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人初乳中含有对产肠毒素大肠杆菌的定居因子I和II有反应的IgA抗体。

Human colostrum contains IgA antibodies reactive to colonization factors I and II of enterotoxigenic Escherichia coli.

作者信息

Corrêa Simone, Palmeira Patricia, Carneiro-Sampaio Magda Maria Sales, Sanae Nishimura Lucília, Guth Beatriz Ernestina Cabilio

机构信息

Department of Immunology, Universidade de São Paulo, São Paulo, Brazil.

出版信息

FEMS Immunol Med Microbiol. 2006 Jul;47(2):199-206. doi: 10.1111/j.1574-695X.2006.00082.x.

Abstract

Diarrhea is an important cause of morbidity and mortality amongst infants of low socio-economic levels in developing countries and in travelers who visit such areas. Enterotoxigenic E. coli strains express two sets of virulence-associated factors: enterotoxins (heat-stable toxins or heat-labile toxins) and colonization factors. Studies have shown that breast-feeding protects infants against infectious diseases, such as diarrhea, as it presents a great variety of immunological components. The aim of this study was to analyze the reactivity of immunoglobulin A from human colostrum to colonization factor antigens I and II. The colostrum ability in preventing enterotoxigenic E. coli adhesion to Caco-2 cells was also evaluated. Colostrum samples were collected from 32 healthy women, and a human colostrum pool was prepared. Enterotoxigenic E. coli strains expressing colonization factor antigens I and II were utilized. The colostrum pool and individual samples showed variable antienterotoxigenic E. coli immunoglobulin A titers, that were reactive with colonization factor antigen I and CS1/CS3 (colonization factor antigen II). The human colostrum pool and individual samples inhibited enterotoxigenic E. coli colonization factor antigen I and II adhesion to Caco-2 cells, at variable levels, and this ability was a result of immunoglobulin A antibodies reactive to these colonization factors. The immunoglobulin A-depleted pool lost this inhibitory ability. As bacterial adhesion is the initial mechanism of enterotoxigenic E. coli infection, breast-feeding could protect the offspring against diarrhea caused by this agent.

摘要

腹泻是发展中国家社会经济水平较低的婴儿以及前往这些地区的旅行者发病和死亡的重要原因。产肠毒素大肠杆菌菌株表达两组与毒力相关的因子:肠毒素(热稳定毒素或热不稳定毒素)和定植因子。研究表明,母乳喂养可保护婴儿免受诸如腹泻等传染病的侵害,因为母乳含有多种免疫成分。本研究的目的是分析人初乳中免疫球蛋白A对定植因子抗原I和II的反应性。还评估了初乳预防产肠毒素大肠杆菌黏附于Caco-2细胞的能力。从32名健康女性收集初乳样本,并制备人初乳混合液。使用表达定植因子抗原I和II的产肠毒素大肠杆菌菌株。初乳混合液和个体样本显示出不同的抗产肠毒素大肠杆菌免疫球蛋白A滴度,这些滴度与定植因子抗原I和CS1/CS3(定植因子抗原II)反应。人初乳混合液和个体样本在不同程度上抑制产肠毒素大肠杆菌定植因子抗原I和II黏附于Caco-2细胞,这种能力是对这些定植因子有反应的免疫球蛋白A抗体的结果。去除免疫球蛋白A的混合液失去了这种抑制能力。由于细菌黏附是产肠毒素大肠杆菌感染的初始机制,母乳喂养可保护后代免受该病原体引起的腹泻。

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