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
In Mendel’s dihybrid cross, cross is made between heterozygous round green seeds(RRyy) to another heterozygous wrinkled yellow seeds(rrYY). In the F1 generation, all plants are heterozygous Round yellow(RrYy).
Parents : RRyy X rrYY
F1: RrYy
Gametes : RY, Ry, rY, ry
|
RY |
Ry |
rY |
ry |
|
|
RY |
RRYY |
RRYy |
RrYY |
RrYy |
|
Ry |
RRYy |
RRyy |
RrYy |
Rryy |
|
rY |
RrYY |
RrYy |
rrYY |
rrYy |
|
ry |
RrYy |
Rryy |
rrYy |
rryy |
Genotypic ratio : 1:2:1:2:4:2:1:2:1
To find the genotypes and phenotypes in the F2 generation, we need to consider the possible combinations of alleles:
Genotypes: RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy, rryy (total 9)
Phenotypes: Round yellow, Round green, Wrinkled yellow, Wrinkled green (total 4)
So, the correct answer is (3)- 9 genotypes and 4 phenotypes.