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Question #45290
The Question
In a human primary spermatocyte, an error occurs during Metaphase I where the kinetochores of both homologous chromosomes of bivalent 21 attach to spindle fibers originating from the same pole. Assuming Meiosis II proceeds normally, which of the following accurately describes the mechanical event at Anaphase I and the final chromosomal constitution of the four resulting spermatids?
Options
- A.
Homologous chromosomes fail to segregate; all four spermatids will be aneuploid, with two being n+1 and two being n-1.
- B.
Sister chromatids fail to separate; two spermatids will be normal (n), one will be n+1, and one will be n-1.
- C.
Homologous chromosomes fail to segregate; resulting in polyploidy where two spermatids are 2n and two lack chromosome 21.
- D.
The centromeres split prematurely; all four spermatids will be aneuploid, with two being n+1 and two being n-1.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Comparative_Analysis
Sharpen your ability to evaluate multiple scenarios and distinguish between competing potential answers.
Concepts Behind This Question
Down/Klinefelter/Turner Syndrome Karyotypes|Homologous chromosomes
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Distinguishing between Anaphase I non-disjunction (failure of homologous chromosomes to separate, yielding 100% aneuploid gametes: two n+1, two n-1) and Anaphase II non-disjunction (failure of sister chromatids to separate, yielding 50% normal gametes).
Students frequently confuse the mechanical events of Anaphase I (separation of homologs without centromere splitting) with Anaphase II (centromere splitting) and conflate the terms Aneuploidy and Polyploidy.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Cell Cycle and Cell Division - Class 11 |
| Concept | Spindle Attachment Geometry |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |