Question

In a Mendelian dihybrid cross, the probability of getting seeds with genotype Rryy, RrYy, rrYy and RrYY in F₂ generation is respectively

• A. $\frac{2}{16}:\frac{4}{16}:\frac{1}{8}:\frac{1}{8}$
• B. $\frac{2}{16}:\frac{2}{16}:\frac{2}{16}:\frac{2}{16}$
• C. $\frac{4}{16}:\frac{4}{16}:\frac{2}{16}:\frac{2}{16}$
• D. $\frac{1}{8}:\frac{1}{4}:\frac{2}{8}:\frac{1}{16}$

Answer 1

Answer: (A)

In a Mendelian dihybrid cross, the genotypic ratio is RRYY:RrYY:RRYy:RrYy:rrYY:rrYy:RRyy:Rryy:rryy is 1:2:2:4:1:2:1:2:1.