Q.49 In a Mendel’s dihybrid experiment, a homozygous pea plant with round yellow seeds
was crossed with a homozygous plant with wrinkled green seeds. F1 intercross
produced 560 F2 progeny. The number of F2 progeny having both dominant
traits (round and yellow) is _________.
In Mendel’s classic dihybrid cross, a homozygous round yellow pea plant (RRYY) crossed with a homozygous wrinkled green plant (rryy) produces F1 hybrids (RrYy) that self-cross to yield a 9:3:3:1 phenotypic ratio in F2. For 560 F2 progeny, the number with both dominant traits (round and yellow) follows this standard ratio.
Dihybrid Cross Ratio Explained
The F1 generation (RrYy × RrYy) results in 9/16 round yellow, 3/16 round green, 3/16 wrinkled yellow, and 1/16 wrinkled green progeny due to independent assortment. Calculate round yellow progeny as (9/16) × 560 = 315. This matches Mendel’s observed 9:3:3:1 ratio from actual experiments with similar totals like 2160 seeds.
Step-by-Step Progeny Calculation
- Total F2 progeny: 560.
- Round yellow fraction: 9/16 = 0.5625.
- Number: 0.5625 × 560 = 315 (exact integer as 9 × 35 = 315).
- Round green: (3/16) × 560 = 105; wrinkled yellow: 105; wrinkled green: 35. The sum confirms 560 total.
Common Options and Misconceptions
| Option Scenario | Calculation | Progeny Count | Why Incorrect |
|---|---|---|---|
| Monohybrid 3:1 ratio | (3/4) × 560 | 420 | Ignores second trait; applies to single gene only. |
| Recombinant only (2/16) | (2/16) × 560 | 70 | Underestimates dominants; confuses with test cross. |
| All dominant (12/16) | (12/16) × 560 | 420 | Wrong ratio; assumes linkage, not independent. |
| Correct 9:3:3:1 | (9/16) × 560 | 315 | Matches Mendel’s law perfectly. |
