Chem153A Spring 2013 Midterm #2 Form D Last Name:_____________________________ Page 1 of 7 Chem 153A Spring 2013 Midterm #2 Form D May 17, 2013 _____ ____ ___ ___ _ Name (Last) (First) Student ID Lecture Page Score Be sure to read all the questions carefully, and answer the question that is asked. Answer questions as concisely as possible. The word limits are more than sufficient to fully answer the question. Nothing after the word limit will be graded. 2 (19 pts) 3 (21 pts) 4 (20 pts) 5 (18 pts) 6 (10 pts) 7(12 pts) Total (100pts):Chem153A Spring 2013 Midterm #2 Form D Last Name:_____________________________ Page 2 of 7 1 1. (4 pts) How do enzymes increase reaction rates? (40 words maximum) Enzymes preferentially bind the transition state, which stabilizes it, lowering the energy of activation for the reaction , and increasing the reaction rate. 2. Consider the mechanism of Deoxyribose-5-phosphate Aldolase diagramed below: a. (10 pts) name all the enzymatic mechanism(s) indicated by each number (if they occur more than once in a picture you only have to list them once): 1. Proximity and Orientation 2. Covalent catalysis Acid catalysis base catalysis 3. Base catalysis b. (2 pts) What amino acid could be represented by A? Justify your answer Asp/Glu, it is negatively charged when deprotonated and neutral when protonated c. (1 pt) What Enzyme class does this enzyme belong to? Lyase d. (2 pts) What can you conclude about the reaction catalyzed by Glucose:NADP+ oxidoreductase? Transfers electrons from glucose to NADP+ 2 3Chem153A Spring 2013 Midterm #2 Form D Last Name:_____________________________ Page 3 of 7 3. You assay 0.005 μM of Enzyme X at two different substrate concentrations in the presence and absence of an inhibitor, with the following results: [substrate] (μM) μM product after 20 sec of reaction without Inhibitor μM product after 20 sec of reaction with 2 mM Inhibitor 1/16.67 = 0.06 66.67/20 = 3.33  1/3.33 = 0.3 28.57/20 = 1.43  1/1.43 = 0.7 1/33.33 = 0.03 100/20 = 5  1/5 = 0.2 50/20 = 2.5  1/2.5 = 0.4 a. (5pts) Construct a Lineweaver-Burke plot of the data. Express Vo in μM/sec. Be sure to label your axes b. (6 pts) Determine each of the following for the enzyme in the absence of inhibitor. Show all your work and include units in the answer. 1/0.1 = 10uM/s -1/-0.03 = 33.33 uM 10uM/s / 0.005uM = 2000/s Vmax______________ KM____________ turnover number (kcat)_______________ c. (3 pts) How would each of these parameters change if you used 1/3 the amount of enzyme? Give a numerical answer. 3.33 uM/s 33.33 uM 2000/s Vmax______________ KM____________ turnover number (kcat)_______________ d. (3pt) Is this an efficient enzyme? Show all Your work Efficiency = kcat/Km = 2000/s /0.000033M = 6x10^7 M-1s-1 YES e. (2pt) What properties would make an enzyme efficient? A fast enzyme, and achieved that fast rate at a low substrate concentration f. (1 pt) What type of inhibition is demonstrated by your results? ___competitive____ g. (1pt) This inhibitor is most likely to bind to the: (circle one) Free enzyme Enzyme-substrate complex Both 00.20.40.60.81-0.04 -0.02 0 0.02 0.04 0.061/V [uM/s] 1/[S] mMChem153A Spring 2013 Midterm #2 Form D Last Name:_____________________________ Page 4 of 7 4. You are studying the utilization of a key metabolite. You have strains containing mutations in all the enzymes involved in the pathway. You grow each strain in the presence of the radiolabeled first metabolite and determine which intermediates can still become radiolabeled. A (+) indicates that when the enzyme to the left is mutated the radiolabel can still be detected in the intermediate at the top of the column. A B C D E F 1 + + - + + + 2 - + - - + - 3 + + - + + - 4 - + - - - - 5 - + + + + + a). (9 pts) Indicate the order of intermediates and the location of the enzymes in the pathway. 4 2 3 1 B  E  D  F  C 5 A b). (5 pts) Describe the strategy you would use to determine if the expression of the enzymes in this pathway are induced by the first metabolite of the pathway. Include your expected results. Grow cells in the presence and absence of compound B. Isolate proteins: Tag sample from the presence of intermediate with light ICAT and the sample in the absence of intermediate with heavy ICAT (or vice versa). Process samples. If it is induced the light ICAT peak will be higher. IF there is no induction the two peaks will be equal Run each sample on a separate gel. Identify the spots that correspond to the enzymes. If it is induced the spots corresponding to the enzymes on the gel with the sample from in the presence of the metabolite will be darker. If there is no induction the spots for the enzyme swill be the same on both gels c. (2 pts) Which enzymes in the pathway in part a would you expect to be regulated? 4 (first enzyme), 3 and 5 (branchpoint enzymes d. (4 pts) How would you expect the intermediates in the pathway to regulate these enzymes? Assume the pathway is under cumulative feedback inhibition. (50 words maximum) C should inhibit enzyme 3 and partially inhibit enzyme 4. A should inhibit enzyme 5 and partially inhibit enzyme 4 C and A together will fully inhibit enzyme 4.Chem153A Spring 2013 Midterm #2 Form D Last Name:_____________________________ Page 5 of 7 5. (5 pts) Most reactions required in glycolysis are thermodynamically unfavorable under standard conditions. What techniques are used by cells to make them thermodynamically favorable so that enzymes can catalyze the reaction in the necessary direction? (50 words maximum) Some reactions are highly unfavorable and the cell could never make these favorable, so they are coupled to highly thermodynamically favorable reactions Other reactions are slightly unfavorable so the cell can change the favorability by manipulating [P]/[R]. Thus by keeping [P] low (or [R] high) this pushes these reactions forward 6. ( 8 pts) Explain all the ways GAPDH is responsible for energy production from glycolysis. The product of GAPDH is 1,3-BPG which directly produces ATP in the next step (PGK) by substrate level phosphorylation GAPDH incorporations a Pi so that there can be a net production