Levitt D G
Biophys J. 1978 May;22(2):209-19. doi: 10.1016/S0006-3495(78)85485-X.
The electrostatic energy profile of one, two, or three ions in an aqueous channel through a lipid membrane is calculated. It is shown that the previous solution to this problem (based on the assumption that the channel is infinitely long) significantly overestimates the electrostatic energy barrier. For example, for a 3-A radius pore, the energy is 16 kT for the infinite channel and 6.7 kT for an ion in the center of a channel 25 A long. The energy as a function of the position of the ion is also determined. With this energy profile, the rate of crossing the membrane (using the Nernst-Planck equation) was estimated and found to be compatible with the maximum conductance observed for the gramicidin A channel. The total electrostatic energy (as a function of position) required to place two or three ions in the channel is also calculated. The electrostatic interaction is small for two ions at opposite ends of the channel and large for any positioning of the three ions. Finally, the gradient through the channel of an applied potential is calculated. The solution to these problems is based on solving an equivalent problem in which an appropriate surface charge is placed on the boundary between the lipid and aqueous regions. The magnitude of the surface charge is obtained from the numerical solution for a system of coupled integral equations.
计算了通过脂质膜的水通道中一个、两个或三个离子的静电能分布。结果表明,此前针对该问题的解决方案(基于通道无限长的假设)显著高估了静电能垒。例如,对于半径为3埃的孔,无限通道的能量为16kT,而对于位于25埃长通道中心的离子,能量为6.7kT。还确定了能量作为离子位置的函数。利用这种能量分布,估算了穿过膜的速率(使用能斯特-普朗克方程),发现其与短杆菌肽A通道观察到的最大电导率相符。还计算了在通道中放置两个或三个离子所需的总静电能(作为位置的函数)。对于位于通道两端的两个离子,静电相互作用较小,而对于三个离子的任何排列,静电相互作用都较大。最后,计算了施加电位在通道中的梯度。这些问题的解决方案基于求解一个等效问题,即在脂质和水区域之间的边界上放置适当的表面电荷。表面电荷的大小通过耦合积分方程组的数值解获得。