Figure Q11–1 Microvilli of intestinal epithelial cells in profile and cross section (Problem 11–7). (Left panel, from Rippel Electron Microscope 11–6 Ion transporters are “linked” together—not physi-Facility, Dartmouth College; right panel, from David Burgess.) cally, but as a consequence of their actions. For example, cells can raise their intracellular pH, when it becomes too acidic, by exchanging external Na+ for internal H+, using 11–8 According to Newton’s laws of motion, an ion a Na+–H+ antiporter. The change in internal Na+ is then exposed to an electric field in a vacuum would experience redressed using the Na+-K+ pump. a constant acceleration from the electric driving force, just a. Can these two transporters, operating together, as a falling body in a vacuum constantly accelerates due to normalize both the H+ and the Na+ concentrations inside gravity. In water, however, an ion moves at constant velocthe cell? ity in an electric field. Why do you suppose that is?
21.4 nm
Figure Q11–2 A “ball” tethered by a “chain” to a voltage-gated K+ channel (Problem 11–9).
11–9 In a subset of voltage-gated K+ channels, the N-terminus of each subunit acts like a tethered ball that occludes the cytoplasmic end of the pore soon after it opens, thereby inactivating the channel. This “ball-andchain” model for the rapid inactivation of voltage-gated K+ channels has been elegantly supported for the shaker K+ channel from Drosophila melanogaster. (The shaker K+ channel in Drosophila is named after a mutant form that causes excitable behavior—even anesthetized flies keep twitching.) Deletion of the N-terminal amino acids from the normal shaker channel gives rise to a channel that opens in response to membrane depolarization, but stays open instead of rapidly closing as the normal channel does. A peptide (MAAVAGLYGLGEDRQHRKKQ) that corresponds to the deleted N-terminus can inactivate the open channel at 100 µM.
Is the concentration of free peptide (100 µM) that is required to inactivate the defective K+ channel anywhere near the local concentration of the tethered ball on a normal channel? Assume that the tethered ball can explore a hemisphere [volume = (2/3)πr3] with a radius of 21.4 nm, which is the length of the polypeptide “chain” (Figure Q11–2). Calculate the concentration for one ball in this hemisphere. How does that value compare with the concentration of free peptide needed to inactivate the channel?
11–10 The giant axon of the squid (Figure Q11–3) occupies a unique position in the history of our understanding of cell membrane potentials and nerve action. When an electrode is stuck into an intact giant axon, the membrane potential registers –70 mV. When the axon, suspended in a bath of seawater, is stimulated to conduct a nerve impulse, the membrane potential changes transiently from –70 mV to +40 mV.
Table Q11–1 Ionic composition of seawater and of the cytosol in the squid giant axon (Problem 11–10).
For univalent ions and at 20°C (293 K), the Nernst equation reduces to where Co and Ci are the concentrations outside and inside, respectively.
Using this equation, calculate the potential across the resting membrane (1) assuming that it is due solely to K+ and (2) assuming that it is due solely to Na+. (The Na+ and K+ concentrations in the axon cytosol and in seawater are given in Table Q11–1.) Which calculation is closer to the measured resting potential? Which calculation is closer to the measured action potential? Explain why these assumptions approximate the measured resting and action potentials.