7.014 Quiz I Solutions

Transcription

1 7.014 Quiz I Solutions Class Average = 64.7 Median = 66 Grade Range % A B C D F

2 Question 1 (32 points) You are a biology specialist on a space exploration team. During one of your trips, you discover life on a new planet, which you name Proteon. Conditions on Proteon are strikingly similar to those of Earth: it is abundant in water and has an oxygen and nitrogen rich atmosphere. You analyze a sample of the alien bacteria from the planet and realize that, in addition to the 20 amino acids found on Earth, this bacteria also uses a new 21st amino acid. You determine the chemical structure of this new amino acid, which you call Neonine (neo): a) Classify neonine as: polar charged (indicate basic or acidic), polar uncharged, or non-polar (hydrophobic). Polar uncharged b) Would you expect to find neonine on the surface or in the core of a protein? Why? On the surface, where it can form hydrogen bonds with water molecules in the environment. If neonine is found on the inside of a protein, we would expect it to be interacting with other polar or charged amino acid side chains. You analyze two mutant strains of Proteon bacteria that both seem to be unable to maintain proper water balance. You suspect that these bacteria have a defect in the protein porin, a membrane channel protein made up of three identical subunits that is responsible for the transport of water across the cell membrane. The ribbon representation of the structure of porin is shown below. A B 2

3 Question 1, continued c) What element of secondary structure is most abundant in porin? Β-sheet d) Circle all that apply: What type(s) of amino acids do you expect to find on the porin polypeptides i) on the inside of each barrel (arrow A on the diagram points into one of three barrels)? Polar Nonpolar Positively charged Negatively charged Why? Porin transports water molecules, so we would expect the amino acids found on the insides of the barrels to be able to interact with the water molecules being transported, i.e. we expect these amino acids to be able to form hydrogen bonds with the water molecules. ii) in the center section of the outside of each barrel (arrow B points to one such section. Note that these are the sections of the protein that are embedded in the cell membrane)? Polar Nonpolar Positively charged Negatively charged Why? Since this part of the protein is embedded in the nonpolar part of the membrane (the lipid part of the phospholipids bilayer), we would expect nonpolar amino acids to be found there. You isolate porin from wild-type and the two mutant strains of Proteon bacteria and run these proteins on the denaturing and native gels. You get the following results (first lane in each gel is molecular size standards, with sizes marked on the sides): Denaturing gel Native gel - 175kDa 150kDa Wildtype M1 M2-175kDa 130kDa Wildtype M1 M2 75kDa 80kDa 35kDa 25kDa 37kDa 30kDa + + 3

4 Question 1, continued e) From the gel data we can conclude that the primary structures of M1 and M2 porin (circle the best choice) are the same are different may or may not be different Justify your answer The native gels show that the two mutants have different quaternary structures, and both are different from the wild-type porin. Porin is a membrane protein, and not a prion. Therefore, because there is a change at higher levels of protein structure, we know that the difference must have started at the level of the sequence of amino acids primary structure. f) Describe the quaternary structure of M2 porin protein. Use gel data to justify your answer. M2 porin protein has two interacting subunits rather than three like the wild-type protein. This is apparent because while the denaturing gel shows that individual subunits of wild-type and mutant proteins are of the same length, the native gel shows that M2 porin is only made up of two subunits while M1 is made up of one subunit. Proper alignment of two particular beta sheets in a porin polypeptide is stabilized by amino acids Lys 363, Gln 366, and Asp 370 located in a loop turn between the two beta sheets. g) What is the strongest possible interaction between i) Lys 363 and Asp 370? Ionic bond ii) Gln 366 and Asp 370? Hydrogen bond h) For each substitution below, indicate whether the resulting mutant would be likely to maintain alignment of beta sheets in this particular loop turn, and give a brief explanation of your choice. Substitution Maintain alignment (choose one)? Explain why Asp 370 Arg likely not likely would disrupt ionic bond; two basic aas would repel Asp 370 Glu likely not likely new aa still acidic, and close in size bonds maintained Gln 366 Pro likely not likely polar aa sub d by a nonpolar likely to disrupt interactions Question 2 (22 points) a) Can any amino acids be essential for an autotroph? Why or why not? No, autotrophs by definition make everything they need from CO 2, water, organic N, and minerals. 4

5 Question 2, continued Below are simplified chemical flowcharts of glycolysis and the citric acid cycle. Citric Acid Cycle Pyruvate In the course of glycolysis, NAD + is reduced to NADH. Fermentation and cellular respiration then recycle NADH back to NAD +. b) What would happen to a cell if it lost its ability to oxidize NADH back to NAD +? Why? Glycolysis would be shut down because of lack of NAD +. With no ATP available, all other cell functions would shut down as well, and cell would die. The cells in our muscles can function aerobically or anaerobically. The aerobic mode is preferred. Often, after a period of strenuous physical activity, such as in the middle of running a marathon, people feel burning in the muscles. That sensation has as its root cause lack of oxygen in the muscles. c) Step by step, trace the molecular events that lead from muscles running out of oxygen to the burning sensation athletes experience. i) What is the role of oxygen in respiration? O 2 is the final electron acceptor in respiration. 5

6 Question 2, continued ii) What happens to the Electron Transport Chain if the cell runs out of oxygen? The chain would begin to back up as first the most electronegative proteins, and then each next protein in the chain would retain the electrons that normally only reside on them transiently. iii) What effect would the events in (ii) eventually have on the cell s ability to recycle NADH through respiration? Why? When all the initial proteins of the ETC in the mitochondria retain the electrons deposited on them by NADH, no new NADH would be able to release its electrons, and the cell would loose its ability to recycle NADH through respiration. iv) Explain why if events in (iii) were to occur, the cell would be unable to continue to execute Citric Acid Cycle. With no free NAD + available, the Citric Acid Cycle would shut down because several steps of the cycle depend on the presence of NAD + (that gets converted to NADH in those steps). v) If the cell is unable to execute Citric Acid Cycle, what alternative pathway would be triggered in a muscle cell? Why? If Citric Acid Cycle is shut down, free pyruvate would begin to accumulate. This would enable a muscle cell to begin executing the lactic acid fermentation reaction. Fermentation would allow the muscle cell to recycle the NADH back to NAD +, and thus enable it to continue performing glycolysis. vi) What product of this alternative pathway causes the muscles to feel as if they are burning? Lactic acid. Question 3 (25 points) Recall that humans can consume alcohol due to the enzyme alcohol dehydrogenase, an enzyme that catalyzes the oxidation of alcohols to aldehydes coupled with reduction of NAD + to NADH. Interestingly, this is the very same enzyme that catalyzes the reverse reaction as the final step of the fermentation pathway in yeast. Alcohol dehydrogenase enzymes in humans and yeast are not identical, but are closely related. reduced a) In the fermentation reaction, acetaldehyde is oxidized and NADH is reduced oxidized b) Is the absolute value of G likely to be big or small for the reaction catalyzed by alcohol dehydrogenase? Why? G absolute value is likely to be small since rxn proceeds spontaneously (although in the presence of enzyme) without additional energy input in both directions in the two organisms above (only one direction in each organism). 6

7 Question 3, continued c) Below, sketch the energy diagram of this reaction without enzyme added in the human liver, where conditions favor production of aldehydes. On the diagram, label the following: E a ( G ) G 0 reactants (name these) products (name these) E a Enzyme lowers E a alcohol + NAD + G 0 aldehyde + NADH d) Alcohol is toxic to the yeast cells. During fermentation, any ethanol produced is quickly transported outside of the cell. i) Explain how that ensures that the fermentation reaction always proceeds in the direction of ethanol synthesis. Exporting ethanol out creates a concentration gradient in the direction of the products of this reaction, and ensures that the reaction proceeds in that direction. ii) Why is it important for the well being of a yeast cell for this reaction to proceed in the direction of ethanol synthesis? Fermentation reactions allow the cell to recycle NADH back to NAD +. Thus, although ethanol is toxic and needs to be exported, the cell needs to produce it in order to recycle NADH. e) i) On the graph above, draw and label the effect that adding alcohol dehydrogenase would have on the reaction. ii) Does alcohol dehydrogenase change the speed of the reaction? Why or why not? Yes, the catalyst changes the rate of the reaction by lowering the activation energy which allows more reactant molecules to reach the new transition state. In humans, alcohol dehydrogenase also catalyzes the conversion of methanol into formaldehyde, a substance that can cause death of cells in the human body. The standard treatment for methanol poisoning used to be to administer ethanol. f) Why do you think this treatment works? Explain on the molecular level. Both molecules apparently are converted in the same active site on the alcohol dehydrogenase enzyme. In addition, ethanol apparently out competes methanol, i.e. the enzyme has higher affinity for ethanol than for methanol. 7

8 Question 4 (21 points) You perform an experiment to measure the base composition of DNA in various animals and tissues. You first determine the base composition for skin cells from a dog. a) Given what you know about the structure of DNA and Chargaff s rule, fill in the missing base composition percentages: A: 34% C: 16% G: 16% T: 34% You also determine the base composition of a sample taken from dog white blood cells. b) Do you expect the base composition of this sample to be the same or different from the sample above? Why? We would expect the base composition to be the same because (almost) all cells in an organism have the same DNA sequences. You then determine the base composition of human skin cells. c) Do you expect the base composition of this sample to be the same or different compared to the original dog skin cells sample? Why? We would expect the base composition to be different because different organisms are likely to have differing base compositions of their genomes. d) Watson initially envisioned DNA as having A-A, C-C, G-G and T-T base pairs. Would Chargaff s rule support this hypothesis? Why or why not? No, because this base pairing model would not guarantee that the ratio of T/A, C/G, and purines/pyrimidines would always be ~1 in all organisms and tissue types. Meselson and Stahl did an elegant experiment to show that DNA replication is semiconservative. In this experiment, they grew bacteria in 15 N (a heavy nitrogen isotope) for many generations, and then switched to 14 N (the light isotope) for a given number of cycles. Using a cesium-chloride gradient they were able to visualize the bands of 15 N- and 14 N-labeled DNA. You dream that you are the one originally doing the Meselson-Stahl experiment. Recall that one of the two alternative hypothesis of replication was the conservative model, which hypothesized that after replication two strands of the old molecule remained together, and two newly synthesized strands formed a new molecule of DNA. The third hypothesis postulated a discontinuous model, in which after replication, each strand would contain some old and some new DNA. 8

9 Question 4, continued You see the following results: light heavy # of replication cycles e) i) Which model of DNA replication (semi-conservative, conservative, or discontinuous) is supported by the data in your dream universe? For each model not supported, explain what piece of data is inconsistent with that model. In this dream, the conservative model is supported by the data because the original HH band remains through the cycles (indicating that molecules consisting entirely of old DNA remain through the cycles), only the light band (LL) appears (indicating that molecules consisting entirely of new DNA appear after replication is performed), while no intermediate band (HL-- which would indicate a DNA molecule that combined old and new DNA) appears. Semi-conservative model is not supported because no intermediate band (HL) appears, indicating that no molecule where one strand is from old and one is from new DNA is created. At the same time, heavy band remains (so no old-old molecule is disrupted) and light band is created. Discontinuous model is not supported because no intermediate band (HL-- which would indicate a DNA molecule that combined old and new DNA) appears. At the same time, heavy band remains (so no old-old molecule is disrupted) and light band is created. ii) Explain why the lines in the figure are of varying thickness. The line thickness indicates the amount of product in each band. As DNA is replicated, the original molecules remain the same in this model, while the amount of newly created DNA (LL) increases with each cycle. This is reflected in the thicknesses of lines in the diagram above. 9