1. The negatively charged glutamate residues mimic the negatively charged phosphoserine or phosphothreonine residues and stabilize the active conformation of the enzyme.

2. No. Phosphoserine and phosphothreonine are considerably shorter than phosphotyrosine.

3. The GTPase activity terminates the signal. Without such activity, after a pathway has been activated, it remains activated and is unresponsive to changes in the initial signal. If the GTPase activity were more efficient, the lifetime of the GTP-bound G_α subunit would be too short to achieve downstream signaling.

4. Two identical receptor molecules must recognize different aspects of the same signal molecule.

5. Growth-factor receptors can be activated by dimerization. If an antibody causes a receptor to dimerize, the signal--transduction pathway in a cell will be activated.

6. The mutated α subunit will always be in the GTP form and, hence, in the active form, which would stimulate its signaling pathway.

7. A G protein is a component of the signal-transduction pathway. GTPγS is not hydrolyzed by the G_α subunit, leading to prolonged activation.

8. Calcium ions diffuse slowly because they bind to many protein surfaces within a cell, impeding their free motion. Cyclic AMP does not bind as frequently, and so it diffuses more rapidly.

9. Fura-2 is a highly negatively charged molecule, with five carboxylate groups. Its charge prevents it from effectively crossing the hydrophobic region of the plasma membrane.

10. G_αs stimulates adenylate cyclase, leading to the generation of cAMP. This signal then leads to glucose mobilization (Chapter 21). If cAMP phosphodiesterase were inhibited, then cAMP levels would remain high even after the termination of the epinephrine signal, and glucose mobilization would continue.

11. If the two kinase domains are forced to be within close proximity of each other, the activation loop of one kinase, in its inactivating conformation, can be displaced by the activation loop of the neighboring kinase, which acts as a substrate for phosphorylation.

12. The full network of pathways initiated by insulin includes a large number of proteins and is substantially more elaborate than indicated in Figure 14.26. Furthermore, many additional proteins take part in the termination of insulin signaling. A defect in any of the proteins in the insulin signaling pathways or in the subsequent termination of the insulin response could potentially cause problems. Therefore, it is not surprising that many different gene defects can cause type 2 diabetes.

13. The binding of growth hormone causes its monomeric receptor to dimerize. The dimeric receptor can then activate a separate tyrosine kinase to which the receptor binds. The signaling pathway can then continue in similar fashion to the pathways that are activated by the insulin receptor or other mammalian EGF receptors.

14. The truncated receptor will dimerize with the full-length monomers on EGF-binding, but cross-phosphorylation cannot take place, because the truncated receptor possesses neither the substrate for the neighboring kinase domain nor its own kinase domain to phosphorylate the C-terminal tail of the other monomer. Hence, these mutant receptors will block normal EGF signaling.

15. Insulin would elicit the response that is normally caused by EGF. Insulin binding will likely stimulate dimerization and phosphorylation of the chimeric receptor and thereby signal the downstream events that are normally triggered by EGF binding. Exposure of these cells to EGF would have no effect.

16. 10⁵

17. The formation of diacylglycerol implies the participation of phospholipase C. A simple pathway would entail receptor activation by cross-phosphorylation, followed by the binding of phospholipase C (through its SH2 domains). The participation of phospholipase C indicates that IP₃ would be formed and, hence, calcium concentrations would increase.

A18

18. Other potential drug targets within the EGF signaling cascade include, but are not limited to, the kinase active sites of the EGF receptor, Raf, MEK, or ERK.

19. In the reaction catalyzed by adenylate cyclase, the 3′-OH group nucleophilically attacks the α-phosphorus atom attached to the 5′-OH group, leading to displacement of pyrophosphate. The reaction catalyzed by DNA polymerase is similar except that the 3′-OH group is on a different nucleotide.

20. ATP-competitive inhibitors are likely to act on multiple kinases because every kinase domain contains an ATP-binding site. Hence, these drugs may not be selective for the desired kinase target.

21. (a) X ≈ 10⁻⁷ M; Y ≈ 5 × 10⁻⁶ M; Z ≈ 10⁻³ M. (b) Because much less X is required to fill half of the sites, X displays the highest affinity. (c) The binding affinity almost perfectly matches the ability to stimulate adenylate cyclase, suggesting that the hormone–receptor complex leads to the stimulation of adenylate cyclase. (d) Try performing the experiment in the presence of antibodies to G_αs.

22. (a) The total binding does not distinguish binding to a specific receptor from binding to different receptors or from nonspecific binding to the membrane.

(b) The rationale is that the receptor will have a high affinity for the ligand. Thus, in the presence of excess nonradioactive ligand, the receptor will bind to nonradioactive ligand. Therefore, any binding of the radioactive ligand must be nonspecific.

(c) The plateau suggests that the number of receptor-binding sites in the cell membrane is limited.

23. Number of receptors per cell =