Biology • Year 12 • Module 5 • Lesson 13
Sources of Genetic Variation — Meiosis, Crossing Over, Fertilisation, Mutation
Apply the four sources of variation to a sibling-genotype scenario, to a real allele-frequency dataset, to a diagram critique, and to a "sort the source" classification task.
1. Sort the source — classify each event
For each event in the table, identify which one of the four sources of variation best explains it (crossing over · independent assortment · random fertilisation · mutation) and give a one-sentence justification using lesson terminology. 12 marks (1 source + 2 justification = 3 marks per row × 4 rows; or 1.5 each as marked)
| # | Event | Source | Justification (1 sentence) |
|---|---|---|---|
| 1.1 | Two non-identical siblings inherit a slightly different mix of grandparents' alleles because their parents' gametes happened to fuse in different combinations. | ||
| 1.2 | A single base in a parent's germ-line DNA changes from G to A, producing a new allele that neither parent previously carried. | ||
| 1.3 | During meiosis in a parent, paired homologous chromosomes physically exchange a segment, producing a recombinant chromosome with both maternal and paternal alleles along its length. | ||
| 1.4 | At metaphase I, the orientation of each homologous pair is independent of every other pair, so each gamete receives a different combination of whole maternal and paternal chromosomes. |
2. Interpret graph — sibling genotype identity vs number of independently assorting chromosome pairs
The graph below plots the maximum number of genetically distinct gametes a single parent can produce through independent assortment alone, against the haploid number n (number of chromosome pairs). The formula is 2n. The dashed line marks the human value (n = 23). 6 marks
Independent assortment alone, 2n distinct gamete types. Crossing over and mutation are not included in this count.
2.1 Using the graph, estimate the number of distinct gametes a human can produce by independent assortment alone. Show how you read this off the graph. 2 marks
2.2 Explain why the curve is steeper on a log axis at higher n, in terms of what each extra chromosome pair contributes to gamete variety. 2 marks
2.3 The 8.4 million figure ignores crossing over, random fertilisation and mutation. Briefly state how each of those three would change the actual number of possible offspring genotypes a couple can produce. 2 marks
3. Diagram critique — what's wrong with this student's diagram?
A Year 12 student has drawn the diagram below to explain "where genetic variation comes from". There are three biological errors. Identify each error and write the correction. 6 marks (2 per error: 1 identify, 1 correct)
3.1 Error 1: What is wrong?
Correction:
3.2 Error 2: What is wrong?
Correction:
3.3 Error 3: What is wrong?
Correction:
4. Apply to a new scenario — sibling pairs and a brand-new allele
A clinical geneticist sequences three siblings (S1, S2, S3) born to the same two parents (P1, P2). She finds that:
- For 99.6% of the genome, every variant in each sibling can be traced back to alleles already present in P1 or P2.
- S2 alone carries a single G→A base change at one position in a gene; this exact variant is present in neither parent.
- The three siblings are clearly genetically similar (consistent with shared parents) but easily distinguishable from one another.
6 marks
4.1 Account for the 99.6% of variation among the siblings — which of the four lesson sources is doing the work, and how? 3 marks
4.2 The geneticist concludes that the G→A change in S2 must be a de novo mutation in a parental germ-line cell. Justify this conclusion using the lesson's distinction between "new allele combinations" and "new alleles". 2 marks
4.3 If S2 went on to reproduce, would this G→A variant be transmissible to S2's children? Justify in one sentence. 1 mark
Q1 — Sort the source (12 marks)
1.1 Random fertilisation. Any one gamete from one parent can fuse with any one gamete from the other parent, producing many possible zygote genotypes — this is what differs from sibling to sibling [3].
1.2 Mutation. A change in DNA base sequence creates a genuinely new allele not present in either parent — only mutation can do this [3].
1.3 Crossing over. Exchange of corresponding segments between paired homologous chromosomes during meiosis produces recombinant chromosomes carrying new combinations of existing alleles [3].
1.4 Independent assortment. Random orientation of homologous pairs at metaphase I distributes whole chromosomes into gametes in different combinations [3].
Q2.1 — Estimate from graph (2 marks)
Read up from n = 23 on the x-axis to the curve; the y-value sits just above the 10⁶ gridline and just below 10⁷, at approximately 8.4 × 10⁶ (about 8 million) distinct gametes [1 estimate + 1 method described].
Q2.2 — Why steeper on log axis at higher n (2 marks)
Each additional chromosome pair doubles the number of distinct gametes (2n), because each pair independently contributes a binary choice [1]. On a log axis the curve is in fact a straight line of constant slope (each unit of n adds a factor of 2), but the absolute gain per extra chromosome rises exponentially — going from n = 22 to 23 adds 4 million extra gamete types where going from 1 to 2 adds only 2 [1].
Q2.3 — Effect of the three other sources (2 marks)
Crossing over multiplies the count further by allowing each chromosome itself to come in many recombinant forms, not just maternal-or-paternal [0.5]. Random fertilisation multiplies the parental gamete counts together (≈ 8.4×10⁶ × 8.4×10⁶ for humans, so ~7×10¹³ zygote types per couple) [0.5]. Mutation contributes extra variation by introducing new alleles not in the existing gamete count at all [1].
Q3 — Diagram critique (6 marks)
3.1 Error 1 ("crossing over creates new alleles"): Crossing over reshuffles existing alleles by exchanging chromosome segments — it does not change DNA base sequence. Correction: redraw the arrow from "crossing over" to "new allele combinations", and have the arrow to "new alleles" come from mutation only. [1 + 1]
3.2 Error 2 (random fertilisation drawn before meiosis): Fertilisation is the fusion of gametes, which can only happen after meiosis has produced those gametes. Correction: redraw the flow so meiosis (with crossing over and independent assortment) sits upstream and random fertilisation sits downstream, combining the meiotic products. [1 + 1]
3.3 Error 3 ("mutation only in body cells, never passes to offspring"): Mutation can occur in either body cells (somatic) or germ-line cells (eggs and sperm). Only germ-line mutations are inherited, but they absolutely do occur and are the source of new heritable alleles. Correction: relabel as "mutation in germ-line cells is heritable; mutation in body cells is not" and keep mutation as a valid source of new alleles in the population. [1 + 1]
Q4.1 — 99.6% of sibling variation (3 marks)
Crossing over during meiosis in each parent produced recombinant chromosomes carrying new combinations of P1's and P2's existing alleles [1]. Independent assortment then randomised the maternal/paternal chromosome mix in each gamete [1]. Random fertilisation finally combined a different gamete-pair to form each sibling, so each sibling carries a different reshuffle of the same parental allele pool — explaining the bulk of the differences [1].
Q4.2 — Why the G→A is a de novo mutation (2 marks)
Crossing over, independent assortment and random fertilisation can only produce new combinations of alleles already carried by the parents — they cannot generate a base that neither parent has [1]. The G→A variant is therefore not present in the parental "pool" to be reshuffled, so it must have arisen by a DNA sequence change (mutation) in a germ-line cell of one parent, since only mutation creates genuinely new alleles [1].
Q4.3 — Heritability to S2's children (1 mark)
Yes — because the variant arose in a germ-line cell of a parent and is therefore present in every cell of S2, including S2's own germ-line, it can be passed to S2's children at the normal Mendelian rate [1]. (Accept also: yes, provided the variant is in S2's germ-line cells.)