FinalExam2015_CellBio Key

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MCB104 Final Exam Student name: 1 PART 1: Cell Biology Total points: 134 1 (12 points). For each of the following sentences, select the best word or phrase from the list below to fill in the blanks. Not all words or phrases will be used; each word or phrase should be used only once. G-protein-coupled receptors (GPCRs) all have a similar structure with __________________ transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of __________________ subunits, becomes activated. __________________ of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the _ subunit is bound to __________________, which is exchanged for __________________ on stimulation. The intrinsic __________________ activity of this subunit is important for inactivating the G protein. Active G protein then stimulates adenylyl cyclase, causing production of the second messenger ______ adenylyl cyclase alpha GTPase AMP diacylglycerol beta ATP five seven ATPase four three Ca2+ GDP twelve cAMP GTP two 2 points per correctly filled blank: G-protein-coupled receptors (GPCRs) all have a similar structure with seven transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of three subunits, becomes activated. Two of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the alpha subunit is bound to GDP, which is exchanged for GTP on stimulation. The intrinsic GTPase activity of the alpha subunit is important for inactivating the G protein. Active G protein stimulates adenylyl cyclase, causing production of the second messenger cAMP. 2 (9 points) . Binding of a G-protein-coupled receptor to its ligand can lead to activation of protein kinase C (PKC) as depicted in the diagram below. You have discovered a novel protein, Tumor Inducing Kinase 1 (TIK1, in red), which is phosphorylated and activated by PKC. When TIK1 becomes phosphorylated, it promotes oncogenic transformation of cells. This is a potentially important finding because it implies that you could use one of the many existing GPCR inhibitors as a novel anti-cancer drug. However, you first want to learn more about how TIK1 becomes activated. In particular, which of the following conditions would lead to signal-independent activation of TIK1? Which one(s) would activate TIK1 in a signal-dependent way? Explain your reasoning. (a) the expression of a constitutively active phospholipase C

MCB104 Final Exam Student name: 2 (b) a mutation in the GPCR that binds the signal more tightly (c) a Ca2+ channel in the endoplasmic reticulum with an increased affinity for IP3 (d) a mutation in the gene that encodes TIK1 such that the enzyme can no longer be phosphorylated by PKC (a) (3 points) A constitutively active phospholipase C will lead to the constitutive production of IP3 and diacylglycerol, leading to activation of PKC in a signal- independent manner; thus, TIK1 activation and the lipid modification will be signal-independent. (b) and (c) (2 points each) will increase activity of the signal transduction pathway in a signal-dependent manner. (d) (2 points) will prevent PKC from activating TIK1 and will thus prevent the oncogenic transformation. 3 (10 points). Would each of the following events increase, decrease or not affect Receptor Tyrosine Kinase (RTK) signaling in a cancer cell? Briefly explain your answer. (a) (2 points) Chromosomal amplification of a ligand-dependent receptor (b) (2 points) Chromosomal amplification of a ligand-independent receptor (c) (2 points) Mutation of a Tyrosine in the intracellular domain to Alanine (d) (2 points) Mutation of a Serine in the intracellular domain to Alanine (e) (2 points) Mutation of an amino acid in the extracellular domain required for receptor dimerization

MCB104 Final Exam Student name: 3 2 points for each correct answer (a) Increase: amplification of a ligand-dependent RTK would increase the signaling as long as some ligand is present. (b) Increase: amplification of a ligand-independent RTK would boost signaling under all circumstances (c) Decrease: cross-phosphorylation of the intracellular tyrosines of an RTK is required for signal propagation (d) No effect: Tyrosine, not serine phosphorylation propagates the signal (e) Decrease: Receptor dimerization is required for cross-phosphorylation of the intracellular domains and activation 4 (4 points). What would be the most obvious outcome of repeated cell cycles consisting of S phase and M phase only? (a) Cells would not be able to replicate their DNA. (b) The mitotic spindle could not assemble. (c) Cells would get larger and larger. (d) The cells produced would get smaller and smaller. (d) (4 points) The cells produced would get smaller and smaller, as they would not have sufficient time to double their mass before dividing 4. (5 points) A mutant yeast strain stops proliferating when shifted from 25°C to 37°C. When these cells are analyzed at the two different temperatures, using a machine that sorts cells according to the amount of DNA they contain, the graphs in Figure Q18-3 are obtained.

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MCB104 Final Exam Student name: 4 Figure 18-3 Which of the following would not explain the results with the mutant? (a) inability to initiate DNA replication (b) inability to begin M phase (c) inability to activate proteins needed to enter S phase (d) inappropriate production of a signal that causes the cells to remain in G1 (b) (4 points) At 37°C, the cells all have one genome-worth of DNA, meaning that they have not replicated their DNA and therefore have not entered S phase. Cells that are unable to begin M phase should have two genomes-worth of DNA, as they would have completed DNA replication and arrested in G2. 6. (10 points) The yeast S. pombe has been used to dissect the mechanism of cell cycle regulation. It has been observed that cells expressing higher than normal levels of active cdc25 protein exhibit a small cell size or wee phenotype, as depicted in the picture below. normal elevated cdc25 levels

MCB104 Final Exam Student name: 5 a. (3 points) Explain this observation in light of what you know about cdc25 activity. Cdc25 is a phosphatase that removes an inhibitory phosphate from the Cdk subunit of the Cdk/cyclin complex (or MPF), promoting the G2 to M transition. If there are elevated levels of cdc25, the phosphate will be removed and Cdk will be activated prematurely, and the G2 to M transition will occur at a smaller cell size, resulting in a small cell or wee phenotype. b. (3 points) If you could express higher levels of another protein to counteract the effect of elevated cdc25 levels and restore cells to their normal size, which protein would you choose and why? Wee1. It is a kinase that phosphorylates Cdk, which inhibits Cdk activity. Its activity would directly counteract that of cdc25. c. (4 points) How does cdc25 contribute to the switch-like increase in Cdk activity at the G2 to M transition? Activation of cdc25 and dephosphorylation of the Cdk subunit of the Cdk/cyclin complex (MPF) is the final step in its activation. Furthermore, cdc25 is itself activated by Ckd. Thus, when a small amount of either cdc25 or Cdk are activated, each can activate the other, setting up a positive feedback loop that leads to rapid MPF activation. 7a (4 points) Progression through the cell cycle requires a cyclin to bind to a Cdk because _________. (a) the cyclins are the molecules with the enzymatic activity in the complex. (b) the binding of a cyclin to Cdk is required for Cdk enzymatic activity. (c) cyclin binding inhibits Cdk activity until the appropriate time in the cell cycle. (d) without cyclin binding, a cell-cycle checkpoint will be activated. Choice (b) is correct. Cyclins have no enzymatic activities themselves [choice (a)], and cyclin binding to Cdk activates the Cdk [choice (c)]. As far as we know, cyclin–Cdk binding is not directly monitored by checkpoints [choice (d)] 7b (4 points) Levels of Cdk activity change during the cell cycle, in part because ________________. (a) the Cdks phosphorylate each other.

MCB104 Final Exam Student name: 6 (b) the Cdks activate the cyclins. (c) Cdk degradation precedes entry into the next phase of the cell cycle. (d) cyclin levels change during the cycle. Choice (d) is correct. Cdks do not phosphorylate each other [choice (a)]. The Cdks do not activate the cyclins [choice (b)], and Cdks are not degraded during specific phases of the cycle [choice (c)]. The cyclins, however, are degraded in a cell- cycle-dependent fashion, and they are required for Cdk activity. 7c (4 points) The concentration of mitotic cyclin (M cyclin) ________________. (a) rises markedly during M phase. (b) is activated by phosphorylation. (c) falls toward the end of M phase as a result of ubiquitylation and degradation. (d) is highest in G1 phase. (c) The concentration of mitotic cyclin rises gradually during G2 and it is ubiquitylated and degraded during late M phase. 7d (4 points) The Retinoblastoma (Rb) protein blocks cells from entering the cell cycle by ______. (a) phosphorylating Cdk. (b) marking cyclins for destruction by proteolysis. (c) inhibiting cyclin transcription. (d) activating apoptosis. (c) is correct 8 (9 points). Using genetic engineering techniques, you have created a set of proteins that contain two (and only two) conflicting signal sequences that specify different compartments. Predict which signal would win out for the following combinations. Explain your answers. A. (3 points) Signals for import into the nucleus and import into the ER. B. (3 points) Signals for export from the nucleus and import into the mitochondria. C. (3 points) Signals for import into mitochondria and retention in the ER. A. The protein would enter the ER. The signal for a protein to enter the ER is recognized as the protein is being synthesized and the protein will end up either

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MCB104 Final Exam Student name: 7 in the ER or on the ER membrane. Cytosolic nuclear transport proteins recognize proteins destined for the nucleus once those proteins are fully synthesized and fully folded. B. The protein would enter the mitochondria. For a nuclear export signal to work, the protein would have to end up in the nucleus first and thus would need a nuclear import signal for the nuclear export signal to be used. C. The protein would enter the mitochondria. To be retained in the ER, the protein needs to enter the ER. Because there is no signal for ER import, the ER retention signal would not function. 9. (4 points) If you remove the ER retention signal from a protein that normally resides in the ER lumen, where do you predict the protein will ultimately end up? Explain your reasoning. The protein would end up in the extracellular space. Normally, the protein would go from the ER to the Golgi apparatus, get captured because of its ER retention signal, and return to the ER. However, without the ER retention signal, the protein would evade capture, ultimately leave the Golgi via the default pathway, and become secreted into the extracellular space. The protein would not be retained anywhere else along the secretory pathway: it presumably has no signals to promote such localization because it normally resides in the ER lumen. ALSO ACCEPTABLE: the protein may be re-routed to the lysosome and degraded there. 10. (4 points) The figure below shows the organization of a protein that normally resides in the plasma membrane. The boxes labeled 1 and 2 represent membrane- spanning sequences and the arrow represents a site of action of signal peptidase. Given this diagram, which of the following statements must be true ? (a) The N-terminus of this protein is cytoplasmic. (b) The C-terminus of this protein is cytoplasmic. (c) The mature version of this protein will span the membrane twice, with both the N- and C-terminus in the cytoplasm. (d) None of the above. (b) The mature version of this protein will span the membrane once, with membrane-spanning segment 2 in the membrane and the C-terminus facing the cytoplasm.

MCB104 Final Exam Student name: 8 11 (10 points). You are curious about the dynamic instability of microtubules and decide to join a lab that works on microtubule polymerization. The people in the lab help you grow some microtubules in culture using conditions that allow you to watch individual microtubules under a microscope. You can see the microtubules growing and shrinking, as you expect. The professor who runs the lab gets in a new piece of equipment, a very fine laser beam that can be used to sever microtubules. She is very excited and wants to sever growing microtubules at their middle, using the laser beam. A. (5 points) Do you predict that the newly exposed microtubule plus ends will grow or shrink? Explain your answer. B. (5 points) What do you expect would happen to the newly exposed plus ends if you were to grow the microtubules in the presence of an analog of GTP that cannot be hydrolyzed, and you then severed the microtubules in the middle with a laser beam? A. The newly exposed microtubule plus ends will most probably shrink if you sever the microtubules in the middle. This is because a microtubule grows by adding GTP-carrying subunits to the plus end. The GTP is hydrolyzed over time, leaving only a cap of GTP-carrying subunits at the plus end, with the remainder of the tubulin protofilament containing GDP-carrying subunits. Therefore, if you sever a growing microtubule in its middle, you will most probably create a plus end that contains GDP-carrying subunits. The GDP-carrying subunits are less tightly bound than the GTP-carrying subunits and will peel away from each other, causing depolymerization of the microtubule and shrinkage. B. If you were to polymerize the microtubules in the presence of a nonhydrolyzable analog of GTP and you then severed the microtubules with a laser, the newly exposed plus end would contain a GTP cap and so would probably continue to grow. 12a (4 points) Which of the following statements regarding dynamic instability is false? Explain your reasoning. (a) Each microtubule filament grows and shrinks independently of its neighbors. (b) The GTP cap helps protect a growing microtubule from depolymerization. (c) GTP hydrolysis by the tubulin dimer promotes microtubule shrinking. (d) The newly freed tubulin dimers from a shrinking microtubule can be immediately captured by growing microtubules and added to their plus end. Choice (d) is untrue. A newly dissociated tubulin dimer will be bound to GDP; this GDP will need to be exchanged for GTP before it can be added to a newly growing microtubule. 12b (5 points) Consider an animal cell that has eight chromosomes (four pairs of homologous chromosomes) in G1 phase. How many of each of the following structures will the cell have at mitotic prophase?

MCB104 Final Exam Student name: 9 A. sister chromatids B. centromeres C. kinetochores D. centrosomes E. centrioles 1 points for each correct number A—16; B—16; C—16; D—2; E—4 12c (4 points) Compared to the normal situation, in which actin monomers carry ATP, what do you predict would happen if actin monomers that bind a nonhydrolyzable form of ATP were incorporated into actin filaments? (a) Actin filaments would grow longer. (b) Actin filaments would grow shorter because depolymerization would be enhanced. (c) Actin filaments would grow shorter because new monomers could not be added to the filaments. (d) No change, as addition of monomers binding nonhydrolyzable ATP would not affect actin filament length. (a) Addition of monomers carrying a nonhydrolyzable form of ATP would stabilize the interactions between the monomers of a filament, stabilizing the filament and inhibiting depolymerization, resulting in longer actin filaments. 13 (5 points) Imagine that you could microinject cytochrome c into the cytosol of both wild-type cells and cells that were lacking both Bax and Bak, which are apoptosis-promoting members of the Bcl2 family of proteins. Would you expect one, both, or neither of the cell lines to undergo apoptosis? Explain your reasoning. The presence or absence of Bak and Bax would not affect whether a microinjection of cytochrome c would promote apoptosis, because Bax and Bak act upstream of cytochrome c by promoting its release from mitochondria. By promoting the formation of the apoptosome and the activation of procaspases, microinjection of cytochrome c bypasses the need for Bax or Bak in promoting apoptosis. 14. (12 points) The following proteins are all implicated in the generation of cancer cells. For each, indicate whether mutations are gain of function (GOF) or loss of function (LOF), and 1 sentence summary of their cellular role. Example: Ras GOF. Ras is a GTPase that in its activated form stimulates cell proliferation through the MAP kinase pathway.

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MCB104 Final Exam Student name: 10 a. p53 (2 points) LOF. p53 is a transcription factor that halts cell cycle in response to DNA damage. b. RasGAP (2 points) LOF. RasGAP is the GTPase activating protein that turns off activated Ras. c. TGF- β (2 points) LOF. TGF- β is a growth factor that inhibits G1-Cdk complexes, thereby keeping Rb active, which inhibits the E2F transcription factor and the G1/S transition. d. p21 (2 points) LOF. p21 is an inhibitor of G1/S-Cdk complexes which halts cell cycle progression and can induce apoptosis. e. EGFR (2 points) GOF. EGFR is a growth factor receptor that when bound to EGF induces cell proliferation through the Ras/MAP kinase pathway. f. RhoC – (2 points) GOF. RhoC is a small GTPase that induces acto-myosin contractility and cancer cell metastasis. 15. (12 points) Imagine that all ATP suddenly disappeared from a cell. For each process, indicate whether it would be severely affected and why a. microtubule elongation (2 points) No. Microtubule elongation occurs via addition of new alpha-beta tubulin dimers loaded with GTP. b. nuclear import of a transcription factor (2 points) No. Nuclear import of cargo such as transcription factors require the Ran GTPases, which utilize GTP for their transport cycle. c. actin filament elongation (2 points)

MCB104 Final Exam Student name: 11 Yes. Actin filaments grow by addition of ATP-loaded actin monomers to the plus end. d. separation of spindle poles (2 points) Yes. Spindle poles separate through the sliding of interpolar microtubules onto one another. This process is mediated by tetrameric kinesin-5, an ATP-powered molecular motor. e. scission of a clathrin coated vesicle (2 points) No. Clathrin-coated vesicles separate from the plasma membrane via constriction of their neck by the dynamin GTPase, a GTP-dependent molecular motor. f. Cytokinesis (2 points) Yes. Cytokinesis is mediated by the formation of a contractile actin ring that constricts and severs the membrane bridge between two daughter cells. Both actin and myosin in the contractile ring require ATP.

FinalExam2015_CellBio Key

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