Calculations 1064 using these average molecular

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CALCULATIONS 10–64 Using these average molecular weights, there are 96 lipid molecules (phos- pholipid + cholesterol) for every protein molecule {[(50,000/800) + (50,000/386)]/2 = 96}. A similar lipid-to-protein ratio is present in many cell membranes. 10–65 A. The calculation for the number of spectrin molecules per red blood cell is shown in detail below. In essence, one first calculates the fraction of total protein that is spectrin and then converts that number into the number of spectrin molecules using the molecular weight of spectrin and Avogadro’s number: The analogous calculation gives values of 9 ¥ 10 5 molecules of band 3 and 2.3 ¥ 10 5 molecules of glycophorin per red cell. The calculated number of glycophorin molecules per cell is too low by a factor of 2.5 because about 60% of the molecular weight of glycophorin is carbohydrate, which is not stained by Coomassie Blue. B. The fraction of the plasma membrane that is occupied by band 3 is the area of the face of a single band 3 molecule ( p r 2 ), times the total number of band 3 molecules per cell (9 ¥ 10 5 ), divided by the total area of the red blood cell (10 8 nm 2 ). Note that the height of the molecule is irrelevant to the calculation. Thus band 3 occupies about 25% of the surface area of a red blood cell. This result is consistent with freeze-fracture electron micrographs of red blood cells, which show a high density of intramembrane particles that are thought to be dimers of band 3. DATA HANDLING 10–66 A. Elimination of sialic acid staining by sialidase treatment indicates that car- bohydrate is exposed on the external surface. Because the carbohydrate is attached to glycophorin, it follows that glycophorin is also exposed on the external surface. This conclusion is supported by the results with pronase digestion, which eliminates sialic acid staining, presumably by clipping the peptide backbone. The results with pronase digestion indicate that band 3 is exposed to the external surface as well. In this case the appearance of the new protein band at about 70,000 daltons allows you to estimate that at least 30,000 daltons of band 3 are exposed on the external surface. In neither MEMBRANE PROTEINS A231 spectrin 5 mg protein 0.25 spectrin mmol spectrin 6 ¥ 10 20 molecules cell 10 10 cells total protein 250,000 mg mmol spectrin = ¥ ¥ ¥ = 3 ¥ 10 5 spectrin molecules/cell band 3 3.14 ¥ (3 nm) 2 9 ¥ 10 5 molecules 1 cell plasma membrane molecule cell 10 8 nm 2 = ¥ ¥ = 0.25
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digestion were the two spectrin bands affected, suggesting that spectrin is not exposed on the external surface. B. One direct experimental approach for testing your colleague’s objection is to break open the red cell ghosts before digesting them with pronase. If spec- trin were resistant to pronase, its mobility on SDS polyacrylamide gels would be unaltered. If spectrin were located on the cytoplasmic surface, its mobility would be altered dramatically. These control experiments have been done; they show that spectrin is sensitive to pronase. Another approach is to make inside-out ghosts and see if it is possible to dissociate spectrin from the membrane by treatments that do not disrupt the mem- brane. This approach also has been successful, confirming that spectrin is on the cytoplasmic side and is not embedded in the membrane.
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