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Unformatted text preview: 7.06 Spring 2004 Exam 1 Name___________________________ 7.06 Cell Biology EXAM #1 February 24, 2004 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; if you use the back of a sheet make it clear which answer is where. And be sure to put your name on each page in case they become separated! Remember that we will Xerox many exams at random. Good luck! F=2.3 x 10^4 cal/(V*mol) R=1.987 cal/(K*mol) Question 1 (35 pts). You have recently isolated a yeast that grows in industrial wastewater, and find that its optimal growth medium contains glucose, glycine, as well as 150 mM KBr (potassium bromide) and 15 mM KCl. Curious as to how such an organism grows under such strange conditions, you sequence its genome. You find only one gene sequence (i.e. one encoded protein) homologous to the mammalian Ca 2+ and K+/ Na+ pump. Curious as to what ions this pump pumps, you produce large amounts of it, incorporate it into liposomes, and find that the protein is an ATP- powered Cl- and Br- pump. For each ATP molecule hydrolyzed, two Br- ions are pumped outward (from the cytosolic to the exoplasmic face) and two Cl- ions are pumped inward. Furthermore, you examine these yeast cells growing in their preferred medium of 150 mM KBr and 15 mM KCl., and find that their cytosol has a different ionic composition: 15 mM KBr and 150 mM KCl. a, 5 pts) To measure binding of radiolabeled Cl- and Br- to the ATP- powered Cl- and Brpump, you want to use a detergent to solubilize the purified protein from your liposomes. Would you use an ionic or nonionic detergent? Explain your answer. You would need to use a nonionic detergent (3 pts) because it will solubilize the protein in its native state, allowing you to study its binding properties in liposomes. An ionic detergent will denature the protein. 2 pts – for mentioning the native state OR that ionic would denature the protein. 0 pts – if nonionic is chosen but completely invalid reasons are given 1 of 9 7.06 Spring 2004 Name___________________________ Exam 1 b, 5 pts) You find that the potential across the plasma membrane of these yeast cells is 59 millivolts, with the inside of the cell negative to the outside. How would you make this measurement? You would insert a microelectrode into the cell and one in the medium and connect the two to a potentiometer. 3 pts – microelectrode 2 pts – potentionometer 0 pts – patch clamp (w/ or w/o microelectrode or volt meter) because patch clamping allows us to investigate the ion conductance of a single ion channel only. c, 5 pts) Would you expect that this potential is generated by open Cl- channels in the plasma membrane, by open Br- channels in the plasma membrane, or by the ATPpowered Cl- and Br- pump? Explain your answer in a sentence or two. The negative potential would be generated by open Br- channels. The negative Brions will flow with their concentration gradient to build up a net negative charge within the cell. The Cl-/Br- pump does not change the charge of the cell, and a Clchannel would allow negative charges to flow out of the cell. +3 pts – Open Br- channels +2 pts – Cl-/Br- pump does not change the potential of the cell +1 pt – Cl- channels would allow negative charges out of the cell (if you said all 3, only 5 pts were awarded) d, 10 pts) Interested in how these cells take up glucose from the culture medium, you find in the plasma membrane a chloride- glucose antiporter – it transports 3 chloride ions in one direction and one glucose in the other. Assuming that this is the sole protein used by the cells to import glucose, calculate the equilibrium concentration of glucose inside the cell if the external concentration is 1 µM (10-6 M). Assume that the temperature is 310 K, the concentrations of K+, Cl-, and Br- in the medium and the cytosol are as stated above, and that the membrane potential is – 59 mV as also stated above. Show all of your work, and in particular show the equations that you use to calculate your answer. 3Clin + 1 Glucose out fi 3Clout + 1 Glucose in [Clin] = 150mM [Clout] = 15mM zCl = -1 [Glucosein] = ??? [Glucoseout] = 10^-6 M zGlucose = 0 DGTotal = 3DGcl + DGglucose 2 of 9 7.06 Spring 2004 Name___________________________ Exam 1 DGcl = zClFE + RT ln(Clout/Clin) DE for moving chloride IN to OUT = E(final) – E(initial) = 0V – (-0.059V) = +0.059 V DGglucose= zGlucoseFE + RT ln(Glucosein/Glucoseout) DE for moving glucose OUT to IN = E(final) – E(initial) = -0.059V – 0V = -0.059 V DGtotal = [zClFE + RT ln(Clout/Clin)] + [zGlucoseFE + RT ln(Glucosein/Glucoseout)] DGTotal is zero at equilibrium, so substituting all known values into the equation for DGTotal: 0 = 3[ (-1)(23000)(0.059) + (1.98)(310) ln (15/150) ] + [ (0)(23000)(-0.059) + 1.98(310) ln (x/10^-6) ] x= 0.74 M The glucose concentration inside the cell is 0.74 M. 8/10 – if there was a single calculation error but everything else was correct 7/10 – if Cl(out)/Cl(in) was reversed, but everything else was correct 0/10 – if the Nernst Equation is used. e 5 pts) Your yeast cells are haploid and are not known to form diploids (that is, they appear not to have a sexual cycle.) But through exhaustive screening of a mutagenized population of these yeast cells you have isolated a mutant that is defective in its ability to transport glucose efficiently into the cell; this mutant requires 1 mM glucose in the medium for growth whereas the wild- type cell grows in glucose concentrations as low as 1 µM. Suspecting that this strain harbors a mutation in the chloride- glucose antiporter gene, you want to clone the gene. How do you do it, assuming that you can use any plasmids or vectors that already have been used for other yeast strains? Answer in just a few sentences. Transfect a genomic or cDNA library from the wild type cells into the mutant strain of yeast, and then select for colonies that will grow in 1mM glucose (in which the mutant yeast cannot normally grow). Sequence the cDNAs that confer the ability to grow on low glucose, and one should be the antiporter. f, 5 pts) Would you expect to find a chloride- glucose antiporter in any human cell? Why or why not? You wouldn’t expect to find a chloride-glucose antiporter in a human cell because the chloride concentration is much great outside of human cells than it is inside. Thus transport of a chloride ion outwards is thermodynamically unfavorable, and will not provide the energy necessary to transport glucose into the cell. 3 of 9 7.06 Spring 2004 Exam 1 Name___________________________ Question 2 (15 pts). You find that the purified chloride- glucose antiporter protein is phosphorylated in a test tube following addition of active yeast Protein Kinase A, and you further find that this yeast Protein Kinase A has a regulatory subunit that functions in a manner similar to its homolog in mammalian cells. Furthermore, you find that addition of the yeast hormone tyrosinol to these yeast cells results in activation of adenylyl cyclase and Protein Kinase A and in inhibition of glucose uptake. Thus you hypothesize that phosphorylation of the chloride- glucose antiporter by Protein Kinase A results in inhibition of its function. Determine whether you would expect the following conditions to increase or decrease glucose uptake by the yeast cells, both in the absence of any hormone and in the presence of tyrosinol, and give a brief explanation for your answer. Note that your answers depended on whether you were comparing the mutant to wildtype or whether you were just comparing within the mutant strain itself. a, 5 pts) A mutation in the Gas subunit that prevents binding of GTP once a bound GDP has dissociated. 1) comparing within the mutant strain: Glucose uptake will remain at the same high level in both the presence and absence of tyrosinol because the G-protein needs to bind GTP in order to activate. Without active G-protein, PKA will not be activated and the transporter will not be downregulated in response to hormone. In the absence of hormone the G-protein pathway is not being activated and so the mutation in the Gas subunit has no effect on levels of glucose uptake. 2) comparing to wildtype: In the absence of hormone the G-protein pathway is not being activated and so the mutation in the Gas subunit has no effect and so glucose uptake will remain at the same high level. In the presence of tyrosinol the G-protein can’t bind GTP and without active G-protein, PKA will not be activated and the transporter will not be downregulated in response to hormone and so glucose uptake should increase relative to wildtype yeast in the presence of hormone. b, 5 pts) A mutation in the regulatory domains of PK-A that prevents binding to the catalytic subunit. 1) comparing within the mutant strain: Glucose uptake will be decreased in the presence or absence of hormone. PKA that cannot bind its regulatory subunit is constitutively active, and so the glucose antiporter will be constitutively downregulated. 4 of 9 7.06 Spring 2004 Exam 1 Name___________________________ 2) comparing to wildtype: In the absence of hormone glucose uptake wil be decreased since PKA that cannot bind its regulatory subunit is constitutively active, and so the glucose antiporter will be constitutively downregulated. In the presence of hormone, levels of glucose uptake will be the same as wildtype or perhaps decreased slightly. This will depend on whether constitutively active PKA gives the exact same level of inhibition of glucose uptake as hormone binding and subsequent G-protein signaling does. c, 5 pts) A mutation to alanine of the single serine residue in the chloride- glucose antiporter protein that is phosphorylated by Protein Kinase A 1) comparing within the mutant: Glucose uptake will remain the same in the presence or absence of hormone. The transporter needs to be phosphorylated in order to be downregulated. The alanine mutation will prevent phosphorylation of the receptor and prevent its downregulation. In the absence of hormone this mutation will be of no consequence since there is no signaling through the pathway to begin with. 2) comparing to wildtype: The transporter needs to be phosphorylated in order to be downregulated. The alanine mutation will prevent phosphorylation of the receptor and prevent its downregulation. In the absence of hormone this mutation will be of no consequence since there is no signaling through the pathway to begin with. In the presence of hormone glucose uptake will be upregulated compared to wildtype since the transporter can’t be inhibited by phosphorylation. Question 3. 25 pts The hormone TGFß1 inhibits the proliferation of many types of body cells, and loss or inactivation of a TGFß receptor or a component of its downstream signaling pathway frequently occurs in human cancers. As an oncologist, you are studying a particularly aggressive type of AML leukemia, and have cloned lines of pure leukemia lymphocyte cells from each of five patients. Each of the cell lines contains a mutation in some component of the TGFß signaling pathway, but you do not know whether the cells themselves are sensitive or resistant to the growth- inhibitory effects of TGFß: i) Patient A has a deletion in one Smad3 allele and a point mutation in the other Smad 3 gene such that the nuclear localization sequence (NLS) is nonfunctional. ii) Patient B has a deletion in one Smad3 allele and a frame shift mutation in the other Smad 3 gene such that the carboxyl- terminal 9 amino acids, including the three serine residues that are phosphorylated by the Type I TGFß receptor, are deleted. 5 of 9 7.06 Spring 2004 Exam 1 Name___________________________ iii) Patient C has a deletion in one Smad4 allele and a mutation in the other Smad 4 gene such that the mutant Smad4 protein cannot bind to Smad3. iiii) Patient D has a deletion in one allele encoding the Type I TGFß receptor, and a mutation in the second allele such that the Type I receptor cannot become phosphorylated by the Type II receptor. (v) Patient E has a deletion in one Smad4 allele and a mutation in the other Smad 4 gene such that the mutant Smad4 protein cannot bind to the Ski protein a, 10 pts) Explain, for each of the five patients, whether or not the indicated mutations would cause the leukemic cells to become resistant to growth inhibition by added TGFß, and for each case explain your answer in one or two sentences. i) ii) iii) iv) v) These tumor cells will be resistant to growth inhibition by TGFb. Without an NLS Smad3 will never enter the nucleus and activate transcription of its target genes. Resistant. Phosphorylation is necessary for Smad3 activation and eventual transport into the nucleus. Thus without these residues Smad3 will not activate transcription of its target genes. Resistant. Smad4 is necessary for TGFb-induced transcriptional activation. Without Smad4, the Smad3/Smad4 complex can never form and enter the nucleus to activate TGFb responsive genes. Resistant. Type I receptors need to become phosphorylated in order to be activated. Inactive Type I receptors will not phosphorylate and activate Smad3, and therefore Smad 3 will not enter the nucleus and activate transcription of its targets. These cells would be hypersensitive to TGFb. Ski downregulates TGFb transcription by Smads. If Ski can’t bind its target, than TFGb-induced transcription will increase. b, 5 pts) How could you determine the location – nucleus or cytosol – of Smad3 in living cultured leukemia cells? Use flourescence microscopy on cells expressing a fusion protein of Smad3 with GFP. Where the GFP signal is present is where Smad3 is localized. Points were not given for antibody staining because that technique involves fixing (i.e. killing) the cells. c, 10 pts) For each of the indicated cell lines, explain whether, in the presence of added TGFß, you would expect the Smad3 protein and the Smad4 protein to be localized to the nucleus or to the cytosol. Use N for nucleus and C for cytosol, and explain each of your answers. 6 of 9 7.06 Spring 2004 Exam 1 Wild-type (nonleukemic) lymphocytes Patient A lymphocytes Patient B lymphocytes Patient C lymphocytes Patient D lymphocytes Patient E lymphocytes Name___________________________ Smad3 N C C N/C C N Smad4 N C C C C N In wild type lymphocytes, Smad3 forms a heterotrimeric complex with Smad4 that enters the nucleus to activate transcription upon TGFb stimulation. In A Smad3 has no NLS and cannot enter the nucleus, Smad4 is dependent on Smad3 for nuclear entry, so it too will remain cytoplasmic. In B Smad3 cannot get activated and cannot enter the nucleus because its NLS remains hidden. Smad4 is cytoplasmic for the same reasons stated above. Points were given for N or C for Smad3 because it was not clear whether or not Smad3 can enter the nucleus without Smad4. In either case, Smad4 has no NLS, so Smad 4 that cannot bind Smad3 is cytoplasmic. In D Smad3 cannot be activated, so both it and Smad 4 will be cytoplasmic just like in patients A and B. In E, both Smad3 and 4 are normal and will be nuclear upon TGFb stimulation. Question 4, 15 pts. You are a brilliant gastroenterologist who has just finished analyzing a collection of mouse mutants, isolated at the Jackson laboratories in Bar Harbor, Maine because they are underweight. You find that two strains have an unusually high stomach pH of 4 rather than the normal pH of 1. The stomach lumen acidification is achieved by secretion of HCl by parietal cells in response to endocrine signals. The H+/K+- ATPase pump, together with Cl- and K+ channels and a Cl-/ HCO3- antiporter, are responsible for HCl secretion into the stomach lumen. a. 5 pts) You suspect that in one of your mouse lines the Cl-/ HCO3- antiporter has a mutation that causes it to become localized to the apical plasma membrane rather than its normal site in the basolateral plasma membrane. Describe, step- by- step, a simple experiment you could to determine where in the plasma membrane this antiporter is located. 7 of 9 7.06 Spring 2004 Exam 1 Name___________________________ You could use immunoflourescence microscopy as in 3b. Fix and permeabilize the cells, add an antiporter antibody, then add a fluorescent secondary antibody which will bind the primary antibody. Visualize the cells using confocal fluorescent microscopy and determine which side of the membrane, apical or basolateral, the antiporter is located. b. 5 pts) Why would mis-localization of the Cl-/ HCO3- antiporter to the apical plasma membrane result in an elevation of the pH of the stomach lumen? Mislocalization of the antiporter will pump HCO3- into the stomach lumen. Because HCO3- is basic, the pH of the lumen will rise. c. 5 pts) In a second line of mutant mice you determine that there is a mutation in the Claudin1 gene such that the number of tight junctions that normally interconnect the epithelial cells lining the stomach is reduced from 6 to 2 per cell pair. Explain why a mutation in this protein would result in an elevation of the pH of the stomach lumen? The reduced number of tight junctions might allow for the diffusion of H+ ions from the stomach lumen into the basolateral space or OH- ions from the basolateral space to the stomach lumen, raising the pH of the lumen. Alternatively, the Cl-/HCO3antiporter could diffuse from the basolateral membrane to the apical membrane, also raising the lumen pH. Question 5, 10 pts. You are interested in signaling by the newly isolated hormones BMP6, BMP7, and BMP8. Knowing nothing about the nature of the cell surface receptors for these hormones, you conduct a binding experiment with 35S labeled BMP6, and determine that your lung cell line contains 50,000 surface receptors with a dissociation constant of 1011 M. You cannot radiolabel either BMP 7 or BMP8, so you decide to carry out a competition assay. The binding reaction contains 2 x 10-12 M 35S labeled BMP6 and the indicated molar concentrations of unlabeled BMP 7 or BMP8, and you measure the fraction of 35S labeled BMP6 bound to the cell surface, relative to the amount bound in the presence of no added unlabeled hormone: 8 of 9 7.06 Spring 2004 Name___________________________ Exam 1 100 80 B MP8 60 B MP7 40 20 10-12 10-11 10-10 10-9 10-8 10-7 unlabeled hormone (Molar) a. 5 pts. ) Which hormone – BMP6 or BMP7, binds with a higher affinity to cell surface receptors? Explain your answer. BMP-6 binds with a higher affinity. From the graph, the Kd of BMP7 is about 10-10, and we know the Kd of BMP6 is 10-11. The lower Kd of BMP6 is indicative of tighter receptor binding. b. 5 pts) Does the data indicate that there are more than one type of cell surface receptor that binds BMP6? Explain your answer. Yes. BMP8 can only compete off about half of the radiolabeled BMP6 from the cell surface, suggesting that BMP6 binds an additional receptor to which BMP8 cannot bind. 9 of 9 ...
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This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore.

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