Cell Division
Level 3 Paper: Production (From-Scratch Reasoning & Explanation)
Time limit: 45 minutes Total marks: 60
Instructions: Answer all questions. Diagrams should be labelled clearly. Where a calculation is required, show all working. "Explain" means give reasoning, not just state facts.
Question 1 — The cell cycle from scratch (10 marks)
(a) From memory, draw a circular diagram of the cell cycle. Label G1, S, G2 and the four sub-stages of M phase in correct sequence. (4)
(b) Explain one key molecular event that occurs in each of G1, S and G2 that must be completed before the next phase begins. (3)
(c) A cell has 12 picograms (pg) of nuclear DNA at the start of G1. State the DNA content (in pg) at the following points and justify each: end of S phase, end of G2, and in each daughter cell after mitosis. (3)
Question 2 — Checkpoints, cyclins and cancer (12 marks)
(a) Explain how cyclin and CDK concentrations change through the cycle and how they together drive a cell past the G2/M checkpoint. (4)
(b) Name the three main checkpoints and state, for each, the specific condition being monitored. (3)
(c) A mutation makes a cyclin permanently active (it is never degraded). Reason out, step by step, what consequence this has for the cell and how it can lead to cancer. (5)
Question 3 — Mitosis walkthrough (10 marks)
(a) A parent cell has a diploid number of . For each stage of mitosis (Prophase, Metaphase, Anaphase, Telophase), explain in your own words what happens to the chromosomes, and state the number of chromosomes present at each pole during Anaphase. (6)
(b) Explain two ways cytokinesis differs between a plant cell and an animal cell, and explain why each difference exists in terms of cell structure. (4)
Question 4 — Meiosis and genetic variation (12 marks)
(a) Explain, from first principles, the difference between what happens in Anaphase I and Anaphase II of meiosis. (4)
(b) Explain how crossing over produces genetic variation. Include the terms homologous chromosomes, chiasmata and recombination. (4)
(c) An organism has . Calculate the number of genetically different gametes possible from independent assortment alone (ignore crossing over). Show your reasoning and give the general formula. (4)
Question 5 — Compare and reason (10 marks)
(a) Construct a comparison table of mitosis vs meiosis with four distinct points of difference (e.g. number of divisions, daughter cell number, ploidy, genetic identity). (4)
(b) Explain out loud (in writing) why meiosis must reduce the chromosome number by half, using the concept of fertilisation to support your reasoning. (3)
(c) Explain what nondisjunction is and predict the chromosome number of a resulting zygote if a normal gamete ( in humans) fertilises a gamete produced by nondisjunction of one chromosome pair. (3)
Question 6 — Chromosome structure (6 marks)
(a) Draw and label a duplicated chromosome, identifying: sister chromatids and the centromere. (3)
(b) Explain the functional importance of the centromere during mitosis. (3)
END OF PAPER
Answer keyMark scheme & solutions
Question 1 (10)
(a) Circle divided: Interphase (G1 → S → G2) then M phase (Prophase → Metaphase → Anaphase → Telophase). (4: 1 for correct interphase order, 1 for M after G2, 1 for PMAT order, 1 for clear circular layout)
(b) (1 each)
- G1: cell grows, synthesises proteins/organelles; passes restriction/G1 checkpoint.
- S: DNA replication — each chromosome copied to form two sister chromatids.
- G2: further growth, DNA proofread/repaired, proteins for mitosis (e.g. tubulin) made.
(c) (1 each)
- End of S: 24 pg — DNA has been fully replicated (doubled). ✔
- End of G2: 24 pg — no replication in G2, content unchanged. ✔
- Each daughter after mitosis: 12 pg — replicated DNA divided equally between two cells. ✔
Question 2 (12)
(a) (4) Cyclin concentration rises and falls cyclically; CDK levels are constant but inactive alone (1). Cyclin binds CDK to form an active cyclin–CDK complex (1). Rising mitotic (M) cyclin activates M-CDK (1); this phosphorylates target proteins triggering entry into mitosis once G2/M checkpoint conditions are met (1).
(b) (1 each)
- G1 checkpoint: cell size, nutrients, growth signals, DNA damage.
- G2/M checkpoint: DNA fully and correctly replicated, no DNA damage.
- Spindle/M (metaphase) checkpoint: all chromosomes attached to spindle at metaphase plate.
(c) (5) Cyclin never degraded → CDK stays permanently active (1) → checkpoints continually driven past/overridden (1) → cell divides continuously without pausing for repair (1) → damaged/mutated DNA passed on and uncontrolled proliferation occurs (1) → forms a tumour/mass of cells = cancer (1).
Question 3 (10)
(a) (6: ~1 per stage description + 2 for anaphase count reasoning)
- Prophase: chromatin condenses into visible chromosomes (each two chromatids), nuclear envelope breaks down, spindle forms.
- Metaphase: chromosomes align single-file at the metaphase (equatorial) plate; spindle fibres attach to centromeres.
- Anaphase: centromeres split, sister chromatids pulled to opposite poles. Since , 8 chromatids move to each pole → 8 chromosomes per pole. ✔
- Telophase: chromosomes decondense, nuclear envelopes reform, spindle disappears.
(b) (2 per difference)
- Plant: cell plate forms in the middle and grows outward → new cell wall — because plants have a rigid cell wall that cannot pinch in.
- Animal: cleavage furrow — contractile ring of actin pinches membrane inward — possible because the flexible membrane has no cell wall.
Question 4 (12)
(a) (4)
- Anaphase I: homologous chromosomes separate; each chromosome (still 2 chromatids) moves to opposite poles → reduction division, ploidy halves. (2)
- Anaphase II: sister chromatids separate (like mitosis); centromeres split → single chromatids to poles. (2)
(b) (4) Homologous chromosomes pair up (synapsis) in prophase I (1). Non-sister chromatids exchange segments at points called chiasmata (1). This is crossing over, producing recombinant chromosomes (1). New allele combinations not present in either parent → genetic variation (1).
(c) (4) Formula: number of combinations where = number of homologous pairs (1). Here (1). So (1). Reasoning: each of 3 pairs orients independently at metaphase I, giving arrangements (1). Answer: 8. ✔
Question 5 (10)
(a) (1 per correct row, max 4)
| Feature | Mitosis | Meiosis |
|---|---|---|
| Divisions | 1 | 2 |
| Daughter cells | 2 | 4 |
| Ploidy of products | diploid (2n) | haploid (n) |
| Genetic identity | identical to parent | genetically varied |
(b) (3) Fertilisation fuses two gametes (1). If gametes were diploid, the zygote would double chromosome number each generation (1). Meiosis halves the number so that fusion restores the correct diploid number and keeps it constant across generations (1).
(c) (3) Nondisjunction = failure of homologous chromosomes (meiosis I) or sister chromatids (meiosis II) to separate, giving gametes with an abnormal chromosome number (1). Abnormal gamete has (one extra) (1). Fertilised by normal : zygote = chromosomes (trisomy) (1). ✔
Question 6 (6)
(a) (3) Duplicated X-shaped chromosome: two identical sister chromatids joined at the centromere (constricted region). 1 mark shape, 1 chromatids labelled, 1 centromere labelled.
(b) (3) Centromere is where the kinetochore forms (1); spindle fibres attach to the kinetochore (1); allows chromatids to be pulled apart accurately to opposite poles ensuring equal distribution (1).
[
{"claim":"DNA doubles from 12pg to 24pg after S phase","code":"g1=12; after_S=g1*2; result = (after_S==24)"},
{"claim":"Daughter cell DNA after mitosis returns to 12pg","code":"after_G2=24; daughter=after_G2/2; result = (daughter==12)"},
{"claim":"Gamete combinations for 2n=6 is 8","code":"n=6//2; combos=2**n; result = (combos==8)"},
{"claim":"Nondisjunction zygote has 47 chromosomes","code":"normal=23; abnormal=24; zygote=normal+abnormal; result = (zygote==47)"},
{"claim":"Anaphase pole chromosome count for 2n=8 mitosis is 8","code":"twoN=8; per_pole=twoN; result = (per_pole==8)"}
]