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Let $n$ count chromosomes in one complete set, and let $C$ denote the nuclear DNA amount in one unreplicated haploid set of this organism. Ploidy counts sets, whereas $C$ units count material; $C$ is not a universal mass shared by all species. Here we count a paired replicated chromosome as one until its sister chromatids separate. A copied diploid genome is therefore still diploid through S phase: $2n$ persists even as $2C$ becomes $4C$.

An original bookkeeping example uses a hypothetical diploid biological cell with $n=3$. Name its three homologous chromosome pairs $A,a$, $B,b$ and $D,d$. Assume ordinary nuclear DNA replication, no chromosome loss, and one continuous double-stranded DNA molecule per unreplicated chromosome or replicated chromatid. Count nuclear material only, excluding mitochondrion DNA.

For example, $A$ becomes the paired copies $A_1,A_2$ during S phase. They remain one replicated chromosome until separation. Each completed daughter receives one $A$ copy and one $a$ copy, and similarly for the other two pairs. Across both daughters, $2\times 6=12$ double-stranded DNA molecules and $2(2C)=4C$: division allocates the copied material without another round of DNA replication.

The whole undivided anaphase biological cell transiently contains $4n$ separately counted daughter chromosomes. This counting change does not establish a four-set ploidy in a completed daughter genome: each pole receives $2n$, and each completed daughter cell nucleus is diploid. Always state whether a number refers to one cell nucleus, one pole, or the entire undivided biological cell. Compare meiotic DNA content: after meiosis I, a haploid cell nucleus can contain $2C$, so amount alone cannot determine ploidy.

A chromosomal crossover is a reciprocal exchange of segments between nonsister chromatids of homologous chromosomes during meiosis. It can change the haplotypes inherited at selected genetic loci; whether it does so depends on the transmitted chromatid and interval.

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.

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).

Ploidy counts sets of chromosomes; DNA amount counts material. Let $n$ be the number of chromosomes in one complete set and $C$ the DNA amount of one unreplicated set. For the ordinary diploid model, read each row per completed cell nucleus, not as a total across all products:

In the illustrated $n=2$ example, the starting biological cell has four chromosomes. DNA replication gives eight chromatids, organized into four replicated chromosomes. Each product of meiosis I has two replicated chromosomes, hence four chromatids; each final product has two unreplicated chromosomes. Across completed products, $2(2C)=4C$ after meiosis I and $4(C)=4C$ after meiosis II: division redistributes the replicated DNA. A haploid cell nucleus can therefore have $2C$ without becoming diploid.

The recombination fraction is the probability that a transmitted gamete carries a recombinant haplotype between two specified genetic loci, relative to the parent's known haplotype phase. Write it as $r$ here; the symbol $\theta$ is also common in genetic linkage analysis. In the ordinary symmetric two-locus model, $0\le r\le 1/2$, with $r=0$ for complete genetic linkage and $r=1/2$ for independent assortment.

For phase-known genotype $AB/ab$, fair Mendelian segregation and equal reciprocal gamete probabilities giveFor example, 1,000 independent informative transmissions with observed counts $400,400,100,100$ in this order have empirical recombination fraction $(100+100)/1000=0.2$. Those illustrative counts match the expected values in the toy model, but actual counts fluctuate. The genetic linkage diagram shows both the parental and recombinant classes, so their separate probabilities are not confused with the total $r$.

The recombination fraction measures the probability of an odd number of origin switches along the transmitted chromatid. It is neither the expected value of the exchange count nor the probability of at least one exchange: even positive counts can be invisible at the two endpoints. Converting an observed recombination fraction to a genetic map distance therefore needs an explicit model such as the Haldane mapping function. The distinction between transmitted-chromatid counts and four-chromatid exchange counts is discussed in Weeks, Tang and Kwon (2009), Casares' map function: No need for a 'corrected' Haldane's map function, Genetica 135, 305–307.

Sister chromatids are the paired chromatids originating from DNA replication of the same chromosome.