Question

A round-yellow (RRYY) seed producing pea plant is crossed with wrinkled-green (rryy) seed producing pea plant. Plants obtained by this cross is
allowed to self-cross. If 165 seeds obtained in F2 generation are round-green, then what should be the number of round-yellow seeds obtained in
this generation?
BIO​

Answer 1

Answer:

495

Explanation:

we know that the ratio for dihybrid cross is 9:3:3:1

given that 165 seeds are round green which is present in the ratio of 3.

then if 165 seeds are present in eating 3,then the number of seeds for ratio 9 will be

(9*165)/3 =495

Answer 2

Answer:

The number of round-yellow seed-producing plants obtained in the F2 generation is 495.

Explanation:

• In the Mendelian dihybrid cross experiment, the round-yellow seed-producing plant is crossed with a wrinkled-green seed-producing plant. In the F2 generation, the phenotypic ratio was 9:3:3:1 for wrinkled-yellow seeds, round-yellow seeds, wrinkled-green seeds, and round-green seeds.
• Thus, it can be concluded that 1 round-green seed-producing plant is obtained per 3 round-yellow seed-producing plants, so if 165 are round-green seed-producing plants, the number of round-yellow seed-producing plants obtained in the F2 generation is:
• 3×165/1
• =495
• Thus, 495 round-yellow seed-producing plants are obtained in the F2 generation.