Stony Brook University BIO 310/BIO310 All Lecture quizzes (answered) study this for a 100% A grade.

Quiz questions based on Lecture 13 (Signaling-1). 1. What is always present in a signaling pathway? A. Signal that originates outside a cell B. Signal that remains outside the responding cell during signaling C. Cell-surface receptor D. A and B are correct 2. Adrenaline is a signal used in the fight-or flight response. Adrenaline signaling is A. Endocrine B. Paracrine C. Autocrine D. Contact-dependent 3. One signal can have different effects on the same cell; for example, it can alter both metabolism and gene expression. This occurs when A. One activated receptor binds several effector proteins B. One effector protein causes several responses C. One intracellular signaling molecule activates several effector proteins D. One effector protein activates several intracellular signaling molecules 4. What is true of signaling molecules that bind intracellular receptors? A. They are linked to enzymes B. They always act in an autocrine manner C. They always act in a paracrine manner. D. They easily diffuse across the plasma membrane 5. What is true of an inactive G protein? A. It is not attached to the plasma membrane B. It is bound tightly to an inactive receptor C. It keeps its target protein inactivated by binding to it D. It is a heterotrimer (3 different polypeptide subunits) 6. What is the first step in activation of a G-protein coupled receptor signaling pathway? A. A ligand (small signaling molecule) binds a cell-surface receptor B. A ligand binds the alpha subunit of a G protein C. A cell-surface receptor binds GTP D. The alpha subunit of a G protein binds GTP 7. What is true of an activated G protein? A. It is released from the plasma membrane into the cytosol B. It is a heterotrimer (3 different polypeptide subunits) C. The alpha subunit is bound to GTP D. The beta subunit is bound to GTP8. What is important in turning off (inactivating) G protein signaling? A. Binding of an inactive G protein to a target protein B. Hydrolysis of GTP to GDP C. Separation of the G protein alpha subunit from the beta-gamma complex D. Release of the G protein from the plasma membrane into the cytosolQuiz on Lecture 14, Signaling-2 (review session 3/29/21) 1. What is true of EGFR (EGF receptor)? A. Only one EGFR monomer in a dimer is ever activated; it phosphorylates several Tyr on the other monomer. B. EGFR phosphorylates substrates (including Ras) on Tyr to activate them. C. EGFR is activated by a cytosolic kinase, which phosphorylates EGFR on Tyr residues. D. EGF binding changes EGFR conformation to allow its dimerization. 2. Two proteins, Grb2 and Cbl, are both recruited to the cytoplasmic domain of EGFR after EGFR is activated. What is true? A. Grb2 and Cbl compete with each other to bind EGFR, since they both bind the same site (P-Tyr) B. Cbl binds directly to EGFR P-Tyr, while Grb2 is recruited indirectly to EGFR by binding to Sos C. Grb2 and Cbl each bind two sites on EGFR: both a specific P-Tyr, and also a second site nearby D. Grb2 binds a specific P-Tyr on EGFR, while Cbl binds to monoubiquitinated EGFR 3. Ras A. Is recruited to the plasma membrane by binding to Sos, a Ras-GEF B. Binds activated EGFR via its SH2 domain C. Initiates a kinase cascade when activated D. Binds to Erk when activated 4. The duration of Ras activation following EGF addition to cells was measured using a FRET assay. What is true of FRET assays in general, and this experiment in particular? A. In FRET experiments in general, one fluorescent protein absorbs light emitted by a second fluorescent protein. B. In this experiment, FRET decreased as GTP was hydrolyzed. C. In this experiment, FRET was used to test whether Ras and Sos (a Ras GEF) bind each other. D. Both A and B are correct. 5. Akt A. Is recruited to the plasma membrane by binding PI(3,4,5)P3 B. Is phosphorylated by PDK1 C. Is phosphorylated by mTOR (in a complex with other proteins) D. A, B, and C are correct 6. You express a mutant Akt that has no kinase activity in cells, while blocking expression of any wildtype Akt. What happens? A. Bad binds an apoptosis-inhibitory protein. B. Bad binds 14-3-3. C. Bad is phosphorylated by PDK-1. D. Both A and B are correct. 7. Receptor tyrosine kinase signaling eventually leads to degradation of the receptor. What is a step in this pathway? A. The ubiquitin ligase Cbl binds to a P-Tyr on the receptor B. An arrestin binds to a phosphorylated tyrosine on the receptor C. Ubiquitination targets a receptor for degradation by the proteasome D. Both A and C are correct 8. What is important in reducing (turning off) GPCR signaling? A. GPCR phosphorylation B. Arrestin binding to GPCR to block G protein binding C. Removal of the GPCR from the plasma membrane by endocytosis D. All of the aboveQuiz on Lecture 15 Cytoskeleton-1 (actin) 1. What is true of actin treadmilling in vitro (no proteins other than purified actin present)? A. Actin filaments get shorter. B. Actin monomers are added to the plus and minus ends of filaments at the same rate. C. ATP is hydrolyzed by the actin-ATP monomers in solution at the same rate that monomers are added to the filament plus ends $D. The actin monomer concentration is lower than the critical concentration of the minus end. 2. You divide a solution of purified actin monomers between 2 tubes, then add pre-formed actin filaments to Tube A only (no filaments added to Tube B), and observe the tubes over time. What is true? $A. For the first few minutes, filament growth rate is faster in Tube A than Tube B. B. Filament growth rate is faster in Tube A at all times (including the first few minutes), until steady state is reached. C. Tube A and Tube B reach steady state at the same time. D. Choices A and B are both correct 3. Arp 2 and Arp 3 A. Can be incorporated into new actin filaments, at the minus ends B. Are present in a protein complex that overcomes the slow lag phase in actin polymerization C. Are both structurally similar to actin $D. A, B, and C are correct 4. What keeps the concentration of actin monomers higher in cells than would be expected from the behavior of purified actin in vitro? $A. Binding to thymosin B. Binding to CapZ C. Continual activity of cofilin D. High rate of ATP hydrolysis by actin-ATP monomers in cells 5. Formins A. Stimulate actin polymerization by binding to and stabilizing filament minus ends $B. Have a domain that forms dimers, which encircle growing actin filaments C. Make branched actin filaments D. A and B are both correct 6. What provides the driving force to move the front of a migrating cell forward? A. Actin-myosin contractile bundles push against the plasma membrane B. Filopodia protrude from the leading edge $C. Arp 2/3-dependent polymerization forms actin filaments that push against the plasma membrane D. Actin treadmilling pushes against the plasma membrane 7. Why are integrins necessary in cell migration? A. Integrins send signals that stimulate actin polymerization at the leading edge. $B. Integrins link actin filaments to the extracellular matrix, anchoring them and allowing them to push the plasma membrane forward. C. Integrins send signals that slow actin disassembly at the leading edge.D. Integrins stimulate actin filament turnover just behind the leading edge 8. Cofilin severs actin filaments to help recycle actin monomers for re-use. What helps cofilin find the appropriate filaments to sever? $A. It binds better to actin-ADP than actin-ATP. B. It only severs filaments made by formins. C. It only severs filaments that are not bound to actin bundling proteins. D. It only severs actin filaments in stress fibers at the rear of the cell.Quiz on Cytoskeleton-2 1. What is true of microvilli? A. They contain actin and myosin filaments that slide past each other when they move. $B. They contact the interior (lumen) of the intestine. C. They are present on the basal surface of epithelial cells. D. They form tight junctions. 2. What’s true of myosin II in muscle and non-muscle cells? A. Myosin II dimers only form in non-muscle cells $B. Only in non-muscle cells do two myosin II dimers join in opposite directions, to form tetramers C. Myosin II only binds directly to actin in muscle cells D. Myosin II only binds ATP in muscle cells 3. Dimers containing one monomer each of α-tubulin and β-tubulin are present in the cytosol. These dimers A. Are added to the plus ends of growing microtubules B. Contain bound GTP C. Are part of tubulin small complexes (TuSC) $D. Both A and B are correct 4. +TIP proteins A. Bind near the plus ends of microtubules B. Can stabilize microtubules C. Can bind best to regions of MT containing only GTP $D. A, B, and C are correct 5. The microtubule organizing complex (MTOC) in fibroblasts in interphase A. Is located very close to the leading edge of the cell B. Anchors the plus ends of MT that spread out through the cell $C. Contains many embedded γ-TURC D. Contains a single centriole 6. What is true of axonemal (ciliary) dynein? A. The tail is stably bound to one microtubule (MT) of a MT doublet in a cilium, while the heads come on and off the other MT of the same doublet. $B. Its activity should cause it to move along MT, but its linkage to other axonemal proteins causes the MT to bend instead. C. Individual dyneins assemble into dynein filaments that are arranged between MT doublets D. Both A and B are correct 7. Nuclear lamins A. Use energy of ATP hydrolysis to move chromosomes in the nucleus B. Use energy of GTP hydrolysis to move chromosomes in the nucleus $C. Can bind chromatin to regulate gene expression D. Provide tracks for transport of chromatin in the nucleus by dynein 8. What is true of desmosomes?$A. They anchor intermediate filaments at the sites of junctions between two adjacent epithelial cells, strengthening a monolayer. B. They anchor belts of contractile actin filaments in each of 2 adjacent cells. C. They link intermediate filaments to the nuclear envelope, strengthening cells. D. They anchor intermediate filaments to microvilli in epithelial cells.Lecture 17 Quiz questions 1. What is true of antigenic shift but not of antigenic drift? A. It changes the influenza genome. B. It is caused by the high error rate of influenza RNA polymerase. C. It can produce more infectious virus particles. $D. It can produce virus particles with novel combinations of genes 2. The influenza M2 protein is an ion channel. How does it function in infection? A. Aids in the conformational change in HA needed for fusion of the viral envelope with an endosome membrane. B. Permeabilizes the endosome membrane to release the virus into the cytosol. C. Enhances binding of HA to sialic acid on the cell surface. D. Enhances cleavage of sialic acid by NA (neuraminidase) to facilitate budding. $E. Disassembles ribonucleoproteins from other viral proteins, allowing them to enter the cytoplasm. 3. What is required for fusion of the influenza viral envelope with the endosome after endocytosis? A. SNARE proteins $B. Insertion of a fusion peptide at the end of an extended intermediate form of the viral HA protein into the endosome membrane C. Conformational change in HA induced by binding to sialic acid on the cell surface D. An amphipathic helix on the HA protein 4. What is true of the 5’ end of influenza mRNA molecules? A. No 5’ cap is present: viral mRNA have sequences that allow translation without a cap B. The same enzyme that adds a 5’ cap to cellular mRNAs also adds caps to viral mRNAs C. A modified 5’ cap, derived from ATP instead of GTP, is added to viral mRNA $D. A 5’ cap is cleaved off of a cellular RNA, and used to prime synthesis of viral mRNA E. The viral RNA polymerase has a separate activity that forms a 5’ cap as it copies the template 5. HIV virus particles $A. Infect macrophages and CD4-positive helper T cells (TH) B. Infect epithelial cells in the urogenital system during sex C. Have a segmented RNA genome, with 8 separate RNA molecules D. Must be exposed to acid pH to fuse with a cellular membrane E. Bind sialic acid linked to glycoproteins on the host cell surface 6. What is true of HIV and retroviral-like retrotransposons? A. Both encode an RNA-dependent RNA polymerase $B. Both make DNA copies of an RNA molecule and insert them into the host cell genome C. Both encode gp120 D. Only HIV encodes an integrase E. Only retroviral-like retrotransposons use a DNA-RNA hybrid intermediate 7. HIV protease inhibitors are used to treat HIV patients. What does the protease do for the virus? $A. Cleaves the gag poly-protein precursor into matrix, capsid, and nucleocapsid proteins B. Cleaves a precursor form of reverse transcriptase to the active form C. Cleaves gp120 from its receptor to release virus particle from the cell during buddingD. A and B are correct E. A, B, and C are correct 8. What is true about coronavirus and influenza (flu) virus? A. Both have + strand single-stranded RNA genomes. B. Coronavirus binds ACE2 on the target cell, while flu binds CD4 $C. Both have membrane proteins made on membrane-bound ribosome on the ER D. Coronavirus RNA is coated with nucleocapsid proteins, while flu RNA does not associate tightly with proteins E. Both have membrane envelopes acquired from the plasma membrane during buddingQuiz questions Lecture 18 Immunity-1 1) What is true of blood cell development? A) All blood cells are derived from a common pluripotent hematopoietic stem cell B) One precursor cell gives rise to lymphocytes, while a different one gives rise to neutrophils, macrophages, and eosinophils C) Neutrophils are precursors of macrophages $D A and B are correct E) A, B, and C are correct 2) What is true of pattern recognition receptors (PRR)? A) They can be present on the cell surface, in endosomes, or in the cytoplasm. B) Peptidoglycan, polysaccharides, and nucleic acids can all be recognized by different PRR. C) Toll-like receptors (TLR) are present in the cytoplasm, and recognize PAMPS derived from intracellular pathogens. *D) A and B only are correct. E) A, B, and C are correct. 3) What is NOT part of the inflammatory response, triggered by entry of a pathogen into a tissue? A) Histamine release. B) Increased blood vessel permeability. C) Neutrophil migration in response to chemokines. D) Cytokine (including chemokine) release by macrophages. *E) All of the above are involved in the inflammatory response. 4) What contributes to kill of internalized bacteria following phagocytosis? A) Phagosomes fuse with lysosomes, and hydrolytic lysosomal enzymes degrade bacterial macromolecules. B) NADPH-oxidase on the phagosome membrane triggers an oxidative burst that releases reactive oxygen species. C) Cytokines released into the phagosome cause the bacterium to undergo apoptosis. *D) A and B only are correct. E) A, B, and C are correct. 5) What is true of PAMPS and C3b? A) Binding of phagocytic cells to either PAMPS or C3b on the pathogen surface triggers phagocytosis B) PAMPS are normally present on the pathogen surface, while C3b is attached to the pathogen during the immune response C) Both PAMPS and C3b are anaphylatoxins that bind GPCR on leukocytes *D) Both A and B are correct E) A, B, and C are correct 6) Cleavage of C3 to C3a and C3b is a central step in complement activation in all 3 activation pathways. What is true? A) In the lectin pathway, MASP proteases cleave C3 in response to MBL binding to PAMPS. B) In the alternative pathway, C3 is cleaved spontaneously at a low rate. C) C3a stimulates the inflammatory response. D) A and B only are correct.*E) A, B, and C are correct. 7) What is true of lymphocytes? A. Bone marrow is the primary lymphoid organ, while the thymus and lymph nodes are examples of secondary lymphoid organs B. Lymphocytes permanently leave the bloodstream after maturing, and circulate between secondary lymphoid organs via the lymphatic system $C. In lymph nodes, B cells concentrate in lymphoid follicles, while T cells concentrate in the paracortex D. Before contacting antigen, B cells express antibodies on their cell surface and also secrete soluble forms of the same antibodies. E. Memory cells are formed early in lymphocyte development, before contacting antigen, to enhance the response upon antigen contact 8) You isolate some mutant lymphocytes that cannot perform L-selectin-mediated adhesion, and thus also cannot perform downstream adhesion and migration events. What is true of these lymphocytes? *A) They can be found in the bloodstream. B) They can be found in the lymphatic vessels C) They can be found in lymph nodes D) A and B only are correct E) A, B, and C are correctLecture 19 Immunity 2 Quiz questions 1. What is true of antibody gene rearrangement events that occur before contact with antigen? $A. DNA between one V segment and one D segment is removed from one H chain locus. B. DNA between one J segment and one C segment is removed from one H chain locus. C. DNA between one D segment and one J segment is removed from one L chain locus. D. A and B only are correct. E. A, B, and C are correct. 2. What is true of BOTH MHC Class I and MHC Class II? $A. They both bind antigenic peptide and present it to T cells. B. Their genes undergo rearrangement during development, to increase diversity. C. They are expressed on most cells in the body. D. A and B only are correct. E. A, B, and C are correct. 3. What is true of the invariant chain? A. It binds antibody H chain during early B cell development, to prevent cell-surface expression in the absence of L chain. B. It binds T cell receptors to block peptide binding. C. It binds Class I MHC protein in the early secretory pathway to block peptide binding. $D. It binds Class II MHC protein in the early secretory pathway to block peptide binding. E. It binds cell-surface antibody molecules on B cells to enhance activation by T cells. 4. How can antibodies stimulate phagocytosis of pathogens by phagocytic cells? A. Fc receptors on phagocytic cells bind antibodies. B. Antibodies bound to the surface of a pathogen bind complement, which binds to complement receptors on phagocytic cells. C. Antibodies bind MHC Class II on the surface of phagocytic cells. $D. A and B only are correct. E. A, B, and C are correct. 5. Helper T cells (TH) can bind to both dendritic cells and B cells. What is true? $A. Only activated TH bind B cells B. The TCR on TH binds to cell-surface antibody (BCR) on B cells C. TH bind dendritic cells via peptide-bound MHC Class II, and B cells via peptide-bound MHC Class I D. TH activate dendritic cells by binding peptide-bound MHC Class II on the dendritic cells E. A and B are correct 6. What is true about peptide loading onto MHC proteins? A. It always occurs in endosomes B. It always occurs in the ER C. Only foreign peptides are loaded onto MHC proteins D. Peptides are loaded onto Class II MHC proteins on the cell surface $E. Peptides are loaded onto Class I MHC proteins in the ER, and onto Class II MHC proteins in endosomes 7. Why are transplanted organs often rejected?A. Their cells usually display peptides seen as foreign by the recipient B. They contain cytotoxic T cells seen as foreign by the recipient C. They contain helper T cells that cannot bind the recipient’s B cells $D. MHC Class I proteins on their cells are often seen as foreign by the recipient E. MHC Class II proteins on their cells are often seen as foreign by the recipient 8. What is required for cytotoxic T cells (TC) to kill their target cells? A. Granzymes enter pores made from perforins secreted by the T cell B. T cell receptor binds MHC Class I (bound to peptide) on the target cell C. Formation of active caspase to induce apoptosis in the target cell D. A and B are correct $E. A, B, and C are correctCancer cont. and Development review questions 1. According to the cancer stem-cell model for tumor growth and propagation, … A. transit amplifying cells are incapable of cell division, which is reserved for the cancer stem cells. B. transit amplifying cells constitute the great majority of the cells in the tumor. C. if each stem cell divides to create one stem cell and one transit amplifying cell, the abundance of stem cells in the tumor will increase exponentially over time. D. tumors are genetically heterogeneous, even though they show high phenotypic homogeneity. E. the transit amplifying cells, even though each acts transiently, carry the whole responsibility for maintenance of the tumor in the long term. 2. Suppose you are studying tumor heterogeneity in a certain type of melanoma. You have used fluorescence-activated cell sorting (FACS) to specifically isolate those melanoma tumor cells that either do (first category) or do not (second category) express a specific marker present in normal stem cells in the tissue of origin (i.e. the melanocyte stem cells). You implant the same number of cells from each of these categories into severely immunodeficient mice and compare the tumor-formation efficiencies after several weeks, which turn out to be significantly higher for the first category. You then analyze the new tumors using FACS, and find out that the majority of the cells in the tumors that originated from the first category of cells harbor the stem-cell marker, whereas the majority of the cells in the tumors that originated from the second cell category lack the marker, just like their respective founder cells. Do these observations support the existence of cancer stem cells? Write down Yes or No as your answer. 3. Mutation in which of the following genes is most prevalent in human colorectal cancer cells? A. K-Ras B. β-Catenin C. Apc D. p53 E. MLH 4. The Rb gene in retinoblastomas is similar to the Apc gene in polyposis colon carcinomas in that both genes … A. are tumor suppressors. B. are mutated in one copy in all cells of patients with a hereditary form of the cancer. C. are in a locus that shows loss of heterozygosity in the hereditary form of the cancer. D. should be inactivated in both copies to cause the nonhereditary form of the cancer. E. All of the above. 5. The genotypes of 400 colorectal cancer tumors are tabulated below, where the number of tumors with or without mutations in each of the two cancer-critical genes, β-catenin andApc, are indicated. Which row (a or b) corresponds to those with a mutant Apc gene? Which column (1 or 2) corresponds to those with a mutant β-catenin gene? β-catenin 1 2 Apc a 351 2 b 13 34 A. a; 1 B. a; 2 C. b; 1 D. b; 2 6. Which of the following is estimated to be the leading cause of death from cancer in the United States? A. UV light B. Obesity C. Smoking D. Alcohol consumption E. Sedentary lifestyle 7. You are studying the rising incidence of a certain subtype of cervical cancer in Oceania, and are curious to know whether environmental factors are the dominant cause of the disease. You collect the incidence statistics from indigenous populations as well as from two different immigrant populations in three different countries, as shown in the following table. Do these data appear to be consistent with a dominant role of environmental risk factors (E) or a genetic background (G) for this type of cancer? Write down E or G as your answer. Populations Annual age-adjusted incidence rate in females (per 100,000 women) Australia (entire population) 2.2 Indigenous inhabitants 2.1 Japanese immigrants (first generation) 2.0 Melanesian immigrants (first generation) 2.1 Japan (entire population) 3.5 Indigenous inhabitants 3.6 Australian immigrants (first generation) 3.7 Melanesian immigrants (first generation) 3.6 Melanesia (entire population) 12.9 Indigenous inhabitants 12.6 Australian immigrants (first generation) 12.9 Japanese immigrants (first generation) 12.98. Most DNA tumor viruses inhibit the products of … A. Apc and Rb B. Brca1 and Rb C. p53 and Rb D. Brca1 and Apc E. Brca1 and p53 9. The simplified drawing below depicts early stages of animal development. Indicate which letter (A to E) in the drawing corresponds to each of the following terms. Your answer would be a five-letter string composed of letters A to E only, e.g. ECDBA. ( ) Blastula ( ) Ectoderm ( ) Endoderm ( ) Mesoderm ( ) Gastrula 10. Indicate whether each of the following organs or tissues arises from ectoderm (C), mesoderm (M), or endoderm (N). Your answer would be a four-letter string composed of letters C, M, and N only, e.g. MMCC. ( ) Blood ( ) Liver and pancreas ( ) Brain ( ) Bone and cartilage 11. Indicate true (T) and false (F) statements below regarding animal development. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) A fertilized egg is totipotent. ( ) Differences in their regulatory DNA can largely explain the differences between animal species. ( ) Inductive signaling is mostly mediated through G-protein-coupled receptors. ( ) A cell’s response to a signal depends on its exposure to other signals at that present time as well as in the past. A B E C D12. Indicate whether each of the following descriptions better applies to these poles established in an early embryo: anterior (A), posterior (P), dorsal (D), ventral (V), animal (N), or vegetal (G). Your answer would be a six-letter string composed of letters A, D, G, N, P, and V only, e.g. ADGPVN. ( ) It defines the part to become internalized in gastrulation. ( ) It defines the parts to remain external. ( ) It defines the location of the future head. ( ) It defines the location of the future tail. ( ) It defines the location of the future belly. ( ) It defines the location of the future back. 13. In classical experiments done half a century ago, the cells of early frog embryos were disaggregated and later reaggregated in desired combinations. The cells managed to rearrange and sort themselves out into an overall arrangement similar to that of a normal embryo. This effect is mainly due to … A. chemotactic cell motility. B. asymmetric cell division. C. cell adhesion. D. convergent extension. E. lateral inhibition. 14. Fill in the blank in the following paragraph describing collective cell rearrangements. Do not use abbreviations. “Cells form lamellipodia and attempt to crawl over one another, essentially pulling their neighbors inward into a narrow zone. This is accompanied by elongation along the long axis of the narrow zone. This process of … depends on Wnt signaling and is observed multiple times during the development of a vertebrate.” 15. Indicate whether each of the following conditions favors a larger (L) or smaller (S) tissue or body size. Your answer would be a four-letter string composed of letters L and S only, e.g. SSSS. ( ) Hippo overexpression ( ) Insulin-like growth factor overexpression ( ) Growth hormone deficiency 16. Indicate true (T) and false (F) statements below regarding Hox genes. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) The order of expression of Hox genes along the body corresponds to their order in the Hox complex. ( ) Generally, the more anterior of the Hox genes dominate (or suppress) posterior Hox genes. ( ) When a posterior Hox gene is artificially expressed in an anterior region of the embryo, the tissue maintains its anterior character.( ) Hox genes control the A-P axis in both vertebrates and invertebrates. 17. Which of the following evolutionary changes better explains morphological differences between different animals despite many common molecular mechanisms governing their development? A. Changes in the types of genes responsible for key developmental processes. B. Changes in regulatory DNA controlling expression of key developmental genes. C. Changes in DNA-binding domains of key transcription regulators involved in development. D. Changes in coding sequences of key developmental genes. E. Changes in copy number of key developmental genes.Answers 1. Answer: B Difficulty: 1 Section: Cancer-critical Genes Feedback: Cancer stem cells can be responsible for tumor growth and maintenance, yet remain only a small part of the tumor cell population. 2. Answer: No Difficulty: 3 Section: Cancer-critical Genes Feedback: Cancer stem cells are expected to constitute a minority of the tumor cell population. Even though the transplantation is more efficient with the subpopulation of cells that expresses the stem-cell marker, this might simply reflect the heterogeneity in the original tumor, with these cells harboring a higher ability to found new tumors. 3. Answer: C Difficulty: 1 Section: Cancer-critical Genes Feedback: The Apc tumor suppressor gene is mutated (and is defective) in over 80% of colorectal cancers. 4. Answer: E Difficulty: 2 Section: Cancer-critical Genes Feedback: Apc and Rb are tumor suppressors, and inactivation of both copies of each gene occurs in colorectal cancers and retinoblastomas, respectively. The hereditary forms, however, are easier to develop since the person receives only one functional copy to begin with, which can become lost or defective. 5. Answer: B Difficulty: 2 Section: Cancer-critical Genes Feedback: The majority of colorectal carcinoma cells have a loss-of-function mutation in Apc. Among those that do not, the majority have a gain-of-function mutation in β- catenin. Having both mutations is not common, implying that the activation of the Wnt pathway, and not the individual mutations, confers a selective advantage to the tumor cells. 6. Answer: C Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: Smoking is estimated to account for about 33% of cancers and to kill near 190,000 people in the US every year as a result. 7. Answer: E Difficulty: 3Section: Cancer Prevention and Treatment Feedback: The incidence rates for this type of cancer in each immigrant population mirrors that of the host country, suggesting an overall dominance of environmental factors. 8. Answer: C Difficulty: 1 Section: Cancer Prevention and Treatment 9. Feedback: Small DNA tumor viruses encode proteins that interfere with the Rb and p53 pathways, allowing the replication of the viral genome. Answer: ACEDB Difficulty: 1 Section: Overview of Development Feedback: Please refer to Figure 21–3. 10. Answer: MNCM Difficulty: 1 Section: Overview of Development Feedback: Ectoderm gives rise to the epidermis and the nervous system; endoderm gives rise to the gut tube and its appendages, such as lung, pancreas, and liver; mesoderm gives rise to muscles, connective tissues (e.g. bone and cartilage), blood, kidney, and various other components. 11. Answer: TTFT Difficulty: 1 Section: Overview of Development Feedback: A fertilized egg is totipotent, able to give rise to all the different cell types in an organism. Changes in regulatory DNA seem to be mainly responsible for the dramatic differences between one class of animals and another, even though the coding DNA has been, for the most part, highly conserved between the classes. Most known inductive signaling events during animal development are mediated by the transforming growth factor-β (TGFβ), Wnt, Hedgehog, and receptor tyrosine kinase (RTK) pathways. During development and in an adult organism, the cellular response to a signal depends not only on the identity of the signal, but also on the other signals that the cell is receiving, as well as on the previous experiences of the cell. 12. Answer: GNAPVD Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Three axes generally have to be established in the early embryo. In most animals, the animal-vegetal (A-V) axis defines which parts are to remain external in gastrulation and which are to become internalized. The anteroposterior (A-P) axis specifies the locations of the future head and tail. The dorsoventral (D-V) axis specifies the future back and belly. 13. Answer: CDifficulty: 2 Section: Morphogenesis Feedback: Cell adhesion molecules such as cadherins play a major role in forcing the cells into their natural arrangement. 14. Answer: convergent extension Difficulty: 1 Section: Morphogenesis Feedback: In convergent extension, the cells in a sheet crawl over one another in a coordinated fashion, causing the sheet to narrow along one axis (converge) and elongate along another (extend). 15. Answer: SLS Difficulty: 2 Section: Growth Feedback: Growth hormone stimulates growth, and its deficiency leads to a smaller body size. It acts by inducing the expression of insulin-like growth factor, overexpression of which may lead to a larger body size in an animal. The Hippo pathway inhibits organ and body growth in general; overexpression of Hippo, an upstream component of this pathway, inhibits growth. 16. Answer: TFFT Difficulty: 1 Section: Mechanisms of Pattern Formation Feedback: The genomic location of Hox genes (which control the anteroposterior [A-P] axis in both vertebrates and invertebrates) corresponds well with their order of expression along the A-P axis. In other words, Hox genes are expressed according to their order in the Hox complex. Generally, the more posterior of the Hox genes dominate the more anterior ones, so that when both are expressed in a given segment, the segment will assume an identity dictated by the more posterior Hox genes. 17. Answer: B Difficulty: 1 Section: Mechanisms of Pattern Formation Feedback: Changes in regulatory DNA controlling expression of key developmental genes are largely responsible for morphological differences between animal species.1. A patient with a tumor arrives at Stony Brook Hospital. Dr. Woodward suggests treating the tumor with a drug that inhibits the VEGF pathway. Dr. Samuel disagrees with this treatment because she thinks that it might have long-term problems even though the short term outcome might look good. What are the pros and cons of this treatment? A. In the short term blocking VEGF will directly cause apoptosis in the tumor but some of the cancer stem cells will survive and seed new tumor formation in the long term. B. In the short term blocking VEGF will cause the tumor to stop growing by limiting the blood supply in the tumor but without oxygen some of the tumor cells will undergo epithelial to mesenchymal transition and metastasize to new sites. C. In the short term blocking VEGF will cause the tumor to stop growing by limiting the blood supply in the tumor but the tumor will rapidly evolve resistance to the drug and then continue growing. D. In the short term blocking VEGF will directly cause apoptosis in the tumor but the adjacent cells that phagocytose the apoptotic bodies will become cancerous. E. None of the above. 2. Which of the following evolutionary changes better explain morphological differences between different animals despite many common molecular mechanisms governing their development? A. Changes in the types of genes responsible for key developmental processes. B. Changes in regulatory DNA controlling expression of key developmental genes. C. Changes in DNA-binding domains of key transcription regulators involved in development D. Changes in coding sequences of key developmental genes. E. Changes in copy number of key developmental genes. 3. A new organism you discover expresses only 4 Hox genes (Hox A, B, C, and D) along the length of its developing body in a sequential fashion, with A being expressed in the most anterior head region, then B, C, and finally D in the most posterior region. Based on Hox gene function in other organisms, what would you expect to happen if you mutated Hox B to eliminate its function. A. Regions normally expressing Hox B would be unchanged. B. Regions normally expressing Hox B would adopt the cell fates of region C. C. Regions normally expressing Hox B would adopt the cell fates of region A. D. Regions normally expressing Hox B would adopt the cell fates of region D. E. Regions normally expressing Hox B would adopt the cell fates of region C and D. 4. When dorsal organizer region of an early frog embryo is transplanted into the ventral region of a host frog embryo, the embryo develops with two fused bodies, the original body plus a secondary body. If you analyze the secondary induced body to determine the content of cells that originated from the original organizer graft, you would see:A. The grafted tissue makes up a small number of cells in the secondary body, most of the secondary body originated from the host tissue. B. The grafted tissue makes up all of the cells in the secondary body, and none of the secondary body originated from the host tissue. C. The grafted tissue is completely absent from the secondary body, all of the secondary body originated from the host tissue. D. None of the above. 5. A loss of function mutation in which of the following genes would cause an increase in organ size in a developing embryo? A. Hippo kinase B. Yap C. Myc D. Cyclin E E. Bantam 6. An inherent problem in cancer treatment is the development of resistance to otherwise effective treatments. How does cancer resistance develop? A. Treatments, such as chemotherapy, induce the cell to produce different proteins that are resistant to the treatment. B. Treating with any one compound places selective pressure on tumor populations, selecting for those cells that are most able to survive, which often contain resistance mutations by chance. C. Cancer cells change their membrane compositions so that drugs can no longer get into the cells, preventing them from reaching their protein targets. D. Cancer cells target drugs for selective degradation by immune cells. E. None of the above. 7. The anti-PD1 antibody is an effective treatment against some cancer types because: A. Blocking PD1 function exposes the cancer specific antigens and allows the cancer cells to be recognized by T-cells B. Blocking PD1 function causes cancer cells to have more phosphatidylserine in their outer leaflet, causing them to be eaten by macrophages C. Blocking PD1 function prevents cancer cells from inhibiting T-cell activation, allowing Tcells to be fully active against the cancer cells D. Blocking PD1 function prevents cancer cells from undergoing EMT E. Blocking PD1 function prevents cancer cells from migrating out of blood vessels 8. During the process of combinatorial signaling, two cells exposed to the same signal (signal X) can end up having different fates. This is due to:A. The cells were differentially exposed to two different factors earlier in their life B. The cells are each receiving a different second signal, causing them to respond differently to signal X C. The cells are receiving different levels of signal X, causing them to respond differently D. The cells are in a different state of the cell cycle than each other, causing them to respond differently to signal XCancer lecture review questions 1. Indicate true (T) and false (F) statements below regarding the properties of cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer cells invade and colonize territories that normally belong to other cells. ( ) Unlike in normal tissues, cell death is extremely rare in tumors. ( ) Cancer cells grow and proliferate in defiance of normal restraints. ( ) Malignant tumors are composed of cells that grow and proliferate, but still have not acquired invasiveness. 2. Indicate whether each of the following cancers can be best classified as a carcinoma (C), sarcoma (S), or neither of the two (N). Your answer would be a four-letter string composed of letters C, S, and N only, e.g. SSNC. ( ) Breast cancer ( ) Lung cancer ( ) Colorectal cancer ( ) Myeloma 3. Among the following cancers, one is currently leading to the most number of deaths in the United States and in the rest of the world. In the United States, it contributes to cancer mortality more than the next three killing cancers combined. Worldwide, it claims more than 1.5 million lives every year. Which cancer is this? A. Lung cancer B. Breast cancer C. Colon cancer D. Pancreatic cancer E. Stomach cancer 4. X-chromosome inactivation in female mammals occurs mostly randomly early in development, resulting in a heterogeneous cell population, with each cell having inactivated one or the other of its X chromosomes and passing on the same X-inactivation choice to its offspring. The inactivated X chromosome is generally hypermethylated and transcriptionally inactive. You are studying a newly discovered type of colon tumor in women that has a morphology distinct from that of other colon adenomas. You extract chromosomal DNA from the tumor cells. You then either keep the DNA untreated, or digest the DNA with a methylation-sensitive restriction enzyme that only cleaves its recognition DNA sequence if the sequence is not methylated. Finally, youamplify by polymerase chain reaction (PCR) a locus on the X chromosome known to be polymorphic in length (i.e. it is expected to be of different sizes in different X chromosomes). The locus has a restriction site for the mentioned enzyme, such that cleavage would prevent PCR amplification. You quantify the amount of PCR products corresponding to shorter and longer versions of the locus, and obtain the results shown in the following table. Do these data appear to be in better agreement with a monoclonal (M) or a polyclonal (P) origin of cancer? A monoclonal origin would mean that all cells in the tumor are the clonal descendants of a single abnormal cell, while a polyclonal tumor is composed of cells from different lineages. Write M or P as your answer. Healthy tissue sample Tumor sample Undigested sample Longer PCR product 49% 49% Shorter PCR product 51% 51% Digested sample Longer PCR product 47% 99% Shorter PCR product 53% 1% 5. Indicate true (T) and false (F) statements below regarding cancer. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer can be induced by infectious agents such as viruses. ( ) The earlier a cancer is diagnosed, the better the chances are for a cure. ( ) Most cancers originate from a single aberrant cell. ( ) A single mutation is NOT enough to turn a normal cell into a cancer cell. 6. The requirement for accumulation of multiple mutations in cancer progression is manifested in the normalized percentage of new cases of cancer diagnosed in different age groups. Which of the curves A to E in the following graph better represents the incidence of human cancers as a function of age?7. Indicate whether each of the following descriptions better applies to a cancer cell (C) or a normal adult cell (N). Your answer would be a four-letter string composed of letters C and N only, e.g. CCNC. ( ) Higher lactate production ( ) Higher oxidative phosphorylation ( ) Contact inhibition ( ) Anchorage independence 8. In medical oncology, PET (positron emission tomography) is used to selectively image tumors in the body and to monitor cancer progression and response to treatment. Before performing a PET scan, the patient should fast for at least several hours for blood glucose to be sufficiently low. At the time of the scan, the positron-emitting glucose analog fluorodeoxyglucose (FDG) is injected into the bloodstream and the patient is asked to wait for up to an hour while avoiding physical activity. Finally, the scanner moves slowly over the body to reveal the location of possible tumors. Why do you think the patient should avoid physical activity before the scan? A. To prevent the Warburg effect B. To accelerate glucose uptake by the tumor cells C. To prevent the absorption of the radioactive tracer by healthy tissues D. To promote fermentation in healthy tissues E. All of the above 9. Compared to cells of a normal tissue, which of the following occurs less frequently in cells within a solid tumor? A. Apoptosis Age Rate of new cancer incidence per 100 individuals Birth 80 years 3 0 A B C D EB. Necrosis C. Cell division D. Mitotic recombination E. Stress 10. Indicate true (T) and false (F) statements below regarding cell proliferation in human somatic cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer cells show replicative cell senescence. ( ) Cancer cells maintain their telomeres by inhibiting the enzyme telomerase. ( ) Some cancer cells do not rely on telomerase for telomere lengthening. ( ) Most cancer cells lack telomeres. 11. Which of the following proteins is NOT encoded by a proto-oncogene? A. Src B. Ras C. EGF receptor D. Myc E. E-cadherin 12. Mutations in two important cancer-critical genes, encoding p53 and Rb, respectively, are commonly found in cancers. What type of mutations are these expected to be? A. Loss-of-function mutations in both genes B. Loss-of-function mutation in p53 and gain-of-function mutation in Rb C. Gain-of-function mutation in p53 and loss-of-function mutation in Rb D. Gain-of-function mutations in both genes 13. Indicate whether each of the following descriptions better applies to proto-oncogenes (P) or tumor suppressor genes (T). Your answer would be a four-letter string composed of letters P and T only, e.g. PPPT. ( ) Cancer mutations in these genes are usually recessive. ( ) Cancer mutations in these genes include gene duplications. ( ) Cancer mutations in these genes are responsible for most hereditary cancers. ( ) Cancer mutations in these genes are commonly in the form of nonsense (truncating) mutations that abort protein synthesis.14. The homologous chromosome pairs in our cells do not carry identical sequences in all loci. This heterozygosity (difference between the two copies) can be altered in cancer: in fact, loss of heterozygosity at many loci is observed in cancer cells, through an increase in either homozygosity (two identical copies) or hemizygosity (i.e. loss of one copy). Researchers can take advantage of this loss of heterozygosity in cancer cells to identify genomic loci that contain cancer-critical genes. What type of gene would you expect to find in chromosomal regions with a loss of heterozygosity? Proto-oncogenes (P) or tumor suppressor genes (T)? Write down P or T as your answer. 15. The immortalized non-malignant mouse cell line NIH-3T3 was derived from normal mouse fibroblasts in the early 1960s. These cells are able to readily take up exogenous DNA and are prone to transformation by cancer-causing agents, including some retroviruses. DNA extracted from a human bladder carcinoma line is able to transform these cells, as judged by a significant increase in the number of foci (cell clumps) in the cell-culture plates when the DNA is added. The malignant cells contain human DNA, and the DNA can be shown by sequence analysis to contain a single mutant gene that is present in the original bladder carcinoma cell line. The gene codes for a monomeric G protein and was one of the first cancer-critical genes to be identified in this way. The protein encoded by this gene is … A. Src B. Ras C. Myc D. p53 E. Apc 16. Retinoblastoma is an early-onset cancer of the retina with a rapid progression, and is mostly diagnosed in children. In its hereditary form, multiple eye tumors usually arise in both eyes, while the nonhereditary form usually causes fewer tumors in only one eye. Treatment may involve a combination of chemotherapy, radiotherapy, and other therapies and the majority of patients can be cured if given the right treatment. However, survivors of one form of retinoblastoma (and not the other form) have a markedly increased frequency of subsequent neoplasms that can lead to other cancers later in life, especially soft-tissue sarcomas. These patients should therefore be closely monitored throughout their lives. Which gene is affected by the primary driver mutation in this cancer as well as the later sarcomas? Which form of retinoblastoma do you think is associated with a higher risk of subsequent neoplasms? A. p53; hereditary B. Rb; hereditaryC. Ras; nonhereditary D. p53; nonhereditary E. Rb; nonhereditary 17. The effect of the deletion of one copy of the gene encoding p53 is different from the effects caused by other p53 mutations. For example, some loss-of-function mutations in the DNA-binding domain of p53 cripple its function as a transcription regulator. Such a mutation in only one copy of the p53 gene can be enough to confer a p53 loss-of-function phenotype, even when the other copy of the gene on the homologous chromosome is wild type. This is because … A. p53 is a proto-oncogene. B. these mutations are recessive. C. p53 is a tumor suppressor. D. p53 forms a tetramer. E. p53 can induce apoptosis. 18. Indicate true (T) and false (F) statements below regarding the mutational landscape of cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) In each cancer, usually there is one driver mutation and a large number of passenger mutations. ( ) It is estimated that about 20% of our genes are cancer-critical. ( ) Cancer-critical genes can encode metabolic enzymes or components of the RNA splicing machinery. ( ) The karyotype is often severely disordered in cancer cells. 19. Myelomas are cancers of blood plasma cells—white blood cells that are normally responsible for producing large quantities of antibodies. In the following “Circos plot” for myelomas in a hypothetical mammalian genome, the interchromosomal rearrangements are indicated by red lines and variations in copy numbers are indicated in blue. The positions of named genes are indicated with arrows. On which chromosome do you expect to find the antibody genes? Write down the chromosome number (1 to 5) as your answer.20. Three fundamental controls seem to have been subverted in essentially every type of cancer. Choose these three among the following regulatory axes. Your answer would be a threeletter string composed of letters A to F only, in alphabetical order, e.g. BDF. (A)Wnt pathway (B) Rb pathway (C) RTK/Ras/PI3K pathway (D)p53 pathway (E) Hippo pathway (F) GPCR/PKA pathway 21. Genetically knocking out both copies of the p53 gene in rats … A. is embryonic lethal. 1 2 3 4 5 FGF receptor Rb p53 Myc Ras WntB. results in a lower malignancy rate, but the rats are otherwise seemingly normal. C. results in a higher rate of cancer onset, but the rats are otherwise seemingly normal. D. increases cell death by apoptosis, leading to developmental defects. E. does not have any effect unless the rats live outside of the laboratory and are exposed to various types of stress. 22. Which of the following can lead to p53 stabilization and activation? A. Hypoxia B. Overexpression of Myc C. DNA damage D. Telomere loss E. All of the above 23. From an immortalized human HeLa cell line with wild-type p53 genes, you have derived a line that lacks both copies of the gene. You then treat the original and derived cells with the anticancer drug doxorubicin, which can activate the p53 pathway in the cell by stalling DNA replication forks and inducing double-strand breaks in DNA. You measure cell proliferation in the presence of different doses of the drug in each of the two cell lines, and plot the results as shown in the graph below. Which cell line (A or B) do you expect to be the original HeLa line? Write down A or B as your answer. 24. Indicate true (T) and false (F) statements below regarding cancer. Your answer would be a four-letter string composed of letters T and F only, e.g. TFF. Concentration of doxorubicin (nM, log scale) Relative cell proliferation 100 1 0 A B 0.1 1 10( ) Cancers become less and less heterogeneous as they progress. ( ) Knocking out Ras or Myc genes individually leads to a higher incidence of cancers in mice, and knocking out both genes simultaneously has an even stronger phenotype. ( ) Wnt signaling is important in colon epithelial cells and, correspondingly, mutations in genes in the Wnt pathway are present in most colorectal cancers. 25. Carcinoma cells that have acquired malignancy and started local invasiveness to begin metastasis … A. decrease the expression of E-cadherin and undergo mesenchymal–epithelial transition. B. increase the expression of E-cadherin and undergo mesenchymal–epithelial transition. C. decrease the expression of E-cadherin and undergo epithelial–mesenchymal transition. D. increase the expression of E-cadherin and undergo epithelial–mesenchymal transition. 26. The Ames test is used to test the mutagenicity of a compound suspected to be a carcinogen. In a simple form of the test, the carcinogen is first mixed with a rat liver extract. A disc of filter paper is soaked with this mixture and placed on a culture of a strain of Salmonella typhimurium that is defective in a gene involved in the synthesis of histidine, an amino acid that is essential for cell growth and proliferation. The strain is thus normally unable to grow into visible colonies when the histidine in the culture medium is depleted. In the presence of a mutagen, however, mutations (often “reverse mutations” in the same gene) can enable the bacteria to produce histidine on their own, and therefore grow into colonies. The results of the Ames test for three compounds A, B, and C—each used at the same concentration—are shown in the schematic diagram below. Colonies are indicated with black dots, and the disc is indicated with a white circle at the center of each plate. Which compound (A to C) appears to be a stronger mutagen in this assay? Write down A, B, or C as your answer. A B CAnswers 1. Answer: TFTF Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: Unrestrained growth and division plus invasiveness are two key features of true malignant cancer cells. Cells within a tumor still die in large numbers. 2. Answer: CCCN Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: It is not surprising that many cancers are derived from epithelial cells (i.e. are carcinomas) as these cells are more proliferative and also more exposed to the environment. 3. Answer: A Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: After the cancers of the respiratory system (lung), the next leading cancers by mortality in the US are those of the digestive organs, reproductive tract, and breast. 4. Answer: M Difficulty: 3 Section: Cancer as a Microevolutionary Process Feedback: The individual is heterozygous in the locus, represented in a ratio close to 1:1 for the PCR products of the longer to shorter alleles when an undigested DNA is used. When fully digested, however, only the methylated DNA is expected to be represented in the PCR products. Whereas the polyclonal healthy tissue sample still has a mostly unbiased 1:1 ratio, the monoclonal tumor cells (all with the same inactive X chromosome) have a skewed ratio. 5. Answer: TTTT Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: A single mutation is not enough to change a normal cell into a cancer cell: most cancers develop gradually from a single aberrant cell by the accumulation of a number of genetic and epigenetic changes over time. Treatment is generally easier if the cancer is diagnosed at earlier stages. Some cancers can be induced by infectious agents. 6. Answer: E Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: The incidence of cancers grows exponentially as a function of age in adulthood, although it is thought to reach a plateau or even decline after the age of 80. 7. Answer: CNNCDifficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: Transformed cells display an altered growth control (e.g. lack of contact inhibition or anchorage dependence) and an altered sugar metabolism (e.g. deemphasized oxidative phosphorylation, increased glucose uptake, and increased lactic acid fermentation). 8. Answer: C Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: When the blood glucose level is low, the added radioactive glucose analog is preferentially and rapidly taken up by tumor cells as a result of the Warburg effect. Physical activity stimulates uptake by muscle cells also (including heart muscle), which interferes with the imaging. 9. Answer: A Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: Although cancer cells tend to avoid apoptosis, they still die on a massive scale by necrosis in solid tumors. 10. Answer: FFTF Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: Mammalian cancer cells avoid replicative cell senescence (which generally depends on telomere shortening) in two ways: they can either maintain telomerase activity to prevent telomere shortening, or evolve an alternative mechanism based on homologous recombination to lengthen their telomeres. 11. Answer: E Difficulty: 1 Section: Cancer-critical Genes Feedback: Loss-of-function mutations in the tumor suppressor E-cadherin promote the epithelial–mesenchymal transition and local invasiveness. 12. Answer: A Difficulty: 1 Section: Cancer-critical Genes Feedback: Both p53 and Rb are coded by tumor suppressor genes, and their inactivation promotes cancer. 13. Answer: TPTT Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: Truncating mutations in tumor suppressor genes can create recessive, loss-offunction mutants associated with most hereditary cancers. A proto-oncogene, in contrast,can be activated by a limited set of dominant, gain-of-function mutations (e.g. missense point mutations or other accidents such as gene duplication, which can produce an abnormally large amount of gene product). 14. Answer: T Difficulty: 3 Section: Cancer as a Microevolutionary Process Feedback: Once the first copy of a tumor suppressor gene is lost or inactivated, the remaining copy is commonly lost by a less specific mechanism, leading to loss of heterozygosity in the gene as well as in its neighboring loci. This represents a way of finding loci that contain tumor suppressor genes. 15. Answer: B Difficulty: 1 Section: Cancer-critical Genes Feedback: Ras was also found earlier to be the oncogene carried by some sarcoma viruses, and the identification of the same gene (with similar mutations) in both cases was an important discovery that advanced our understanding of the molecular biology of cancer. 16. Answer: B Difficulty: 2 Section: Cancer-critical Genes Feedback: Due to a germ-line mutation in the Rb gene, people with hereditary retinoblastoma are at a higher risk of cancers in older age. Note that this is different from “relapse” of partially treated retinoblastomas, which can be accounted for by the survival of a few original cancer cells after the initial treatment. 17. Answer: D Difficulty: 2 Section: Cancer-critical Genes Feedback: The dominant negative effect of p53 mutations is due to the fact that defective p53 monomers can block the function of the tetramers in which they participate. 18. Answer: FFTT Difficulty: 2 Section: Cancer-critical Genes Feedback: The number of driver mutations in most cancers is estimated to be on the order of 10. About 300 genes (less than 2% of our genes) are strongly suspected to be cancercritical. They encode proteins of various functions. In addition to DNA sequence changes, chromosomal breaks and translocations are common in cancer cells. 19. Answer: 5 Difficulty: 3 Section: Cancer-critical GenesFeedback: Translocation between immunoglobulin loci and proto-oncogene loci activates the proto-oncogenes and can transform the cell. These translocations are common in myelomas, where each one takes advantage of the powerful transcription potential of the immunoglobulin genes. 20. Answer: BCD Difficulty: 1 Section: Cancer-critical Genes Feedback: Genes in each of these pathways are mutated in most cancer cells that have been analyzed. 21. Answer: C Difficulty: 2 Section: Cancer-critical Genes Feedback: Most rats (or mice) with a homozygous knockout of the gene encoding p53 develop cancer within a few months after birth but appear normal otherwise, suggesting that p53 function is required only in special circumstances. 22. Answer: E Difficulty: 1 Section: Cancer-critical Genes Feedback: Cells raise their concentration of p53 in response to a whole range of conditions. 23. Answer: B Difficulty: 3 Section: Cancer-critical Genes Feedback: Cells in the original HeLa line better respond to the drug since they can trigger apoptosis through the p53 pathway. Knocking out p53 makes the cells more resistant to this effect of the drug. 24. Answer: FFT Difficulty: 2 Section: Cancer-critical Genes Feedback: Cancer risk is increased by gain-of-function mutations in Ras and Myc, not by their knockout. Cancer progression coincides with increasing heterogeneity of the tumor cell population. 25. Answer: C Difficulty: 1 Section: Cancer-critical Genes Feedback: The transition in cancer cells to become invasive involves shifting to a less adhesive and more motile character, resembling the epithelial–mesenchymal transition in normal development. A key part in this process involves switching off the expression of E-cadherin, which is a cell adhesion molecule. 26. Answer: BDifficulty: 3 Section: Cancer Prevention and Treatment Feedback: More colonies (revertants) appear in the presence of a stronger mutagen, with the distribution of colonies reflecting the dose–response relationship for the mutagen (stronger mutagens will affect bacteria further away from the disc).1. HPV can lead to the formation of cervical cancer by the sustained expression of viral protein E7 or E6. How does E7 lead to the eventual development of cancer? A. E7 binds to p53 and inactivates it, allowing for the accumulation of mutations in the DNA B. E7 binds to Rb and prevents it from inhibiting E2F, causing the cells to cycle more frequently C. E7 activates the transcription of Myc, which causes cells to cycle more frequently D. E7 binds to p21 and prevents it from inhibiting Cdk4, causing the cells to cycle more frequently E. E7 binds to p16 and prevents it from inhibiting Cdk4, causing the cells to cycle more frequently 2. Which of the following genetic changes cannot convert a proto-oncogene into an oncogene? A. A mutation that introduces a stop codon immediately after the codon for the initiator methionine. B. A mutation within the coding sequence that makes the protein hyperactive. C. An amplification of the number of copies of the proto-oncogene, causing overproduction of the normal protein. D. A mutation in the promoter of the proto-oncogene, causing the normal protein to be transcribed and translated at an abnormally high level. 3. Ras is a GTP-binding protein that is often defective in cancer cells. A common mutation found in cancer cells causes Ras to behave as though it were bound to GTP all the time, which will cause cells to divide inappropriately. From this description, the normal Ras gene is _______. A. a tumor suppressor. B. an oncogene. C. a proto-oncogene. D. a gain-of-function mutation. 4. Your studies in cell biology have revealed how APC, a protein encoded by a tumor suppressor gene that is frequently inactivated in people with colorectal cancer, functions in the Wnt signaling pathway. This has inspired you to study Wnt signaling. You would like to design a treatment for people with colorectal cancer caused by APC mutations. Assuming you have drugs that target and inactivate each of the following proteins, which one would be best for treating these patients? A. The Wnt receptor B. APC C. Glycogen synthase kinase 3 beta (GSK3) D. The TCF transcription factor5. Which of the following is not a common property of a transformed cell? A. Increased cell size B. Aneuploidy C. Reduced requirement for growth factors D. Failure of normal differentiation E. Immunodeficiency 6. One of the properties of a transformed cell is that it displays the Warburg effect. The Warburg effect is best described as: A. The increased utilization of aerobic respiration to meet the needs of increased ATP consumption. B. The decreased utilization of aerobic respiration to divert carbon for use as cellular building blocks. C. The increased migration of cells to distant sites within the body. D. The increase in aneuploidy caused by non-disjunction. E. The decreased rate of apoptosis. 7. Mutations in the Rb gene often lead to retinoblastoma. You are examining a patient at Stony Brook Hospital that has retinoblastoma. You examine tissues collected from other parts of the body as well and observe that in many of the cell types there is also a homozygous loss of Rb function, but there is