= Mitotic daughter chromosome accounting
{title2=$2n=4$}
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.
|| Stage and compartment
|| <Chromosomes>
|| Double-stranded <DNA> <molecules>
|| Nuclear <DNA> amount
| One <G1 phase> parent
| 4 unreplicated
| 4
| $2C$
| One <prophase> or <metaphase> <biological cell>
| 4 replicated
| 8 in 4 pairs of <sister chromatids>
| $4C$
| Whole undivided <anaphase> <biological cell>
| 8 separated daughter <chromosomes>
| 8
| $4C$ total
| One <anaphase> pole's allocation
| 4 daughter <chromosomes>
| 4
| $2C$ allocated
| Each daughter after <mitosis> and <cytokinesis>
| 4 unreplicated
| 4
| $2C$ each
\Image[/mitosis-daughter-chromosome-dna-accounting.png]
{title=Four replicated <chromosomes>, eight <DNA> <molecules>, then four of each per daughter}
{description=Original accounting diagram for $2n=4$, counting nuclear <DNA> only. Each colored stroke represents one double-stranded <DNA> <molecule>; the small bridge in a replicated pair indicates association, not a new <DNA> <molecule>. The eight separate strokes in the whole <anaphase> <biological cell> are already partitioning into two groups. The final row states counts per daughter, not their sum.}
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.
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