Problem Set 8 Key 1 of 8

7.06 2003 Problem Set 8 Key 1 of 8 7.06 2003 Problem Set 8 Key 1. As a bright MD/PhD, you are interested in questions about the control of cell number in the body. Recently, you've seen three patients with abnormally high amounts of T-cells and concurrent autoimmune disease, where the body's immune system begins to attack itself. To understand why these patients are ill, you isolate T-cells from them and begin to study the cells in the lab to determine what particular mutations are causing disease. a. Why might the patients have abnormally high levels of T-cells? overproliferation or defect in apoptosis b. You begin to study the cells in culture and first decide to look at the protein Caspase-9 by Western blotting, since you have a good antibody available. You make protein extracts from all cell lines in the presence or absence of trophic factors, run the extracts out on an SDS-PAGE gel, transfer to a membrane and probe the membrane with anti-caspase-9. Your results are shown below: WT Mut 1 Mut 2 Mut 3 Cell Strain + - + - + - + - Trophic Factor What can you conclude about mutant cell lines 1 and 2? the apoptotic pathway is not being activated properly in the absence of trophic factor, as caspase-9 is not cleaved into its active state. furthermore, the mutations in patients 1 and 2 are in the apoptotic pathway upstream of caspase 9 activation (so, loss of function in Bax, Apaf1, Cytochrome C, Bad, a constitutively active mutation in either PI-3 or AKT kinase, or even a dominant negative mutation in Bcl-2/Bcl-xl could be responsible for the lack of apoptosis) c. You find that the cells isolated from patient 3 never undergo apoptosis, even under extreme conditions of hypoxia and gamma irradiation. It seems that patient 3 has a mutation in their apoptotic pathway, so which gene

7.06 2003 Problem Set 8 Key 2 of 8 might you conclude the mutation affected, given the data in b? How could you test this hypothesis experimentally? the mutation may affect Caspase-3, as Caspase-9 is properly cleaved and presumably activated in these cells in the absence of trophic factor. to test this, you would perform a rescue experiment in which you transfected the gene encoding a wt copy of caspase-3 into the mutant cells, and make wt protein in the cells (not overexpression!). then you would check to see if they were able to do apoptosis in the absence of trophic factor. if the mutation affected caspase-3, the cells with the new wt gene should now be able to apoptose when growth factor was removed. d. You return to mutants 1 and 2 and check to see if the Bax protein is affected. The experiment you perform is a Co-IP, where you use anti-bax antibodies to do the immunoprecipitation and then do a Western blot to look at Bcl-2 binding to Bax. You first probe your blot with anti-bax to make sure the IP worked and then strip the anti-bax antibody from it and reprobe the blot with anti-bcl-2. Both blots are shown below: WT Mut 1 Mut 2 Cell Strain + - + - + - Trophic Factor Bax Bcl-2 What genes could be mutated in patients 1 and 2, using this data and the data in b? Both mutants seem to undergo activation of the apoptotic pathway just fine up to the point where Bax is released from inhibition by Bcl- 2 binding, but then fail to cleave and activate Caspase-9 (from b), so the mutation could affect any genes acting in the pathway between those Bax and Caspase-9 (Bax itself, Cytochrome c, Apaf1, Caspase- 9 itself). e. You discover that the mutant 2 cells (but not the mutant 1 cells) have a dramatically decreased respiration rate in addition to the defect in apoptosis. What gene is the mutation affecting?

7.06 2003 Problem Set 8 Key 3 of 8 Cytochrome c is mutated. This protein would give both the apoptotic defect of not cleaving Caspase-9, since it is needed to help Apaf1 activate the cleavage of Caspase-9, and the respiration defect, since it is necessary for the electron transport in the mitochondria during cellular respiration. f. Propose an experiment to determine which gene is affected in mutant 1. We have the possible genes narrowed to Bax, Apaf1, or Caspase-9 at this point, so we could transfect into the mutant cells the wt gene for each of these and then ask if they undergo apoptosis when trophic factor is removed, as we did in c above. 2. Besides there being induction of apoptosis when trophic factors are removed, there are pathways in which a death ligand (e.g. CD95/FasL) binds to a death receptor and signals the activation of regulator Caspases like Caspase-9, which then activate other Caspases like Caspase-3. You isolate some leukemic cells from several patients which you think are defective in apoptosis and treat them with the anti-cancer drug doxorubicin, which promotes the production of death ligand. Your results follow: WT Mut A Mut B cell strain + - + - + - doxorubicin yes no yes no no no Undergo Apoptosis? a. What part of the "death pathway" might be affected in Mut A? there may be not enough ligand produced and doxorubicin stimulates production of enough ligand to get apoptosis to occur. b. What part of the "death pathway" might be affected in Mut B? the receptor or the downstream components like regulator caspases (like caspase-9), or even effector caspases like caspase-3. 3. What would you expect to happen in a normal cell line when the following proteins were expressed at the designated levels: Protein expressed + Trophic Factor - Trophic Factor Overexpress Bcl-2 that No Apoptosis No Apoptosis, acts as will not release Bax a dominant negative Overexpress Bad that doesn't bind Bcl-2 No Apoptosis Apoptosis, as some wildtype Bad remains

7.06 2003 Problem Set 8 Key 4 of 8 Express at normal levels a Bad protein with S to D mutations at its phosphorylation sites Express at normal levels a truncated Caspase-9 that looks like the cleaved form of the protein Express at normal levels a Caspase-9 mutant that has an aspartate to alanine mutation Apoptosis Apoptosis, this seems to be a constitutive form of Caspase-9 that can activate Caspase-3 all the time No Apoptosis and can inhibit Bcl-2 No Apoptosis, as this protein acts as phosphorylated Bad, which cannot bind Bcl-2 and is probably sequestered by 14-3- 3 Apoptosis Apoptosis, even though this mutant cannot be activated because its cleavage site is mutated, there is still WT caspase-9 to do the job. 4. You are a gifted grad student, pursuing four different thesis projects, which requires you to use four different cell lines. Feeling ever so slightly sleep deprived, you mislabel your flasks of cells one night. No problem, I can fix this like I always do, you think, especially since these cells are of four pretty different types: tumor cells, primary fibroblasts, mouse embryonic stem cells, and an immortalized cell line from mouse intestines. a. What characteristics of these cells might you check to distinguish between them? you would ask: do they display density dependent inhibition, are they anchorage dependent (require ECM), do they display loss of apoptotic controls, have they acquired telomerase activity, can they mimic metastasis (degrade ECM), do they require growth factor to divide, can they be induced to differentiate b. What experiments might you perform to determine if each cell line displays these characteristics in a? take a fraction of your cells to perform the following experiments, making sure to keep them growing to do your real experiments when you get this sorted out...

7.06 2003 Problem Set 8 Key 5 of 8 density dependent inhibition: grow cells to confluency, check to see if they begin to grow on top of each other or if they stop growing when they begin to touch each other. anchorage dependence: grow cells with and without ECM components (fibroblast helper cells which secrete ECM), cells which don't require ECM will live, those which do require ECM will die loss of apoptotic controls: grow cells under normal conditions and condition like hypoxia, gamma irradiation, removal of trophic factor, which would induce apoptosis normally. then perform an assay like TUNEL or fluorescent antibody staining for the cleaved form of caspase-3, and look at under the microscope to determine if any of the cells undergo apoptosis under the apoptosis-inducing conditions. have they acquired telomerase activity: perform a telomerase assay using extracts from each of the cell lines (don't need to know the details of the assay, just that it measures telomerase activity) degradation of ECM: incubate the cells with ECM components, then recover the extracellular medium with ECM components, run SDS- PAGE and Western Blot for breakdown of laminins, fibronectins, collagen require growth factor to divide: grow the cells in the absence of growth factor and see if they survive can they be induced to differentiate: grow cells in the presence of differentiation factors like Epo and see if they begin to differentiate c. What might you expect from each cell line in terms of these characteristics? Tumor Cells: density independent growth, may not require ECM, may degrade ECM, grow in absence of growth factor, would not differentiate, lost apoptotic controls, may have telomerase activity Primary Fibrobalsts: density dependent growth, make their own ECM, don't degrade ECM, require growth factor, may differentiate, have apoptotic controls intact, no telomerase activity Embryonic Stem Cells:

7.06 2003 Problem Set 8 Key 6 of 8 require ECM, don't degrade ECM, require growth factors, differentiate, have apoptotic controls inact, telomerase active, density independent growth Immortalized Intestine Cells: density dependent inhibition, require ECM, don't degrade ECM, may not require growth factors, active telomerase likely, don't differentiate, may have apoptotic controls intact. 5. You are performing a screen for mutations in genes that could cooperate with the c-myc oncogene. To do this, you make use of a transgenic mouse that contains a c-myc oncogene under the control of a strong promoter. These mice express high levels of c-myc and develop several types of tumors only after a variable latency period. You then infect these mice with a slow-acting carcinogenic retrovirus, which integrates itself into tht host genome, and observe the following: 100% % tumor-free mice 75% 50% 25% 0% myc + + retrovirus myc + 50 100 150 200 250 Age in days a. What is the effect of the retrovirus infection in myc + mice? The retrovirus infection reduces the latency period of the onset of tumors in the myc + mice. b. What is the mechanism by which the retrovirus allow you to identify other oncogenes that synergize with the c-myc oncogene? The retrovirus inserts itself in the myc + mice genome more or less at random. If the virus is inserted near a promoter or enhancer region of a proto-oncogene, this could lead to this proto-oncogene to become overexpressed turning it into an oncogene. WT

7.06 2003 Problem Set 8 Key 7 of 8 You cloned the genomic DNA from a tumor flanking a proviral insertion. This same DNA sequence was found to be overexpressed in 21/29 tumors and codes for a novel gene which you call ccg for c-myc cooperating gene. c. How would you test in vitro if ccg functions as an oncogene that could synergize with c-myc? We can do a COLONY FORMATION ASSAY or SOFT AGAR ASSAY. In order to do this, we overexpress c-myc, ccg or c-myc + ccg in primary cells from wild-type mice and plate them on soft agar at low density. After several days of incubation, we count the number of colonies formed for each sample. If c-myc cooperates with ccg, c-myc + ccg should give a significantly higher number of colonies than c-myc or ccg alone. d. What would be a good positive control for the experiment proposed in c? Since ras is a known oncogene which cooperates with c-myc in the formation of tumors as well as colonies in soft agar, a c- myc + ras would be a good positive control. This sample should give you a significant number of colonies in this assay. Some of your mice develop melanomas which correlate with a very poor prognosis. These mice die shortly after and upon dissection, you discover that they had many tumors in different part of their bodies. You suspect that the original melanoma became metastatic and led to the accumulation of many other tumors. e. What would you do to test your hypothesis? We could isolate samples from the melanomas as well as the other tumors found in these mice. If these other tumors are due to metastasis of the original melanoma, we would expect that the melanoma has gained additional mutations which are the same as in the secondary tumors. However, the secondary tumors should contain more additional mutations which would confer them with their metastatic ability. 6. Two brothers are affected with a rare childhood cancer that can be cured. They are both succesfully treated and reach adulthood. One

7.06 2003 Problem Set 8 Key 8 of 8 brother moves to the countryside whereas the second brother moves to an industrial city which is known to be a chemical waste dump. a. What would explain the difference in cancer incidence among the offspring of the two brothers? Both brothers are heterozygous for a germline tumor suppressor mutation which is passed on to roughly half of their offspring. The family that lives in the industrial city is exposed to more mutagens and therefore the offspring that inherited the mutant allele is much more likely to undergo loss of heterozygosity. b. Is it likely that the mutation has occurred in this pedigree? If so, in which individuals? Yes, because the two brothers are affected but neither of their parents are. The mutation could have arisen in any of their parents. c. Explain why a loss of function mutation of a tumor suppressor gene is recessive at the cellular level yet it shows dominant inheritance in families. In order to achieve loss of function of a tumor suppressor, two mutational events are required. In hereditary cancers, one mutation is inherited through the germline. This means that only one additional mutation is required, increasing its probability such that the penetrance of the dissease approaches 100%. So, even though the mutation is recessive, loss of heterozygosity makes the inheritance appear dominant.