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MCMP 208 Exam III - 1 Examination III MCMP 208 – Biochemistry for Pharmaceutical Sciences I April 14, 2014 Correct answers in multiple choice questions are indicated in RED and underlined. Correct answers to essay questions are indicated in RED in comic book font. In some cases and explanation is provided in BLUE/BLUE MATCHING. For problems 1 and 2, a set of numbered answers is provided immediately below. For each problem, select from the list of answers the single choice that best matches the item described in the problem. Mark that answer on your answer sheet. An answer may be used more than once or not at all. [3 points each] 1. Cholesterol Cholic acid 17β-Estradiol Leukotriene Prostaglandin α-Linolenic acid Stearic acid Sphingomyelin Phosphatidylserine Phosphatidylcholine The molecule has the sphingosine moiety with the phosphocholine head group. 2. The molecule has the cholesterol-like ring structure with multiple polar functional groups (three hydroxyl groups and a carboxylic group). MCMP 208 Exam III - 2 MULTIPLE CHOICE. For problems 3 to 22, select from the list immediately following each question the single most correct choice to complete the statement, solve the problem, or answer the question. Mark that answer on your answer sheet. [3 points each] 3. _____________ are the largest lipoproteins and are produced in the intestine to deliver lipids in diet to other organs. Chylomicrons Very-low-density lipoproteins Intermediate-density lipoproteins Low-density lipoproteins High-density lipoproteins 4. Which of the following statements is INCORRECT about integral membrane proteins? The surface of transmembrane domains is mostly hydrophobic. Hydropathy plots can be used to predict the putative transmembrane helices. Integral membrane proteins can be removed from the membranes easily without disrupting the membranes. Peripheral membrane proteins can be removed from the membranes without disrupting the membranes. Integral membrane proteins are removed from the membrane only after the membrane is dissolved by a detergent. Integral membrane proteins do not change their orientation in the membrane. Integral membrane proteins may have hydrophilic interior for transport functions. 5. Acetylcholine receptor is ______________________. an ATPase an acetylcholine pump an acetylcholine transporter a voltage-gated channel a ligand-gated channel 6. Which of the following statement is INCORRECT about ATP-ADP transporter? ****** Note: Either of Choice 1 or choice 3 will be scored as correct ****** It is a symporter. It is an antiporter. ATP and ADP move in different directions. It exchanges ATP and ADP. It functions at the mitochondrial outer membranes. It is on the inner mitochondrial membrane. The membrane potential promotes its action. As one net charge moves across the membrane, the transport is not electrically neutral. The exchange of ATP and ADP results in the transport of one net negative charge. ATP has 4 negative charges and ADP has 3 negative charges. 7. The main form in which triglycerides are stored inside human cells is in an VLDL Not used for storage inside cells in a chylomicron Not used for storage inside cells in a droplet in an endosome Not used for storage in a complex with albumin Not used for storage inside cells as thioesters with coenzyme A Not a storage mechanism in a bilayer membrane Not a storage mechanism – membranes have an essential barrier function MCMP 208 Exam III - 3 8. Longer chain fatty acids such as palmitate or stearate enter the mitochondria for oxidation in what form? monoacylglycerol diacylglycerol triacylglycerol phosphatidic acid free fatty acids thioesters with Coenzyme A carnitine esters cholesteryl esters thioester with acyl carrier protein 9. The regulated step in fatty acid synthesis is catalyzed by acyl-coenzyme A synthase citrate lyase glucose-6-phosphate dehydrogenase acetyl-coenzyme A carboxylase pyruvate dehydrogenase PEP carboxykinase glycerol kinase carnitine acyl transferase thiolase none of the above 10. Why does the liver make ketone bodies? To reduce glucose utilization by the brain To increase glucose production by the liver To reduce fatty acid mobilization in adipocytes To increase fatty acid mobilization in adipocytes To provide muscle with a third type of circulating energy molecule To reduce insulin levels in the blood To metabolize glycerol that had been freed from triacylglycerol during fatty acid mobilization To carry out glyceroneogenesis 11. Some phosphoglycerolipds are made directly from other phosphoglycerolipids. Other phosphoglycerolipids are made directly (in a single enzyme-catalyzed step) from diacylglycerol phosphatidic acid Not made directly into PL triacylglycerol Not made directly into PL ceramide This is a sphingolipid and is not used to make glycerol containing PL CDP-diacylglycerol diacylglycerol or CDP diacylglycerol Some use one, some use the other diacyglycerol or ceramide diacylglycerol or triacylglycerol diacylglycerol, CDP diacylglycerol, or ceramide MCMP 208 Exam III - 4 12. The activated 5 carbon unit that is added one or more time during the second-stage oligomerization that produces the isoprenoids (terpenoids) is β-hydroxy-β-methylglutaryl-CoA 6 carbons β-hydroxy-β-methylglutarate 6 carbons mevalonate Needs reduction and activation β-hydroxybutyrate 4 carbons L-methylmalonyl-coA 4 carbons isopentenylpyrophosphate α-ketoisovaleryl-CoA geranylpyrophosphate 10 carbons farnesylpyrophosphate 15 carbons 13. Before it is secreted, acetoacetate mostly is reduced to β-hydroxybutyrate. What is this advantageous? Acetoacetate is unstable and breaks down spontaneously. β-hydroxybutyrate is stable. β-hydroxybutyrate contains more energy than acetoacetate. It has one additional reducing equivalent. Acetone is not metabolically useful. Acetoacetate decomposes to acetone and carbon dioxide All of the above None of the above 14. The metabolite that (a) is a major ammonia donor in metabolic reactions and (b) is regenerated by a reaction that involves free ammonia is aspartate Aspartate is a frequent ammonia donor but it is always regenerated using glutamate. glutamine asparagine Asparagine is generated using ammonia, but is not used as an ammonia donor 5-phosphoribosylpyrophosphate 5-phosphoribosylamine This does not donate ammonia tetrahydrofolate S-adenosylmethionine carbamoylphosphate 15. Two amino acids are considered semi-essential because they cannot be made in adequate amounts by children but can be made in adequate amounts by adults cannot be made in people with certain hereditary disorders of amino acid metabolism are made metabolically from non-essential amino acids are made metabolically from essential amino acids Tyr from phe and cys from met are recycled back to their original structures after they have been metabolized require one carbon metabolism for their synthesis These are semiessential because they are only needed in the diet if Phe or Met are inadequate in the diet to meet the needs for Phe AND Tyr or Met AND Cys. MCMP 208 Exam III - 5 16. Hydroxylation of aromatic amino acids on their aromatic ring always occurs in metabolic reactions that involve molecular oxygen and NADPH molecular oxygen and NADH molecular oxygen and ascorbic acid molecular oxygen and tetrahydrobiopterin molecular oxygen and tetrahydrofolate molecular oxygen and FADH2 molecular oxygen and ATP molecular oxygen only, without any other reactant hydration of an olefin with a molecule of water hydrogen peroxide 17. The nucleotide salvage reaction produces ribose-1-phosphate 5-phosphoribosylpyrophosphate monophosphonucleotides nucleosides free bases (purines and pyrimidines) 18. The vitamin that is essential for the reaction that regenerates methionine from homocysteine is choline biotin cobalamin folate While methyl-THF can serve as methyl donor for this reaction, it is not essential because a betaine such as choline or N,N,N-trimethylglycine can also serve as a methyl donor (but cobalamin is still needed for this methyl transfer reaction. tetrahydrobiopterin pyridoxine pantothenic acid thiamine niacin riboflavin 19. The molecule that is used to tag proteins for degradation by the proteasome is ubiquinone albumin cobalamin ubiquitin thioredoxin glutathione xanthine geranylgeranol 4-phosphopantathiene MCMP 208 Exam III - 6 20. Of the pyrimidines that are in nucleotides used to make nucleic acids, which one(s) has/have an amino group outside the pyrimidine ring? orotate This has no exo-cyclic amine (and is not made into nucleotides used to make nucleic acids). thymidine This has no exo-cyclic amine. uracil This has no exo-cyclic amine. cytosine cytidine This is a pyrimidine nucleoside, not a pyrimidine base adenine This has an exo-cyclic amine, but it is a purine guanine This has an exo-cyclic amine, but it is a purine uridine This is a pyrimidine nucleoside, not a pyrimidine base (and it has no exocyclic amines) uracil and cytosine uracil and thymidine 21. The interconversions of purine nucleotides refers to the catabolism of nucleotides and their resynthesis by the salvage pathway the catabolism of purines to urate and their resynthesis from urate the generation of higher phosphorylated states of purine nucleotides and their eventual loss of phosphates in metabolism the synthesis of AMP, GMP, or IMP from AMP, GMP, or IMP the synthesis of ATP, GTP, or ITP from ATP, GTP, or ITP the synthesis of two molecules of NDP from one NMP and one NTP where N is a purine the synthesis of purine deoxyribonucleotides from purine ribonucleotides 22. The common indication both in gout and Lesch-Nyhan syndrome is developmental neurological abnormalities Lesch-Nyhan only fatal at young age Lesch-Nyhan only urine that is discolored or has an unusual smell Not related to either hypercholesterolemia Not related to either hyperuricemia anemia Not related to either obesity Not related to either acetone presence in exhaled breath Not related to either MCMP 208 Exam III - 7 ESSAY PROBLEMS. Write your answers to problems 23 to 29 in the space immediately below each problem. 23. [4 points] The action of Na+/K+-ATPase is shown below. a. [2 points] Indicate which side of the membrane is extracellular and cytosolic in the blanks below. ___ Extracellular ___ __ Cytosolic _____ This transporter maintains the low Na+ concentration and the high K+ concentration in the cytosol by moving Na+ out of the cytosol and bringing K+ into the cytosol. b. [2 points] Explain briefly what happens to the membrane potential by the action of this transporter. The extracellular side becomes more positive. As the transporter exchanges 3 Na+ and 2 K+ simultaneously, the transporter’s action results in the net movement of one positive charge from the cytosol to the extracellular space. The transporter, therefore, generates the positive membrane potential on the extracellular side (or the negative membrane potential on the cytosolic side). 24. [2 points] Arachidonic acid (20:4Δ5,8,11,14 or 20:4 ω-6) is a polyunsaturated fatty acid and the most abundant precursor for the synthesis of eicosanoids. Using the given abbreviated formula, draw the structure of arachidonic acid. 20-carbon fatty acid with 4 double bonds at 5, 8, 11, 14th carbon from the carboxylic end. 25. [3 points] The transverse movement (flip-flop) of membrane lipids is very slow unless catalyzed by enzymes. Explain briefly why the transverse movement is so slow. Membrane lipid molecules are amphipathic with hydrophobic tails and a hydrophilic head group. During the transverse movement, the hydrophilic head group needs to move across the lipid bilayer to the other side of the membrane, which is energetically costly and makes the movement very slow. MCMP 208 Exam III - 8 26. [6 points] During β-oxidation of saturated linear fatty acids in the mitochondria, there are four steps, each catalyzed by a different enzyme, that are repeated over and over in the same order. Starting with a generic fatty acyl-CoA, describe below (using words only) what happens in each of these four steps. Specifically describe (words only) the type of reaction, how the fatty acyl-CoA product differs from the fatty acyl CoA substrate, and which (if any) other reactants are involved. For the other reactants you may use acronyms/abbreviations to refer to them. Do not provide the name of the enzyme. Step 1: The single bond between the alpha and beta carbons is oxidized (using FAD) to a trans olefin bond. Step 2: The alpha-beta trans olefin bond is hydrated (using water) to become the betahydroxy fatty acyl-coA Step 3: The beta-hydroxyl group is oxidized (using NAD+) to become a beta ketone Step 4: Coenzyme A’s thiol attacks the beta-ketone carbon and the first two carbons of the beta-keto fatty acyl CoA are separated as acetyl CoA. The beta ketone carbon becomes a carboxylate in a thio ester linkage to Coenzyme A. 27. [6 points] Answer the following three questions about the urea cycle. [2 points for each question.] a. The step that forms L-citrulline involves forming an amide. Why is ATP not required for this? The carboxyl group (in carbamoyl phosphate) that is used to from the amide (with the side chain amino group in ornithine) is in an anhydride linkage to another acid (phosphoric acis). The anhydride is already “activated” in that it does not require that a water molecule be removed in order for it to acylate an amine. b. Why is urea not simply hydrolyzed from L-citrulline? While the resulting primary alcohol could be converted to an aldehyde and then back to ornithine to complete the cycle, this would not allow arginine to be synthesized. Since the reactions that produce urea also are used to produce arginine for protein synthesis, the pathway must produce urea from arginine, not from ornithine. c. Why is transamination not possible for adding ammonia to L-citrulline? Because the ketone in citrulline would become reduced during a transamination, and in the guanadino group of arginine, the carbon is a ketimine, which is the same level of oxidation as the ketone in citrulline. 28. [7 points] Starting with farnesylpyrophosphate, there are three major phases (with two key intermediates at the points where one phase ends and the next phase begins) in the biosynthesis of cholesterol. Using the outline below provide the names of these two key intermediates and for each phase provide a general description (using words) of the chemical changes that happen during that phase. (Simply stating that MCMP 208 Exam III - 9 farnesylpyrophosphate is converted to key intermediate #1 is not sufficient! You must describe the changes that occurred to make that happen.) [Hint: Both key intermediates are very stable compounds with highly distinct structures from eachother and from farnesylpyrophosphate.] Farnesylpyrophosphate Phase I: Two farnesol groups are joined (condensed) with a single C-C bond between them. This requires that the pyrophosphates and alcohols are removed and there is a reduction. Key intermediate #1: _squalene__ Phase II: Squalene is folded (while it is bound to a protein). Then the four fused rings of the sterol nucleus are produced in a chain reaction that is initiated by a reaction involving an epoxide group. Key intermediate #2: __lanosterol__ Phase III: Several methyl groups are removed and some double bonds are moved or removed. Cholesterol The exact wordings used above (Q28) are not required. What is required that there is a general description provided which explicitly includes those events described above. 29. [6 points] Methotrexate prevents utilization of the vitamin folate in metabolism, yet it is an effective antiproliferative used to treat rapidly growing cancers. This means that at doses where it stops the growth of cancer cells it has minimal or no impact on cells growing at a normal rate. a. [3 points] What are the two ways that the active form of folate used in the synthesis of nucleotides? i) Two formyl-THF are used for each IMP made (in de novo purine biosynthesis) ii) One methylene-THF is used for each dTMP made from dUMP by thymidylate synthase. [Either the reaction or the enzyme name are required, but not both.] b. [3 points] The inhibition by methotrexate of the utilization of the vitamin-form of folate has little impact on cells because the very small amounts of the vitamin-form of folate are needed by cells. The key to understanding why it is effective for cancer therapy is a unique detail of one of the two ways the active form of folate is used in the synthesis of nucleotides (your answer to part a). Which of these two ways is the key, and what is the detail of this that makes methotrexate a selective and effective anti-cancer (anti-proliferating cell) agent? The key reaction is the one catalyzed by thymidylate synthase. It results in the oxidation of THF (from methylene-THF) to DHF. This DHF must be reduced back to the active form (THF). This is what is inhibited by methotrexate.
