These tools will no longer be maintained as of December 31, 2024. Archived website can be found here. PubMed4Hh GitHub repository can be found here. Contact NLM Customer Service if you have questions.
Pubmed for Handhelds
PUBMED FOR HANDHELDS
Search MEDLINE/PubMed
Title: Effect of eradication of Helicobacter pylori infection on gastric epithelial cell proliferation. Author: Cahill RJ, Xia H, Kilgallen C, Beattie S, Hamilton H, O'Morain C. Journal: Dig Dis Sci; 1995 Aug; 40(8):1627-31. PubMed ID: 7648960. Abstract: Helicobacter pylori infection has been linked with gastric carcinoma. Epithelial cell proliferation is an indicator of cancer risk. The aim of this study was to assess gastric epithelial cell proliferation before and after eradication therapy and to assess the efficacy of treatment of H. pylori infection using lanzoprazole and clarithromycin. Twenty-three patients with H. pylori-associated gastritis were treated with lanzoprazole 30 mg daily for four weeks and clarithromycin 500 mg three times a day for two weeks. Antral mucosal biopsies were taken for gastric epithelial cell proliferation analysis using the in vitro bromodeoxyuridine (BrdU) immunohistochemical technique before and four weeks after eradication therapy. Labeling index percent (LI%) was calculated as the percent ratio of proliferating cells to the total number of cells in the gastric pit. Efficacy of treatment was assessed in 16 subjects. Eight were negative for H. pylori infection 28 days after therapy and in eight patients H. pylori infection was not eradicated. The eradication rate for the regime was 50%. Cell kinetics were assessed in 19 subjects who completed treatment. Patients with H. pylori infection had a significantly higher LI% compared to normal (N = 19, LI%: 5.01 +/- 0.3 vs 3.2 +/- 0.2, N = 29). Eradication of H. pylori infection significantly reduced epithelial cell proliferation (N = 9, LI%: 5.2 +/- 0.4 to 3.2 +/- 0.8, P < 0.001), whereas it was unaltered in those whose infection was not eradicated (N = 10, LI%: 4.8 +/- 0.4 to 5.5 +/- 0.5, P = 0.18).(ABSTRACT TRUNCATED AT 250 WORDS)[Abstract] [Full Text] [Related] [New Search]