CH 10 Membrane Structure: Lipid Bilayer

If a lipid raft is typically 70 nm in diameter and each lipid molecule has a diameter of 0.5 nm, about how many lipid molecules would there be in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules per protein molecule (50% protein by mass), how many proteins would be in a typical raft? (Neglect the loss of lipid from the raft that would be required to accommodate the protein.)

A raft 70 nm in diameter would have an area of 3.8 × 103 nm2 (3.14 × 352), and a lipid molecule 0.5 nm in diameter would have an area of 0.20 nm2 (3.14 × 0.252). Thus, there would be about 19,000 lipid molecules per monolayer of raft (3.8 × 103/0.20 = 19,000), and about 38,000 molecules in the raft bilayer. At a ratio of 50 lipids per protein, a raft would accommodate about 760 protein molecules. The true ratio of lipids to proteins in a raft is unknown.

Predict which of the following organisms will have the highest percentage of unsaturated fatty acid chains in their membranes. Explain your answer. A. Antarctic fish B. Desert iguana C. Human being D. Polar bear E. Thermophilic bacterium

A. Antarctic fish, which are cold-blooded, live in freezing waters. In order to maintain an appropriate fluidity of their membranes under such extreme conditions, they require a higher proportion of unsaturated fatty acid chains in their membranes to keep them from freezing solid like a stick of margarine. Polar bears also live in extreme cold, but they are warm- blooded and maintain a high internal temperature; thus, they have no special requirement for unsaturated fatty acids in their membranes.

A classic paper studied the behavior of lipids in the two monolayers of a membrane by labeling individual molecules with nitroxide groups, which are stable free radicals (Figure 10-3). These spin-labeled lipids can be detected by electron spin resonance (ESR) spectroscopy, a technique that does not harm living cells. Spin-labeled lipids are introduced into small lipid vesicles, which are then fused with cells, thereby transferring the labeled lipids into the plasma membrane. The two spin-labeled phospholipids shown in Figure 10-3 were incorporated into intact human red blood cell membranes in this way. To determine whether they were introduced equally into the two monolayers of the bilayer, ascorbic acid (vitamin C), which is a water-soluble reducing agent that does not cross membranes, was added to the medium to destroy any nitroxide radicals exposed on the outside of the cell. The ESR signal was followed as a function of time in the presence and absence of ascorbic acid as indicated in Figure 10-4A and B. A. Ignoring for the moment the difference in extent of loss of ESR signal, offer an explanation for why phospholipid 1 (Figure 10-4A) reacts faster with ascorbate than does phospholipid 2 (Figure 10-4B). Note that phospholipid 1 reaches a plateau in about 15 minutes, whereas phospholipid 2 takes almost an hour.

A. The difference in rate of loss of the ESR signals is due to the location of the nitroxide radical on the two phospholipids. The nitroxide radical in phospholipid 1 is on the head group and is therefore in direct contact with the external medium. Thus, it can react quickly with ascorbate. The nitroxide radical in phospholipid 2 is attached to a fatty acid chain and is therefore partially buried in the interior of the membrane. As a consequence, it is less accessible to ascorbate and is reduced more slowly.

The asymmetric distribution of phospholipids in the two monolayers of the plasma membrane implies that very little spontaneous flip-flop occurs or, alternatively, that any spontaneous flip-flop is rapidly corrected by phospholipid translocators that return phospholipids to their appropriate monolayer. The rate of phospholipid flip-flop in the plasma membrane of intact red blood cells has been measured to decide between these alternatives. One experimental measurement used the same two spin-labeled phospholipids described in Problem 10-23 (see Figure 10-3). To measure the rate of flip-flop from the cytoplasmic monolayer to the outer monolayer, red cells with spin-labeled phospholipids exclusively in the cytoplasmic monolayer were incubated for various times in the presence of ascorbate and the loss of ESR signal was followed. To measure the rate of flip-flop from the outer to the cytoplasmic monolayer, red cells with spin-labeled phospholipids exclusively in the outer monolayer were incubated for various times in the absence of ascorbate and the loss of ESR signal was followed. From the results in Figure 10-6, estimate the rate of flip-flop from the cytoplasmic to the outer monolayer, and from the outer to the cytoplasmic monolayer. A convenient way to express such rates is as the half-time of flip-flop—that is, the time it takes for half the phospholipids to flip-flop from one monolayer to the other.

A. The half-time for flip-flop in these experiments is the point at which 50% of the ESR signal is lost. For cells labeled in the cytoplasmic monolayer, the curve in Figure 10-6 suggests a half-time for flip-flop of about 7 hours. For cells labeled in the outer monolayer, the half-time of flip-flop is much longer but cannot be estimated reliably. These data indicate that the rate of flip-flop of phospholipids between the two monolayers of the plasma membrane in red cells is extremely low. Similar experiments using synthetic bilayers have given even longer times; in fact, in the best experiments, when great care was taken not to allow oxidation or other damage to the lipids, the rate of flip-flop was immeasurably low (less than once per month).

ganglioside

Any glycolipid having one or more sialic acid residues in its structure; especially abundant in the plasma membranes of nerve cells.

liposome

Artificial phospholipid bilayer vesicle formed from an aqueous suspension of phospholipid molecules.

Predict the properties of a lipid bilayer in which all of the hydrocarbon chains were saturated. What would be the properties if all of the hydrocarbon chains were unsaturated?

Bilayers formed by lipids with saturated hydrocarbon tails would be much less fluid. Whereas a normal lipid bilayer has the viscosity of olive oil, a bilayer made of lipids with saturated hydrocarbon tails would have the consistency of bacon fat. In contrast, bilayers formed by lipids with unsaturated hydrocarbon tails would be much more fluid. Also, because the lipids would pack together less well, there would be more gaps and the bilayer would be more permeable to small water-soluble molecules.

Using the information in this problem, propose a method to generate intact red cells that contain spin-labeled phospholipids exclusively in the cytoplasmic monolayer, and a method to generate cells spin-labeled exclusively in the outer monolayer.

C. One can make intact red cells with spin-labeled phospholipids exclusively in the cytoplasmic monolayer by introducing phospholipid 2 into the membrane and then incubating the red cells for 1 hour in the presence of ascorbate. Ascorbate reduces the lipids in the outer monolayer, leaving red cells that are labeled only in the cytoplasmic monolayer. Similarly, one can make intact red cells with spin-labeled phospholipids exclusively in the outer monolayer by introducing phospholipid 1 into the membrane and then incubating the red cells for 15 minutes in the absence of ascorbate. In this case, the spin-labeled lipids in the cytoplasmic monolayer are reduced by agents in the cytoplasm, leaving red cells that are labeled only in the outer monolayer.

Which one of the phospholipids listed below is present in very small quantities in the plasma membranes of mammalian cells? A. Phosphatidylcholine B. Phosphatidylethanolamine C. Phosphatidylinositol D. Phosphatidylserine E. Sphingomyelin

C. Phosphatidylinositol is a minor component of the phospholipids in the plasma membrane, yet it plays a very important role in cell signaling. All the other phospholipids are common components of plasma membranes and play important structural roles in membrane integrity.

Fluorescence resonance energy transfer (FRET) has been used to investigate the existence of lipid rafts in living cells. To test for the presence of lipid rafts by FRET, you use two different cell lines: one that expresses a glycosylphosphatidylinositol (GPI)-anchored form of the folate receptor, and one that expresses the transmembrane-anchored form. Folate receptors can be made fluorescent by addition of a fluorescent folate analog. Cells tagged in this way show variation in fluorescence intensity over their surface because of chance variations in the density of labeled receptors. This allows different densities of receptors to be analyzed by examining different places in the same cell. The proximity of labeled receptors can be determined by FRET, which depends on the distance between receptors. The ratio of FRET to direct fluorescence gives different expectations for dispersed receptors versus receptors that are clustered together, as depicted in Figure 10-5. Explain the basis for the difference between the graphs of these expectations.

For randomly dispersed receptors (see Figure 10-5A), FRET will depend critically on the concentration of the receptors in the membrane. At high density, there will be efficient FRET, but low FRET at low density. For receptors that are confined to microdomains such as lipid rafts (see Figure p10-5B), the overall fluorescence intensity will decrease with decreasing density of the rafts, but FRET, as a fraction of direct fluorescence, will remain constant.

amphiphilic

Having both hydrophobic and hydrophilic regions, as in a phospholipid or a detergent molecule.

What is meant by the term "2D fluid"?

In a two-dimensional fluid, the molecules are constrained to move in a plane; the molecules in a normal fluid can move in three dimensions.

The lipid bilayers found in cells are fluid, yet asymmetrical in the composition of the monolayers. Is this a paradox? Explain your answer.

It is not a paradox. The fluidity of the bilayer is strictly confined to one plane. The lipid molecules can diffuse laterally, but do not readily flip from one monolayer to the other. Specific types of lipid molecule remain in the monolayer they are inserted into, unless they are actively transferred by an enzyme—a phospholipid translocator (a flippase).

Five students in your class always sit together in the front row. This could be because (1) they really like each other or (2) nobody else in your class wants to sit next to them. Which explanation holds for the assembly of a lipid bilayer? Explain your answer. If the lipid bilayer assembled for the opposite reason, how would its properties differ?

Lipid bilayers assemble because the surrounding water molecules exclude the component lipids; thus, analogy (2) is the correct one. If bilayers formed because of attractive forces among the lipids—analogy (1)—the properties of the bilayer would likely be quite different. Molecules "attract" one another by forming specific bonds that hold them together. Such bonding among lipids would make the bilayer less fluid, perhaps even rigid, depending on the strength of the interaction

cholesterol

Lipid molecule with a characteristic four-ring steroid structure that is an important component of the plasma membranes of animal cells

Many snake venoms contain enzymes that cause red blood cells to lyse. Imagine that you have purified such an enzyme. When you add the purified enzyme to red blood cells, you find that in addition to cell lysis, choline with a phosphate group attached to it is released, as well as diacylglycerol (glycerol with two fatty acid chains attached). What molecule is cleaved by the enzyme to cause cell lysis?

Phosphatidylcholine is the phospholipid that is cleaved by your snake venom enzyme.

From what you learned about the behavior of the two spin-labeled phospholipids in Problem 10-23, deduce which one was used to label the cytoplasmic monolayer of the intact red blood cells, and which one was used to label the outer monolayer.

Phospholipid 2 was used to label the cytoplasmic monolayer, and phospholipid 1 was used to label the outer monolayer. As shown by the experiments in Figure 10-4B, phospholipid 2 in the cytoplasmic monolayer is not reduced by the cytoplasm of red cells; when it is present in the outer monolayer, it can be reduced by ascorbate. Thus, phospholipid 2 is appropriate for measuring the rate of flip-flop from the cytoplasmic to the outer monolayer. As shown by the experiments in Figure 10-4A, phospholipid 1 in the cytoplasmic monolayer is reduced by red cell cytoplasm, but it is stable in the outer monolayer in the absence of ascorbate. Thus, phospholipid 1 is appropriate for measuring the rate of flip-flop from the outer to the cytoplasmic monolayer.

lipid raft

Small region of the plasma membrane enriched in sphingolipids and cholesterol.

To investigate the difference in extent of loss of ESR signal with the two phospholipids, the experiments were repeated using red cell ghosts that had been resealed to make them impermeable to ascorbate (Figure 10-4C and D). Resealed red cell ghosts are missing all of their cytoplasm, but have an intact plasma membrane. In these experiments, the loss of ESR signal for both phospholipids was negligible in the absence of ascorbate and reached a plateau at 50% in the presence of ascorbate. What do you suppose might account for the difference in extent of loss of ESR signal in experiments with red cell ghosts (Figure 10-4C and D) versus those with normal red cells (Figure 10-4A and B).

The key observation is that the extent of loss of ESR signal in the presence and absence of ascorbate is the same for phospholipids 1 and 2 in resealed red cell ghosts, but different in red cells. These results suggest that there is an undefined reducing agent in the cytoplasm of red cells (which is absent from red cell ghosts). Like ascorbate, this cytoplasmic agent can reduce the more exposed phospholipid 1, but not the less exposed phospholipid 2. Thus, in red cells, phospholipid 2 is stable in the absence of ascorbate; in the presence of ascorbate, the spin-labeled phospholipids in the outer monolayer are reduced, causing loss of half the ESR signal. Phospholipid 1, on the other hand, is not stable in red cells in the absence of ascorbate because the phospholipids in the cytoplasmic monolayer are exposed to the cytoplasmic reducing agent, which destroys half the ESR signal. When ascorbate is added, labeled phospholipids in the outer monolayer are also reduced, causing loss of the remaining ESR signal.

phosphoglyceride

The main type of phospholipid in animal cell membranes, with two fatty acids and a polar head group attached to a three-carbon glycerol backbone.

Phosphatidylserine, which is normally confined to the cytoplasmic monolayer of the plasma membrane lipid bilayer, is redistributed to the outer monolayer during apoptosis. How is this redistribution accomplished?

The redistribution of phosphatidylserine from the cytoplasmic to the outer monolayer of the plasma membrane occurs by two mechanisms: (1) the phospholipid translocators that normally transport this lipid from the noncytoplasmic monolayer to the cytoplasmic monolayer are inactivated in apoptotic cells; and (2) a "scramblase" that transfers phospholipid nonspecifically in both directions between the two monolayers is activated.

Were the spin-labeled phospholipids introduced equally into the two monolayers of the red cell membrane?

The results in Figure 10-4 indicate that the labeled phospholipids were introduced equally into the two monolayers of the red cell plasma membrane. Phospholipid 2 was 50% sensitive to ascorbate, indicating that half the label was present in the outer monolayer, and 50% insensitive to ascorbate, indicating that half was present in the cytoplasmic monolayer. Similarly, phospholipid 1 was 50% sensitive to the cytoplasmic reducing agent and 50% sensitive to ascorbate, indicating an even distribution between the cytoplasmic and outer monolayers.

The actual experiments showed that transmembrane-anchored folate receptors followed the expectations shown in Figure 10-5A, whereas the GPI-anchored folate receptors followed those in Figure 10-5B. Do these experiments provide evidence for the existence of lipid rafts in the plasma membrane? Why or why not?

The results suggest that transmembrane-anchored folate receptors are randomly dispersed in the membrane, while GPI-anchored receptors are clustered in microdomains. Although such microdomains are likely to be lipid rafts, these experiments do not prove that point.

When a lipid bilayer is torn, why does it not seal itself by forming a "hemi-micelle" cap at the edges, as shown in Figure 10-2?

The same forces that dictate that certain lipids will form a bilayer, as opposed to micelles, operate in the repair of a tear in the bilayer. The tear will heal spontaneously because a bilayer is the most energetically favorable arrangement. The lipids that make up a bilayer are cylindrical in shape and therefore do not readily form a micelle (or a hemi-micelle), which would require cone-shaped lipids.

If lipid rafts form because membrane components such as sphingolipids and cholesterol molecules preferentially associate with one another, why do you think it is that they aggregate into multiple tiny rafts instead of into a single large one?

The size of a lipid raft depends on the affinity of its components for one another. If sphingolipids and cholesterol molecules, for example, bound one another sufficiently tightly, they would aggregate into a single domain in the membrane. If they bound one another with the same affinity as they bind to other species of lipid molecules, they would remain dispersed. The small size of the lipid rafts indicates that sphingolipids and cholesterol molecules, for example, have only a slightly higher affinity for one another than for other lipids. Presumably, at the typical size of a raft, the aggregated lipid components, including sphingolipids and cholesterol molecules, are in equilibrium with their free forms, so that they are added to and leave a raft at equal rates.

Glycolipids are never found on the cytoplasmic face of membranes in living cells.

True. Glycolipids are synthesized in the lumen of the Golgi apparatus, which is topologically equivalent to the outside of the cell, and cannot flip-flop across the bilayer.

T/F: Although lipid molecules are free to diffuse in the plane of the bilayer, they cannot flip-flop across the bilayer unless enzyme catalysts called phospholipid translocators are present in the membrane.

True. The hydrophobic interior of the lipid bilayer acts as a barrier to the passage of the hydrophilic lipid head groups that must occur during flip- flop. The energetic cost of this movement effectively prevents spontaneous flip-flop of lipids, so that it occurs extremely rarely in the absence of specific catalysts known as phospholipid translocators.

T/F: All of the common phospholipids: phosphatidylcholine phosphatidylethanolamine phosphatidylserine sphingomyelin carry a positively charged moiety on their head group, but none carry a net positive charge.

True. The positively charged moieties in all cases are balanced by the negative charge on the phosphate group; thus, none of the common phospholipids carries a net positive charge.

Margarine is made from vegetable oil by a chemical process. Do you suppose this process converts saturated fatty acids to unsaturated ones, or vice versa? Explain your answer.

Vegetable oil is converted to margarine by reduction of double bonds (by hydrogenation), which converts unsaturated fatty acids to saturated ones. This change allows the fatty acid chains in the lipid molecules to pack more tightly against one another, increasing the viscosity, turning oil into margarine.

Hydrophobic solutes are said to "force the adjacent water molecules to reorganize into icelike cages" (Figure 10-1). This statement seems paradoxical because water molecules do not interact with hydrophobic solutes. How could water molecules "know" about the presence of a hydrophobic solute and change their behavior to interact differently with one another? Discuss this seeming paradox and develop a clear concept of what is meant by an "icelike" cage. How does it compare to ice? Why would such a cagelike structure be energetically unfavorable relative to pure water?

Water is a liquid, and thus hydrogen bonds between water molecules are not static; they are continually formed and broken again by thermal motion. When a water molecule happens to be next to a hydrophobic solute, it is more restricted in motion and has fewer neighbors with which it can interact because it cannot form any hydrogen bonds in the direction of the hydrophobic solute. It will therefore form hydrogen bonds to the more limited number of water molecules in its proximity. Bonding to fewer partners results in a more ordered water structure, which constitutes the icelike cage in Figure 10-1. The true cage of water molecules exists in three dimensions, forming a pentagonal dodecahedron (like a soccer ball) or clusters of them that enclose the hydrophobic solute. The structure is similar to ice, although it is a more transient, less organized, and less extensive network than even a tiny ice crystal. The formation of any ordered structure decreases the entropy of the system, which is energetically unfavorable.