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Question #45254
The Question
A diploid microspore mother cell (2n = 24, initial G1 DNA content = 2C) is treated with colchicine, which completely inhibits spindle fiber formation during Anaphase I. Assuming the cell forms a restitution nucleus and fails to undergo cytokinesis, what will be the chromosome number and DNA content of this cell at the end of Meiosis I, and what is the primary agricultural consequence?
Options
- A.
12 chromosomes, 2C DNA; leads to normal haploid gametes.
- B.
24 chromosomes, 4C DNA; leads to the formation of unreduced diploid (2n) gametes.
- C.
48 chromosomes, 4C DNA; leads to the immediate formation of tetraploid (4n) gametes.
- D.
24 chromosomes, 2C DNA; leads to the formation of aneuploid gametes due to partial nondisjunction.
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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
Cell Cycle|Aneuploidy vs Polyploidy|Spindle Apparatus
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often blindly memorize that 'colchicine doubles chromosome number' based on its mitotic application.
However, in Meiosis I, colchicine prevents the reduction of chromosomes (maintaining 2n instead of reducing to n).
It does not double the chromosome count to 4n because homologous chromosomes simply fail to separate; centromere splitting (which actually doubles the chromosome count) does not occur until Anaphase II or Mitosis.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Cell Cycle and Cell Division - Class 11 |
| Concept | DNA vs Chromosome Content |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |