Interpret conditional mitotic lineage counts without inventing a growth rate
Invented full-round totals are 3, 6, 12 and 24 cells, requiring 21 completed events. If only three of eight cells divide once, the new total is eleven; the chart is not a time-based growth forecast.
Section 17.2 · Mitosis
Extended only (Supplement); Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked
Before you begin
- Recall chromosomes, genes, chromosome sets, selective gene expression and the distinction between a nucleus and a whole cell.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Three starting cells complete three full no-loss doubling rounds. What is the final count?
What you will learn
- Count completed events and whole-population rounds
- Handle a partially dividing population
Original teaching model. Counts and records are stipulated; drawings are schematic.
| Feature | Meaning |
|---|---|
| Model rounds | 0, 1, 2 and 3; not hours |
| Cell totals | 3, 6, 12 and 24 |
| Completed events | 3 + 6 + 12 = 21 |
| Partial round | 8 cells + 3 net additions = 11 |
Count completed events and whole-population rounds
A mitotic lineage model counts completed cell divisions, not only nuclear snapshots. One parent replaced by two daughter cells adds one cell. If every cell in a population completes one division in each stated round and no cells are lost or added from elsewhere, the population doubles per round. Those assumptions make a calculation possible; mitosis alone does not supply them.
An original record starts with three cells. Whole-population rounds zero, one, two and three contain three, six, twelve and twenty-four cells. The final twenty-four is eight times the initial three, and the net increase is twenty-one. The individual division-event counts are three, six and twelve in the three rounds, also totalling twenty-one events.
Handle a partially dividing population
A partial round gives a different result. If eight cells are present and only three complete division, the other five remain single cells while the three parents become six daughters. The final population is eleven, not sixteen. Equivalently, add one net cell for each of the three completed events: eight plus three. A round label does not imply that every cell divided.
The chart's horizontal axis counts model rounds, not hours, and its vertical axis counts cells. It therefore gives no measured division speed, time to repair or real tissue growth forecast. Cells in organisms differ in division behaviour, and loss or migration changes population totals. Preserve the supplied assumptions and use comparable per-cell chromosome counts when discussing the genetic outcome.
Pause and explain
Eight cells are present; only three complete division. What is the new count with no other changes?
Worked example
Three cells complete three full doubling rounds with no loss. Count the final cells and events.
Show the worked solution
- Use three, six, twelve, twenty-four.
- Sum three plus six plus twelve events.
Answer There are twenty-four cells after twenty-one completed division events, a net increase of twenty-one.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the three-cell lineage after three complete no-loss rounds.
Give me a hint
Double the entire population at each stated round.
Compare my reasoning
- Calculate three to six to twelve to twenty-four.
- Subtract the initial three from the final total.
The final population is twenty-four cells, with a net increase of twenty-one.
Look for these in your work
- Use the stated assumptions
- Separate final total and increase
Independently calculate the individual completed events across those three rounds.
Give me a hint
Count the parents dividing in each round.
Compare my reasoning
- Count three, six and twelve events.
- Add the three event counts.
The rounds contain three, six and twelve completed division events, totalling twenty-one events.
Look for these in your work
- Count events rather than rounds
- Show the addition
Eight cells are present and exactly three divide once; calculate the final population and state why it does not double.
Give me a hint
Five stay single and three become six daughters.
Compare my reasoning
- Count five undivided cells plus six daughters.
- Compare with whole-population doubling.
The final population is eleven cells. It does not double because only three of the eight original cells divide.
Look for these in your work
- Include the undivided cells
- Retain the partial-division condition
Common mistakes
- A round is not an individual division event.
- Only the stated cells divide in a partial-round model.
What can you explain now?
Eight cells are present; only three complete division. What is the new count with no other changes?
Compare with the explanation
Eleven
Each of the three events adds one net cell.
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Make it stick
Without your notes, count completed events and whole-population rounds and handle a partially dividing population. Distinguish nuclear division, chromosome copies and completed daughter cells before opening the solutions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.