7.06_2004_PS2key - 7.06 Spring 2004 PS 2 KEY 1 of 7 7.06...

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Unformatted text preview: 7.06 Spring 2004 PS 2 KEY 1 of 7 7.06 Problem Set 2 Answer Key 1) You have started working in lab that investigates signaling pathways downstream of TGFß receptors. One of the senior graduate students has been working on a screen for novel TGFß’s and has one that she thinks might interest you—she calls it TGFß57. She’s already cloned the gene so you can make pure TGFß57. You decide to take up the challenge of characterizing this new growth factor. a) You would first like to find some high affinity receptors that bind your novel growth factor. Design an experiment to identify the cell-surface TGFß57 receptors. Answer: There is no one right way to do this, however, a common method is to radiolabel the purified growth factor with iodine-125 (125I) and culture the labeled growth factor with cultured cells (you would probably use the same fibroblast cell line that’s been used in earlier experiments with TGFß57). Then you’d treat the culture with a chemical crosslinker to covalently bind the growth factor to any receptors it bound. You can then use standard purification methods to purify the labeled receptors from cell membranes (nonionic detergents to separate the receptors from the cell membrane and chromatography or electrophoresis to separate your receptor from other proteins). You would now like to investigate the binding properties of TGFß57. b) You decide to look at the maximal physiological response of TGFß57R to TGFß57. Based on the graph below determine the Kd (in relative concentration) of TGFß57 binding, the percentage of receptors occupied at 50% of maximal physiological response, and the percentage of maximal physiological response when half the receptors are occupied. What does this tell you about the percentage of bound receptors needed to achieve maximal response? *** NOTE: y-axis label: Fraction of maximum cellular response Answer: Kd is about 1.2. At half-maximal physiological response, only about 26% of the receptors are bound. When 50% of receptors are bound, about 77% of the maximal physiological response is elicited. This tells you that cells do not nee to have 100% of the receptors bound by ligand in order to achieve a maximal response. 7.06 Spring 2004 PS 2 KEY 2 of 7 c) You would now like to determine the binding affinity of TGFß57 for the TGFß57 receptor that you have been working with. Design an experiment to measure the Kd of TGFß57 binding high- affinity receptors in a cultures fibroblast cell line. Explain how you would deal with TGFß57 binding nonspecifically to low-affinity receptors. Answer: You’d want to design some kind of binding assay to measure receptor binding. What is commonly done is to take a sample of cells and incubate them with increasing concentrations of 125I-labeled TGFß57. After incubation, you separate the cells from unbound TGFß57 (usually by centrifugation) and the amount of radioactivity bound to them is measured. This gives you the total binding of TGFß57 to receptors. In order to determine specific binding to high-affinity receptors you have to measure nonspecific binding. This is determined by repeating the binding assay in the presence of a 100-fold excess of unlabeled TGFß57, which saturates all the specific high-affinity sites. In this case all the labeled V binds to nonspecific sites. To determine specific binding you simple calculate the difference between total binding and nonspecific binding. d) Using the specific binding curve below, determine the Kd of TGFß57 for high-affinity receptors. Also, find the total number of receptors per cell. *** NOTE: y-axis label: [I125] TGFb-57 Bound (molecules per cell) Answer: The Kd is determined by finding the concentration of TGFß57 at half-maximal binding. Kd= 10nM. The total number of receptors per cell is 19,000. e) You’d now like to determine the sensitivity of these cells to TGFß57. You know that the Kd for binding TGFß57 to its receptor is about 1X10-8 and you know there are a total of 19,000 receptors per cell. You have previously determined that only 13% of the 19,000 receptors must be bound to obtain the maximal cellular response. Determine the TGFß57 concentration needed to induce the maximal response. What would be the concentration of TGFß57 required to obtain maximal response if there were only 5000 receptors per cell, what is the fold difference 7.06 Spring 2004 PS 2 KEY 3 of 7 compared to the normal concerntration you just calculated, and would this make the cell more or less sensitive to TGFß57? Why might a cell want to regulate the number of receptors to a given external signal? Answer: To calculate the concentration of TGFß57 to induce maximal response, you use the equation [L] = Kd/(RT/[RL])-1. [RL] is the number of TGFß57-occupied receptors needed to induce maximal response—which is 13% of 19,000. Plugging all the values in to the equation gives you [TGFß57] = 1X10-9M. If the total number of receptors was reduced to 5000, the [RL] would still be 2470, then [TGFß57] = 5X10-9M—a 5-fold increase in the Kd of TGFß57. This decrease makes the cell less sensitive to TGFß57 because there needs to be a higher concentration of TGFß-57 released in order to achieve the maximal response. A cell might want to regulate the number of receptors in order to direct physiological or even developmental events. f) If TGFß57 binding to its receptor had occurred with a low affinity (a Kd higher than 1X-7M), you couldn’t have measured the Kd by performing the binding assay from above. How could you have determined the Kd if this had been the case? Answer: You could perform a competition assay with another ligand that binds to the same receptor with high affinity. In this type of assay, increasing amounts of an unlabeled, low-affinity ligand are added to a cell sample with a constant amount of the radio-labeled high-affinity ligand. Binding of unlabeled competitor blocks binding of the radioactive ligand to the receptor. Consequently the concentration of competitor required to inhibit binding of half the radioactive ligand approximates the Kd value for binding if the competitor to the receptor. 2. A colleague in your lab discovers that the binding of TGFb57 to its receptor turns on expression of a gene that inhibits cell growth, SloGro. SloGro is frequently mutated in cancers, and you suspect that components involved in the TGFb57 signaling pathway might also be relevant to cancer research. You decide to set up a cell culture screen to identify proteins involved in TGFß57 signaling. Your screen is a success, and you isolate a number of cell lines that have mutations in the TGRb57 signaling pathway. A) Explain how the following mutations would affect TGFb57 signaling. Assume that the basic components of the b57 pathway are identical to the canonical TGFb signaling pathway. i)The MH1 domain of Smad 3 is missing. The MH1 domain contains a NLS sequence. Thus Smad 3 will not localize to the nucleus even in the presence of TGFb57 and will not activate transcription of SloGro. ii)R1 has a truncated cytoplasmic tail and cannot be phosphorylated. The R1 kinase is activated by phosphorylation of its cytoplasmic tail by RII. A truncated version of R1 that cannot be phosphorylated will be unable to phosphorylate Smad3, and will not activate the signaling cascade. 7.06 Spring 2004 PS 2 KEY 4 of 7 iii) Smad 7 is constitutively active. Smad 7 blocks the ability of R1 to phosphorylate Smad 3, which prevents it from entering the nucleus and activating transcription. B) You also isolate a version of Smad3 that cannot bind DNA, and find that this mutant version exerts a dominant negative effect on the TGFß57 pathway. Give one possible mechanistic explanation for how mutant Smad3 could act dominant negatively on the pathway. One possible scenario is as follows: Smad 3 can’t bind DNA, and therefore the complex between mutant Smad3 and Smad4 can’t activate transcription of target genes. Because the complex that binds DNA and activates transcription involves two Smad 3 proteins, the mutant version gets incorporated into complexes with wild type Smad 3, effectively inactivating complexes with wild type protein as well. 3) Epineprhine increases the concentration of glucose 1-phosphate in wild type liver cells. Explain how the following mutations would affect the glucose 1-phosphate concentration within an epinephrine stimulated liver cell. i) The cell contains a PKA with a nulceotide binding site that is unable to bind cAMP. cAMP binding liberates the active version of PKA from its psuedosubstrate. A PKA that is unable to bind cAMP will therefore not be activated upon epinephrine stimulation and glucose 1-phosphate concentration will not increase. ii) The cell contains an unphosphorylatable version of Inhibitor of Phosphoprotein Phosphatase. Inhibitor of Phosphoprotein phosphatase will not inactivate phosphoprotein phosphatase in its unphosphorylated form. Thus phosphoprotein phosphatase will be active even in the presence of active PKA, and glucose 1-phosophate concentration will not increase. iii) The cell contains a mutant version of Gsa that cannot hydrolyze GTP . Gsa that cannot hydrolyze GTP will remain active once turned on. This will constitutively activate adenylyl cylase and all the downstream targets of PKA, which will lead to increased glucose 1-phosphate concentration. 4) a) Given the following values, calculate the electrochemical equilibrium potential of each ion (F=23, 062cal/(mol V), R = 1.987 cal/K mol, T = 293K). K+out = 3 mM Na+out = 117 mM Cl-out = 120 mM 7.06 Spring 2004 K+in = 90 mM PS 2 KEY Na+in = 30 mM 5 of 7 Cl-in = 4 mM E = (RT/zF) ln [ionout/ionin] EK = - 85.9 mV ENa = 34.4 mV ECl = - 85.9 mV b) What does the electrochemical equilibrium potential reflect/what is true at the electrochemical equilibrium potential? At the electrochemical equilibrium potential, the concentration gradient for the ion is balanced by the electrical gradient. Hence, there is not net flux of ions across the membrane. That is, the rate (notice rate, not concentration) of movement of the ion in both directions is equal. c) If one reduces the magnitude of the K+ concentration gradient across the cell, one will increase/decrease (circle one) the equilibrium potential. Why? One would decrease the equilibrium potential. Remember that at the electrochemical equilibrium potential, the concentration gradient is balanced by the electrical gradient. If we decrease the magnitude of the concentration, the electrical gradient will also necessarily decrease. Another way to think about it, is as follows: by increasing extracellular K+, we are decreasing the flow of K+ out of the cell because the concentration gradient that makes this process energetically favourable has decreased. Assuming the Na+ concentration remains the same, its flow in the cell remains the same. So, we have decreased the flow (current) of K+ out of the cell while not changing the flow of Na+ into the cell. As a result, the cell should depolarize. d) Assume that the resting membrane of a cell is determined primarily by the equilibrium potential of potassium. If you insert an electrode into this cell and clamp the membrane voltage to +20 mV, would K+ be moving into or out of the cell through the potassium channels (K+out = 3 mM, K+in = 90 mM)? From above, the equilibrium potential of potassium is –87.2 mV. The ratio of K+ (out/in) is 1/30. At +20 mV, the ratio of K out/in would be: E = 0.02V, E = 0.059log (out/in), (out/in) = 10(E/0.059) = 891.3. Since the present K+ concentration ratio of 1/30 is less than that required at +20 mV, the potassium will leave the cell until the concentration outside is nearly 3 orders of magnitude greater than the one on the inside. 7.06 Spring 2004 PS 2 KEY 6 of 7 e) You isolate cells that have only K+ channels. You also have a drug that causes this channel to open. With a voltage clamp apparatus, you find, as expected, that in the presence of the drug the membrane potential is dependent on the concentration of extracellular K+ used in the experiment. At one concentration of K+ of 5 mM, you find that the membrane potential is -80.5 mV. From this information, calculate the intracellular K+ concentration. E = - 0.0805 V, E = 0.059 log (Kout/Kin) Kout/Kin = 10(E/0.059) = 0.043 Kin = Kout / 0.043 = 115.7 mM 5) You are doing a summer internship in a lab that studies a recently discovered receptor called RKR. It is a G-protein coupled receptor that is expressed in kidney cells and found to be mutated in patients with a kidney disease. You sequence the receptor in 100 patients and find that the mutations fall into two classes, A and B. Unlike wildtype RKR, both mutants fail to increase cellular levels of cAMP. (i)The ligand for the RKR receptor is called MC. How could you find out if the mutant receptors are expressed on the surface of kidney cells? First you would transfect cDNAs for wt and mutant RKRs into cultured human kidney cells. These receptors would have been tagged e.g. using GFP, and then you would incubate the ligand with the cells and then look under the microscope to see whether the receptor is localized at the membrane and you would examine the relative levels of expression between mutant and wild type. (ii)You find that mutant A is localized to the membrane at approximately the same levels as wildtype but mutant B has very low levels of RKR receptor on the surface. Mutation A lies in the conserved C3 loop, which is known to be important for binding and activation of the Ga subunit. How could you determine whether Ga becomes activated in cells expressing mutant A? One way to do this would be to use the FRET system mentioned in class. You would tether a YFP tag to the Gb subunit and a CFP tag to the Ga subunit. In the absence of ligand both subunits are close together and so when CFP is excited at 440nm FRET occurs between CFP and YFP and so light of 527nm is emitted. When ligand is bound to wildtype receptor the Ga subunit binds GTP and dissociates from the Gbg subunits. Now when CFP is excited, light emits at 490nm since no FRET can occur. (iii) While trying to understand the basis for mutant B you look at receptor localization in the presence and absence of ligand using the GFP tagged versions of the receptors from part (i). What do you expect to see in the wildtype cells after extended exposure (30mins) to ligand? You would expect to see the receptor localized at the plasma membrane when ligand is not present. After 30 mins of ligand exposure you would expect to see some of the GFP 7.06 Spring 2004 PS 2 KEY 7 of 7 fluorescence in the cytoplasm in a punctate pattern, since upon extended interaction with the ligand the receptors are internalized into endosomes using b-arrestin. In the cells expressing mutant B the receptor is found to localize predominantly to the cytoplasm in the presence or absence of ligand. How could you explain this? How could you demonstrate this experimentally? b–arrestin could be constitutively binding to the mutant RKR receptor in the presence and absence of ligand, resulting in the receptor being internalized and hence there is insufficient receptor on the surface to activate the required signaling pathways. To demonstrate this you could express a version of b-arrestin that is tagged with YFP and look for colocalization with mutant RKR-GFP in the endosomes in the absence of the MC ligand. (iv) The RKR receptor has a conserved motif in the cytoplasmic portion of the protein that contains Ser-Ser-Thr. You want to show that if you prevent the constitutive b-arrestin mediated endocytosis of the mutant B version of the RKR receptor that it will then be capable of signaling Your advisor suggests that you mutate all three residues to alanine in the mutant receptor? Why does he choose these three residues? Why does he suggest mutating them to alanine? After extended exposure to ligand, GPCRs become phosphorylated and then b-arrestin can bind to the phosphorylated residues and this results in endocytosis of the receptor and termination of signaling. Serine, threonine and tyrosine are all residues that can be phosphorylated and hence if they are converted to another residue then you won’t get internalization of the receptor. He chose alanine because it is small and so probably won’t affect the secondary structure of the protein and also it is used in mutagenesis studies because it allows you to determine the functional importance of the side chain of an amino acid – which is where phosphorylation occurs on serine and threonine. ...
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