sp07 first exam

sp07 first exam - Metabolic Biochemistry / BIBC 102 First...

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Metabolic Biochemistry / BIBC 102 First Exam / Spring 2007 I. (25 points) Fill in all of the enzyme catalyzed reactions which convert glucose to ethanol . Draw the correct structure for each intermediate molecule, showing the correct position of the hydrogen atoms at each carbon atom. Include the names or abbreviations of any other reactants and products (eg, ATP, ADP, Pi, NAD+, NADH, etc), and any required prosthetic groups (eg, biotin, TPP, etc.). (Note:You do not need to give the correct name for each intermediate molecule or the correct name of the enzyme to receive full credit.) Glycogen (draw the 2 terminal sugar subunits of the polymer only) lactate (draw structure)
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2 II. (18 points: answer any 3 of the following 4 problems A, B, C, and D, and cross out the other; if you don’t, we’ll automatically cross out the first problem). (A) In the space below, please sketch the Vo versus [S] plot for an enzyme obeying Michaelis-Menten kinetics in the presence and absence of an irreversible inhibitor that inactivates exactly ½ of the enzyme molecules. Label the two lines, write out the Michaelis-Menten equation , and indicate on your graph Km and Vmax for each condition. (B) In the space below, sketch the double reciprocal plot (Lineweaver-Burk plot) for an enzyme obeying Michaelis-Menten kinetics. Give the values of the x and y axis intercepts in terms of Km and Vmax. Also show on this plot what the graph would look like if you repeated the kinetic analysis in the presence of a competitive inhibitor. (C) In the space below, please sketch the Vo versus S plot for an enzyme which shows positive cooperativity in substrate binding, in the presence and absence of an allosteric activator ; label each line . What form of the allosteric enzyme binds the activator better, the R state or the T state ?
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5 (C). Consider the reaction: ATP + pyruvate phosphoenol pyruvate + ADP Given that the G °′ for ATP hydrolysis is –30.5 kJ/mol, the G °′ for phosphoenolpyruvate hydrolysis is –61.9 kJ/mol, RT is 2.48 kJ/mol, and the concentration are: (ATP) = 2.24 mM, (ADP) = 0.25mM, (pyruvate) = 0.051mM, (phosphoenolpyruvate) = 0.023mM, and (phosphate) = 8mM, what will the G be for this reaction? Do not calculate a numerical answer. Full credit will be given for correct set up of the
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sp07 first exam - Metabolic Biochemistry / BIBC 102 First...

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