Life Sciences 1a. Practice Problems 4

Transcription

1 Life Sciences 1a Practice Problems 4 1. KcsA, a channel that allows K + ions to pass through the membrane, is a protein with four identical subunits that form a channel through the center of the tetramer. A cartoon representation of KcsA in the membrane is illustrated on the left, with each of the four subunits numbered. The structure of the amino acids lining the neck of the channel is shown on the right (carbons are colored yellow, oxygens red and nitrogens blue; hydrogens are not shown). The two polypeptides shown are from two facing subunits; the other two adjacent subunits are omitted for clarity. K + K + utside of cell Subunit 1 Subunit 3 Inside of cell a) What specific structural feature facilitates the passage of K + ions through the channel? b) KcsA is a selective channel that allows K + but not Na + ions to pass through. Based on the structure shown, and the fact that K + ions are larger than Na + ions, propose an explanation for this selectivity. c) In order to determine the structure of membrane proteins such as KcsA, researchers must first purify them from membranes into aqueous solution. This can be difficult because membrane proteins are often insoluble in aqueous solutions. Why are membrane proteins often insoluble in aqueous solution?

2 d) To make membrane proteins more soluble in aqueous solution, researchers often add detergents to the solution. The structure for one detergent is shown below. Based on the chemical properties of this detergent, explain how it can make membrane proteins soluble in aqueous solution. H (Triton X-114 detergent) 2. A bacterium is expelled from a warm human intestine into the cold world outside. Which of the following adjustments might the bacterium make to maintain the same level of membrane fluidity? (a) Increase the length of the hydrocarbon tails in its membrane phospholipids. (b) Increase the proportion of unsaturated hydrocarbon tails in its membrane phospholipids. (c) Increase the proportion of hydrocarbon tails with no double bonds in its membrane phospholipids. 3. Some antibiotics act as carriers that bind an ion on one side of a membrane, diffuse through the membrane, and release the ion on the other side. The conductance of a lipid-bilayer membrane containing a carrier antibiotic decreased abruptly when the temperature was lowered from 40 C to 36 C. In contrast, there was little change in conductance of the same bilayer membrane when it contained a channel-forming antibiotic. Why? 4. McMurry 24.31: Why does the presence of double bonds lower the Tm of a fatty acid? Would cis fatty acids have a higher or lower Tm than the corresponding trans fatty acids? Why?

3 5. Proteins that form a ß-barrel pore in the membrane have several ß-strands that span the membrane. The amino acid side chains facing the inside of the pore would be hydrophilic whereas the amino acid side chains facing the lipid bilayer would be hydrophobic. Which of the three 10-amino acid sequences listed below is the most likely candidate for a transmembrane ß-strand in a ß-barrel protein? Explain your choice. (a) A D F K L S V E L T (b) A F L V L D K S E T (c) A F D K L V S E L T 6. Fluorescence recovery after photobleaching (FRAP) is a technique that allows visualization of diffusion within the membrane. You set out to perform FRAP using four different samples of cells. In sample 1, you have fluorescently labeled a phospholipid. In samples 2, 3, and 4 you have fluorescently labeled membrane proteins X, Y, and Z, respectively. You photobleach an area of the membrane in each sample and record the rate of recovery of fluorescence. The data you obtain are shown in the graphs below a) Using the data from these graphs, list the proteins in order of their ability to diffuse in the membrane, from fastest to slowest. b) Given the data on the graphs above, is the following statement a good hypothesis: protein X and protein Z are always present in the cell as part of the same protein complex? Explain your answer.

4 1. a) The polar carbonyl groups from the polypeptide backbone line the inside of the channel. The partially negatively charged oxygens from the four poylypeptide subunits can interact with the positively charged K + ions to facilitate their passage. b) Since electrostatic interactions are very dependent on distance, a small increase in distance between interacting groups greatly weakens the interaction. The carbonyl oxygens lining the channel are precisely spaced at a distance that optimizes their electrostatic interactions with the potassium ion. Since the sodium ion is smaller, it forms less favorable electrostatic interactions with the carbonyl oxygens because they are too far apart. c) Membrane proteins have a hydrophobic region that crosses the phospholipid membrane. This region of the protein would not make favorable interactions with water, but would cause water to form a cagelike structure around it. Membrane proteins would tend to clump together or aggregate in aqueous solution in order to group their hydrophobic regions together and maximize the entropy of water. d) The detergent is small amphipathic molecule. It has a hydrophobic ring and hydrocarbon chain (on the left of the molecule above) as well as a hydrophilic portion (the right hand side of the molecule beginning with the first oxygen and ending with the hydroxyl group). When mixed with membranes the hydrophobic ends of the detergent interact with the hydrophobic regions of the protein. Since the other end of the detergent molecule is polar, it can interact favorably with water. The detergent coats the hydrophobic region of the protein so the protein-detergent complex is soluble in water. 2. (b) is the correct answer. At colder temperatures, the membrane will be less fluid. Hence, in order to maintain the status quo, the bacterium will have to take measures to increase membrane fluidity. Both choices (a) and (c) would decrease membrane fluidity. 3. The membrane underwent a phase transition from a highly fluid to a nearly frozen state when the temperature was lowered. A carrier can shuttle ions across a membrane only when the bilayer is highly fluid. A channel former, in contrast, allows ions to traverse its pore even when the bilayer is quite rigid. 4. Saturated fatty acids are straight and can easily order themselves to make the maximum number of Van der Waal s interactions between chains. A double bond in a fatty acid produces a kink, which makes it more difficult for the molecules to pack together so they make less Van der Waal s interactions and thus a double bond lowers the Tm of the fatty acid. A cis fatty acid has a lower Tm than the corresponding trans fatty

5 acid because a cis double bond produces a larger kink in a fatty acid hydrocarbon chain than a trans double bond produces. 5. (a) is the correct answer. In a β-sheet, the amino acid side chains project alternately above and below the plane of the sheet. Therefore, every other amino acid side chain will face the same side of the strand. If a β- sheet were part of a β-barrel pore, that would mean that one side of the sheet would face the lipid bilayer while the other side would face the hydrophilic pore. This would necessitate an alternation between hydrophilic and hydrophobic amino acid side chains so that one side of the sheet would be hydrophobic while the other side was hydrophilic. The amino acids in choice (a) alternate between amino acids with nonpolar (hydrophobic) side chains and amino acids with polar (hydrophilic) side chains. 6. a) X diffuses the fastest, followed by Z, with Y barely diffusing at all. (The faster the recovery of fluorescence in the bleached area, the greater the diffusion coefficient of the protein and the faster the protein diffuses.) b) It is unlikely that X and Z are part of the same protein complex because then the rate of diffusion of X and Z should be the same.