Figure 11–23 The ionic basis of a membrane potential. A small flow of inorganic ions through an ion channel carries sufficient charge to cause a large change in the membrane potential. The ions that give rise to the membrane potential lie in a thin (< 1 nm) surface layer close to the membrane, held there by their electrical attraction to their oppositely charged counterparts (counterions) on the other side of the membrane. For a typical cell, 1 microcoulomb of charge (6 × 1012 monovalent ions) per square centimeter of membrane, transferred from one side of the membrane to the other, changes the membrane potential by roughly 1 V. This means, for example, that in a spherical cell of diameter 10 μm, the number of K+ ions that have to flow out to alter the membrane potential by 100 mV is only about 1/100,000 of the total number of K+ ions in the cytosol. This amount is so minute that the intracellular K+ concentration remains virtually unchanged.
618 Chapter 11:  Membrane Transport of Small Molecules and the Electrical Properties of Membranes
Figure 11–24 The structure of a bacterial K+ channel. (A) The transmembrane α helices from only two of the four identical subunits are shown. From the cytosolic side, the pore (schematically shaded in blue) opens up into a vestibule in the middle of the membrane. The pore vestibule facilitates transport by allowing the K+ ions to remain hydrated even though they are more than halfway across the membrane. The narrow selectivity filter of the pore links the vestibule to the outside of the cell. Carbonyl oxygens line the walls of the selectivity filter and form transient binding sites for dehydrated K+ ions. Two K+ ions occupy different sites in the selectivity filter, while a third K+ ion is located in the center of the vestibule, where it is stabilized by electrical interactions with the more negatively charged ends of the pore helices. The ends of the four short “pore helices” (only two of which are shown) point precisely toward the center of the vestibule, thereby guiding K+ ions into the selectivity filter (Movie 11.8). (B) Peptide bonds have an electric dipole, with more negative charge accumulated at the oxygen of the C=O bond and at the nitrogen of the  N–H bond. In an α helix, hydrogen bonds (red) align the dipoles. As a consequence, every α helix has an electric dipole along its axis, resulting from summation of the dipoles of the individual peptide bonds, with a more negatively charged C-terminal end (δ–) and a more positively charged N-terminal end (δ+). (A, adapted from D.A. Doyle et al., Science 280:69–77, 1998.) the cell where K+ ions exit from the channel (Figure 11–24). The polypeptide chain that connects the two transmembrane helices forms a short αhelix (the pore helix) and a crucial loop that protrudes into the wide section of the cone to form the selectivity filter. The selectivity loops from the four subunits form a short, rigid, narrow pore, which is lined by the carbonyl oxygen atoms of their polypeptide backbones. Because the selectivity loops of all known K+ channels have similar amino acid sequences, it is likely that they form a closely similar structure.
The structure of the selectivity filter explains the ion selectivity of the channel. A K+ ion must lose almost all of its bound water molecules to enter the filter, where it interacts instead with the carbonyl oxygens lining the filter; the oxygens are rigidly spaced at the exact distance to accommodate a K+ ion. A Na+ ion, in contrast, cannot enter the filter because the carbonyl oxygens are too far away from the smaller Na+ ion to compensate for the energy expense associated with the loss of water molecules required for entry (Figure 11–25).