7.06 Problem Set #3, Spring 2005

Transcription

1 7.06 Problem Set #3, Spring The Drosophila compound eye is composed of about 800 units called ommatidia. Each ommatidium contains eight photoreceptor neurons (R1 through R8), which develop in a fixed order. Mutations have been isolated that perturb the differentiation of the R7 cells. Two such mutations are in genes called Sev (Sevenless) and Boss (Bride Of Sevenless). R7 progenitor cells in fly eyes that do not express either (or both) of these genes differentiate into non-neural cone cells (instead of R7 cells). Too many R7 cells develop in fly eyes that are overexpressing either BOSS or SEV (or both). a. Overexpression of SEV in fly eyes that do not express functional Boss leads to ommatidia with too many R7 cells. Overexpression of BOSS in fly eyes that do not express functional Sev generates ommatidia with no R7 cells. What do these results imply about the relationship between Sev and Boss? This epistasis analysis allows one to order genes into a genetic pathway. Epistasis is inferred from the phenotype of double mutant cells or organisms. One analyzes the phenotype of the double mutant organism, and compares it to the two single mutant phenotypes. Whichever phenotype is displayed by the double mutant is the phenotype of the mutant that is further downstream in the pathway. Overexpression of SEV alone causes too many R7 cells to develop, and a loss of function in Boss alone leads to no R7 cells. The double mutant leads to too many R7 cells developing. Thus one would place the Sev gene downstream of the Boss gene in the pathway for the development of the R7 cell. (Downstream means closer to the end of the pathway, or the output. Upstream means closer to the signal for the pathway, or the input.) b. You have isolated a gain of function mutation in Sos; you call this allele Sos*. You find that flies that are expressing Sos* in their eyes have high numbers of R7 cells. What two strains would you make (given that you have gain of function and loss of function alleles in SEV and BOSS available to you) if you didn t already know the order of Sos in the fly eye developmental pathway with Sev and Boss, and what phenotype would each strain have? Sos is downstream of Sev and Boss in the fly eye developmental pathway. (BOSS is the ligand for SEV (the RTK receptor), and SOS is a downstream effector in this RTK-Ras signaling pathway. To determine the order (Boss Sev Sos an R7 cell), you could have made: Flies with eyes expressing Sos* that do not express functional Boss; these would have extra R7 cells. AND 1

2 Flies with eyes expressing Sos* that do not express functional Sev; these would have extra R7 cells. Note that you have to use double mutants that contain two single mutations that yield opposite phenotypes. That is, you couldn t use a fly that was expressing Sos* and overexpressing SEV in its eyes, for instance, because both single mutant phenotypes = too many R7 cells. c. Given what is known now about the RTK/Ras pathway and fly eye development, how do you think the SOS* protein operates molecularly that makes it different from the wildtype SOS protein? Any mutation that would make SOS constitutively active would be acceptable. Examples include: -- a mutant version of SOS that is always at membrane (regardless of where GRB-2 is) so that SOS* is always proximal to RAS (the protein upon which it acts as a GEF) -- a mutant form of SOS that can always interact with GRB-2 s SH3 domains so that SOS* is always in the conformation that is active for being a GEF for RAS (regardless of whether GRB-2 is bound to phosphotyrosines on SEV or not) d. To learn about the expression pattern of your genes Sev, Boss and Sos, you perform in situ hybridizations (a molecular technique for detecting where in a cell or an organism a specific mrna is expressed). In your in situ hybridization experiment, you incubate three different sets of fly eye cells with antisense probes that are complementary to the Sev, Boss, and Sos mrnas. You can then use a colormetric assay to detect where in the fly eyes these mrnas are being expressed. Given what is known about fly eye development, what do you predict that you will see? You would find that Sos and Sev mrnas are expressed in the precursors of R7 cells, while Boss is expressed in the precursors of R8 cells. This is because R8 cells stimulate R7 cell development by producing the ligand (BOSS) for the receptor (SEV) found on R7 cells that triggers the RTK/Ras signaling pathway that leads to R7 cell development. e. You next perform a mosaic experiment. In this experiment, you make mosaic flies, or flies that are a mix of two cell types that differ in their genotypes. In one line of mosaic flies, the R1, R2, R3, R4, R5, R6 and R7 progenitor cells have one functional copy of Boss, while the R8 progenitor cells have no functional copies of Boss. In another line of mosaic flies, the R1-R7 progenitor cells have no functional copies of Boss, but the R8 progenitor cells have one functional copy of Boss. For each of these flies, predict whether their ommatidia will form normally. 2

3 In the flies in which the R8 cells have no functional copies of Boss, you would observe that normal ommatidia fail to form, that is no R7 cell will form. Normal ommatidia will form in the flies that contain a wild-type copy of Boss in the R8 cells. This is because only the R8 cell needs to produce BOSS in order to induce R7 cells, and R7 cells do not need to make their own BOSS ligand. (R7 cells instead need to make SEV and all of the downstream components of the pathway, so that they can respond to the signal being emitted from the R8 cells.) f. Your colleague is studying GRB-2. She creates two strains of mutant flies that are expressing an aberrant form of GRB-2 called GRB-2a. GRB-2a mutants lack the SH3 domains of the protein. Strain #1 expresses both Grb-2 and Grb-2a in its eyes. Strain #2 only expresses Grb-2a in its eyes. What eye phenotype would you expect for these two mutant strains of flies? Explain your predictions. GRB-2 is an adaptor protein that binds to phosphorylated tyrosines on the intracellular domain of the RTK (SEV). Its SH3 domains are required for the binding and activation of SOS (a GEF for Ras). Because GRB-2a s SH3 domains are missing, SOS would not be recruited to GRB-2 and would never be activated. The linear flow in the RTK pathway would be interrupted and no signal would be transduced. Thus fly eyes that are only expressing Grb-2a would have abnormal ommatidia lacking R7 cells. However, GRB-2a can still bind to phosphotyrosine residues on SEV when it is activated by its ligand BOSS, because GRB-2a still has an intact SH2 domain. Thus GRB-2a will soak up activated SEV, preventing wild-type GRB-2 from binding, and will not transmit the signal. Thus Grb-2a is a dominant negative form of Grb-2. For this reason, fly eyes that are expressing both Grb-2 and Grb-2a would also have abnormal ommatidia lacking R7 cells. 2. A tumor is a group of cells that are growing and dividing when they are not supposed to. All of the cells in the tumor originate from a single cell that accumulates enough mutations in genes controlling cell growth and division that this cell no longer responds to the control signals that normally keep cell growth in check. Thus this cell grows and divides out of control, forming a tumor. The mutations that lead to such uncontrolled cell proliferation are usually found in three general classes of genes: -- proto-oncogenes, which are genes that produce proteins in non-cancerous cells that promote cell growth and division of these cells only when it is appropriate for them to divide (e.g. to heal up a wound) 3

4 -- tumor-suppressor genes, which are genes that produce proteins in non-cancerous cells that inhibit cell growth and division of these cells when it is inappropriate for them to divide (e.g. when cells are terminally differentiated and are therefore performing specific specialized functions in the brain, lung, etc.) -- genes that regulate apoptosis (programmed cell death), which we will discuss later in the course a. Is the gene encoding Raf a proto-oncogene or a tumor suppressor gene? Describe one kind of mutation you might find in tumors containing mutations in Raf. Would one or both copies of the Raf gene have to be mutated to cause cells to proliferate? Explain your answers. The gene that encodes Raf is a proto-oncogene, because Raf normally promotes cell proliferation in wild-type cells that are receiving the signal to grow and divide. One type of mutation that would transform the Raf proto-oncogene into an oncogene would be a mutation that causes Raf to be constitutively active. An example would be a mutation of the serine in Raf that normally gets phosphorylated (allowing for Raf to bind the protein and be inhibited) to an alanine, which cannot be phosphorylated. Such a mutation is a gain of function (GOF) mutation that requires only a single mutant copy in a diploid cell to show its effect, because constitutively activating Raf, and thereby constitutively activating the RTK/Ras signaling pathway, is a dominant effect. Another possible mutation that is found in tumors is a version of Raf that is consitutively found at the plasma membrane. This finding tells you that Ras s important function appears to be the recruitment of Raf to the membrane. b. Is the gene encoding a proto-oncogene or a tumor suppressor gene? Describe one kind of mutation you might find in tumors containing mutations in Would one or both copies of the gene have to be mutated to cause cells to proliferate? Explain your answers. The gene encoding the protein is a tumor suppressor gene, because the protein normally binds to Raf and inhibits its proliferation-inducing function by preventing it from phosphorylating MEK. One type of mutation that would allow Raf to be constitutively active (and thus constitutively signaling for proliferation to occur) would be a null mutation in the gene encoding Mutations in tumor suppressor genes need to exist in both copies of the gene, because the presence of a single wild-type copy of the gene encoding is enough to allow for production of the protein that will suppress proliferation by binding and inhibiting Raf. 4

5 c. You are studying two different cell lines with the enhanced PDGF receptor signaling properties that can lead to carcinogenesis. One of your cell lines contains a single point mutation in the gene encoding the PDGF receptor, while the other cell line contains an abnormally high number of wild-type receptors in the plasma membrane. What are two different possibilities of where the point mutation might be in the open reading frame that encodes the mutant receptor? Also, outline how you would determine experimentally which of the two cell lines was which if you forgot to properly label them. Two possibilities of where the point mutation might be are: -- in the extracellular domain. There could be a mutation which leads to constitutive dimerization of receptor monomers, thereby causing constitutive activation of the pathway. -- in the cytoplasmic kinase domain. There could be a mutation there which leads to constitutive kinase activation or constitutive interaction with Grb-2. The overexpressing cell line can be distinguished from the other cell line by using Western blotting on cell extracts and probing for the PDGF receptor using PDGF receptor-specific antibodies. You could also sequence the gene encoding the receptor in both cell lines and check if there is a mutation in the open reading frame or not. d. Ras is found linked to the plasma membrane by a farnesyl fatty acid membrane anchor that is covalently attached to the Ras protein. Predict what would happen if you added serum (a cocktail of growth factors) to a cell that only expressed a mutant form of Ras that lacked the amino acid to which the farnesyl group is normally covalently attached, and discuss how your prediction relates to using farnesyl transferase inhibitors in cancer therapy. Without its farnesyl group, Ras cannot reach the membrane and be activated by Sos (which is activated in the presence of growth factors). Adding serum would thus have no Ras-mediated proliferative effects on these cells. Because Ras requires a farnesyl group to be able to promote the proliferation-inducing RTK/Ras pathway signaling, farnesyl transferase inhibitors are currently being used to treat some cancers. 3. You are studying cell differentiation of gastric parietal cells. To do this you study canine gastric parietal cells in primary cell culture. In gastric parietal cells, the α-subunit of the H + /K + -ATPase is considered a marker of parietal cell differentiation. This protein plays a key role in gastric acid production. It is known that the addition of Epidermal Growth Factor (EGF) to parietal cells induces expression of the α-subunit of the H + /K + - 5

6 ATPase. However, the exact signaling pathway that leads to the H + /K + -ATPase α- subunit expression is unknown. Sonic hedgehog (Shh) is a member of the hedgehog family of proteins. You want to determine if Shh plays a role in parietal cell differentiation. You know that EGF induces the expression of the gene encoding the H + /K + -ATPase α-subunit in canine parietal cells. You now want to know if this induction is dependent on the Shh signaling pathway. a. To this end, you examine whether the gene encoding the H + /K + -ATPase α-subunit is induced upon addition of both EGF and the drug cyclopamine to cells. (Cyclopamine is a specific inhibitor of Smo.) You conclude from the results of your experiment that the induction of gene expression by EGF is dependent on Shh signaling. What result did you get from this experiment? Give a brief explanation of your answer. You found that expression of the gene encoding the H + /K + -ATPase α-subunit was not induced. The proposed pathway is that EGF signaling causes Shh release. Shh, in turn, binds Ptc, which then ceases its inhibition of Smo. This allows Smo to act on downstream proteins, leading to changes in gene expression. However, cyclopamine inhibits Smo. Therefore, the pathway is inhibited and there is no change in H + /K + -ATPase α-subunit expression. b. Through experiments involving cyclopamine, you determine that the Shh signaling pathway is involved in inducing expression of the H + /K + -ATPase α-subunit. You want to find out what other genes expression levels are affected by Shh signaling. Describe two experiments you could do to identify such genes. One approach is to use a microarray analysis of canine parietal cells exposed to Shh compared to parietal cells that were not exposed to Shh. In this experiment, the mrna from both sets of cells is isolated and the corresponding fluorescently-labeled cdna is synthesized using a different color for each condition (e.g. green labeling for cells without Shh stimulation, red labeling for cells with Shh stimulation). This cdna is then incubated with a microarray chip containing oligonucleotide probes complementary to most of the genes in the canine genome. The color of fluorescence on each spot (each spot on the array corresponds to one canine gene) indicates whether the expression level of each gene is higher prior to or after stimulation with Shh. (Green spots = higher expression without Shh; Red spots = higher expression with Shh; Yellow spots = equal expression under both conditions). 6

7 A second approach is to use subtractive hybridization. In this approach, you collect mrna from cells that have been incubated with Shh and cells that have not been exposed to Shh. Only the cdna corresponding to the mrna from the untreated cells is synthesized. Both the mrna from the treated cells and the cdna from the untreated cells are heated to denature any secondary structures, and the cdna is then allowed to hybridize with the mrna of the treated cells. Those double-stranded hybrid molecules composed of one strand of cdna and one strand of mrna are then separated from the single-stranded cdna by hydroxylapatite column chromatography. The single-stranded mrna is then isolated and identified by sequencing and BLAST searches. Any single-stranded mrna that is left must be from genes that were expressed only when cells were treated with Shh, because these mrnas did not have any cdna partners from untreated cells to bind to via hydrogen bonds between complementary DNA bases. c. After identifying a group of genes affected by Sonic Hedgehog signaling, you use a motif-search program to analyze the DNA sequences found upstream of the open reading frames whose expression is affected by Shh. In this way, you discover a DNA sequence motif that is found upstream of many of the open reading frames you identified. Describe an approach you could use to determine whether this upstream DNA sequence motif is necessary for Shh-induced gene expression. Make a reporter gene system using an assayable gene product (such as GFP or luciferase) in a vector. Insert different portions of the upstream DNA sequence motif from a gene known to be regulated by Shh into the vector, and transfect the vectors into the canine parietal cells. After exposure to Shh, only those cells containing vectors harboring the necessary upstream control sequences will show induction of GFP or luciferase. By inserting smaller and smaller segments of upstream sequence into the vector, the sequence that is necessary to form a functional Shh-responsive element can be determined. d. You determine that the nucleotide motif that you discovered is indeed an Shhresponsive control sequence. You are now curious about which transcription factors bind this sequence. You suspect that one of the members of the Gli family of proteins is involved. There are 3 Gli proteins (Gli1, Gli2, Gli3), which are the mammalian equivalents of the Drosophila Ci protein. Describe an approach that you could use to detect the binding of one or more of these Gli proteins to the Shh-response element upon the addition of Shh to the canine parietal cells. One technique you could use in ChIP (Chromatin Immunoprecipitation). ChIP is a lot like co-immunoprecipitation, except that, in co-ips, you are looking for other proteins that immunoprecipitate along with a protein, and in ChIP, you are looking for DNA sequences that immunoprecipitate along with a protein. 7

8 For this approach, canine parietal cells are exposed to Shh. The live cells are then incubated with formaldehyde. Formaldehyde crosslinks proteins to other proteins, and crosslinks proteins to nucleic acids. The cells are then harvested and lysed, and the nuclei are isolated by centrifugation. The nuclei are lysed, and the chromatin is then sheared. An immunoprecipitation is performed on the chromatin with an antibody against the protein you are studying (in this case, one of the Gli proteins). The Gli protein and any chromatin to which it is bound is then eluted, and the crosslinking is reversed by heat. This releases any DNA sequences to which Gli was bound. You then attempt to amplify the Shhresponse element by PCR. If the response element has been immunoprecipitated along with the Gli protein you re studying, then there should be an amplification of this specific sequence by the PCR reaction. You can then detect the amplified Shh-response element as a band in an agarose gel. If the response element has not been immunoprecipitated with the Gli protein, then there will be no amplification of the sequence. Therefore, you can determine which protein binds the response element in response to Shh signaling. Several important negative controls must be done during this experiment, such as immunoprecipitating Gli from uninduced cells, and attempting to amplify a DNA site other than the Shh-responsive element from Gli that has been immunopreicipitated from Shhinduced cells. 4. Your first task as new UROP in a yeast lab is to look at the cell cycle profiles of three temperature-sensitive mutant haploid yeast strains using FACS (fluorescence activated cell sorting). a. You incubate wild-type cells and your three temperature sensitive mutant cells at the permissive temperature, and then shift to the restrictive temperature for at least one cell cycle. You then stain the DNA of these cells with a fluorescent DNA stain. You then perform FACS analysis on the strains, and obtain the FACS profiles below. Which profile is from wild-type cells, and what are the defects in the three mutant strains? Explain your answers. A B 1C 2C DNA content 1C 2C DNA content 8

9 C D 1C 2C 1C 2C DNA content DNA content B shows wild-type cells. In a wild-type haploid yeast population, about half of the cells will be in G1 with a 1C DNA content. The remaining cells will be in G2 or M with their DNA replicated (2C DNA content) or in S phase (currently replicating their DNA so that they have between a 1C and 2C DNA content). A shows cells that are delayed in S phase. Thus a greater proportion of cells will be in S phase at any one time, as is indicated by the lack of complete separation between the 1C and 2C peaks. The cells in between 1C and 2C are currently in the process of replicating their DNA. C shows G2/M-arrested cells. These cells have already replicated their DNA, but are arrested before or during chromosome segregation and the subsequent cell division. D shows G1-arrested cells. These cells have arrested in G1 and are unable to proceed into S phase. b. You grow the four strains described above at the permissive temperature, and then incubate the strains at the restrictive temperature for the length of at least one cell cycle. You then stain the DNA and examine the cells under a microscope. Draw the terminal phenotypes (what the cells look like under a microscope once they are frozen at the point in the cell cycle at which they become arrested or delayed) you would expect to see for each of the strains based on the defects you assigned above. B (wild-type) cells won t have a terminal phenotype. Cells in this population will be at all stages of the cell cycle because they do not contain a temperature sensitive mutation that arrests them somewhere in the cell cycle when incubated at the restrictive temperature. 9

10 D cells are arrested in G1 or at the G1/S transition and so they will look like this: G1 C cells are arrested in G2, M, or at the G2/M transition and may look like the following: M M G2 (anaphase) (telophase) A cells are arrested or delayed in S phase and would look like this: G1/S or S phase c. Some of the FACS profiles above could also have represented wild-type yeast strains that were treated with different compounds, such as α-factor, HU (hydroxyurea) or nocodazole. Which profiles in part a could represent such treated wild-type strains, and why? Profile D could represent α-factor-treated wild-type cells. α-factor arrests cells in G1 before budding. Profile D could also represent HU-treated wild-type cells. HU will block cells in very early S phase with a small bud because HU (hydroxyurea) depletes cells of free nucleotides that they would use to incorporate into the new DNA strands that they would be synthesizing during S phase. The HU block is so early in S phase that the amount of DNA that these cells do synthesize is not detectable by FACS. Profile C could represent nocodazole-treated wild-type cells. Nocodazole arrests cells in M phase by depolymerizing microtubules, which are necessary for chromosome segregation. d. You identify the gene in which one of your temperature sensitive mutations lies. Based on homology, you suspect that the gene encodes a kinase. The grad student you are working with has already purified the wild-type protein product of this gene. She tells you that many kinases will phosphorylate a protein called Histone H1 in vitro 10

11 (in a test tube), regardless of what the actual substrate of the kinase is inside a cell. How would you test whether your protein has kinase activity using Histone H1? Kinases transfer the γ-phosphate from ATP to their substrates. To track this transfer in vitro, you can incorporate radioactive phosphorous ( 32 P) into ATP molecules and use this radiolabeled ATP as the phosphate source for the kinase. You would therefore incubate your purified protein and Histone H1 with radioactive ATP in a test tube. If Histone H1 becomes radiolabeled, you would conclude that your protein acts as a kinase. e. You want to know when the wild-type kinase you have discovered is expressed during the cell cycle. Briefly describe two experiments you could do to figure this out. Two possible techniques are: Synchronize your cells in G1 (with α-factor), release them (ie. wash away the α-factor so that the cells enter the cell cycle in a synchronous manner), and take samples of the cells at different time-points during the cell cycle. Run these samples on an SDS-PAGE gel and use Western Blotting to detect your protein. To be sure of which stages of the cell cycle each of your samples came from, you can also do Western Blotting against previouslycharacterized cell cycle stage markers like cyclins. You can also tag your protein with GFP so you can visualize it in live cells. Then, as above, synchronize and release your cells and follow the location of your GFP-tagged protein throughout the cell cycle under the microscope. This method reveals when your protein is expressed and where your protein localizes in the cell. You can be sure of the cell cycle stage each cell is in by using, for example, fluorescently-tagged tubulin, which forms cell cycle stage-dependent spindle structures. Alternatively, you can do immunofluorescence on fixed cells to answer the same question, by using one primary antibody against the tubulin (to figure out what stage of the cell cycle the cells are in) and another primary antibody against your protein. 11