These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. No constants are needed. Ignore crossing over in calculations unless a question says otherwise. All organisms marked fictional, and all data sets, are invented.
Question 1 (multiple choice · foundation)
Which statement correctly describes crossing over?
- (A) Non-sister chromatids of a homologous pair exchange matching segments during prophase I.
- (B) Sister chromatids of one chromosome exchange segments during prophase II.
- (C) Chromosomes from two different homologous pairs exchange segments during metaphase I.
- (D) Sister chromatids exchange segments during S phase, as the DNA is copied.
Answer and explanation
Answer: (A). Crossing over needs the homologues to be paired, which happens only in prophase I. The exchange is between a chromatid of one homologue and a chromatid of the other.
- (B) is wrong twice: homologues are no longer paired in prophase II, and sister chromatids are identical, so swapping between them would make no new combination.
- (C) describes an exchange between non-homologous chromosomes. That is not crossing over; non-homologous chromosomes carry different genes, so the segments would not match.
- (D) puts the event in S phase. Crossing over happens after replication, during synapsis in meiosis I.
Question 2 (multiple choice · core)
A fictional plant species has 2n = 14. Ignoring crossing over, how many different combinations of chromosomes can its gametes contain?
- (A) 14
- (B) 49
- (C) 128
- (D) 16 384
Answer and explanation
Answer: (C). 2n = 14, so n = 7 homologous pairs. Each pair assorts independently: 2⁷ = 128.
- (A) uses 2n, the diploid number, as the answer.
- (B) uses n² = 7² = 49 instead of 2ⁿ.
- (D) uses 2²ⁿ = 2¹⁴. That counts the combinations in a zygote from two parents (128 × 128), not in one parent’s gametes.
Question 3 (multiple choice · core)
In a species with 2n = 10, one homologous pair fails to separate during meiosis I. Meiosis II is normal. What are the chromosome numbers of the four gametes?
- (A) 6, 6, 4, 4
- (B) 5, 5, 6, 4
- (C) 10, 10, 0, 0
- (D) 5, 5, 5, 5
Answer and explanation
Answer: (A). After meiosis I, one cell has n + 1 = 6 chromosomes and the other has n − 1 = 4. Each divides normally in meiosis II, giving two gametes of 6 and two of 4.
- (B) is the result of nondisjunction in one cell during meiosis II, not meiosis I.
- (C) is what would happen if all five pairs went to the same pole in meiosis I, not just one pair.
- (D) is normal meiosis, with no nondisjunction.
Question 4 (model · core)
The diagram shows a pair of homologous chromosomes in prophase I. The chromosome from the mother carries alleles F, G and H. The chromosome from the father carries f, g and h. One crossover happens at the chiasma shown.
(a) List the combination of alleles on each of the four chromatids at the end of meiosis II. (b) Identify which chromatids are recombinant. (c) Explain why alleles F and G stay together on every chromatid, but G and H do not. (d) Explain why the two cells produced from one cell in meiosis II are genetically different in this example, even though meiosis II separates sister chromatids.
Worked solution and suggested Marlbridge rubric
(a) F G H, F G h, f g H and f g h.
(b) F G h and f g H are recombinant. F G H and f g h are parental.
(c) The crossover happened between G and H. Everything on one side of the crossover point (F, G and the centromere) stays on its original chromatid; the segment on the other side (H or h) is swapped. F and G are on the same side, so they are never separated by this crossover.
(d) Before crossing over, the two sister chromatids of the maternal chromosome were identical (F G H and F G H). After it, one is F G H and the other is F G h. So when the sister chromatids separate in anaphase II, the two cells receive different alleles of gene H.
| Point | What earns it |
|---|---|
| 1 | All four combinations correct |
| 1 | The two recombinants correctly identified |
| 1 | Crossover point between G and H, so only the segment beyond it is exchanged |
| 1 | Sister chromatids no longer identical after crossing over, so meiosis II gives different cells |
Question 5 (investigation · core)
A fictional freshwater snail can reproduce asexually (offspring are clones of the mother, made by mitosis) or sexually. A biologist notices that in ponds where a parasite is common, a higher proportion of the snails come from sexual reproduction.
(a) Pose a testable question based on this observation. (b) Describe an experiment to answer it. Give the independent variable, the dependent variable and two controlled variables. (c) Predict the result, and justify it using the processes of meiosis and fertilization. (d) The biologist exposes 50 clonal offspring of one mother and 50 sexual offspring of the same mother to each of three parasite strains. The numbers that survive:
| Parasite strain | 1 | 2 | 3 |
|---|---|---|---|
| Clonal offspring surviving (of 50) | 48 | 0 | 1 |
| Sexual offspring surviving (of 50) | 31 | 27 | 22 |
Calculate the percentage survival of each group against each strain, and the mean for each group. Use the data to evaluate the claim: “Sexual reproduction always gives higher survival.”
Model answer and suggested Marlbridge rubric
(a) “Do snails produced by sexual reproduction survive infection by the parasite better than snails produced asexually?”
(b) Independent variable: the way the offspring were produced (sexual or asexual). Dependent variable: the percentage of offspring that survive, or that remain uninfected, after exposure to the parasite. Controlled variables (any two): age and size of the snails, same mother (or mothers of the same population), dose and strain of parasite, temperature and water quality, length of the trial. Use many snails per group.
(c) Sexual offspring should survive better as a group. Crossing over, independent assortment and random fertilization give each sexual offspring a different combination of alleles, so some are likely to carry combinations that resist the parasite. Clonal offspring are genetically identical: if the mother’s genotype is susceptible, they all are.
(d) Clonal: 48/50 = 96.0%, 0/50 = 0.0%, 1/50 = 2.0%; mean = (96.0 + 0.0 + 2.0) ÷ 3 = 32.7%. Sexual: 31/50 = 62.0%, 27/50 = 54.0%, 22/50 = 44.0%; mean = (62.0 + 54.0 + 44.0) ÷ 3 = 53.3%.
The claim is not supported as written. Against strain 1, the clones survived better (96.0% against 62.0%): this clonal genotype happens to resist strain 1. But the clones almost all died with strains 2 and 3, while some sexual offspring survived every strain. The data support a narrower claim: sexual reproduction gives more consistent survival across different parasite strains, because the offspring are genetically varied. Evidence is limited to one mother and three strains.
| Point | What earns it |
|---|---|
| 1 | A testable question that names reproduction mode and a measurable outcome |
| 1 | IV, DV and two controlled variables correct |
| 1 | Prediction justified by variation from meiosis and fertilization and identical clones |
| 1 | All six percentages and both means correct |
| 1 | Claim evaluated using strain 1 as a counter-example and the consistency of the sexual group |
Question 6 (prediction · stretch)
A fictional mammal has 2n = 8. In one parent, the sister chromatids of chromosome 2 fail to separate in one cell during meiosis II. Meiosis I was normal.
(a) Give the chromosome numbers of the four gametes produced from that original cell. (b) What percentage of these gametes are abnormal? How does this compare with nondisjunction in meiosis I? (c) Each gamete fuses with a normal gamete from the other parent. Give the chromosome number of each resulting zygote and name the abnormal ones. (d) An offspring is found with three copies of chromosome 2. The parent’s two copies of chromosome 2 carry different versions of a DNA marker close to the centromere, where crossing over is rare. Explain how the markers in the offspring could show whether the extra chromosome came from an error in meiosis I or meiosis II.
Worked solution and suggested Marlbridge rubric
(a) n = 4. The normal cell from meiosis I divides into two gametes of 4. The faulty cell gives one gamete of 5 and one of 3. Gametes: 4, 4, 5, 3.
(b) 2 of 4 = 50% abnormal. After nondisjunction in meiosis I, all four gametes (100%) are abnormal (5, 5, 3, 3), because the error happens before the cell splits into two.
(c) 4 + 4 = 8 (normal), 8 (normal), 5 + 4 = 9 (2n + 1, trisomy of chromosome 2), 3 + 4 = 7 (2n − 1, monosomy of chromosome 2).
(d) In meiosis I, the two homologues fail to separate, so the faulty gamete carries both different versions of the marker. In meiosis II, two sister chromatids fail to separate, so the gamete carries two identical copies of one version. Compare with the other parent’s marker to find which two copies came from the parent with the error. If those two copies differ, the error was in meiosis I; if they are the same, it was in meiosis II. Choosing a marker near the centromere matters because crossing over farther along the arm could make sister chromatids differ there.
| Point | What earns it |
|---|---|
| 1 | Gametes 4, 4, 5, 3 |
| 1 | 50% compared with 100% for meiosis I, with the reason |
| 1 | Zygotes 8, 8, 9, 7 with trisomy and monosomy correctly named |
| 1 | Meiosis I gives two different homologues, meiosis II gives two identical sister copies |
Question 7 (calculation · stretch)
A fictional fish species has 2n = 10. Ignore crossing over unless told otherwise.
(a) Calculate the number of chromosome combinations possible in one fish’s gametes. (b) Calculate the probability that a gamete contains only chromosomes the fish inherited from its father. Give your answer as a fraction and a percentage. (c) Calculate the number of chromosome combinations possible in the offspring of one pair of fish. (d) A student claims: “Two offspring of the same pair of fish have a 1 in 1024 chance of being genetically identical.” Evaluate this claim.
Worked solution
(a) n = 5, so 2⁵ = 32.
(b) (1/2)⁵ = 1/32 = 0.03125, which is 3.1% to 1 decimal place.
(c) 32 × 32 = 1024.
(d) The claim treats the second offspring as having a 1 in 1024 chance of getting exactly the same combination as the first. That is right only if there is no crossing over. In reality, crossing over in prophase I creates recombinant chromatids, so gametes with the same set of whole chromosomes still differ in their combinations of alleles. The true chance of two (non-twin) offspring being genetically identical is far smaller than 1 in 1024. The claim also assumes every homologous pair differs in its alleles; if some pairs were identical, offspring could match more often for those chromosomes.
Suggested mark points (3): 1 for 32 and 1/32 (3.1%); 1 for 1024; 1 for evaluating the claim with crossing over as the reason the true chance is lower.
Common error: writing 2 × 5 = 10 or 5² = 25 for (a). Each pair doubles the number of combinations, so multiply 2 by itself n times.
How did you do?
- Q1 or Q4 wrong: re-read “Crossing over” and Figure 1 in the study guide.
- Q2 or Q7 calculations wrong: rework Worked example 1; check you used 2ⁿ with n = the number of pairs.
- Q3 or Q6 wrong: rework parts (a) to (c) of Worked example 2 and the nondisjunction table.
- Q5 incomplete: re-read part (d) of Worked example 2 on posing a testable question, then check your answer named a variable to change and one to measure.
Then tick off the topic checklist.