Biology • Year 12 • Module 5 • Lesson 14b
Dihybrid Crosses and Independent Assortment
Build HSC Band 5–6 extended-response technique on dihybrid crosses: linking the 9:3:3:1 ratio to its meiotic mechanism, evaluating data that breaks the ratio, and analysing a fresh stimulus on linked genes.
1. Extended response — link the 9:3:3:1 ratio to the meiotic mechanism (Band 5–6)
7 marks Band 5–6
Q1. Explain how Mendel's law of independent assortment produces the 9:3:3:1 phenotypic ratio in the F2 of an AaBb × AaBb cross. In your response you must:
- State and define Mendel's law of independent assortment.
- Identify the specific cellular event in meiosis that produces independent assortment.
- Explain how this event generates the four equally frequent gamete types (AB, Ab, aB, ab) from a single AaBb parent.
- Use a 4×4 Punnett square (or equivalent reasoning) to derive the 9:3:3:1 ratio from the gamete frequencies.
- State the condition under which the ratio would NOT hold.
2. Stimulus-based extended response — Bateson & Punnett's sweet pea anomaly (Band 5–6)
8 marks Band 5–6
Stimulus. In 1905, William Bateson and Reginald Punnett crossed sweet pea plants of genotype PpLl × PpLl, where P (purple flower) is dominant to p (red flower) and L (long pollen grain) is dominant to l (round pollen grain). They expected a 9:3:3:1 phenotypic ratio, but their F2 of 6952 plants gave:
| Phenotype | Observed | Expected (9:3:3:1) |
|---|---|---|
| Purple, long (P_L_) | 4831 | 3911 |
| Purple, round (P_ll) | 390 | 1304 |
| Red, long (ppL_) | 393 | 1304 |
| Red, round (ppll) | 1338 | 434 |
Q2. Analyse and evaluate this data set with reference to Mendel's law of independent assortment.
In your answer:
- Describe the discrepancy between observed and expected for each phenotypic class.
- Identify which two classes are overrepresented and explain what this reveals about the chromosome arrangement of the P and L alleles in the parents.
- Explain why this result is incompatible with independent assortment, and what meiotic mechanism would be required for the data to fit 9:3:3:1.
- Reach a justified conclusion about whether the two genes are on the same or different chromosomes.
3. Evaluate this claim (Band 5–6)
6 marks Band 5–6
"Mendel's 9:3:3:1 ratio is a universal law of genetics. Any cross of two heterozygotes for two genes will always produce offspring in this ratio, because independent assortment is a property of every chromosome pair in every species."
Q3. Evaluate this claim. Identify the parts that are correct, the parts that are biologically wrong, and reformulate it into a defensible statement using the lesson's framing of independent assortment as a chromosome-dependent mechanism.
Q1 — Sample Band 6 response (7 marks), annotated
Mendel's law of independent assortment states that the alleles of two different genes segregate independently of each other during gamete formation. [1 — law stated and defined]
The cellular event responsible is the random orientation of bivalents (paired homologous chromosomes) on the metaphase plate during meiosis I. Either homologue of each pair can face either pole, and the alignment of one bivalent has no effect on the alignment of another bivalent on a different chromosome. [1 — correct meiotic mechanism identified]
For an AaBb parent, the A/a homologue pair and the B/b homologue pair (assumed to be on different chromosomes) align independently. This produces four equally likely combinations in the resulting gametes — AB, Ab, aB, ab — each at a frequency of 1/4. [1 — four equal gamete types derived from independent alignment]
When two AaBb parents are crossed, the 4×4 Punnett square contains 16 equally probable cells. Grouping by phenotype: nine cells show A_B_ (both dominant), three cells show A_bb, three cells show aaB_, and one cell shows aabb. [1 — 4×4 grid grouping explicit] The resulting phenotypic ratio is therefore 9 : 3 : 3 : 1. [1 — ratio derived from gamete frequencies]
The 9:3:3:1 ratio is therefore a direct mathematical consequence of independent assortment, given two dominant–recessive genes. [1 — ratio explicitly linked back to mechanism]
The ratio holds only when the two genes lie on different chromosomes. Genes on the same chromosome are linked: their alleles travel together during meiosis, the four gamete types are no longer equally frequent (parental combinations outnumber recombinants), and the F2 deviates significantly from 9:3:3:1. [1 — exception correctly stated]
Q1 — Marking criteria
- 1 mark — States Mendel's law of independent assortment correctly.
- 1 mark — Identifies random orientation of bivalents at the metaphase plate of meiosis I as the cellular mechanism.
- 1 mark — Explains how this produces all four gamete types (AB, Ab, aB, ab) at equal frequency.
- 1 mark — Sets up a 4×4 Punnett square (or equivalent multiplication of 3/4 × 3/4 reasoning) showing the 16 cells.
- 1 mark — Correctly groups the 16 cells into the 9 : 3 : 3 : 1 phenotypic classes.
- 1 mark — Explicitly links the ratio to the gamete-frequency / independent-assortment mechanism (not just stated).
- 1 mark — Identifies that the ratio fails when the two genes are linked (on the same chromosome).
Q2 — Sample Band 6 response (8 marks), annotated
Comparing observed and expected counts, two phenotypic classes are massively overrepresented: purple-long (4831 observed vs 3911 expected, +920) and red-round (1338 observed vs 434 expected, +904 — over three times the expected count). The two intermediate classes — purple-round (390 vs 1304) and red-long (393 vs 1304) — are correspondingly under-represented by roughly a factor of three. [1 — describes discrepancies quantitatively]
The two overrepresented classes (P_L_ and ppll) correspond to the parental allele combinations. This reveals that in the original parents, the dominant P allele and the dominant L allele lay on the same chromosome, while the recessive p and l alleles lay together on the homologue. [1 — identifies overrepresented classes as parental] [1 — infers chromosome arrangement]
For the data to fit a 9:3:3:1 ratio, the four gamete types (PL, Pl, pL, pl) would need to be produced in equal frequency from each AaBb-equivalent parent. This in turn requires the random orientation of bivalents at meiosis I to assort the P/p homologues independently of the L/l homologues — which is only possible if the two genes are on different chromosomes. [1 — links 9:3:3:1 to required meiotic mechanism]
Because the parental gametes (PL and pl) are overrepresented and the recombinant gametes (Pl and pL) are underrepresented, the two alleles must travel together more often than they assort independently. This is incompatible with Mendel's second law as applied to two independently assorting genes. [1 — explicitly rules out independent assortment]
The biological explanation is genetic linkage: the genes for flower colour (P) and pollen-grain shape (L) lie on the same chromosome in sweet peas. Because they travel together during meiosis, the parental combinations are inherited as a unit and recombinant combinations occur only when crossing-over physically separates them. [1 — concludes linkage on same chromosome]
In conclusion, the data is incompatible with independent assortment as predicted by Mendel; the genes are linked on the same chromosome. Bateson and Punnett's anomaly is historically the first demonstration that Mendel's second law is conditional on chromosome location, not universal. [1 — justified conclusion with historical link] [1 — overall coherence + precise lesson terminology used throughout]
Q2 — Marking criteria
- 1 mark — Describes the discrepancy quantitatively for at least two classes (with figures from the table).
- 1 mark — Identifies P_L_ and ppll as the overrepresented (parental) classes.
- 1 mark — Infers correctly that P and L are on one chromosome while p and l are on its homologue.
- 1 mark — States that 9:3:3:1 requires equal-frequency gametes via random bivalent orientation in meiosis I.
- 1 mark — Explicitly states the data is incompatible with independent assortment.
- 1 mark — Concludes the genes are linked on the same chromosome.
- 1 mark — Reaches a justified, integrated conclusion linking observation → mechanism → exception.
- 1 mark — Uses precise lesson terminology throughout (independent assortment, bivalent, parental vs recombinant, linkage).
Q3 — Sample Band 6 response (6 marks)
The claim is partly correct but importantly flawed. [1 — overall judgement]
What is defensible. Independent assortment is a real biological mechanism that operates during meiosis I in every species with paired chromosomes, and a cross of two heterozygotes for two unlinked genes will generate a 9:3:3:1 phenotypic ratio under simple dominance. [1 — concedes correct elements]
What is wrong. First, independent assortment applies only to genes on different chromosomes. Genes on the same chromosome are linked and their alleles travel together, so the four gamete types are not equally frequent and 9:3:3:1 does not hold. [1 — refutes "universal" by invoking linkage] Second, the ratio also assumes simple dominant–recessive inheritance for both genes — incomplete dominance, codominance, lethal alleles or epistatic interactions can change both the genotype-to-phenotype mapping and the observed ratio, even with unlinked genes. [1 — refutes "always" by invoking non-Mendelian inheritance] Third, real data deviates from 9:3:3:1 even when genes are unlinked, because finite samples are subject to chance / sampling variation; Mendel's own 556 pea data are very close to 9:3:3:1 but not exact. [1 — refutes "always" by invoking statistical variation]
Defensible reformulation. "When two heterozygous parents are crossed for two genes that lie on different chromosomes and each shows simple dominant–recessive inheritance, the random orientation of bivalents at meiosis I produces four equally frequent gamete types and an expected F2 phenotypic ratio of 9:3:3:1. The ratio is conditional, not universal: genetic linkage, non-Mendelian inheritance patterns and small sample size can all cause real F2 data to deviate from this prediction." [1 — biologically defensible reformulation invoking chromosome-dependent mechanism]
Q3 — Marking criteria
- 1 mark — States an overall evaluative judgement (e.g. "the claim is partly correct but flawed").
- 1 mark — Correctly identifies the defensible element (9:3:3:1 does arise from independent assortment in the standard case).
- 1 mark — Refutes "universal" by invoking linkage / same-chromosome exception.
- 1 mark — Refutes "always" by invoking non-Mendelian inheritance (incomplete dominance, codominance, epistasis).
- 1 mark — Refutes "always" by invoking sampling variation in finite F2 populations.
- 1 mark — Reformulates the claim into a defensible, conditional statement that uses precise lesson terminology (independent assortment, bivalent, linkage).