7.06 Problem Set #3, 2006

Size: px
Start display at page:

Download "7.06 Problem Set #3, 2006"

Transcription

1 7.06 Problem Set #3, You are studying the EGF/Ras/MAPK pathway in cultured cells. When the pathway is activated, cells are signaled to proliferate. You generate various mutants described below. (Assume that none of the mutations affect the ability of the proteins to fold properly.) Assume that, when you add ligand (EGF), it is present in abundance. You perform each of your studies in two ways: (i) Overexpression of the mutant while the wild-type endogenous gene is present (ii) Replacement of the endogenous gene with the mutant gene In each case, predict the outcome and explain your reasoning. Your choices are: constitutive signaling (+/- ligand), no signaling (+/- ligand), or no effect (proliferation in the presence of ligand and no proliferation in the absence of ligand). a) Mutation of the EGF receptor so that dimerization is not possible. (i) No effect. The mutant EGF receptors cannot dimerize and signal, but the wild-type receptors retain the ability to signal. (ii) No signaling. The mutant EGF receptors cannot dimerize and signal b) Mutation of the proline-rich domain of Sos to an alanine-rich domain. (i) No effect. Grb2 is unable to bind to Sos via its SH3 domain, because GRB2 binds to the proline rich domain of Sos. However, wild-type Sos can still bind to GRB2 and trigger Ras activation. (ii) No signaling. Grb2 is unable to bind to Sos via its SH3 domain, because GRB2 binds to the proline rich domain of Sos. c) Mutation of the SH2 domain of Grb2 so that it is non-functional. (i) No effect. Mutant Grb2 is unable to bind to the activated phosphorylated form of the EGF receptor, but the wild-type Grb2 retains this ability. (ii) No signaling. Mutant Grb2 is unable to bind to the activated phosphorylated form of the EGF receptor d) Mutation of the SH3 domain of Grb2 such that it is non-functional. (i) No signaling. If GRB2 has no SH3 domain, then it cannot bind and activate Sos. This would lead to the inability of the Ras pathway to signal, regardless of whether wild-type GRB2 is present. This is because this form of GRB2 is dominant negative, as it retains its ability to bind to RTKs, and thus occupies the RTK and prevents wild-type GRB2 from binding. 1

2 (ii) No signaling. If GRB2 has no SH3 domain, then it cannot bind and activate Sos. e) A serine-to-alanine mutation at each of the serines in the N-terminal regulatory domain of Raf to which binds. (i) Constitutive. This is a dominant mutant. Alanines cannot be phosphorylated, and thus cannot bind and regulate the mutant Raf, as it is never phosphorylated. Thus, the mutant Raf is always active, regardless of the presence or absence of ligand. (ii) Constitutive. Alanines cannot be phosphorylated, and thus cannot bind and regulate the mutant Raf, as it is never phosphorylated f) A serine-to-aspartate mutation at each of the serines in the N-terminal regulatory domain of Raf to which binds. (i) No effect. The aspartate mimics phosphorylated serine. Thus, the mutant Raf is constitutively bound and inhibited by However the wild-type Raf can be released from as usual, and thus can allow for normal signaling to occur. (ii) No signaling. The aspartate mimics phosphorylated serine. Thus, the mutant Raf is constitutively bound and inhibited by g) Mutation of the two most C-terminal cysteine residues in Ras to alanine. (Hint: Think back to problem set one.) (i) No effect. The mutant Ras cannot be farnesylated and thus is not recruited to the membrane. However the wild-type Ras that is present can be at the membrane as usual. (ii) No signaling. The mutant Ras cannot be farnesylated and thus is not recruited to the membrane. Thus, Sos cannot perform its GEF activity and mutant Ras remains GDP bound (inactive). 2. You are working on a hormone named Insulin-like growth factor 1 (IGF-1). Without IGF-1, the pancreatic cells you are working with undergo apoptosis (programmed cell death). This apoptotic response can be prevented by adding IGF-1 to the cell culture medium. Immediately, you realize that IGF-1 turns on an anti-apoptotic (i.e. cell survival) signaling pathway and you are very excited to understand more about this pathway. Using affinity labeling (remember back to Unit One), you identify the receptor for IGF-1 and name this receptor IGF-1R. You discover that, upon IGF-1 stimulation, IGF-1Rs are phosphorylated on several tyrosine residues. a) Describe how you would perform a genetic screen to identify other proteins involved in this anti-apoptotic (or cell survival) pathway. 2

3 1)Use mutagen to treat your cells heavily, so that both copies of the chromosome would be mutated. You could then screen through the mutagenized cells looking for temperature sensitive mutant cell lines because they would undergo apoptosis even in the presence of IGF-1, as described below: Genotype Temperature Phenotype Phenotype with IGF-1 without IGF-1 WT Permissive (33 0 C) Apoptosis die Cells survive WT Non-permissive (37 0 C) Apoptosis die Cells survive Mutants Permissive (33 0 C) Apoptosis die Cells survive Mutants Non-permissive (37 0 C) Apoptosis die Apoptosis die 2) Use cloning by complementation to identify which genes are mutated in your cells. b) After performing the experiment described in part a), you discover that you have isolated loss-of-function mutants in two genes involved in the cell survival pathway: IRS-1 and Pi3K. You plug the sequences of these two proteins into the BLAST search program, and it tells you that IRS-1 contains a PTB domain, whereas Pi3K contains an SH2 domain. You know from class that both of these domains can bind to phospho-tyrosine residues, and you wonder if both, one, or neither of these proteins interact with activated IGF-1R. How would you design experiment(s) to test for these potential physical interactions? Assume that you have generated antibodies to both of these proteins. One possible experiment: Co-IP (co-immunoprecipitation) 1)Stimulate cells with IGF-1 and lyse cells in buffer containing phosphatase inhibitors; 2)Immunoprecipitate with anti-igf-1r antibody; 3) Do Western Blot analysis using an anti-irs-1 antibody (for example, but here you could also use an anti-pi3k antibody). If activated IGF-1R interacts with IRS-1, then you can see the band on the Western using anti- IRS-1 antibody, even though you were using an immunoprecipate that was isolated by using an anti-igf-1r antibody. c) You suspect that IRS-1 is also tyrosine phosphorylated upon IGF-1 stimulation. How would you design an experiment to test this hypothesis? (Hint: Use an antibody that recognizes phosphotyrosine residues.) Answer: Add IGF-1 to your cells, make cell lysate in buffer containing phophatase inhibitors and immunoprecipitate IRS-1. Do a Western blot on your immunoprecipitate, and blot with antiphospho-tyrosine specific antibody. If you see antibody binding +IGF-1 but not IGF-1, then IRS-1 is phosphorylated upon IGF-1 signaling. d) You also isolated a loss of function mutant version of the AKT gene in your screen, and you want to determine if Pi3K is upstream or downstream of AKT activation. Both proteins are kinases and you have made constitutively active mutant versions of these two proteins. How would you design experiments to test the order of these two genes in this pathway? List the possible results of these experiments, and how you would interpret each possible result. 3

4 You need to do an epistasis test using a double mutant strain that contains mutations that cause two opposite mutant phenotypes. You can over-express a constitutively active AKT kinase in cells that have loss-of-function mutations in Pi3K. If Pi3K is upstream of AKT, you will see the following: Genotype of cells Phenotype without IGF-1 Phenotype with IGF-1 Pi3K kinase dead mutant (Pi3K-/-) Die of apoptosis Die of apoptosis Constitutively active AKT (AKT*) survive survive Both Pi3K-/- and AKT* Survive Survive If Pi3K is downstream of AKT, the last line of the chart will be different: Both Pi3K-/- and AKT* Die of apoptosis Die of apoptosis e) You want to find out which genes are regulated by this pathway. You read from the cell biology book about DNA microarray experiments. How would you perform such an experiment to determine which genes are regulated by your cell survival pathway? Take two sets of cells, and add IGF-1 to one set but don t add IGF-1 to the other set. After a certain time, wash away IGF-1, lyse the cells and isolate mrna from these two sets of cells. Label one set of mrna with green fluorescence (say cells stimulated with IGF-1) and the other (no stimulation) with red fluorescence. This is actually done by making cdnas from the mrnas and then labeling the cdnas. Mix equal amounts of labeled cdna from each set of cells, and allow that cdna mixture to hybridize on microarray glass slides. These slides contain a series of spots, and each spot is a probe for one gene in the genome. After washing, scan slides using a scanner that can detect both wavelengths of light to visualize how much fluorescence of each color is bound to each spot of probe. Spots that have mostly green mrna bound to them are genes upregulated upon IGF-1 stimulation, and spots that have mostly red mrna bound to them are downregulated by IGF-1. f) Now you have a long list of genes that are potentially regulated by IGF-1. How should you confirm if they are truly regulated by this pathway or not? Try to name as many ways as you can. You can generate reporter gene constructs for each of your candidates. You would insert the promoter region of each candidate gene in front of the open reading frame for GFP to make the reporter constructs. Then you would transfect each of these reporter constructs into your cells and either add or not add IGF-1. True candidates would either be repressed (GFP is expressed without IGF-1 but not +IGF-1) or activated (GFP is expressed + IGF-1 but not without IGF-1) by the presence of IGF-1. Other ways to confirm changes in expression of genes in response to the IGF-1 signal are to measure gene expression +/- IGF-1 by Northern blotting or by RT-PCR. 4

5 3. You have recently joined a laboratory that studies the Hedgehog (Hh) signaling pathway. Before you undertake your own experiments, you would like to reproduce some of the work of a previous graduate student that elucidated the function of PKA in this pathway. a) PKA was identified as an inhibitory component of the Hh pathway. Compare what you would see from the results of the following experiments, depending on whether you did or did not express a constitutively active form of the PKA catalytic subunit in wild-type cells that are being incubated in the presence of the Hedgehog ligand: (i) You purify Cubitis interuptus from cells and run the purified protein through a gel filtration column. You then load each fraction that comes off the column (fractions 1-10) into a different lane of an SDS-PAGE gel and stain with Coommassie Blue. Constitutively active PKA means the pathway is always inhibited, so you would get the smaller repressor sized fragment of Ci. Thus Ci would elute later from your gel filtration column (i.e. fraction 8 for example). If PKA were not consititutively active, Hh would stimulate the pathway, Ci would be its larger 155 kda activator form, and would elute earlier from the column (be in lane 2 for example). The size of the fragment in the gel WOULD ALSO CHANGE because the Ci polypeptide that had been run through the column in the absence of mutant PKA would be longer, whereas the Ci polypeptide that had been run through the column in the presence of PKA would be shorter. (ii) You isolate total RNA from cells and run a Northern blot. Your probe is specific to an RNA that is normally induced by hedgehog. There would be decreased signal (or no signal) in the lane with constitutive PKA. This is because PKA inhibits the Hh signaling pathway, so if PKA is always active, the Hh target genes will be repressed. b) How would you demonstrate that PKA phosphorylates specific residues of Cubitus interuptus (Ci)? Assume that you can use the recombinant PKA catalytic subunit that your lab mate has already prepared. Suggest some possible controls that you would do as part of your experiment. You would perform an in vitro kinase assay. You would first have to purify Ci from cells somehow (either by immunoprecipitation or by some other purification technique such as column chromatography). You would then incubate the Ci with recombinant PKA protein in the presence of [γ- 32 P]ATP. Next, you would run the sample on an SDS-PAGE gel, transfer to nitrocellulose, and expose to x-ray film. You would look for a dark band on the film at the size of the Ci protein. Some possible controls to perform would be to: - Incubate the immunoprecipitated Ci with an unrelated kinase in the presence of [γ- 32 P]ATP and look for the absence of band at the size of the Ci protein. - Purify a form of Ci from cells that you have engineered to lack all of the PKA phosphorylation sites (assuming that you know the phosphorylation consensus sequence) 5

6 to alanines. Repeat the in vitro kinase assay and look to see that you have eliminated (or at least reduced) phosphorylation. c) You make a transgenic Drosophila mutant S2 cell line into which you have introduced a copy of the Ci gene with all of its PKA phosphorylation sites mutated to alanine. You have made this transgenic line in the background of a Ci null mutation. You expect to see constitutive activation of Hedgehog target genes even in the absence of Hedgehog signaling. Your reasoning is that PKA will not be able to phosphorylate Ci, and Ci will fail to undergo Slimb dependent proteolytic cleavage to the 75 kda repressor form. As you expect, when you perform a Western blot using cell lysate and an antibody against Ci, you only detect the 155 kda full length protein. However, you fail to see constitutive activation of Hh target genes in the absence of Hh ligand. Suggest some possible explanations for these results. It is possible that the ability of Ci to act as a functional transcriptional activator is not only dependent on the protein being spared from proteolysis but also on the full-length protein undergoing an activation or modification step dependent on Hh signaling. Indeed, the transmembrane protein Smoothened (Smo) must indirectly lead to the phosphorylation of Fu and Cos2 so that the larger form of Ci can be released into the nucleus to activate transcription. Fu and Cos2 only get phosphorylated when Hh is present, and you have not added Hh to cells. Thus Ci is still in a complex with Fu and Cos2, such that Ci is sequestered in the cytoplasm. d) Your colleague is confused that the Cubitus interuptus protein can have two opposing functions, transcriptional repressor or transcriptional activator, depending on whether Ci has undergone proteolytic cleavage. Provide an explanation as to how two protein products of the same gene can result in contrasting outputs. One explanation is that only the full-length protein contains the transcriptional activator domain, because this domain is the one that is cleaved off in the 75 kda form. It is also possible that only the full-length protein can bind to the co-activators (such as CBP) that are necessary to recruit RNA polymerase and activate transcription. e) You are interested in studying the post-translational processing of the Hedgehog protein, specifically cholesterol modification. You overexpress a mutant version of Hh that does not undergo cholesterol modification (but is otherwise identical in protein sequence) in the normal place where Hh is expressed in the Drosophila wing imaginal disc. You see a gain of function phenotype in which there is an abnormally wide range of Hh signaling and activation of Hh target genes. What conclusions can you draw about the function of cholesterol modification in Hh signaling? The cholesterol moiety targets Hh so that its activity is restricted to certain areas. Cholesterol modification of Hh is required to confer the correct range of Hh action, and it thereby controls the range of action of the inductive signal. Also, you can conclude that cholesterol is NOT required for the signaling activity of the protein. 6

7 f) Smoothened is a 7-transmembrane protein involved in the Hedgehog signaling pathway. You purify Smoothened from cells and then reincorporate it into liposomes. You treat the liposomes with protease, and then wash the protease away. You then solubilize the liposomes and perform SDS-PAGE on the solubilized material. Draw what you expect to see on your Coommassiestainined SDS-PAGE gel. You should see a lane with four different protein fragments in it. This is because the protease you treated with cannot cross the membrane, so it chews up all of the parts of Smo that are exposed outside the liposomes, but all of the parts of Smo inside the liposomes are protected. Smo crosses the membrane 7 times, so before and after protease treatment, that protein would look like this: Protease 4. You re interested in making mutations in Gene X, which is highly conserved from yeast to man but has an unknown function. You compare the sequences of potential translation products of Gene X from different eukaryotic organisms and discover two serine residues that appear to be especially conserved. Using your favorite model organism, S. cerevisiae, you construct mutants in which one of the two serine residues is replaced by alanine. While growing your yeast cultures, you quickly notice that one of your mutant strains (Mutant #1) appears to grow at a normal rate, while the other (Mutant #2) has a slow-growing phenotype. You stain an asynchronously-growing population of cells from each mutant with a fluorescent dye that binds DNA, and then perform FACS analysis. You see the following profiles. 7

8 a) Based on the FACS profiles above, describe the general defect (if any) in cell cycle progression for each mutant. Mutant #1 exhibits wild-type growth roughly half of the cells are in G1 with a 1C DNA content, and the remaining cells are in S phase (between 1C and 2C DNA content) or G2/M (2C DNA content). Mutant #2 appears to be delayed in S phase a larger proportion of cells have between 1C and 2C DNA content. b) Which different cell morphologies you would expect to see if you were to look at the each of the mutant cell populations under a fluorescence microscope? Compare the results you would get between the two populations of cells. Neither mutant will have a terminal phenotype, so you would expect to see all of the different possible morphologies in each mutant cell population. Mutant #1 will have a normal distribution of cells in different stages of the cell cycle, whereas a larger proportion of Mutant 2 cells will have the following S-phase morphology: c) Your colleague has just completed a screen for mutants defective in DNA replication elongation, and pulled out two independent mutants that share the same phenotype with your Mutant #2. The two of you perform complementation tests with your three mutants to see if your Mutant #2 is mutated in the same gene as either of your colleague s mutants. You perform the crosses listed in the labels above each plot drawn below. You then do FACs analysis on the resulting strains after staining them with a fluorescent DNA stain, and see the following profiles. What do you conclude and why? 8

9 Crossing your Colleague s Mutant #1 to your Mutant #2 leads to complementation of the delayed S phase phenotype of both mutants, so the two mutations are in separate genes. Crossing your Mutant #2 to your Colleague s Mutant #2 does not lead to complementation of the delay in S-phase, so the two mutations are in the same gene. d) Why are the axes different in part c) than they were before? Explain the significance of the difference in the axes. In part c, where you mate two haploids to do the complementation test, your resulting strains are diploid. Diploid strains have 2C DNA content in G1, and 4C DNA content in G2. In part b, you were dealing with haploid strains. Haploid strains have 1C DNA content in G1, and 2C DNA content in G2. e) You now take the strain that you constructed in part d) from the plot on the left, and sporulate it to isolate haploids that contain both of the two mutations. You again perform FACS analysis on this haploid stained with the same DNA stain, and see the following profile. What do you conclude? What is your interpretation of this result? The delay in S-phase looks even more severe in the double mutant, suggesting that there is a synergistic defect in S-phase progression in the double mutant. It is possible that the two genes mutated in each of the single mutant strains act in the same pathway, and reducing function in both genes makes the pathway even less able to occur than if only one component is reduced in function. Alternatively, the two genes could act in parallel pathways that are redundant in function. f) It turns out that your slightly absent-minded colleague mistook the diploid that you constructed for complementation analysis (the plot on the left in part c) for the haploid that you made in part e). She attempts to synchronize the complementation test strain with alpha factor, but she could not get the cells to arrest properly. Explain why your colleague is unable to arrest the diploids, and suggest an alternative way to synchronize the diploids (assuming your colleague still wants to continue her work on the diploid strain). 9

10 Alpha factor arrests haploid strains upon the beginning of the mating process. Only haploid yeast can mate. Diploid yeast do not mate, and so alpha factor has no effect on diploid yeast. Diploid yeast cells do not respond to alpha factor because they do not synthesize the receptors necessary to bind and recognize the pheromone. In order to synchronize the diploid cells, you would have to use drugs such as hydroxyurea, nocodazole, and taxol to induce arrest (depending on which stage you want the arrested cells to be in). 10