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PHRM 836 Exam II Key - 1 Examination II Key PHRM 836 – Biochemistry for Pharmaceutical Sciences II October 24, 2016 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 PHRM 836 Exam II Key - 2 MULTIPLE CHOICE. For problems 1 to 15, 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] 1. Which choice does NOT accurately describe TATA binding protein? 2. has favorable electrostatic interactions with DNA has favorable hydrophobic interactions with DNA distorts the DNA structure away from B-form DNA has non-sequence-specific interactions with DNA has sequence-specific interactions with DNA interacts only with DNA forms part of the preinitiation complex recognizes a DNA region with the sequence TATAAAA A number of DNA binding motifs were discussed in lecture. Which phrase does NOT apply to some type of DNA binding motif found in proteins? one, two, or three tandem repeats of the motif. a dimerization region to form either a homodimer or heterodimer with another transcription factor one of several domains in eukaryotic transcription factors can bind a specific DNA sequence must have non-sequence-specific interactions in order to bind DNA amino acid sidechains hydrogen bond to GC and AT basepairs of DNA a good steric fit between an -helix and the major groove a long helix with Leu residues spaced periodically along the chain. 3. The domain structure of the transcription factor Sp1 is shown below. Sp1 binds GC box regions and is known to require Zn2+ to enhance the transcription rate. Which choice accurately describes Sp1? Sp1 binds DNA as a homodimer. Sp1 binds DNA as a heterodimer. The Ser/Thr rich domain is a transmembrane region. After the DNA binding domain binds a GC box, the Gln rich domain interacts with DNA to enhance transcription. The DNA binding domain has at least two Cys residues. The DNA binding domain likely has Asp residues to interact with the DNA phosphate backbone. Sp1 binds the minor groove of DNA. Sp1 distorts DNA structure by coordinating Zn2+ to the phosphate backbone. The DNA binding domain is the activation domain. PHRM 836 Exam II Key - 3 4. The ABC transporters are primary, active transporters because 5. the protein is phosphorylated during the transport cycle. the NBD has the highly conserved Walker A and Walker B sequence. the electrochemical gradient across the membrane drives transport. the concentration of transported substances, including drug molecules, is not higher inside the cell. the transported substance diffuses across the membrane. these transporters, along with cytochrome P450’s, account for most pharmacokinetics of a drug. Which statement correctly describes RNA polymerase? RNA polymerase comprises one polypeptide chain of over 3500 residues. RNA polymerase I and RNA polymerase II transcribe mRNA. RNA polymerase II opens DNA for transcription, but DNA regains the double stranded structure upon exiting the RNA polymerase complex. RNA polymerase II transcribes rRNA. DNA and RNA bind in a pocket on the surface of RNA polymerase. RNA polymerase IV transcribes tRNA and other small RNAs. 6. What is circled in the structure of the Na+/K+ exchanging ATPase shown to the right? the transmembrane region the extracellular domain of the protein the region where ATP binds the region that transports K+ the region that transports Na+ the region that has the ATP binding cassette the region that is phosphorylated choices and choices and choices , and 7. In a segment of linear dsDNA with anchored ends but without any protein bound to it there is a single stable negative plectonemic supercoil. If the ends of this segment are moved further apart such that it was impossible for this DNA to writhe, then what would happen? The DNA would have one negative toroid supercoil The DNA would have one additional helical turn than normal for B-DNA The DNA would have several additional helical turns compared to normal B-DNA The DNA would have one fewer helical turn than normal for B-DNA The DNA would have several fewer helical turns compared to normal B-DNA The DNA would have exactly the number of helical turns expected of B-DNA Both strands of the DNA would break One strand of the DNA would break PHRM 836 Exam II Key - 4 8. Which statement about nucleosomes is true? The core histones in nucleosomes are held together by the DNA that is wrapped around them. The DNA in a nucleosome has the form of a negative toroid supercoil. The generation of nucleosomes requires the action of DNA gyrase. Nucleosomes form only during mitosis. The core histones are histones 1, 2a, 2b, and 3. Only AT-rich DNA can form into a nucleosome. The DNA in nucleosomes is so tightly bound to the core histones that the DNA cannot move at all. The only function of nucleosomes is to compact DNA so it fits inside the cell. Nucleosomes are responsible for all the compaction needed of nuclear DNA in all organisms. 9. How much compaction of DNA happens via nucleosome formation? 2-fold 8-fold between 83 and 84-fold between 8 and 84-fold depending on how closely spaced the nucleosomes are it depends on whether the nucleosomes involve positively or negatively supercoiled DNA 10. Unlike passive helicases, active helicases unwind DNA only if it is supercoiled are molecular machines take advantage of DNA that has unwound due to locally higher thermal energy separate dsDNA to make two ssDNA move along the ssDNA can move either direction along ssDNA rather than in just one direction 11. Signaling by NO is an exception to broad generalizations about signaling by cell permeable signal molecules because signaling by NO involves cAMP involves a gated ion channel involves a GPCR does not involve a second messenger does not involve ligand-activated transcriptional regulation does not involve a protein kinase cascade 12. STAT proteins signal in a manner that is uncommon for signaling proteins that carry out similar functions in cell signaling because STAT proteins are transcriptional regulators that also are heterotrimeric G-proteins are transcriptional regulators that are directly activated by binding the primary signal molecule are transcriptional regulators that are activated at the plasma membrane of the cell are transcriptional regulators that are regulated by second messengers are transcription regulators that function entirely at the plasma membrane of the cell PHRM 836 Exam II Key - 5 13. A src-homology 2 domain (SH2 domain) in a protein is known to always bind phosphorylated tyrosine in proteins bind phosphorylated serine and phosphorylated threonine in proteins have serine/threonine protein kinase activity have tyrosine protein kinase activity be a pseudo-substrate domain be a key location where phosphorylation results in the activation of the protein bind to PIP3 at the cytosolic surface of the plasma membrane bind to DNA in a sequence-specific manner bind to diacylglycerol bind NO 14. A “licensing factor” is any protein that triggers the start of replication of DNA by phosphorylating various nuclear proteins is made in G1 and degraded after completing its regulatory or functional role in DNA replication is degraded quickly by the proteosome is made in a specific phase of the cell cycle for the sole purpose of regulating protein kinases that regulate the progression of the cell cycle regulates transcription initiation unwinds DNA for replication or transcription regulates DNA repair regulates DNA recombination 15. Gleevec (imatinib) treatment produces a prolonged remission (but not a cure) of most cases of CML. In all the responsive cases, the Abl tyrosine protein kinase activity in the cancer cells has been activated by mutational activation of the activation loop aberrant recombination that caused loss of the inhibitory domain from the catalytic domain increased transcription mutation of its dimerization domain mutational inactivation of a specific protein inhibitor mutational activation of an upstream activating protein kinase gene duplication PHRM 836 Exam II Key - 6 ESSAY PROBLEMS. Write your answers to problems 16 to 21 in the space immediately below each problem. 16. [6 points] From the information about cell surface receptors provided in class and the required readings, answer the following questions, fill in the blanks, or circle the correct term. A. [1.5 points] An example of a ligand where its receptor has enzyme activity is __any one of the following: EGF, PDGF, VEGF, TGFBeta, ANP, insulin_. B. [1.5 points] For the receptor for the ligand you specified as your answer to part A of this problem, what kind of enzyme activity does the receptor have? _Guanulate cyclase for ANP, serine/threonine kinase activity for TGFBeta, and tyrosine protein kinase for the rest_ C. [1.5 points] An example of a ligand where its receptor is dimeric but has NO enzyme activity is _TNFalpha, integrin ligand (cell matrix or RDG-sequence), IgG, innate immunity/TOLL/TOLL-like receptor antigen (since these are also known as pattern or pathogen pattern receptors, indicating the antigen for these is also acceptable), T-cell antigen, growth hormone, interferon, erythropoietin, prolactin_. The intent with this question was to ask about receptors that use a scaffold mechanism, but without using the word scaffold. This would have allowed for dimeric and trimeric (TNFalpha) but not gated ion channels or GPCRs. Also, nuclear/intracellular recepors are excluded by the stipulation at the top of the question that this is referring to cell surface receptors. Unfortunately, the wording was not done in a way consistent with the intent. So, if this problem (as written) is answered correctly, then part D is trivial (dimeric.) D. [1.5 points – no partial credit] The receptor for the ligand specified as your answer to part C of this problem is monomeric/dimeric/trimeric in structure. (circle one) Dimeric for all correct answers for part C. If part C is not correct it still may be possible to get part D correct or wrong, depending on what was specified in part C. If in part C the answer was TNFalpha the correct answer here is trimeric. If in part C, a GPCR ligand or nuclear/intracellular ligand (including NO) was given, then the correct answer here is monomeric. If the ligand in C is stated to be acetylcholine, GABA, glycine, glutamate or serotonin, the type of receptor is a gated ion channel and there is no choice here that is correct since these ligand-gated channels are typically pentameric or tetrameric. PHRM 836 Exam II Key - 7 17. [4 points] The structure of an eukaryotic ABC transporter is shown below. A. [1 point] Draw on the figure the direction for exporting xenobiotics out of the cell. B. [1 point] Circle the NBD part of the protein C. [2 points] P-glycoprotein, a member of the MDR family of ABC transporters, is found on the luminal surface of epithelial cells in the small intestine but not on the basal (blood) side. Efflux by Pglycoprotein therefore works against the absorption of substances into the blood by transporting them back into the lumen of the intestine. Digoxin is a drug that is a P-glycoprotein substrate, while erythromycin is a drug that is an inhibitor of P-glycoprotein. The co-administration of digoxin and erythromycin has been noted to result in a change in the serum levels of digoxin. Would the serum levels of digoxin increase or decrease, and why? ANS C. An inhibitor of P-gp will prevent efflux of a substrate back into the lumen and increase the bioavailability of a P-gp substrate. Therefore, the serum levels of digoxin will increase when given with erythromycin b/c P-gp is blocked by erythromycin and not available to efflux digoxin back to the intestine. 18. [6 points] Contrast type 1 topoisomerases with type 2 topoisomerases in this problem. A. [2 points] What is the one main difference between the structure that is common to all type I topoisomerases compared with the structure that is common to all type 2 topoisomerases? Type 1 topoisomerases are monomeric while type 2 are never monomeric. B. [4 points] What are the TWO differences between the substrate(s) of all type 1 topoisomerases compared to the substrate(s) of all type 2 topoisomerases? Difference 1: Type 1 topoisomerases do not utilize ATP as a substrate, while all Type 2 topoisomerases use ATP. Difference 2: Type 1 topoisomerases work on a single segment of dsDNA, while type 2 topoisomerases work on two segments of dsDNA. PHRM 836 Exam II Key - 8 19. [4 points] One reason why cellular signaling is so complex is that a single signaling pathway often splits to become several separate pathways. From the material presented in class and the assigned readings, indicate one specific example of where a signaling pathway splits to become two or more independent signaling pathways. There are many examples. While this list may not include all those presented in class and readings, these were definitely presented in class and each one is an acceptable answer: MAPK cascade (activation of AP1 and two other kinases: MNK and RSK, each of which activates different TFs) EGF receptor – its own Tyr PK activity and scaffold activity (GRB2) that actiates ras Integrins (a dimeric scaffold receptor) signal via cytoskeleton (actin) and via FAK, a nonreceptor tyrosine protein kinase Ras signaling activates Raf, ralGDS and PI3K PI3K (via PIP3) activates Akt, PDK1 (also other proteins including rac-GEFs) GPCRs activate different heterotrimeric G-proteins that signal differently GPCRs activate both G-alpha subunit and G-beta/G-gamma subunit complex Gαi signals via inhibiting Adenylate cyclase and regulating ion channels, phospholipases, and phosphodiesterases cAMP activates PKA and some ion channels and GEFs for small G-proteins Calcium activates calmodulin and also activates PKC Calmodulin activates CAM-PK and myosin light chain kinase Some phosphodiesterases hydrolyze both cAMP and cGMP, affecting both signaling paths PKC phosphorylates many target proteins – specifics are not required PKA phosphorylates many target proteins – specifics are not required PKG phosphorylates many target proteins – specifics are not required CAM-PK phosphorylates many target proteins – specifics are not required (The phrase “phosphorylates many target proteins – specifics are not required” applies to many protein kinases described in class, including the above and raf, non-receptor tyrosine protein kinases, and some receptor tyrosine protein kinases) 20. [4 points] PKA and PKG are regulated in a very similar manner with two very important differences. Answer the questions below about the regulation of PKA compared to the regulation of PKG. A. [2 points] What is the common way in which both PKA and PKG are activated? Both PKA and PKG are regulated by a regulatory part of the protein that inhibits the catalytic part of the protein in the absence of the cyclic nucleotide, and the cyclic nucleotide removes the inhibition by this inhibitory part of the protein B. [2 points] What are the TWO differences in how PKA and PKG are activated? Difference 1: cAMP activates PKA while cGMP activates PKG Difference 2: cAMP causes dissociation of a regulatory subunit of PKA, while cGMP causes only a change in shape/conformation of the regulatory domain of PKG. This is because both catalytic and regulatory domains are part of the PKG polypeptide while they are two separate polypeptides with PKA. PHRM 836 Exam II Key - 9 21. [6 points] Germ-line mutations in tumor suppressor genes dramatically increase the risk of cancer. A. [4 points] Name TWO different examples of human tumor suppressor genes that were presented in class or the assigned readings. For each example you give, describe the cancer or syndrome that results from germ-line mutations in that tumor suppressor gene. ***Important note*** The information about the cancer or syndrome from germ line mutations was on a slide that was not presented on Oct 18 and was carried over to the October 25 lecture. Therefore the two examples of TS genes each will be worth 2 points, and there will be no expectation for the cancer/syndrome part of the question to be completed. The TS answers provided for this problem may be any tumor suppressor gene given in lecture or readings, incuding the slide that was not covered on Oct 18 (this was on slide 22). The following TS genes were presented in class (others possible from the readings): BRCA1, p53 (also known as TP53), APC, Rb, MSH2, WT1, NF1, PTEN Shown on slide 22 (but also acceptable for this answer) are BRCA2, MLH1, XP genes (XP1 through XP8), BLM, ATM, ATR. B. [1 point – no partial credit] Mutations in a tumor suppressor gene that elevate cancer risk have what impact on the activity (or abundance) of the gene product of that tumor suppressor gene? Mutations always eliminate or reduce activity or abundance of the TS gene product. C. [1 point – no partial credit] The products of tumor suppressor genes are involved in cellular process that can be grouped into a few groups sharing a common purpose. Of these functional (purpose) groupings of tumor suppressor gene functions, the largest group shares what common purpose of the affected cellular process? Maintenance of genomic stability (or genome integrity). All TS gene products maintain normal cell phonotype, so this cannot be the common purpose of “the largest group”.
