In a dihybrid cross between RRYY and rryy parents, the number of RrYy genotypes possible in F $_{2}$ generation will be

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Solution

The correct option is A 4
This could be calculated by using Punnet square.
Genotype: RRYY X rryy
Gametes: RY ry
F $_{1}$ Generation: RrYy
Selfing of F $_{1}$
Genotype: RrYy X RrYy
Gametes: RY, Ry, rY and ry X RY, Ry, rY and ry
F $_{2}$ Generation:
RRYY RRYy RrYY RrYy
RRYy RRyy RrYy Rryy
RrYY RrYy rrYY rrYy
RrYy Rryy rrYy rryy
Thus, it is clear that RrYy occurs 4 times in F $_{2}$ generation.

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In a didybrid cross between RRYY and rryy, the number of RrYy $F_{2}$ genotypes will be

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