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An ancestral recombination graph records ancestral coalescences and genetic recombination events for sampled chromosomal material. Backwards in time, each pair of ancestral lineages merges at rate one in standard diploid coalescent units, while a lineage splits at a recombination breakpoint at rate $\rho /2$ times the genetic span of the material it carries. The breakpoint splits that lineage's ancestral material into two parental contributions. Waiting times are exponential with the sum of the current event rates. At each genomic position, retaining the ancestry at that position gives a marginal phylogenetic tree with the neutral Kingman's coalescent distribution. Different positions have dependent trees because they share parts of the graph.

// Target: biology.bigb

Association mapping localizes genetic influences on a phenotype by testing population genetic association with markers. Linkage disequilibrium allows a typed marker to tag a nearby causal variant; historical genetic recombination can narrow the associated interval. Population stratification can produce association without a local causal relation.

// Target: biology.bigb

Evolution is descent with inherited modification. Within a population, changes in allele frequencies provide a precise genetic description; over longer histories, branching descent connects populations through common descent. An individual growing larger is biological development, whereas an inherited variant becoming more frequent across generations is evolution. See UC Museum of Paleontology: the definition of evolution.

Mutation supplies new alleles, genetic recombination reshuffles inherited material, natural selection favors variants with greater reproductive contribution in a specified environment, and genetic drift changes frequencies through chance. Gene flow moves variants between populations. Consequently, an increasing allele frequency alone does not identify natural selection. See Clark, Douglas and Choi, Biology 2e: population genetics.

Evidence includes ordered fossils, shared inherited structures and similarities in DNA; phylogenetics uses such observations to infer relationships. A phylogenetic tree is an inferred historical account, with uncertainty about particular branches. See Clark, Douglas and Choi, Biology 2e: evidence of evolution. Evidence also comes from experiments: Blount, Borland and Lenski reported an Escherichia coli lineage able to use citrate with oxygen present by 31,500 generations, and tested historical dependence by replaying evolution from stored ancestors. See Blount, Borland and Lenski, 2008: original experimental report. Such observations constrain explanations; no single frequency equation captures every evolutionary process.

Under an infinite sites mutation model, two variable sites on one unrecombined rooted tree have nested or disjoint derived-allele clades. At most three of the four biallelic haplotypes can then occur. Observation of all four haplotypes proves that the single-tree, single-mutation-per-site model cannot hold: some ancestral genetic recombination is necessary if recurrent mutation is excluded. The test gives no proof of absence of recombination when only three haplotypes occur.

// Target: biology.bigb

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.

A haplotype is the set of alleles on one chromosome across specified genetic loci. Closely linked markers without intervening genetic recombination are transmitted as one haplotype; they are not independent observations of inheritance.

A haplotype block is a chromosomal region with limited historical genetic recombination, strong local linkage disequilibrium and comparatively few common haplotypes. Its definition is operational and population-dependent; it does not mean that recombination never occurs within it.

// Target: biology.bigb

An identical-by-descent segment is a chromosomal tract inherited from the same ancestral chromosome without an intervening breakpoint in the tracked ancestry. With a specified total number $m$ of independent meioses and a uniform recombination map, the probability that a fixed tract of genetic length $d$ survives without genetic recombination is approximately $e^{-md}$. Recent common ancestry therefore tends to give longer shared tracts. Lengths of tracts selected by covering a random point are length-biased relative to tracts sampled by their boundaries.

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.

Meiosis reduces two sets of chromosomes to one. In the ordinary diploid life cycle, this makes it possible for fertilization of two haploid gametes to restore the diploid state rather than double the number of sets in every generation. Its two divisions perform different tasks: meiosis I separates homologous chromosomes, whereas meiosis II separates sister chromatids. One round of DNA replication precedes both divisions; there is no second round between them. See ^[ref].

Two homologous chromosomes carry corresponding genetic loci, potentially with different alleles; two sister chromatids originate by DNA replication of one chromosome. After chromosomal crossover between nonsister chromatids, the sister chromatids need not carry identical alleles everywhere. Separating the two relationships explains why a haploid product of meiosis I still contains replicated chromosomes.

The two divisions explain Mendelian segregation; variable orientation of different homologous chromosome pairs connects it to independent assortment. Genetic recombination adds another source of allele combinations. These mechanisms reshuffle inherited material without requiring a new mutation.

The symmetric diagram is useful for counting, but human oogenesis produces one functional ovum rather than four. The polar bodies receive discarded chromosomes and little cytoplasm. See Human Reproductive Anatomy and Gametogenesis (§43.3, Oogenesis and Figure 43.13: unequal divisions and polar bodies).

In ordinary diploid meiosis, the two copies at a genetic locus separate: each gamete receives one copy. With fair transmission, a heterozygote $Aa$ produces gametes carrying $A$ or $a$ with probability $1/2$ each. This is a transmission mechanism conditional on a parental genotype, rather than a statement about how common either allele is in a population.

For the one-locus cross $Aa\times Aa$, assume independent parental transmissions, no new mutation, and no differences in fertilization or survival between the resulting genotypes. Multiplying the two gamete probabilities gives four equally likely ordered outcomes: $AA,Aa,aA,aa$, each with probability $1/4$. Combining the two orders of the heterozygote givesOnly under the additional ideal model of complete dominance of $A$, complete penetrance, and no environmental changes to the scored characteristic do $AA$ and $Aa$ share the dominant phenotype. Its probability is then $0.25+0.50=0.75$. The inheritance calculation alone does not determine a phenotype or predict a generic disease risk.

These are probabilities, not a requirement that every four offspring contain one $AA$, two $Aa$ and one $aa$. For 100 independent offspring under the ideal model, the expected values of the three counts are $25,50,25$; the $aa$ count has a binomial distribution with standard deviation $\sqrt{100(0.25)(0.75)}\approx 4.33$. Each new transmission keeps the same conditional probabilities given the parental genotypes.

In his paper published in 1866, Mendel reported 5,474 round and 1,850 wrinkled pea seeds: $5474/7324\approx 0.7474$ round, close to the ideal $0.75$. Following descendants also distinguished two inherited classes within the round phenotype, illustrating why genotype cannot always be read from appearance.^[ref] At several genetic loci, transmission patterns additionally depend on genetic linkage and genetic recombination.