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Question #45589
The Question
A geneticist performs a test cross on a plant that is trisomic for a specific autosome, possessing the genotype AAa. Assuming all gametes are equally viable, the progeny exhibit a 5:1 phenotypic ratio of dominant to recessive, completely deviating from the classic 1:1 Mendelian test cross ratio. Which of the following best explains the cytological basis for this altered genetic ratio?
Options
- A.
During Anaphase I, the three homologous chromosomes segregate randomly, producing 'n' and 'n+1' gametes in equal proportions, leading to an unbalanced transmission of alleles.
- B.
The extra chromosome undergoes secondary non-disjunction during Anaphase II, resulting in diploid gametes that skew the phenotypic expression.
- C.
Dosage compensation mechanisms inactivate one of the dominant 'A' alleles during embryogenesis, mimicking a standard monohybrid F2 cross.
- D.
The presence of a trivalent during Pachytene completely suppresses recombination, causing the recessive 'a' allele to be lost during gametogenesis.
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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
Down/Klinefelter/Turner Syndrome Karyotypes|Test Cross vs Back Cross Ratios
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often memorize the 1:1 test cross ratio without deeply connecting it to its cytological basis (the equal segregation of exactly two homologues).
This question tests the synthesis of Aneuploidy (Trisomy) with Mendelian segregation, requiring the student to deduce how three homologous chromosomes segregate during Anaphase I to produce unbalanced 'n' and 'n+1' gametes, thereby breaking the classic 1:1 prediction.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Principles of Inheritance and Variation - Class 12 |
| Concept | Test Cross vs Back Cross Ratios |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |