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The topomer-sampling model of protein folding

机译:蛋白质折叠的拓扑异构体采样模型

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Clearly, a protein cannot sample all of its conformations on an in vivo folding timescale (<1 s). To investigate how the conformational dynamics of a protein can accommodate subsecond folding time scales, we introduce the concept of the native topomer, which is the set of all structures similar to the native structure (obtainabIe from the native structure through local backbone coordinate transformations that do not disrupt the covalent bonding of the peptide backbone). We have developed a computational procedure for estimating the number of distinct topomers required to span all conformations (compact and semicompact) for a polyPeptide of a given length. For 100 residues, we find approximately equals 3 x 10(7) distinct topomers. Based on the distance calculated between different topomers, we estimate that a l00-residue polyPeptide diffesively samples one topomer every approximately equals 3 ns. Hence, a 100-residue protein can filld its native topomer by raudom sampling in just approximately 100 ms. These results suggest that subsecond folding of modest-sized, single-domain proteins can be accomplished by a two-stage process of (i) topomer diffesion: random, diffusive sampling of the 3 x 10(7) distinct topomers to find the native topomer ( approximately equals 0.1 s), followed by (ii) intratopomer ordering: nonrandom, local conformational rearrangements within the native topomer to settle into the precise native state.
机译:显然,蛋白质无法在体内折叠时间尺度(<1 s)上取样其所有构象。为了研究蛋白质的构象动力学如何适应亚秒级折叠时间尺度,我们引入了天然拓扑异构体的概念,它是与天然结构相似的所有结构的集合(通过进行局部骨干坐标转换从天然结构中获得不会破坏肽骨架的共价键)。我们已经开发出一种计算程序,用于估算跨越给定长度的多肽的所有构象(紧凑和半紧凑)所需的不同拓扑异构体的数量。对于100个残基,我们发现大约等于3 x 10(7)个不同的拓扑异构体。根据不同拓扑异构体之间的距离计算,我们估计一个100残基的多肽大约每3 ns扩散地采样一个拓扑异构体。因此,一个100残基的蛋白质可以在大约100毫秒内通过raudom采样来填充其天然拓扑异构体。这些结果表明,中等大小的单域蛋白的亚秒级折叠可以通过(i)拓扑异构体扩散的两个阶段过程来完成:对3 x 10(7)个不同的拓扑异构体进行随机扩散采样以找到天然拓扑异构体(约等于0.1 s),然后进行(ii)拓扑异构体内部排序:在天然拓扑异构体中进行非随机,局部构象重排,以沉降为精确的天然状态。

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