(考试资料下载)中南大学 生物化学期末考试试卷2006a卷修改

中南大学考试试卷20005-20006学年 二 学期 时间110分钟 生物化学 课程 64 学时 4 学分 考试形式： 闭 卷 专业年级：生物工程/生物技术04级 总分100分，占总评成绩70 % 注：1-4页不作答题纸，请将答案写在答题纸上Part I Filling Blankets (0.5 point each, 25 pts)Please fill each blanket with suitable word(s) either in English or Chinese1. Glycolysis provides the organism ( ) as well as energy.2. The two typical photosystems PSI and PSII for plant cells or cyanobacteria （蓝细菌） are ( ) and ( ), where the original electron donor is ( ) and the final electron acceptor is ( ). Besides ATP, the other substance produced by the photosystems is ( ).3. The two basic amino acids are ( ) and ( ), the two acidic amino acids are ( ) and ( ), and the three aromatic （芳香）amino acids are ( ), ( ) and ( ).4. The two epimers （异头体）of glucose in pyrano-forms are ( ) and ( ).5. The typical storage lipids are named ( ), the typical membrane lipids are ( ).6. For a typical membrane, when dipalmitoyl phosphtidylcholine （二棕榈酸磷脂酰胆碱）increases in the membrane, Tm value of the membrane will （ ）.7. Melting temperature of DNA may ( ) with increase in electrolyte concentration or GC content.8. The two major products of pentose phosphate pathway are ( ) and ( ).9. Proline can occur in the ( ) of an a-helix of polypeptide.10. The two standard amino acids that have not chairal (手性) carbon atoms are ( ) and ( ).11. The membrane proteins can be classified into three types, they are ( ), ( ) and ( ).12. The amino acids that link to the sugar units either by O- or N-linkage in proteoglycans （蛋白多糖）or glycoproteins （糖蛋白）, typically include ( ), ( ), and ( ). 13. a-keratins have a-helix, b-keratins have ( ). 14. The typical five sugars repeat unit in the core structure of oligosaccharides （寡聚糖） in N-linked glycoproteins include two molecules of ( ) and three molecules of ( ).15. In the reversible competitive inhibition, the inhibitor binds to the ( ) site of the enzyme, and in the reversible noncompetitive inhibition, the inhibitor binds to the ( ) site of the enzyme. Both these two inhibitions are by way of ( ) binding. 16. The two mobile electron carriers in the electron transport chain are ( ) and ( ).17. In the K-type of inhibition, the apparent Km will ( ), though the Vmax keeps unchanged.18. The anticodon of AUG is ( ).19. The general procedure for the transformation of atmosphere N2 into organic substance is (1) ( ), (2) assimilation of NH3 into amino acids, and (3) biotransfer of amino acids (normally beginning with glutamate) into other organic substance. 20. The Calvin cycle is responsible for the fixation of ( ).21. DNA is always synthesized in the ( ) direction.22. In a typical cycle of b-degradation of C16 saturated fatty acids, it includes two steps of oxidation, in which the electron carriers or cofactors are respectively ( ) and ( ).23. Carboxylation （羧化）enzymes often require the prosthetic group (辅基) ( ).24. Transamination (转氨作用) with tansaminase needs an prosthetic group ( ).25. The most representative molecules with high energy include ( ), and ( )，besides the NTPs.26. The common substance entering into TCA cycle for b-oxidation of fatty acids and glycolysis is ( ).Part II Structural details for some typical biomolecules (15 pts)1. If a physiological phosphatidylcholine （磷脂酰胆碱）consists of palmitic and linoleinic acids（棕榈酸和亚麻酸）, please write the physiologically correct structure, if you take P and L as palmitic and linoleinic acids, respectively. （2分）2. Please give the name for the following oligosaccharide (寡糖)，in which you should give the glycolinkage forms, steric configuration (构型) as well as the names of the monosaccharide residues （单糖残基）.（3分）3. 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)4. Please give the nucleotide sequence of the DNA template strand from which the following mRNA segment was transcribed: 5-UAGUGACAGUUGCGAU-3. （2分）5. Please write the open reading frame for mRNA of typical prokaryotic cells. （2分）6. Please write the typical structure of Lac operon （乳糖操纵子）. （3分）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 ? (3分）2. Please compare the differences in characteristics between passive transport and active transport. （4分）3. What are the differences in physiological functions between NADH and NADPH? （4分）4. What are major fates of the final product of glycolysis? （3分）5. 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）6. What are the typical types of the secondary structures of proteins, and what force(s) for the secondary structures of proteins? （5分）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)8. 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)Part 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)（2） What are the entry and exit substance, respectively, for this pathway. （1分）（3） What specific enzymes are used for steps 2 and 3, respectively? （1分）（4） Why citrate needs to be transferred into isocitrate before it is oxidized ? （1分）（5） In step 6, why the cofactor FAD is used as the electron acceptor. （1分）（6） What are the major control steps? （1分）（7） How do the high concentrations of ATP and NADH regulate the pathway, why ? （1分）（8） Explain the important role for this pathway. （1分）（9） How many ATPs can be formed from one molecule of pyruvate by way of TCA cycle and oxidative phosphorylation. (2分)2. Please explain the mechanism of covalent modification and allosteric regulation of glycogen phosphorylase （糖原磷酸水解酶）according to the figure below, in which the phosphorylase a is more active than the phosphorylase b . （5分）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 doe