Bioc385-Exam2-Spr16-KEY

Bioc385-Exam2-Spr16-KEY - € BIOC 385 – EXAM 2-­ March...

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Unformatted text preview: € BIOC 385 – EXAM 2 -­ March 8 2016 KEY Student Name________________________ Preceptor/Team___________ Energy Charge (EC): Change in Free Energy: [ product]actual ΔG = ΔGº '+RT ln [substrate]actual [ATP] + 0.5[ADP] [ATP] + [ADP] + [AMP] ATP exchange ratios: 2 NADH = 5 ATP 2 FADH2 = 3 ATP € 1 Acetyl-­‐CoA = 10 ATP USEFUL CONSTANTS: R = 8.3 x 10–3 kJ/mol•K F = 96.5 kJ/V•mol 25 °C = 298 K ΔGº’ = -RTln Keq ΔGº' = -nFΔEº' ΔG = RT·ln(C2/C1) + ZFΔV ΔE º '= ( E º ' e − acceptor) − ( E º ' e − donor) ΔG = ΔpH + ΔΨ € Physiological Standard Free Energy (ΔGº’) changes for selected glycolytic reactions. REACTION ΔGº' (kJ/mol) Glucose + ATP -­‐-­‐> glucose6-­‐P + ADP -­‐16.7 Glucose-­‐6P <-­‐-­‐> fructose-­‐6P +1.7 ATP -­‐-­‐> ADP + Pi -­‐30.5 Fructose-­‐1,6BP <-­‐-­‐> fructose-­‐6-­‐P + Pi -­‐16.3 Fructose-­‐1,6-­‐BP <-­‐-­‐> DHAP + glyceraldehyde-­‐3P +23.8 1,3-­‐BPG +ADP <-­‐-­‐> 3-­‐PGA + ATP -­‐18.8 3-­‐PGA <-­‐-­‐> 2-­‐PGA +4.6 2-­‐PGA <-­‐-­‐> PEP +1.7 PEP + ADP <-­‐-­‐> pyruvate + ATP -­‐31.4 Table of Standard Reduction Potentials Half-­‐reactions + α-­‐Ketoglutarate + CO2 + 2 H + 2 e-­‐ -­‐-­‐> isocitrate + NAD+ + H + 2 e-­‐ -­‐-­‐> NADH + -­‐ 1,3 BPG + 2H + 2 e -­‐-­‐> GAP + FAD(free) + 2 H + 2 e-­‐ -­‐-­‐> FADH2(free) + Oxaloacetate 2 H + 2 e-­‐ -­‐-­‐> malate + Fumarate + 2 H + 2 e-­‐ -­‐-­‐> succinate + Q + 2 H + 2 e-­‐ -­‐-­‐> QH2 + FAD(enz) + 2 H + 2 e-­‐ -­‐-­‐> FADH2(enz) + ½ O2 + 2 H + 2 e-­‐ -­‐-­‐> H2O Eº' (V) -­‐0.38 -­‐0.32 -­‐0.29 -­‐0.22 -­‐0.16 +0.03 +0.04 +0.05 +0.82 1 NO Notes, Books, Smart Phones, or Pre-­Programmed Calculators may be used. Put your name on this Exam Copy and on the ScanTron sheet. Record your answers on BOTH the provided ScanTron sheet and the EXAM, in case ScanTron is destroyed in machine or lost. You must turn in your EXAM with your ScanTron to be graded;; One exam and One ScanTron. The answers are graded objectively as correct or incorrect, there is no partial credit. ScanTron is the FINAL answer. BIOC 385 – EXAM 2 -­ March 8 2016 KEY Student Name________________________ Preceptor/Team___________ This page is purposely left blank 2 BIOC 385 – EXAM 2 -­ March 8 2016 KEY Student Name________________________ Preceptor/Team___________ 1. (6 pts) Blood glucose levels are initially high after breakfast, but then return to normal by 2 hours. Moreover, despite that you have not eaten for eight hours before breakfast, your blood glucose levels are near normal. Which statements below best describe how blood glucose levels are normal before breakfast, and 2 hours after breakfast, given that 8 hours of fasting is followed by a meal with carbohydrate (sugar). 1. Glucagon hormone stimulates glycogen degradation and gluconeogenesis before breakfast. 2. Insulin hormone stimulates glycogen degradation and gluconeogenesis before breakfast. 3. Insulin and glucagon are responsible for the level of red blood cells in the serum, hence glucose levels. 4. Glucagon hormone stimulates glucose export before breakfast to keep blood glucose levels steady. 5. Glucagon hormone stimulates glucose import before breakfast to keep blood glucose levels steady. 6. Insulin hormone stimulates glycogen synthesis and glycolysis after breakfast to lower blood glucose. 7. Insulin hormone inhibits glycogen synthesis and glycolysis after breakfast to raise blood glucose levels. 8. Glucagon hormone stimulates glycogen degradation before breakfast to lower blood glucose levels. A. 2, 5, and 8 are correct. B. 3, 7, and 8 are correct. C. 1, 4, and 6 are correct. D. 2, 3, and 7 are correct. E. None of these statements are correct. 2. (5 pts) Do plants and animals have similar metabolism at night? Explain. Choose one answer. A. Yes. They both metabolize nutrients using aerobic metabolism to generate ATP in mitochondria. B. No. Plants use photosynthesis at night to generate ATP and animals use glycolysis at night. C. No. Only animals metabolize nutrients at night but plants actually export nutrients to the soil at night. D. Yes. They both metabolize nutrients using anaerobic metabolism to generate ATP in mitochondria. E. Yes. Animals and plants both respond to glucagon (before breakfast) and degrade glycogen to sugar. 3. (6 pts) Calculate the ΔGº’ for the conversion of glucose-­6-­phosphate to fructose-­1,6-­bisphosphate in which the net reaction is described as glucose-­6-­P + ATP <-­> Fructose-­1,6-­BP + ADP using the table of ΔGº’ values on the front page of the exam. Choose the BEST answer. A. -­45.1 kJ/mol B. +12.5 kJ/mol C. -­43.4 kJ/mol D. -­12.5 kJ/mol E. -­28.4 kJ/mol 4. (6 pts.) Calculate the ΔG value (actual change in free energy) at 25 ºC for the aldolase reaction, which converts fructose-­1,6-­bisphosphate (F-­1,6-­BP) to dihydroxyacetone-­phosphate (DHAP) and glyceraldehyde-­3-­phosphate (GAP), given the ΔGº’ values on front page of exam and the concentrations of these metabolites at steady state;; F-­1,6-­BP = 15 mM, DHAP = 4 x 10-­5M, GAP = 0.04 mM. A. +1.2 kJ/mol ΔG = +23.8 kJ/mol + (8.3 x 10–3 kJ/mol•K) • (298K) • ln (4 x 10-­5M) (4 x 10-­5M) B. -­15.8 kJ/mol 1.5 x 10-­2M C. -­2702 kJ/mol ΔG = +23.8 kJ/mol + (8.3 x 10–3 kJ/mol•K) • (298K) • (-­16.0) D. -­20.1 kJ/mol ΔG = +23.8 kJ/mol + (-­39.6 kJ/mol)) E. +20.5 kJ/mol ΔG = -­ 15.8 kJ/mol 3 BIOC 385 – EXAM 2 -­ March 8 2016 KEY Student Name________________________ Preceptor/Team___________ 5. (6 pts) What is the metabolic advantage of having glucokinase activity limited to the liver and pancreas? 1. It provides a way to target glycogen to tissues with ATP synthesis during rest. Which statements are 2. It provides a way to increase glycogen stores in liver cells, which is important to correct? maintain safe blood glucose levels in between meals. 3. It provides a way to link blood glucose levels to glucagon release from A. 1 and 4 pancreatic beta cells. B. 2 and 6 4. It changes the equilibrium constant for glucose dephosphorylation in liver and C. 3, 5 and 7 pancreas cells. D. 2, 6 and 7 5. It prevents diabetes by increasing expression of glucagon receptors on E. None are correct. pancreatic beta cells. 6. It provides a way to stimulate insulin release from pancreatic beta cells by activating ATP synthesis and calcium-­mediated vesicle fusion. 7. It provides a way to stimulate insulin release from pancreatic beta cells in response to high blood glucose levels (after breakfast). 6. (6 pts) Select the answer below that has the correct underlined words in this passage. Glycolysis is an anaerobic / aerobic pathway, which metabolizes dietary hexose sugars such as fructose / lactase to yield 2 net ATP, 2 FADH2 / 2 NADH, and 2 pyruvate per mole of hexose sugar. When energy charge is low, flux through glycolysis is stimulated / inhibited by the allosteric effector ATP / AMP through activation of phosphofructokinase-­1 / hexokinase-­1 activity. A. aerobic, lactase, 2 NADH, stimulated, ATP, phosphofructokinase-­1 B. anaerobic, fructose, 2 NADH, stimulated, AMP, phosphofructokinase-­1 C. anaerobic, fructose, 2 FADH2, stimulated, AMP, hexokinase-­1 D. anaerobic, lactase, 2 NADH, stimulated, ATP, phosphofructokinase-­1 E. None of these combinations are correct. 7. (6 pts.) Calculate the standard free energy change (ΔGº') for the combined reactions in glycolysis converting 1,3-­bisphosphoglycerate to pyruvate using some, none, or all of the reactions listed below. Glyceraldehyde-­3P + NAD+ + Pi <-­-­> 1,3-­bisphosphoglycerate + NADH + H+ ΔGº' +6.3 kJ/mol 1,3-­bisphosphoglycerate + ADP <-­-­> 3-­phosphoglycerate + ATP ΔGº' -­18.8 kJ/mol 3-­phosphoglycerate <-­-­> 2-­phosphoglycerate ΔGº' +4.6 kJ/mol 2-­phosphoglycerate <-­-­> phosphoenolpyruvate + H2O ΔGº' +1.7 kJ/mol + phosphoenolpyruvate + ADP + H <-­-­> pyruvate + ATP ΔGº' -­31.4 kJ/mol pyruvate + NADH + H+ <-­-­> lactate + NAD+ ΔGº' -­25.1 kJ/mol A. -­106.6 kJ/mol B. -­100.3 kJ/mol C. -­43.9 kJ/mol D. -­68.9 kJ/mol E. -­50.2 kJ/mol 8. (6 pts) The disease beriberi is caused by a nutritional deficiency in vitamin B1 (thiamin). (Q1) What key mitochondrial enzyme uses thiamin as a coenzyme in a reaction that generates acetyl-­CoA?. (Q2) What would be the best way to treat someone with disease symptoms resulting in beriberi? A. (Q1) Pyruvate carboxylase, (Q2) feed them a high cornmeal diet. B. (Q1) Citrate synthase, (Q2) feed them a meal with brown rice C. (Q1) Pyruvate dehydrogenase, (Q2) feed them a meal with white rice D. (Q1) Phosphofructokinase, (Q2) feed them a sushi meal E. None of these (Q1) and (Q2) answer combinations are correct. 4 BIOC 385 – EXAM 2 -­ March 8 2016 KEY Student Name________________________ Preceptor/Team___________ 9. (6 pts) A muscle biopsy from an individual who was incapable of carrying out prolonged intense exercise was found to contain a deficiency in the enzyme lactate dehydrogenase. What is the biochemical explanation for the inability of this individual to run up three flights of stairs? Choose the one best answer. A. The deficiency prevented the synthesis of ATP because calcium levels in the muscle are dependent on the reduction of NADH needed to replenish NAD+ for the glycolytic pathway. B. The deficiency provides another mechanism for ATP depletion through the dephosphorylation of phosphoenolpyruvate by the pyruvate kinase, so no ADP, no muscle contraction. C. The deficiency prevented the oxidation of NAD+ needed to replenish NADH for the citrate cycle and reduced production of ADP following muscle contraction under aerobic conditions. D. The deficiency prevented the oxidation of NADH needed to replenish NAD+ for the glycolytic pathway and ongoing production of ATP for muscle contraction under anaerobic conditions. E. The deficiency prevented the reduction of NADH needed to replenish FAD for the glycolytic pathway and ongoing production of ATP for muscle contraction under aerobic conditions. 10. (6 pts) Match the reactants and products of the pyruvate Reactants Products dehydrogenase reaction using the numbers given as the correct 1. NADH 7. Pyruvate reactants (choose from 1-­6) and correct products (choose from 7-­12). 2. CoA 8. CO2 3. Acetyl-­CoA 9. NADH A. Reactants;; 1, 3, 4 and Products;; 8, 10, 12. 4. CO2 10. NAD+ B. Reactants;; 1, 3, 5 and Products;; 7, 10, 11. 5. NAD+ 11. Pyruvate C. Reactants;; 2, 5, 6 and Products;; 8, 9, 12. 6. Pyruvate 12. Acetyl-­CoA D. Reactants;; 3, 4, 5 and Products;; 10, 11, 12. E. Reactants;; 2, 4, 6 and Products;; 7, 8, 9. 11. (6 pts) Choose the correct standard reduction potentials (Eº') from the table on page 1 to calculate the change in standard free energy (ΔGº') for the citrate cycle reaction that converts isocitrate to α-­ ketoglutarate, and in the process reduces NAD+ to generate NADH + H+. A. -­36.7 kJ/mol ΔEº’ = (-­0.32 V) – (-­0.38 V) = +0.06 V B. -­73.3kJ/mol ΔGº' = -­nFΔEº' C. +11.6 kJ/mol ΔGº' = (-2) • (96.5 kJ/mol•V) • (+0.06 V) D. -­5.8 kJ/mol ΔGº' = -­11.6 kJ/mol E. -­11.6 kJ/mol 12. (6 pts) The complete oxidation of glucose by combined metabolic reactions is described by: Glucose (C6H12O6) + 6 O2 → 6 CO2 + 6 H2O. What enzymatic reactions account for the 6 CO2 ? A. Pyruvate dehydrogenase, isocitrate dehydrogenase, α-­ketoglutarate dehydrogenase B. Phosphofructokinase-­1, citrate synthase, α-­ketoglutarate dehydrogenase C. Pyruvate dehydrogenase, citrate synthase, isocitrate dehydrogenase D. Malate dehydrogenase, pyruvate carboxylase, citrate dehydrogenase E. Pyruvate hydrogenase, citrate dehydrogenase, α-­ketoglutamate dehydrogenase 13. (5 pts) What explains the ΔG value of ~0 kJ/mol for the citrate cycle reaction that is catalyzed by malate dehydrogenase considering that the ΔGº’ value for the reaction is ~30 kJ/mol? Choose the one best answer. A. The reason is that the ΔGº’ and ΔG values are affected by the level on enzyme, high enzyme, low ΔG. B. The reason why the ΔGº’ = ~30 kJ/mol but the ΔG = ~0 kJ/mol is because the citrate cycle is blocked. C. Malate is at high concentrations after breakfast, but it decreases with insulin, hence ΔG = ~0 kJ/mol. D. Oxaloacetate concentration is very low under normal conditions because it is quickly converted to citrate. E. None of these answers are correct. 5 BIOC 385 – EXAM 2 -­ March 8 2016 KEY Student Name________________________ Preceptor/Team___________ 14. (6 pts) What is the ΔG value in kJ/mol to transport 10 H+ across the mitochondrial membrane at 25ºC in yeast cells given that the concentration of H+ in the mitochondrial matrix is ~5% that of the concentration of H+ in the intermembrane space, and the membrane potential is 160 mV? A. +22.8 kJ/mol ΔG = (8.3 x 10-­3 kJ/mol· K)·(298 K)·ln(20) + [(+1)·(96.5 kJ/mol·V)·(0.16V)] B. +19.4kJ/mol ΔG = 7.4 kJ/mol + 15.4 kJ/mol C. +162 kJ/ ΔG = 22.8 kJ/mol for 1 H+ D. +228.0 kJ/mol ΔG = 228 kJ/mol for 10 H+ E. -­218.0 kJ/mol 15. (6 pts.) Two villages depend on corn as their primary food source. Corn contains niacin. Pellagra was prevalent in village A, but no pellagra was found in village B, even though the amount of dietary corn was equal in both villages. The one major difference was that village A cooked their fresh corn in water, whereas, village B first soaked the fresh corn in mineral lime (calcium hydroxide) before cooking. What explains the presence of pellagra in village A? Choose the one best answer. A. Village A ate corn that was grown in soil contaminated with calcium hydroxide, so niacin was absent. B. Niacin is bound to corn proteins and it needs to be released by treating with lime before cooking. C. It is likely that the reason for pellagra in village A is that they harvested fresh corn instead of aged corn. D. Resident of both villages had pellagra but the residents in village B only ate a little bit of corn daily. E. None of these answers are correct. 16. (6 pts.) Pyruvate carboxylase is a mitochondrial enzyme that converts pyruvate to oxaloacetate. Explain why a mutation in this enzyme, which blocks its ability to be allosterically-­activated by acetyl-­CoA, results in lower rates of energy conversion via the citrate cycle. Choose the one best answer. A. Because energy charge and energy conversion are two different things, therefore acetyl-­CoA is critical. B. Because mutations occur in the non-­coding DNA and therefore histone acetylation by CoA is blocked. C. Because the enzyme is degraded when it cannot be allosterically activated by a metabolite. D. Because energy conversion via the citrate cycle depends on the synthesis of pyruvate from citrate. E. Because conversion of pyruvate to oxaloacetate is unable to balance the input of acetyl-­CoA. 17. (6 pts). Energy charge, levels of oxidized and reduced coenzymes, and feedback inhibition by downstream metabolites, provide regulatory strategies for key metabolic enzymes. For example, if ATP levels are high, then enzymes that catalyze flux through ATP generating reactions are inhibited, and similarly, if NAD+ levels are high, then enzymes that promote NAD+ reduction are activated. Circle the ONE word that best describes the effect of each regulatory molecule on the given enzyme activity. 1. Hexokinase I is activated / inhibited / unregulated by glucose-­6P. + 2. Pyruvate dehydrogenase is activated / inhibited / unregulated by NAD and ADP. 3. Phosphofructokinase I is activated / inhibited / unregulated by ATP and citrate. 4. Isocitrate dehydrogenase is activated / inhibited / unregulated by ADP. 5. Citrate synthase is activated / inhibited / unregulated by NADH and citrate. 6. Glucokinase is activated / inhibited / unregulated by glucose-­6P. A. inhibited(1) B. unregulated(1) C. inhibited(1) D. inhibited(1) E. activated(1) activated(2) activated(2) inhibited(2) inhibited(2) activated(2) inhibited(3) activated(4) inhibited(5) inhibited(3) inhibited(4) inhibited(5) unregulated(3) activated(4) inhibited(5) inhibited(3) activated(4) activated(5) activated(3) unregulated(4) inhibited(5) 6 unregulated(6) unregulated(6) activated(6) unregulated(6) inhibited(6) ...
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