7.06 Cell Biology EXAM #2 March 20, 2003

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1 7.06 Cell Biology EXAM #2 March 20, 2003 This is an open book exam, and you are allowed access to books, a calculator, and notes but not computers or any other types of electronic devices. Please write your answers in pen (not pencil) to the questions in the space allotted; if you use the back of a sheet make it clear which answer is where. And be sure to put your name on each page in case they become separated! Remember that we will Xerox many exams at random. Good luck! Question 1 Question 2 Question 3 Question 4 Question 5 TOTAL / 20 points / 25 points / 15 points / 20 points / 20 points / 100 points 1

2 Question 1 (20 pts.) Bristle development in Drosophila is controlled by the Notch signaling pathway, with Delta being the ligand, and is initiated by transcription of Achaete and Scute in proneural cells. This group of cells is called the proneural cluster. Only one or two proneural cells maintain Achaete and Scute expression and differentiate into bristles. These cells emit an inhibitory signal that inhibits bristle development in their neighbors. The inhibition is initiated by the interaction of the Delta ligand produced by the differentiating bristle cell with the extracellular domain of the Notch proteins on neighboring cells. Proneural cluster Bristle cell (black) expressed Delta and Inhibits bristle development in neighboring Notch-expressing cells You are interested in understanding the Notch signaling pathway using bristle development in flies as a system. Through a mutagenesis screen, you have already identified four nuclear proteins that are necessary for normal Notch signaling: mutations muta, mutb, mutc, that constitute loss of function mutations and mutation MUTD that constitutes a constitutive active mutation. You want to start by ordering them into a genetic pathway by conducting an epistasis analysis. You make use of the mutant flies obtained from your screen and cross them to generate double mutants. These are the bristle phenotypes you observe: Genotype muta mutb mutc MUTD mutb, mutc muta, mutc muta, mutb muta MUTD mutb MUTD mutc MUTD Phenotype bristle formation in neighboring cells (no inhibition) bristle formation in neighboring cells (no inhibition) no bristle formation in neighboring cells (total inhibition) no bristle formation in neighboring cells (total inhibition) no bristle formation in neighboring cells (total inhibition) bristle formation in neighboring cells (no inhibition) bristle formation in neighboring cells (no inhibition) no bristle formation in neighboring cells (total inhibition) no bristle formation in neighboring cells (total inhibition) no bristle formation in neighboring cells (total inhibition) A) (8 pts.) Order the genes in the pathway and explain the relationships between the genes. MutB I MutC I MutA ! MutD MutB functions upstream of MutC and MutC upstream of MutA. MutB must function to negatively regulate MutC and MutC must be a negative regulator of MutA. This is the only manner in which the observed double mutant phenotypes can be explained. MutD functions 2

3 downstream of all other genes, is positively regulated by MutA and is a positive regulator of the signaling pathway as a constitutive active mutation in this gene causes constitutive inhibition of bristle formation. Having established the epistatic relationship amongst your genes you want to try and understand the molecular details of this pathway. To this end you create a cell line that expresses Notch. In this cell line Notch is the receptor for the ligand Delta. Upon Delta binding, Notch undergoes two proteolytic cleavages that release the intracellular domain. You suspect that this fragment must interact with the first member of the pathway to initiate the signal. Because MutA, MutB, MutC and MutD are all nuclear proteins, if the Notch fragment interacts with any of them, it has to translocate into the nucleus. B) (5 pts.) Describe an experiment, without detailing the experimental steps, to demonstrate that the Notch fragment translocates into the nucleus only upon Delta binding knowing that you can purchase the Delta ligand from TLC, also known as the Thousand Ligand Company. Also, give the expected result. By immunofluorescence, one could determine the location of the Notch protein. To show that the translocation into the nucleus occurs in response to Delta signaling, one should compare cells in the presence and absence of Delta ligand. In the cells that have been exposed to Delta, the fluorescence should be detected in the nucleus; whereas in the cells that have not been exposed to Delta, the fluorescence should be localized to the plasma membrane. Encouraged by the finding that Notch indeed translocates into the nucleus you now want to determine whether the Notch fragment interacts with the first member of the pathway, whose identity you determined using the epistasis analysis in (A). (We will designate this pathway member X to not give away the answer.) To do so, you decide to do a co-immunoprecipitation assay and obtain the following results: Delta ligand added: Immunoprecipitation Ab against: Western blot Ab against: Lane: no no Notch Notch Notch Notch Notch Notch X X Notch X 3

4 C) (7 pts.) Interpret the results indicating which lanes are negative and positive controls for the experiment and why they were necessary. Lane 1 Positive control: Lane 1 - This experiment is important to ensure that we indeed immunoprecipitated Notch in our experiment. Lane 2 Positive control: Lane 2 - This experiment is important to ensure that we indeed immunoprecipitated Notch in our experiment. Lane 3 Negative control: Lane 3 absence of Delta signaling. This is necessary to make sure that the Notch-X interaction is specific to Delta ligand addition. Lane 4 Experiment: Lane 4 - Notch and X interact upon Delta signaling. This is indicated by the ability of Notch to co-immunoprecipitate X only in the presence of ligand. 4

5 Question 2 (25 pts.) You obtained a colon carcinoma cell line from your collaborator at MGH and want to characterize this cell line further with the aim to eventually find a cure for this disease. A) (3 pts.) You remember that a key characteristic of cancer cells is that they continue to proliferate in the absence of serum growth factors. You therefore want to test whether your cell line indeed enters S phase when serum is removed. Describe one method with which you could test this. I remove the serum and then do a FACS analysis on these cells or measure 3H-Thymidine incorporation or BrdU incorporation to detect DNA replication. Having established that your cell line indeed is growing in the absence of serum and because you attended 7.06 you know that the Ras/MAPK pathway is often hyperactive in cancer cells. You, therefore, want to investigate whether this is true for your cell line. B) (6 pts.) Before further analyzing your mutant cell line you first decide to examine the effect of various Ras/MAPK pathway mutations on growth of normal colon cells. To this end you overexpress the various mutant proteins. What do you expect to see? Fill in the chart: Overexpressed Mutant protein MEK with a S->D mutation at the Raf phosphorylation site Grb2 that binds unphosphorylated tyrosine MAPK that fails to dimerize Mutation in Ras that fails to bind GTP SOS that lacks its C-terminus Constitutively active GAP Growth without serum /no no no no C) (6 pts.) While you were characterizing the effects of your mutations on the normal colon cell line, your collaborator at MGH determined that the Ras/MAPK pathway was indeed hyperactive in the colon cancer cell line and that it was Raf that carried a constitutive active mutation. He now suggests to test your constructs in the carcinoma cell line. Because you paid a lot of attention in 7.06 you tell him you do not need to do the experiment, in fact you can predict the outcome and send him the following table by e- mail. Fill in the chart before you send it!: Overexpressed Mutant protein MEK with a S->D mutation at the Raf phosphorylation site Grb2 that binds unphosphorylated tyrosine MAPK that fails to dimerize Mutation in Ras that fails to bind GTP SOS that lacks its C-terminus Constitutively active GAP Growth without serum /no no 5

6 D) (10 pts.) Being impressed with your wits your collaborator now wants to know how he could indeed prove that Raf is constitutively active. In your back you suggest the following experiment. First immunoprecipitate the Raf with a Raf specific antibody bound to beads. Then incubate the sample with radioactively labeled ATP and a substrate to allow incorporation of the radioactivity into the substrate. Spin down the beads, run the supernatant out on a gel, and visualize on a autoradiogram. For a substrate could use MEK since it is a biological substrate for Raf. To show that the protein is constitutively active you show that Raf activity is high even in the absence of serum. 6

7 Question 3 (15 pts.) You have been recruited into a lab to study an epithelial cell line to further the understanding of epithelium formation. You have been working for months to culture this epithelial cell line but have not been having much luck. The cells do not form an epithelial layer in your dish. A) (4 pts.) You suspect that their failure to form an epithelium is due to the absence of ECM. How could you remedy the situation? I could either add fibroblasts to the culture as these cells secrete ECM or I could coat the dishes with laminin or fibronectin in order to mimic presence of ECM. Then the cells should organize into an epithelium and grow happily. You perform the steps you described in the answer to part A and now your cells are organized into a sheet. Your cells continue to do well until you change to a new bottle of media after which all the cells seem to just fall apart and lose their highly ordered structure. Looking for an explanation you find out that the kitchen added EDTA, which chelates divalent cations, to the latest batch of media. B) (4 pts.) How might this offer an explanation for what you observed with your cells? EDTA is a chelator and will complex all the Ca++ and other metals. Thus there will be very little free Ca++ in the media. Tight junctions, adherens junctions, and desmosomes are all disrupted in the absence of Ca++ so the cells will lose almost all cell-cell adhesion and thus the epithelial sheet will lose its structure and organization. You have the kitchen make up new media without EDTA, and you finally have your cell line stable and doing well. Now you can test for the presence of different types of cell-cell junctions in your epithelial sheet. First you suspect that the cells are forming GAP junctions. You perform an immunofluorescence experiment that shows that connexin (the protein that makes up GAP junctions) is present and in the proper position. C) (7 pts.) What experiment would you do next to prove that the GAP junctions are functional and that these junctions can close and open? Inject a colored dye into one cell and observe its movement under the microscope. If the GAP junctions are functional then the dye will quickly move into the adjacent cells and eventually, throughout the epithelial sheet. To make sure the GAP junctions can close properly simply repeat the injection of the dye along with injecting high levels of Ca++. The Ca++ will block the channel and the dye will not spread to neighboring cells. 7

8 Question 4 (20 pts.) It is difficult to understand how the inactivation of CDK inhibitors (CKIs) leads to chromosome loss, which has been implicated in cancer formation. Because Prof. Amon told you that S. cerevisiae cell cycle regulation is very similar to that in human cells you decide to analyze the CKI Sic1 in budding yeast. Indeed you discover that deletion of SIC1 causes chromosome loss and you decide to investigate the reasons for this chromosome loss. Your first experiment is to arrest both wild type and sic1 mutant yeast in the G1 stage of the cell cycle then release the cells from the arrest and observe S phase progression by measuring BrdU incorporation as cells progress through the cell cycle in a synchronous manner. This is what you observe. 90 % BrdU Incorporation Time (min) Wild type Sic1 A) (4 pts.) What two things can you conclude about the dynamics of S phase in the sic1 mutants? DNA replication occurs delayed and it is prolonged in the mutant. B) (2 pts.) The BrdU incorporation profile indicates that deleting SIC1 has an effect on S phase. In order to further establish that sic1 mutants lose chromosomes, you assay both sic1 mutants and wild type cells for the loss of a mini-chromosome (Mini-chromosomes are minimal chromosomes that are stably maintained in wild type yeast). In order to be stably maintained in yeast cells, what features must the mini-chromosome possess? Origin of replication, centromere, optional: telomere C) (4 pts.) Your mini-chromosome showed that SIC1 mutants lose mini-chromosomes at a high rate. You know that Sic1 inhibits all Clb-CDKs in yeast, which include the mitotic cyclins Clb1, Clb2, Clb3 and Clb4 and the S phase cyclins Clb5 and Clb6. You hypothesize that loss of sic1 allows Clb-CDK to run amuck and affect S phase. But, you don t know which group of Clbs, Clb1-4 or Clb5 and 6, is affecting S phase in the sic1 mutant. You create sic1 clb4 and sic1 clb5 double mutants and assess their chromosome loss phenotype at the restrictive temperature. 8

9 8 % Minichromosome 6 loss 4 2 Wt sic1 clb4 clb5 sic1, sic1, clb4 clb5 Which Clb-CDK is allowed to run amuck and affect S phase in the sic1 mutant yeast and explain how you came to this conclusion? Clb5-CDK because deletion of CLB5 suppresses the mini-chromosome loss phenotype of sic1 mutants. D) (5 pts.) You noticed while analyzing your sic1 mutant using FACS analysis that the mutant is also delayed in G2. You suspect that this is due to activation of the DNA damage checkpoint. To test this hypothesis you analyze the FACS profile of sic1 rad53 double mutants. What FACS profile would you expect in the double mutant if the G2 delay were indeed due to activation of the DNA damage checkpoint and provide an explanation for why you drew the shape of the FACS profile in this manner. %Cells %Cells 1C 2C sic1 mutant 1C 2C sic1 rad53 mutant Inactivation of SIC1 causes the activation of the DNA damage checkpoint, which delays cells in G2. If the DNA damage checkpoint pathway is inactivated due to the inactivation of RAD53, cell cycle progression is no longer delayed in G2 and cells continue to divide despite DNA damage. Thus the FACS profile will look like that of exponentially growing wild-type cells. However, these cells will eventually die. 9

10 E) (5 pts.) How would inactivation of RAD53 affect the frequency of chromosome loss in sic1 mutants if the G2 delay in sic1 mutants were indeed due to activation of the DNA damage checkpoint? It would lead to an increase in mini-chromosome loss because the checkpoint is no longer active. Thus, DNA damage can not be repaired and mini-chromosomes are lost at a higher frequency during mitosis. 10

11 Question 5 (20 pts.) You are studying the metaphase to anaphase transition in the budding yeast S. cerevisiae. You examine the cell-cycle phenotypes of a yeast strain carrying a temperature sensitive mutant version of the CDC20 gene (cdc20-1) and a strain carrying a temperature sensitive mutant version of the gene encoding the Securin protein (securin-38). Both of these mutant yeast strains are haploids, and therefore do not carry a wild-type copy of the mutated genes. In order to study the cell cycle defects in these mutants, you arrest wild type, cdc20-1, and securin-38 cells in G1 using mating pheromone at 25 C, the permissive temperature, and then release the cells from the arrest at 37 C, the restrictive temperature, allowing the cells to progress synchronously through the cell cycle at 37 C. You take samples every 15 minutes for 90 minutes and in each time point, you examine the protein levels of the cyclin Clb5 and of Cohesin at each time point for each yeast strain. Below are the Western blots and the graph showing the data you obtained from each yeast strain during the time course. Wild type securin (time) (time) Clb5 Clb5 Cohesin-full length Cohesin-full length Cohesin cleavage product Cohesin cleavage product Western blot Western blot cdc (time) Clb5 Cohesin-full length Cohesin cleavage product Western blot 11

12 A) (5 pts.) Based on the western blot data, what cell cycle regulator is responsible for the cycling of the Clb5 protein and during what stage of the cell cycle does this protein get degraded? The APC-Cdc20 degrades Clb5 at the metaphase to anaphase transition. This is evident because in the wild type cells, Clb5 is degraded concomitant with cohesin cleavage and in the cdc20-1 mutant, Clb5 is not degraded. B) (5 pts.) What process specifically is defective in the securin-38 cells? In this mutant, separase is not active since cohesin is not cleaved. To further investigate the securin-38 phenotype, you decide to examine the localization of both Separase and Securin-38 mutant protein in the securin-38 cells during S phase and metaphase at 25 C and at 37 C. Your localization studies reveal the following: Protein 25 C S-phase 25 C 37 C S-phase 37 C metaphase metaphase Separase cytoplasmic nuclear cytoplasmic cytoplasmic Securin-38 cytoplasmic nuclear cytoplasmic nuclear C) (10 pts.) Why do you think that Separase is mis-localized in the securin-38 cells at 37 C but not at 25 C? Propose an experiment, without giving experimental detail, to test your hypothesis. Securin normally binds to Separase and inhibits its function until the metaphase to anaphase transition. However, the binding of Separase to Securin may also be necessary for the localization of Separase to the nucleus, where it can cleave cohesin. Perhaps the Securin-38 protein cannot bind to Separase efficiently at 37 C because the mutant securin protein becomes mis-folded and therefore, Separase cannot localize to the nucleus. You could perform a co-ip experiment to see whether the mutant Securin-38 and Separase interact at 25 C but not at 37 C. 12

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