Cell Division

Question 1 (12 marks)

(a) (6 marks) — G1 = 6 pg, 8 chromatids (8 single-stranded chromosomes).

• (i) End of G2: 12 pg; 16 chromatids. DNA was replicated during S phase, doubling the content; each chromosome now has 2 sister chromatids (8 chromosomes × 2). (2)
• (ii) End of mitotic anaphase (per pole): 6 pg; 8 chromatids. Sister chromatids separated; each pole receives one chromatid of each chromosome = 8 chromatids = 6 pg (same as G1). (2)
• (iii) End of meiosis II (per daughter): 3 pg; 4 chromatids. Two divisions halve chromosome number to n = 4 with each now single-chromatid; DNA = quarter of G2 = 3 pg. (2)

(b) (4 marks)

• Gradual rise = S phase — DNA replicated progressively as many replication origins fire over time. (2)
• Sharp drop = mitosis / cytokinesis (end of M) — the doubled DNA is instantly partitioned between two cells in a single event, so content per cell falls abruptly. (2)

(c) (2 marks) In meiosis I, homologous chromosomes (not sister chromatids) separate; two cells form, each receiving half the chromosome number. DNA per cell halves because chromosomes are distributed, not because DNA is degraded. (2)

Question 2 (11 marks)

(a) (2 marks) — (1) DNA has been correctly and completely replicated; (2) DNA is undamaged (no breaks/errors). (1 each)

(b) (4 marks) — Any two, with mechanism:

• Chromosome breaks/mutations passed on — damaged DNA is not repaired before mitosis, so daughters inherit mutations. (2)
• Aneuploidy / unequal chromosome distribution — incompletely replicated DNA leads to breakage during anaphase and mis-segregation. (2) (Accept: genomic instability, non-viable cells.)

(c) (3 marks) Normally when DNA is damaged, an inhibitory phosphate is added to the CDK (by Wee1-type kinase), keeping the cyclin–CDK complex inactive (1). The active mitosis-promoting cyclin–CDK complex is required to trigger events like nuclear-envelope breakdown/chromosome condensation (1); without its activity the cell cannot enter M phase and is held at G2 until repair is complete (1).

(d) (2 marks) Bypassed checkpoints allow cells with DNA damage/mutations to keep dividing (1); accumulated mutations and uncontrolled proliferation produce a tumour/cancer (1).

Question 3 (10 marks)

(a) (2 marks) n = 3, so number of combinations = $2^n = 2^3 = 8$ gamete types. (working: 1, answer: 1)

(b) (3 marks) Actual variation is greater because of crossing over (recombination at chiasmata) during prophase I (1) and random fertilisation combining any two gametes (1); these create new allele combinations beyond the fixed $2^3$ (1).

(c) (3 marks) Stage = metaphase I (1). Bivalents (pairs of homologues joined) align, not individual chromosomes; there are half as many structures (3 bivalents) because homologues pair up (1). In mitotic metaphase all 6 individual chromosomes align separately — no pairing occurs (1).

(d) (2 marks) Any two: same size/length and same centromere position; carry the same genes in the same order but may carry different alleles (1 each) — one is maternal, one paternal in origin.

Question 4 (9 marks)

(a) (4 marks)

• Cell X: mitosis, anaphase. Single chromatids of every chromosome separate; each pole gets one copy of all four → daughters remain diploid (2n=4). (2)
• Cell Y: meiosis I, anaphase I. Whole chromosomes (each still two chromatids) move to poles; homologous pairs separate — indicates reduction division. (2)

(b) (3 marks) Plant cell: a cell plate / new cell wall forms from vesicles at the centre, growing outward to divide the cell (2). Animal cells lack a cell wall so use a cleavage furrow (constriction of the plasma membrane) instead; the plant's rigid wall prevents furrowing (1).

(c) (2 marks) Daughter cells from cell Y are haploid (n = 2) (1); halving the chromosome number ensures that fertilisation restores the diploid number, maintaining a constant chromosome number across generations (1).

Question 5 (8 marks)

(a) (4 marks)

• (i) Meiosis I: homologous chromosomes fail to separate — both daughter cells of meiosis I are affected, giving 2 gametes with n+1 and 2 with n−1. (2)
• (ii) Meiosis II: sister chromatids fail to separate — only one meiosis-II product pair is affected, giving 1 (n+1), 1 (n−1) and 2 normal (n) gametes. (2)

(b) (2 marks) Zygote has 47 chromosomes (1); condition = Down syndrome (trisomy 21) (1).

(c) (2 marks) Nondisjunction in meiosis I affects the segregation before the second division, so all four resulting gametes are abnormal (2 with extra, 2 missing) (1); in meiosis II only one of the two secondary cells is affected, so only half its products (2 of 4) are abnormal (1).

[
  {"claim":"G2 DNA content is double G1 (6 pg -> 12 pg)","code":"g1=6; g2=g1*2; result = g2==12"},
  {"claim":"Meiosis II daughter DNA is quarter of G2 (12 pg -> 3 pg)","code":"g2=12; m2=g2/4; result = m2==3"},
  {"claim":"Gamete types from independent assortment with n=3 is 8","code":"n=3; combos=2**n; result = combos==8"},
  {"claim":"Trisomy 21 zygote chromosome number is 47","code":"normal=23; abnormal_gamete=24; zygote=normal+abnormal_gamete; result = zygote==47"}
]