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Asexual reproduction produces offspring without fusion of gametes. An operational species concept for such organisms needs evidence other than interbreeding.

Fertilization is fusion of gametes. Fusion of two haploid gametes restores the diploid state in the inheritance model used for Mendelian segregation.

A gamete is a reproductive biological cell that can fuse with another gamete during fertilization. In the ordinary diploid inheritance model, a haploid gamete carries one copy at each genetic locus.

The plant life-cycle generation that produces gametes. In an angiosperm, male gametophytes are pollen grains and the female gametophyte is the embryo sac.

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 Hardy-Weinberg principle describes the genotype proportions produced by independently combining two parental gametes from the same allele pool. For alleles $A,a$ with allele frequencies $p,q$, where $p+q=1$, the Hardy-Weinberg proportions areThe two ordered heterozygote outcomes contribute $pq+qp$. These proportions sum to $(p+q{)}^2=1$, and their implied $A$ allele frequency is $p^2+(2pq)/2=p$. Reshuffling copies therefore preserves their frequencies.

For stable Hardy-Weinberg equilibrium across generations, use a large population with random mating, equal allele frequencies in the parental pools, ordinary Mendelian segregation, and no natural selection, mutation, gene flow or genetic drift. Negligible genetic drift is an idealization of large population size. The one-generation random-mating equilibrium needs independent union of gametes; persistence additionally requires processes that change allele frequencies to be absent.

At $p=0.70,q=0.30$, the predicted genotype proportions are $0.49,0.42,0.09$. For 100 sampled individuals the expected values of the counts are $49,42,9$, rather than the $25,50,25$ of a specified $Aa\times Aa$ cross. The latter fixes both parental genotypes; Hardy-Weinberg proportions average over population mating.

An observed population need not have these proportions. For example, $70AA$, no $Aa$, and $30aa$ among 100 individuals still gives $p=0.70$, but has no heterozygotes. Random mating predicts $0.42$ heterozygosity in the next generation under the stated model. A finite sample can also depart by chance, so a discrepancy is a starting point for investigating inbreeding, population stratification or genotyping error, rather than proof of one cause.

Hardy's 1908 note made the conceptual payoff explicit: a dominant phenotype does not by itself make its allele spread. Complete dominance describes expression; changes in frequency require a separate mechanism such as natural selection. See Hardy, Mendelian Proportions in a Mixed Population, Science 28 (1908), 49–50.

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.

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).