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Unformatted text preview: LAST NAME: __________________ FIRST NAME: __________________ Exam #4 BIO 320 There are 30 multiple-choice questions worth 2 points each. Use a #2 pencil to answer the multiple-choice questions. WRITE AND MARK OFF THE BUBBLES FOR YOUR LAST NAME AND UT EID ON THE SCANTRON FORM There are 10 fill-in-the-blank questions, 2 points each. There are 3 essay questions worth a total of 20 points. Use a pen if you want a regrade for these questions. Write legibly and please do not ramble. Multiple choice questions, 2 points each 1) In a folded protein, most of the nonpolar amino acids are buried inside the protein fold through: a. Ionic interaction. b. hydrophobic interaction. c. apolar interaction. d. hydrophilic interaction. e. hydrogen bonding. 2) The resolving power of a light microscope is dependent on: a. the intensity of the light source. b. the wavelength of the light source. c. the thickness of the specimen being observed. d. the magnifying power of the objective lens. e. None of the above. 3) The isopeptide bond involved in ubiquitination links which of the following pairs of amino acids? a. Glycine and lysine. b. Cysteine and glycine. c. Serine and lysine. d. Asparagine and glycine. e. Lysine and cysteine. 4) The average lengths of the nonpolar hydrocarbon tail for the following lipid molecules are in the descending order of: a. Cholesterol > sphingolipid > phosphoglyceride (i.e., longest for cholesterol). b. Sphingolipid > phosphoglyceride > cholesterol. c. Sphingolipid > cholesterol > phosphoglyceride. d. Phosphoglyceride > cholesterol > sphingolipid. e. Phosphoglyceride > sphingolipid > cholesterol. 5) Which of the following statements concerning degradation of misfolded proteins is CORRECT: a. misfolded cytosolic proteins bind E3 in the cytosol and misfolded ER luminal proteins bind E3 in the ER lumen . b. misfolded cytosolic and ER luminal proteins spend excessive amount of time being associated with molecular chaperones. c. a mannose molecule is removed from misfolded cytosolic and ER luminal proteins d. specific E3 proteins bind cytosolic and ER luminal molecular chaperones. e. b and d. 6) Which of the following associations does NOT occur in a cell? a. b. c. d. e. Clathrin and cis-Golgi. Clathrin and plasma membrane. Dynamin and clathrin. COPI and cis-Golgi. COPII and ER. 7) Which of the following does NOT occur in the ER lumen? a. Glucose trimming. b. Disulfide bond formation. c. N-linked glycosylation. d. Protein folding. e. O-linked glycosylation. 8) The figure below shows the predicted organization of a newly discovered protein. The boxes labeled 1 and 2 represent stretches of hydrophobic sequences (of ~20 amino acids) and the arrow represents a site of action of signal peptidase. Which of the following statements is TRUE if this protein ends up on the plasma membrane? a. The N-terminus of this protein is in the cytoplasm. b. The C-terminus of this protein is in the cytoplasm. c. Both the N and C-termini of this protein are in the cytoplasm. d. Both the N and C-termini of this protein are outside the cell. e. None of the above. N 1 2 C 9) Which of the following statements about secretion is CORRECT? a. Upon reaching the trans-Golgi, a protein does not need additional signal to reach the plasma membrane (or be secreted). b. Vesicles for the regulated secretory pathway will not bud off the trans-Golgi network until the appropriate signal has been received from the cell. c. Proteins destined for the constitutive secretory pathway have signal sequences that ensure their packaging into the correct vesicles at the trans-Golgi. d. Proteins destined for the constitutive secretory pathway aggregate as a result of the acidic pH of the trans-Golgi network. e. Clathrin is required for the fusion of secretory vesicles with the plasma membrane. 10) What allows for the dissociation of KDEL-containing proteins from their receptor in the ER lumen? a. A greater affinity between the receptor and a competing molecule. b. Removal of phosphate in the ER lumen. c. The more basic pH in the ER lumen. d. The more acidic pH in the ER lumen. e. GAP activity in the ER lumen. 11) Receptor-mediated endocytosis is mediated by which protein(s) a. COPI. b. Caveolin. c. Clathrin. d. Sar1. e. a and b. 12) For proteins that are encoded by the mitochondrial genome, which of the following proteins is required for their localization to the inner mitochondrial membrane? a. TOM complex. b. TIM22 complex. c. TIM23 complex. d. OXA complex. e. None of the above. 13) A drug that blocks the ability of Ran to exchange GDP for GTP will most likely affect nuclear import of cargo protein in which of the following ways? a. Import receptors would be unable to bind cargo. b. Import receptors would be unable to interact with the nuclear pore fibrils. c. Import receptors would interact irreversibly with the nuclear pore fibrils. d. Import receptors would be unable to release their cargo in the nucleus. e. None of the above. 14) Which of the following molecules is not a component of lipoproteins? a. Cholesterol. b. Apoprotein. c. Phospholipid. d. Triglyceride. e. They are all components. 15) Which of the following statements is CORRECT regarding critical concentration Cc for actin? a. Cc of plus end is lower than Cc of minus end. b. Cc of plus end is higher than Cc of minus end. c. Treadmilling occurs at concentration between Cc for the two ends. d. Treadmilling occurs at concentration below Cc for the minus end. e. a and c. 16) Which of the following statements regarding dynamic instability is INCORRECT? a. Each microtubule end grows and shrinks independently of its neighbors. 17) 18) 19) 20) 21) b. The GTP cap helps protect a growing microtubule end from depolymerization c. GTP hydrolysis by the tubulin dimer promotes microtubule shrinking. d. The newly freed tubulin dimers from a shrinking microtubule end can be immediately captured by growing microtubules and added to their ends. e. Microtubules in a cell are more dynamic during M phase. Regarding only the ends where formin and the Arp complex work, a plus-end capping protein: a. has no effect on the ability of formin to promote actin assembly. b. prevents/diminishes the ability of formin to promote actin assembly. c. has no effect on the ability of the Arp complex to promote actin assembly. d. prevents/diminishes the ability of the Arp complex to promote actin assembly. e. a and c. In a living cell, rescue can occur at a depolymerizing microtubule end because: a. of a sudden dramatic rise in tubulin dimer concentration. b. there are islands of GTP-tubulin within the microtubule lattice. c. of the binding of a capping protein. d. the conformational change of tubulin dimers. e. none of the above. Which of the following proteins is responsible for the branched arrangement of some actin filaments? a. Formin. b. Gelsolin. c. profilin. d. Arp complex. e. fimbrin. Kinesin-1 and myosin II are similar in that: a. they are both processive motors. b. they both move towards the plus end of the filaments that they work on. c. they can both form thick bundles. d. All of the above. e. None of the above. The step size of a myosin motor is: a. inversely proportional to the duration of the nucleotide hydrolysis cycle. b. inversely proportional to the angle of movement for the lever arm. c. proportional to the length of the lever arm. d. proportional to the length of the coiled-coil tail. e. None of the above 22) p53 is an example of? a. a transcription factor. b. a protein that is required for cell cycle progression. c. a protein that shuttles into and out of the nucleus. d. All of the above. e. a and c. 23) You have isolated a mutant strain of yeast that enters M phase prematurely (i.e., sooner than normal). You learned that this mutant has a mutated form of M-Cdk. Which of the following changes is consistent with the behavior of this mutant? a. The site on M-Cdk normally phosphorylated by CAK is mutated. b. The catalytic function of M-Cdk is mutationally abolished. c. The site on M-Cdk normally phosphorylated by Wee1 is mutated. d. The binding site on M-Cdk for cyclin is mutationally destroyed. e. None of the above. 24) __________________ can remain associated with a depolymerizing microtubule plus end: a. The +TIP EB1. b. A ring-like complex that encircles the microtubule. c. A standard plus end-directed kinesin. d. A standard minus end-directed kinesin. e. Dynein 25) Metaphase to anaphase transition is triggered by: a. degradation of M-cyclin. b. activation of aurora-B kinase. c. phosphorylation of cohesin. d. degradation of securin. e. degradation of condensin. 26) Which of the following statements regarding steroid hormone and it receptor is INCORRECT? a. The receptor undergoes conformational change upon ligand binding. b. The receptor is often found in the cytosol. c. The ligand-bound receptor is found in the nucleus. d. The steroid hormone is hydrophobic e. All of the above are correct. 27) Which of the following processes is NOT commonly used to terminate (or down-regulate) GPCR signaling? a. Receptor endocytosis. b. Receptor phosphorylation. c. Synthesis of RGS proteins. d. Binding of protein to GPCR to prevent its association with trimeric G protein. e. They are all commonly used. 28) Scaffold proteins are important for MAP kinase cascade signaling because: a. MAP kinases are not active when not bound to scaffold proteins. b. Scaffold proteins prevent cross-talk between parallel MAP kinase cascades. c. Scaffold proteins increase the efficiency of signal transmission within the cascade. d. All of the above. e. b and c. 29) Cadherins and selectins are similar in that: a. they are both linked to the actin cytoskeleton. b. they are both dependent on calcium for function. c. they both bind oligosaccharides. d. a and b. e. a and c. 30) Occludin is part of: a. adherens junctions. b. gap junctions. c. tight junctions. d. desmosomes. e. connexons. Fill in the blank with one or a small number of appropriate words, 2 points each. 1. SDS-PAGE is often used to separate proteins according to __size/molecular weight/length_______________________. 2. A phospholipid membrane is more fluid when the phospholipid molecules have more __(cis) double bonds_______________________ and are _____shorter______________ in length. 3. The signal sequence that directs a protein into the mitochondria forms an -helix that is _____hydrophobic_______________________ on one side and _____positively charged___________________________ on the opposite side. 4. The binding of ___v-SNARE (or Rab protein)______________ to ___t-SNARE (or Rab effector)_______________ is one way to obtain specificity in the fusion of donor vesicle to appropriate target membrane. 5. The rate-limiting step in the assembly of microtubules is ______nucleation_______________________. 6. The conformation of GTP-bound tubulin dimer is more ________straight_______________ than that of GDP-bound tubulin dimer. 7. In motile cells, membrane protrusion at the leading edge is caused by actin filament assembly that is driven by __the Arp complex_____________________________. 8. The M-Cdk complex phosphorylates _______Cdc25 and Wee1__________________________ in a positive feedback loop that leads to abrupt activation of the M-Cdk complex (i.e., abrupt G2 to M transition). 9. Chromokinesin that provides the polar/astral ejection force to move chromosome arms is _____plus_____________ end-directed. 10. PLC catalyzes the formation of _______diacylglycerol/DAG___________ and ________IP3______________. Assay questions. 1. Immunofluorescence is a method commonly used when trying to detect specific proteins in a cell. Typically, unlabeled primary antibodies and then fluorescently labeled secondary antibodies are used. Briefly explain the reasoning behind using fluorescently labeled secondary antibodies rather than just fluorescently labeled primary antibodies. Use simple diagrams to help in your answering if you want. (4 points) Multiple molecules of secondary antibodies can recognize a single molecule of primary antibody (that is bound to the protein of interest). Therefore, there is amplification of fluorescent signal (that one does not get if only labeled primary antibodies are used). 2. If you were to add a nonhydrolyzable form of ATP (instead of ATP) to myosin, how would you affect its movement on actin filaments? Very briefly explain your answer. (5 points) Myosin has very weak affinity for actin filaments when bound to ATP. Hence, it will fall off actin filaments and be unable to move along actin filaments at all. How is this similar to or different from adding a nonhydrolyzable form of ATP to kinesin? Briefly explain your answer (5 points) Kinesin binds microtubule tightly when bound to ATP. Hence, when ATP cannot be hydrolyzed, the kinesin head cannot dissociate from the microtubule. It will not move on the microtubule since the kinesin head cannot take any steps. 3. In the absence of pheromone, yeast cells exhibit normal growth. During the mating process, pheromones bind GPCRs on the surface of yeast cells and lead to the activation of trimeric G proteins. As a result, the cells arrest growth as a mating response. If yeast cells do not undergo the appropriate response after binding of pheromone to GPCRs, they are considered sterile (and continue normal growth). Yeast cells that lack one or more components of the G protein have characteristic phenotypes in the absence and presence of the pheromone, which are listed below. mutation None (wild type) a subunit deleted b subunit deleted g subunit deleted a and b deleted a and g deleted b and g deleted minus pheromone normal growth arrested growth normal growth normal growth normal growth normal growth normal growth phenotype plus pheromone arrested growth, mating response arrested growth, sterile normal growth, sterile normal growth, sterile normal growth, sterile normal growth, sterile normal growth, sterile Which of the following models is consistent with the data from the analysis of these mutants? Explain your answer. (6 points) (a) (b) (c) activates the mating response but is inhibited when bound to . activates the mating response but is inhibited when bound to . Either free or free complex is capable of activating the mating response. (Note: line 4 of the table should read: subunit deleted arrested growth arrested growth, sterile mating response) (b) is consistent. With (a), when subunit is deleted alone or incombination with or subunit, there should be no growth arrest (i.e., no mating response) with or without pheromone. With (c), when , or , or subunit is deleted alone, there should be growth arrest (i.e., mating response) with or without pheromone. With (b), when or subunit is deleted alone or in combination with subunit, there should be no growth arrest (i.e., no mating response) with or without pheromone. ...
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This note was uploaded on 02/06/2012 for the course BIO 320 taught by Professor Staff during the Spring '08 term at University of Texas at Austin.
- Spring '08
- cell biology