= Meiotic DNA content
{title2=$(2n,2C)\to(2n,4C)\to(n,2C)\to(n,C)$}
<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:
|| Stage
|| <Ploidy>
|| <DNA> amount
|| <Chromosome> organization
| Before <DNA replication>
| $2n$
| $2C$
| Two unreplicated sets
| After <DNA replication>
| $2n$
| $4C$
| Two sets; each <chromosome> has two <sister chromatids>
| After <meiosis I>
| $n$
| $2C$
| One replicated set
| After <meiosis II>
| $n$
| $C$
| One unreplicated set
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>.
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