Cell division

What the exam asks

Expect a diagram of a dividing cell, a table of DNA content against time, or a described gamete, and a question that is really about counting. The recurring items are: identify the stage from what is separating, state the chromosome and chromatid number at a labelled point, work out which meiotic division a non-disjunction happened in, and predict where a described drug arrests the cycle. The trap that costs most is anaphase I. Candidates see chromosomes moving apart, call it 'chromatids separating', and lose both the count and the difference between the two divisions. Look for whether the moving structures still have two arms each: if they do, homologues are separating and this is meiosis I. The second trap is reading doubled DNA as doubled chromosome number after S phase, which comes from counting arms instead of centromeres.

Two divisions are examined, and the exam is almost never interested in the names of the phases. It is interested in bookkeeping: how many chromosomes, how many chromatids, how much DNA, and what separated to produce that count. Get the counting rule right and most of this topic becomes arithmetic.

The rule is that a chromosome is one centromere. Replication in S phase doubles the DNA and gives every chromosome a second chromatid, but it does not make new chromosomes, because no new centromeres appear. A human cell after S phase has 46 chromosomes and 92 chromatids, not 92 chromosomes. Chromosome number changes only when centromeres are pulled apart, which happens in anaphase of mitosis and in anaphase II, or when whole chromosomes are separated from their partners, which happens in anaphase I.

Meiosis differs from mitosis in one structural act: homologous chromosomes pair in prophase I and are pulled apart in anaphase I. Everything else follows from it. Pairing allows crossing over, so chromatids stop being identical. Pairing allows the two homologues of each pair to face either pole independently, which is independent assortment. And separating homologues rather than chromatids is what halves the count, which is why meiosis I is the reductional division and meiosis II is just a mitosis on a haploid cell.

Checkpoints and non-disjunction are the two applied ends of the topic. A checkpoint is a question the cell asks before committing, and knowing which question each asks lets you predict where any described drug or defect will stall the cycle.

What to hold

• A chromosome is counted by its centromere, so replication doubles the DNA and the chromatid count while leaving the chromosome number unchanged.
• S phase takes a human cell from 46 chromosomes and 46 chromatids to 46 chromosomes and 92 chromatids.
• Mitosis produces two diploid cells genetically identical to the parent; meiosis produces four haploid cells, each genetically distinct.
• Anaphase I separates homologous chromosomes and leaves sister chromatids attached; anaphase II separates sister chromatids.
• Meiosis I is reductional because it halves the chromosome number; meiosis II is equational and is mechanically a mitosis performed on a haploid cell.
• Crossing over occurs in prophase I, between non-sister chromatids of paired homologues, and is what makes sister chromatids no longer identical.
• Independent assortment happens at metaphase I, because each homologous pair orients towards the poles independently of every other pair.
• The G1 checkpoint asks whether conditions and the genome are fit to commit; the G2 checkpoint asks whether replication finished without damage; the spindle checkpoint asks whether every chromosome is properly attached before anaphase.
• A drug that prevents microtubule function arrests cells at metaphase, because the spindle checkpoint never receives the attachment signal it is waiting for.
• Non-disjunction in meiosis I gives a gamete carrying both homologues of a pair; non-disjunction in meiosis II gives a gamete carrying two copies of the same homologue.
• Non-disjunction during mitosis after fertilisation produces a mosaic, an individual with two cell lines rather than one abnormal karyotype throughout.
• Loss of checkpoint control, rather than fast division alone, is what lets a damaged genome be copied and passed on.