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α-乳白蛋白纤维化的构象先决条件。

Conformational prerequisites for alpha-lactalbumin fibrillation.

作者信息

Goers John, Permyakov Sergei E, Permyakov Eugene A, Uversky Vladimir N, Fink Anthony L

机构信息

Department of Chemistry and Biochemistry, University of California, Santa Cruz, California 95064, USA.

出版信息

Biochemistry. 2002 Oct 15;41(41):12546-51. doi: 10.1021/bi0262698.

Abstract

Bovine alpha-lactalbumin, a small acidic Ca(2+)-binding milk protein, formed amyloid fibrils at low pH, where it adopted the classical molten globule-like conformation. Fibrillation was accompanied by a dramatic increase in the beta-structure content and a characteristic increase in the thioflavin T fluorescence intensity. S-(Carboxymethyl)-alpha-lactalbumin, a disordered form of the protein with three out of four disulfide bridges reduced, was even more susceptible to fibrillation. Other partially folded conformations induced in the intact protein at neutral pH, either by the removal of Ca(2+) or by the binding of Zn(2+) to the Ca(2+)-protein complex, did not fibrillate, although Zn(2+)-loaded alpha-lactalbumin precipitated out of solution as amorphous aggregates. Our data suggest that the transformation of a protein into an essentially unfolded (thus, highly flexible) conformation is required for successful fibril formation, whereas more rigid (but still flexible) species may form amorphous aggregates.

摘要

牛α-乳白蛋白是一种酸性的小分子量钙结合乳蛋白,在低pH值下会形成淀粉样纤维,此时它呈现出经典的类熔球构象。纤维形成过程伴随着β-结构含量的显著增加以及硫黄素T荧光强度的特征性增强。S-(羧甲基)-α-乳白蛋白是该蛋白的一种无序形式,其四对二硫键中有三对被还原,它甚至更易形成纤维。在中性pH值下,通过去除Ca(2+) 或使Zn(2+) 与Ca(2+)-蛋白复合物结合,在完整蛋白中诱导产生的其他部分折叠构象不会形成纤维,尽管负载Zn(2+) 的α-乳白蛋白会以无定形聚集体的形式从溶液中沉淀出来。我们的数据表明,蛋白质成功形成纤维需要转变为基本未折叠(因此高度灵活)的构象,而更刚性(但仍灵活)的物种可能形成无定形聚集体。

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