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PHRM 836 Exam I - 1 Examination I PHRM 836 – Biochemistry for Pharmaceutical Sciences II September 30, 2014 Name:______________________________________________ Instructions 1. Check your exam to make certain that it has 10 pages including this cover page. Ask for a new copy of the exam if you are missing any pages. 2. Use a pencil for filling in the answer sheet for computerized grading. 3. Write your name on the line above (this page) AND on the answer sheet for computerized grading. Also place your student ID number on the answer sheet. Fill in the circles underneath where you write your name and student ID number. Do NOT place any other identification (name or ID number) anywhere else on this exam or on the answer sheet. Also, do NOT fill in anything for “section number”. 4. Your answers to problems 1 through 24 will be graded by computer. All computer-graded questions will be worth three points each. Put your answers for these problems on the answer sheet and fill in the appropriate circle using a pencil. It is your responsibility to be certain that the answers for these problems are correctly marked on the answer sheet at the time you hand in your completed exam. You are strongly encouraged to double check this at the time you submit the answer sheet, as no points will be given for having mistakenly marked your answer sheet. The computer grading answer sheets will not be returned to the students, so you are encouraged to record your answers to the computer-graded questions on this exam as well as on the answer sheet. However, the computerized grading of your answers to these questions will be done solely on the basis of the optically scanned answer sheet, not on what is recorded on your exam. The course policy is that there will be no re-grading of the computer answer sheets. 5. Examination problems 25 through 28 are to be answered with short answers, brief essays, or drawings. Put your answer to each of these directly on this exam, in the space provided below each of these questions. 6. After completing your exam, both the exam and the answer sheet must be submitted for grading and both must be identified only with your name. 7. This is a closed-book, closed-notes exam. Calculators and electronic devices of any kind are not allowed to be used during this exam. Any electronic device that is not in a backpack, bookbag, purse, pocket, etc, will be considered to be in use, so put away all electronic devices including cell phones and calculators. Anyone observed to be using a calculator, any book, handouts, notes, printed or handwritten material of any kind, or the exam or answer sheet of any other student will be considered to have committed an act of academic dishonesty. Students that do so will not be allowed to complete the exam and will be given a zero for this exam. Also remember that the first instance of academic dishonesty also results in a grade of "F" in this course and reporting this episode to the Pharmacy Dean and the Dean of Students office. PHRM 836 Exam I - 2 MATCHING. For problem 1, 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. An enzyme assay measuring initial velocities at different substrate concentrations was run in the absence of compound X and in the presence of compound X. The following Michaelis-Menton values were determined from the data. Match the plot that best indicates the effect of compound X. absence of X presence of X -1/KM (μM)-1 -0.14 -0.08 1/Vmax (μ mol/min)-1 0.8 0.8 ANS: choice #1 The changes indicate the compound X is an enzyme inhibitor that increases KM. but does not affect Vmax, as shown by the effect on the Michaelis-Menton curves shown in choice 1 (slope but not the Y-intercept). The other plots indicate changes in Vmax or are related to cooperativity and not relevant. PHRM 836 Exam I - 3 MULTIPLE CHOICE. For problems 2 to 24, 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] 2. All of the following statements are an accurate description of cytochrome P450s except they activate molecular oxygen through coordination with iron of a heme and thus oxidize substrates. they catalyze a diverse set of chemical reactions including hydroxylation, hydrolysis, dealkylation, and N-oxidation. they add oxygen to different sites of the substrate but exhibit preferences for certain sites NADPH-cytochrome P450 reductase passes electrons to them. they are heme-containing enzymes. none of the above (i.e. all of the above statements are correct) P450s do not hydrolyze substrates. All other statements properly describe P450s. 3. A zymogen is a pro-drug. is metabolized by cytochrome P450. is an enzyme named with the suffix –ogen. cleaves a peptide bond. is always activated to be a serine protease. is allosteric. Zymogens are inactive enzymes sometimes named with the suffix –ogen, e.g. trypsinogen. Serine proteases sometimes are activated by serine proteases, but not always. The inactive form does not cleave a peptide bond, only the active form does so. Other choices are not sensible. 4. The effectiveness of a small-molecule inhibitor of an enzyme is determined from a comparison of the enzyme activity in the presence and absence of the inhibitor. The apparent Vmax and/or KM differ from the actual Vmax and/or KM by a factor that is independent of inhibitor concentration. is independent of the equilibrium dissociation constant for inhibitor is independent of the equilibrium association constant for inhibitor increases with higher inhibitor concentrations can be either greater or less than 1.0 choices and choices and Enzyme inhibitors are detected by measuring the apparent KM and/or Vmax as [I] is varied. The values in the apparent KM and/or Vmax vary by a factor that depends on the binding constant for inhibitor, which is always greater than one: 1 + KI. The dissociation constant, KI, is the inverse of the association constant for inhibitor. 5. Which of the following phrases is not an accurate description of an allosteric effector? it binds at a site other than the protein’s active site binding often results in a conformational change it can enhance the enzyme’s activity it can decrease the enzyme’s activity it cannot be the same molecule as one of the substrates it binds at a site other than where the substrate binds PHRM 836 Exam I - 4 An allosteric inhibitor can be the same as a substrate, a homotropic effector. 6. The lowering of cholesterol by statins is due to inhibition of HMG-coA reductase. In class, the crystal structure of a complex of HMG-coA reductase with an HMG-CoA analogue and NADP was compared to the binary complex of HMG-coA reductase with Rosuvastatin. Atorvastatin is another highly effective statin with an IC50 similar to that of Rosuvastatin. What type of inhibitor is Atorvastatin most likely to be assuming both statins bind HMG-CoA reductase in a similar fashion? competitive with HMG-CoA because Atorvastatin would interact with NADPH when bound to HMG-CoA reductase. competitive with HMG-CoA because Atorvastatin would bind in the same sites as both HMG-CoA and NADPH. competitive with HMG-CoA independent of NADPH binding. competitive with NADPH. would not be competitive with HMG-CoA but might compete with NADPH. would be competitive with neither HMG-CoA nor NADPH. We saw in class that Rosuvastatin bound in a site that overlapped with the site of binding HMG-CoA, but did not overlap that for binding NADPH. As such, atorvastatin would be competitive with HMG-CoA and likely not with NADPH. The best choice is therefore #3 as the other choices are inaccurate. 7. Misregulation of an enzyme can be the cause for disease. A mutation can lead to improper regulation of an allosteric enzyme for the following reasons, excluding: the accumulation of the enzyme product as a result of increased kcat. a mutation in the binding site of an activating effector. a mutation in the binding site of an inhibitory effector. a mutation that increases KD of an inhibitory effector. a mutation that alters the interactions at a domain-domain interface but does not change the KD of the effector. a mutation that alters the interactions at a domain-domain interface and increases the KD of the effector. none of the above All of the answers are valid reasons for why a mutation in an allosteric enzyme could lead to disease. The stated effects alter the activity, and therefore the amount of the product present in the cell, by altering the allosteric regulation either directly (effect KD or kcat) or indirectly (domain-domain interactions). The amount of product can lead to disease. 8. Unlike enzymes that are not allosteric, an allosteric enzyme can be regulated by product inhibition. over a narrow range in substrate concentration. by a competitive inhibitor. PHRM 836 Exam I - 5 by pH. by BPG. by substrate concentration. choices and choices and choices and The sigmoidal dependence on [S] of activity is the hallmark of allostery and results in a large change in activity over a small range of [S] relative to a hyperbolic dependence. Product inhibition, pH, competitive inhibition and [S] effect non-allosteric enzymes. Regulation by BPG is relevant only to hemoglobin, not allosteric enzymes in general. 9. Which of the following conditions promotes release of oxygen from hemoglobin? increased CO2 increased H+ decreased 2,3-BPG the partial pressure of oxygen in the lungs increased pH choices and choices and choices and all of the above Hb affinity for oxygen depends on H+ (Bohr effect) and therefore on CO2 because of the equilibrium with bicarbonate and H+. Higher [H+] (lower pH) shifts the deoxy/oxy Hb equilibrium toward the deoxy form. BPG also shifts this equilibrium toward the deoxy form so decreased amounts of BPG would promote the oxy form, not deoxy form. The partial pressure of oxygen is high in the lungs so that promotes binding, not release. 10. Hemoglobin transports molecular oxygen to the peripheral tissues by directly binding O2 to a proximal histidine. by switching O2 with CO2 to coordinate to the heme. through binding O2 at the αβ dimer interface by coordination of O2 with Fe(II) by cooperatively binding O2 to the heme. choices and choices and choices and O2 fills the 6th coordination site of Fe(II) in the heme, and binds Hb cooperatively, as described by the sigmoidal binding curves of Hb. The proximal histidine of Hb also coordinates Fe(II) but does not contact O2. None of the other answers are correct. 11. CYP2E1 metabolism of acetaminophen causes acetaminophen-induced liver toxicity. Which statement is NOT an accurate description of the drug interaction between alcohol and acetaminophen? An interaction exists because alcohol and acetaminophen are both substrates for CYP2E1. This interaction can be protective if alcohol is consumed at the same time acetaminophen is taken. This interaction affects the metabolism of acetaminophen by Cyp2E1 to the toxic substance NAPQI. Chronic alcohol consumption induces CYP2E1 so that greater amounts of NAPQI are produced. This interaction always increases the risk of liver damage. PHRM 836 Exam I - 6 choice and choice and choice and Choices 1 to 4 are true, and because choice 2 is true, choice 5 cannot be true. Choice 2 is true because alcohol can ‘occupy’ the pool of CYP2E1 and make less of it available to metabolize acetaminophen. 12. Of the three isoforms of nitric oxide synthase (NOS), the inducible NOSII has been targeted for drug intervention because of its implicated role in septic shock. Which of the following approaches is most likely to succeed for designing an inhibitor that is selective for NOSII over NOSI and NOSIII? screening for a suicide inhibitor that will hit only NOSII making an inhibitor that competes with arginine binding designing a molecule that looks like the transition state for NOSII screening for a molecule that has a higher KI value for NOSII than for either NOSI or NOSIII making the inhibitor more hydrophobic looking at the crystal structures of the three isoforms to search for differences in amino acid types in the active site of the NOS isoforms Achieving selectivity among isoforms cannot come from the general approaches of designing inhibitors (suicide inhibitors, substrate analogues, transition state analogues) as these target features common to all isoforms. A rational approach is to exploit specific amino acid differences, as in the example given in lecture. A higher KI for NOSII is the opposite of what is wanted to selectively inhibit NOSII. 13. Both aspirin and the NSAID Celebrex bind the COX enzymes in a similar region and block access of the substrate to the active site. Aspirin differs from Celebrex and other NSAIDS that inactivate COX because the other NSAIDS are not selective for COX-2. because aspirin is a mechanism-based inhibitor. because aspirin inhibits the role of COX in the production of prostaglandins. because aspirin acetylates COX and is therefore an irreversible inhibitor. because aspirin is a competitive inhibitor while other NSAIDS are uncompetitive inhibitors. because aspirin is a transition-state analogue Unlike Celebrex and other NSAIDS, aspirin acetylates a residue (i.e. is irreversible) that is along the entry pathway for the substrate so both are competitive with substrates but for different reasons. Some NSAIDS are selective for COX-2. 14. An accurate description of the structure of hemoglobin excludes the phrase an α-helical protein two αβ dimers four heme groups four iron atoms a rigid structure positive cooperativity Hill coefficient greater than 1 two polypeptide chains pH dependent activity Hemoglobin cooperativity is based on a change in structure so the only incorrect statement is choice 5. PHRM 836 Exam I - 7 15. The serine protease called thrombin is involved in blood clotting and is highly selective for cleaving the peptide bond between Arg and Gly, for example the sequence Leu-Val-Pro-Arg-Gly-Ser. Choose the sentence or phrase that accurately describes thrombin. The P1’ residue is Arg. The active site includes a linear sequence of the residues SerHisAsp. It is reasonable that the S1 site of thrombin would include a Glu residue. Prothrombin is proteolytically cleaved to form the inactive form of thrombin. an all helical protein Arg (positively charged) is the P1 residue (the one immediately N-terminal to the cleaved peptide bond), and thus the protein S1 site reasonably has a Glu residue (negatively charged). The catalytic triad of Ser, His and Asp do not occur as a linear sequence in the polypeptide chain. Choices 4 and 5 are incorrect statements. 16. The phrases accurately describe DNA binding proteins except they bind to DNA with the same types of interactions that occur in other biochemical complexes. they can deform the DNA double helix. they may or may not recognize a specific DNA sequence. there is always a helix that binds the major groove. electrostatic interactions are generally involved. they can bind other proteins in addition to DNA. Some, but not all, of the DNA binding motifs include a helix binding to the major groove. All other choices are correct phrases. PHRM 836 Exam I - 8 17. Zinc finger domains coordinate zinc through interactions with the backbone atoms of His residues bind to TATA boxes of the DNA can be designed to bind specific regions of DNA and thus have therapeutic potential bind to DNA with high sequence specificity but do not bind to RNA bind to DNA with no sequence specificity ZF domains have Cys and His residues that coordinate Zn through sidechains. ZFs bind to specific DNA sequences, but not to TATA boxes. ZFs also bind to molecules other than DNA including RNA. 18. Which DNA binding motif occurs in the DNA-protein complex shown here? helix-turn-helix Zn-finger basic leucine zipper helix-loop-helix beta-alpha-beta TATA binding protein The drawing shows two helix-turn-helix motifs in close contact with the major groove, not the other motifs for the other choices. 19. In eukaryotic RNA polymerase II, there are over 3,000 amino acids in the elongation complex. the elongation complex includes protein, DNA and RNA. multiple proteins must somehow work together to open the DNA helix, and elongate the RNA strand. the work to elucidate transcription at a molecular level was recognized by an award of a Nobel prize. the structure shown in lecture is part of the pre-initiation complex for transcription. all of the above All of the choices are accurate descriptions as covered in lecture. 20. ATP-binding cassette transporters have both a membrane-spanning domain that recognizes ATP and a cytoplasmic domain that binds substrate. are all P-glycoproteins. are secondary active transporters. bind ATP but do not hydrolyze ATP. involve a mechanism for coupling ATP hydrolysis to transport, but the details of this mechanism are unknown. include the multidrug resistance family of ABC transporters, which cause the genetic disease cystic fibrosis. PHRM 836 Exam I - 9 ABC transporters bind substrate in the membrane region and have an ATP binding domain that is in the cytoplasm and hydrolyzes ATP during transport. P-glycoproteins (also called MDR transporters) are a subset of ABC transporters. The ABC transporter that causes cystic fibrosis is not an MDR transporter. PHRM 836 Exam I - 10 21. A doctor found that a patient’s cholesterol levels were not responding to a usually effective statin drug she had prescribed. Upon querying the patient, the doctor learned that the patient was taking St. John’s wort to improve his mood. St. John’s wort is an herbal remedy not requiring a prescription, and known to induce Cyp3A4. Which statement is a reasonable explanation regarding the patient’s condition? The statin is a substrate for Cyp3A4 and is being inactivated more rapidly in this patient. As a xenobiotic, the statin is oxidized by Cyp3A4 in order to increase its pharmacological activity. A drug interaction between St. John’s wort and statin prevents the induction of Cyp3A4. Because statins inhibit HMG-CoA reductase, the use of St. John’s wort cannot be the cause for a poor response in cholesterol levels. choices and choices and The lack of patient response is explained by an inactivation of the statin drug through Cyp3A4 metabolism, and thus St. John’s wort can cause the poor response. An increase in activity and prevention of Cyp3A4 expression do not explain the lack of response. 22. The following accurately describe the movement of Na+ and K+ across the plasma membrane by Na+K+-ATPase EXCEPT which statement? utilizes ATP hydrolysis to induce a conformational change in the protein. must work against a concentration gradient of Na+ and K+. is a major source for the synthesis of ATP by muscle cells. even though Na+ and K+ are simple ions, the transmembrane domain requires many transmembrane helices for translocation. is a symport mechanism choices and choices and Na+K+-ATPase, an integral membrane protein with many helices, maintains the concentration gradient of Na+ and K+ and must work against an existing gradient to do so. The process involves substantial conformational changes in the protein and consumes, not synthesizes, enormous amounts of ATP. 23. Membrane transporters are peripheral membrane proteins. are open to both sides of the membrane simultaneously. can use a concentration gradient. can be voltage-gated. require either a concentration gradient or an electrochemical gradient. only transport molecules into the cell choices , and choices , and Transporters are integral membrane proteins that switch between open and closed forms for each side of the membrane. Some use a concentration gradient, others use ATP hydrolysis to drive transport. Transport can be either into or out of the cell. 24. Nitric oxide synthase (NOS) catalyzes the synthesis of NO. The isoforms of NOS form NO from molecular oxygen and histidine by their production of NO, have a role in regulation of blood pressure, neurotransmission, or play a part in our defense against parasites PHRM 836 Exam I - 11 are either constitutive or inducible depending on calcium levels have multiple protein components and are allosteric choices and choices and choices , and NO is produced from Arg (not His) and molecular oxygen and used in various processes include those listed in choice 2. NOSII is induced by cytokines, not calcium. NOS is a redox system, not allosteric. PHRM 836 Exam I - 12 ESSAY PROBLEMS. Write your answers to problems 25 to 28 in the space immediately below each problem. 25. [12 points] The structure of an IgG antibody is illustrated with the two figures below; on the left is a schematic showing the full molecule, while on the right is a ribbon diagram showing only part of an IgG molecule. a. [2] Circle and label the part of the ribbon diagram that corresponds to the CH2 region. b. [3] Indicate where the antigen binds on BOTH figures. Hint: there are 3 total sites. c. [3] Circle and label one Ig domain in the left figure. What type of secondary structure does this domain have and what is the approximate size in either the number of amino acids or molecular weight? ANS: beta sheet ~100 amino acids, ~10 kDa d. [4] In 2 to 3 sentences, explain the structural basis for the remarkable specificity and highaffinity binding to a given antigen by an Ig molecule. ANS: the hypervariable loops, 3 in each of the N-terminal Ig domains (arrows), have sequences that complement the antigen both chemically and spatially to form strong interactions and thus high affinity. Only the antigen for which the antibody is selected will have the correct structure to bind with optimal interactions. PHRM 836 Exam I - 13 26. [5 points] In pregnant women, there is a 30% increase in intracellular 2,3-BPG to allow more oxygen to be offloaded to the fetus in the maternal uterine arteries. The fetus has a low sensitivity to 2,3-BPG, so its hemoglobin has a higher affinity for oxygen. Draw in one plot a binding curve for oxygen binding to maternal Hb and a second curve for oxygen binding to fetal Hb. 27. [8 points] Based on your knowledge about the sodium-dependent transport of glucose across the plasma membrane by the Na+/glucose cotransporter described in lecture, answer the following questions about a Na+/amino-acid cotransporter that also uses a symport mechanism. a. What is the source of energy for the transport of Na+ by the Na+/amino-acid cotransporter? ANS: sodium transport is passive and driven by the concentration gradient across the plasma membrane (from high concentration outside the cell to low concentration inside the cell). b. What is the source of energy for the transport of the amino acid by the Na+/amino-acid cotransporter? ANS: the amino acid is co-transported with each sodium, and thus it is a secondary active transport being driven in effect by the [Na+] gradient. 28. [3 points] The plot below is a velocity curve for an allosteric enzyme in the absence of any allosteric effector. Draw a curve on the plot below to show what the reaction velocity as a function of [S] would be in the presence of an activating effector that alters affinity. The curve should be to the left of the black solid line to indicate a lower effective KM.
