A pea plant with yellow and round seeds (YYRR) is crossed with a pea plant having green and wrinkled (yyrr) seeds, then in $F_{2}$ generation of this dihybrid cross 320 plants are produced. Out of which 180 plants have some phenotypic characters. Identify this phenotype.

Yellow ad wrinkled seeds

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Yellow and round seeds

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Green and round seeds

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Green and wrinkled seeds

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Solution

The correct option is D Yellow and round seeds
When the pea plant with yellow and round seeds (YYRR) is crossed with a pea plant having green and wrinkled (yyrr) seeds, the phenotypic ratio is obtained as 9:3:3:1.
There is a total of 320 seeds produced out of which 180 have some phenotype.
There are total of 16 chambers in the Punnett square. So, 320/16=20. There are 20 seeds in one chamber.
If there are 180 seeds that have the same phenotype, these seeds are present in the 180/20=9 chambers. So, the seeds will have the phenotype as yellow and round seeds.
Thus, the correct answer is option B.

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Q. A pea plant with yellow and round seeds (YYRR) is crossed with a pea plant having green and wrinkled (yyrr) seeds, then in

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A pea plant with yellow and round seeds (YYRR) is crossed with a pea plant having green and wrinkled (yyrr) seeds, then in F2 generation of this dihybrid cross 320 plants are produced. Out of which 180 plants have some phenotypic characters. Identify this phenotype.

A pea plant with yellow and round seeds (YYRR) is crossed with a pea plant having green and wrinkled (yyrr) seeds, then in $F_{2}$ generation of this dihybrid cross 320 plants are produced. Out of which 180 plants have some phenotypic characters. Identify this phenotype.