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Question #45833
The Question
A female heterozygous for a severe X-linked recessive disorder (X^R X^r) unexpectedly exhibits the full clinical disease phenotype. Karyotyping confirms a normal 46, XX complement with no structural deletions (ruling out Turner syndrome). Epigenetic analysis reveals a severe skewing in her dosage compensation mechanism. Which of the following molecular events best explains this 'pseudohomozygous' expression?
Options
- A.
Preferential transcription of the XIST gene on the X^R chromosome across her somatic cells.
- B.
Hypermethylation of the XIST gene promoter on the X^R chromosome.
- C.
Failure of Barr body formation, leading to simultaneous euchromatinization of both X chromosomes.
- D.
Hypomethylation of the XIST gene promoter on the X^r chromosome.
Check Your Answer
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
Barr Bodies|X-inactivation|Lyons Hypothesis
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often misunderstand the function of the XIST gene, assuming its transcription 'activates' the chromosome's functional genes.
The trap relies on knowing that XIST acts in cis to INACTIVATE the chromosome.
Therefore, to get a recessive disease in a heterozygote, the XIST gene must be highly active on the chromosome carrying the DOMINANT allele, effectively silencing the protective wild-type gene.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Principles of Inheritance and Variation - Class 12 |
| Concept | Sex-Linked Recessive Inheritance |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |