Study questions for Bio 230_Exam_2_2022_answers

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Study questions for Bio 230, exam 2 1. What must be regenerated in glycolysis in order for reactions to proceed? NAD+ 2. What does an enzyme do that speeds up a reaction, in terms of energetics? An enzyme lowers the activation energy of the transition state for a reaction. 3. The  G of a reaction is -5 kcal/mol. An enzyme is added at high concentration to catalyze the reaction. What is the  G now? The  G of the reaction is still -5 kcal/mol. Enzymes do not change the free energy of the initial and final states of a reaction. 4. Describe a mechanism by which the ES complex promotes the transition to EP. Substrate orientation, altering the local charge environment at the active site, and putting mechanical strain on the substrate were mentioned as way to promote the transition to EP. 5. In the above diagram, what is Vmax? Draw a line showing where Km falls. What is the reaction velocity at Km? For the catalyzed reaction (non-linear line), Vmax = 200. So the reaction velocity at Km would be 100. 6. In the above diagram, which line is the catalyzed reaction (solid or dashed)? The dashed line is the catalyzed reaction. The y-intercept of a

Lineweaver-Burke plot is 1/Vmax. Since the solid line has a larger value of 1/Vmax than the dashed line, this means that its Vmax is smaller and represents the uncatalyzed reaction. 7. On the same plot, draw a line showing a) competitive inhibition b) classic non-competitive inhibition c) mixed non-competitive inhibition. A competitive inhibitor would increase Km but not Vmax, so the y-intercept would be the same and the slope of the line, given by Km/Vmax, would increase. The x-intercept at -1/Km would decrease. A classic non- competitive inhibitor would decrease Vmax without changing Km. Therefore, the y-intercept at 1/Vmax would increase and the slope of the line would increase. The x-intercept would not change. For a mixed non-competitive inhibitor, Vmax decreases and Km increases. This would increase the slope, increase the x-intercept and decrease the y-intercept. NOTE that we did not talk about mixed inhibition, so you will not have to know that. 8. What is pyruvate converted to in order to begin the TCA cycle? Acetyl-CoA 9. Suppose a toxin prevents the dissociation of CoA from succinyl-CoA. How many NADH molecules have been produced from each pyruvate in this scenario? 3. How many FADH molecules have been produced? 0. Assuming a continuing supply of pyruvate, what would have to be added to the medium or into the cell to continue producing NADH? Any of the following could be added to continue the reaction: succinate, fumarate, malate or oxaloacetate. If one wanted to get all of the electron carriers, it would be best to add succinate. After a long time, HS-CoA would also need to be added. 10. Why are fatty acids able to ultimately produce so much more ATP than glucose molecules? Two reasons. First, HS-CoA binds to the end of the fatty acid chain. Then fatty acid enters the fatty acid cycle, during which FADH2 and NADH are produced. These can enter the electron transport chain. Second, the two terminal carbons are cleaved with CoA to form acetyl-CoA and a fatty acid chain that is two carbons shorter. The acetyl-CoA enters the TCA cycle, producing another 3 NADH, 1 FADH2, and 1 GTP. So for a 16-carbon fatty acid, we run through the fatty acid cycle 7 times (the last cycle will just be acetyl-CoA that does not need to be cleaved) and the TCA cycle 8 times (16 carbons/2 carbons for each acetyl-CoA). 11. What is the source of energy used to drive the formation of ATP? The proton gradient across the inner mitochondrial membrane.

12. If the solid line represents standard Michaelis Menten kinetics, is the Hill coefficient of the dashed line in the above plot >1, <1, 1, or 0? What type of cooperativity does this plot depict? Positive and >1 . 13. What advantage, if any, does positive cooperativity provide at high [S]? At moderate [S]? For high [S], positive cooperativity does not influence v because it is near Vmax. At moderate [S], positive cooperativity can allow a major change in v over a small [S] range. 14. What kind of inhibitor binds very tightly to an enzyme often forming a covalent bond with an amino acid in the active site? A) irreversible B) reversible C) uncompetitive D) reversible and uncompetitive E) None of these are correct. 15. In what ways can pyruvate and NADH be metabolized in anaerobic respiration? Pyruvate can be converted by lactate or ethanol in anaerobic respiration and NADH is regenerated to NAD+. In aerobic respiration? Pyruvate is converted to Acetyl CoA, and the electrons from NADH are transported to the mitochondria. 16. What molecule is responsible for conveying 2 carbons from pyruvate to the Krebs cycle. Acetyl CoA 19. In which of the compartments or membranes would the following things be found? a) Glycolytic enzymes. Cytoplasm b) Electron transport system. Inner mitochondrial membrane/cristae c) Krebs Cycle enzymes. Mitochondrial matrix (mostly) and inner mitochondrial membrane (succinate dehydrogenase) d) ATP synthase. Cristae

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20. Given the table above, will  -ketoglutarate + CO2 + 2H+ + 2e- spontaneously form isocitrate? Why or why not? No. ΔG’° = -nF ΔE’ 0 . Since ΔE’ 0 is negative, ΔG’° will be positive and the reaction will not proceed spontaneously. 21. Using the same table above, name one molecule that NADH would donate electrons to and one molecule that NAD+ would accept electrons from. NADH will donate electrons to any molecule whose E’ 0 for accepting electrons (reactions going from left to right in table) is more positive than -0.32 V. NADH will accept electrons from any molecule whose E’ 0 is more negative than -0.32 V, because the reverse reaction (right to left on table) will be more positive than 0.32 V and can overcome the free energy barrier for NAD+ to accept electron. Thus, NADH will donate electrons to any of the molecules in the left half of the equation from acetaldehyde downward. Also, we know that oxygen and most of the molecules in the bottom half of the chart are electron acceptors in the electron transport chain. NADH will accept electrons from any molecule in the right half of the equation from cysteine upward, including isocitrate, H2, acetaldehyde, or alpha-ketoglutarate. 22. Which steps in the electron transport chain would be blocked by limiting motion within the inner mitochondrial membrane? Movement of UQH2(ubiquinol) from complex I or complex II to complex III. Movement of cytochrome c from complex III to complex 4. Everything taking place in complex III and complex IV would be blocked, because those steps depend on reduced ubiquinone to move through the membrane from complexes I and II to complex III. 23. Describe the three conformations of the beta subunit in ATP synthase. Open state (O). Low affinity for ATP. If bound, ATP is released. Loose state (L). ADP+Pi are loosely bound. Tight state (T). ADP+Pi are tightly bound (high affinity for them) or after spontaneous conversion to ATP, ATP is tightly bound. 24. Would a mutated ATP synthase without a gamma subunit be able to produce ATP? Why or why not? No, it would not. Rotation of the c subunits by proton influx causes the gamma subunit to rotate. The orientation of the gamma subunit with respect to the beta subunit determines the beta subunit’s

conformation. If the mutated gamma subunit were non-functional, it would not be able to change the beta subunit conformations to produce ATP. (Don’t need to remember delta subunit) If the delta subunit were mutated, the ATP head could be disconnected from the base. ATP synthase could still produce ATP as long as the alpha and beta subunits of the ATP synthase head could be held in the proper place, but otherwise it would not produce ATP. 25. What is the role of H+ translocation in generating ATP by ATP synthase? Proton translocation causes the c subunits in the base of the ATP synthase to rotate. This in turn rotates the gamma subunit, whose orientation of the gamma subunit with respect to the beta subunit determines the beta subunit’s conformation. 26. What are the components of the proton motive force? Mitochondrial membrane potential psi and the proton gradient. 27. What is the mobile electron carrier in the electron transport chain? Ubiquinone/Ubiquinol 28. Describe one way that the kinetics of the TCA cycle are regulated. High concentrations of ATP and NADH both inhibit different steps of the TCA cycle (both inhibit pyruvate to Acetyl CoA, isocitrate to alpha-ketoglutarate, and alpha ketoglutarate to succinyl Coa). High concentrations of ADP upregulate the isocitrate to alpha ketoglutarate step. 29. If a toxin inserts sodium-permeable channels into the inner mitochondrial membrane, what will happen to the proton motive force and why? The proton motive force will decrease significantly because the mitochondrial membrane potential will largely collapse. Recall that the voltage gradient produced by Na and K charge imbalance is more responsible for the proton motive force than the pH gradient. 30. What types of DNA are represented by the highly repeated, moderately repeated and non- repeated fractions of eukaryotic DNA? Highly repeated are things like satellite and microsatellite DNA and are relatively short, moderately repeated are things like critical and common proteins such as histones and proteins and RNA needed for ribosomal RNA, and non-repeated are most genes and their nearby regulatory regions. 31. How does DNA fingerprinting work? Many highly repeated DNA segments have variable copy numbers that can be used to identify individuals. DNA samples can be obtained from individuals. Each sample is then broken into fragments at specific sequences by restriction enzymes. Fragments are sorted and then amplified by PCR. Then the fragments are run out on a gel. Individual differences will show up as bands of different weights that will be consistent across samples from an individual but unique across individuals. 32. What does semi-conservative replication mean? This will not be on exam 2. From original (parent) double stranded DNA, two new copies are made that consist of one parent strand and one strand of new DNA. 33. Name three ways that a genome can change over generations. Insertion, deletion, duplication, transposition, mutation, inversion.

34. If you were to boil down your research article to a single hypothesis, what would it be? Draw a graph or figure that best summarizes how the article’s results address the hypothesis. This is specific to the research article. 35. If the  -clamp of the replisome is damaged, will the leading strand or lagging strand, both or neither be affected and why? If the beta clamp is damaged, both strands will be affected. The beta clamp attaches to DNA polymerase III and aligns the parent DNA strand. The gamma clamp is used To load the beta clamp and DNA polymerase onto the lagging strand for each new Okazaki fragment. 36. Draw and label the key components of replication on the leading and lagging strands. In addition to These, could also add the beta clamp and DNA polymerase, along with the gamma clamp loader. 37. How is DNA unwound, and then how are strands separated? Gyrase unwinds the DNA in prokaryotes, and helicase separates the DNA strands. 38. How is a nucleotide triphosphate added to the DNA chain? A polymerase adds a nucleotide triphosphate to the open 3’ OH group of a growing chain. 39. Which components are found in both DNA replication and DNA repair? Ligase and DNA polymerase I 40. During replication, which strand would contain more RNA and why? The lagging strand would contain more RNA because there would be an RNA primer for each Okazaki Fragment. 41. What are the functions of DNA polymerase I? It has three functions – removing the RNA primer 5 ’ to 3’ exonuclease, 3’ to 5’ exonuclease for error

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Correction, and DNA polymerase. 42. How are endonucleases used in nucleotide excision repair versus base excision repair, and how many bases are removed in each? In nucleotide excision repair, the endonucleases cut out a section of DNA. In base excision repair, The AP endonucleases cut the sugar phosphate at a single site where the nucleotide has been Removed by DNA glycosylases. 43. Describe the steps in base excision repair and nucleotide excision repair. Base excision repair: DNA glycosylase removes mismatched nucleotide AP endonucleases break sugar phosphate DNA polymerase beta inserts the correct nucleotide Ligase seals the nick in the backbone Nucleotide excision repair: UVRA/B sense the damage UVRB/C are endonucleases that cut out the section containing the mismatch or damage UVRD is a helicase to separate the DNA DNA polymerase I fills in the appropriate nucleotides DNA ligase seals the nick 44. Name a post-transcriptional modification in eukaryotes and state what role that it serves. 5’ methyl guanosine cap or 3’ polyadenylation tail, both of which prevent the mRNA from being Degraded in the cytoplasm. 45. Why are immature mRNAs much longer than mature mRNAs? Immature mRNAs contain exons, which will appear in the mature mRNA, and introns, which will not. 46. Where does RNA polymerase bind to initiate transcription in prokaryotes or eukaryotes? RNA polymerase binds where it is directed to by the sigma factor or another transcription factor. This is the start site. The RNA polymerase and sigma factor together are the holoenzyme. 47. Describe the process of transcription in prokaryotes The holoenzyme (RNA polymerase and sigma factor) binds at the DNA promoter region via the sigma Factor. mRNA polymerization begins, and after about 10 nucleotides, the sigma factor falls off. RNA polymerase continues until it reaches a termination sequence and then it dissociates. 48. What is a consensus sequence? A consensus sequence is a short sequence of nucleotides that is upstream of the start site that has a Specific sequence that can be recognized by the sigma factor. In bacteria, there is a sequence about 10 nucleotides upstream and another about 35 upstream, both of which are bound by the sigma Factor.

49. What process would be affected by alteration of the sigma factor? If the sigma factor has been altered, RNA polymerase would not bind specifically for transcription, but Randomly instead. The sigma factor identifies the promoter regions in prokaryotes.

Study questions for Bio 230_Exam_2_2022_answers

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