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Q.

If pea plant produces 2560 seeds in F2 generation when a dihybrid cross is done between homozygous Round and Yellow seeded plant and wrinkled and green seeded plant, how many of them have the phenotype wrinkled yellow seeds?

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a

160

b

320

c

480

d

1280

answer is C.

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Detailed Solution

The phenotypic ratio of F2 generation Round Yellow 9 : Round green 3 : wrinkled yellow 3 : wrinkled green 1. Hence the number of wrinkled yellow seeded plants would be 2560 × 316 = 480.
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