Biology • Year 12 • Module 5 • Lesson 8
Meiosis — Reduction Division and Continuity Across Generations
Apply meiosis to real chromosome-number scenarios, sequence the events, interpret quantitative data on genetic variation, and predict what happens when meiosis goes wrong.
1. Sequence the steps — one diploid cell through meiosis to gametes
The eight events below describe what happens between a diploid germ-line cell entering meiosis and four haploid gametes leaving it. They are listed in a shuffled order. Write the correct order (1–8) in the right-hand column. 8 marks
| # | Event (shuffled) | Order |
|---|---|---|
| 1.1 | Four haploid daughter cells are produced. | |
| 1.2 | Homologous chromosome pairs line up at random (independent assortment). | |
| 1.3 | DNA replication: each chromosome becomes two sister chromatids joined at a centromere. | |
| 1.4 | Sister chromatids separate (meiosis II). | |
| 1.5 | The diploid parent cell enters meiosis. | |
| 1.6 | Homologous chromosomes pair up and exchange segments (crossing over). | |
| 1.7 | Two haploid daughter cells exist briefly between meiosis I and meiosis II. | |
| 1.8 | Homologous chromosomes are separated into different cells (meiosis I — reduction division). |
2. Chromosome arithmetic across four species
The table below gives the diploid chromosome number (2n) for four species. Use it to answer the questions. 8 marks
| Species | Diploid number (2n) | Haploid number (n) — fill in | Zygote number — fill in |
|---|---|---|---|
| Human (Homo sapiens) | 46 | ||
| Fruit fly (Drosophila melanogaster) | 8 | ||
| Domestic dog (Canis lupus familiaris) | 78 | ||
| Garden pea (Pisum sativum) | 14 |
2.1 Fill in the haploid number (n) and the zygote number for each species in the empty columns above. 4 marks (½ each)
2.2 Calculate how many different chromosome combinations can theoretically be produced by independent assortment alone in human gametes. Show your working using 2n where n is the haploid number. (Ignore crossing over.) 2 marks
2.3 Explain why a fruit fly (2n = 8) still produces gametes that vary, even though it has only 4 homologous pairs. Refer to both sources of variation in your answer. 2 marks
3. Interpret data — recombination frequency along a chromosome
The graph below shows the mean number of crossover events per meiosis measured along human chromosome 1 (modelled, simplified after Kong et al., 2010, Nature 467: 1099–1103). Each peak corresponds to a region with a higher frequency of crossing over (a "recombination hotspot"). 7 marks
Figure 3.1. Mean crossover events per meiosis along human chromosome 1. Modelled curve simplified after Kong et al. (2010), Nature 467: 1099–1103.
3.1 Describe the overall shape of the curve in 1–2 sentences. 2 marks
3.2 Estimate the mean crossover frequency at the position of hotspot 2 (≈145 Mb), and compare it with the trough at ≈75 Mb. 2 marks
3.3 A student claims: "Crossing over produces new alleles, which is why these hotspots create such high variation." Use the lesson's framing to correct the student in 2–3 sentences. 3 marks
4. Predict-and-justify — what happens when meiosis fails?
In a small percentage of human meioses, a homologous chromosome pair fails to separate during meiosis I (non-disjunction). One daughter cell receives both homologues of that pair while the other receives neither. 5 marks
4.1 Predict the chromosome number of each of the four gametes produced when this happens to chromosome 21 in a human cell (2n = 46). Show your working. 2 marks
4.2 If one of the gametes carrying an extra chromosome 21 fertilises a normal gamete, predict the zygote's chromosome number and justify your prediction by reference to the role of meiosis as a reduction division. 3 marks
Q1 — Correct order
1.5 (1) → 1.3 (2) → 1.6 (3) → 1.2 (4) → 1.8 (5) → 1.7 (6) → 1.4 (7) → 1.1 (8). One mark per event placed in the correct position; partial credit if the broad block-order is right but two events within a block are swapped.
Q2.1 — Haploid number and zygote number
Human: n = 23, zygote = 46. Fruit fly: n = 4, zygote = 8. Domestic dog: n = 39, zygote = 78. Garden pea: n = 7, zygote = 14. (½ mark per correct cell, 8 cells = 4 marks.)
Q2.2 — Independent-assortment combinations in human gametes
Humans have n = 23 homologous pairs. Independent assortment gives 223 combinations [1] = 8 388 608 ≈ 8.4 million possible gamete chromosome combinations from independent assortment alone [1]. (Accept any working that shows 223.)
Q2.3 — Variation in fruit fly gametes
Independent assortment of 4 homologous pairs gives 24 = 16 different chromosome combinations [1]. Crossing over during meiosis I exchanges segments between homologous chromosomes within each pair, producing additional new combinations of existing alleles — so the actual variation between fruit-fly gametes is much greater than 16 [1].
Q3.1 — Shape of curve (2 marks)
Crossover frequency along chromosome 1 is not uniform: the curve oscillates with three distinct peaks ("hotspots") around 60 Mb, 145 Mb and 220 Mb, separated by lower-frequency troughs [1]. The mean crossover number rises to roughly 3 per meiosis at the hotspots and falls to about 0.5 between them [1].
Q3.2 — Estimate at hotspot 2 vs trough at ≈75 Mb (2 marks)
At hotspot 2 (≈145 Mb), the curve peaks at approximately 3.2 crossovers per meiosis (accept 2.8–3.4) [1]. At the trough near 75 Mb, the curve is approximately 0.5 crossovers per meiosis, so hotspot 2 is roughly 6× higher than the nearby trough [1].
Q3.3 — Correct the student (3 marks)
The student is wrong on the mechanism. Crossing over exchanges corresponding segments between homologous chromosomes and produces new combinations of existing alleles, not new alleles [1]. New alleles arise through mutation, not crossing over [1]. The hotspots do increase variation, but they do so by increasing the rate at which existing alleles are reshuffled into new combinations on chromosomes that then enter different gametes [1].
Q4.1 — Predict gamete chromosome numbers (2 marks)
Normal gametes from a 2n = 46 cell would each carry 23 chromosomes. With non-disjunction of chromosome 21 in meiosis I, the two cells produced after meiosis I have 24 and 22 chromosomes respectively (one has both homologues of chromosome 21, the other has neither) [1]. After meiosis II separates sister chromatids, the four gametes have chromosome counts 24, 24, 22, 22 [1].
Q4.2 — Zygote chromosome number (3 marks)
If one of the n = 24 gametes fuses with a normal n = 23 gamete, the zygote will contain 24 + 23 = 47 chromosomes [1] — three copies of chromosome 21, a condition known as trisomy 21. This is direct evidence that meiosis must work as a reduction division: meiosis I normally separates homologous chromosomes so each gamete carries one homologue of each pair, and fertilisation then restores 2n [1]. When meiosis I fails to halve chromosome 21 properly, fertilisation cannot restore the correct 2n number, and the zygote has the wrong chromosome count from generation 1 [1].