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在肝细胞中,葡萄糖必须被磷酸化以激活糖原合酶,但不是为了使糖原磷酸化酶失活。

Glucose has to be phosphorylated to activate glycogen synthase, but not to inactivate glycogen phosphorylase in hepatocytes.

作者信息

Carabaza A, Ciudad C J, Baqué S, Guinovart J J

机构信息

Departament de Bioquímica i Biologia Molecular, Universitat Autònoma de Barcelona, Spain.

出版信息

FEBS Lett. 1992 Jan 20;296(2):211-4. doi: 10.1016/0014-5793(92)80381-p.

Abstract

2-Deoxyglucose and 5-thioglucose, in the same fashion as glucose, cause the inactivation of the rat hepatocyte glycogen phosphorylase and the activation of glycogen synthase. However, 6-deoxyglucose and 1,5-anhydroglucitol inactivate phosphorylase without increasing the activation state of glycogen synthase. With 3-O-methylglucose no changes in the activity of these enzymes occurred. These results prove that while glucose is the molecule that triggers the inactivation of phosphorylase, glucose 6-phosphate is the signal for glucose synthase activation and that a metabolite control of the activation state of glycogen synthase is operative in hepatocytes.

摘要

2-脱氧葡萄糖和5-硫代葡萄糖与葡萄糖一样,可导致大鼠肝细胞糖原磷酸化酶失活并激活糖原合酶。然而,6-脱氧葡萄糖和1,5-脱水葡萄糖醇可使磷酸化酶失活,而不会增加糖原合酶的激活状态。3-O-甲基葡萄糖对这些酶的活性没有影响。这些结果证明,虽然葡萄糖是触发磷酸化酶失活的分子,但6-磷酸葡萄糖是葡萄糖合酶激活的信号,并且糖原合酶激活状态的代谢物控制在肝细胞中起作用。

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