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Question #46771
The Question
During Prophase I of meiosis, CDK1-Cyclin B activation normally synchronizes chromatin condensation with nucleolar disassembly. In an experiment on plant meiocytes, a drug selectively inhibits the condensin complex. This delays chromatin condensation, leaving the nucleolus intact and functional. This temporal mismatch completely disrupts meiotic spindle formation. If the cell bypasses the spindle assembly checkpoint and fails to undergo cytokinesis, which of the following represents the ultimate genetic consequence?
Options
- A.
Aneuploidy, due to the non-disjunction of a single homologous chromosome pair during Anaphase I.
- B.
Polyploidy, due to the complete failure of the meiotic reductional division and retention of the entire chromosome set.
- C.
Nullisomy, because the prolonged nucleolar activity degrades the uncondensed homologous chromosomes.
- D.
Normal haploid gamete formation, as the intact nucleolus acts as a secondary microtubule organizing center.
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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
Nucleolar Functionality|Chromatin Condensation States|Meiotic Reductional Division
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Distinguishing between the scale of non-disjunction.
Students often reflexively link 'non-disjunction' exclusively to Aneuploidy (e.g., Down's syndrome).
However, a complete failure of the meiotic spindle and subsequent cytokinesis affects the entire chromosomal set, leading to Polyploidy, a critical distinction emphasized in the NCERT Principles of Inheritance and Variation chapter.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Cell – The Unit of Life - Class 11 |
| Concept | Nucleolar Functionality |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |