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Question #45313
The Question
In a typical diploid angiosperm, a meiotic nondisjunction event during microsporogenesis leads to the formation of unreduced, diploid (2n) pollen grains. If this pollen successfully germinates on the stigma of a normal diploid plant of the same species and undergoes double fertilization, what will be the ploidy of the resulting zygote and primary endosperm nucleus (PEN), and the most likely developmental fate of the seed?
Options
- A.
Zygote: 3n, PEN: 4n; Seed abortion due to maternal-paternal genomic imbalance in the endosperm.
- B.
Zygote: 3n, PEN: 5n; Development of a viable but sterile triploid plant.
- C.
Zygote: 4n, PEN: 6n; Seed abortion due to extreme polyploidy in the embryo.
- D.
Zygote: 3n, PEN: 4n; Normal seed development leading to a fertile tetraploid plant.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Regulatory_Failure
Understand how NOT, EXCEPT, and logical constraints are used to manipulate exam results.
Concepts Behind This Question
Microsporogenesis Ploidy Shift|Outbreeding Device Logic|Embryo Sac Nuclei Count
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often rote-memorize that 'zygote is 2n and endosperm is 3n' without understanding the additive mechanics of double fertilization.
This question forces them to recalculate ploidy based on an unreduced male gamete (2n + n = 3n; 2n + 2n = 4n).
Furthermore, it tests the 'NCERT-Plus' evolutionary concept that endosperm viability relies on a specific maternal:paternal genomic ratio, and deviations act as a reproductive barrier leading to seed abortion.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Sexual Reproduction in Flowering Plants - Class 12 |
| Concept | Outbreeding Device Logic |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |