Biologyhard · Past Paper
In a dihybrid cross ($RRYY \times rryy$), what is the probability of getting an $RrYy$ genotype in the F2 generation?
A1/16
B2/16
C4/16
D9/16
✓ Correct Answer: C — 4/16
In the $9:3:3:1$ dihybrid cross, the double heterozygote $RrYy$ occurs in 4 out of 16 boxes of the Punnett square.
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