In a dihybrid cross in garden pea in ,F2 90 plants were obtained Thai had yellow wrinkles seeds . How many plants will have yellow round seeds? (1) 390 (2)270 (3)810 (4)180?

Question

Answer

Given; yellow wrinkled seeds=90...we know that;phenotypic ratio of dihybrid cross=9(round yellow):3(round green):3(wrinkled yellow):1(wrinkled green)...in question it asked about round yellow....9 round yellow->3 wrinkled yellow...?.round yellow->90 wrinkled yellow..no. of plants with round yellow = 90 × 9/3=270...option b )270 is true..

Answer

270 hoga answer bbbb

Answer

Calculation of Plants with Yellow Round Seeds in Dihybrid Cross

Given Information:
- F2 generation from a dihybrid cross in garden pea
- 90 plants with yellow wrinkled seeds were obtained

Explanation:
- In a dihybrid cross of garden peas, the genotype for seed shape and seed color follows a 9:3:3:1 ratio in the F2 generation.
- This means that out of 16 possible combinations, 9 plants will have yellow round seeds, 3 plants will have yellow wrinkled seeds, 3 plants will have green round seeds, and 1 plant will have green wrinkled seeds.

Calculation:
- If 90 plants have yellow wrinkled seeds, then the total number of plants in the F2 generation is 90 * 16 = 1440.
- Out of these 1440 plants, 9/16 will have yellow round seeds.
- Therefore, the number of plants with yellow round seeds = (9/16) * 1440 = 810.

Conclusion:
- Therefore, out of the 1440 plants in the F2 generation, 810 plants will have yellow round seeds.
Therefore, the correct answer is:
- (3) 810

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