Midterm1 (Spring 10) - Chem 135: Exam I 11 February 2010...

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Chem 135: Exam I 11 February 2010 Please provide all answers in the space provided. Extra paper is available if needed. You may not use calculators for this exam. Including the title page, there should be 12 single sided pages in this exam booklet. ¡ Good luck ! Name: KEY (1) (7 points) (2) (18 points) (3) (20 points) (4) (10 points) (5) (10 points) (6) (18 points) (7) (17 points) TOTAL (100 points)
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2 1. a. Draw the peptide: RTDQM at pH 7.0 (with the correct stereochemistry at ALL C α -carbons). + H 3 N H N O O N H O H N O N H O O - NH OH NH 2 + H 2 N O O - O NH 2 S b. What is the total charge of the peptide at pH 1? +2, N-terminus and R are positively charged, there are formal no negative charges c. What is the total charge of the peptide at pH 14? -2, C-terminus and D are negatively charged, the are no formal positive charges 2. While sampling for unique microbial activity in the “fountain” in front of Lewis Hall, you find a new protein and take it back to the laboratory to study. Luckily this new protein is very small so you name it Tiny. The first thing you do is to analyze the primary structure of Tiny by reducing and cleaving it with different reagents in separate tubes. You are then able to separate and sequence the different peptides in each tube and find the following: Tube 1 : PGCGAKLISTCKFGNWEDG, M, EERHCVLRLWM Tube 2 : FGNWEDG, LWMPGCGAK, HCVLR, LISTCK, MEER Tube 3 : MEERHCVLRLW, MPGCGAKLISTCKF, GNW, EDG a. What cleaving agent did you use for each tube? Describe the specificity of each cleaving agent. 1. Cyanogen Bromide, cleaves after M 2. Trypsin, cleaves after R and K (positively charged residues) 3. Chymotrypsin, cleaves after F, W, and Y (aromatic residues) b. What is the sequence of Tiny using single-letter codes for the amino acids? MEERHCVLRLWMPGCGAKLISTCKFGNWEDG
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3 c. Using a technique called gel electrophoresis, you can determine the size of Tiny under reducing and non-reducing conditions. In the gel below, Lane 1 contains the molecular weight marker as a reference. Lane 2 contains Tiny under reducing conditions and Lane 3 contains Tiny under non-reducing conditions. Draw the reaction that occurs on Tiny when you add a reducing agent and describe the quaternary structure of Tiny. When a reducing agent is added the disulfide bond is broken. S S SH HS Gel electrophoresis shows that Tiny is a homotetramer. Under reducing conditions, when the monomers cannot form quaternary structure the size is 3000 Da while under non-reducing conditions when quaternary structure forms the size is 12000 Da (four times larger).
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4 d. After Tiny has been reduced, you monitor the pH dependence of the formation of the high-molecular weight band seen in Lane 3 of the gel from 2c. Please explain briefly showing a chemical equation for the protonation/deprotonation event and explain if the equilibrium is shifted. If there is a shift, please give a possible explanation for the observed p K a . In order to form the disulfide bond the Cys must become deprotonated, see
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Midterm1 (Spring 10) - Chem 135: Exam I 11 February 2010...

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