Institute of Molecular and Cellular Biology, Leeds University, Leeds, United Kingdom.
J Phys Chem B. 2011 Oct 27;115(42):12315-24. doi: 10.1021/jp208585r. Epub 2011 Oct 3.
A fundamental problem in the analysis of protein folding and other complex reactions is the determination of the reaction free energy landscape. The current experimental techniques lack the necessary spatial and temporal resolution to construct such landscapes. The properties of the landscapes can be probed only indirectly. Simulation, assuming that it reproduces the experimental dynamics, can provide the necessary spatial and temporal resolution. It is, arguably, the only way for direct rigorous construction of the quantitatively accurate free energy landscapes. Here, such landscape is constructed from the equilibrium folding simulation of FIP35 protein reported by Shaw et al. Science 2010, 330, 341-346. For the dynamics to be accurately described as diffusion on the free energy landscape, the choice of reaction coordinates is crucial. The reaction coordinate used here is such that the dynamics projected on it is diffusive, so the description is consistent and accurate. The obtained landscape suggests an alternative interpretation of the simulation, markedly different from that of Shaw et al. In particular, FIP35 is not an incipient downhill folder, it folds via a populated on-pathway intermediate separated by high free energy barriers; the high free energy barriers rather than landscape roughness are a major determinant of the rates for conformational transitions; the preexponential factor of folding kinetics 1/k(0) ∼ 10 ns rather than 1 μs.
蛋白质折叠和其他复杂反应分析中的一个基本问题是确定反应自由能景观。当前的实验技术缺乏构建此类景观所需的空间和时间分辨率。只能间接地探测景观的性质。模拟,假设它再现了实验动力学,可以提供必要的空间和时间分辨率。可以说,这是直接严格构建定量准确的自由能景观的唯一方法。在这里,从 Shaw 等人报道的 FIP35 蛋白的平衡折叠模拟中构建了这样的景观。Science 2010, 330, 341-346。为了使动力学能够准确地描述为自由能景观上的扩散,反应坐标的选择至关重要。这里使用的反应坐标是这样的,即投影在其上的动力学是扩散的,因此描述是一致和准确的。获得的景观提出了对模拟的另一种解释,与 Shaw 等人的解释明显不同。特别是,FIP35 不是一个初始的下坡折叠物,它通过由高自由能势垒隔开的充满途径的中间产物折叠;高自由能势垒而不是景观粗糙度是构象转变速率的主要决定因素;折叠动力学的指数前因子 1/k(0)∼10 ns 而不是 1 μs。