Vitamin D3 (cholecalciferol) undergoes sequential hydroxylations. The hydroxylation that produces the most physiologically active form (1,25-(OH)2D3 / calcitriol) and is tightly regulated by PTH, FGF-23, and calcium is performed by which enzyme in which location?
- A 1-alpha-hydroxylase (CYP27B1) in kidney proximal tubule mitochondria ✓
- B 25-hydroxylase in liver microsomes
- C 24-hydroxylase in kidney (inactivation step)
- D 7-dehydrocholesterol reductase in skin
Correct answer: A. 1-alpha-hydroxylase (CYP27B1) in kidney proximal tubule mitochondria
Explanation
CYP27B1 (1-alpha-hydroxylase) in the proximal tubule mitochondria converts 25-OH D3 (calcidiol) to 1,25-(OH)2D3 (calcitriol). This step is the tightly regulated one: PTH and low plasma phosphate stimulate it; FGF-23, high calcium, and 1,25-(OH)2D3 itself (negative feedback) inhibit it. 25-Hydroxylation in liver is constitutive and substrate-driven. CYP24A1 (24-hydroxylase) in the kidney inactivates calcitriol to calcitroic acid. Renal failure reduces CYP27B1 activity, causing secondary hyperparathyroidism.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
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