BS11 Midterm 2 (1999)

Transcription

1 Question points. A. (4 pts) Briefly explain the difference in melting points between trans-oleic acid (18:1, 9)(44.5C) and cis-oleic acid (18:1, 9) (13.4C). The cis double bond puts a kink or bend in the hydrocarbon chain, while the trans-oleic acid is more or less fully extended. The trans configuration allows better packing in a crystal than the cis arrangement, and hence has a higher melting point. B. (4 pts) Would you expect to see an amphipathic helix in the transmembrane domain of a single-pass transmembrane protein or a channel protein? Why? Answer: A channel protein. Amphipathic helices have one side that is polar or charged, which would be unfavourable in the nonpolar membrane. In a channel protein this charged or polar side of the helix face the channel, and other polar residues, and away from the nonpolar lipids. C. (4 pts) How many amino acids would be the minimum to cross the lipid bilayer? Show your calculations for full credit. The minimum residues required to cross the lipid bilayer of 3-5 nm thickness, given 15 Å per aa in a helix 3/1.5 or 5/1.5 =20-35 aa 2

2 Question points. A. (8 pts) The difference in ph between the internal and external surfaces of the inner l membrane is 1 ph unit (external side acidic). If the membrane potential is 0.18V(negative inside), calculate G for the transport of H + from outside to inside the matrix. Answer: G = 2.3RT(pH out - ph in ) + zf Ψ =2.3(0.6kcal/mol)(-1) + (1)(23kcal/V*mol)(-0.18V) = kcal/mol 4.14 kcal/mol = kcal/mol B. (15 pts). Assume it takes 3 moles of H + to synthesize 1 mole of ATP. What is the largest ratio of [ATP]/[][Pi] under which ATP synthesis can occur? ( G 0 ATP = 7.5 kcal/mol) Answer: G H+ = G 0 ATP + RT ln[atp]/[][p I ] (5.52 x3) kcal = 7.5 kcal ln [ATP]/[][P I ] [ATP]/[][P I ] = e 9.06/0.6 = 3.6 x10 6 /1 3

3 Question points Inhibitors of l function have provided extremely useful tools for analyzing how work. The figure below shows three distinct patterns of oxygen electrode traces obtained using a variety of inhibitors. In all experiments, were added to a phosphate-buffered solution containing succinate as the sole source of electrons for the respiratory chain. After a short interval and then an inhibitor were added, as indicated in the figure. The rate of O 2 consumption at various times during the experiment is shown by downward sloping lines, with faster rates shown by steeper lines. A. (10 pts) From the description of the inhibitors in the list that follows, assign each inhibitor to one of the O 2 traces in the figure. All these inhibitors stop ATP synthesis. Inhibitor Action Trace (Answer) 1. Dinitrophenol (DNP) Makes membranes permeable to protons b 2. Malonate Prevents oxidation of succinate a 3. Cyanide Inhibits cytochrome oxidase a 4. Atractylate Inhibits -ATP exchange c 5. Oligomycin Inhibits ATP synthase c a. b. inhibitor inhibitor c. inhibitor 4

4 B. (15 pts) Using the same experimental protocol, sketch the O 2 traces that you would expect for the SEQUENTIAL addition of the pairs of inhibitors listed below. 1. DNP followed by cyanide DNP cyanide 2. DNP followed by oligomycin DNP oligomycin 3. Oligomycin followed by DNP oligomycin DNP 5

5 Question points. A student just started to study protein translocation with the eventual goal of isolating the individual components involved. His first task was to isolate the different subcellular fractions that are required for this process. Unfortunately, his labels fell off and he could not tell which fraction was what. Since he had the in vitro translation system (IVT) ready to go, he decided to use that to help him figure out what the fractions were. Shown below are the pertinent reactions. The IVT contains all the components necessary for cell-free translation, an mrna that encodes a protein known to translocate to the ER, and 35 S-labelled methionine to visualize the translated polypeptide. Translation was carried out in the absence (lane 1) or presence of fraction C (lane 2), fraction E (lane 3), or both (lanes 4-7). Endo-glycosidase H (endo H) or protease was added after translation was complete. The samples were analyzed by SDS-PAGE, followed by exposure of the dried gel to X-ray film. Results are shown below. 1. IVT 2. IVT + fraction C 3. IVT + fraction E 4. IVT + fraction C + fraction E 5. IVT + fraction C + fraction E + detergent + endo H 6. IVT + fraction C + fraction E + protease 7. IVT + fraction C + fraction E + detergent + protease a) 5 pts. What is fraction C? Briefly explain. The cytoplasmic fraction containing SRP. Addition to the IVT causes translational arrest seen in lane 2. b) 5 pts. What is fraction E? What are the components in this fraction that our student would be able to isolate? This is the ER, or microsomal fraction. It contains SR, Sec61p (translocon/channel), glycosylase and signal peptidase. BiP, calnexin and other proteins that assists protein folding are also present, though the experiment does not implicate them. 6

6 c) 5 pts. What is the purpose of the endoglycosidase H treatment? What accounts for the size of the band in lane 5? Endonuclease removes N-linked sugars that are added in the ER. The polypeptide is smaller than that in lanes 3 and 4 because sugars and the signal peptide have been removed. d) 5 pts. Based on the data, can you tell whether the test protein is lumenal or transmembrane? Briefly explain. The protein is probably transmembrane. A lumenal protein will not be digested with protease alone, since it is protected inside the microsome. There are probably cytoplasmically exposed regions to which the protease have access, leaving transmembrane and ER fragments seen in lane 6. 7

7 Question points. In most organisms, the citric acid cycle begins with the reaction of acetyl-scoa with oxaloacetate, as shown on the next page. However, in Bacillus logotenens, a bacterium found near ocean vents, the initial reaction occurs between acetyl-scoa and pyruvate (shown below). CH 3 C=0 C00 - Draw a cycle for the oxidation of acetate to 2 CO 2 that might occur in this organism. The rules for making and breaking C-C bonds and for oxidation reactions are the same in B. logotenens as in other organisms. 8