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Question #45280
The Question
A male with blood group AB and a female with blood group O produce a child whose erythrocytes possess both A and B surface antigens. Assuming no novel mutations or epistasis (e.g., Bombay phenotype), this anomalous inheritance is best explained by a meiotic non-disjunction event. Identify the specific meiotic stage where this failure occurred, the parent involved, and the fundamental Mendelian principle directly violated.
Options
- A.
Anaphase I in the male; Law of Segregation
- B.
Anaphase II in the male; Law of Independent Assortment
- C.
Anaphase I in the female; Law of Segregation
- D.
Anaphase II in the male; Law of Segregation
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Evolutionary_Exception
Tackle problems where standard rules evolve or change based on specific environmental constraints.
Concepts Behind This Question
Multiple Allelism Limitation|Meiotic Reductional Division|Mendelian Segregation
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students frequently confuse the genetic consequences of Anaphase I versus Anaphase II non-disjunction.
Anaphase I failure results in gametes containing both homologous chromosomes (heterozygous alleles, I^A and I^B), whereas Anaphase II failure results in gametes containing unseparated sister chromatids (homozygous alleles, I^A I^A or I^B I^B).
Additionally, linking a chromosomal aberration directly to the violation of Mendel's First Law (Segregation) tests the synthesis of Mendelian Genetics with the Chromosomal Theory of Inheritance.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Cell Cycle and Cell Division - Class 11 |
| Concept | Meiotic Reductional Division |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |