CH395 G Exam 2 Multiple Choice - Fall 2004

Transcription

1 C395 G Exam 2 Multiple Choice - Fall Which of the following fatty acids has the lowest melting point? a. Fatty acids with sites of unsaturation with cis double bonds b. Fatty acids with sites of unsaturation with trans double bonds c. Fatty acids with no sites of unsaturation d. Fatty acids with longer hydrophobic tails e. None of these will make a difference 2. What is the major lipid component of biological membranes? a. triacylglycerols b. phosphoglycerides c. fatty acids d. cholesterol e. waxes 3. A membrane is impermeable to charged molecules and does not allow passage of ions because of: a. the thickness of the membrane b. the hydrophobic nature of the membrane c. the hydrophilic nature of the membrane d. all of the above e. none of the above 4. The carbohydrate portion of a glycoprotein is found: a. on the interior side of the cell membrane b. spanning the cell membrane c. on the exterior side of the cell membrane d. all of the above e. none of the above 5. Which of the following is not a specific advantage of enzymes? a. Enzymes can be quickly turned off by allosteric control. b. Enzymes can function at large ranges of p and temperature. c. Enzymes exhibit very high substrate specificity. d. Enzymes can catalyze reactions several orders of magnitude faster than chemically-catalyzed reactions. e. Enzyme activity can be enhanced through covalent modifications.

2 6. Which of the following statements about cofactors/coenzymes is not true? a. Many vitamins serve as coenzyme precursors. b. Metal ions could serve as cofactors. c. NAD is a coenzyme that catalyzes redox reactions. d. Coenzymes can be either transiently or permanently associated with an enzyme. e. Once a coenzyme is chemically changed in a reaction, it must be regenerated by the same enzyme. 7. Which of the following equations accurately describes enzyme-catalyzed conversion of substrate to product? a. E + S ES E + P b. E + S ESP E + P c. E + S EP E + P d. E + P E + S e. none of the above 8. What is an accurate velocity equation for the following reaction if [B] is in excess? A + B C a. v = -d[a] b. v = -d[b] c. v = -d[c] d. v = d[a] e. v = d[b] 9. Competitive inhibitors have the following effects on K M and V max : a. They alter the value of K M but not V max b. They alter the value of V max but not K M c. They alter the value of K M and V max d. none of the above e. all of the above 10. Mixed inhibitors have the following effects on K M and V max : a. They alter the value of K M but not V max b. They alter the value of V max but not K M c. They alter the value of K M and V max d. none of the above e. all of the above

3 11. A system at its transition state a. is at its lowest free energy value. b. requires a large amount of energy to proceed to product. c. has an equal probability of proceeding to product or decomposing back to reactants if G reaction = 0. d. describes a system in which the reactants have not yet come into contact with one another. e. describes a system in which products have been produced. 12. One way in that an enzyme can carry out the catalysis of a reaction is by a. preferentially binding the transition state analog of a reaction. b. carrying out redox reactions. c. bringing substrates close to one another. d. aligning substrates in the proper orientation for a reaction. e. all of the above. 13. In gluconeogenesis, the initial reaction converts pyruvate to a. oxaloacetate b. acetyl-coa c. phosphoenolpyruvate d. lactate e. malate 14. The enzyme in gluconeogenesis that reverses the reaction of phosphofructokinase in glycolysis is a. glucose-6-phosphatase. b. fructose-1,6-bisphosphatase. c. pyruvate carboxylase. d. hexokinase. e. none of the above 15. If a reaction has a negative value and a negative S value, it is a. spontaneous at all temperatures. b. unspontaneous at all temperatures. c. spontaneous at high temperatures and unspontaneous at low temperatures. d. unspontaneous at high temperatures and spontaneous at low temperatures. e. none of these are correct. 16. The conversion of dihydroxyacetone phosphate to give glyceraldehydes-3-phosphate is a(n) a. phosphorylation b. cleavage c. phosphate transfer d. isomerization e. dehydration

4 17. Glucose residues are cleaved from a glycogen branch until is about three units long. What enzyme is required to continue glycogen breakdown? a. debranching enzyme b. branching enzyme c. glycogen phosphorylase d. glycogen synthase e. phosphoglucomutase 18. In glycogen synthesis, what happens when UTP concentrations get low? a. UTP levels have no effect on glycogen synthesis. b. Glycogen synthesis stops. c. ATP is used instead of UTP. d. ATP replenishes UTP supply through nucleoside diphosphate kinase. e. Another branch point is formed and synthesis continues from there. 19. Pyruvate carboxylase a. is located in the cytosol. b. has acetyl-coa as a negative allosteric effector. c. uses biotin as a coenzyme. d. converts pyruvate to malate. e. None of these are true. 20. Which of the following is not an activator of PFK? a. ADP b. camp c. F2,6BP d. F1,6BP e. citrate

5 C395G Exam 2 - Fall Small lipid vesicles that are designed to contain drugs or other water-soluble molecules for targeted delivery are called. liposomes / micelles In animal and plant tissues the energy storage oils and fats are typically triacylglycerides. / triglycerides / triacylglycerols A type of lipid found in bilayers that is a precursor to steroid hormones is. cholesterol In fatty acids found in bilayers, the double bond is typically in the cis configuration. (+1/2 point for each fill-in-the-blank answer) One method of lysing (breaking open) bacterial cells involves the use of the enzyme lysozyme. a) What does the lysozyme do to allow it to break open bacteria? (+1) The cell walls of bacteria are made up of a mesh network containing polysaccharides with alternating NAG-NAM units. Lysozyme cleaves the β(1-4) linkage between the NAG-NAM units thus opening up the cell wall and allowing lysis. b) Why isn t a protease a good alternative choice for this task? (+1) Proteases are ineffective because the cell wall is primarily polysaccharide. Common, inexpensive proteases do not cleave the peptide portions of the cell wall. c) Why could a triacylglyceride lipase not be used successfully for this job? (+1) The cell membranes of bacteria do not contain triacylglycerides. Cholesterol can alter the fluidity of cell membranes. Explain briefly in what manner it does this and what properties of cholesterol suit it to this task. (+2) Cholesterol decreases the fluidity of cell membranes because the rigid structure of the steroid ring interferes with motion of the fatty acid tails of the phospholipids. In addition, hydrophobic interactions between the tails and cholesterol restrict movement. Cholesterol is suited to this task because it has a fused ring structure, which is essentially planar and rigid. The single hydrophobic -O group orients cholesterol in the membrane with the ring structure perpendicular to the surface.

6 2. The progress of a reaction, measured as the disappearance of substrate or the appearance of product over a specific time period, is called the. velocity The plot of substrate concentration versus velocity results in a hyperbolic curve. In rate equations, the symbol k refers to the rate constant. In a second-order reaction, the rate is dependent on the concentration of reactant(s). two In an enzyme-catalyzed reaction that is zero order with respect to substrate, the rate is dependent on the concentration of. enzyme The velocity at the start of the reaction, or v o, is called the. initial velocity When [S] is equal to K M, the reaction velocity is equal to half the maximum velocity. At very high substrate concentrations, the enzyme becomes saturated with substrate. (+1/2 per fill-in-the-blank) The Michaelis-Menton equation is used to describe the behavior of very simple enzyme catalyzed reactions. a) Briefly describe two of the assumptions utilized in the derivation of the equation. (+1) i) k -1 >>k 2 so that the first step E + S k 1 k 2 ES achieves equilibrium ii) [ES] maintains a steady state or d[es] = 0 b) Name one class of enzyme for which the Michaelis-Menton equation is poorly suited. (+1) allosteric/ multi-subunit reactions c) On the graph below draw a comparison between an enzyme in the presence of no inhibitor and the enzyme in the presence of two different concentrations of an inhibitor that affects the enzyme in a classical uncompetitive manner. Please be sure to clearly label your lines. (+2) See Fig (a) has [I] (b) has higher [I] 1/v o (b) (a) no inhibitor 1/[S]

7 3. The enzyme aspartyl transcarbamoylase carries out the reaction carbamoyl phosphate + aspartate N-carbamoylaspartate + phosphate This enzyme regulates pyrimidine synthesis. (+2) CTP inhibits this enzyme in an allosteric fashion. On the graph below draw an approximation of the enzyme behavior in the presence and absence of this inhibitor. (+2) no inhibitor see p 465 Fig 13-5 v o + CTP [S] Very briefly describe, using diagrams if you choose, what happens when the substrates for this enzyme bind to the enzyme. (+2) See pp , especially Fig In the absence of substrate, the catalytic subunits are in the T-state with the carbamoyl phosphate and aspartate binding sites separated by steric interfence. Upon substrate binding, the subunits move apart and rotate, forming the R-state and allowing the substrate sites to closely approach each other and thereby promote reaction to give product. ow might the inhibitor CTP affect these changes? (+2) In the presence of CTP, this inhibitor binds to the regulatory subunits and stabilizes the T-state of the enzyme (with substrate binding sites far apart). This T-state stabilization lessens the chance that a transition to the active R-state will occur and thus causes inhibition of enzymatic activity.

8 4. (+8) On the graph below, draw (to approximate scale using a solid line) the reaction coordinate diagram for a simple, uncatalyzed reaction for the conversion of S P where K eq = 8.5 x 10 5 and G act = 24.6 kj/mol (T = 25 o C) On the same graph, draw (dashed line) the reaction coordinate diagram for a simple M- M enzyme catalyzed reaction for the conversion of S P where the enzyme destabilizes the ES and EP complexes by 3.2 kj/mol and the enzyme stabilizes the transition state by 13.9 kj/mol. What is the extent of rate enhancement for the enzyme catalyzed reaction? ~1000 fold increase a) Keq = 8.5 x 10 5 ; G o = -RT lnkeq = kj/mol c) kcat / kuncat = exp( G)/RT = exp( )/2.48 = 988 or ~ 1000 fold rate enhancement

9 5. Identify the role or explain the meaning of each of the following terms as they relate to the catalytic mechanism of trypsin, a serine protease. (+2 each) a) tetrahedral intermediate: The tetrahedral intermediate is formed at the carbonyl carbon of the scissile bond by the nucleophilic attack on the carbonyl C by the Ser-O to form the oxyanion.. O - Ser-O R N R' b) acyl-enzyme: The acyl-e is a covalently modified form of the enzyme whereby the new C-terminus of the N-terminal portion of the polypeptide chain being cleaved is temporarily in covalent (ester) linkage with the attacking Ser-O- of the enzyme. c) catalytic triad: The catalytic triad refers to the three, -bonded, active-site residues in serine proteases (Asp is Ser195) that cooperate to make the O of Ser195 a better nucleophile to attack the carbonyl C of the scissile bond to be cleaved. d) oxyanion hole: The carbonyl C is in a trigonal form before attack by Ser195. During the formation of the tetrahedral intermediate an oxyanion is formed. This shifts the oxygen (oxyanion) into a new position (oxyanion hole) where the charge is stabilized in part by new -bonds to backbone =N- groups. 6. BLOSUM50 is a substitution matrix. What is a substitution matrix? ow is it generated? ow is it an aid when searching for similar sequences doing a BLAST search? (+5) A substitution matrix is a matrix that contains values proportional to the probability of one amino acid substituting (mutating into) another amino acid. The BLOSUM50 (BLOcks SUbstitution Matrix) is constructed by first assembling a large and diverse (no more than 50% sequence identity in this case) sample of verified, pairwise sequence alignments of amino acids. With a large enough sample, the observed mutations will reflect the probability in nature that each amino acid has for being substituted by each of the other amino acids. Such a substitution matrix with its values enables a numerical score to be assigned when two sequences are aligned in a pairwise fashion. The significance of this score is then tested against random permutations of the same sequences to ascertain an Expect value of the relative significance that alignment.

10 7. a) Consider the glycolytic pathway as illustrated below. Draw the structures of the missing compounds B, C, and E: (+2 each) B = F-1,6-BP C = 1,3 BPG E = Pyruvate -- O3 PO O OPO 3 -- O O C OPO 3 -- C O -- C OPO 3 - O O C C O C 3 Write out a balanced equation for the reversible reduction of NAD +. You need only to show the detailed structure of the nicotinamide portion of the coenzyme. (+3) See p 461, Fig 13-2 CON 2 CON 2 N + R NAD [] + + (= hydrogen atom) N R NAD or : - (hydride ion) no + or e - + +

11 8. Our study of glucose catabolism illustrated a role for some of the B vitamins, namely niacin and thiamine. Name or describe one enzyme or metabolic step in alcoholic fermentation that relies on each of these vitamins. (+1 each) a) niacin i) Glyceraldehyde 3-phosphate D (GAP or G3P 1,3 BPG) ii) Alcohol D (acetaldehyde ethanol) b) thiamine Pyruvate decarboxylase ( pyruvate acetaldehyde ) The hormone glucagon has a major effect on glycogen and glucose metabolism. Briefly but clearly, in words or with the aid of a diagram, describe its action, subsequent resulting actions and what effects glucagon has on these pathways. (+5) The hormone glucagon is released from the pancreas (α islets cells) in greater amounts under low [glucose] conditions. Glucagon stimulates the liver to release glucose both by promoting gluconeogenesis and by breaking down its liver glycogen stores. The steps involved are summarized below: 1. Released glucagon binds to a glucagons receptor. 2. Binding at receptor stimulates adenylate cyclase to convert ATP into c-amp. 3. c-amp acts as an allosteric effector to activate PKA which leads to increased phosphorylation. 4. PKA leads to phosphorylation of phosphorylase kinase and PFK-2 5. Phosphorylase kinase phosphorylates glycogen phosphorylase-b into glycogen phosphorylase-a (more active form) breakdown of glycogen to glucose-6-p 6. Phosphorylation of PFK-2 leads to inactivation of PFK-2 and activation of FBPase- 2, decrease in [F-2,6-BP] 7. Lower [F-2,6-BP] leads to activation of FBPase and lower activity of PFK. slow down of glycolysis and increase in gluconeogenesis more glucose-6-p. 8. Glucose-6-P from gluconeogenesis and glyogenolysis can be converted in the liver into Glucose for export to increase blood glucose levels.