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WRITE YOUR NAME AND I.D. NUMBER LEGIBLY ON EVERY PAGE – PAGES WILL BE SEPARATED FOR GRADING! CHECK TO BE SURE YOU HAVE 6 PAGES , NAME (print): ANSWERS INCLUDING COVER PAGE. 5-digit course ID # I swear/affirm that I have neither given nor received any assistance with this exam. Signature: Date: BIOCHEMISTRY 460 FIRST HOUR EXAMINATION FORM A (yellow) ANSWER KEY May 30, 2007 A NON-PROGRAMMABLE CALCULATOR MAY BE USED ON THIS EXAM. No programmable calculators are permitted, and no sharing of calculators. We have a couple of spare calculators to lend in an emergency. SHOW YOUR WORK FOR ALL CALCULATIONS , AND BE SURE TO STATE UNITS OF ANY NUMERICAL ANSWERS. If the reasoning, calculations, or answer are shown anywhere other than in the space provided, make a note in the space provided and put answer on BACK OF SAME PAGE so the grader for that page will have it. USEFUL CONSTANTS : R (gas constant) = 8.315 J •mol –1 •Kelvin –1 = 8.315 x 10 –3 kJ •mol –1 •Kelvin –1 If temperature = 25 °C, absolute temperature T = 298 K (Assume this temperature unless problem states otherwise.) Use these “generic” pK a values only when no precise pK a for a specific group is given. Ionizable group in peptides and proteins Approximate ("generic") pK a (from Berg, Tymoczko & Stryer, Biochemistry , 6th ed., 2007) Potentially Useful Equations Δ G' = Δ H' T Δ S' Δ G' = Δ G°' + RTln(mass action ratio) α -carboxyl 3.1 side chain carboxyl 4.1 imidazole 6.0 α -amino 8.0 thiol 8.3 aromatic hydroxyl 10.9 ε -amino 10.8 guanidino 12.5 Michaelis-Menten Equation: V o = V max [S] [S]+K m
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Biochemistry 460, Exam #1 Form A ANSWERS NAME May 30, 2007 Last 4 digits of SID: ___________________ page 2 1. (9 pts, 3 pts each) Fill in the blanks below with terms that fit the definitions: 2,3-bisphosphoglycerate the ligand that binds in the central cavity of hemoglobin in humans competitive the type of reversible enzyme inhibitor that cannot be bound to the enzyme at the same time as the substrate – binding of S and I is mutually exclusive. disulfide the type of bond that results from oxidation of two cysteine side chains 2. (12 pts) Consider the following reaction: Be sure to show your calculations/reasoning, and state units where appropriate. A. (3 pts) Suppose that for the uncatalyzed reaction, k F(uncatalyzed) = 4.5 x 10 –4 s –1 , and K eq(uncat) = 3.0 x 10 3 . What is k R(uncat) ? k R(uncat) = k F(uncat) /K eq(uncat) = 1.5 x 10 –7 s –1 . B. (9 pts) Consider the same reaction as in part A, with rate constants and equilibrium constant given or calculated in part A for the uncatalyzed reaction. Suppose that in the presence of an enzyme, k F(catalyzed) = 1.8 x 10 8 s –1 . 1) (3 pts) What is K eq for the catalyzed reaction? Catalyst doesn’t change K eq , so K eq(cat) = K eq(uncat) = 3.0 x 10 3 . 2) (3 pts) What is the rate enhancement brought about by the enzyme?
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This note was uploaded on 04/07/2008 for the course BIOC 460 taught by Professor Ziegler during the Spring '07 term at Arizona.

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