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]