Interpret potential gamete combinations in a bounded chromosome model
Count possible combinations across a model, without changing the complement per nucleus.
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 and alleles, complete chromosome sets, mitosis and fusion of gamete nuclei.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does n count in this pathway?
What you will learn
- Enumerate combinations from independent stated choices
- Separate possible outcomes from observed proportions
Original teaching model. Counts, source combinations and sample records are stipulated; drawings are schematic.
| Feature | Meaning |
|---|---|
| Two type choices | AA, AB, BA, BB: four possibilities |
| Three independent choices | 2 × 2 × 2 = eight possibilities |
| Complement remains | n = number of chromosome types per complete product |
| Ten-record sample | 4, 2, 1, 3 counts give 40%, 20%, 10%, 30% |
Enumerate combinations from independent stated choices
A simplified model may stipulate one choice from each of two corresponding chromosome pairs. For type one choose A or B; for type two also choose A or B. If the model permits these choices independently, the four possible source combinations are AA, AB, BA and BB. The first position identifies type one and the second type two. Each combination contains two chromosomes, one complete two-type set.
These four possibilities are considered across repeated meiotic events under the model assumptions. They are not a statement that every one event produces all four combinations, and they do not represent every source of biological genetic variation. Exchange between corresponding chromosomes is omitted. The assumptions make the enumeration useful while keeping its interpretation narrower than a complete real-gamete model.
Separate possible outcomes from observed proportions
For three independently chosen pairs, each with two stated source choices, there are two times two times two, or eight, possible source combinations. The eight is a number of combinations across possible events, not the chromosome count in a nucleus. Each normal product in this three-type model still contains three chromosomes. Combinations and complement size use different denominators and describe different properties.
Observed samples need not contain equal numbers of each possible outcome. An invented sample of ten model products records AA four times, AB twice, BA once and BB three times: forty, twenty, ten and thirty per cent. Those sample proportions do not establish future probabilities or prove a biased biological process. Equal theoretical probabilities would require explicit assumptions; a small supplied record alone cannot prove them.
Pause and explain
Three independent pairs with two source choices each give how many potential combinations?
Worked example
Enumerate two independent A/B choices and interpret sample counts 4/2/1/3 out of ten.
Show the worked solution
- Fix the first position as A and list AA and AB.
- Fix it as B and list BA and BB.
- Divide each observed count by ten: 40%, 20%, 10% and 30%.
Answer There are four potential source combinations; the observed proportions need not be equal, and possibility alone does not prove future probability.
Try a topic test
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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.
List all combinations from independently choosing A or B for type one and A or B for type two. State the chromosomes per product.
Give me a hint
Use an ordered first and second position.
Compare my reasoning
- List both second-position choices for first-position A.
- Repeat for first-position B.
- Count one of each of two types in a product.
AA, AB, BA and BB: four possible source combinations. Each product has two chromosomes, one of each type, in this model.
Look for these in your work
- Lists four distinct ordered combinations
- Two chromosomes per complete product
A three-type model independently offers A or B at each type. Calculate potential combinations and distinguish that result from n and 2n.
Give me a hint
Multiply choices; then count chromosome types.
Compare my reasoning
- Calculate 2 × 2 × 2.
- Use three types for n.
- Use two sets for 2n.
Eight possible source combinations, n = three chromosomes per complete product, and 2n = six in the diploid precursor.
Look for these in your work
- Eight combinations
- n three and 2n six
An invented sample of ten products records AA:4, AB:2, BA:1, BB:3. Calculate each sample percentage and assess the claim that each possible combination must appear exactly 25% of the time.
Give me a hint
Use ten as the observed denominator; possibility is not an observed frequency guarantee.
Compare my reasoning
- Divide each count by ten and multiply by one hundred.
- Compare the observed proportions with the claim.
- Keep future probabilities separate from sample data.
AA 40%, AB 20%, BA 10%, BB 30%. The record does not show equal proportions, and a list of possible outcomes does not guarantee exactly 25% in each sample or establish future probabilities.
Look for these in your work
- Four correct percentages
- Rejects guaranteed equal observed frequencies
Common mistakes
- Calling four potential combinations four chromosomes per nucleus
- Assuming every single division gives every potential combination
- Reporting an invented sample as a measured universal probability
What can you explain now?
Three independent pairs with two source choices each give how many potential combinations?
Compare with the explanation
Eight
Two multiplied by two multiplied by two is eight.
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Make it stick
Without your notes, enumerate combinations from independent stated choices and separate possible outcomes from observed proportions. State the chromosome number per nucleus and distinguish chromosome-set reduction from fertilisation.
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.