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Membrane Transport 15 Membrane Transport To understand how small The cytoplasmic membrane is a hydrophobic barrier that limits the entry Goal molecules move across and exit of molecules from the cell. Nonetheless, living cells require the membranes. exchange of many types of molecules between the cytoplasm and the outside environment. Cells need nutrients and inorganic ions, and they Objectives must remove waste products and expel other ions. This chapter focuses on After this chapter, you should be able to membrane impermeability and on membrane proteins that facilitate and • explain why particular molecules can control the transport of molecules across the lipid bilayer with a special or cannot cross the membrane. emphasis on ions (e.g., K+ and Na+). • distinguish between passive and active transport and between channels and Membrane bilayers are permeable to small, uncharged molecules carrier proteins. • explain how the bacterial potassium Mome molecules can freely diffuse across membranes (Figure 1). Such channel discriminates against sodium molecules are generally small and uncharged. These include small ions. hydrophobic molecules composed mostly of nonpolar bonds. It is • define the electrochemical gradient and energetically favorable for such molecules to enter the membrane. Larger explain how it influences the movement hydrophobic molecules can also enter the membrane, but due to their larger of ions across the membrane. surface areas, they are unlikely to cross the membrane and be released into • explain how membrane potential is the aqueous environment on the opposite side. Gaseous molecules, such as established by K+ leak channels and the O2, N2, and CO2, can also diffuse freely across lipid bilayers. Na+/K+ ATPase. Polar molecules are less free to diffuse across membranes. In order for a • describe the function of membrane polar molecule to cross a membrane, it must break its interactions with potential in transporting glucose, maintaining osmolarity, and in water as it enters the hydrophobic interior of the bilayer. Breaking these generating an action potential in interactions is energetically unfavorable and requires an input of energy. neurons. Chapter 15 Membrane Transport 2 Figure 1 The membrane imposes small small, uncharged large, uncharged hydrophobic ions a barrier to the diffusion of large polar molecules polar molecules polar molecules and all charged molecules - + molecules amino acids HCO3 Na O2 CO2 H2O glycerol glucose H+ K+ Ca2+ N2 benzene ethanol nucleotides Cl- Mg2+ impermeable impermeable movement across membranes requires freely diuse through lipid bilayer specialized transport proteins This energetic cost creates a kinetic barrier that impedes polar molecules from crossing the membrane. Amino acids, glucose, and nucleotides are examples of polar molecules that are too large to diffuse across membrane bilayers. Molecules such as these require specialized transport proteins that facilitate their transport through membranes. If a polar molecule is small enough, however, the kinetic barrier is not great enough to prevent diffusion from occurring on a time scale relevant to that of living systems. Consequently, small polar molecules, such as water, ammonia, and ethanol, are capable of diffusing across membranes. Even though water can diffuse through membranes, the kinetic barrier associated with its diffusion slows the process. As a result, some cells use specialized membrane proteins called aquaporins to accelerate the rate at which water crosses the membrane. Despite their small size, charged ions cannot easily diffuse through membrane bilayers. Ions form ion-dipole interactions with water that are both strong and numerous. For example, both sodium (Na+) and potassium (K+) ions associate with six water molecules at a time in aqueous solutions. Hydrated ions such as these are simply too big to cross the membrane. Consequently, the ion must become dehydrated (i.e., lose its water molecules) to cross the membrane. However, breaking the interactions between an ion and its water molecules is highly unfavorable (Figure 2), creating an insurmountable kinetic barrier that prevents ions from freely diffusing through membranes. Since ions cannot freely diffuse through membranes, cells use specialized membrane proteins to desolvate ions and facilitate transport, as we will explain. Transport via diffusion requires a concentration gradient The free diffusion of a molecule through a membrane is bidirectional, meaning that the molecule can both enter and exit a cell. As a consequence, a concentration gradient across the membrane is needed to produce a net change in the concentrations of molecules inside and outside of the Chapter 15 Membrane Transport 3 Figure 2 Ion dehydration is highly unfavorable hydrated ion water dehydrated ion K+ K+ dehydration ∆Grxn >> 0 cell (Figure 3). Consider a cell in which a diffusible molecule is at equal concentrations on both sides of the membrane. Diffusion of molecules into the cell will occur at the same rate as diffusion out of the cell. Such a system is at equilibrium; consequently, the concentrations of the molecule on both sides of the membrane will remain the same. In contrast, consider a cell in which a diffusible molecule is at a higher concentration outside the cell than inside. The molecule will still diffuse in both directions, but since the molecule is more concentrated outside the cell, the rate at which molecules diffuse into the cell will be greater than the rate at which they diffuse out. Because the rates in and out are different, the concentrations of the molecule on both sides of the membrane will change over time. The concentration of molecules outside of the cell will decrease and the concentration inside will increase. The concentration gradient provides the energy (entropic energy) that drives the net movement of molecules from one side of the membrane to the other. Some molecules cannot freely diffuse across the membrane because they are too large or carry a charge. Nonetheless, and for certain molecules, diffusion is still possible due to a process known aspassive transport. Net diusion into cell No net diusion Net diusion out of cell ratein > rateout ratein = rateout ratein < rateout extracellular space cytoplasm Figure 3 Net diffusion requires a concentration gradient Net transport via diffusion only occurs when there is a concentration gradient. The left and right panels show cases in which the solute (represented by a blue sphere) is at different concentrations on opposite sides of the membrane (i.e., there is a concentration gradient). In the left panel, net diffusion is possible because the rate of diffusion into the cell is faster than the rate of diffusion out of the cell, and the concentration of the solute outside the cell will decrease over time as the concentration of solute inside the cell increases. In the right panel, net diffusion out of the cell is faster than the rate of diffusion into the cell, and the concentration of the solute outside the cell will increase over time as the concentration of solute inside the cell decreases. The center panel shows a system at equilibrium in which there is no concentration gradient; consequently, the rate of diffusion into the cell equals the rate of diffusion out of the cell, and the concentration of solute on both sides of the membrane remains constant. In other words, in the left and right panels molecules move spontaneously because energy is released as disorder increases; the middle panel shows the case with maximal disorder of the molecules. Chapter 15 Membrane Transport 4 Passive transport involves specialized membrane proteins that facilitate movement across the membrane. One kind of passive transport is mediated by channel proteins, which create protein-lined pores through which ions that are the right size and charge can freely diffuse. Another kind is mediated by carrier proteins. Carrier proteins are also pores, but the pores have gates that can open and close on one or the other side of the membrane. These carrier proteins also have binding sites for specific molecules. The binding of a molecule to such a site triggers a conformational change in the carrier protein that opens the gate and releases the molecule on the opposite side of the membrane from which it entered the pore. As with diffusion through a channel, carrier protein-mediated passive transport requires a concentration gradient for the net movement of molecules. The structure of the potassium ion channel reveals how it can be fast and selective A well-understood example of a channel protein is the potassium ion channel of bacteria, whose structure was solved by X-ray crystallography. The potassium channel operates at great speed and with high selectivity. It allows the movement of tens of millions of ions per second while discriminating against sodium, which has very similar properties (being a singly charged ion), by a factor of 1,000. This potassium channel is composed of four identical polypeptide subunits, each consisting of three α-helices (Figure 4). Two long helices in each subunit are transmembrane helices, whereas the third and shorter pore helix is inside the bilayer but does not span the entire membrane. (A) (B) Figure 4 The potassium channel is composed of four polypeptides that assemble to form a pore at their center Shown is the structure of a bacterial potassium channel (PDB 1BL8). (A) Side view. (B) Top view. Each of the four polypeptide chains is individually colored. Chapter 15 Membrane Transport 5 Figure 5 The potassium channel extracellular selectivity selectivity transports potassium ions quickly space – – – – – loop lter – and selectively – – – – – – – – – – The potassium channel is a tetramer, but – – for simplicity only two subunits are shown. The exterior of the channel is coated with O K+ O negatively charged amino acid residues that O O pore help attract positive potassium ions. These O K+ O helix ions enter the aqueous vestibule, where they remain hydrated. The selectivity loops of the four subunits come together to create a narrow passage called the selectivity filter. To pass through the selectivity filter, ions K+ must become dehydrated.
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