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通过随机卷曲或α-螺旋中的带电残基实现兰尼碱受体1(RyR1)C末端的同型二聚化。

Homo-dimerization of RyR1 C-terminus via charged residues in random coils or in an alpha-helix.

作者信息

Lee Eun Hui, Allen Paul D

机构信息

Department of Physiology, College of Medicine, The Catholic University of Korea, Seoul 137-701, Korea.

出版信息

Exp Mol Med. 2007 Oct 31;39(5):594-602. doi: 10.1038/emm.2007.65.

Abstract

To investigate the mechanism by which the C-terminus (4,938-5,037) of the ryanodine receptor 1 (RyR1) homo-tetramerizes, forming a functional Ca(2+)-release channel, the structural requirements for the tetramerization were studied using site-directed mutagenesis. Alanine-substitutions at five charged residues, E4976, H5003, D5026, E5033 and D5034, significantly decreased the formation of homo-dimers (reduced by >50%). Interaction between the C-terminus and cytoplasmic loop I (4,821-4,835) required two positively charged residues, H4832 and K4835. Based on the predicted protein secondary structures, all seven charged residues are located in random coils. Paired alanine-substitutions at six negatively charged residues (E4942A/D4953A, D4945A/E4952A and E4948A/ E4955A) of the alpha-helix (4,940-4,956) in the C-terminus increased homo-dimerization. Therefore, the homo-tetramerization of RyR1 may be mediated by intra- and/or inter-monomer electrostatic interactions among the C-terminal charged residues in random coils or in an alpha-helix.

摘要

为了研究兰尼碱受体1(RyR1)同四聚体的C末端(4938 - 5037)形成功能性Ca(2+)释放通道的机制,我们使用定点诱变研究了四聚化的结构要求。对五个带电荷残基E4976、H5003、D5026、E5033和D5034进行丙氨酸取代,显著降低了同二聚体的形成(减少超过50%)。C末端与胞质环I(4821 - 4835)之间的相互作用需要两个带正电荷的残基H4832和K4835。根据预测的蛋白质二级结构,所有七个带电荷残基都位于无规卷曲中。对C末端α螺旋(4940 - 4956)中的六个带负电荷残基(E4942A/D4953A、D4945A/E4952A和E4948A/E4955A)进行成对丙氨酸取代增加了同二聚化。因此,RyR1的同四聚化可能由无规卷曲或α螺旋中C末端带电荷残基之间的单体内部和/或单体间静电相互作用介导。

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