Good M, Hollenstein R, Sadler P J, Vasák M
Biochemisches and Organisch-Chemisches Institut, Universität Zürich, Switzerland.
Biochemistry. 1988 Sep 6;27(18):7163-6. doi: 10.1021/bi00418a074.
The formation of two metal-thiolate clusters in rabbit liver metallothionein 2 (MT) has been examined by 113Cd NMR spectroscopy at pH 7.2 and 8.6. The chemical shifts of the 113Cd resonances developing in the course of apoMT titration with 113Cd(II) ions have been compared with those of fully metal occupied 113Cd7-MT. At pH 7.2 and at low metal occupancy (less than 4), a cooperative formation of the four-metal cluster (cluster A) occurs. Further addition of 113Cd(II) ions generates all the resonances of the three-metal cluster (cluster B) in succession, suggesting cooperative metal binding to this cluster also. In contrast, similar studies at pH 8.6, at low metal occupancy (less than 4), reveal a broad NMR signal centered at 688 ppm. This observation indicates that an entirely different protein structure exists. When exactly 4 equiv of 113Cd(II) are bound to apoMT, the 113Cd NMR spectrum changes to the characteristic spectrum of cluster A. Further addition of 113Cd(II) ions again leads to the cooperative formation of cluster B. These results stress the determining role of the cluster A domain on the overall protein fold. The observed pH dependence of the cluster formation in MT can be rationalized by the different degree of deprotonation of the cysteine residues (pKa approximately 8.9), i.e., by the difference in the Gibbs free energy required to bind Cd(II) ions to the thiolate ligands at both pH values.
通过在pH 7.2和8.6条件下的¹¹³Cd NMR光谱研究了兔肝金属硫蛋白2(MT)中两个金属硫醇盐簇的形成。将脱辅基MT用¹¹³Cd(II)离子滴定过程中¹¹³Cd共振的化学位移与完全被金属占据的¹¹³Cd₇-MT的化学位移进行了比较。在pH 7.2和低金属占有率(小于4)时,四金属簇(簇A)协同形成。进一步添加¹¹³Cd(II)离子会依次产生三金属簇(簇B)的所有共振,这也表明金属与该簇的结合具有协同性。相比之下,在pH 8.6和低金属占有率(小于4)下的类似研究显示,在688 ppm处有一个宽的NMR信号。这一观察结果表明存在一种完全不同的蛋白质结构。当恰好4当量的¹¹³Cd(II)与脱辅基MT结合时,¹¹³Cd NMR光谱变为簇A的特征光谱。进一步添加¹¹³Cd(II)离子再次导致簇B的协同形成。这些结果强调了簇A结构域在整个蛋白质折叠中的决定性作用。MT中观察到的簇形成对pH的依赖性可以通过半胱氨酸残基不同程度的去质子化(pKa约为8.9)来解释,即通过在两个pH值下将Cd(II)离子与硫醇盐配体结合所需的吉布斯自由能差异来解释。