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Name: _____________KEY__________________ 7.06 Cell Biology EXAM #1 KEY February 28, 2006 This is an OPEN BOOK exam, and you are allowed access to books, a calculator, and notes BUT NOT computers or any other types of electronic devices. Please write your answers in PEN (not pencil) to the questions in the space allotted. Please write only on the FRONT SIDE of each sheet, as we will Xerox all of the exams and thus only grade writing on the front. And be sure to put your NAME ON EACH PAGE in case they become separated! There are EIGHT pages including this cover sheet. Question 1. 23 pts __________ Question 2. 27 pts __________ Question 3. 21 pts __________ Question 4. 9 pts __________ Question 5. 20 pts __________ TOTAL 100 pts __________ 1 Name: _____________KEY__________________ Question 1. (23 pts) You are studying transepithelial transport of glucose from the intestinal lumen to the extracellular medium surrounding the basal surface in cells that are growing in culture. You are using a system identical in set-up to that described for MDCK cells in the textbook, copied below: (a, 3 pts) If you performed immunofluorescence on permeabilized intestinal epithelial cells using an antibody against GLUT2, where would you find the fluorescent signal? You would see signal only in the basolateral surface of the membrane, as that is the only portion of the cell in which GLUT2 protein is found. GLUT2 is the uniporter that allows glucose to flow down its concentration gradient, from the inside of the cells to the blood. (b, 5 pts) If you performed the same experiment as above on intestinal epithelial cells that had been depleted of occludin and claudin, where would you see the fluorescent signal? Justify your answer. You would see the signal all over the plasma membrane (both the basolateral and apical sides), because these two proteins are essential for proper formation of tight junctions. One of the very important functions of tight junctions is to disallow proteins from diffusing freely between the basolateral and the apical sides of the membrane. (c, 8 pts) You are trying to design a new artificial transporter that would transport glucose from the intestinal lumen into intestinal epithelial cells. You have the ability to create novel transporters and target them to the apical membrane. You make the two new transporters listed below, and to test them you use recombinant DNA techniques learned in 7.02 and 7.03 to express these in the MDCK cells. For each new transporter below, state whether it could allow for movement of glucose across the apical membrane at all, and if yes, whether you think that glucose would move into or out of cells. (i) A 2 Ca2+/1 glucose symporter Yes, it would allow movement of glucose inside of the cell. Glucose is higher inside these cells than outside, and calcium is higher outside than inside. Thus the free energy 2 Name: _____________KEY__________________ released from moving calcium down its concentration gradient could be used to drive the movement of glucose into the cells (up its concentration gradient). (ii) A 2 Ca2+/1 glucose antiporter Yes, it would allow movement of glucose outside of the cell. Glucose is higher inside these cells than outside, and calcium is higher outside than inside. Thus it would be impossible for this antiporter to move calcium outside (against its gradient) and glucose inside (against its gradient). It would be possible, however, for the antiporter to move calcium inside (down its gradient) and glucose outside (down its gradient). (d, 7 pts) You expose wild-type mice to a modified version of ouabain. Ouabain inhibits the Na+/K+ ATPase in all body cells, but your version inhibits the Na+/K+ ATPase only in intestinal epithelial cells. You think this might be useful as an antidiabetic drug. Predict the following: (i) What would happen over time to glucose uptake from the intestinal lumen to the blood and why? Glucose uptake would cease. The uptake of glucose from the lumen is powered by the Na+ concentration gradient, because sodium flowing down its gradient (from the lumen to inside cells) releases free energy that is used to power the movement of glucose up its gradient (from the lumen to inside cells). The Na+/glucose symporter is the protein that allows for this movement. If the Na+/K+ ATPase were inhibited, then the concentration gradients of Na+ and K+ would eventually be dissipated, and so there would no longer be a concentration gradient of Na+. Thus the Na+/glucose symporter would no longer be powered. (ii) What would happen over time to the magnitude of the membrane potential of the intestinal epithelial cells? It would decrease in magnitude. If the Na+/K+ ATPase were inhibited, then the concentration gradients of Na+ and K+ would eventually be dissipated, and so there would no longer be a concentration gradient of K+. Thus the movement of K+ through the open resting potassium channels would cease. This movement of K+ is what creates the membrane potential of cells, so therefore the magnitude of the membrane potential would decrease. Question 2. (27 pts) You are studying the TGF-β pathway in mouse cells growing in culture. Describe how the results obtained from each of the two following experiments would be different, depending on whether or not TGF-β had been added the cells for an hour. Also describe how you would interpret that difference in results (i.e. what the different results in the presence or absence of TGF-β tell you about the effect that TGF-β has). (a, 7 pts) You perform immunofluorescence on cells using an antibody against Smad-3. (Assume you performed the immuofluorescence correctly, using standard procedures.) 3 Name: _____________KEY__________________ Result without TGF-β: You would see staining in the cytoplasm. Result + TGF-β: You would see staining in the nucleus. Interpretation of difference in results: TGF-beta pathway activation leads to the translocation of the the transcription factor Smad-3 into the nucleus. This is because TGF-beta binding to its receptor results in receptor dimerization and activation, and subsequent phosphorylation of Smad-3. Phosphorylated Smad-3 complexes with Smad-4, and Smad-3’s now exposed NLS sequences allow it to be translocated into the nucleus. (b, 7 pts) You dissolve the cells in a nonionic detergent, remove all remaining membrane fragments by centrifugation, and run the cell lysate over a gel filtration column. You then do Western blotting on each fraction that comes off the column using an antibody against Smad-3. Result without TGF-β: Smad-3 would elute in a relatively late fraction, so you would see a band corresponding to Smad-3 in a lane of your Western in which you loaded a relatively late fraction. Result + TGF-β: Smad-3 would elute in a relatively early fraction, so you would see a band corresponding to Smad-3 in a lane of your Western in which you loaded a relatively early fraction. Interpretation of difference in results: TGF-beta pathway activation leads Smad-3 forming a complex. Without TGF-beta, it is a monomer. This is because TGF-beta binding to its receptor results in receptor dimerization and activation, and subsequent phosphorylation of Smad-3. Phosphorylated Smad-3 complexes with Smad-4 and another monomer of Smad-3. (c, 4 pts) You overexpress in wild-type cells a mutant version of Smad-3 in which its three critical serine residues at the C- terminus – the ones normally phosphorylated by the TGF-β receptor - were replaced with aspartic acid. These cells (in contrast to wild-type cells) rapidly stop growing in culture whether or not TGF-β is added to the cells. Explain this result. 4 Name: _____________KEY__________________ This is a constitutively active mutant version of Smad-3, because you are replacing the amino acid that is normally phosphorylated with an amino acid that has a negatively charged R group that mimics phosphorylation. Thus this Smad-3 will always be activated, and will always form complexes and translocate into the nucleus, where it will turn on genes that inhibit the cell cycle. Thus cells are always inhibited from growing rapidly in culture. (d, 3 pts) Would the result from part (c) change if the cells had not also been expressing endogenous wild-type Smad-3? If so, how would it change? No. This is a mutant that is constitutively active, which is why you see the mutant phenotype even though there is wild-type Smad-3 present in addition to mutant Smad-3. (e, 6 pts) Would cells in culture rapidly stop growing if you over-expressed (in wild-type cells) a mutant version of Smad-3 that lacked an NLS… …in the absence of TGF-β? …in the presence of TGF-β? Explain. No in the absence and no in the presence. If Smad-3 lacked an NLS, it could never go into the nucleus, and thus it could never turn genes off that inhibited the cell cycle. Thus the cells would always proliferate. This is a dominant negative mutation, which is why you see the mutant phenotype even though there is wild-type Smad-3 present in addition to mutant Smad-3. Mutant Smad-3 will bind to any wild-type Smad-3, and prevent the entire complex from translocating into the nucleus. Question 3. (21 pts) You form a liposome in buffer #1 that contains 5mM KCl and ATP at pH7. You then place the formed liposome into buffer #2 that contains 50mM KCl and ATP also at pH7. For each subpart below: Describe what would happen very shortly after liposome formation, and then eventually (after a long period of time). (a, 7 pts) There are no proteins incorporated into the liposome. (i) Will K+ move across the membrane, and if so, in which direction does it move? No. Ions cannot pass through the membrane unless there is a channel, pump, or transporter that allows movement of the ion. (ii) What happens to the concentration gradient across the liposome bilayer over time? Nothing. Ions cannot pass through the membrane unless there is a channel, pump, or transporter that allows movement of the ion. If no ions can flow, then the concentration gradient cannot change. 5 Name: _____________KEY__________________ (iii) How does the magnitude of the membrane potential across the lipid bilayer change over time? It doesn’t. It remains at zero the whole time. Ions cannot pass through the membrane unless there is a channel, pump, or transporter that allows movement of the ion. Ions traveling through channels is what creates a membrane potential in the first place. (b, 7 pts) You now repeat the experiment from the introduction, but using liposomes into which the H+/K+ ATPase (isolated from stomach epithelial cells) is incorporated in the same orientation as it is in the cell, with the ATP-binding site inside the liposome. (i) Will K+ move across the membrane, and if so, in which direction does it move? Yes, inward. The H+/K+ ATPase brings K+ into the cell and pumps H+ out of the cell. (ii) What happens to the K+ and H+ concentration gradients across the liposome bilayer over time? The H+/K+ ATPase brings K+ into the cell and pumps H+ out of the cell. Thus, an H+ gradient will be created so that H+ is higher outside the liposome. The K+ gradient will initially dissipate (as it starts off higher outside than inside), but once the concentration gradient of K+ reaches zero, the pump keeps pumping, thus creating a new gradient where K+ becomes higher inside. (iii) How does the magnitude of the membrane potential across the lipid bilayer change over time? It does not change. No net charge is flowing across the membrane. The outside and inside of the liposome remain neutral. (c, 7 pts) You now repeat the experiment from the introduction, but using liposomes into which only a K+ channel is incorporated. (i) Will K+ move across the membrane, and if so, in which direction does it move? Yes, inward. K+ begins higher inside than outside. When the channels are first open, K+ will want to flow down its concentration gradient (from the outside to the inside). This results in the inside of the liposome becoming positive. Thus, eventually, K+ will want to stop flowing inside because it is positive and thus repelled by the + charge inside the liposome. Thus eventually there will be no net flow of K+ because an equilibrium has been reached (where the membrane potential across the liposome membrane is equal to the Nernst potential for potassium). (ii) What happens to the concentration gradient across the liposome bilayer over time? 6 Name: _____________KEY__________________ Very little. It may decrease slightly because K+ is flowing inward, but then it reaches an equilibrium, and the number of ions that must flow to create the membrane potential at that equilibrium is miniscule compared to the number of ions that create the concentration gradient. (iii) How does the magnitude of the membrane potential across the lipid bilayer change over time? At first, the inside of the liposome will be becoming more and more positive, but eventually the membrane potential will stabilize at the Nernst potential for K+. K+ begins higher inside than outside. When the channels are first open, K+ will want to flow down its concentration gradient (from the outside to the inside). This results in the inside of the liposome becoming positive. Thus, eventually, K+ will want to stop flowing inside because it is positive and thus repelled by the + charge inside the lipsome. Thus eventually there will be no net flow of K+ because an equilibrium has been reached (where the membrane potential across the liposome membrane is equal to the Nernst potential for potassium). Question 4. (9 pts) You are studying muscle contraction. When muscles receive the signals from neurons to contract, calcium is released into the cytosol from an organelle called the sarcoplasmic reticulum (SR), which stores calcium. The sarcoplasmic reticulum contains abundant amounts of a P-class Ca2+ ATPase. (a, 5 pts) What would be the effect over time on muscle contraction if you added a drug that inhibited the Ca2+ ATPase in the SR, and why? Initially, contractions would be continual. This is because the SR calcium pump pumps calcium from the cytosol into the lumen of the SR. If you inhibit this pump, when calcium rushes into the cytosol, it will not be cleared from the cytosol back into the SR. Thus the cell will feel like it is constantly being stimulated to contract. However, eventually the SR would be depleted of calcium, and thus the muscle cell would lose the ability to contract. This is because the calcium that initially flows into the cytosol can be cleared into the extracellular medium by the calcium pump that is present in the plasma membrane of all cells. (b, 4 pts) There are calcium channels in the membrane of the SR. Do you think that those channels are non-gated or gated? Justify your answer. Gated. If the channels were non-gated, then calcium would always be flowing into the cytosol, because calcium wants to flow down its concentration gradient (from the SR to the cytosol). 7 Name: _____________KEY__________________ Question 5. (20 pts) Adipocytes (fat cells) store fatty acids as triacylglycerol – a molecule of glycerol (HOCH2 – CHOH- CH2OH) esterified to three long chain fatty acids. Lipase, the enzyme that hydrolyzes triacylglycerols (TAGs) to one molecule of glycerol and three long chain fatty acids, is found mainly in adipocytes. It is phosphorylated by Protein Kinase A (PK-A) and is catalytically active only when phosphorylated. You are studying a line of cultured adipocytes that express the G-protein-coupled receptor for adrenaline that is coupled to Gsα. For parts (a) – (d) below, state whether the following conditions would result in either: constitutively high levels of TAGs, constitutively low levels of TAGs, high levels of TAGs in the absence of adrenaline but low levels in its presence, OR low levels of TAGs in the absence of adrenaline but high levels in its presence. Explain your choice. (a, 5 pts) In addition to the wild-type β-adrenergic receptor, your cells express a mutant β-adrenergic receptor (containing three amino acid changes in the cytosol-facing Cterminal segment) that is a GEF for Gsα regardless of the presence or absence of adrenaline. Constitutively low levels of TAGs. If the receptor always allows for the exchange of GDP for GTP, then G alpha will always be bound to GTP. Thus it will always activate adenylate cyclase, leading to constitutively high cAMP levels. Thus PKA will always be active and lipase will always be phosphorylated and thereby activated. Thus TAGs will always be cleaved. (b, 5 pts) A mutation in the catalytic domain of PK-A that prevents dissociation from the regulatory domains even if 3’-5’-cAMP is present. (No wild-type PKA is present.) Constitutively high levels of TAGs. If the catalytic subunit of PKA can never dissociate from the regulatory subunit, then PKA can never be an active kinase. Thus it will never phosphorylate and activate lipase, so TAGs will never be hydrolyzed. (c, 5 pts) A mutation in IP (the inhibitory subunit of phosphoprotein phosphatase) that prevents its phosphorylation by PK-A. (No wild-type IP is present.) Constitutively high levels of TAGs. PP is a phosphatase that takes off the phosphate groups put on substrate proteins by PKA. (PKA and PP are opposites of each other.) PP is inhibited by phosphorylated IP. If IP cannot be phosphorylated, then IP will never inhibit PP, and the phosphatase will always be active. Thus the phosphatase will always remove any phosphorylation put on substrate proteins by PKA. Thus the substrates will 8 Name: _____________KEY__________________ never phosphorylated. Thus lipase will never be active and will never be able to cleave TAGs. (d, 5 pts) You incubate the cells in growth medium to which non-hydrolyzable GTP has been added. Normal levels (i.e. high in the absence of adrenaline and low in its presence). GTP of any form cannot cross the plasma membrane, because it is charged, and charged molecules can only cross the membrane if there is a channel, pump, or transporter to move that molecule. 9