A certain plant homozygous for yellow seeds and red flowers was crossed with a plant homozygous for green seeds and white flowers. The F1 plants had yellow seeds and pink flowers. The F1 plants were selfed to get F2 progeny. Assuming independent assortment of the two characters, how many phenotypic categories are expected for these characters in the F2 generation?
Answer: (B) 6. Answer: (B) 6 Solution: Seed colour (yellow dominant over green) shows complete dominance, so its F2 has 2 phenotypes in a 3:1 ratio.
- A.4
- B.6✓
- C.9
- D.16
Correct Answer
(B) 6
Solution & Explanation
Answer: (B) 6 Solution: Seed colour (yellow dominant over green) shows complete dominance, so its F2 has 2 phenotypes in a 3:1 ratio. Flower colour shows incomplete dominance (red x white gives pink F1), so its F2 has 3 phenotypes in a 1:2:1 ratio. With independent assortment the number of combined phenotypic categories is 2 x 3 = 6, which is option (2). NCERT Reference: NCERT Class XII, Ch 4, p.60, lines 13-18: "the F2 resulted in the following ratio 1 (RR) Red: 2 (Rr) Pink: 1 (rr) White. Here the genotype ratios were exactly as we would expect in any mendelian monohybrid cross, but the phenotype ratios had changed from the 3:1 dominant : recessive ratio."