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Question #46260
The Question
In the family Liliaceae, the androecium is characteristically epiphyllous. A geneticist identifies two distinct, tightly linked genes on the same autosome: Gene P controls perianth formation and Gene E controls stamen adhesion. If a heterozygous plant (PpEe) with alleles in a cis-arrangement is test-crossed, which of the following statements correctly describes the expected genetic outcome?
Options
- A.
The progeny will exhibit a 1:1:1:1 phenotypic ratio, as crossing over during the pachytene stage ensures the independent assortment of the P and E genes.
- B.
The progeny will predominantly display parental phenotypes, demonstrating an exception to Mendel's Law of Independent Assortment due to a recombination frequency approaching zero.
- C.
The F2 generation of a self-cross will yield a 9:3:4 ratio, indicating that the gene for perianth formation is supplementary to the gene for stamen adhesion.
- D.
The test cross will yield exactly 50% recombinant types because the genes are located on the same chromosome, validating Morgan's chromosomal theory of inheritance.
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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
Epiphyllous|Perianth|Gene_Linkage
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often mechanically associate dihybrid test crosses with a 1:1:1:1 ratio or assume that being on the same chromosome automatically yields 50% recombinants.
The trap lies in failing to recognize that 'tight linkage' implies a recombination frequency near zero, which fundamentally breaks Mendel's Second Law (Independent Assortment) and preserves the diagnostic epiphyllous trait of Liliaceae across generations.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Morphology of Flowering Plants - Class 11 |
| Concept | Androecium Cohesion/Adhesion |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |