Two Anticancer Drugs and Two Very Different Arrests
Three flasks of human cells (46 chromosomes in G1) are grown in identical medium.
- Flask 1, untreated: mitotic index 4%. Dividing cells show 46 chromosomes at metaphase, each composed of two sister chromatids; anaphase and telophase figures are present; cell number doubles about every 24 hours.
- Flask 2, drug X for 24 h: mitotic index 35%. Nearly all mitotic cells are arrested at metaphase with 46 chromosomes, each with two chromatids. No anaphase figures are seen. Total cell number is unchanged from the start of treatment.
- Flask 3, drug Y for 24 h: mitotic index 2%. Cells contain 46 chromosomes and a DNA content of 2C, the G1 value; flow cytometry shows they are stuck before S phase.
The lab must decide which drug to pursue and must explain each arrest before presenting the data.
1. Where does drug X arrest the cycle, and why do the arrested cells show 46 chromosomes each with two chromatids?
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At metaphase. In G1 the cell has 46 chromosomes; during S phase each is replicated into two sister chromatids, so at metaphase the count is still 46 chromosomes but 92 chromatids, joined at their centromeres.
The spindle assembly checkpoint will not release a cell into anaphase until every kinetochore is attached to spindle fibres and the chromosomes are under tension. If drug X prevents microtubules from assembling or binding, that checkpoint stays unsatisfied, so cells accumulate in the metaphase configuration and the sisters never separate. Note what has not happened: the chromosome number has not doubled from 46. The DNA content has, from 2C to 4C. Counting chromosomes alone hides that.
2. Drug Y leaves cells with 46 chromosomes and 2C DNA. Why does that also lower the mitotic index?
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2C DNA with 46 chromosomes means the DNA has not been replicated, so the cells are sitting in G1 and failing to enter S phase. If they never pass the G1/S checkpoint they never reach G2, never duplicate their chromatids, and never appear as mitotic figures at all. The mitotic index falls to 2% not because mitosis is directly blocked but because the supply of cells reaching mitosis has dried up.
That is the contrast worth presenting: drug X acts inside M phase and inflates the mitotic index, while drug Y acts before M phase and deflates it. Reading the mitotic index alone, without the DNA content, would confuse the two mechanisms.
3. Predict the outcome if drug X is washed out after two hours, and if it remains for 48 hours.
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Two hours: the block is reversible. Microtubules repolymerize, kinetochores attach, the checkpoint is satisfied and anaphase proceeds normally, giving two daughters with 46 chromosomes each, like untreated cells.
Forty-eight hours: cells cannot stay arrested indefinitely. Many will undergo apoptosis; others undergo mitotic slippage, exiting mitosis without dividing and becoming a single tetraploid cell with 92 chromosomes, which may then die or continue cycling abnormally. The reason the answer depends on duration is that the checkpoint is a reversible pause, not a permanent stop, and a prolonged arrest is itself damaging. That is also why dosing schedules, not just drug choice, matter in the clinic.