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. Useful relationships: mitotic index = (cells in mitosis ÷ total cells counted) × 100; time in a stage ≈ (cells in that stage ÷ total cells) × length of the cell cycle. A calculator is assumed. Keep full calculator values until the last step. All data sets are fictional.
Question 1 (multiple choice · foundation)
Which event happens during the S phase of the cell cycle?
- (A) The cell grows and makes more mitochondria and ribosomes.
- (B) The DNA of each chromosome is replicated, producing two sister chromatids joined at a centromere.
- (C) Chromosomes condense and the mitotic spindle begins to form.
- (D) The sister chromatids of each chromosome separate and move to opposite poles.
Answer and explanation
Answer: (B). “S” stands for synthesis: the DNA, still spread out as chromatin, is copied. The two copies stay joined at the centromere as sister chromatids until anaphase.
- (A) describes G1, when the cell grows and copies organelles before replicating its DNA.
- (C) describes prophase. Condensing happens after replication, not during it.
- (D) describes anaphase. The chromatids made in S phase stay together until then.
Question 2 (multiple choice · core)
A plant cell has 14 chromosomes in G1. How many DNA molecules (chromatids) does it contain at the end of G2, and how many chromosomes will each daughter cell receive after mitosis and cytokinesis?
- (A) 28 DNA molecules; 14 chromosomes
- (B) 14 DNA molecules; 7 chromosomes
- (C) 28 DNA molecules; 28 chromosomes
- (D) 14 DNA molecules; 14 chromosomes
Answer and explanation
Answer: (A). In S phase each of the 14 chromosomes is copied, giving 14 × 2 = 28 chromatids (DNA molecules), still counted as 14 chromosomes. In anaphase one chromatid of each pair goes to each pole, so each daughter cell gets 14 chromosomes, the same as the parent.
- (B) ignores replication and then halves the number, which is what meiosis does, not mitosis.
- (C) counts each chromatid as a chromosome in the daughter cells. The 28 chromatids are shared between two cells, 14 each.
- (D) forgets that DNA is copied in S phase, so G2 would have the same DNA as G1.
Question 3 (multiple choice · core)
A chemical stops vesicles from fusing with each other at the middle of a dividing plant cell. Mitosis is not affected. Which result is most likely?
- (A) The chromosomes fail to line up at the equator, so mitosis stops at metaphase.
- (B) A cleavage furrow forms instead, so the cell divides normally.
- (C) Two nuclei form, but no cell plate is built, so the cell is left with two nuclei in one cytoplasm.
- (D) The cell skips S phase, so each daughter cell has half the normal DNA.
Answer and explanation
Answer: (C). In plant cells, Golgi vesicles fuse at the middle of the cell to form the cell plate, which becomes the new wall. If the vesicles cannot fuse, cytokinesis fails. Mitosis has already produced two nuclei, so the cell ends up with two nuclei.
- (A) The chemical acts on vesicle fusion, not on the spindle; the question says mitosis is not affected.
- (B) The rigid cell wall prevents a plant cell from pinching in; plant cells cannot switch to a cleavage furrow.
- (D) S phase happens long before cytokinesis, and vesicle fusion has nothing to do with DNA replication.
Question 4 (data · core)
A student counts cells in two regions of a garlic root, using stained squashes.
| Region | Interphase | Prophase | Metaphase | Anaphase | Telophase |
|---|---|---|---|---|---|
| A: 0–2 mm from the tip | 150 | 28 | 12 | 6 | 4 |
| B: 5–7 mm from the tip | 192 | 5 | 2 | 1 | 0 |
(a) Calculate the mitotic index of each region. (b) Calculate the percentage change in mitotic index from region A to region B, and describe the pattern. (c) The cell cycle in region A takes 16 hours. Estimate the time a cell in region A spends in metaphase, in minutes. (d) Suggest a biological explanation for the difference between the regions.
Worked solution and suggested Marlbridge rubric
(a) Region A: cells in mitosis = 28 + 12 + 6 + 4 = 50; total = 200; mitotic index = 50 ÷ 200 × 100 = 25%. Region B: in mitosis = 5 + 2 + 1 + 0 = 8; total = 200; mitotic index = 8 ÷ 200 × 100 = 4.0%.
(b) Percentage change = (4.0 − 25) ÷ 25 × 100 = −84%. The mitotic index is more than six times higher near the tip (25 ÷ 4.0 = 6.25): cell division is concentrated in the first 2 mm.
(c) Fraction in metaphase = 12 ÷ 200 = 0.060. Time = 0.060 × 16 h = 0.96 h = 0.96 × 60 = 58 minutes (57.6).
(d) Cells near the tip are actively dividing, adding new cells for root growth. Farther back, most cells have stopped dividing: they have left the cycle (G0) and are growing longer and specialising.
| Point | What earns it |
|---|---|
| 1 | Both mitotic indices correct (25% and 4.0%), with interphase cells included in the total |
| 1 | −84% (accept “an 84% decrease”) with a description of the pattern |
| 1 | About 58 minutes, using 12 ÷ 200 of the 16-hour cycle |
| 1 | Explanation linking region A to active division for growth and region B to cells that have stopped dividing (G0) |
Common error: using only the dividing cells as the total in (c) (12 ÷ 50) gives about 3.8 h, which is wrong because the fraction must be of the whole cycle.
Question 5 (constructed response · core)
The table shows the mass of DNA in one cell, measured every 2 hours. Mitosis and cytokinesis were completed between 16 h and 20 h; the last value is for one of the two daughter cells. The cell has 10 chromosomes in G1.
| Time / h | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 20 |
|---|---|---|---|---|---|---|---|---|---|---|
| DNA per cell / pg | 6.0 | 6.0 | 6.0 | 7.5 | 9.0 | 10.5 | 12.0 | 12.0 | 12.0 | 6.0 |
(a) Identify the time interval for S phase and for G2, using the data. (b) Calculate the rate of DNA synthesis during S phase. (c) State the number of chromosomes and the number of chromatids in the cell at 14 h. (d) Explain why the daughter cell at 20 h contains 6.0 pg of DNA.
Model answer and suggested Marlbridge rubric
(a) S phase: 4 h to 12 h, when the DNA mass rises steadily from 6.0 to 12.0 pg. G2: 12 h to 16 h (or until mitosis starts), when the mass stays at 12.0 pg.
(b) Rate = (12.0 − 6.0) pg ÷ (12 − 4) h = 6.0 ÷ 8 = 0.75 pg h⁻¹. Check: the rise is 1.5 pg every 2 h, and 1.5 ÷ 2 = 0.75.
(c) 10 chromosomes, 20 chromatids. Replication has doubled the DNA, but each chromosome is now two sister chromatids joined at one centromere.
(d) The 12.0 pg of replicated DNA was shared equally in mitosis: one chromatid of every chromosome went to each pole. After cytokinesis each daughter cell has one complete set, 12.0 ÷ 2 = 6.0 pg, which is the same as the parent cell had in G1.
| Point | What earns it |
|---|---|
| 1 | S phase 4–12 h and G2 12–16 h, each justified from the data |
| 1 | 0.75 pg h⁻¹ with units |
| 1 | 10 chromosomes and 20 chromatids |
| 1 | Equal sharing of replicated DNA in mitosis and division of the cell at cytokinesis |
Do not award (d) for “the DNA is broken down” or “half the DNA is lost”: the copies are divided between two cells.
Question 6 (constructed response · stretch)
A gardener grows many new African violet plants from leaf cuttings taken from one parent plant. New plants grow from the cut leaf by mitosis.
(a) Explain why the new plants are genetically identical to the parent. (b) In one dividing cell of a different organism with 8 chromosomes, the two sister chromatids of one chromosome fail to separate in anaphase and both move to the same pole. Predict the number of chromosomes in each daughter cell, and explain your answer. (c) Explain why the separation of sister chromatids in anaphase is essential for passing a complete genome to the next generation of cells.
Model answer and suggested Marlbridge rubric
(a) In S phase every chromosome is copied to make two identical sister chromatids. In mitosis, one chromatid of each chromosome goes to each pole, so each new nucleus receives a complete copy of every chromosome. With no change to the DNA (apart from rare copying errors), the cells of the new plant carry the same genes as the parent.
(b) One daughter cell receives 9 chromosomes and the other 7. The 7 other chromosomes separate normally, so each cell gets 7 from them. The faulty pair stays together, so one cell gets both chromatids (7 + 2 = 9) and the other gets none (7 + 0 = 7). The total is still 16, but it is shared unequally.
(c) Each sister chromatid is a complete copy of the chromosome’s DNA. If, and only if, the two copies are pulled to opposite poles, both daughter cells receive one copy of every chromosome. A cell that is missing a chromosome lacks the genes on it; a cell with an extra one has extra copies of those genes. Both can disrupt how the cell works.
| Point | What earns it |
|---|---|
| 1 | Replication makes identical sister chromatids and one of each goes to each daughter cell |
| 1 | 9 and 7 chromosomes |
| 1 | Working or reasoning showing why: one cell gains the extra chromatid, the other lacks it |
| 1 | Separation gives each cell one copy of every chromosome; a missing or extra chromosome means missing or extra genes |
Question 7 (calculation · stretch)
A fictional tissue culture starts with 1000 cells. In this culture, 30% of the cells are in G0 and do not divide during the experiment. The other cells complete one full cell cycle every 22 hours. Assume no cells die and no G0 cells re-enter the cycle.
(a) Calculate the number of cells after 66 hours. (b) Calculate how many cells there would have been if every cell had kept dividing, and state the percentage of the final culture that is G0 cells in part (a). (c) After 66 hours, a signal is added that causes the G0 cells to re-enter the cycle. Explain what this tells you about G0, and give one example of when this happens in the body.
Worked solution and suggested Marlbridge rubric
(a) 66 h ÷ 22 h = 3 cycles. G0 cells: 0.30 × 1000 = 300, which stay at 300. Dividing cells: 1000 − 300 = 700; after 3 cycles 700 × 2³ = 700 × 8 = 5600. Total = 300 + 5600 = 5900 cells.
(b) If all divided: 1000 × 8 = 8000 cells. Percentage of the final culture in G0 (part a) = 300 ÷ 5900 × 100 = 5.1% (from 30% at the start).
(c) G0 is a non-dividing state, not a dead end: cells in G0 are alive and can return to the cycle when they receive appropriate cues. Example: after part of the liver is damaged or removed, many liver cells leave G0 and divide until the tissue is replaced. (Accept another valid example, such as cells near a wound beginning to divide during repair.)
| Point | What earns it |
|---|---|
| 1 | 3 cycles and 5600 dividing-cell descendants |
| 1 | 5900 total, with the 300 G0 cells added back |
| 1 | 8000 and about 5.1% |
| 1 | G0 cells are alive and can re-enter in response to cues, with a valid example |
Common error: multiplying all 1000 cells by 8 (8000), which ignores the G0 cells.
How did you do?
- Q1 or Q3 wrong: re-read “Interphase: G1, S and G2” and “Cytokinesis: two new cells” in the study guide, with Figure 1.
- Q2, Q5 or Q6 counts wrong: rework Worked example 2 and look again at Figure 2: count centromeres for chromosomes and chromatids for DNA molecules.
- Q4 calculations wrong: rework Worked example 1, and check that interphase cells are in your total.
- Q7 wrong: re-read “G0: leaving the cycle”, then redo the calculation with G0 cells kept separate.
Then tick off the topic checklist.