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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.

A genetic marker is an observed inherited genomic state used to track chromosomes, test genetic association, or perform genetic linkage analysis. Genetic marker alleles need not themselves cause the phenotype. Several genetic markers on one nonrecombining region form a single observed haplotype.

Ascertainment in a genetic study specifies how individuals or families enter a genetic study. If $A$ is the event of inclusion, then $P(X\mid A)=P(X)P(A\mid X)/P(A)$. Selection on disease phenotype can change an unconditional pedigree likelihood. When comparing genetic marker probabilities conditional on a specified disease pattern, selection depending only on that pattern cancels from the linkage likelihood ratio.

A genotyping error is an incorrect observed allele or genotype call. A probabilistic observation model replaces the deterministic genetic marker factor by $P(X=x\mid G=g)$, allowing nonzero probability of an observed call inconsistent with the latent genotype. Such error can bias family transmission tests and linkage likelihoods.

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]

An observed genetic marker tags an unobserved causal genetic variant when linkage disequilibrium correlates their allele indicators. Their squared allelic correlation $r^2$ measures prediction strength. Under standard local-alternative score-test approximations, using the marker rather than the causal allele reduces the noncentrality parameter by roughly $r^2$, suggesting a sample-size multiplier about $1/r^2$. This is an approximation requiring the same disease model and sampling design, not a universal power identity.

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.

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