7.06 Cell Biology Exam #3 April 23, 2002

Transcription

1 RECITATION TA: NAME: 7.06 Cell Biology Exam #3 April 23, 2002 This is an open book exam and you are allowed access to books, notes, and calculators. Please limit your answers to the spaces allotted after each question. If you write on the back of a page, be certain to indicate clearly which answer is which, otherwise additional writing will be ignored. You may find it useful to read each question completely before you begin answering; often you will get clues from the later parts of a question. Be sure to put your name on each page in case they become separated. You will have 90 minutes to complete the exam. Good luck! Question 1 : / 20 Question 2 : / 20 Question 3 : / 10 Question 4 : / 20 Question 5 : / 10 Question 6: / 20 TOTAL: / 100

2 Question 1 (20 points) You have isolated the cdna that encodes cerebrin, an important cell surface membrane protein normally expressed by a set of cerebral nerve cells. The sequence of the protein predicted from the cdna is mostly hydrophilic, but there are two stretches of 23 amino acids that are hydrophobic enough to form a membrane- spanning segment. However, because there are two charged amino acids in the middle of each of these stretches you are not certain whether or not either or both of these actually spans the plasma membrane. To explore the membrane orientation of cerebrin you use the cerebrin cdna in an in vitro transcription system to synthesize cerebrin mrna. Then you translate cerebrin mrna into protein in a cell-free protein synthesizing system (containing ribosomes, trna, initiation and elongation factors, radioactive amino acids, etc.) in the absence or presence of microsomes (rough endoplasmic reticulum vesicles stripped of their own ribosomes.) You then prepare samples from these translation reactions in four different ways: (1) no treatment, (2) add trypsin (a protease), (3) add trypsin and detergent, and (4) treat reaction with detergent and add endo-glycosidase H (endo H), which removes N-linked sugars that are added in the ER. You analyze these samples by SDS polyacrylamide gel electrophoresis and autoradiography; the results are indicated in the table below. Sample number Microsomes present? Treatment of the reaction products Results 1 NO no treatment One radioactive protein band of mw 42,000 2 NO trypsin No radioactive protein 3 NO trypsin and detergent No radioactive protein 4 NO detergent and endo H One radioactive protein band of mw 42,000 5 YES no treatment One radioactive protein band of mw 43,000 6 YES trypsin Two radioactive protein bands, one of mw 18,000 and the other of mw 21,000 7 YES trypsin and detergent No radioactive protein 8 YES detergent and endo H One radioactive protein band of mw 40,000 You conducted one additional experiment: you removed the two radioactive protein bands, from the experiment in sample 6, treated them with endoh, and reran them on an SDS gel. You found that the molecular weight of the mw 21,000 polypeptide was unaffected, while the mw 18,000 polypeptide migrated considerably faster on the gel, at about mw 15,000. PAGE: 2 OF EXAM #3

3 N C cytosol (=outside of microsome) protease sensitive in expt exoplasm (=inside of microsome) protease protected in expt A. 5 pts.) Is the protein normally glycosylated? Explain why or why not. YES, this protein is normally glycolsylated. The best lanes to compare are lanes 5 & 8. In 5, the mature protein translated in the presence of microsomes weighes 43,000. In 8, the mature protein translated in the presence of microsomes and treated with endo H and detergent weighs only 40,000. Detergent will not change the weight of a protein. Endo H cleaves off sugars, therefore the protein normally is glycosylated with sugars that weigh about 3,000Daltons. If you said YES and compared lanes 1 & 5 (saying 5 weighs more, so sugars must have been added to the protein) and noted the reduction in weight following endo H treatment, then you still got full credit. A YES without reasoning received 2 pts; YES with mention of endo H but no further explanation 3pts, a NO 0 pts. B. 5 pts.) Is cerebrin synthesized with a cleaved signal sequence? Explain why or why not. YES, cerebrin is synthesized with a cleaved signal sequence. The molecular weight of the protein in lane 8 is less than that in lanes 4 or 1. The proteins in each of these lanes are all non-glycosylated either due to lack of microsomes or endo H treatment. A comparison of molecular weights tells you that a 2,000 Dalton fragment was lost. This is presumably the ER signal sequence that is cleaved upon entry into the ER. YES with wrong reasoning or no reasoning received 2pts. NO received 0 pts. C. 10 pts.) Draw the structure of cerebrin in the plasma membrane. Be sure to indicate where the N- and C- termini of the protein are located. And explain why your diagram is consistent with the experimental data stated above. The keys to this problem were to realize (1) that a cleaved signal sequence causes the N-terminus of the protein to end up in the lumen of the microsomes (= the exoplasmic side of the plasma membrane) and (2) that when the protein is treated with trypsin, the intracellular loop (exposed outside of the microsomes) is digested and that two protein segments remain. These two protein segments must be inside of the microsomes (= outside the cell), and the only way for such a protein to have two extracellular segments is for it to have two transmembrane segments. This is consistent with the PAGE: 3 OF EXAM #3

4 information given in the problem which suggests that there are 2 possible transmembrane helices. Either the N or C terminus is glycolsylated, but there is not enough information given for you to determine specifically which one is glycosylated. A perfect drawing and explanation that included how the trypsin fragments related to the drawing received 10pts. Partial credit was given for having the correct # of transmembrane segments, having the N-terminus on the exoplasmic side and stating that this was due to the protein s cleaved signal sequence, etc.. 1 point was deducted for each piece of incorrect information. There was no penalty for drawing the protein in the membrane of the microsome instead of the plasma membrane of a cell so long as you clearly indicated what was on either side of the membrane (exoplasm and cytosol for the cell s membrane or cytosol and lumen for the microsome s membrane). Question 2 (20 points) Many transcription factors shuttle into and out of the nucleus. One such factor that has been widely studied is STAT5. In unstimulated erythroid cells STAT5 is present exclusively in the cytosol. After binding of erythropoietin (Epo) to cell surface Epo receptors, the protein tyrosine kinase JAK2 becomes activated. JAK2 then phosphorylates STAT5 on a tyrosine near its C- terminus. STAT5 then moves into the nucleus, where it activates expression of a set of genes. After Epo is removed from the cell the phosphate residue on the tyrosine of nuclear STAT5 is hydrolyzed, and the (non- phosphorylated) STAT5 moves immediately (i.e. within 5 min.) from the nucleus back into the cytosol. A. 5 pts) Assume you can add segments of amino acids or even sequences encoding entire proteins to the C- terminus of STAT5 and still have a fully functional STAT5 protein. How could you monitor experimentally the movement of STAT5 into and out of the nucleus in response to Epo addition or removal in living cells? ANSWER: The best method for following protein localization in living cells is by making a fluorescent protein fusion. One could fuse the coding region for GFP onto the end of the STAT5 gene, creating a C-terminal fusion. The protein's localization could then be followed using flourescence microscopy, both in the presence and absence of Epo. B. 5 pts) You treat erythroid cells with Epo for 15 min. and observe that about half of the STAT5 is phosphorylated on tyrosine while half is not. You treat these cells with a nonionic detergent which dissolves all cell membranes, including the nuclear membrane. Then you incubate the cell lysate at 4 C for an hour with small plastic beads onto which are immobilized an antibody to Importin alpha PAGE: 4 OF EXAM #3

5 protein. You then recover the beads by gentle centrifugation, wash them to remove all proteins not bound specifically, and then analyze the bound proteins. You find STAT5 in this mixture of bound proteins. Would you expect the bound STAT to be tyrosine phosphorylated, unphosphorylated, or a mixture of tyrosine phosphorylated and unphosphorylated protein? Explain your answer. ANSWER: The bound STAT5 would be phosphorylated. Importin alpha is responsible for carrying STAT5 into the nucleus, which occurs only when STAT5 is phosphorylated. Importin alpha will only bind to this phosphorylated form. C. 5 pts) You take the same detergent lysate of Epo- treated cells as in Part B, and incubate it at 4 C for an hour with small plastic beads onto which are immobilized an antibody to the Ran protein. You then recover the beads by gentle centrifugation, wash them to remove all proteins not bound specifically, and then analyze the bound proteins. Again, you find STAT5 in this mixture of bound proteins. Would you expect the bound STAT5 to be tyrosine phosphorylated, unphosphorylated, or a mixture of tyrosine phosphorylated and unphosphorylated protein? Explain your answer. ANSWER: In this case, the bound STAT5 will be unphosphorylated. Ran only binds cargo during export from the nucleus. STAT5 is unphosphorylated when it is exported. Ran will only be associated with STAT5 (via exportin) in its unphosphorylated form. D. 5 pts) You made a mutant of STAT5 in which amino acids are replaced by alanines. When expressed (by recombinant DNA in erythroid cells) this mutant, called 6A, STAT is present exclusively in the cytosol as long as Epo is not added. Epo addition causes the protein to move normally into the nucleus PAGE: 5 OF EXAM #3

6 (and activate transcription normally). However, after Epo removal 6A becomes dephosphorylated normally but remains in the nucleus for at least 30 min. Provide an explanation for the function of amino acids in STAT5. ANSWER: These 6 amino acids normally function as the nuclear export sequence, or the place where exportin will recognize it to bind. With these amino acids mutated, the Ran/exportin complex will no longer be able to easily bind STAT5 to carry it out of the nucleus. Question 3 (10 points) Plastocyanin is a protein localized to the thylakoid space in chloroplasts. Newly made Plastocyanin contains at its N- terminus a typical stromal import sequence that is cleaved by a stromal protease. This sequence is followed by a thylakoid targeting sequence that is cleaved off by a protease in the thylakoid space. Citrate synthase is a typical protein encoded by nuclear DNA and is transported from the cytosol into the mitochondrial matrix. Suppose you grafted on to the C- terminus of the citrate synthase gene the sequence encoding the Plastocyanin thylakoid targeting sequence. You then expressed this chimeric protein in cultured petunia (a plant) cell that contain both functional mitochondria and chloroplasts. Where would the chimeric citrate synthase/ Plastocyanin thylakoid targeting sequence protein be localized. Explain. The protein would be localized to the mitochondria. The N-terminal sequence of the citrate synthase protein would still direct it to the mitochondria. The thylakoid targeting sequence would not be functional unless preceeded by the stromal import sequence (remember the order of these sequences is essential for them to be functional) and would also need to be located on the N-terminus of the protein. In addition once the protein entered the mitochondria (as directed by the N-terminal sequence) it would fold there and be unable to leave the matrix. PAGE: 6 OF EXAM #3

7 Question 4 (20 points) A. 7 pts) How would you alter the citrate synthase gene (i.e. delete or change some sequences, add others, etc.) such that the citrate synthase protein would become a protein resident in the lumen of the rough endoplasmic reticulum and not targeted to the mitochondrial matrix? Include in your answer a diagram showing the relevant segments of the normal citrate synthase protein and in the one you construct. N matrix targeting sequence C normal N ER signal sequence KDEL C construct Replace the mitochondrial matrix targeting sequence (1 pt) with an ER signal sequence (2 pts) at the N-terminus (1 pt). Add a KDEL sequence (2 pts) at the C-terminus (1 pt) for retention of the protein in the ER. B. 7 pts) How would you alter the citrate synthase gene (i.e. delete or change some sequences, add others, etc.) such that the citrate synthase protein would be directed to the peroxisome and not to the mitochondrial matrix? Include in your answer a diagram showing the relevant segments of the normal citrate synthase protein and in the one you construct. N matrix targeting sequence N C C SKL normal construct Delete the matrix targeting sequence at the N-terminus (2pts). Add a peroxisome targeting sequence (SKL) (3 pts) at the extreme C-terminus (2pts). PAGE: 7 OF EXAM #3

8 C. 6 pts) You are successful in your attempts in Part B and you have generated a cultured CHO cell in which the citrate synthase protein indeed is found in peroxisomes. But it has no enzymatic activity. Provide an explanation for this. Some of the possible explanations are listed below: 1) The protein is not folded properly. Folding of mitochondrial proteins happens in the mitochondria matrix with the help of matrix chaperones and chaperonins --these are not present in the peroxisome. 2) The oxidizing environment of the peroxisome may prevent the normal functioning of the protein. 3) The difference in ph may also affect its activity. 4) Cofactors required for protein activity are not present 5) Many of you suggested that the SKL signal sequence may interfere with protein activity. This is possible but not very likely since this short 3 amino acid long sequence is located at the extreme C-terminus of the enzyme. Partial credit was awarded if you stated this as your sole explanation. Question 5 (10 points) Cathepsin D is a protease that is found in the lysosome lumen; it is targeted there by the mannose 6-phosphate pathway. Cathepsin D has one asparagine linked carbohydrate chain. Using a complex mutagenesis scheme you have identified a clonal line of CHO cells in which all of the Cathepsin D is secreted rather than localized to the lysosomes. The content of all other lysosomal enzymes in the lysosome is normal. You then sequence the mutant gene and find that the protein has a single point mutation that changes one amino acid into another. Suggest two ways in which a single point mutation in Cathepsin D could cause it to be secreted and describe a simple test to tell you which of these two possibilities might explain your particular point mutant. Possible Point Mutations (+3 pts each): 1. ASN -> another amino acid (or possibly an adjacent amino acid mutation). 2. Point mutation in a portion of the protein that blocks the GlcNac- Phosphotransferase from interacting with Cathepsin D Could be an O- linked mutation. Although a mutation in a disulfide may prevent GlcNac interaction, it is not a good answer since most misfolded proteins are not secreted but instead degraded. Tests which are good for full credit (+4 pts): 1. Look for glycosylation of Cathepsin D in the ASN fragment (change in size, glycosylation of fragment, Endo H with SDS-PAGE, are all acceptable answers). 2. Perform immunoppt and look for interactions with GlcNacphosphotransferase. 3. Look for phosphorylation of the mannose group. 4. Add exogenous protein and look for uptake into the lysome. Test which are good for partial credit (+3/+2 pts): 1. Pulse-chase with 32P and look for hot phosphate on the protein. 2. If you are looking for some kind of protein interactions, however, the approach you suggest was bogus. PAGE: 8 OF EXAM #3

9 Question 6 (20 points) You are studying a cultured line of neuroblastoma cells; these are tumors of neuronal cells that grow well in culture. When placed in a certain medium they generate axons and dendrites and produce the neurotransmitter serotonin. Experimentally, you can induce an action potential in these cells by adding the chemical acetylcholine; arrival of the action potential at the axon termini causes these cells to release the neurotransmitter serotonin into the medium. From this line of cells you isolate several mutants which grow normally and normally generate axons and dendrites. These neurons also generate and conduct action potentials normally. These cells also synthesize serotonin normally and contain a normal amount of serotonin in their axon termini, and they have a normal complement of synaptic vesicles docked at the plasma membrane of the axon termini. All of the V-SNAREs and T-SNAREs are also normal. Nonetheless, arrival of the action potential at the axon terminus does not cause serotonin secretion. A. 10 pts) Suggest two different genes in which the mutation could have occurred, and explain how each of these mutations would cause the observed phenotype. There are actually three genes in which the mutation could have occurred. 1) Synaptotagmin a mutation in this gene which caused the protein to be unable to bind calcium or to be unable to change conformation upon calcium binding would result in a permanent block to fusion. 2) The serotonin antiporter a mutation in this gene would mean that although serotonin is made properly it will not be loaded into the vesicles. 3) The calcioun channel A mutation that rendered the calcium channel unable to open upon membrane depolarization would mean that synaptotagmin again would never bind calcium and change conformation. NOTE: SNAPs are a type of SNARE and thus you were told in the question they are normal but you still got partial credit. Synapsin mutation would result in abnormal vesicle localization and you were told vesicles are docked normally. I don t know where people got synaptophysin from but we certainly never addressed it. B. 10 pts) For each of your mutant cell lines describe a simple experiment that would allow you to confirm the defect you proposed. PAGE: 9 OF EXAM #3

10 Synaptotagmin Transfect your cell line with a plasmid carrying the WT gene from which you could OVEREXPRESS the WT protein and see if you get vesicle fusion and secretion of serotonin. (Mutant protein would still be present to block fusion so you need to overexpress). Serotonin antiporter Transfect cells with plasmid carrying WT gene. Or use a fluorescent antibody to seretonin to see if it is contained in vesicles (punctate staining pattern) or whether it is in the cytoplasm (diffuse staining). Calcium Channel Microinject calcium into the cell and see if seretonin is secreted (look for its presence in the culture media). This particular experiment would distinguish between calcium channel mutaion and synaptotagmin mutation. You could also transfect in a plasmid with the WT gene. NOTE: These are just some examples. There are other experiments that would work, but they must test what mutation in present in your cell line. PAGE: 10 OF EXAM #3