Show answerA 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) microscopy, 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 induced 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 the moment of the difference in extent of loss of ESR signal, offer an explanation for why phospholipid 1 (10-4A) reacts faster with ascorbate than does phospholipid 2 (10-4B). Note that phospholipid 1 reaches plateau in about 15 minutes, whereas phospholipid 2 takes almost an hour.
B. 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 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 (10-4C and D), versus those with normal red cells (10-4A and B).
C. Were spin-labeled phospholipids introduced equally into the two monolayers of the red cell membrane?
We will go number-by-number.
A. I think that the speed of ESR signal extinction is linked with probability of reaction between ascorbate and nitroxide group. Thus, speed of phospholipids 2 signal extinction should be significantly less than for phospholipids 1, because it is not as easy for ascorbate to hit side chain than head group.
B. Simple comparison of ghosts membranes and intact cells tells us that there should be kind of agent – reducing agent – in the red cells' membranes that can attack head groups but not side chains. Of course red cell ghosts lack that agent, so only ascorbate acts. That is why loss of ESR signal in 10-4A is 50% in abscence and 100% in presence of ascorbate. Also this idea explain ESR signal reaction in ghosts also.
C. Yes, they were introduced equally in the interior and exterior layers of membrane. This is right both for intact cells ("natural" 50% signal decrease) and for ghosts (50% decrease in presence of ascorbate).
Now lets check the right answer.
A. I think that the speed of ESR signal extinction is linked with probability of reaction between ascorbate and nitroxide group. Thus, speed of phospholipids 2 signal extinction should be significantly less than for phospholipids 1, because it is not as easy for ascorbate to hit side chain than head group.
B. Simple comparison of ghosts membranes and intact cells tells us that there should be kind of agent – reducing agent – in the red cells' membranes that can attack head groups but not side chains. Of course red cell ghosts lack that agent, so only ascorbate acts. That is why loss of ESR signal in 10-4A is 50% in abscence and 100% in presence of ascorbate. Also this idea explain ESR signal reaction in ghosts also.
C. Yes, they were introduced equally in the interior and exterior layers of membrane. This is right both for intact cells ("natural" 50% signal decrease) and for ghosts (50% decrease in presence of ascorbate).
Now lets check the right answer.
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.
B. The key observation is that the extent of loss of ESR signal in the presence and absence of ascorbate is the same 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.
C. 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.
