Biochemistry · Mineral and Trace Element Metabolism (Iron, Copper, Zinc, Calcium-Phosphate)

A 55-year-old man with chronic kidney disease (stage 4) has serum phosphate 6.2 mg/dL, calcium 8.1 mg/dL, and markedly elevated intact PTH. Which step in vitamin D metabolism is primarily impaired, leading to decreased intestinal calcium absorption and secondary hyperparathyroidism in this patient?

  • A 7-Dehydrocholesterol conversion to cholecalciferol in the skin
  • B 1-Alpha-hydroxylation of 25-hydroxyvitamin D in the proximal renal tubule
  • C 25-Hydroxylation of cholecalciferol in the liver
  • D 24-Hydroxylation of 25-hydroxyvitamin D in the kidney
Correct answer: B. 1-Alpha-hydroxylation of 25-hydroxyvitamin D in the proximal renal tubule

Explanation

The biologically active form of vitamin D, 1,25-dihydroxyvitamin D (calcitriol), is produced by 1-alpha-hydroxylase (CYP27C1) in the proximal renal tubule. In chronic kidney disease, loss of functional renal mass impairs this conversion, reducing calcitriol synthesis. This decreases intestinal calcium absorption, causing hypocalcemia, which together with hyperphosphatemia drives secondary hyperparathyroidism. Option A describes cutaneous vitamin D synthesis, not the rate-limiting renal step. Option C (25-hydroxylation) occurs in the liver and is generally preserved until advanced malnutrition. Option D (24-hydroxylation) is the catabolic inactivation pathway, not the activation step.

Reference: Ganong's Review of Medical Physiology, 26th ed.

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