midterm key

midterm key - BIBC 102 Metabolic Biochemistry Summer...

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BIBC 102 Metabolic Biochemistry KEY Summer Session 2 2011 Midterm Exam (400 pts.) Exam A Section 1—Do this section of the exam first: 1) (24 pts.) Using the axes below, sketch a plot of Gibbs free energy vs. the reaction coordinate for an endergonic chemical reaction. On the plot, indicate the standard free energy change for the reaction, and the standard free energy of activation for both the forward and reverse reactions. Label each of these with the correct symbol. -3 pts. each if correct symbol is not used. 2) (6 pts.) For a particular chemical reaction, K eq = 6.3 x10 -6 . The standard free energy change for this reaction would be: a. large and positive X b. large and negative____ c. close to zero____ 3) (12 pts.) What compound is the starting point for glycolysis (the substrate for hexokinase), and how many carbon atoms does it contain? Glucose 6 carbon atoms 4) (12 pts.) What compound is the end point for the oxidation of a carbohydrate in glycolysis (the product of pyruvate kinase), and how many carbon atoms does it contain? Pyruvate 3 carbon atoms 5) (6 pts.) What is the net gain in the number of ATP for the oxidation of a carbohydrate molecule through all 10 reactions of glycolysis? 2 ATP G Reaction Coordinate ΔG’° [6 pts.] ΔG° forward [6 pts.] ΔG° reverse [6 pts.] [6 pts. for correct curve]
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6) (12 pts.) What is the equilibrium constant (K eq ) for the following reaction, where k 1 = 0.04 M -1 sec -1 and k -1 = 833 M -1 sec -1 ? acetate + ethanol ethyl acetate + H 2 0 K eq = k 1 /k -1 = 0.04 M -1 sec -1 / 833 M -1 sec -1 = 6 pts. for correct set up, 6 pts. for correct answer 4.8 x10 -5 7) (12 pts.) The hydrolysis of sucrose to glucose and fructose has a standard free energy change of -29.3 kJ/mol. What is the equilibrium constant for this reaction? G’° = -RT ln K’ eq therefore ln K’ eq = G’°/-RT (3 pts.) = -29.3 kJ/mol = -29.3 kJ/mol -8.315 J/mol•K(310 K) -2.58 kJ/mol (3 pts.) ln K’ eq = 11.36 (3 pts.) K’ eq = 8.55 x10 4 (3 pts.) 8) (14 pts.) Draw a generic amino acid structure with the R-group indicated as simply “R.” Make this amino acid at neutral or physiologic pH. (Partial credit at grader’s discretion) 9) (12 pts.) A new enzyme inhibitor is developed and its effect on the kinetic properties of the enzyme are studied. C CO - O H 3 N + H R k 1 k -1 1/V 0 1/[S] enzyme alone + inhibitor
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Based on the Lineweaver-Burke reciprocal plot above (check one for each category): This inhibitor is X competitive _____noncompetitive This inhibitor is X reversible _____irreversible 10) (6 pts.) The enzyme P7 peroxidase has broad substrate specificity in that it can oxidize a variety of small organic molecules that have a similar double-ring structure. Some of these substrates are listed below along with their respective K M values. Which one of these compounds is the best Luminol—lowest K M substrate for P7 peroxidase? Compound
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This note was uploaded on 03/27/2012 for the course BIBC BIBC 102 taught by Professor Price during the Summer '08 term at UCSD.

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midterm key - BIBC 102 Metabolic Biochemistry Summer...

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