Chem153A Spring 2012 Midterm #2 Form D Last Name:_____________________________ Page 1 of 7 Chem 153A Spring 2012 Midterm #2 Form D May 23, 2012 _____ ____ ___ ___ _ 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 over the word limit will be graded. 2 (17 pts) 3 (24 pts) 4 (22 pts) 5 (15 pts) 6 (10 pts) 7(12 pts) Total (100pts):Chem153A Spring 2012 Midterm #2 Form D Last Name:_____________________________ Page 2 of 7 1. (4 pts) What is meant by stabilization of the transition state? How does this allow enzymes to increase the rate of chemical reactions? (40 words maximum) The Enzyme makes additional contacts to the transition state (that aren’t makes to the substrate or product), stabilizing it/lowering its free energy. This lowers the free energy of activation of the reaction, increasing the reaction rate. Consider the mechanism of Lysozyme diagramed below: (9pts) name all the enzymatic mechanism(s) indicated by each number: 1. Proximity and Orientation 2. Electrostatic Catalysis Preferential binding of the TS 3. Covalent catalysis (3 pts) What amino acid could be represented by A? Justify your answer His/Lys/Arg, are/can be protonated under physiological pH so they have a proton to donate. Glu/Asp -1 (1 pt) What Enzyme class does Lysozyme belong to? Hydrolase A A -A -A -A 1 2 3Chem153A Spring 2012 Midterm #2 Form D Last Name:_____________________________ Page 3 of 7 2. You assay 0.00005 μM of Enzyme X at two different substrate concentrations in the presence and absence of an inhibitor, with the following results: [substrate] mM [product] after 10sec of reaction without Inhibitor [product] after 10sec of reaction with 2 mM Inhibitor 16.67  1/[S] = 0.06 50/10 = 5 uM/s 0.2 12.5/10 = 1.25 uM/s  0.8 50 1/[S] = 0.02 100/10 = 10uM/s  0.1 25/10 = 2.5uM/s  0.4 a. (6pts) 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. Vmax_1/0.05 = 20uM/s_ KM_-1/-0.02 = 50 mM__ turnover number (kcat)_20uM/s / 0.00005 uM = 400000/sec_ c. (3 pts) How would each of these parameters change if you doubled the concentration of enzyme? Vmax___40uM/s_________ KM_No Change___ turnover number (kcat)__No Change______ d. (4 pts) Explain your Answer in part c. Vmax depends on the enzyme concentration, so if you double the amount of enzyme you double Vmax. Km and Kcat are constants so chaging the enzyme concentration will not change their value. e. (3pt) Is this an efficient enzyme? Show all Your work. 400000/sec / 0.05 M = 8 x 106 Yes f. (1 pt) What type of inhibition is demonstrated by your results? _Non-Competitive________________ g. (1pt) This inhibitor is most likely to bind to the: (circle one) Free enzyme Enzyme-substrate complex Both Can’t tell without math 00.20.40.60.81-0.04 -0.02 0 0.02 0.04 0.061/V [uM/s] 1/[S] mMChem153A Spring 2012 Midterm #2 Form D Last Name:_____________________________ Page 4 of 7 3. Compounds H and F are the products of a biosynthetic pathway, including intermediates A through H, that branches after intermediate D. H is an essential metabolite, whereas F is not. You have 7 individual strains containing a mutation in an enzyme involved in the overall pathway and assess them for their ability to grow on each metabolite. The data collected is presented in the table below. (a + means the strain with a mutation in the enzyme to the left is capable of growing when fed the metabolite at the top of the column) a. (12pts) Indicate the possible orders of the metabolites and the enzymes in the pathway. 3 2 5 7 1 E or G  B  C  D  A  H 6 or 4 E or G F 6 or 4 b. (4 pts) Given that there are 2 intermediates and 2 enzymes whose place in the pathway you cannot determine. How could you distinguish between these possibilities? 30 words maximum. Assess mutants 4 and 6 for intermediate accumulation. The branch point will only accumulate 1 of E/G and that is the first intermediate in the pathway. c. (2 pts) Which enzymes in this pathway would you expect to be regulated? 3, 7, 6or4 (the other branch point enzyme d. (4 pts) Why? (40 words maximum) The first committed step (3) is regulated to ensure biological efficiency and flexibility. Branch points (7 and 6 or 4) are regulated to proper use of of the shared metabolite (D) A B C D E F G H 1 + 2 + + + + 3 + + + + + 4 Not Auxotrophic 5 + + + 6 Not Auxotrophic 7 + +Chem153A Spring 2012 Midterm #2 Form D Last Name:_____________________________ Page 5 of 7 4. (5 pts) Briefly explain how fructose-2-6-bisphosphate is a regulatory molecule. Include how it reflects the overall conditions of both the individual cell and the whole organism. (50 words maximum) Fructose 2-6-bisphosphate activates glycolysis via PFK-1 when glucose concentrations are high. It is produced from F-6-P whose build up indicates there are high glucose levels in the cell. When organism wide glucose concentrations are high PFK-2 is dephosphorylated by hormone signaling, and will phosphorylate F-6-P to F-2,6-bP. 5. (5 pts) Why is hexokinase not suitable as the first step in glycogen production? How does the liver compensate for this? (50 words maximum) Hexokinase is inhibited by its product G-6-P. So at high glucose concentrations when we need to make glycogen hexokinase is inactive. The liver has glucokinase, which is active at high glucose because it is not inhibited by G-6-P 6. (5 pts) What two methods does the cell use in order to drive thermodynamically unfavorable reactions in the desired direction? Give one example of each method. (50 words maximum) Coupling it to a highly thermodynamically favorable reaction i.e hydrolysis of ATP, as in hexokinase or phosphofructokinase, or GAPDH and phosphoglycerate kinase Maintaing very low product concentrations i.e. phosphoglucose isomerase is driven by low F-6-P