opportunities for an unusually large score somewhere. The genome-wide interpretation therefore differs from an analysis of a single prespecified comparison. Marker density, inheritance model, trait structure and the analysis procedure can affect calibration.
The research discussion All LODs Are Not Created Equal is useful for understanding why a number’s label does not make every implementation statistically interchangeable. The point for this article is to preserve the distinction between the score’s definition and a study-specific decision threshold.
A classroom problem may explicitly supply a threshold and ask whether the calculated statistic exceeds it. Apply that instruction while stating its role. A research report must justify why its threshold is appropriate for the actual procedure and scope of the search.
A score just above a convention does not identify a causal allele or establish a mechanism. A linked region can contain several candidate loci, and further evidence may be needed to resolve them. Evidence of co-inheritance and evidence of biological function answer different questions.
An effective reading habit is to locate the methods statement beside the headline result. Ask how the score was defined, what was searched, which model was used and how the threshold was obtained. This keeps a familiar number from acquiring more authority than the procedure supports.
Known phase is a major simplification in our example. A double heterozygote with genotype AaBb could carry AB/ab or Ab/aB. The same offspring marker combination can be classified differently depending on that phase. If phase is uncertain, a real analysis must incorporate or resolve that uncertainty appropriately.
Choosing whichever phase produces the largest score without accounting for the selection step can overstate evidence. So can dropping inconvenient observations, treating uncertain genotypes as certain or classifying an uninformative transmission as non-recombinant.
Genotyping errors can create apparent recombinants, including patterns that suggest extra exchanges over short intervals. Missing data can reduce information or alter which observations contribute. A good analysis retains the raw observations and documents quality control rather than silently editing counts to fit a preferred map.
The three-point mapping guide offers a useful neighbouring example: a surprising offspring class may be biologically informative, misclassified or affected by data quality. Investigation should precede a confident mechanism claim.
For a classroom audit, put every observation into one of three categories: confidently non-recombinant, confidently recombinant or unresolved under the available information. Only the first two enter the simple count formula as demonstrated. The unresolved group is not automatically discarded from every possible analysis; it needs a model appropriate to its information.
This diagnostic sequence follows the foundation-first approach illustrated in the Clementi mathematics guide. Repair the observation-to-category step before increasing sample size or adding a more sophisticated statistical transformation.
Compare equal fractions at different sample sizes
Consider two invented datasets with the same observed fraction of 0.2. The first has four non-recombinants and one recombinant. The second has forty non-recombinants and ten recombinants. At θ = 0.2, the first LOD is about 0.418540 and the second is about 4.185399.
The parameter estimate is unchanged, while the evidence against θ = 0.5 is much stronger with the larger independent sample. This comparison repairs a common confusion between effect size or parameter location and evidence strength. One describes the pattern’s magnitude; the other also reflects how much information supports it.
Now keep n fixed at twenty and compare sixteen-and-four with eleven-and-nine. The second fraction is near 0.5 and provides much less support for a smaller shared recombination probability. A larger score is not generated merely by calling the majority category parental; the actual balance of categories matters.
Inspect a score contribution that opposes the preferred model
Suppose the tested θ is 0.1, but a new independent family supplies five non-recombinants and five recombinants. Its score is 5 log10(0.9/0.5) + 5 log10(0.1/0.5), approximately −2.2185. Adding it at that same θ reduces any earlier positive joint score.
The negative contribution is scientifically useful. It says that these observations fit the specified θ = 0.1 poorly relative to the reference. It does not justify deleting the family, relabelling observations or switching its θ independently while retaining the claim of a shared-parameter model.
A next step could inspect data quality, model assumptions or genuine heterogeneity, depending on the study. Those possibilities require evidence. The arithmetic identifies a tension; it does not choose a convenient explanation for removing it.
Build a minimal reproducible calculation
A small calculation record can include the tested θ, N, R, the non-recombinant log contribution, the recombinant log contribution and their sum. Keeping these pieces visible makes it possible to reconstruct the result without depending on a software label.
For the opening example, enter 0.2, 16 and 4. Compute log10[(1 − θ)/0.5] once and multiply by N. Compute log10[θ/0.5] once and multiply by R. Check that the sum becomes zero at θ = 0.5 and falls if θ becomes implausibly small given the observed recombinants.
If a logarithm is undefined, inspect the boundary and count combination. A model assigning zero probability to an observed recombinant has zero likelihood for that dataset. A software error message can therefore reflect a meaningful incompatibility rather than a missing calculator trick.
The final audit separates the observation record from the model grid. Changing θ changes the likelihood assigned to the fixed observations. It does not change which offspring were seen. This simple discipline protects both reproducibility and interpretation when a more complicated pedigree model replaces the classroom count model.
Problem A gives eight non-recombinants and two recombinants, known phase, and θ = 0.2. Calculate the LOD against θ = 0.5. The answer is 8 log10(1.6) + 2 log10(0.4), approximately 0.837080. The corresponding likelihood ratio is about 6.87.
Problem B gives a score of −1 at a specified θ. Interpret it without inventing a posterior probability. The linked parameter value makes the observed data one tenth as likely as the no-linkage reference under the stated model. It does not prove that every possible linked parameter value has been excluded.
Problem C supplies two independent datasets evaluated at the same θ with scores 1.2 and 0.8. Their joint score is 2.0 and their likelihood ratios multiply: 10^1.2 × 10^0.8 = 100. The operation is addition on the log scale because it is multiplication on the likelihood-ratio scale.
Now alter Problem C: the two numbers are separately maximised scores at incompatible θ values, while the joint model requires a common θ. You no longer have enough information to calculate the joint maximum from those two peak heights alone. You need the curves or sufficient data to evaluate a common parameter.
For a delayed check, ask which observation would most directly repair each mistake. A wrong logarithm base needs a numerical-method correction. An inflated n needs a data-structure correction. A correct score described as a probability of causation needs an interpretation correction. These are different learning problems.
The task is complete when a learner can calculate the number, explain its denominator, name its comparison and state one inference it does not establish. Fluent arithmetic and restrained interpretation belong together.
Does a larger LOD always mean a smaller recombination fraction?
No. A score reflects evidence strength under a model, while the estimated fraction describes a parameter. A large sample with a moderate fraction can provide strong evidence; a tiny sample with no recombinants can remain weakly informative. Report the estimate and evidence separately.
Can a negative LOD be added to positive evidence?
Yes, when the datasets are independent and evaluated at the same compatible parameter and models. Negative contributions reduce the joint score. Omitting them simply because they disagree would alter the evidence being combined.
Does linkage identify the causal variant?
No. It can support co-inheritance or localise a region under the analysis. Functional interpretation requires additional evidence. Our marker-count examples make no claim about phenotype, disease or a mechanism of action.
Why use logarithms when a ratio is easier to explain?
Logarithms make multiplication manageable, keep very small likelihoods numerically tractable and allow compatible evidence to be accumulated through addition. The interpretation can still be translated back with 10^LOD so that the numerical scale remains understandable.
What should accompany a reported score?
State the data structure, informative observations, model, phase treatment, tested parameter or maximisation procedure and reference model. If a threshold is used, explain its calibration and search scope. A peak number without those details is difficult to reproduce or compare.
A useful final language check is to replace “proof” with the actual result: the data favour one stated model over another by the reported likelihood ratio. That sentence is precise without pretending the calculation settles every biological question.
The NHGRI LOD-score glossary introduces the term. Its genome-analysis teaching handout provides linkage-analysis context, and All LODs Are Not Created Equal discusses the importance of statistical definition and calibration. Our count tables and comparisons are original fictional teaching exercises.
Continue with Haldane mapping to convert a recombination fraction under a stated crossover model, and D′ and r² to distinguish population association from family transmission. Return to the Genetics subject root to place the methods in the wider inheritance framework.
For retrieval tomorrow, rebuild the likelihood ratio from a new pair of counts without looking at the equation. Explain why the binomial coefficient cancels. Then combine two compatible datasets at one specified θ. A week later, introduce different family-specific peaks and decide why adding their maxima changes the model.
This sequence turns a statistical label into a reusable reasoning tool. The evidence chain runs from informative observation to category count, from count to likelihood, from likelihood to a named comparison, and from comparison to a carefully limited conclusion.
The Gold Standard of Education and How X Works library share that goal: independent use when the wording changes. Here, independence means recognising which quantities may be combined, which assumptions support them and when the data do not answer the stronger question being asked.
Continue the connected explanation
Connect inheritance evidence with the stages that produce observed ratios. Read The Gold Standard of Genetics | Segregation Distortion and the Survival Filter Behind an Inheritance Ratio.