Explain separation of chromosome copies into complete daughter complements
A copied A B C complement separates into A B C in each daughter. Equal totals do not rescue a distribution with a missing or duplicated chromosome identity.
Exam scope and assessment notes
Extended only (Supplement) · Biology 0970, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0970 has the same subject content as IGCSE 0610, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
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
What is the expected normal distribution of copied A B C chromosomes?
What you will learn
- Distribute every chromosome identity
- Recognise a missing or duplicate product
Original teaching model. Counts and records are stipulated; drawings are schematic.
| Feature | Meaning |
|---|---|
| Normal daughter 1 | A, B and C |
| Normal daughter 2 | A, B and C |
| Faulty counterexample | A A B / B C C: missing identities |
| Limit | Three labels alone do not establish ploidy |
Distribute every chromosome identity
During mitosis, the chromosome copies separate so each daughter nucleus receives the complete complement. In the original three-label model, chromosomes A, B and C have each been copied before division. One copy of A, one of B and one of C must go to each daughter. It is not enough for the total across both products to contain six objects if their distribution is incomplete.
The two products A B C and A B C retain the original three-chromosome list. A hypothetical faulty distribution A A B and B C C also totals six objects, but neither product contains the complete original list. That record is a counterexample to normal accurate mitosis, not an alternative normal method of chromosome maintenance.
Recognise a missing or duplicate product
This model specifies three labelled chromosomes without supplying a diploid or haploid set classification. The number three alone cannot establish its ploidy. It simply lets you check identity, completeness and count. The later diploid model supplies the chromosome-set information needed to show maintenance of both sets as well as the individual chromosome number.
Copies that have separated are counted as daughter chromosomes. The completed daughter nuclei provide the comparable endpoint for the parent-to-product count. Within one not-yet-divided cell during separation, the combined chromosome count is a different question. Do not mix that temporary combined quantity with the maintained number in each final daughter nucleus.
Pause and explain
A A B and B C C total six chromosome objects. Does that establish normal distribution?
Worked example
Which daughter lists preserve A B C: A B C / A B C or A A B / B C C?
Show the worked solution
- Compare every identity in each product.
- Do not rely only on the combined total.
Answer A B C in both daughters preserves the complete complement; the other distribution has missing and duplicate identities.
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.
Explain the normal separation of copied A B C chromosomes.
Give me a hint
Give every identity to each daughter.
Compare my reasoning
- Identify two copies of each original chromosome.
- Assign one of each to both products.
One A, one B and one C copy go to each daughter nucleus, giving A B C in both and maintaining the original number three.
Look for these in your work
- Distribute all identities
- State the per-daughter number
Compare correct A B C / A B C with faulty A A B / B C C distributions.
Give me a hint
Check completeness as well as totals.
Compare my reasoning
- Count three in each product.
- Compare each list with A B C.
Both distributions total six objects, but only A B C / A B C gives every identity to each daughter. A A B lacks C; B C C lacks A.
Look for these in your work
- Check both daughters
- Identify the missing entries
A three-chromosome diagram has no chromosome-type or set key. Can its number alone show that it is haploid?
Give me a hint
A set must be defined.
Compare my reasoning
- Identify that only a total is supplied.
- State the absent set or type information.
No. The total three does not establish haploid status without information about a complete set or the chromosome types.
Look for these in your work
- Distinguish number and ploidy
- State the evidence limit
Common mistakes
- A combined total can conceal faulty distribution.
- A count alone does not specify chromosome sets.
What can you explain now?
A A B and B C C total six chromosome objects. Does that establish normal distribution?
Compare with the explanation
No; each daughter has a missing and a duplicated identity
Check every complement separately.
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Make it stick
Without your notes, distribute every chromosome identity and recognise a missing or duplicate product. 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.