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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.

Chromosome segregation allocates chromosomes to the products of division. Mitosis separates sister chromatids, preserving ploidy in completed daughter cell nuclei. Meiosis I separates homologous chromosomes with their sister chromatids still associated, reducing diploid to haploid ploidy; meiosis II separates the sister chromatids. Only the first of the two divisions reduces the number of complete sets.^[ref]

For example, correct chromosome segregation of the stipulated three homologous chromosome pairs gives each meiosis I product one $A$, one $B$ and one $D$, each replicated. Giving one product both $A$ members and the other none would preserve the total DNA across products while failing to give each a complete set. Conservation of material alone does not prove correct chromosome segregation.

A haploid cell nucleus has one complete set of chromosomes. For the stipulated $n=3$ example, both products after meiosis I and products after meiosis II are haploid; their Nuclear DNA contents differ because the former retain associated sister chromatids. Haploid therefore does not mean “unreplicated”.^[ref]

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.

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.

Count the same hypothetical diploid animal biological cell in three ways: chromosomes, double-stranded DNA molecules, and nuclear DNA amount. Assume one continuous double-stranded DNA molecule per unreplicated chromosome or replicated chromatid, complete S phase, and no losses or mutations. Exclude mitochondrion DNA. Here $n=2$ is the size of one haploid set, and $C$ is its unreplicated nuclear DNA amount, as defined in cell-cycle DNA content. A double-stranded DNA molecule counts as one, despite containing two strands.

The allocation can be checked copy by copy. DNA replication turns $A$ into $A_1,A_2$, $a$ into $a_1,a_2$, and similarly for $B,b$. One daughter receives $A_1,a_1,B_1,b_1$ and the other $A_2,a_2,B_2,b_2$. Each therefore inherits the same four chromosome identities and the same gene and allele content as the unreplicated parent, under the stated assumptions. Across both daughters there are $4+4=8$ DNA molecules and $2C+2C=4C$: partition conserves the material present after S phase.

The transient whole-cell count of eight does not mean that each daughter's ploidy has doubled. Each completed daughter is diploid, containing both homologues of each homologous chromosome pair. This differs from separating homologous chromosomes during meiosis I. Faithful mitosis preserves a nuclear genome through growth and renewal; equal inherited genes need not imply equal gene expression or equal cell fate. Mutations or chromosome segregation errors can break the ideal inheritance pattern.

Nuclear DNA content is the amount of DNA in a specified cell nucleus. Define $1C$ relative to one unreplicated haploid set of the chosen species. It includes coding and non-coding DNA and excludes DNA outside the cell nucleus, such as a mitochondrion's genome. $C$ is an amount, whereas $n$ is a chromosome count; neither fixes the other across different species.

For an original numerical model, assign the three chromosome types unreplicated DNA amounts of 3, 2 and 1 arbitrary units. Then $C=3+2+1=6$ units: diploid G1 phase contains 12 units, G2 phase 24, each completed meiosis I product 12, and each completed meiosis II product 6. The totals $2\times 12=24$ and $4\times 6=24$ check conservation after copying. Equal chromosome counts alone would not justify equal DNA amounts if the sets had different sequence lengths.

How can an experiment measure the amount independently of chromosome counting? UCL's Cell Cycle Analysis by Propidium Iodide Staining, pp. 1–2, specifies a flow cytometric DNA-content assay. Washed biological cells receive cold 70% ethanol dropwise while vortexing, then at least 30 minutes on ice. After centrifugation and two washes, ribonuclease A removes RNA that would also bind the dye. Propidium iodide stains the preparation. A flow cytometry instrument such as the listed BD LSR II records individual-event fluorescence on a linear scale. Signal area versus height or width excludes doublets and clumps; acquisition is below 400 events per second, retaining at least 10,000 single-cell events. These are protocol settings, not measured results.^[ref]

For a ploidy comparison, add a known diploid G1 phase reference processed with the same staining and detector settings, and an unstained background control. These are proposed comparison controls, beyond the protocol's clump-exclusion procedure. Proportional fluorescence then estimates relative Nuclear DNA content; it does not count chromosomes. A $2C$ event could be a diploid G1 phase product or a replicated haploid product; $4C$ could be replicated diploid material or an unreplicated four-set cell nucleus. Peaks alone therefore cannot establish ploidy or cell cycle stage without a reference and biological context. Uneven staining, residual RNA and unresolved doublets limit this inference; two $2C$ particles can mimic one $4C$ event.

Gregor Mendel's first law is Mendelian segregation: the two alleles at a genetic locus in a diploid individual separate into different gametes, so each gamete receives one. In ordinary meiosis, separation of homologous chromosomes at meiosis I provides the chromosomal mechanism; their sister chromatids separate at meiosis II.

The second law is independent assortment: segregation at one genetic locus is independent of segregation at another when the loci are unlinked. Random orientation of different pairs of homologous chromosomes at meiosis I supplies the mechanism for genes on different chromosomes. Genetic linkage is an essential qualification: nearby loci on one chromosome do not generally assort independently, although sufficiently frequent genetic recombination can make their observed combinations approach independence. Genetic dominance concerns the phenotype of a heterozygote and is not one of these two segregation laws.

Ploidy counts complete sets of chromosomes in a cell nucleus. Write $n$ for the number of chromosomes in one set: haploid means one set, diploid two. DNA replication produces sister chromatids without changing ploidy.^[ref]

Consider an illustrative species with $2n=6$, hence $n=3$. One set contains three distinguishable chromosome types, $A,B,D$; the diploid cell nucleus contains two homologous chromosomes of each type. Labels describe chromosome identity, not particular alleles. Let $C$ be the Nuclear DNA content of one unreplicated set. All entries below refer to one completed product of division, or the starting cell nucleus, rather than the sum over products. Here “chromatid count” includes the single DNA copy in each unreplicated chromosome.

The arithmetic follows the allocation rules. DNA replication replaces each of six single copies with two associated sister chromatids: $6\times 2=12$ copies still grouped as six chromosomes. Meiosis I allocates one homologous chromosome of every type to each product: three replicated chromosomes, with $3\times 2=6$ copies. Meiosis II allocates one copy from each pair of sister chromatids to each final product: three single-copy chromosomes. Compare meiotic DNA content for the general symbolic bookkeeping. Chromosomal crossover can change the alleles on a chromatid without changing these counts when exchanges preserve material.

Keeping ploidy, chromosome count and Nuclear DNA content separate makes the beauty of meiosis explicit: copying supplies material for two divisions, while sorting determines which complete sets each product inherits. It also prevents interpreting a doubled DNA signal as evidence of an extra homologous chromosome set.