Biology • Year 12 • Module 5 • Lesson 14b
Dihybrid Crosses and Independent Assortment
Apply the 9:3:3:1 ratio to predict offspring numbers, build a Punnett square from a fresh scenario, interpret a real dihybrid data set, and reason about linkage when the ratio breaks down.
1. Build a Punnett square from scratch — pea pod colour × plant height
Two pea plants, both PpTt, are crossed. Allele P (green pod) is dominant to p (yellow pod). Allele T (tall plant) is dominant to t (dwarf plant). Assume independent assortment. 10 marks
1.1 List the four gamete types each parent can produce. 1 mark
1.2 Complete the 4×4 Punnett square for PpTt × PpTt. 4 marks
| ♀ / ♂ | PT | Pt | pT | pt |
|---|---|---|---|---|
| PT | ||||
| Pt | ||||
| pT | ||||
| pt |
1.3 Count the offspring in each phenotypic class and write the simplified ratio. 3 marks
Green tall (P_T_) ____ : Green dwarf (P_tt) ____ : Yellow tall (ppT_) ____ : Yellow dwarf (pptt) ____
Simplified ratio: ____ : ____ : ____ : ____
1.4 Which one of the 16 genotypes can ONLY be produced from one specific gamete combination (i.e. appears in just one cell)? Give its genotype and explain why. 2 marks
2. Predict offspring numbers
A horticulturist crosses two pea plants, both YyRr, where Y (yellow seed) and R (round seed) are dominant. They harvest a total of 640 F2 seeds. Assume the 9:3:3:1 ratio holds. 5 marks
2.1 Predict the expected number of seeds in each phenotypic class. Show your working. 4 marks
(a) Yellow round (Y_R_) = ____/16 × 640 = ____
(b) Yellow wrinkled (Y_rr) = ____/16 × 640 = ____
(c) Green round (yyR_) = ____/16 × 640 = ____
(d) Green wrinkled (yyrr) = ____/16 × 640 = ____
2.2 Check: do your four numbers add to 640? If not, what does the discrepancy tell you about the 9:3:3:1 ratio? 1 mark
3. Interpret Mendel's real F2 data
Mendel's actual dihybrid cross of YyRr × YyRr pea plants yielded the F2 counts shown below. 6 marks
| Phenotype | Observed count (Mendel) | Expected count (9:3:3:1 of 556) |
|---|---|---|
| Yellow round | 315 | |
| Yellow wrinkled | 101 | |
| Green round | 108 | |
| Green wrinkled | 32 | |
| Total | 556 | 556 |
3.1 Calculate the expected count for each phenotype out of 556 (round to whole seeds) and fill the right-hand column above. 2 marks
3.2 Express Mendel's observed yellow-round : green-wrinkled ratio as a simplified fraction (divide both by the smaller). How close is this to 9:1? 2 marks
3.3 Mendel's observed counts do not match the expected counts exactly. Give two biological reasons why real F2 data deviates a little from the 9:3:3:1 prediction even when the genes are unlinked. 2 marks
4. When the ratio breaks — linkage scenario
A genetics student crosses two plants both AaBb and harvests 800 F2 offspring. The phenotype counts are shown below. 6 marks
| Phenotype | Expected (9:3:3:1 of 800) | Observed |
|---|---|---|
| A_B_ | 450 | 610 |
| A_bb | 150 | 50 |
| aaB_ | 150 | 40 |
| aabb | 50 | 100 |
4.1 The student claims the result confirms independent assortment. Do you agree? Justify your answer using two specific data points. 2 marks
4.2 Which two phenotypic classes are overrepresented compared with expectation? What does this suggest about the parental allele combinations on the chromosomes? 2 marks
4.3 Propose the most likely biological explanation for this deviation, in one sentence. 2 marks
Q1.1 — Gametes
Each PpTt parent produces four gamete types in equal frequency: PT, Pt, pT, pt. [1]
Q1.2 — Completed Punnett square
| PT | Pt | pT | pt | |
|---|---|---|---|---|
| PT | PPTT | PPTt | PpTT | PpTt |
| Pt | PPTt | PPtt | PpTt | Pptt |
| pT | PpTT | PpTt | ppTT | ppTt |
| pt | PpTt | Pptt | ppTt | pptt |
Mark allocation: 4 marks for fully correct grid (allow one slip for 3 marks; two for 2; three for 1).
Q1.3 — Phenotype counts and ratio
Green tall (P_T_) = 9 • Green dwarf (P_tt) = 3 • Yellow tall (ppT_) = 3 • Yellow dwarf (pptt) = 1. Simplified ratio = 9 : 3 : 3 : 1. [3]
Q1.4 — Single-cell genotype
Two double-homozygous genotypes appear in only one cell each: PPTT (only from PT × PT) and pptt (only from pt × pt). Either answer earns full marks. Reason: a genotype that requires two specific gametes — and where each gamete is itself only one of four — can occur in only 1 / (4 × 4) = 1/16 of the boxes; no other gamete combination produces it. [2]
Q2.1 — Expected offspring numbers
(a) Yellow round = 9/16 × 640 = 360. (b) Yellow wrinkled = 3/16 × 640 = 120. (c) Green round = 3/16 × 640 = 120. (d) Green wrinkled = 1/16 × 640 = 40. [4 — 1 per row]
Q2.2 — Total check
360 + 120 + 120 + 40 = 640. The four fractions of the 9:3:3:1 ratio (9/16 + 3/16 + 3/16 + 1/16) must equal 16/16 = 1 by construction, so the predicted counts always sum to the total offspring. [1]
Q3.1 — Expected counts out of 556
Yellow round = 9/16 × 556 ≈ 312.75 → 313. Yellow wrinkled = 3/16 × 556 ≈ 104.25 → 104. Green round = 3/16 × 556 ≈ 104.25 → 104. Green wrinkled = 1/16 × 556 ≈ 34.75 → 35. (Whole-seed rounding accepted; total may differ by ±1.) [2]
Q3.2 — Yellow round : green wrinkled
315 / 32 ≈ 9.84 : 1. The expected ratio is 9 : 1; Mendel's observed ratio is very close (about 9% higher than predicted), which strongly supports independent assortment. [2]
Q3.3 — Why real F2 data deviates from 9:3:3:1
Any two valid reasons earn full marks: (i) sampling / chance variation — with only 556 seeds, random fertilisation events will not produce exact 9:3:3:1 proportions; (ii) slight differences in viability between phenotypic classes (e.g. some genotypes survive to germination slightly better than others); (iii) human error in scoring phenotypes; (iv) very rare mutation or environmental modifiers. [2]
Q4.1 — Does the data confirm independent assortment?
No. [1] The observed A_B_ count is 610 vs an expected 450, and the observed aabb count is 100 vs an expected 50 — both are far outside reasonable sampling variation, so the ratio is not 9:3:3:1. [1]
Q4.2 — Over-represented classes and what they suggest
A_B_ (both dominant) and aabb (both recessive) are overrepresented. [1] This suggests that on the chromosomes of the parents, alleles A and B travelled together (as did a and b), and were inherited as parental combinations rather than being shuffled independently. [1]
Q4.3 — Most likely biological explanation
The two genes are linked — they lie on the same chromosome, so the parental allele combinations are inherited together and the four gamete types are no longer equally frequent. [2]