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Capítulo 15
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2. Inactivation of phosphatase 2 would lead to the permanent activation of kinase 3, which would lead to the continued unregulated activation of the transcription factor and likely the unregulated growth of the cell. Inactivation of protein kinase 3 might be expected to stop the cell from progression through the cell cycle because the transcription factor could not be activated.

3. Inactivation of cAMP phosphodiesterase would be expected to cause prolonged responses to hormonal stimulation and excessive glucose mobilization. An inactive glucagon receptor would prevent the cell from responding to this hormone, which would be expected to lead to a problem maintaining adequate blood glucose levels. Cells with an inactive phosphorylase kinase would be unable to mobilize glucose from glycogen because the phosphorylase molecules could not be activated. A G protein unable to hydrolyze GTP would be expected to continue to send stimulatory signals downstream, leading to a constitutively active adenylyl cyclase.

4. All three act by binding to a protein, but Ca 2+ binds to a protein (calmodulin) that activates other proteins, IP3 binds to and directly opens an ion channel, and cAMP binds to and directly activates an enzyme.

5. Amplification occurs when activated MAPKKK (which is Raf) phosphorylates MAPKK molecules; when activated MAPKK (MEK) phosphorylates MAPK molecules; when activated MAPK (ERK) phosphorylates transcription factors; and when activation of a single DNA sequence leads to the production of large numbers of gene products.

6. You could see if they competed with one another for the same binding site. This could be determined, for example, by radioactively labeling the two hormones and determining the amount of bound radioactivity when cells were incubated with the two hormones separately or together over a range of hormone concentrations.

7. The cardiac muscle cells responded to FSH by contracting, which they would not normally do. That the second messenger (in this case cAMP) was diffusing through the gap junctions between the cells.

8. The GTP analogue would presumably bind to the G protein and remain unhydrolyzed, causing the stimulation to be prolonged. The liver cell would continue to break down glycogen, while the epithelial cell treated with EGF would continue to proliferate. The cholera toxin would be expected to have a similar effect as the GTP analogue on the glucagon treated liver cell, but would not have an effect on proliferation of an epithelial cell since this response is not mediated by a heterotrimeric G protein.

9. For example, incubate the cells with radioactively labeled choline to label the PC of the membrane, then chase the cells for a period to remove labeled choline from the cell, then stimulate the cells, homogenize the cells, discard the membrane fraction, and determine whether there is an increase in soluble label over that found in unstimulated controls. Subsequent experiments could analyze the nature of the labeled compound(s) that appear after stimulation to determine if the label is in choline phosphate.

10. Cells possess high concentrations of calcium-binding proteins that are able to "soak up" excess calcium ions, and they also possess large numbers of calcium pumps in their plasma and cytoplasmic membranes that removes the ions from the cytosol. If you injected the calcium solution, you would expect to see a very localized and transient rise in Ca 2+ levels, with a quick return to normal levels.

11. The wave is the result of calcium-induced calcium release in which free calcium released at one site binds to calcium channels at nearby sites, causing them to open and release additional calcium. As a result, a wave of calcium release sweeps across the cell.

12. No. Calmodulin is a small protein whose structure is highly conserved throughout eukaryotes. It is much more likely that such a molecule would have only one or two recognizable binding sites and that the various effectors have similar sites that are capable of binding to calmodulin.

13. A deficiency in insulin production, a deficiency or alteration in the insulin receptor so that cells do not respond to insulin, a deficiency or alteration in an IRS or a component downstream in the pathway.

14. Yes, because the response to EGF requires that EGF receptors, which exist as monomers within the plasma membrane, must collide with one another following ligand binding if they are to form active dimers. The insulin receptor already exists within the membrane as a dimer.

15. Dominantly, because the creation of a single mutant RAS gene in a cell would cause the encoded protein to act constitutively to signal proliferation.

16. Cells become malignant when they develop mutations in genes whose products are involved in cell growth and division. If these mutations were to trigger an apoptotic response, then the cell would be eliminated before it had a chance to develop into a life-threatening disease.

17. Protein kinase A may be mutated so that it no longer recognizes Raf, which it normally phosphorylates; or Raf may be altered so that it is no longer a substrate for protein kinase A.

18. Both are self-propagating. A rise in calcium concentration in one region triggers opening of calcium channels in neighboring regions, just as a depolarization at one site in a nerve cell membrane triggers a depolarization at an adjacent site.

19. They are almost universally bitter tasting and the rodents find them inedible.

20. The inhibitor would stop the cell from undergoing apoptosis. This is required for reproduction of the virus. If this didn't occur, the viral DNA would be degraded along with the host cell DNA, stopping the progress of the infection.

21. Use of the intermediate provides a major, early amplification step because each activated receptor can phosphorylate numerous IRS substrates. In addition, the IRSs are soluble proteins that can diffuse to other parts of the cell to activated downstream effectors.

22. Yes. Both PI3K and PKB are key upstream components of the signaling pathways that lead to the major insulin responses, i.e., glucose uptake, glycogen synthesis, and protein synthesis. Therefore inhibition of PI3K would be expected to block these responses and overexpression of PKB would be expected to induce them.

23. They have an enlarged brain due to lack of death of excess neurons normally generated in the embryonic nervous system. Cytochrome c knockouts would be expected to die at a very early stage of development due to lack of ability to carry out electron transport.

24. Those who do not find PROP to have a bitter taste are presumed to lack a G protein-coupled gustatory receptor capable of binding to the compound.







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