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Biology  ·  Year 12  ·  Module 5  ·  Lesson 14b

HSC Exam Practice

Dihybrid Crosses and Independent Assortment

8 questions / 3 sections / 30 marks total
Section 1

Short answer

1.Short answer

1.1

Define independent assortment.

2marks Band 3
1.2

Identify the four gamete types produced by a parent of genotype AaBb, assuming the two genes are on different chromosomes.

2marks Band 3
1.3

Outline the cellular event in meiosis that produces independent assortment.

3marks Band 4
1.4

A cross of two heterozygous pea plants YyRr × YyRr produces 480 F2 seeds. Calculate the expected number of green wrinkled (yyrr) seeds, showing your working.

2marks Band 4
1.5

Distinguish between the 9:3:3:1 phenotypic ratio and the genotypic ratio produced by an AaBb × AaBb cross.

3marks Band 4
Section 2

Data response

2.Data response — dihybrid cross in tomato plants

2.1

A geneticist crossed two heterozygous tomato plants RrTt × RrTt, where R (red fruit) is dominant to r (yellow fruit) and T (tall plant) is dominant to t (dwarf plant). The F2 phenotype counts from 1600 offspring are shown in the table below.

Phenotype Expected (9:3:3:1) Observed
Red, tall (R_T_)900912
Red, dwarf (R_tt)300287
Yellow, tall (rrT_)300305
Yellow, dwarf (rrtt)10096
Table 2.1. Observed vs expected F2 phenotypes from a tomato dihybrid cross (n = 1600). Source: simulated data based on standard tomato genetics teaching materials.

(a) Calculate the expected number of red, tall offspring from a total of 1600 F2 plants. Show your working.

(b) Assess whether the observed data is consistent with the 9:3:3:1 ratio predicted by independent assortment. Refer to the data.

(c) A second cross of the same two genes in a different population gave 600 red-tall, 75 red-dwarf, 80 yellow-tall and 445 yellow-dwarf out of 1200 plants. Account for the deviation from 9:3:3:1.

8marks Band 4–5
Section 3

Extended response

3.Extended response

3.1

Explain how the 9:3:3:1 phenotypic ratio observed in the F2 of an AaBb × AaBb dihybrid cross provides evidence for Mendel's law of independent assortment, and identify the conditions under which this ratio would not be observed. In your response, refer to the meiotic mechanism that produces independent assortment.

10marks Band 5–6

Biology · Year 12 · Module 5 · Lesson 14b

Answer Key & Marking Guidelines

1.1

Section 1 · Short answer · 2 marks · Band 3

Sample response. Independent assortment is Mendel's second law. It states that the alleles of two (or more) different genes segregate independently of each other during gamete formation, so the inheritance of one allele does not influence the inheritance of an allele at a different locus.

Marking notes. 1 mark for identifying independent assortment as a process of allele segregation during gamete formation / meiosis; 1 mark for stating that alleles of different genes segregate independently of each other (i.e. the inheritance of one does not depend on the other).

1.2

Section 1 · Short answer · 2 marks · Band 3

Sample response. AB, Ab, aB, ab — all four produced in equal frequency (1/4 each).

Marking notes. 2 marks for all four correct gamete types. 1 mark for two or three correct.

1.3

Section 1 · Short answer · 3 marks · Band 4

Sample response. During metaphase I of meiosis, paired homologous chromosomes (bivalents) align along the metaphase plate. The orientation of each bivalent is random — either homologue may face either pole — and the orientation of one bivalent is independent of the orientation of any other bivalent. When two genes are located on different chromosome pairs, this random alignment produces all possible combinations of maternal and paternal alleles in the resulting gametes at equal frequency.

Marking notes. 1 mark — identifies the event as the alignment of bivalents at the metaphase plate of meiosis I. 1 mark — states the orientation is random / each homologue has equal chance of facing either pole. 1 mark — explains that independent alignment of different bivalents produces all possible allele combinations at equal frequency in gametes.

1.4

Section 1 · Short answer · 2 marks · Band 4

Sample response. Green wrinkled (yyrr) is 1/16 of offspring in the 9:3:3:1 ratio. Expected number = 1/16 × 480 = 30 seeds.

Marking notes. 1 mark for identifying that yyrr is 1/16 of the F2 ratio; 1 mark for correctly calculating 1/16 × 480 = 30 (no working shown loses 1 mark).

1.5

Section 1 · Short answer · 3 marks · Band 4

Sample response. The phenotypic ratio of 9:3:3:1 describes the four observable trait combinations (both dominant : A dominant only : B dominant only : both recessive). The genotypic ratio is more complex because nine genotypes are produced from an AaBb × AaBb cross, occurring in the ratio 1:2:2:4:1:2:1:2:1 (AABB : AABb : AaBB : AaBb : AAbb : Aabb : aaBB : aaBb : aabb). The phenotypic ratio groups multiple genotypes together when they produce the same observable phenotype.

Marking notes. 1 mark for correctly identifying the phenotypic ratio as 9:3:3:1 with the four observable classes; 1 mark for identifying that nine distinct genotypes are produced; 1 mark for explaining that the phenotypic ratio results from grouping genotypes that share a phenotype (e.g. A_B_ groups AABB, AABb, AaBB, AaBb).

2.1

Section 2 · Data response · 8 marks · Band 4–5

Sample response (a). Red-tall is the A_B_ class (9/16 of offspring under 9:3:3:1). Expected = 9/16 × 1600 = 900 plants.

Sample response (b). Yes, the data is consistent with 9:3:3:1. Each observed count is very close to its expected count: red-tall 912 vs 900 (+1.3%); red-dwarf 287 vs 300 (-4.3%); yellow-tall 305 vs 300 (+1.7%); yellow-dwarf 96 vs 100 (-4%). All four classes are within roughly 5% of expectation, which is well within the range of sampling variation for n = 1600. The ratio supports independent assortment of the R/r and T/t genes — they are on different chromosomes.

Sample response (c). The second cross gives a strongly distorted ratio — red-tall (600) and yellow-dwarf (445) together account for over 87% of the 1200 offspring, whereas under 9:3:3:1 they should account for 10/16 ≈ 62.5%. The two "parental" combinations (R with T together, and r with t together) are massively overrepresented; the "recombinant" combinations are heavily underrepresented. The most likely explanation is that the R and T alleles (and r and t alleles) are linked on the same chromosome, so they tend to be inherited together. Linkage violates Mendel's law of independent assortment and produces a phenotypic ratio that deviates predictably from 9:3:3:1.

Marking notes. Part (a) — 1 mark for identifying 9/16 as the relevant fraction; 1 mark for the calculation 9/16 × 1600 = 900. Part (b) — 1 mark for stating the data is consistent with 9:3:3:1; 1 mark for referring to specific paired observed/expected counts; 1 mark for linking small deviations to sampling variation. Part (c) — 1 mark for identifying that the parental combinations are overrepresented (or recombinants underrepresented); 1 mark for naming linkage as the explanation; 1 mark for connecting linkage to genes lying on the same chromosome and the resulting violation of independent assortment.

3.1

Section 3 · Extended response · 10 marks · Band 5–6

Sample response. Mendel's law of independent assortment states that the alleles of two different genes segregate independently of each other during gamete formation. The 9:3:3:1 phenotypic ratio observed in the F2 of an AaBb × AaBb cross provides strong evidence for this law because it can only arise mathematically if the four gamete types (AB, Ab, aB, ab) produced by each parent occur in equal frequency. The cellular mechanism that produces this equal frequency is the random orientation of bivalents on the metaphase plate during meiosis I. When the A/a homologous pair and the B/b homologous pair are located on different chromosomes, their alignments are independent — either homologue of one pair can face either pole, irrespective of how the other pair is oriented. This independence generates all four allele combinations with equal probability, so each AaBb parent produces AB, Ab, aB and ab gametes at a frequency of 1/4 each. When two such parents are crossed, the 4×4 Punnett square contains 16 equally probable cells. Grouping by phenotype under simple dominance: nine cells show A_B_ (both dominant phenotypes), three cells show A_bb (only A dominant), three cells show aaB_ (only B dominant), and one cell shows aabb (both recessive). The phenotypic ratio is therefore 9:3:3:1 — a ratio that is a direct mathematical consequence of independent assortment, given simple dominant–recessive inheritance for both genes. The 9:3:3:1 ratio is conditional, not universal. It would not be observed under at least three circumstances. First, if the two genes are linked (on the same chromosome), the alleles travel together during meiosis, the four gamete types are no longer equally frequent (parental combinations outnumber recombinant combinations), and the F2 will show predictable deviation in favour of the parental phenotypes — exactly as in Bateson and Punnett's 1905 sweet pea data. Second, if either gene shows non-Mendelian inheritance — for example incomplete dominance, codominance, lethal alleles or epistatic interaction — the genotype-to-phenotype mapping changes and a different observable ratio is produced even when the genes are unlinked. Third, even when the two genes are unlinked and show simple dominance, small samples are subject to chance variation, so real F2 data only approximate 9:3:3:1 rather than match it exactly. Mendel's own 556 F2 pea seeds gave a 315:101:108:32 split — extremely close to the predicted 313:104:104:35, but not identical. The 9:3:3:1 ratio therefore both provides evidence for independent assortment and constrains the conditions under which Mendel's second law can be applied.

Marking notes. 1 mark — defines independent assortment correctly. 1 mark — states that the 9:3:3:1 ratio requires the four gamete types from each parent to be equally frequent. 1 mark — identifies random orientation of bivalents at the metaphase plate of meiosis I as the cellular event responsible. 1 mark — explains why independent alignment of bivalents on different chromosomes generates four equally frequent gamete types. 1 mark — sets up the 4×4 Punnett square (or equivalent reasoning) explicitly. 1 mark — correctly derives the 9:3:3:1 phenotypic grouping from the 16 cells under simple dominance. 1 mark — identifies that the ratio is not observed when the two genes are linked (on the same chromosome). 1 mark — identifies that the ratio is not observed under non-Mendelian inheritance patterns such as incomplete dominance or codominance. 1 mark — recognises that the ratio is also an expectation that real data only approximate due to sampling variation. 1 mark — overall coherence and use of precise lesson terminology (independent assortment, bivalent, metaphase plate, parental/recombinant, linkage) throughout.