Biochemistry · Heme Synthesis and Porphyrias

Lead poisoning inhibits two enzymes in the heme biosynthetic pathway: delta-aminolevulinic acid dehydratase (ALAD) and ferrochelatase. ALAD converts 2 molecules of ALA to porphobilinogen (PBG). Ferrochelatase inserts Fe2+ into protoporphyrin IX. Which laboratory findings would be EXPECTED in a child with significant lead exposure?

  • A Elevated urine ALA and PBG, elevated RBC zinc protoporphyrin (ZPP)
  • B Elevated urine ALA, normal PBG, elevated RBC zinc protoporphyrin (ZPP), microcytic anemia with basophilic stippling
  • C Elevated urine PBG, normal ALA, elevated serum porphyrins, normocytic anemia
  • D Normal ALA and PBG, elevated urine coproporphyrin III, sideroblastic anemia
Correct answer: B. Elevated urine ALA, normal PBG, elevated RBC zinc protoporphyrin (ZPP), microcytic anemia with basophilic stippling

Explanation

Lead inhibits ALAD (ALA → PBG step) causing ALA accumulation in urine, but because the block is upstream of PBG, urine PBG remains normal (not elevated). Lead also inhibits ferrochelatase (final step), so iron cannot be inserted into protoporphyrin IX; instead, zinc (which doesn't need ferrochelatase regulation) is inserted, forming zinc protoporphyrin (ZPP), which accumulates in RBCs — measured by hematofluorometer. This ZPP elevation is a sensitive screening test for lead exposure. The combined ALAD and ferrochelatase inhibition causes sideroblastic-pattern microcytic anemia. Basophilic stippling occurs because lead inhibits pyrimidine 5'-nucleotidase (a separate effect), causing aggregation of ribosomal RNA and granules. The pattern: elevated ALA, normal PBG, high ZPP, basophilic stippling.

Reference: Harper's Illustrated Biochemistry, 32nd ed.

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