Level 5 — MasteryCell Division

Cell Division

⏱ 75 minutes60 marksprintable — key stays hidden on paper

Level 5 Mastery Examination: Quantitative & Analytical Cell Division

Time limit: 75 minutes Total marks: 60 Instructions: Answer ALL questions. Show full reasoning, derivations, and calculations. Where code is requested, use clear pseudocode or Python. Use $...$ notation for any mathematical expressions.

Question 1 — Combinatorics of Genetic Variation (22 marks)

Meiosis generates variation through independent assortment, crossing over, and (in fertilisation) random fusion.

(a) For an organism with diploid number $2n$ , derive from first principles the formula for the number of genetically distinct gametes produced by independent assortment alone. Explain what "independent" means in the context of homologous chromosome alignment at metaphase I. (5)

(b) Humans have $2n = 46$ . Compute the number of distinct gametes from independent assortment alone, and the number of distinct zygote combinations from random fertilisation of two such individuals (ignoring crossing over). Express both as powers of 2 and give the fertilisation value in scientific notation to 3 significant figures. (5)

(c) Crossing over adds variation. Suppose each of the $n=23$ chromosome pairs undergoes on average exactly one crossover, and each crossover recombines the chromosome into one of 2 distinguishable recombinant forms (in addition to the 2 parental forms, giving 4 distinguishable arrangements per pair). Derive a modified expression for distinct gamete types and evaluate it as a power of 2. Compare its order of magnitude to the answer in (b). (6)

(d) Write a short function distinct_gametes(n, forms_per_pair) that returns the number of distinct gamete types, then state the two calls that reproduce your answers in (b) and (c). Explain why the model is a lower bound on true variation. (6)

Question 2 — Cell Cycle Kinetics & Checkpoint Control (20 marks)

A population of $10^4$ cultured cells is cycling asynchronously. The cycle lengths are: $G_1 = 11$ h, $S = 8$ h, $G_2 = 4$ h, $M = 1$ h.

(a) Under the standard assumption that the fraction of cells in a phase is proportional to that phase's duration in an exponentially growing steady-state population, the exact fraction in phase of length $t_i$ within total cycle $T$ is $f_i = \frac{\ln 2 \cdot 2^{-\,t_{\text{start},i}/T}\left(1 - 2^{-t_i/T}\right)}{\ln 2 \cdot (1 - 2^{-1})}\cdot\frac{1}{1}$ For a simpler examiner-accepted linear approximation, $f_i \approx t_i/T$ . Using the linear model, compute the number of cells expected in each of $G_1$ , $S$ , $G_2$ , $M$ . (6)

(b) A drug blocks the $G_2/M$ checkpoint, arresting all cells at that transition. Explain molecularly, in terms of cyclins and CDKs, what this checkpoint normally verifies and why blocking it prevents entry into mitosis. (6)

(c) After 24 h of drug exposure, describe qualitatively (with reasoning) how the phase distribution changes and what happens to total DNA content per arrested cell relative to a $G_1$ cell. (4)

(d) Relate a loss-of-function mutation in this checkpoint to cancer. Identify the class of gene involved and explain the consequence of unchecked progression despite DNA damage. (4)

Question 3 — Nondisjunction: Modelling & Consequence (18 marks)

(a) Define nondisjunction and distinguish the outcomes when it occurs in meiosis I versus meiosis II for a single chromosome pair. State the resulting chromosome numbers in the four gametes for each case (starting from $2n$ with the pair of interest). (8)

(b) A gamete produced by nondisjunction (n+1 for chromosome 21) fuses with a normal gamete. Compute the total chromosome number of the zygote and name the resulting condition. Classify it as monosomy/trisomy and as aneuploidy/euploidy. (4)

(c) In a large sperm population, suppose the per-meiosis probability of nondisjunction at a given pair is $p = 3\times10^{-3}$ , independent across the $n=23$ pairs. Derive the probability that a random gamete is aneuploid for at least one pair, and evaluate it to 3 significant figures. (6)

Answer keyMark scheme & solutions

Question 1

(a) (5)

At metaphase I each homologous pair aligns independently; either maternal or paternal homolog can face a given pole (1).
Each pair has 2 orientations, choices are independent across pairs (1).
With $n$ pairs, total combinations $= 2 \times 2 \times \dots = 2^n$ (2).
"Independent" = the orientation/segregation of one bivalent does not influence any other (1).
Result: $2^n$ distinct gametes.

(b) (5)

Gametes per individual: $2^{23} = 8\,388\,608$ (2).
Fertilisation combines one gamete from each parent independently: $2^{23}\times2^{23}=2^{46}$ (2).
$2^{46} = 7.037 \times 10^{13}$ (3 s.f.) (1).

(c) (6)

With 4 distinguishable arrangements per pair instead of 2: replace base 2 by 4 (2).
Distinct gametes $= 4^{n} = 4^{23} = 2^{46}$ (2).
$4^{23}=2^{46}\approx 7.04\times10^{13}$ ; this equals the fertilisation magnitude of (b), i.e. crossing over raises single-gamete variation to the same order (~ $10^{13}$ ) as random fertilisation without crossing over (2).

(d) (6)

def distinct_gametes(n, forms_per_pair):
    return forms_per_pair ** n

Call reproducing (b): distinct_gametes(23, 2) → $2^{23}$ (1).
Call reproducing (c): distinct_gametes(23, 4) → $4^{23}=2^{46}$ (1).
Correct function structure (2).
Lower bound because: crossover position is continuous (not just 2 forms), crossover number varies (0,1,2…), gene-level mutation and variable chiasma sites add further diversity — so real variation exceeds the model (2).

Question 2

(a) (6) $T = 11+8+4+1 = 24$ h.

$f_{G_1}=11/24 \Rightarrow 10^4\times11/24 \approx 4583$ cells (2).
$f_S=8/24 \Rightarrow 3333$ cells (1).
$f_{G_2}=4/24 \Rightarrow 1667$ cells (1).
$f_M=1/24 \Rightarrow 417$ cells (1).
Total ≈ 10 000 (rounding) (1).

(b) (6)

$G_2/M$ checkpoint verifies DNA fully and correctly replicated and DNA damage repaired before mitosis (2).
Driven by mitotic cyclin (cyclin B) binding CDK1 to form Maturation-Promoting Factor (MPF) (2).
Blocking checkpoint = MPF kept inactive (CDK1 phosphorylated/inhibited, or cyclin not accumulated), so cells cannot trigger mitotic entry and arrest at $G_2/M$ (2).

(c) (4)

Cells continue until they hit the block, then pile up at $G_2/M$ ; $M$ (short) and later $G_1$ / $S$ deplete as cells accumulate in $G_2$ (2).
Arrested cell has completed S, so DNA content is $2\times$ that of a $G_1$ cell (replicated, $4n$ worth vs $2n$ ) (2).

(d) (4)

Loss of checkpoint control involves tumour-suppressor genes (e.g. p53) (2).
Cells with DNA damage progress into mitosis without repair → accumulation of mutations, genomic instability, uncontrolled proliferation → cancer (2).

Question 3

(a) (8)

Nondisjunction = failure of chromosomes/chromatids to separate properly during anaphase (1).
Meiosis I: homologues fail to separate; both go to same pole. After normal MII → all 4 gametes abnormal: two n+1 and two n−1 (3).
Meiosis II: sister chromatids fail to separate in one secondary cell; the other cell divides normally → gametes: one n+1, one n−1, two normal (n) (3).
Clear statement of counts (1).

(b) (4)

Zygote $= (n+1) + n = 47$ chromosomes (2).
Trisomy 21 = Down syndrome (1).
Classification: trisomy, and aneuploidy (not euploidy) (1).

(c) (6)

Probability a pair segregates normally $= 1-p$ (1).
All 23 pairs normal $=(1-p)^{23}$ (2).
At least one aneuploid $= 1-(1-p)^{23}$ (1).
$=1-(0.997)^{23}=1-0.9331=0.0669$ → 0.0669 (6.69%) to 3 s.f. (2).

[
  {"claim":"2^46 approx 7.04e13 (fertilisation combinations)","code":"val=2**46; result = abs(val - 7.037e13)/7.037e13 < 0.001"},
  {"claim":"4^23 equals 2^46","code":"result = (4**23 == 2**46)"},
  {"claim":"distinct_gametes(23,2)=2^23=8388608","code":"result = (2**23 == 8388608)"},
  {"claim":"At-least-one aneuploid prob = 0.0669","code":"p=Rational(3,1000); val=1-(1-p)**23; result = abs(float(val)-0.0669) < 0.0005"},
  {"claim":"G1 cell count ~4583 of 10000","code":"result = abs(10000*11/24 - 4583.33) < 1"}
]