Q.65. A double homozygous mutant develops green and wrinkled seeds. When it was crossed with a
true-breeding plant having yellow and round seeds, all the F1 plants developed yellow and round seeds.
After self-fertilization of F1, the calculated percentage probability of plants with green and wrinkled
seeds in the F2 population (round off to 2 decimal places) is ____
Yellow and round seeds are dominant traits in pea plants, while green and wrinkled are recessive. In this dihybrid cross, the F2 generation follows a standard 9:3:3:1 phenotypic ratio, making the probability of green wrinkled seeds 1/16 or 6.25%.
Genetic Cross Breakdown
The double homozygous mutant has green wrinkled seeds, denoted as aabb (aa for green, bb for wrinkled). The true-breeding yellow round parent is AABB (A dominant for yellow, B dominant for round). All F1 plants are AaBb, showing yellow round phenotype due to dominance.
Self-fertilization of F1 (AaBb × AaBb) produces F2 with independent assortment of alleles.
Punnett Square Analysis
A 4×4 Punnett square yields 16 combinations. Phenotypes appear in 9:3:3:1 ratio:
-
9 A_B_ (yellow round)
-
3 A_bb (yellow wrinkled)
-
3 aaB_ (green round)
-
1 aabb (green wrinkled)
Green wrinkled (aabb) occurs in 1 out of 16 plants.
Calculated Probability
Probability = (1/4 for aa) × (1/4 for bb) = 1/16 = 0.0625. Rounded to two decimal places: 6.25%.
Common Exam Options Explained
| Option | Probability | Explanation |
|---|---|---|
| 6.25% | Correct | Matches 1/16 from 9:3:3:1 ratio for double recessive. |
| 25.00% | Incorrect | Single trait recessive (3:1 ratio, 1/4). |
| 12.50% | Incorrect | One dominant, one recessive (e.g., 3/16). |
| 3.13% | Incorrect | Inverse of 9/16, not standard. |
| 56.25% | Incorrect | Dominant class (9/16). |
This confirms 6.25% as the precise answer for F2 green wrinkled seeds.
