Meiosis and gamete formation explained: HSC Biology Module 5

Top marks (6) require correctly identifying the type of fertilisation in each group, an extensive comparison of internal and external fertilisation, AND incorporating the data. Lower bands drop the data use and depth (4 = correct identification + sound understanding + limited data reference).

Sample answer (marking guidelines):

• Group A are terrestrial animals using internal fertilisation; Group B are aquatic animals using external fertilisation.
• In both, fertilisation involves fusion of male and female gametes.
• In Group A this occurs inside the female body; fewer gametes are needed because fertilisation is more probable, so far fewer eggs are produced (e.g. 1-100 in snakes) and they can reproduce on land as internal fertilisation protects the gametes.
• In Group B, fertilisation occurs in water so gametes do not dry out; success is much lower as gametes are spread over a large volume, explaining the huge egg numbers (e.g. 100 000 to 2 million in sea urchins).

Markers penalised confusing external fertilisation with asexual reproduction or parental care.

Source: NESA 2025 HSC Biology examination and marking guidelines.

3 marks for comparing the cell division processes of cells R and S; 2 for describing a cell division process; 1 for some relevant information.

Sample answer (marking guidelines): Cell R undergoes mitosis, which results in two genetically identical daughter cells. Cell S undergoes meiosis, with half the chromosome number, producing gametes. Both mitosis and meiosis require DNA replication, where the genetic content doubles.

Markers cautioned students to compare characteristics of the division process (e.g. number/ploidy of products, genetic identity) rather than just listing outcomes.

Source: NESA 2025 HSC Biology examination and marking guidelines.

2 marks for clearly outlining the significance of crossing over in meiosis; 1 mark for some relevant information.

Sample answer (marking guidelines): Crossing over increases genetic variation, which gives the Jack Jumper ant a better chance to survive environmental change.

A full-mark response links crossing over (exchange of alleles between homologous chromosomes) to increased genetic variation in gametes/offspring AND to a survival/adaptation advantage.

Source: NESA 2024 HSC Biology examination and marking guidelines.

3 marks for explaining the processes in meiosis that lead to genetic variation; 2 for explaining one process; 1 for identifying processes or some relevant information.

Sample answer (marking guidelines): In meiosis, homologous chromosomes line up in Metaphase I in random order and orientation (independent assortment). They separate in Meiosis I, resulting in different combinations of parental chromosomes in the gametes. Crossing over is the exchange of genetic material between the chromatids of homologous chromosomes during Meiosis I, leading to new combinations of alleles on each chromatid.

Markers stressed differentiating independent assortment / random alignment / random segregation from crossing over, rather than blurring them together.

Source: NESA 2020 HSC Biology examination and marking guidelines.

(a) Somatic cell (1 mark): diploid, 2n = 46 chromosomes.

(b) Gamete (1 mark): haploid, n = 23 chromosomes.

One mark is awarded for each correct ploidy-and-number pairing. A bare "46" or "23" with no ploidy term earns the mark only if the number is correct; markers want the term (haploid/diploid) and the number together for full clarity.

(a) Muscle cell = 8 (1 mark)

A muscle cell is a somatic cell produced by mitosis, so it keeps the full diploid number, 2n = 8.

(b) Sperm cell = 4 (1 mark)

Sperm are gametes produced by meiosis, which halves the chromosome number to haploid, n = 4.

(c) Zygote = 8 (1 mark)

Fertilisation fuses a haploid sperm (n = 4) with a haploid egg (n = 4) to restore the diploid number, 2n = 8.

One mark per correct value. The examiner is checking that the student applies "mitosis maintains, meiosis halves, fertilisation restores" rather than memorising one number.

(a) Crossing over - Prophase I (1 mark)

Homologous chromosomes pair (synapsis) and exchange segments at the chiasmata.

(b) Independent assortment - Metaphase I (1 mark)

Homologous pairs line up at the equator in a random orientation.

(c) Halving of chromosome number - Anaphase I (1 mark)

Homologous chromosomes (not sister chromatids) are pulled to opposite poles, so each pole receives one of each pair - the cell becomes haploid.

One mark per correct stage. A common error is naming Anaphase II for the halving; the reduction happens at Anaphase I when homologues separate.

Meiosis halves the number (1 mark)

In the germ-line cells of a 2n = 16 organism, meiosis reduces the diploid number to haploid gametes of n = 8.

Fertilisation restores the number (1 mark)

At fertilisation a haploid sperm (n = 8) fuses with a haploid egg (n = 8) to form a diploid zygote of 2n = 16.

Why halving is necessary (1 mark)

If gametes stayed diploid (2n = 16), fertilisation would double the number to 32, then 64, and so on each generation, which is non-viable.

The alternation keeps it constant (1 mark)

Because meiosis halves and fertilisation doubles in every generation, the species-specific number (2n = 16) is restored each cycle and remains constant.

Full marks require the worked numbers (8 and 16), the two opposing processes, AND the consequence of failing to halve. Stating only "meiosis halves it" without fertilisation restoring it caps the response at 2 marks.

A correct comparison table (1 mark per valid compared feature, to a maximum of 3):

"Compare" requires both similarities and differences be drawn out feature by feature; a table satisfies this if each row addresses the SAME feature for both processes. Listing facts about mitosis in one block and meiosis in another (no shared feature axis) is the classic way students lose marks here.

Crossing over - what it is (1 mark)

During Prophase I, homologous chromosomes pair and exchange segments of DNA at the chiasmata.

Crossing over - effect (1 mark)

This produces new combinations of alleles on a single chromatid (recombinant chromatids) that differ from either parental chromosome.

Independent assortment - what it is (1 mark)

During Metaphase I, each homologous pair lines up and orients at the equator independently of every other pair, so either member of a pair can go to either pole.

Independent assortment - effect and distinction (1 mark)

This randomises which combination of whole parental chromosomes ends up in each gamete; unlike crossing over it does not change the chromosomes themselves, only how they are distributed.

The "explain" command word needs cause and effect for both, and the "distinguish" requirement means the answer must make clear that crossing over rearranges alleles WITHIN chromosomes (Prophase I) while independent assortment rearranges WHOLE chromosomes BETWEEN gametes (Metaphase I).

Set the diploid context (1 mark)

Each parent is diploid (humans 2n = 46, 23 homologous pairs). Each child receives one haploid gamete from each parent, so variation arises in how those gametes are formed and combined.

Crossing over (1 mark)

In Prophase I, homologous chromosomes exchange segments at the chiasmata, producing recombinant chromatids with new allele combinations not present in either parent.

Independent assortment (1 mark)

In Metaphase I, the 23 homologous pairs orient randomly and independently, so each gamete receives a random mix of maternal and paternal chromosomes.

Calculation (1 mark)

Independent assortment alone gives $2^{23}$ (about 8.4 million) chromosome combinations per gamete.

Random fertilisation (1 mark)

Any of the $\sim 2^{23}$ egg types can fuse with any of the $\sim 2^{23}$ sperm types, giving about $2^{46}$ (about 70 trillion) possible zygotes per couple, before even counting crossing over.

Synthesis (1 mark)

Because each sibling is the product of independently formed gametes plus random fertilisation, each inherits a unique chromosome and allele combination, so siblings differ.

This is a Band 5/6 extended response. Markers award full marks only for a logically sequenced answer that names all three sources of variation (crossing over, independent assortment, random fertilisation), ties each to the correct meiotic stage, AND includes a correct quantitative calculation ($2^{23}$ per gamete, $2^{46}$ per couple). Omitting the calculation or naming only one source of variation caps the response in the middle band.

Normal baseline (1 mark)

Normally, Meiosis I separates the two homologues of a pair, and Meiosis II separates the sister chromatids, so all four gametes receive exactly one copy of the chromosome (n).

Non-disjunction in Meiosis I (1 mark)

Both homologues of the pair travel to the SAME pole in Anaphase I. As a result, after Meiosis II, all four gametes are abnormal: two carry two copies (n+1) and two carry none (n-1).

Non-disjunction in Meiosis II (1 mark)

Meiosis I is normal, but in one of the two cells the sister chromatids fail to separate in Anaphase II. This affects only that cell's products: one gamete is n+1, one is n-1, and the other two gametes (from the normal cell) are normal (n).

Key difference (1 mark)

Meiosis I non-disjunction affects all four gametes (homologue separation fails earlier); Meiosis II non-disjunction affects only two of the four (the sister-chromatid failure happens later, after the cells have already separated).

Consequence for the zygote (1 mark)

If an n+1 gamete is fertilised by a normal gamete, the zygote has 2n+1 (trisomy); an n-1 gamete gives 2n-1 (monosomy).

Assess - significance (2 marks)

A correct "assess" weighs the outcome: most aneuploid zygotes are non-viable and miscarry, but some (e.g. trisomy 21) survive with altered phenotypes, showing that an error in a single meiotic division can change the chromosome number of an entire organism. The earlier (Meiosis I) error is "more severe" only in that it affects more gametes, but per-gamete the chromosomal imbalance is the same.

This is a top-band response. Full marks require: correct contrast of which gametes are affected (all four vs two), correct ploidy labels (n+1 / n-1, then 2n+1 / 2n-1), AND an evaluative judgement for the "assess" verb. Merely describing the two errors without assessing consequences caps the answer around 4-5 marks.