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结合苯酚的三方晶系R6胰岛素六聚体的结构。

The structure of a rhombohedral R6 insulin hexamer that binds phenol.

作者信息

Smith G D, Dodson G G

机构信息

Medical Foundation of Buffalo, Inc., New York 14203.

出版信息

Biopolymers. 1992 Apr;32(4):441-5. doi: 10.1002/bip.360320422.

DOI:10.1002/bip.360320422
PMID:1623140
Abstract

Different hexameric forms of insulin have been crystallized from a variety of conditions. In the presence of 1% phenol, 1.0 M sodium chloride, and at a pH of 8.5, a rhombohedral form is produced with two monomers in the asymmetric unit, space group R3, a = 79.92 A and c = 40.39 A. The structure has been solved and refined, using data between 8.0 and 2.5 A resolution, to a residual of 0.157. Each of the monomers adopts an R conformation, that is residues B1-B8 are alpha-helical. As a result of the T to R transition, an elliptical cavity is created between symmetry-related monomers and is occupied by a phenol molecule. A region of density within bonding distance to one of the zinc ions has been interpreted as an additional phenol molecule.

摘要

不同的六聚体形式的胰岛素已在多种条件下结晶。在1%苯酚、1.0 M氯化钠存在且pH为8.5的情况下,会产生一种菱面体形式,其不对称单元中有两个单体,空间群为R3,a = 79.92 Å,c = 40.39 Å。利用8.0至2.5 Å分辨率的数据对该结构进行了解析和精修,残余误差为0.157。每个单体都采用R构象,即残基B1 - B8为α螺旋。由于T到R的转变,在对称相关的单体之间形成了一个椭圆形空腔,该空腔被一个苯酚分子占据。与其中一个锌离子距离处于键合距离内的一个密度区域被解释为另一个苯酚分子。

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