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葡萄糖和膳食维生素E对大鼠红细胞中过氧化氢酶失活的保护作用。

Glucose and dietary vitamin E protection against catalase inactivation in the red cells of rats.

作者信息

Chow C K

出版信息

Int J Vitam Nutr Res. 1980;50(4):364-9.

PMID:7203846
Abstract

One-month-old male Sprague-Dawley rats were fed a basal vitamin E deficient diet with or without 50 ppm vitamin E supplementation for 7 months. The washed red cells were suspended in a saline-phosphate buffer, pH 7.4, that contained either 0, 0.011 or 0.055 M glucose and were incubated at 37 C with constant shaking. Catalase activity in the red cells of vitamin E deficient rats was decreased 74% (P less than 0.001) at the end of the 22-hour incubation, and only 9% of the initial value was retained at the end of 46 hours. In the red cells of the vitamin E supplemented group, 82% and 48% of catalase activity was retained at the end of 22 and 46 hours, respectively. Glucose in the medium significantly increased catalase activity during the early hours of incubation and retarded the enzyme inactivation at the end of 22 and 46 hours in both groups of animals. The activities of superoxide dismutase and glutathione peroxidase were not significantly altered by the presence of glucose or by the status of dietary vitamin E during the incubation. The results suggest that both glucose and dietary vitamin E provide protection against inactivation of catalase under the experimental conditions.

摘要

将1月龄雄性斯普拉格-道利大鼠用基础维生素E缺乏饮食喂养7个月,饮食中添加或不添加50 ppm维生素E。洗涤后的红细胞悬浮于pH 7.4的磷酸盐缓冲盐溶液中,该溶液含有0、0.011或0.055 M葡萄糖,并在37℃下持续振荡孵育。在22小时孵育结束时,维生素E缺乏大鼠红细胞中的过氧化氢酶活性降低了74%(P<0.001),在46小时结束时仅保留初始值的9%。在补充维生素E组的红细胞中,在22小时和46小时结束时分别保留了82%和48%的过氧化氢酶活性。培养基中的葡萄糖在孵育早期显著增加了过氧化氢酶活性,并在两组动物的22小时和46小时结束时延缓了酶的失活。在孵育过程中,葡萄糖的存在或饮食中维生素E的状态对超氧化物歧化酶和谷胱甘肽过氧化物酶的活性没有显著影响。结果表明,在实验条件下,葡萄糖和饮食中的维生素E均能提供保护,防止过氧化氢酶失活。

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