706-2000-Exam 4 Answer Key 1) The question asks you to explain peaks A and B in the top graph. The other two graphs were there to give you hints. The question did not ask for these other two graphs to be explained, so no points were awarded for simply repeating what mating pheromone caused or explaining how nocodozole causes arrest. A) Peak A is the population of cells in G1 of the cell cycle. Peak B is the population of cells in G2/M of the cell cycle. These cells have twice the DNA content of Peak A because they have completed S phase (DNA replication). B) mut1 was "BLASTed" against the complete genome of S. pombe. (No other organisms were used.) p34 and 350 other genes were found to be homologous. The relationship of these 350 other genes to p34 is that they are all kinases. They are not cyclins and they are not all cdc genes. Anwers that simply restated the question or said that they were all involved in the cell cycle received no credit (even if it was said that they were all kinases). C) The graph should look exactly like the graph on the left (Mut1 cells + Mating Pheromone for 3 hours at 25 degrees). The cells arrest in G1. D) The graph should look exactly the same as the answer to part C. Nocodozole is a microtubule depolymerizing drug and therefore does not inhibit the cell cycle genes directly. (It affects the spindle apparatus.) Mut1 would still have performed all of its functions in G2/M. The cell would have been unable to divide because of the nocodozole. When the nocodozole is removed and the temperature is shifted, the cells divide but the mut1 protein is now defective so the cells arrest in G1. E) Mut1 is a cdc2-like kinase with analogous functions to the S. pombe cdc2 (p34). It is not the cdc2 protein itself. Wildtype Mut1 functions throughout the cell cycle, promoting entry and progression through S and M. F) The most likely function of MBP1 is as a G1 (start) cyclin. No other answers received credit. G) The curve should have a single peak either at G1/M or S. It is a Mitotic cyclin (or alternatively an S-Phase cyclin). It is not another G1 (start) cyclin because the homology sugguests that S. sockiae is very similar to S. Pombe which has a single G1 (start) cyclin. 2A) Since antibodies have two antigen binding sites, they will dimerize the 706 receptors just like the ligands do. The dimerization will activate 706-R. (3 points) 2B) The cytoplasmic region is necessary for the trans-phosphorylation of the receptors to activate the downstream pathway (1 point). The excessive tailless receptors will dimerize with wildtype receptors (function as dominant negative) so that the transphosphorylation will not happen. The downstream pathways are blocked. (3 points)
2C) SH2 (2 points) 2D) Higher. GTP-bound Ras is constitutively active. (2 points) 2E) Higher. Ras functions downstream of the receptor. Activated Ras can bypass the function of the receptor. (3 points) 2F) Yes, you would expect Pks transcription to be blocked (+2). The Map kinase pathway is upstream of Pks, so if MAPKKK, MAPKK, and/or MAPK are inhibited, you would not get Pks transcription (+2). -2: if you had the answer and also said that just one kinase in the Map Kinase Pathway is a S/T kinase. All three of them could be inhibited. -2: if you had the answer and also said that MAPKKK is a tyrosine kinase. 2G) Many answers are accepted. In general, the point break down per example (and two were needed) is: +1: for a valid protein +2: for a correct description of the mutation and its possible effect on the pathway (i.e. dominant/recessive, gain or loss of function, etc.) +1: for describing how the mutation could be tumorgenic -4: One type of proteins that was not accepted as a relevant answer was the MAPK phosphatases. Loss of function in the phosphatases is highly unlikely, and if one phosophatase is inactivated (through loss of two good copies), the other phosphastase could still function to block the Map Kinase Pathway. -1: Ras-GTP is NOT dephosphorylated; it is HYDROLYZED to GDP. -2: Correct examples supplemented by incorrected examples/mutations. 3A) tsg +/- mutation is rececessive in cells, but in the organism, it appears dominant due to LOH: Loss of Heterozygosity. The cell favors the loss of the good copy of the gene through de novo mutation, loss of good chromosome, or gene conversion. -2: if failed to explain about the loss of the good copy. -2: if used p53 to describe the phenomenon. In the problem, we mentioned the mutation as recessive. If p53 was a likely candidate, then the mutation would appear dominant. -2: if answer said that the loss of the second copy would lead directly to cancer. Multiple mutations are needed for cancer to occur. -1: if confused LOH with the methods it takes to get LOH (de novo, gene conversion, and loss of good chromosome). 3B) PCB appears to be a mutagen (+2), which accelerate the accumulation of mutations, leading eventually to cancer (+2). Cancer is a multistep process, and tsg1 +/- mice are one step closer to cancer due to their inherent mutation. Therefore, we observe tsg+/- animals to develop tumors more rapidly when exposed to PCB s.
3C) (2 Pts) - + - -/- -/+ + -/+ +/+ (3 Pts) % Mice Tumor Free +/+ +/- -/- Time - /- Curves drawn flat with 0% mice tumor free only got 2 points. This is just mutation of ONE oncogene. Cancer is a multistep process that requires multiple other mutations, which takes time. 3D) (3 Pts) The progeny would develop tumors at a similar rate as the first set of progeny from part C. Cancerous cells are almost always derived from a single cell that was able to acquire the 6-8 mutations needed for malignancy. The mutations in the tumor cells are not passed on to other somatic and germ cells. Because such cancer causing mutation are extremely rare, the germ cells probably have the same genotype as the parental cells BEFORE they got cancer. (-1 for not explaining why the rates are similar) 3E) (8 pts) This question asks for GENERAL properties of ALL cancer cells, not just specific mutations that could contribute to the cancer. Each part received 1 point for the property, and 1 point for a gene that is mutated and the type of mutation (gain or loss of function) These are the answers that would have received full credit, with some of the more common answers:
1) Loss of Cell Division Control: Includes loss off Cell Cycle regulation and independence from growth factors. Ex: Rb knock out, Constitutive Ras. 2) Overcome Anchorage Dependence: (exception is blood cells) No longer need signaling through focal adhesions to ECM for growth. Ex: Constitutive signal from integrin w/o focal adhesion. 3) Overcome Contact Inhibition: Uninhibited by neighbor cell-cell adhesions. Ex: Loss of junction proteins which contact neighbors and signal growth arrest. 4) Immortality: Overcome Replicative Senesence. Normal cells can divide a limited number of times before they reach a restriction on replicative ability. Ex: constitutive activity of telomerase. 5) Overcome Apoptotic Signals: Cells don t signal when extracellular conditions are unfavorable or if there is severe DNA damage. Ex: loss of p53, constitutive Bcl2 6) Angiogenic: Cells able to recruit blood vessels for nutrients and oxygen. Ex: constitutive secretion of VEGF 7) Ability to Metastasize: Cancer cells can degrade the ECM so that the cells can invade other tissues and blood vessels. Ex: constitutive secretion of proteases. -1 for any wrong information -2 for giving two examples for the same concept on two parts -1 for incomplete explanation -1 for mutations without specification of loss or gain of function -2 for saying Genomic instability is required. It can help cancer development, but random mutations are also frequently the cause of cancer. If you gave loss of p53 or DNA replair genes as an example, you got 1 pt. 4) Patient 1: p53 Deletion Patient 2: p19 Deletion Patient 3: p19 Point Mutation Patient 4: MDM2 Point Mutation 3pts per patient. -2 for either wrong gene or wrong sort of mutation. -3 for both wrong. Patient 5: MDM2 Deletion OR p53 Point Mutation (Dominant Negative). 2pts for first correct gene + correct mutation 1 pt for second. 1 pt for both correct genes but wrong mutations.
5) 1) A Kinase(6): MAPKKK, RTK, PKC, Cdc2, Src, PKA 1pt for any 3. 2pts for all. 2) Ras binds to (6): MAPKKK, SOS, GTP, GDP, GAP, GEF 1pt for any three. 2pts for all. 3) Interacts with 7-TMR (3): Ligand, Trimeric G Protein, PKA 1 pt for any 2. 2 points for all. 4) Dimerization necessary for activity (2): RTK, Caspase9 1 pt a piece. 5) A Second Messenger (3): IP3, Ca++, camp 1 pt for any 2. 2points for all. 6) GTP binds to (3): Trimeric G-Protein, Ras, Tubulin 1 pt for any 2. 2 points for all.