In a Mendelian dihybrid cross between a pea plant that has seeds with yellow color and round shape (RRYY) and another one that has seeds with green color and wrinkled shape (rryy), select the possible phenotypes obtained on self pollination after $F_1$ generation.

(A) Round yellow
(B) Round green
(C) Wrinkled yellow
(D) Wrinkled brown

Choose the correct answer from the options given below.

(A), (B) and (C) only

The correct answer is Option (2) → (A), (B) and (C) only     

Parental cross:

• Parent 1 = RRYY (Round Yellow)
• Parent 2 = rryy (Wrinkled Green)

F1 generation:

• All offspring = RrYy (heterozygous, Round Yellow, since R and Y are dominant).

F2 generation (selfing RrYy × RrYy):
This is the classic Mendelian dihybrid cross → phenotypic ratio = 9 : 3 : 3 : 1

Gametes from each parent

Each RrYy can produce 4 types of gametes by independent assortment:

• RY, Ry, rY, ry

Step 2: Punnett square (4 × 4 = 16 combinations)

Step 3: Phenotypic ratio

Now group them by phenotype:

• Round Yellow (dominant R, dominant Y) = 9
• Round Green (dominant R, recessive yy) = 3
• Wrinkled Yellow (recessive rr, dominant Y) = 3
• Wrinkled Green (rryy) = 1

So the final ratio is 9 : 3 : 3 : 1

So, the possible phenotypes are:
 Round yellow (A)
Round green (B)
Wrinkled yellow (C)
Wrinkled brown (D) → this trait doesn’t exist in Mendel’s pea characters.