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2005_Exam1_key - Name_KEY 7.06 Cell Biology EXAM#1 This is...

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Name: _________ KEY _________________ 1 7.06 Cell Biology EXAM #1 February 24, 2005 This is an open book exam, and you are allowed access to books, a calculator, and notes but not computers or any other types of electronic devices. Please write your answers in pen (not pencil) to the questions in the space allotted. Please write only on the FRONT SIDE of each sheet. And be sure to put your name on each page in case they become separated! Remember that we will Xerox all of the exams. Good luck! Question 1. 42 pts __________ Question 2. 18 pts __________ Question 3. 29 pts __________ Question 4. 11 pts __________
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Name: _________ KEY _________________ 2 Question 1. (42 pts) You have isolated from the human genome a segment of DNA that encodes a novel protein. Your protein contains seven stretches of hydrophobic amino acids 21 to 26 residues in length, and thus you immediately suspect that it is a typical seven-spanning transmembrane G-protein-coupled receptor. In fact, you suspect that it is the long-sought receptor for Professor Lodish’s new hormone, adiponectin, which plays major roles in regulating fat and sugar metabolism. (a, 6 pts) You show the sequence to your project lab instructor. She tells you that it is unlikely that your protein is a seven-spanning transmembrane G-protein-coupled receptor because, in the middle of each of the 7 presumed membrane-spanning helices, there are charged amino acids. [Helices 1 and 4 (numbering from the amino terminus) contain two arginines each; helices 2 and 7 contain two glutamates each; helices 3 and 5 each contain a glutamine as well as a lysine residue, and helix 6 contains two aspartic acid residues. Otherwise, all of the amino acids in these 7 putative transmembrane helices are hydrophobic.] What can you say to your instructor to explain to her that it is possible for such a protein to be an integral membrane protein? You would say that this protein could still be an integral membrane protein because it could fold into a multi-membrane spanning protein that arranges itself in the membrane such that its seven alpha- helices can bundle together by interacting with each other. The helices would have to bundle such that the outer face of each helix had its hydrophobic R groups sticking out, and that the inner face of each helix contained the charged R groups. Thus the charged R groups would be excluded from contacting the hydrophobic hydrocarbons that make up the inside of a lipid bilayer. Please note that you could not have answered that the charged R groups could be hidden inside each helix, because all R groups in an alpha helix face outwards. (The polar peptide bonds are the parts of the amino acids that are contained within the inside of each helix, thus shielding these polar peptide bonds from the hydrophobic lipid tails.) Also, please note that is not valid to assume that, because this protein has charged amino acids in the middle of its transmembrane domains, that this protein must be a channel. GPCRs are not channels (which allow the passage of small molecules through central pores). They are receptors, which bind
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