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

Question

In a Mendelian dihybrid cross, the probability of getting seeds with genotype Rryy, RrYy, rrYy and RrYY in F2 generation is respectively (A) (2/16):(4/16):(1/8):(1/8) (B) (2/16):(2/16):(2/16):(2/16) (C) (4/16):(4/16):(2/16):(2/16) (D) (1/8):(1/4):(2/8):(1/16)

Tardigrade Admin · Accepted Answer

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.

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

$\frac{2}{16} : \frac{4}{16} : \frac{1}{8} : \frac{1}{8}$

$\frac{2}{16} : \frac{2}{16} : \frac{2}{16} : \frac{2}{16}$

$\frac{4}{16} : \frac{4}{16} : \frac{2}{16} : \frac{2}{16}$

$\frac{1}{8} : \frac{1}{4} : \frac{2}{8} : \frac{1}{16}$

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.

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

162​:164​:81​:81​

162​:162​:162​:162​

164​:164​:162​:162​

81​:41​:82​:161​

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.

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

$\frac{2}{16} : \frac{4}{16} : \frac{1}{8} : \frac{1}{8}$

$\frac{2}{16} : \frac{2}{16} : \frac{2}{16} : \frac{2}{16}$

$\frac{4}{16} : \frac{4}{16} : \frac{2}{16} : \frac{2}{16}$

$\frac{1}{8} : \frac{1}{4} : \frac{2}{8} : \frac{1}{16}$