7.06 EXAM b)(5 points) How would you show that this protein specifically interacts with the minus ends of microtubules?

Transcription

1 7.06 EXAM : /20 2: /20 3: /20 4: /20 5: /20 TOTAL /100 Question 1 (20 pts) You are studying microtubule (MT) associated proteins. I) (12 points) a)(5 points) You identify a gene encoding a protein which genetically interacts with microtubules and wish to prove that its protein product does indeed physically interact with microtubules. Describe one way you would show this. ithe technique mentioned in class was: to radiolabel you protein of interest. Add the protein to a solution containing purified microtubules in the presence of a microtubule stabilizing drug. As a control add the radiolabeled protein to a buffer without MTs. Centrifuge the samples and run the supernatant versus the pellet out on a SDS-PAGE. If the protein associates with MTs then you should see a band by autoradiography corresponding to the protein in the pellet of the sample with the MTs. (Since the MTs are found in the pellet and they have "pulled down" the protein of interest with them). For your control without MTs, you should see a band in the supernatant but not in the pellet lane. Other acceptable answers included: immunofluoresence to show colocalization of your protein and the MTs. However, this is not rigorous enough a test to claim a direct physical interaction since there may be some protein which binds MTs which your protein binds in turn, making it LOOK like its binding the MTs DIRECTLY when it really is not. b)(5 points) How would you show that this protein specifically interacts with the minus ends of microtubules? The key to this question is establishing the polarity of the MT. The technique highlighted both in class and on the problem sets to address this issue was the use of an axenome to determine MT polarity: Label tubulin subunits with a fluor and allow them to polymerize around an axenome. In this case the axenome functions as a MTOC and you can visualize the + ends as long strands coming off one end of the axenome, while the - ends will be very short and on the opposite side of the axenome from the + ends. Label your protein and add it to the axenome/mt complex. Using microscopy you can then visualize whether your protein is binding the +, - or length of the microtubule depending on its placement on the MTs. Half credit was given to a number of schemes which clearly attempted to establish some sort of polarity. For example, some people mentioned using immunofluoresence to visualize whether the protein of interest colocalizes to the MTOC. The problem with this experiment is that again it does not prove DIRECT physical interaction. This was graded more harshly here since the use of the axenome was highlighted in the course quite often. c)(2 points) You find that this protein does indeed bind the minus ends of microtubules. What could this imply about its function? II)(8 points) The protein may function to regulate MT dynamics by stabilizing the - end. Perhaps it is part of a MTOC d)(5 points) You are studying vesicle transport along the MT. You find that in a certain cell type a class of vesicles are transported from the center of the cell OUT towards the periphery. You wish to clone the gene responsible for this movement and suspect it is a kinesin related motor. Given what you know about kinesin's affinity for MTs in the presence of ATP, how would you biochemically isolate this protein?

2 We know that in the presence of ATP the motor binds the microtubule and walks down its length. However, this association is not as stable as we would like since as the ATP is hydrolyzed the motor protein loses some of its contacts with the MT. Therefore we need to use a non-hydrolyzable ATP analogue to allow binding to the microtubule but nothing else. So: Make a column containing microtubules and a MT stabilizing drug. In the presence of non-hydrolyzable ATP add the fractions of cell lysate that you think contains the motor protein. (The fractions could have been separated by charge, size, or whatever else you want. These fractions may be pre-tested for your activity by looking for mobility of vesicles on MTs in vitro. I.e. the fraction that contains an activity that allows vesicle movement in the right direction on MTs in an ATP dependent fashion is the one you want to add to your MT column). Elute anything bound by the MT column off and run an SDS-PAGE. Microsequence any bands of interest. Clone the cdna for the proteins. Test recombinant versions of your protein in the vesicle mobility test to make sure it has the right kind of activity. e)(3 points) You isolate a kinesin family protein and find that it moves from the + end of MTs to the - end. Do you think you've isolated the protein you were looking for? Why or why not? No. The microtubules in the cell are oriented with the - ends in the center of the cell and the + end directed out from the nucleus associated MTOC. We wanted a motor that moves from - to + (from the inside of the cell out) Some people wrote that since the protein that was isolated moves + to - it must be a dynein. This is incorrect. There s a small family of kinesins that can move in that orientation as well. Question 2 (20 pts) The human genetic disorder, junctional epidermolysis bullosa with pyloric atresia (PA-JEB), is characterized by blistering on the skin surface. PA-JEB patients share a common deletion in the gene coding for the cytoplasmic 17 C-terminal amino acids of an integrin b subunit (integrins can function as the trans-membrane component of hemidesmosomes). Other than this deletion, the rest of the gene is just like the wild-type gene. Since the skin is made of an epithelial cell layer, it is reasonable to expect that the blistering is due to lack of attachment to the extracellular matrix (ECM) caused by the mutated integrin subunit. This is confirmed by making a knockout mouse with a deletion in the b subunit gene. These mice die within 8 days of birth and exhibit a lack of epithelial attachment to the ECM, mimicking PA-JEB. Epithelial cells can be cultured and form monolayers if laminin, an ECM component, is used to coat the bottom of the Petri dish. Interestingly, the pentapeptide, YIGSR, induces detachment of epithelial cells from the bottom of the lamininlined Petri dish. If YIGSR is fluorescently labeled, it is seen, through microscopy, to be associated with the membrane of detached cells. a) (5 points) What is the simplest explanation for how YIGSR inhibits epithelial attachment to laminin-lined plates? It is likely that YIGSR is a competitive inhibitor for the integrin receptors ECM binding domain. The localization of YIGSR to the detached cell membrane suggests that it disrupts the laminin binding site of the integrin and that it does not bind to laminin. Epithelial cells from PA-JEB patients can be cultured. However, they do not form epithelial layers on lamininlined plates; instead, they are detached. b)(5 points) How could you use the fluorescently labeled YIGSR peptide to determine if PA-JEB cultured cells can bind to laminin (remember the answer to A)? What would the expected result be? Why? Mix as in (A) and see if YIGSR associates with the cell membrane. If YIGSR acts as a competitive inhibitor, YIGSR binding would imply laminin binding. You would expect that it should since PA-JEB cells have a deletion in the cytoplasmic region suggesting that ECM binding should remain intact. This was not meant to be a trick question, but it seems to have worked out that way. Cell detachment can mean EITHER loss of laminin binding OR detachment from the cytoskeleton. In this case, it means detachment from the cytoskeleton. Plectin is a cellular protein known to bind to keratin (a cytoplasmic intermediate filament). Fluorescently labeled antibodies which recognize plectin show that in normal epithelial cells, plectin localizes to hemidesmosomes. However, in PA-JEB cells, plectin is disperse and not specifically associated with hemidesmosomes. Plectin is a cellular protein known to bind to keratin a (cytoplasmic intermediate filament ). Fluorescently labeled antibodies which recognize plectin in normal epithelial cells, plectin localizes to hemidesmoseomes. However,, in PA-JEB cells, plectin is disperse and not specifically associated with hemidesmosomes. c)(5 points) Propose a model for the defect in plectin localization to hemidesmosomes in human PA-JEB cells. The 17 aa deletion to the integrin b subunit eliminates the plectin binding domain and, therefore, eliminates the ability of this integrin to bind plectin and cause hemidesmosome localization. On this model, plectin is the linker between the cell-ecm attachment protein (the integrin), and the cytoplasmic IF's.

3 d) On the following diagram, label plectin, keratin, laminin, and the 17 amino acid PA-JEB deletion. EPITHELIAL CELLS APICAL SURFACE DESMOSOMES BASAL SURFACE ATTACHED ECM

4 EPITHELIAL CELLS APICAL SURFACE kerat in DESMOSOMES BASAL SURFACE 17amino acid delet ion plectin ATTACHED ECM laminin Question 3 (20 pts) The gene BNI1 encodes a formin-like protein that is implicated in mating projection (shmoo) formation of the yeast Saccharomyces cerevisiae. While a wild-type MATa cell forms a shmoo upon exposure to the pheromone alpha-factor, the mutant deleted for the BNI1 gene (bni1) cannot form a shmoo and remains a round/oval cell upon exposure to pheromone. a) (5 points) You (the wünderkind yeast cell biologist) hypothesize that BNI1 is required for the regulation of actin cytoskeleton organization during shmoo formation. If this hypothesis is correct, what would you expect the actin cytoskeleton of the bni1 mutant to look like? Illustrate (with drawings) the actin cytoskeleton structure of the bni1 mutant cell compared to that of the wild-type cell upon exposure to pheromone. In your answer, provide a brief explanation of how the actin cytoskeleton is visualized. Yor picture should show: i)wild-type: polymerized actin (cables), concentrated in shmoo tip (2 pts) i) bni1d mutant: dispersed actin, i.e., no polymerized actin in shmoo (2 pts) (iii) actin cytoskeleton is visualized by staining with rhodamine-conjugated phalloidin and observing by fluorescent microscopy (1 pt.) Accepted answers: (ii) "no actin at all" in mutant got full credit, although note that you will see actin structure, it's just not organized (iii) "immunofluorescence", "fluorescent-tagged actin antibody", "fluorescent actin monomers", "electron microscopy", "decoration with myosin heads" all got full credit. Note that myosin decoration is usually used with actin filaments that are assembled in vitro; immunofluorescence is a much easier technique to visualize actin structure inside a cell. "Use dye" or "fluorescent microscopy" are along the right idea but not specific enough; those answers received 1/2 pt. b) (5 points) Previous data suggests that Bni1 protein localizes to the shmoo tip, at which Cdc42, a small GTPase, is also known to localize during shmoo tip formation. You wonder whether Bni1 binds to the activated form of Cdc42, thereby linking the activation of mating signal with actin organization. What experiment addresses this question, and what is the expected result? (i) immunoprecipitation/affinity chromatography (2 pts)

5 (ii) add GTPgS (non-hydrolyzable analog of GTP) (1 pt) to Cdc42 protein; <add also GDP to Cdc42>; mix each with Bni1 protein (iii) immunoprecipitate with Cdc42 antibody or affinity chromatography with Cdc42 (1 pt) (iv) Western blot with Bni1 antibody (1 pt) (v) If the two proteins bind, should observe a Bni1 band on Western (1 pt) <no band with Cdc42-GDP and Bni1> Accepted answers: -- gel mobility shift assay with two proteins - full credit gel shift (2 pts) GTPgS-Cdc42 (1 pt) and Bni1 (1 pt), one is radiolabeled Cdc42-Bni1 binding is represented by slower mobility in gel (1 pt) Note that you need to run a non-denaturing (native) gel for this experiment to work. Those who went out of their way to say "SDS-PAGE gel" lost 1/2 pt. -- co-immunolocalization - full credit co-immunolocalization (2 pts) add pheromone to cell to trigger shmoo response (1 pt) add Bni1 antibody and Cdc42 antibody (1/2 pt), each tagged with a different chromophore (1/2 pt) Cdc42-Bni1 binding represented by colocalization (overlap) of the two fluorescent signals in the cell (1 pt) Note that while co-immunolocalization is consistent with the notion that Cdc42 and Bni1 are present in the same part of the cell during shmoo formation, the experiment does not demonstrate direct binding between the two proteins. -- two-hybrid assay- full credit two-hybrid assay (2 pts) Bni1 on activating construct, Cdc42 (GTPgS or dominant active Cdc42) on DNA-binding construct (or vice versa) (2 pts) Cdc42-Bni1 binding represented by activation of the reporter (1 pt) Partial credit: -- mention use of antibody - 1 pt -- described part of an experiment, no result - 2 pts -- described full (correct) experiment, no result - 3 pts -- described full (correct) experiment, said "expected result is that the proteins will bind" - 4 pts -- stating "active Cdc42" without reference to GTP/GDP status - 1/2 pt out of 1 Also some suggested experiments that address different questions. experiment the question it answers Delete CDC42 gene, see if mutant Is CDC42 required for shmoo shmoos formation? Immunodepletion of Cdc42 Is interaction of Cdc42 with another protein required for shmoo formation? Immunodepletion of Cdc42, Is Cdc42-Bni1 interaction with an antibody that blocks required for shmoo Cdc42-Bni1 association formation? The last approach must be based on the presumption that Cdc42 and Bni1 do associate. Note that in the question, you do not know this (in fact that's what you want to know). Since you don't know this, you won't know that by simply adding an Cdc42 antibody, you block the interaction between Cdc42 and Bni1. Also, immunodepletion is a technique that can't really be used for yeast cells... c) (5 points) Further research reveals that Bni1 protein binds Cdc42 as well as profilin, the actin-monomer-binding protein. Bni1 is a protein of approximately 2000 amino acids with four identifiable domains. You express each domain separately and do protein-protein interaction assays of each domain with Cdc42 and profilin with the following results: Bini 1 Protein Amino acid l l l l l l domain I l domain II l domain III l domain IV l l l l l l Protein protein interaction assay results a.acids expressed domain profilin binding Cdc42 binding

6 domain I domain II domain III comain IV - - What can you conclude from these results? (i) domain I binds Cdc42 (2 1/2 pts) (ii) domain II binds profilin (2 1/2 pts) (iii) separate domains of Bni1 bind the two proteins. Partial credit: -- mentioned binding - 1 pt -- mentioned separate domain binds each protein - 2 pts Some wrote that Cdc42 binding to Bni1 triggers profilin binding by Bni1, but this is a model, not something that can be concluded from this data. d) (5 points) You make a construct in which the first N-terminal 1200 amino acids of the BNI1 gene is deleted, and you call this truncated gene BNI1-NT (for N-terminal truncation). When you transform the plasmid with the BNI1-NT gene under a strong promoter into a wild-type cell, you observe that its phenotype is exactly like that of the bni1 deletion mutant. Based on the data from part C, provide an explanation for the phenotype of the wild-type + BNI1-NT strain. (i) BNI1-NT acts as a dominant negative construct (1 pt) (ii) Bni1-NT is unable to bind Cdc42 because domain I is deleted (1 pt) (iii) but Bni1-NT can still bind profilin (1 pt) (iv) and Bni1-NT, overexpressed and more abundant, competes away profilin binding from wild-type Bni1 protein (1 pt) (v) therefore the link between mating signal and cytoskeletal organization is lost, resulting in the mutant phenotype (1 pt) Some answered that domain I/Cdc42 binding is required for profilin binding. Note that data in part C shows that domain II by itself is sufficient for profilin binding. And even if the full-length Bni1 protein in vivo does require Cdc42 binding prior to profilin binding, that model by itself cannot explain why the wild-type Bni1 protein also cannot perform its function in the wild-type + BNI1-NT strain. Question 4 (20 pts) a) (8 points) You infect a strain of mice, Strain 1, with a virus that contains Protein X, Protein Y and Protein Z. This allows you to generate CTL's from Strain 1, CTL-1. How would you determine if indeed these CTL's recognize the virus? If they are against the virus, how would you determine which protein from the virus CTL-1 recognizes? Design experiments to address these questions along with hypothetical answers. You would purify target cells (i.e. macrophages) from Strain 1 +/- viral infection. Then you would mix the CTL-1's with the macrophages from the uninfected mouse, and separately from the infected mouse. You would assay for killing (i.e., chromium release). Only the infected cells would be killed if the CTL's are restricted to viral epitopes. Once you determined that viral infection was required for killing ( the CTL's are against viral peptides), you would infect mice with virus that were missing either protein x, protein y or protein z, and see which target cells the virus could kill. The ones that it couldn't kill (i.e. virus minus protein x) would be the protein that the CTL's recognize. You could also purify the target cells from an uninfected Strain 1 mouse and then infect them with the mutant virus to then assay for CTL mediated killing. Alternatively you could inject purified proteins into the target cells and assay for killing. EXPT CTL's Target Cells Killing? CTL-1 macrophage - virus no CTL-1 macrophage+ virus yes CTL-1 macrophage +virus -protein X no CTL-1 macrophage +virus -protein Y yes CTL-1 macrophage +virus -protein Z yes Therefore CTL-1 is viral specific and recognizes an epitope in protein X. b) (4 points) You now isolate another strain of mouse, Strain 2, that has different MHC alleles than Strain 1. Can CTL-1 cells kill virally infected cells from Strain 2? Why or why not?

7 No, CTL's are MHC class I restricted. Therefore CTL's from strain 1 can only kill target cells from strain 1 or any other matched MHC class I strains. c) (4 points) You mate a mouse from Strain 1 to a mouse from Strain 2 and produce progeny, Strain1/2. Can CTL-1 cells kill infected cells from Strain1/2? Why or why not? CTL-1 can only recognize MHC alleles from Strain 1. However, the progeny of Strain 1 contain the same alleles ( if strain1 is homozygous) as Strain 1 and therefore could kill virally infected target cells from Strain 1/2. Allelic exclusion is exclusive to TCR and BCR's. The polymorphism at the MHC locus exists but that does not mean there is recombination at a high enough frequency to make the progeny of a mouse different from the parent. d) (4 points) You find another strain of mice, Strain 3, that has a mutation in MHC class I such that it produces no protein for that receptor. Would Strain 3 produce CD4 T cells (helper) and/or CD8 T cells (CTLs)? Why or why not? (remember thymic development) Positive selection requires that there be an interaction between a T cell and the appropriate MHC class. CD4 cells require an interaction between the T cell and Class II MHC, which is still possible in this mouse. Therefore CD4 development is normal. CD8 cells require an interaction between the T cell and Class I MHC, which is not possible in this mouse. Therefore CD8 development is blocked. The mouse would be missing the CD8 (CTL) compartment. Question 5 (20 pts) You are studying the llama humoral immune response, and isolate large amounts of antibody through biochemical purification. Certain fractions of your antibody prep have the expected profile on SDS page gels: a50 kd band ans a 25kD band. However, you are surprised to find that some of your fractions show only the 50 kd component, and no matter how careful you are with your purification you can t find any trace of a 25 kd component. a) (2points) Draw a stick diagram of the most likely structure of this antibody, indicating the variable domains. Using some sequence information from your biochemical work, you perform some cloning from a llama genomic DNA library and sequence the gene of this strange antibody. Regions IV and V are highly homologous to the constant region of human and mouse g chains. The overall structure is the following: Region 1 Region II Region III Region IV Region V (250 repeats) (10 repeats) (6 repeats) (1 repeat) (1 repeat) a)(3 points) What do regions I, II, and III correspond to with human and mouse antibody genes?

8 The V, D, and J regions. b)(3 points) What is the expected potential diversity of this antibody (i.e., how many different antibodies can be produced)? 250 x 10 x 6 = 1.5 x 104 different possible sequences. If we assume that with this heavy chain there is also imprecise joining and N-regions as with human heavy chains, then the potential diversity increases around a thousand-fold to 1.5 x 107. c)(4 points) How does this number for potential diversity compare with that for the expected diversity of human or mouse antibodies? The diversity for human and mouse antibodies is on the order of The main reason for the much larger potential diversity of human and mouse antibodies is that they have a light chain, which increases diversity by around a thousand-fold. When you look at the cdna of this antibody from plasma cells, you find two different mrna forms, one lacking and the other including region V. You notice that region V has a very hydrophobic sequence. Based on what you know about human and mouse antibodies: d)(2 points) What is region V? Region V codes for the transmembrane region of the membrane-bound form of the antibody. e)(3 points) Why is region V on one form of the message and not on the other? Antibodies are expressed first in a membrane-bound form on B cells, then in a soluble form in plasma cells. f)(3 points) Briefly, what is the most likely mechanism of exclusion or inclusion of V from the message? Poly A Site Selection