Practice exam 2 key

Practice exam 2 key - Exam 2 Answer Key 1. A) 4 pts When...

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Exam 2 Answer Key 1. A) 4 pts When Adenylyl cyclase becomes activated, it begins to produces high levels of cAMP. cAMP is a small molecule that can rapidly diffuse through the cytosol, acting as a second messenger that affects many downstream targets. B) 12 pts The data shows that in cells expressing WT phosphodiesterase (PDE), activated AC yields high level of cAMP, which is lowered when PKA is activated. Knowing that WT PDE requires phosphorylation to be active, the data suggests that PKA activity may be the kinase what activates PDE and thus leads to breakdown of cAMP. The mutant PDE mimics the active, phosphorylated state. As a result, there is no peak of cAMP upon AC stimulation because PDE immediately begins to break it down as it is being produced. The maximum level of cAMP observed in the mutant is the same observed in cells expressing WT, activated PDE (after PKA stimulation). 2. 9 pts In the resting state of a cell expressing the three types of TGFbeta receptors: Type III is monomeric, Type II is a dimer with constitutive kinase activity, and Type is a dimer that is kinase active only after forming a complex with II and III. Two separate dimers of TGFbeta bind to the Type III receptor monomer and the Type II dimers. This binding event brings about the formation of an active receptor complex containing all three types of receptors and results in the phosphorylation of the Type I kinase domain, activating it. The active Type I receptor dimer phosphorylates Smad3.
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This note was uploaded on 04/07/2008 for the course MCB 54 taught by Professor Lue during the Fall '06 term at Harvard.

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Practice exam 2 key - Exam 2 Answer Key 1. A) 4 pts When...

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