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.
Answer & explanation
Correct answer: option 2
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)
|
RY |
Ry |
rY |
ry |
|
|
RY |
RRYY (Round Yellow) |
RRYy (Round Yellow) |
RrYY (Round Yellow) |
RrYy (Round Yellow) |
|
Ry |
RRYy (Round Yellow) |
RRyy (Round Green) |
RrYy (Round Yellow) |
Rryy (Round Green) |
|
rY |
RrYY (Round Yellow) |
RrYy (Round Yellow) |
rrYY (Wrinkled Yellow) |
rrYy (Wrinkled Yellow) |
|
ry |
RrYy (Round Yellow) |
Rryy (Round Green) |
rrYy (Wrinkled Yellow) |
rryy (Wrinkled Green) |
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.