7.06 Spring 2004 PS 6 KEY 1 of 14

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1 7.06 Spring 2004 PS 6 KEY 1 of 14 Problem Set 6. Question 1. You are working in a lab that studies hormones and hormone receptors. You are tasked with the job of characterizing a potentially new hormone that is released in response to caffeine and binds its receptor to promote cell proliferation. This amazing hormone stimulates cells to proliferate (in culture) when caffeine is added to the growth media. You have cloned the gene for this hormone, which you call cgh for caffeine growth hormone, and you have antibodies against the protein (CGH). The first thing you d like to discover is how cgh is mediating its effect. To this end you have generated mutants that do not respond to caffeine stimulation. You now want to figure out where the defects occur in these mutations. You begin by analyzing the first mutant, MutA. MutA is curious because the hormone is produced but seems to be nonfunctional. A western blot for CGH shows that the cells produce CGH, and an Elisa for CGH shows that the cells secrete CGH in response to caffeine stimulation. However, the cells fail to proliferate. Determined to find the defect in MutA, you decide to explore the secretory pathway for defects. You decide to test for glycosylation modification in CGH because you know that the N-linked Man 8 (GlcNAc) 2 of CGH gets modified in the ER by a CGH specific glucotransferase that was recently discovered in your lab. The CGH glucotransferase adds an acetylglucosamine to the Man 8 (GlcNAc) 2 in CGH. From having taken 7.06, you know that the enzyme endoglycosidase D will cleave unmodified Man 8 (GlcNAc) 2, but once modified the N-linked oligosaccharide becomes resistant to cleavage by endoglycosidase D. The cleaved and uncleaved product can be differentiated by running an SDS PAGE gel of your protein. You decide to isolate CGH form both wild type and MutA cells and test for the acetylglucosamine modification by using endoglycosidase D. The results of this experiment show that CGH from wild type cells are resistant to endoglycosidase D cleavage BUT CGH from MutA cells get cleaved by endoglycosidase D. A) What does this experiment suggest about MutA? ANSWER: This experiment shows that CGH is sensitive to endoglycosidase D in MutA cells, so it seems that acetylglucosamine modification is not occurring in the ER. This could be due to a defective CGH glucotransferase, or the absence of modification signal or an incorrect sorting. Based on the above experiment you now believe there is something wrong with the ER glycosylation modification pathway of CGH. Your first thought is to test for a defective CGH glucotransferase. Luckily since this enzyme was discovered and cloned in your lab, you have antibodies against it. You therefore decide to do an immuno-localization experiment using gold secondary antibodies and electron microscopy (your lab is very

2 7.06 Spring 2004 PS 6 KEY 2 of 14 rich and happens to have an electron microscope). You prepare cell sections for electron microscopy from both wild type and MutA cells and stain them with antibodies against CGH glucotransferase. What you find is that in wild type cells CGH glucotransferase is found in the golgi, in vesicles, and in the ER. In MutA, however, you only see CGH glucotransferase in the golgi. B) What does this experiment suggest about CGH glucotransferase in MutA? Explain your answer. ANSWER: This localization experiment suggests that in MutA CGH glucotransferase is unable to travel back to the ER after being modified in the golgi. This suggests a defect in retrograde vesicular transport. You now want to test for the presence of COPI vesicles in MutA cells. You decide to repeat the immuno-localization experiment with electron microscopy. Only this time you stain with antibodies against coatomers (the coat protein found in COPI vesicles). What you find is that COPI vesicles do in fact form in both the wild type and MutA cells. C) What does this experiment suggest? Explain your answer. ANSWER: This result suggests that the defect of MutA is probably not in COPI vesicles, since you can still detect COPI vesicles in MutA. Your possible target for the MutA defect is the cargo-receptor protein of COPI vesicles. Lucky for you, you have a friend that works in another lab that works on CGH glucotransferase. They actually have an antibody specific for the membrane cargoreceptor protein (in COPI vesicles) specific for CGH glucotransferase. You decide to use this cargo receptor antibody to run a co-immuno-precipitation experiment. You IP the CGH glucotransferase and run a western blot for the CGH glucotransferase COPI cargo receptor. The western shows that wild type cells have the cargo protein receptor, but MutA cells don t. When you do immuno-fluorescence for the cargo protein you find that in wild type cells it localizes in the golgi, COPI vesicles and at very low levels in the ER. However, MutA cells show no trace of the cargo protein. D) What does this experiment suggest and how would you show this directly? ANSWER: This experiment suggests that the MutA cells are missing the CGH glucotransferase COPI cargo receptor. One way of more directly showing this is to do a complementation experiment. You could transfect a cdna library into MutA cells and look for cells that are able to transport CGH glucotransferase back to the ER. Then you could test whether the gene responsible for the complementation is the CGH glucotransferase COPI cargo receptor. E) Based on these experiments, come with a model to explain why CGH is nonfunctional in MutA cells.

3 7.06 Spring 2004 PS 6 KEY 3 of 14 ANSWER: CGH is nonfunctional because it does not have the acetylglucosamine modification on its N-linked Man 8 (GlcNAc) 2. This modification which normally occurs in the ER fails to happen because the CGH specific glucotransferase that adds the acetylglucosamine is absent in the ER of MutA cells. The CGH glucotransferase is absent in MutA cells because the CGH glucotransferase COPI cargo receptor is deleted in MutA cells. Therefore CGH glucotransferase, which gets modified into a mature functional enzyme in the golgi, cannot be loaded into the COPI retrograde transport vesicles in order to be transported back to the ER. Question 2. You are working in a lab that is investigating the molecular basis for a rare autosomal recessive disorder known as Inclusion cell (I-cell) disease. This syndrome presents itself in early infancy and causes facial and skeletal abnormalities with retardation of physical and mental development. Previous studies have shown that fibroblasts from I-cell patients have a build up of fatty substances and complex carbohydrates within intracytoplasmic inclusions (hence the name I-cell). You suspect that this metabolic disorder may be due to the inability of lysosomes from I-cell patients to degrade macromolecules. A) When you culture fibroblasts from these patients, you find that they secrete an excess of lysosome-specific hydrolases and proteases into the extra-cellular medium. Which mutations could lead to this phenotype? ANSWER: Genes involved in targeting vesicles to lysosomes, genes encoding olgi receptors that recognize lysosomal protein cargo, genes for GlcNAc phosphotransferase, or phosphodiesterase (enzymes involved in adding the M6P marker on lysosomal proteins). This phenotype suggests that lysosomal enzymes are mis-sorted. The most likely candidate genes are those that produce the M-6-P epitope and the gene for the M-6- P receptor. B) If you culture fibroblasts from I-cell patients with normal fibroblasts, you can no longer find any cells with cytoplasmic inclusions. What does this suggest? Provide an explanation. ANSWER: Normal fibroblasts may be secreting some factor that complements the mutation in the I-cell fibroblasts. I-cell fibroblasts may be able to take up this factor and incorporate it into their lysosomes or Golgi-sorting machinery (WT fibroblasts may be secreting lysosomal enzymes). Alternatively, cell-cell interactions (i.e. gap junctions) between these two cells allows complementation. C) How could you test your hypothesis from (c)?

4 7.06 Spring 2004 PS 6 KEY 4 of 14 ANSWER: Add extra-cellular medium from WT fibroblasts to I-cell fibroblasts and see if they lose their inclusions. A better experiment would be to add purified lysosomal enzymes. This would test whether WT cells are secreting the complementing factor. D) I-cell fibroblasts take up and retain normal lysosomal enzymes, but lysosomal enzymes secreted by I-cell fibroblasts are not taken up by normal fibroblasts. How could you experimentally show this? ANSWER: Purify radiolabeled enzymes from WT cells (grow cells in the presence of 35-S methionine, use an antibody to purify your enzyme(s) of interest) and add them to the medium of I-cells. After some time, wash away the medium, isolate the cells and look for retention of the radioactivity in the cell pellet. When you repeat this experiment with radiolabeled enzymes from I-cell fibroblasts, there should be little/no retention of the radioactive enzymes in the WT cell pellet. E) Can you construct a model to explain these data and the secretion abnormality in (b)? Be sure to explain the higher level of lysosomal enzymes in the media from I-cells and the inability of WT cells to take these enzymes up. ANSWER: Because I-cell fibroblasts have normal M-6-P Receptors on their cell surface they can take up enzymes released into the medium by WT cells. WT cells, on the other hand, do not recognize enzymes from I-cells, even though these cells release most of their lysosomal enzymes into the medium. Therefore, the lysosomal enzymes from I-cells lack the M6P targeting signal, causing them to be mis-sorted as secretory proteins and released into the extra-cellular medium. The genetic defect of I-cell patients lies in the enzymes involved in M6P addition: GlcNAc phosphotransferase, or Phosphodiesterase we don t know which. Both WT and I-cell fibroblasts constitutively secrete some of their lysosomal proteins, most likely because the M6P receptor cannot sequester and sort all lysosomal proteins with 100% efficiency. Some will inevitable escape the receptor s notice and get packaged by default into secretory vesicles. The level of secretion is much higher in I-cell fibroblasts because they cannot sequester any lysosomal cargo to lysosome-destined vesicles in the TGN. As previously stated, the lysosomal proteins secreted by the WT cells have the M6P epitope and can be recognized by M6P receptor on the plasma membrane of I-cells. After endocytosis, the vesicles fuse with endosomes bringing, in a round-about way, the right enzymes to the right compartment. WT fibroblasts cannot take up enzymes from I-cells because those cells fail to mark lysosomal enzymes with the M6P tag. Therefore, although WT cells also have M6P receptor on their cell surface, it cannot recognize M6P-deficient I cell enzymes. F) Although I-cell fibroblasts lack lysosomal enzymes, some cell types of I-cell patients, such as hepatocytes, do not manifest a storage disorder and contain normal levels of some

5 7.06 Spring 2004 PS 6 KEY 5 of 14 lysosomal enzymes. What does this suggest? ANSWER: This suggests that there is an M6P-independent targeting and transport pathway that sends some lysosomal enzymes to lysosomes. This pathway is either not the predominant pathway in fibroblasts or is not active. Because hepatocytes do not manifest a storage disorder, they must rely on an M-6-P independent pathway. Question 3. A postdoc that you are working with is attempting to characterize the mode of action of several toxins in the hope that he can use the knowledge to develop antidotes. Each toxin (A, B, C) leads to paralysis in animals and based on this and other data you suspect that they are affecting acetylcholine release at neuromuscular junctions. You have at your disposal cdnas encoding each of the toxins, along with antibodies against each of them. A) Your first task is to develop a cell culture assay to monitor toxicity of each toxin. Explain how you would do this. ANSWER: You would culture a neuronal cell line or isolate neurons from an organism and add the toxin to the cells. Then you could monitor for acetylcholine release into the medium after stimulation of neurons (e.g. by depolarization). B) You find that all of the toxins, apart from sample C, inhibit acetylcholine release. Using GFP-fusion proteins you find that toxins A and B are internalized via receptormediated endocytosis and then released into the cytosol of neurons. Toxin A localizes to the presynaptic membrane and so you think it might bind t-snares such as SNAP-25 or syntaxin. You carry out an IP-Western using an extract from the cultured neurons that had been exposed to toxin A and you get the following result. Interpret the data and describe a potential mode of action for toxin-a. IP: Toxin-A SNAP-25 Syntaxin Western Ab: anti-toxin A

6 7.06 Spring 2004 PS 6 KEY 6 of 14 ANSWER: Toxin A interacts with SNAP-25 but not syntaxin. If the toxin binds SNAP-25 then the v-snares, such as VAMP, on the synaptic vesicles can t dock at the membrane in preparation for release of acetylcholine. However, you want to confirm the interaction and so you carry out some in vitro transcription/translation using the toxin A cdna and a SNAP-25 cdna, incubate the proteins together and do an IP-Western. The results are shown below. IP: Toxin-A SNAP-25 Syntaxin Western Ab: anti-toxin A Confused you decide to run both samples of toxin A, one isolated from the neurons and the other from the in vitro transcription/translation reaction, on a native gel. Note that unlike the SDS-PAGE gel shown above, a native gel is run in the absence of SDS and hence the mobility of a protein is determined by charge and conformation. The result is shown below: Source of toxin A: in vitro neuronal extract Western Ab: anti-toxin A You take the in vitro translated form of toxin A and add either buffer A or buffer B, detailed below, to it before incubating it with SNAP-25. You then carry out an IP- Western as before. C) How can the results shown below explain the results of the previous gels?

7 7.06 Spring 2004 PS 6 KEY 7 of 14 Buffer A: 50 mm Tris-HCl ph 8.0, 150 mm NaCl, 1% NP-40 Buffer B: 10 mm Tris-HCl ph 7.5,10 mm NaCl, 10 mm EDTA, 0.5% Triton-X100, DTT

8 7.06 Spring 2004 PS 6 KEY 8 of 14 IP SNAP-25 Buffer A Buffer B Western Ab: anti-toxin A ANSWER: Toxin A exists in an inactive and active form. It normally has disulfide bonds which result in it being in a conformation that can t bind SNAP-25. When it is released from endosomes into the cytosol these disulfide bonds are reduced and the protein can take up its active conformation which binds SNAP-25. The in vitro transcription/translation reaction is carried out in a test tube under oxidizing conditions and so this form of the toxin will be disulfide-bonded and unable to bind SNAP-25 which explains the initial IP-Western. When a native gel is run the toxin A isolated from neuronal cells runs at a different mobility to that produced in vitro because the former doesn t have disulfide bonds. In vitro translated toxin A interacts with SNAP-25 when the reaction is carried out in the presence of DTT because DTT is a reducing agent and so toxin A no longer contains disulfide bonds. D) You carry out a series of IP-Westerns to look for interactions between toxin B and other proteins involved in acetycholine release such as synaptotagmin, VAMP and syntaxin. Toxin B binds none of these proteins. By electron microscopy you find that synaptic vesicles accumulate at the presynaptic membrane when neurons are treated with toxin B. You think that toxin B may inhibit the influx of calcium that stimulates vesicle fusion. How could you test for this? ANSWER: Use a calcium-sensitive dye such as fura-2 and compare calcium fluxes in treated and untreated cells. E) Your results seem to suggest that calcium can t enter cells that have been treated with toxin B. Suggest a possible mode of action for toxin B and how this can explain the results with fura-2. ANSWER: Toxin B may act by preventing the opening of the voltage-gated Ca 2+ channels in response to a nerve impulse, which then prevents synaptic vesicle fusion with the plasma membrane and release of acetylcholine. F) The mode of action of toxin C is still a mystery. You think that it might be affecting the muscle cells that the acetylcholine acts on. Unlike wildtype cells, toxin-treated cells

9 7.06 Spring 2004 PS 6 KEY 9 of 14 don t show an increase in cytosolic calcium levels (monitored using fura-2) when treated with acetylcholine. You think that perhaps the toxin is binding to the nicotinic acetylcholine receptor and so carry out some immunofluorescence studies. You get the following surprising results: (i) the toxin treated cells have no nicotinic acetylcholine receptor on the cell surface (ii) toxin C and the nicotinic acetylcholine receptor colocalize in endosomes (iii) an IP-Western shows that the receptor and toxin are found in a complex together Based on this data propose a possible mode of action for toxin C. ANSWER: Based on the immunofluorescence data toxin C seems to bind, either directly or indirectly, to the nicotinic acetylcholine receptor and causes it to be internalized into endosomes and not recycled to the cell membrane. Since the receptor is internalized it can t respond to acetylcholine, no influx of Na + occurs and so there is no depolarization and no release of calcium from the sarcoplasmic reticulum. Question 4. Answer the following questions about action potentials and ion channels. A) Why are action potential propagated uni-directionally if the depolarization caused by the opening of voltage gated sodium channels spreads bi-directionally? ANSWER: After the Na+ channel has been open for a short while (<0.1 milliseconds) the voltage sensing helices return to their resting positions and the channel inactivating segment blocks the passage of ions through the pore for a few milliseconds after the action potential has passed. This refractory period prevents a channel from opening more than once in a short time span and ensures unidirectional propagation. In 1975 Lily and Yuh Nung Jan were recording electrophysiological measurements on strains of Drosophila with behavioral defects, and discovered that a mutant called Shaker displayed abnormalities. The Shaker flies were identified because they shake uncontrollably under ether anesthesia. The Shaker gene encodes a voltage-sensitive K+ channel that is necessary for repolarization of the membrane during the propagation of an action potential. Its inactivation is very similar to the inactivation of the closely related voltage-gated sodium channel. B) What would happen to the rate of inactivation of the Shaker channel if it contained an insertion of 20 amino acids in the chain segment of its ball and chain? ANSWER: The longer the chain segment is, the more slowly the K+ channel will inactivate, because it will take longer for the inactivating ball enter to the channel and block the passage of ions.

10 7.06 Spring 2004 PS 6 KEY 10 of 14 C) How would adding a chemical inhibitor of the Na+/K+ ATPase immediately affect the nerve cell s ability to propagate an action potential (remember that the Na+/K+ ATPase is responsible for establishing the concentration gradient of sodium and potassium ions within all cells)? What would be the effect of continuous application of the inhibitor? ANSWER: The chemical inhibitor would have no immediate effect on the nerve cell s ability to fire. The change in concentration of sodium and potassium ions that occurs during the propagation of an action potential is about 1/150 th of the total intracellular concentration of each ion. Thus a nerve cell can fire a number of times before the concentration gradient of sodium and potassium is functionally affected. Said again, each ion channel that opens during an action potential only has to let in a few sodium ions to change the local concentration of ions enough that the localized depolarization is propagated to the next channel. Also, the number of potassium ions that need to be pumped out to locally repolarize the membrane are very small. Thus, because the opening and closing of channels results in such a small change in the total sodium and potassium gradients, each neuron can fire a number of times without changing the either gradient all that much. Under prolonged exposure to the drug, however, the gradients of sodium and potassium would be abolished and the nerve cell would no longer be able to fire. D) What type of signal anchor sequence is on the N-terminus of the voltage-gated ion channel an SA-II or an SA-III--and how do you know? ANSWER: The N-terminus of the channel is cytoplasmic so it must contain a type II signal anchor sequence. A type IiI signal anchor would put the N-terminus outside of the cell.

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