Exam3Key - Biochemistry I (CH 339K) Fall 2008 Unique #54715...

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Unformatted text preview: Biochemistry I (CH 339K) Fall 2008 Unique #54715 Professor Adrian Keatinge-Clay Exam #3 October 6, 2008 Do not begin work on this exam until you are instructed to begin. The core values of The University of Texas at Austin are learning, discovery, freedom, leadership, individual opportunity, and responsibility. Each member of the University is expected to uphold these values through integrity, honesty, trust, fairness, and respect toward peers and community. The work that you hand in will be considered to be your own original work prepared without assistance, I will uphold the honor code on this exam. Signed Name: Printed Name: kE Y EID: l. True/False. Mark T or F. (24 points) _F_ Transition states are often stable for longer than a nanosecond. _T_ RNA hairpins primarily exist in “A form” helices. _T_ A decrease in activation energy usually corresponds to an increase in reaction rate. _T_ It is difficult to determine the difference between monosaccharide epimers by mass spectrometry. _F_ For an enzyme to operate at the diffusion-controlled limit, the associated Km must be below 1 mM. _T_ Cofactors are usually inorganic ions. _T_ In general, the longer a plasma glycoprotein retains a terminal sialic acid, the longer it resists degradation. _F_ A reaction will go to completion if the associated AG is negative. _F_ Chitin is a glycosaminoglycan. _T_ A lysine residue can participate in general acid/base chemistry. _FH Chymotrypsin cleaves peptides at the C—terminal end of lysine and arginine residues. _T_ G and U can pair together in folded RNAs. 2. Why is disadvantageous for an enzyme to have a high affinity for its substrate? (5 points) The energy between the ES complex and the transition state (the activation energy) may not have been decreased by the enzyme. Consequently, the reaction rate may not be increased by the enzyme. 3. What is at “steady-state” in steady-state kinetics? (3 points) The ES complex is at “steady-state”. 4. An enzyme transforms substrate, A, to product, B, via the reversible reaction: k f krB ls it possible for the enzyme to evolve a way of increasing the rate of product formation, kf‘, while not changing the reverse rate, k,? Explain. (6 points) No. K“, is set by the thermodynamic ratio set by the free energy difference between A and B; it cannot be changed by an enzyme. Since KN=kp/kr, if the enzyme increases kf, kr must also increase. 5‘ You are studying a remarkable enzyme, UTase, and measure the initial velocities of reactions at various substrate concentrations. For all measurements [UTase] = 10 nM. .—l 'l l s _ f . MM/Wm) MM 9 43] [81mm V0(MM/mm) M “l 0750 0.500 “‘25 01‘” 0,0000 WW 0 07,17- (H20 a) What plot will provide the best measurement of Km and Vmax? (2 points) The Lineweaver—Burk (or double-reciprocal) plot. b) Use this method to determine, as accurately as possible, Km and Vmax- (6 points) c) What is the turnover number of UTase (in s")? (3 points) kc... = me/IEJ = (12 uM/min)/10 nM = 1200 min'1 = 20 s" 6. An inhibitor binds in the active site of an enzyme, preventing binding of the natural substrate. a) What kind of inhibition is this? (2 points) (3) / Competitive inhibition 0 b) How does the inhibitor affect Vmax? (2 points) F (31 It doesn’t. c) How does the inhibitor affect the apparent Km? (2 points) The apparent K... increases with [I]. 7‘ Crocin is a natural carotenoid found in some flowers. It is primarily responsible for the color of saffron. Notice the four glucose residues that help to solubilize crocin in water. ON a) Would crocin test positive in Fehling’s reaction w OH (the qualitative test for the presence of reducing Ho/ sugar)? (2 points) a N0. OH / [is OH x l I ? j \1/ "Oxi/ 1 \ W0” ‘ ‘ H HoA/U " ° 9'” no 1 b) What is the linkage between the monosaccharide units? Y/T (hint: the linkage in amylose is “(11 -)4”) (4 points) K... \V .. /OH HO 5 1-) 6 8. When reacting with penicillin, what catalytic step does B-lactamase perform that transpeptidase does not? (3 points) Hydrolysis of the acyl-enzyme 9. Why is RNA less stable in solution than DNA? (4 points) A 2’ OH group, unique to RNA, can attack a neighboring phosphodiester moiety. This results in cleavage/destruction of the RNA. 10. Show the ring-closure mechanism that yields a B-pyranose. Draw the product using a Haworth proj ectioni (6 points) n [-7\ /H ‘5 k» . Hvo—ou H (3 H a H HO --é--H _—~—§ ll. Shown below are the major purines and pyrimidine bases of nucleic acids. Each R group represents the monosacchan'de residue to which each base is attached within a polynucleotide. In the space below, redraw the DNA bases to show how they hydrogen bond (i.e. how they form base pairs) within the double helix. Label each DNA base (8 points) H ,H 9 “N‘ ‘ r l H ..\ k /H ' IN V/ N/ ‘ ‘N ’N__\/,k\\N "—x L i .. a :1 a / / \ I [V N/ H \N/K\O \N&A\H R H R R Hx /H ‘N. OI “\{/'»:§~N H \ /L\_N/H ‘ . ll I /‘\ /L / \ H ‘N \‘o H N 0 . R n H /l H l (N N‘fk 2“ W: 2 43W“ ‘é‘ 12. The enzyme F abD catalyzes the transfer of a malonyl group from one thiol-containing carrier molecule to another. SUBSTRATES PRODUCTS Rf/‘~\/S‘_H ° 0 FabD # A ,s\ ,o' RéA/S\H "2’ WW 0 O A schematic of the FabD active site reveals some of the catalytic residues: a) Suggest a mechanism for this reaction. (3 points) Through covalent catalysis similar to the chymotrypsin mechanism, the serine acquires the malonyl group. The malonyl group can then be transferred to the second substrate. (This reaction mechanism is referred to as Ping-pong or double- displacement.) b) In the FabD active site above. draw how the malonyl-bound substrate binds to FabD Using arrows, show the electronic rearrangements that lead to the first intermediate. (5 points) l3. What happens to ATP in the presence of hexokinase and xylose? Why does this occur? (4 points) ATP is hydrolyzed. Xylose activates hexokinase through induced fit to catalyze phosphoryl transfer from ATP; however, the xylose molecule is not as well-positioned as the natural substrate, glucose, to attack the terminal phosphoryl group of ATP, and a water molecule acts as the nucleophile. I4. Enzymes increase reaction rates. Use a free energy diagram to explain how they do this. Compare curves for the uncatalyzed and catalyzed reactions, making sure to label the important sections of the curves. (6 points) Enzymes stabilize the transition state to decrease the activation energy for a reaction. fare/*7; 6‘ Free Rpuc f/cvi COC'CJ' “4*: ...
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This note was uploaded on 10/24/2009 for the course CH 369K taught by Professor Techniquesofresearch during the Spring '08 term at University of Texas at Austin.

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Exam3Key - Biochemistry I (CH 339K) Fall 2008 Unique #54715...

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