(考試資料下載)中南大學(xué) 生物化學(xué)期末考試試卷2006a卷修改（答案)

中南大學(xué)2005-2006學(xué)年第二學(xué)期考試試卷(A)答案Exam paper for the 2nd term of 2005-2006 at Central South UniversityCourse: Biochemistry；Specialties：Bioengineering, BiotechnologyPart I Filling Blankets (0.5 point each, 25 pts)Please fill each blanket with suitable word(s) either in English or Chinese1. (some intermediates)2. ( PS700 ) ( PS680 ) （H2O） ( NADP+ ) ( NADPH or O2 ).3. ( lysine 賴氨酸 ) ( arginine 精氨酸 ) ( glutamic acid 谷氨酸 ) ( aspartic acid 天冬氨酸 ) ( phenylalanine 苯丙氨酸 ) ( tryptophan 色氨酸 ) ( Tyrosine 酪氨酸 ) 4. ( a-D-pyranoglucose，a-D-吡喃葡萄糖 ) (b-D-pyranoglucose b-D-吡喃葡萄糖) 5. ( triacylglycerols 三酰甘油 ) ( phosphatidylglycerols 磷脂酰甘油 ) 6. （increase 增加 ） 7. ( increase 增加 ) 8. ( ribose-5-phosphate 核糖-5-磷酸 ) ( NADPH ).9. ( terminal 末端 ) 10. ( glycine 甘氨酸 ) ( proline 脯氨酸 ) 11. T ( membrane integral proteins 膜整合蛋白 ) ( membrane peripheral proteins 膜周邊蛋白 ) (membrane anchoring proteins 膜錨定蛋白 ) 12. ( serine 絲氨酸 ) ( threonine 蘇氨酸 ) ( asparagine 天冬酰胺 ) 13. ( b-sheet b-折疊片) 14. T ( N-acetylglucose or GlcNAc, N-乙酰葡萄糖胺 ) ( mannose 甘露糖 ) 15. ( active 活性 ) ( Non-active, or allosteric 變構(gòu)) ( non-covalent 非共價(jià) ) 16. ( ubiquinone or UQ ) ( Cytochrome c , or Cyt c) 17. ( change ) 18. ( CAU ) 19. (N2 fixation to NH3 ) 20. ( CO2 ) 21. ( 5-3 ) 22. ( FAD ) ( NAD+ ) 23. ( biotin 生物素 ) 24. ( pyridoxal phosphate, pyridoxal or vitamin B6吡哆醛或維生素B6, 答對(duì)任一項(xiàng)均算對(duì)) 25. ( acyl-CoA 酰基-輔酶A, or acetyl-CoA 乙酰輔酶A) ( creatine phosphate 肌氨酸磷酸) 26. ( acetyl-CoA 乙酰輔酶A ) Part II Structural details for some typical biomolecules (15 pts)1. If a physiological phosphatidylcholine consists of palmitic and linoleinic acids, please write the physiologically correct structure, if you take P and L as palmitic and linoleinic acids, respectively. （2分）Answer: where, P: palmitic acid, L: linoleinic acid2. Please give the name for the following oligosaccharide (寡糖)，in which you should give the glycolinkage forms, steric configuration (構(gòu)型) as well as the names of the monosaccharide residues （單糖殘基）.（3分）Answer: Gal b(16) Glc a(14) Glc3. Please complete the following reaction by adding the fitful reactant(s), product(s), enzyme and cofactor on the suitable side of the equation. Known that this reaction is the oxidation step of glycolysis (糖酵解). (3分)Answer: the enzyme for this reaction is glyceroaldehyde 3-phosphate dehydrogenase, it require Mg2+, NAD+ and phosphate for the reaction.4. Please give the nucleotide sequence of the DNA template strand from which the following mRNA segment was transcribed: 5-UAGUGACAGUUGCGAU-3. （2分）Answer: 3-ATCACTGTCAACGCTA-55. Please write the open reading frame for mRNA of typical prokaryotic cells. （2分）Answer: Begin with AUG, end with UAA, UAG or UGA6. Please write the typical structure of Lac operon （乳糖操縱子）. （3分）Answer: Part III Short essay questions (30 pts)Please concisely answer the following short essay questions in English or Chinese.1. What is the key enzyme included in the Calvin cycle, and how does it respond to the light imposed on the photosynthetic organelles ? (2分）Answer: the key enzyme is ribulose-1,5-bisphosphate carboxylase (Rubisco), it can be activated by light imposing on the chloroplasts.2. Please compare the differences in characteristics between passive transport and active transport. （4分）Answer: Passive transport: two modes that are simple diffusion and facilitated diffusion, no need of energy, transport along the gradient of the solute concentration, specific transporter for the facilitated diffusion; Active transport: needs energy and specific transporter, transport against the gradient of solute concentration3. What are the differences in physiological functions between NADH and NADPH? （4分）NADH is the reductive electron donor, which occurs normally in the catabolic pathways, oxidation of which can produce energy by oxidative phosphorylation; NADPH is reductive electron donor that occurs in anabolic pathways and provides the cells reducing environment, it is produced by the pentose phosphate pathway and photosynthetic pathway.4. What are major fates of the final product of glycolysis? （3分）Answer: By anaerobic pathways to form ethanol + CO2, or lactic acid; by aerobic pathway, it goes into the TCA cycle after it is degraded into acetyl-CoA5. An oligopeptide (寡肽) with a sequence of NH3+-SIKDYEFRMP-COO-, please point out the sequence of the peptide in amino acids names. （5分）Answer: the amino acids sequence is：Ser-Ile-Lys-Asp-Tyr-Glu-Phe-Arg-Met-Pro6. What are the typical types of the secondary structures of proteins, and what force(s) for the secondary structures of proteins? （5分）Answer: Typical secondary structures: a-helix, b-sheet, b-turn, b-bulge; and the super secondary structures: aa, bb, bab. The major force for these structures is hydrogen bonding.7. Determine the number of ATP equivalents needed to form palmitic acid from acetyl-CoA. (Assume for this calculation that each NADPH is worth 3.5ATP.)? （3分）(you should describe the reason for that determination)Accoding to the general reaction: 8 acetyl-CoA + 7ATP4- + 14 NADPH + 7H+ Palmitoyl-CoA + 14NADP+ +7 CoASH + 7ADP3- + 7Pi2-The ATPs required for synthesis 1mole of palmitoyl-CoA is 14*3.5+7=568. What are the fates for oxaloacete (or what pathways require oxaloacetate 草酰乙酸)? （4分）(you should describe the relationship between some pathways that require or are related to oxaloacetate)Answer: oxaloacetate in included directly in gluconeogenesis, TCA cycle, glutamate-dependent transamination for synthesis amino acids; and included indirectly in urea cyclePart IV Complex essays ( 15 pts)1. Please answer the following questions according to the figure below (10分). （1） This pathway is TCA cycle, What does TCA mean ? (1分)Answer: TCA: Tricarboxylic acid cycle, or the citric acid cycle（2） What are the entry and exit substance, respectively, for this pathway. （1分）Answer: acetyl-CoA, CO2（3） What specific enzymes are used for steps 2 and 3, respectively? （1分）Answer: aconitase （順烏頭酸酶），isocitrate dehydrogenase(異檸檬酸脫氫酶)（4） Why citrate needs to be transferred into isocitrate before it is oxidized ? （1分）Answer: because citrate is much more difficult to be dehydrogenated.（5） In step 6, why the cofactor FAD is used as the electron acceptor. （1分）Answer: because breakage of C-H have enough energy only for formation of FADH2, instead of NADH.（6） What are the major control steps? （1分）Answer: the major control steps are step 1, 3, 5 and the entry step（7） How do the high concentrations of ATP and NADH regulate the pathway, why ? （1分）Answer: high concentration of ATP and NADH inhibit the pathway, because NADH is the product of this cycle, and ATP is the final products of the degraded pathway.（8） Explain the important role for this pathway. （1分）Answer: This pathway is the key metabolic pathway which links degradation pathways such as glycolysis, b-oxidation of fatty acids and degradation of peptides with oxidative phosphorylation to produce energy and intermediates for the organism.（9） How many ATPs can be formed from one molecule of pyruvate by way of TCA cycle and oxidative phosphorylation. (2分)Answer: By way of TCA cycle, 1pyruvate produce 4 NADH, 1FADH2, 1GTP, so the total ATPs that can be produced are 4*2.5+1*1.5+1=12.5ATPs (considering the consumption of H+ gradient due to ADP/ATP transport through the inner membrane of MT) or 4*3+1*2+1=15ATPs (no consideration of the consumption of H+ gradient due to ADP/ATP transport through the inner membrane of MT)2. Please explain the mechanism of covalent modification and allosteric regulation of glycogen phosphorylase according to the figure below, in which the phosphorylase a is more active than the phosphorylase b . （5分）Answer: The phosphorylase can be regulated by both covalent modification and allosteric regulation, where the allosteric regulation is more effective for the less active phosphorylase b, and when the less active phosphorylase b is phosphorylated, it become more active phosphorylase a, which can directly transferred from the tight state into relax state, high concentration of glucose can allosterically inhibit the phosphorylase a.Part V Calculations and Sequence Analysis (15 pts)1. For a Michaelis-Menten reaction, k1=7107/(mol/L sec), k-1=1103/sec, and k2=2104/sec. What is the value of Km? Does substrate binding approach equilibrium or does it behave more like a steady-state system? If Vmax=100 mmol/mL sec, what is the velocity of the reaction when S=20mmol/L ? (6分) Resolution: Km=(k-1+k2)/k1=(0.1+2)*104/(7*107)=3*10-4 (mol/L)It behaves like a steady-state system, because k2 is 20 times greater than k-1.V=VmaxS/(Km+S)=1