Name: TF: Section Time: LS1a ICE 5. Practice ICE Version B

Transcription

1 Name: TF: Section Time: LS1a ICE 5 Practice ICE Version B

2 1. (8 points) In addition to ion channels, certain small molecules can modulate membrane potential. a. (4 points) DNP ( 2,4-dinitrophenol ), as shown in the three equilibria described below, has the rare ability to cross biological membranes when either protonated or deprotonated. The pk a of DNP is near that of physiological ph. An artificial cell containing no ion channels has a negative membrane potential due only to the electrochemical gradient of hydronium ions across its membrane (such that the ph is lower outside the cell than inside the cell). The cell is treated with protonated DNP. Shown below is a graph of the cell s membrane potential as a function of time. DNP was added at the indicated time. Briefly explain why the addition of DNP makes the membrane potential less negative. Since DNP can cross the membrane when it is either protonated or deprotonated, it can react with a proton to become protonated outside the cell membrane where the concentration of protons is high, it can then diffuse across the membrane to bring the proton inside the cell. Once inside the cell, protonated DNP can dissociate since the proton concentration is low. Since DNP can cross the membrane as an ion, it can exit the cell while deprotonated to repeat the process. As deprotonated DNP leaves the cell, a positivelycharged hydrogen ion is left inside the cell, and a negatively-charged DNP returns to the extracellular space, leaving the inside of the cell more positive and making the outside of the cell more negative. 2

3 b. (4 points) A different artificial cell with no ion channels is produced, and the initial ion concentrations inside and outside of its membrane are listed below: Inside Outside NaCl 10 mm 100 mm HCl 0 mm 50 mm KCl 150 mm 5 mm Amphotericin B, a small molecule, makes cellular membranes permeable to ions, including Cl -, Na +, K +, and H +. Which molecule, DNP or amphotericin B, would produce the largest membrane potential, in terms of magnitude, when added to this artificial cell with the initial conditions described above? What would be the sign of this membrane potential produced by adding the molecule you chose (inside with respect to outside)? Briefly explain your answer. Amphotericin B will yield a neutral membrane since making the membrane permeable to all four ions will result in an equal distribution of cations (positively-charged ions) and anions (negatively-charged ions) on both sides of the membrane. Amphotericin B allows all of the ions that are present in the solution to freely enter and exit the cell, but since both positive and negative ions are allowed to enter, there is no net accumulation of charge on either side of the membrane, which is required for the generation of a membrane potential. By making the membrane selectively permeable to only hydrogen ions, DNP will generate a positive potential (inside with respect to outside), and this positive membrane potential is greater in magnitude than the zero membrane potential generated by amphotericin B. 3

4 2. (6 points) To study how various mutations cause a novel plasma membrane protein, BeatYale, to go to the wrong cellular compartments, you engineer the following DNA construct that fuses your gene of interest (beatyale) with the gene for the protein GFP such that the gfp gene is read in the frame of the beatyale gene and the proteins are covalently linked by the ribosome. A diagram of the fusion protein is shown below. Whereas the wild-type ( un-mutated ) BeatYale-GFP fusion protein localized to the plasma membrane, two different mutations in the beatyale portion of the fusion gene yielded the following results: Mutant 1: Mutant 2: For each mutant, identify the protein targeting error and explain where in the secretory pathway the error must have occurred. (If you do not want to print in color, we suggest you look at this image on a computer monitor). Mutant 1 is supposed to be inserted into the plasma membrane, but is instead secreted. The error in its mistargetting must be that this mutant was translated into the lumen of the ER rather than being inserted into the ER membrane. Mutant 2 never enters the secretory pathway, as it is found throughout the cytosol rather than in a membrane or an organelle. The error must be that it was never brought to the ER membrane but was instead translated in the cytosol. 3. (6 points) A functional potassium leak channel consists of four identical polypeptides. Each polypeptide is a transmembrane protein, and therefore must be synthesized at the endoplasmic reticulum membrane. Below is an image of each of the four individual subunits of the potassium channel on the left before they assemble into a functional channel on the right. A three-amino acid sequence (with sequence RKR) is exposed on the surface of each individual subunit, but this RKR sequence buried between subunits upon subunit assembly. The process of subunit assembly occurs in the endoplasmic reticulum. 4

5 a. (2 points) As the potassium channel assembles in the ER, does the region of the selectivity pore destined to face the extracellular space face the ER lumen or the cytosol? Briefly explain your answer. ER lumen, because the relative orientation of the membranes is conserved throughout the secretory pathway, and the cytosolic face of each organelle/vesicle is maintained throughout, as is the lumenal/extracellular face. b. (4 points) To determine the location of the potassium channel subunits within the cell, the potassium channel gene is replaced with one in which GFP is fused to one of the potassium channel s flexible loops. Having a cell express this fusion gene instead of the normal gene results in the normal membrane potential, indicating the GFP-fusion does not affect the function of the channel. Initially the fluorescence appears both in the plasma membrane and in the ER membrane. You then choose to test the role of the RKR amino acids present in each subunit by mutating them all to AAA (three alanines). This results in a cell in which the GFP fluorescence is almost exclusively located at the plasma membrane, with almost no fluorescence in the ER membrane, and the membrane potential is more positive ( less negative ). Briefly explain why mutating the RKR sequence to AAA changed the GFP fluorescence pattern in the cell and caused the membrane potential to become more positive ( less negative ). The wild-type potassium channels fluorescing at the PM were fully assembled, whereas those in the ER membrane were not fully assembled. The exposure of the RKR sequence in the yet-to-be-fully assembled channels allowed them to be retained in the ER as an aspect of its quality control. Mutating the RKR sequence to AAA allowed potassium channels that were not fully functional to leave the ER and arrive at the membrane, where they are nonfunctional. The insertion of nonfunctional potassium leak channels in the PM causes the membrane potential to rise since the flow of potassium ions through functional channels is what primarily generates the negative membrane potential. 5

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