Site MapHelpFeedbackCapítulo 4
Capítulo 4
(See related pages)

1. For example, integral proteins involved in cell-cell adhesion and cell-cell communication. Epithelial cells form a tightly cohering sheet that requires such proteins, whereas erythrocytes occur in the body as single cells. Epithelia would also be expected to have different types of transport proteins than erythrocytes whose primary transport activity is the exchange of bicarbonate and chloride ions.

2. Because steroids are relatively small hydrophobic compounds, they would be expected to diffuse through the plasma membrane. Accordingly, the receptors for these hormones are found in the cytoplasm. In contrast, receptors for nearly all other types of hormones, including insulin, are found at the surface of the plasma membrane. Being a protein, insulin would not be expected to penetrate the plasma membrane.

3. It was presumed that the outer electron dense layers of the trilaminar image corresponded to proteins that had bound the heavy metal stains. Since they were seen as continuous layers, it was presumed that they reflected the presence of continuous layers of proteins situated outside the lipid bilayer.

4. Incorporate a component, such as an antibody, into the lipid bilayer of the liposome that will recognize and bind to a specific component on the outer surface of the target cell membrane.

5. Polysaccharides, such as starch and glycogen, are monotonous polymers of a single subunit, whereas the oligosaccharides at the cell surface contain a variety of different sugar subunits whose arrangement within the complex is highly defined. This is illustrated by the fact that an antiserum against the type A antigen will precipitate blood cells from a person with type A blood, but will not precipitate those from a person with type B or type O blood, and vice versa.

6. See Figures 4.16 and 4.31c.

7. The flattened biconcave shape allows diffusion of O 2 from the external medium to all parts of the cell (it is only about 1 µm to the cell center).

8. It would be expected to cause the level of saturation of the fatty acids of the membrane to increase (the number of polyunsaturated fatty acids to decrease), which would raise the transition temperature of the lipid bilayer. The activity of membrane desaturases would be expected to decrease.

9. Those with the smallest or most hydrophobic head groups. A flippase is actively moving PC from one leaflet to the other. The integral protein should show no detectable flip-flop since movement of the hydrophilic segments of the protein across the membrane would be extremely unfavorable.

10. A two-dimensional representation (as shown in Figure 4.39a) fails to describe the positions of the various parts of the protein relative to one another as described in a three-dimensional representation (as in Figure 4.15). The former is obtained from determination of amino acid sequences and hydropathy plots that pinpoint transmembrane segments, whereas the latter is obtained from X-ray crystallography. The large numbers of two-dimensional structures are the result of advances in gene sequencing from which amino acid sequences can be deduced. The scarcity of three-dimensional structures stems from the difficulty in obtaining crystals of hydrophobic integral proteins.

11. Both ions would be expected to move into the medium when the axon is at rest; the K + would be diffusing through leak channels, while the Na + would be pumped out by active transport. The firing of an action potential would be accompanied primarily by the movement of labeled K + out of the membrane during the repolarization phase.

12. It makes it difficult to assay for water passage through the channels because of the high baseline created by diffusion. It is difficult to distinguish the two. You can use a substance that inhibits proteins. Mercury-containing compounds have been effective. Frog oocytes exist in a hypertonic medium and are highly impermeable to water entry. With aquaporins in the membrane, the oocytes gain volume and rupture.

13. Lipids are less likely to form large complexes or to be linked to cytoskeletal or extracellular materials of the type that inhibit the movement of integral proteins.

14. Na + would be expected to move into the cell more rapidly because its influx is favored by the electrical component of the gradient as well as the difference in concentration. In contrast, K + would be moving down its concentration gradient as it exited the cell, but against its electrical gradient (inside negative).

15. The internal fluids of marine invertebrates have a much higher osmotic pressure (solute concentration) than those of marine vertebrates. It is thought that marine fishes evolved from species that arose in fresh water and had salt concentrations considerably less than that of sea water.

16. The intracellular concentrations would be expected to be hypertonic to the extracellular fluids, which would cause the cells to gain water and maintain turgor pressure. The intracellular fluids of animal cells would be expected to have similar solute concentrations to the extracellular fluids so that cells would neither gain nor lose water.

17. Action potentials might be able to move in both directions from sites along the axon.

18. From p. 167, E K = 2.3 RT/zF olog10 200mM/10 mM E K = 59 log 10 20 = +77 mV at 25 o C At 37 o C, the term 2.3 RT/zF is equal to 61 mV rather than 59 mV (310K/298K •59), so that E K = +79 mV

19. Would be based on ΔG of 3.1 rather than 6.2. Substituting in the equation on p. 164, the value for X becomes -2.21 and 10 -2.21 = 1/162. The transporter could only work against a gradient about 160 fold, rather than 23,000 fold if two ions were transported per glucose.

21. The equation on p. 167 can be used to determine the membrane potential (V x ) that would be generated by each of the five ions whose concentration is given in the table. The equation can be rewritten

None of these ions are at equilibrium when the membrane potential (V m ) is -70 mV. The distance each is away from equilibrium is equal to V x - V m . Under these conditions, K + and Cl - would move out of the axon through an open channel for that ion, whereas Na + , Ca 2+ , and H + would move into the axon through an open channel.

22. V K+ = 59 mV•log 10 5/140 = -85 mV
V Na+ = 59 mV• log 10 150/10 = +69 mV
V m = (-85 + 69)/2 = -8 mV

23. An alpha helix is able to from internal hydrogen bonds, whereas a beta strand cannot; it must form hydrogen bonds with the residues on an adjacent backbone.

24. Most likely a stack of carbonyl oxygens that are closer together so that K+ ions cannot fit into pore.

25. A channel is much more rapid, because it doesn't depend on a catalytic reaction. Each open channel can conduct millions of ions per second at near diffusion limiting rate, whereas each Na + /K + -ATPase, for example, can only move 2 or 3 ions per catalytic cycle (a fixed stoichiometry).







Biología celular y molecularOnline Learning Center

Home > respuestas preguntas 2 > Capítulo 4