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For independent affected sib pairs with IBD counts $n_0,n_1,n_2$, the likelihood function of the recombination fraction is proportional to $s^{2n_0+n_1}(1-s{)}^{2n_2+n_1}$, with $s=2\theta (1-\theta )$. At fixed sample size, the total shared-copy count $2n_2+n_1$ is a sufficient statistic by the Fisher-Neyman factorization theorem. Thus a one-sided likelihood-ratio test has the same ordering as a test for excess mean IBD sharing. This equivalence depends on the stated fully penetrant recessive model and does not extend to arbitrary disease models.

// Target: biology.bigb

A centimorgan is one hundredth of a morgan. For a sufficiently short genetic map interval, one centimorgan corresponds approximately to a $1\%$ recombination fraction, rather than an exact identity for all intervals.

Complete linkage in a two-locus model means recombination fraction zero: only parental haplotype combinations are transmitted. Multiple genetic markers in a nonrecombining block represent one haplotype segregation observation.

Why can two individually fair allele transmissions fail to be independent? Genetic loci on the same chromosome can travel together through meiosis. This tendency is genetic linkage; in an ordinary two-locus model, detectable genetic linkage means a recombination fraction below $1/2$. A genetic marker can therefore help track a neighboring genetic locus through a pedigree. See Clark, Douglas and Choi, Biology 2e: chromosomal theory and genetic linkage.

Take a diploid individual heterozygous at two genetic loci, with phase-known genotype $AB/ab$: one homologous chromosome carries alleles $A,B$, the other $a,b$. The parental haplotypes are $AB,ab$; the recombinant haplotypes are $Ab,aB$. Assume fair Mendelian segregation, equal probabilities within each reciprocal pair of gametes, and no allele-dependent differences in gamete recovery. If the recombination fraction is $r=0.2$, the gamete probabilities are $0.4,0.4,0.1,0.1$ in that order. Each genetic locus still transmits its two alleles equally, but $P(AB)=0.4$ exceeds $P(A)P(B)=0.25$: fairness at each genetic locus does not imply independence between genetic loci.

At $r=0$, only the parental haplotypes occur, each with probability $0.5$: this is complete genetic linkage. At $r=0.5$, all four haplotypes have probability $0.25$, giving independent assortment. A recombination fraction near $0.5$ alone cannot establish whether the genetic loci lie on different chromosomes or far apart on one chromosome.

Genetic linkage concerns joint transmission through meiosis. Linkage disequilibrium concerns haplotype frequencies in a population, while genetic association compares allele or genotype frequencies with a phenotype. These questions require different observations; pedigrees directly reveal co-transmission.

A genetic map turns observations of inherited haplotypes into an ordered account of genetic loci along a chromosome. Its distances describe genetic recombination, which makes the map useful for locating a genetic locus relative to known genetic markers. A genetic map measures inheritance separation; separation measured in base pairs describes positions along DNA. Recombination hotspots help explain why equal numbers of base pairs need not give equal genetic map distances. See T. A. Brown, Genomes, 2nd edition (2002), chapter 5: Mapping Genomes.

Write $d$ for the expected value of the number of chromosomal crossovers encountered along one transmitted chromatid between the chosen genetic loci, measured in morgans; $100d$ is the distance in centimorgans. This counts exchanges along that transmitted chromatid, rather than all exchanges in the four-chromatid meiotic bundle. For a short interval with very few multiple exchanges, the recombination fraction satisfies $r\approx d$: about $1\%$ recombinant gametes then corresponds to about one centimorgan.

For a longer interval, hidden even numbers of chromosomal crossovers make that approximation inaccurate. Under the Haldane mapping function, $d=0.2$ morgans gives $r\approx 0.164840$; conversely $r=0.2$ gives $d\approx 0.255413$ morgans, or $25.5413$ centimorgans. Applying the inverse formula to an already specified genetic map distance would confuse a count with a probability.

Genetic map distances add along ordered adjacent intervals because the total exchange count is the sum of its interval counts. For independent random variables describing counts in disjoint intervals of the Haldane mapping function model, the endpoint recombination fraction instead obeysExactly one interval must change parental origin. Thus $r_1=r_2=0.2$ gives $r_{12}=0.32$, although the two genetic map distances add to approximately $0.510826$ morgans. The simple composition formula requires the stated independence; crossover interference can invalidate it.

Genetic recombination rearranges inherited genetic material into new combinations. In the meiosis example considered here, chromosomal crossover exchanges segments between nonsister chromatids of homologous chromosomes. A transmitted chromatid can consequently contain DNA from both parental homologues. This produces new haplotypes without requiring a new mutation at either scored genetic locus. See Meiosis.

For phase-known genotype $AB/ab$, a single chromosomal crossover between the selected genetic loci can yield $Ab$ and $aB$ on the participating chromatids; the other two chromatids can remain $AB$ and $ab$. Thus an exchange in the four-chromatid bundle does not mean every resulting gamete is recombinant. Only a chromosomal crossover encountered along the transmitted chromatid within the selected interval affects its two-locus score.

Follow parental origin along the transmitted chromatid from the first genetic locus to the second. Each exchange switches that origin. An odd number of such chromosomal crossovers changes the endpoint combination; an even number restores parental phase. Two exchanges can therefore leave the scored haplotype as $AB$ even though an internal segment came from the other parental homologous chromosome. Recombinant status is relative to the chosen genetic loci, not a complete record of exchanges along the chromosome.

This integer parity distinction connects the physical process of genetic recombination to the observed recombination fraction. It also explains why a long genetic map interval can accumulate many exchanges while its endpoint recombination fraction approaches $1/2$ in the Haldane mapping function model.

The Haldane mapping function connects additive genetic map distance to endpoint recombination fraction in a toy model with no crossover interference. Specify the model directly along a transmitted chromatid: the exchange count $N$ in the interval has a Poisson distribution with expected value $d$, where $d$ is the distance in morgans. Counts in disjoint intervals are independent random variables. This is the transmitted-chromatid convention; $d$ is not the mean number of exchanges across all four chromatids in meiosis.

An odd exchange count changes parental phase. Summing the odd terms of the Poisson distribution givesFor $0\le r<1/2$, the inverse and short-interval approximation areHere $\log$ is the natural logarithm. At $d=0.2$, the recombination fraction is $0.16483997698$; at $r=0.2$, the distance is $0.2554128119$ morgans, or about $25.5413$ centimorgans. The linear approximation $r\approx d$ is useful only near zero. The recombination fraction approaches $1/2$ as $d$ grows; $r=1/2$ has no finite inverse in this model.

The model is a useful baseline, not a universal rule for every interval during meiosis. Crossover interference can change the relationship between distance and recombination fraction. For the count convention and assumptions, see Weeks, Tang and Kwon (2009): genetic map functions and exchange-count conventions. Haldane's original paper is bibliographically recorded as The combination of linkage values, and the calculation of distances between the loci of linked factors, Journal of Genetics 8 (1919), 299–309, in Selected Genetics Papers of J. B. S. Haldane, edited by K. Dronamraju (1990), contents.

Independent assortment means that transmission at different genetic loci has conditional independence given the parental genotype. With fair Mendelian segregation, a parent heterozygous at two genetic loci, $AaBb$, produces $AB,Ab,aB,ab$ gametes with probability $1/4$ each. For example,Here $AB$ labels an allele combination; it does not assert that the two genetic loci lie on the same chromosome.

The physical model is stochastic orientation of different homologous chromosome pairs toward the two poles of the spindle apparatus in meiosis I. One pair's orientation does not determine another's. For two pairs, one division can send $A,B$ together and $a,b$ together; another can send $A,b$ together and $a,B$ together. See The Process of Meiosis (§11.1, Metaphase I and Figure 11.5: random orientation of chromosome pairs).

With $k$ distinguishable homologous chromosome pairs, there are $2^k$ choices of parental homologous chromosome origin for a product at a specified pole, each with probability $2^{-k}$ under this model. For $k=3$, enumerate $MMM,MMP,MPM,MPP,PMM,PMP,PPM,PPP$, where M and P mark parental origins. Whole-pole reversal pairs complementary choices, leaving $2^{k-1}$ unordered partitions for $k\ge 1$. For $k=2$ there are four choices for a specified pole but only the two relative orientations drawn. This counts possible products across meioses, not eight different products of one division when $k=3$. Identical alleles at scored genetic loci can also make different parental-origin choices observationally indistinguishable.

Independent assortment of different chromosome pairs does not imply independence of every pair of genes. Nearby genetic loci on the same chromosome can show genetic linkage. Chromosomal crossover can create new haplotypes, but does not automatically erase genetic linkage. See Laws of Inheritance (§12.3, Law of Independent Assortment and Linked Genes Violate the Law of Independent Assortment). In the ordinary fair two-locus model with parental haplotypes $AB/ab$ and recombination fraction $r$,Thus $r=0$ gives only the two parental haplotypes, while $r=1/2$ gives four equal probabilities. A recombination fraction of $1/2$ alone does not establish that the genetic loci lie on different chromosomes.

A motivating experiment appears in Experiments in Plant Hybridization by Gregor Mendel (§9, “The Reproductive Cells of the Hybrids”). In peas, hybrids for round versus wrinkled seeds and yellow versus green seeds were crossed reciprocally with constant wrinkled-green plants. The readout was visible seed shape and color: hybrid seed parents gave 31 round-yellow, 26 round-green, 27 wrinkled-yellow and 26 wrinkled-green; the reciprocal cross gave 24, 25, 22 and 27.

The apparatus was controlled pollination: opening immature flowers, removing stamens with forceps, and applying selected pollen to the stigma. Two-year trials established constant parental characters; greenhouse pots checked interference from insects in outdoor beds. Reciprocal crosses tested both transmission directions. These procedures are described in Experiments in Plant Hybridization by Gregor Mendel (§§2–3).

With complete dominance at both genetic loci, this test cross makes each phenotype report one hybrid gamete combination. The ideal expected values are $110/4=27.5$ and $98/4=24.5$ per class. The observations support four transmitted combinations, approximately equally frequent. They do not directly image chromosome orientation or prove independence for all genes; failed development and seed germination also limit the inference, as §9 notes.

Linkage disequilibrium means that knowing an allele at one genetic locus changes the probability of an allele at another genetic locus in population haplotypes. It compares their joint probability distribution with the product of their marginal distributions. This is useful because a genetic marker can carry information about another genetic locus: tag SNPs exploit this redundancy, without establishing a causal genetic association.

Consider two genetic loci with alleles $A,a$ and $B,b$. Sample one haplotype, rather than an unphased diploid genotype. Write its four haplotype frequencies as $f_{AB},f_{Ab},f_{aB},f_{ab}$, summing to one, and set the marginal allele frequencies to $p=f_{AB}+f_{Ab}$ and $q=f_{AB}+f_{aB}$. The linkage disequilibrium coefficient isThe second expression follows by substituting $f_{ab}=1-f_{AB}-f_{Ab}-f_{aB}$. The entire two-locus joint probability distribution is thenEvery row and column sum is unchanged when $D$ varies. Linkage equilibrium means $D=0$: all four haplotype frequencies equal products of allele frequencies, so the two allele indicators are independent random variables. For $p>0$, the conditional probability $P(B\mid A)=q+D/p$ shows directly how positive $D$ enriches $B$ among copies carrying $A$, while negative $D$ depletes it. Positive $D$ also enriches $ab$; negative $D$ enriches $Ab$ and $aB$. Swapping the labels $A$ and $a$ reverses the sign of $D$, so neither sign is intrinsically favorable.

For example, fix the allele frequencies at $p=0.60,q=0.40$. At linkage equilibrium, the haplotype frequencies in the order $AB,Ab,aB,ab$ are $(0.24,0.36,0.16,0.24)$. Positive linkage disequilibrium $D=0.12$ instead gives $(0.36,0.24,0.04,0.36)$, with $P(B\mid A)=0.60$; negative linkage disequilibrium $D=-0.12$ gives $(0.12,0.48,0.28,0.12)$, with $P(B\mid A)=0.20$. Counting each allele separately would miss the difference.

Nonnegative haplotype frequencies constrain $D$. Define the nonnegative bounds $L=\min (pq,(1-p)(1-q))$ and $U=\min (p(1-q),(1-p)q)$. The four entries above giveThus $|D|\le 1/4$ globally, but a particular pair of allele frequencies usually permits a smaller, asymmetric interval. The standardized linkage disequilibrium convention used here retains the sign:This convention requires $0<p,q<1$; if either genetic locus has only one allele in the population, $D=0$ but normalized measures are undefined. Some presentations report only $|D^{\prime }|$. At $D^{\prime }=1$ or $-1$, at least one of the four haplotypes has zero frequency; this alone does not prove that genetic recombination has never occurred.

For the indicator random variables $X=1_A$ and $Y=1_B$, their covariance is $D$, and their variances are $p(1-p)$ and $q(1-q)$. The allelic correlation measure is their squared correlation coefficient,The subscript distinguishes it from the recombination fraction $r$ in linkage-disequilibrium decay under recombination. Squaring loses the association's sign. In the illustrated positive and negative examples, $r_{\mathrm{LD}}^2=0.25$ in both, whereas $D^{\prime }=0.75$ and $-0.50$, respectively. For another original example, take $p=0.10,q=0.50,f_{AB}=0.10$: then $D=0.05,D^{\prime }=1$, yet $r_{\mathrm{LD}}^2=1/9$. Every $A$ carries $B$, but most copies carrying $B$ lack $A$. The frequency constraints and the distinction between these normalizations are analyzed in Mathematical properties of the r² measure of linkage disequilibrium by Jenna M. VanLiere and others (sections 2.1 and 2.5, pp. 131–132 and 135–136).

Genetic linkage describes transmission through meiosis; linkage disequilibrium describes a population's haplotype frequencies. Nearby genetic loci on a chromosome can be in linkage equilibrium. Conversely, mixing populations can create linkage disequilibrium even between genetic loci on different chromosomes. For example, mix equal numbers of independently sampled haplotypes from two populations, each internally at linkage equilibrium, with $p=q=0.80$ in one and $p=q=0.20$ in the other. The pooled allele frequencies are $p=q=0.50$, but $f_{AB}=(0.64+0.04)/2=0.34$, giving $D=0.09$. This is a consequence of population stratification, not a measurement of the recombination fraction.

Measured haplotype frequencies also depend on the sample. Finite sample size, related individuals, genetic-study ascertainment, genotyping error and uncertain haplotype estimation can change an estimate. A missing rare haplotype can give empirical $|D^{\prime }|=1$ without demonstrating its absence from the population. Report the sample definition and allele frequencies alongside the association measure; a nonzero sample estimate alone does not establish nonzero population linkage disequilibrium.

A concrete dataset is the 2005 A haplotype map of the human genome by The International HapMap Consortium. The International HapMap Consortium used genotyping platforms including Illumina BeadArray to obtain SNP genotype calls from 269 DNA samples across four collections: 30 parent–offspring trios from Ibadan, 30 trios from Utah, 45 unrelated individuals from Beijing and 44 from Tokyo. Parent–offspring trios and the PHASE algorithm supplied haplotype estimation, rather than direct observation of every chromosome. In Figure 7, 36 SNPs resolve seven haplotypes in a chromosome 2 region in the Utah panel; seven tag SNPs capture the 36 SNPs at $r_{\mathrm{LD}}^2\ge 0.8$. The readout demonstrates reusable allelic correlation, not causal effects.

Controls required over 80% complete calls, at most one discordance among five duplicate genotypes, at most one Mendelian segregation error per trio panel, and a Hardy-Weinberg equilibrium test with $P>0.001$; blind reassessment checked performance. Minor allele frequencies of at least 0.05 were targeted, the collections were not population-representative, and inferred haplotypes retain phase uncertainty. These limits restrict extrapolation of linkage disequilibrium.^[ref]

Genetic recombination erodes linkage disequilibrium by reshuffling alleles between haplotypes, while preserving marginal allele frequencies in the idealized model. Let $f_{AB,t}$ denote the haplotype frequency in the gamete pool of generation $t$, let $p,q$ denote its marginal allele frequencies, and let $r\in [0,1/2]$ be the constant recombination fraction between two autosomal genetic loci. Assume random mating within one population, nonoverlapping generations, fair Mendelian segregation, equal reciprocal gamete production, the same $r$ in both sexes, and a deterministic population large enough to neglect genetic drift. Exclude natural selection, mutation and gene flow. ThenHere $p$ and $q$ remain constant: recombining changes which alleles occur together, not their total numbers. The other haplotype frequencies change by $+r{D}_t$ for $Ab,aB$ and $-r{D}_t$ for $ab$. Consequently the joint probability distribution stays within its allowed bounds and approaches the product of its marginal distributions when $r>0$.

For the original positive example $p=0.60,q=0.40,D_0=0.12$, a recombination fraction $r=0.10$ gives $D_1=0.108$ and $D_{10}\approx 0.04184$. The negative example $D_0=-0.12$ has exactly the opposite sequence: genetic recombination reduces the magnitude without reversing the sign. With $r=0$, the initial linkage disequilibrium persists in this model; even independent assortment at $r=1/2$ only halves $D$ each generation. A population association does not vanish after one generation merely because the genetic loci are unlinked.

The geometric sequence gives a generation scale: for $r>0$, the continuous interpolation reaches $|D_0|/2$ at $t_{1/2}=\log (1/2)/\log (1-r)$. Actual generations are integers; the first at or below half is $\lceil t_{1/2}\rceil$. For small $r$, $t_{1/2}\approx (\log 2)/r$. Because the denominator of allelic correlation stays fixed, $r_{\mathrm{LD},t}^2=(1-r{)}^{2t}{r}_{\mathrm{LD},0}^2$; its decay is faster than the magnitude of $D_t$.

A short derivation uses random mating to assign frequency $2f_{AB}{f}_{ab}$ to the double heterozygote $AB/ab$ and $2f_{Ab}{f}_{aB}$ to $Ab/aB$. Relative to $r=0$, each loses or gains an $AB$ gamete fraction $r/2$, respectively. Their net contribution is $r(f_{Ab}{f}_{aB}-f_{AB}{f}_{ab})=-rD$. The other genotypes produce no net change in $f_{AB}$, proving the recurrence and, by iteration, its solution.

This temporal model does not identify elapsed generations from physical separation alone. The recombination fraction depends on genetic recombination, including recombination hotspots, while population stratification, natural selection, gene flow or genetic drift can create or maintain linkage disequilibrium. Repeated sampling across time would also require accounting for sample size and haplotype estimation. A cross-sectional plot of association against distance between genetic loci is therefore a different observation from the idealized trajectory plotted here.

A morgan is a unit of genetic map distance: an expected value of one exchange along the transmitted chromatid in the interval. One morgan equals 100 centimorgans; it does not mean a recombination fraction of one.

Parametric linkage analysis forms a pedigree likelihood from a specified disease model, including penetrance, allele frequencies and recessive inheritance or another transmission model. It estimates or tests the recombination fraction between a disease locus and a genetic marker.

// Target: biology.bigb

DNA repair preserves usable inherited information, while genetic recombination can improve the combinations in which that information is transmitted. Their evolution need not have the production or prevention of diversity as an ultimate goal. natural selection favors consequences for survival and reproduction in particular contexts; novelty is neither automatically beneficial nor something an organism must maximize.

mutation creates new sequence variants; genetic recombination usually rearranges variants already present. A chromosomal crossover can produce a new haplotype without changing the allele frequency at either of two genetic loci. Conversely, DNA repair can restore sequence integrity after damage without eliminating the inherited differences between individuals. In particular, a normal alternative allele is not an error merely because it differs from another individual's sequence. These operations address different features of inheritance.

Unrepaired damage can block DNA replication or transcription, break a chromosome, or generate a harmful mutation. DNA repair thus has immediate benefits to a biological cell and can reduce the future burden of deleterious changes in descendants. In multicellular organisms, somatic mutation can compromise tissue function, while germline mutation changes what offspring inherit. The relevant balance includes the biochemical costs of DNA repair, the probability and severity of damage, and the possibility of DNA repair errors. natural selection for increased fidelity can become weak relative to genetic drift when an additional improvement is extremely small; perfect fidelity need not evolve. Some error-prone damage responses can also improve immediate survival even if surviving descendants carry more mutations.

The biochemical relationship is complementary. homologous recombination repairs a DNA double-strand break by using a related intact sequence as a template. In meiosis, programmed breaks can be repaired using a homologous chromosome, sometimes producing chromosomal crossovers and sometimes noncrossover products. Biochemical identification of Spo11 linked a component of the meiotic machinery to programmed break formation, demonstrating that meiotic diversification can begin with a controlled DNA-damage-and-repair process.^[ref] Crossovers also help establish the physical connections needed for reliable segregation in many meiotic systems. This immediate transmission function can favor the machinery independently of any later benefit from a novel haplotype. DNA repair with sister chromatids in another context can preserve sequence without the same exchange between parental homologues.

There are also population-level benefits of reshuffling. Suppose two favorable alleles, $A$ and $B$, occur in different lineages as $Ab$ and $aB$. With no exchange, their lineages compete and either favorable allele can be lost when the other lineage prevails. genetic recombination can create $AB$, allowing selection to combine their benefits; it can also separate a favorable allele from linked harmful changes. This is the core of the Fisher-Muller hypothesis and relates to Hill-Robertson interference, the reduction in the efficacy of natural selection caused by associations among selected genetic loci. Recreating low-load genotypes can also slow Muller's ratchet in finite biological populations.

The bookkeeping illustrates why this is not simply increasing all diversity. Let $x_{AB},x_{Ab},x_{aB},x_{ab}$ be haplotype frequencies and $D=x_{AB}{x}_{ab}-x_{Ab}{x}_{aB}$ be linkage disequilibrium. Under the standard random-mating model with recombination fraction $r$ and no other evolutionary forces,For $x_{Ab}=x_{aB}=1/2$ and the other two zero, $D=-1/4$. At $r=0.1$, the next frequencies are $x_{AB}^{\prime }=x_{ab}^{\prime }=0.025$ and $x_{Ab}^{\prime }=x_{aB}^{\prime }=0.475$. Each favorable allele still has frequency $1/2$: the process creates both the favorable combination and its unfavorable counterpart. Subsequent natural selection determines their fates. If the original associations are already advantageous, breaking them can instead impose a recombination load.

In controlled yeast experimental evolution, sexual and asexual treatments differed in their rates and genomic routes of adaptation, with the sexual treatment more efficiently separating beneficial from deleterious changes.^[ref] A mechanistic test should control the costs and bottlenecks of the sexual cycle, compare otherwise matched genetic backgrounds, and use reconstructed changes to distinguish genetic linkage from intrinsically different effects of a mutation. Such results establish a benefit under the tested population regime, not an unconditional advantage of sex in every environment. genetic recombination can disrupt favorable epistasis, and the costs of mating are distinct from those of crossover formation.

For an explanation of the evolution of the machinery itself, the association with its bearer matters. A recombination modifier must on average increase the representation of its own descendants, perhaps through improved segregation or through association with combinations that selection favors. Merely increasing long-run adaptability of the species is insufficient. Similarly, a mutator allele can spread in an asexual lineage by hitchhiking with a beneficial mutation it helped generate despite producing many harmful ones; exchange can weaken that association. The advantages of fidelity and of exchange consequently depend on genetic linkage, population size, environmental change and the current supply of beneficial variants.

DNA repair can regulate which exchanges are permitted as well. DNA mismatch repair corrects copying errors and can reject exchange between sufficiently diverged sequences. In bacteria, disabling mismatch-repair functions greatly relaxed a barrier to genetic recombination between divergent lineages.^[ref] An experiment separating these roles should compare DNA repair of copying errors with exchange between templates of controlled divergence, using a restored DNA repair gene as a rescue. DNA repair and genetic recombination can therefore be parts of the same evolved solution: preserve functional sequence, transmit chromosomes reliably, and reshuffle existing alleles when doing so improves reproductive success.

The recombination fraction is the probability that a transmitted gamete is recombinant between the two loci in the ordinary two-locus genetic linkage model. Its conventional range isZero corresponds to complete linkage and $1/2$ to independent assortment. Multiple physical crossovers can restore parental phase, so the recombination fraction is not the expected number of crossovers or an unrestricted crossover probability.

Genetic linkage analysis asks whether a genetic marker and a disease genetic locus cosegregate within families more often than would be expected under independent assortment. It uses pedigrees and Mendelian segregation to estimate the recombination fraction $\theta$, with evidence for linkage arising when the data favor $\theta <1/2$. Even when founders are in linkage equilibrium, linked loci can show familial cosegregation.

Genetic association analysis instead asks whether observed allele or genotype frequencies differ with disease status in a sampled population, often comparing cases with controls. It can detect a causal variant or a correlated genetic marker through linkage disequilibrium; population structure and selection can also create association. Linkage is about familial transmission; association is about population dependence. Neither an association alone nor a low family recombination estimate establishes that the marker itself causes disease.

Linkage disequilibrium means that alleles at different genetic loci are associated in population haplotypes. For two biallelic loci,The first quantity measures departure from independent allele frequencies; the second is the squared allelic correlation. Standardized linkage disequilibrium $D^{\prime }$ instead compares $D$ to its possible range at the given frequencies. High $|D^{\prime }|$ can coexist with low $r^2$, so it need not make one genetic marker a good predictor of another. Genetic linkage is a different notion: it describes within-family transmission and the recombination fraction.

The chromosome-21 study of Patil and colleagues, Science 2001 found consecutive single-nucleotide polymorphisms organized into regions with comparatively few common haplotypes; typically three haplotypes accounted for over 80% of the sampled chromosomes within a block. Such haplotype blocks can be represented efficiently by tag SNPs. Strong association within blocks and weaker association across boundaries give a useful description, though block boundaries depend on allele frequencies, population history, sampling and the operational definition of a block. They do not divide chromosomes into regions where genetic recombination is respectively impossible and certain.

The ancestry explanation uses an ancestral recombination graph. Going backwards, ancestral lineages coalesce, while a genetic recombination event splits one lineage's ancestral material into contributions from two parents. At a fixed genomic position, retaining only its ancestry gives a phylogenetic tree with the appropriate coalescent law. Nearby positions often retain much of the same tree; mutations on shared branches then produce correlated allele states. Breakpoints can change the local tree, weakening that shared history.

Under a simple random-mating two-locus model, linkage-disequilibrium decay under recombination gives $D_{t+1}=(1-r)D_t$. Actual populations also undergo genetic drift, population bottlenecks, mutation and migration, which can create or preserve association. Recombination rates themselves vary along chromosomes: McVean and colleagues, Science 2004 inferred strong local variation from sequence data. These observations motivate an ancestry model with spatially varying recombination, rather than treating physical distance as the only determinant of linkage disequilibrium.

For fine mapping, a typed genetic marker can show genetic association because it is correlated with a nearby causal genetic variant. Linkage-disequilibrium marker tagging makes a sparse first-stage scan possible, but several variants in one block can produce nearly indistinguishable association patterns. Historical recombination and diverse local genealogies can narrow the interval when denser genotyping is available.

An ancestral recombination graph turns that intuition into a likelihood calculation. One can place a proposed disease-origin event on a local ancestral branch and integrate over possible graphs, mutation histories and unobserved haplotypes. The local phylogenetic tree then relates disease-bearing chromosomes and predicts which markers should remain associated. In a Bayesian inference, uncertainty in ancestry, mutation location and demographic parameters enters the posterior distribution rather than being replaced by one guessed tree. Whole-graph inference is computationally demanding, and approximate or pairwise methods often supply practical estimates.

The useful distinction is between finding an associated region and identifying its causal variant. Strong linkage disequilibrium improves marker coverage but can limit resolution inside the associated region. Population stratification, ascertainment, rare variants and misspecified population history can also mislead an association analysis. Combining fine mapping with independent biological evidence is therefore stronger than interpreting an association peak as proof of causation.

For the Mendelian identity by descent count $J$, the expected shared proportion isThe corresponding expected number of shared copies is one. These are different scales for the same random variable.

With genetic linkage to a disease locus, selecting two affected full siblings preferentially selects transmissions of disease-associated ancestral copies. A nearby genetic marker may then show excess identity by descent: the expected shared proportion can exceed $1/2$, which is the basis of affected-sib-pair analysis. The precise distortion depends on penetrance, inheritance and the recombination fraction; disease status alone does not prescribe a universal alternative distribution. In the fully penetrant recessive model derived below, setting $s=2\theta (1-\theta )$ givesThus the expected shared proportion rises from $1/2$ at $\theta =1/2$ to one at complete genetic linkage, $\theta =0$.

Parametric linkage analysis specifies a disease-transmission model: the mode of inheritance, disease-allele frequency and penetrance of each genotype. Combining these with Mendelian segregation and the recombination fraction gives a pedigree likelihood. One can estimate the recombination fraction or compare a linked model, $\theta <1/2$, with the unlinked model, $\theta =1/2$, using a likelihood-ratio test or LOD score. Incomplete penetrance and unobserved genotypes are accommodated by summing over the possible underlying inheritance states.

Nonparametric linkage analysis instead tests whether relatives with the same disease phenotype share marker alleles identical by descent more often than expected from their pedigree. For example, affected-sib-pair analysis compares sharing among affected full siblings with its Mendelian null distribution. It avoids specifying a complete disease-penetrance model and disease-allele frequency. “Model-free” still assumes an inheritance model for the marker, correct family relationships and a justified treatment of genetic-study ascertainment; it does not mean assumption-free.

Put $N=n_0+n_1+n_2$ and $s=2x=2\theta (1-\theta )$. The usual recombination fraction range $0\le \theta \le 1/2$ gives $0\le s\le 1/2$, and the preceding likelihood function becomesTo see its relation to nonparametric linkage analysis, use Bayes' theorem and the constant disease probability $1/16$ to obtain the affected-pair marker-sharing probabilitiesTheir multinomial likelihood has exactly the same $s$-dependent kernel, since its extra factor $2^{n_1}$ is constant. Set $A=2n_0+n_1$ and $B=2n_2+n_1$. Then $A+B=2N$, and $B$ is the total number of shared ancestral copies. At fixed $N$, the Fisher-Neyman factorization theorem makes $B$ a sufficient statistic. The observed mean shared proportion is $\hat{p}=B/(2N)$, and the binomial likelihood kernel givesThe unlinked null is $s=1/2$, whereas linkage corresponds to $s<1/2$, or expected shared proportion $1-s>1/2$. If $\hat{p}\le 1/2$, the constrained likelihood is maximized at the unlinked null. For $\hat{p}>1/2$, its log likelihood ratio iswhose derivative in $\hat{p}$ is $2N\log [\hat{p}/(1-\hat{p})]>0$. Hence it orders samples exactly by excess IBD sharing. Under this fully penetrant recessive model, parametric linkage testing and a one-sided test of the overall IBD-sharing proportion use the same information. This is allele-sharing sufficiency for recessive linkage, not a general equivalence for all disease models. Use limiting values for terms such as $0\log 0$, and assume $N>0$ for the estimators.

Take the affected sib pairs from independent families with the same disease model. Conditional on their marker sharing states, multiply the recessive affected-sib-pair likelihoods:Write $x=\sqrt{L_0}=\theta (1-\theta )$, so that $\sqrt{L_2}=1/2-x$ and $L_1=x(1/2-x)$. ConsequentlyFor an unordered count sample there can also be a multinomial likelihood coefficient, and the Mendelian probabilities of the observed marker sharing states supply further factors. All these factors are independent of $\theta$.

If genetic-study ascertainment explicitly conditions on both children being affected, the normalizing disease probability is constant:Thus the conditional sharing likelihood differs from the displayed likelihood function only by factors independent of the recombination fraction. Multiple pairs drawn from the same larger sibship would not, in general, justify treating these pair likelihoods as independent.