What will be the number of genotypes and phenotypes obtained in F2 generation when male parent RRyy is crossed with female parent rrYY?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 and 4 respectively
RRyy × rrYY gives F1 RrYy. Selfing RrYy is a dihybrid cross.
Each gene gives 3 genotypes (RR, Rr, rr and YY, Yy, yy), so there are 3 × 3 = 9 genotypes.
Each gene gives 2 phenotypes (dominant, recessive), so there are 2 × 2 = 4 phenotypes, in the ratio 9 : 3 : 3 : 1.
Option (1) 16 and 4 respectively — Incorrect. 16 is the number of boxes in the Punnett square; several boxes share a genotype.
Option (2) 4 and 16 respectively — Incorrect. There cannot be more phenotypes than genotypes.
Option (3) 9 and 4 respectively — Correct. RRyy × rrYY gives an F1 that is RrYy, a dihybrid. Selfing a dihybrid gives 9 genotypes (3 for each gene, 3 × 3) and 4 phenotypes in the ratio 9 : 3 : 3 : 1.
Option (4) 4 and 9 respectively — Incorrect. The numbers are reversed.