中山大学生化上考试往年试题完整版

生物化学I （2007年秋季学期, A 卷）专业(班级)： 姓名： 学号：任课教师：郑利民 李国富 阅卷教师：《中山大学授予学士学位工作细则》第六条：“考试作弊不授予学士学位”。

I.填空题（共15小题，每小题2分，共30分）1. pK 1、pK 2、pK R 分别代表氨基酸的α羧基、α氨基和侧链基团的解离情况，则Lys 的pI ＝ ，Glu 的pI ＝ .2. 维持蛋白质高级结构的非共价作用力包括氢键、 、 等.3. 根据蛋白质的二级结构组成，可将其分为all β、all α、 、 四类.4. α角蛋白的二级结构是 ，含有大量带疏水侧链的氨基酸；同时，α角蛋白还含有丰富的极性侧链氨基酸 .5. 蛋白质体内折叠需要的辅助分子有分子伴侣、 、 等.6. 被解析晶体结构的第一个蛋白是 ，其辅基是 .7. 在抗体基本结构的标示中，H 和L 代表 ，V 和C 代表 .8. 生物大分子结构测定的主要方法有 、 、电子显微镜等.9. 酶在体内的活性调节方式有 、 、酶的切割激活等.10. 列举两个蛋白水解酶家族 、 .11. 糖原和淀粉由葡萄糖通过 糖苷键连接；纤维素由葡萄糖通过 糖苷键连接.12. 寡糖与蛋白的O 连接指 ，N 连接指 ；13. 指示某段DNA 处于功能活动状态的特殊结构有Z-DNA 、 、 等.14. 在Sanger 法DNA 自动测序中，荧光基团可以标记在 上或标记在 上.15. 结构性脂类包括磷脂、糖脂和 ，有些磷脂、糖脂的骨架不是甘油而是 ，因此这部分脂又被称为鞘脂.II. 简答题（每小题6分，共30分，1-5题选做3题，6-9题选做2题）1.有那些氨基酸残基在蛋白质功能中发挥重要作用？列举三例. 2.为什么酶能降低反应的活化能？ 3.解释Ach 受体通道的工作机制（可结合草图说明）. 4.细胞膜表面的寡（多）糖有何生物学功能? 5.物质跨膜运输的类型有那些？ 6.根据肌肉收缩的分子机制，解释“尸僵”现象. 7.朊病毒引起的疾病又称为“蛋白质构象病”，请从蛋白质折叠的热力学角度解释它的致病机理. 8.有一多肽，其一级结构可能是直线式的也可能是闭环式的. 你如何尽可能简单地做出鉴别? 9. 一级结构决定三维结构，那么如何理解三维结构相似（同）的同源蛋白一级结构的较大差别？ III. A protein of unknown structure has been purified and is subjected to gel filtrationchromatography under different conditions, with the following results: 1) Chromatography under native conditions suggests a molecular weight of 240,000 from comparison to standards of known molecular weight chromatographed on the same column; 2) Chromatography in the presence of 5 M urea yields a single protein peak corresponding to a M.W. of 60,000 from comparison to standards of known molecular weight chromatographed on the same column in urea.1. What information do the results of (1) and (2) give about the structure of the protein? Describe thestructure as completely as the data permit;2. On an SDS polyacrylamide gel, the pure proteinshows two stained bands. The relative mobilities ofthose 2 "unknown" bands are shown on the plotbelow as open circles (O), with the mobilities of 5known polypeptide standards of M.W. 15,000;20,000; 30,000, 40,000 and 50,000. Given the resultsof the SDS gel, how would you alter your descriptionof the protein structure from the description you gavein part 1? Be as specific as you can about yourrevised description of the structure, and explain thereason for the altered description.IV. Lysozyme destroys the cell wall of bacteria by catalyzing the hydrolysis of 1,4-β-linkages between N-acetylmuramic acid (NAM) and N-acetyl-D-glucosamine (NAG) residues in a peptidoglycan.Six hexoses of the substrate fit into the active site of lysozyme, a deep cleft across part of the globular protein surface this cleft. Figure A shows the hydrolytic process of 1,4-β-linkages between NAM and NAG and figure B shows the pH-activity profile of lysozyme. Please calculate the ratio of the enzyme molecules which really work at pH 3.8, 5.2, 6.6 to all enzyme molecules, and explain figure B according to your calculation (pH = pKa + lg([A-]/[HA])；pKa Glu = 5.9, pKa Asp =4.5；102.1≈126，100.7≈5).A B生物化学I（2006年秋季学期, A卷）专业(班级)：姓名：学号：任课教师：郑利民李国富阅卷教师：题号I II III IV V VI 合计《中山大学授予学士学位工作细则》第六条：“考试作弊不授予学士学位”。

I.简答题（共25小题，每小题4分，超过50分以50分计入总分）1．哪些标准氨基酸含两个手性碳原子？写出它们的Fisher 投影式。

（4′）2．为什么SDS 变性电泳可以估算蛋白质的分子量？（4′）3．Mass spectroscopy 在生物大分子研究中有什么应用？（4′）4． -Keratin 、Collagen 、Silk fibroin 的二级结构和氨基酸组成有何特点？（4′）5．根据二级结构的组成和排列可将蛋白质结构分为哪些类型？（4′）6．蛋白质in vivo 条件下的折叠有哪些辅助因子？（4′）7．蛋白质在与其他分子的相互作用中发挥作用，这种相互作用有何特点？（4′）8．哪些因素可以导致Hemoglobin 与O 2的亲和力增加？（4′）9．图示抗体的一般结构（包括结构域及可能的二硫键）。

（4′）10．构成肌肉粗丝和细丝的蛋白分子有哪些，各有什么功能？（4′）11．写出蛋白质氨基酸测序的基本步骤。

（4′）12．什么是酶催化的过渡态理论？列出你所知道的支持证据？（4′）13．酶的可逆抑制剂有几类？它们的抑制动力学各有何特点？（4′）14．酶在体内的活性是如何受到调控的？（4′）15．淀粉与纤维素的结构有何区别？（4′）16．细胞膜表面的Glycoprotein 和Glycolipid 有哪些生物学功能？（4′）17．细胞内的DNA 可能会出现哪些异常结构，他们有什么生物学意义？（4′）18．膜脂的分类及其结构特点（4′）19．生物膜的功能有哪些？（4′）20．膜蛋白以哪些方式参与生物膜的构成？（4′）21．图示secondary active transport ，uniport ，symport ，antiport. （4′）22．Acetylcholine receptor 通道是如何控制开关的？（4′）23．K ＋离子通道是如何对通过的离子进行选择的？（4′）24．膜电位是怎样控制Na ＋通道开关的（可图示并说明）？（4′）25．Ca 2+泵和Na +-K +泵的异同点。

（4′）II. You have a crude lysate sample containing a mixture of six proteins (1, 2, 3, 4, 5, andβ-galactosidase). Some characteristics of these proteins are shown in the table below. Give an available procedure of purifying β-galactosidase. (10 pts)III. You are working on a compound as a potential drug that inhibits an overproducedenzyme whose uncontrolled activity leads to uncontrolled cell proliferation (cancer). A steady-state kinetic analysis of the enzyme-catalyzed reaction was carried out, with velocities measured as a function of the concentration of the substrate in the absence and in the presence of 10 nM inhibitor. Theresults of the kinetics experiment are plotted for you below. Please answer the following questions. (10 pts)1. What type of inhibitor is this compound (be as specific as you can)? Show thediagram of the inhibitor ’s mechanism (by E denoting enzyme, S denoting substrate, I denoting inhibitor; for example: ).2. Calculate the K m , K cat in the absence of inhibitor and K m app , K cat app in the presenceof inhibitor (app denoting apparent. The total concentration of enzyme is 2nM).3. Calculate the inhibitor constant K I .IV. Lysozyme destroys the cell wall of bacteria by catalyzing the hydrolysis of1,4-β-linkages between N-acetylmuramic acid (NAM) and N-acetyl-D-glucosamine (NAG) residues in a peptidoglycan (figure A). Figure B shows the binding of the peptidoglycan to the active site of lysozyme, a deep cleft across part of the globular protein surface. Six hexoses of the substrate fit into this cleft. Figure C shows the hydrolytic process of 1,4-β-linkages between NAM and NAG and figure D shows the pH-activity profile of lysozyme. Please answer the following questions. (10 pts)1. What can you talk about the figure B in accord with the enzymatic mechanism?2. What can you talk about the figure C in accord with the enzymatic mechanism?3.Calculate the ratio of the enzyme molecules which really work at pH3.8, 5.2, 6.6 toall enzyme molecules, and explain figure D according to your calculation (pH = pKa + lg([A-]/[HA])；pKa Glu = 5.9, pKa Asp = 4.5；102.1≈126，100.7≈5).Figure CFigure AFigure BFigure DName Class ScoreBiochemistry (I) Final Exam (fall, 2004)NOTE: You must write your answer on the answer sheet!Part I: For the following multiple choice questions, one answer is most appropriate (2.5 18 = 45 points).1. With regard to amino acids, which statement is false?A) Amino acids can act as proton donors and acceptors.B) All amino acids discovered in organism are L enantiomers.C) An L amino acid can be Dextrorotary.D) A conjugate acid/base pair is at its greatest buffering capacity whenthe pH equals its pK.E) Non-standard amino acids can be found in the hydrolysis product of aprotein.2. A mixture of Ala, Arg, and Asp in a pH 5.5 buffer was placed on a cationexchange column (the column is negatively charged) and eluted with the same buffer. What is the order of elution from the column? Use these pK a values: terminal COOH - 2, terminal NH3+- 9, R-amino - 10, R-COOH - 3A) Arg, Ala then AspB) Arg, Asp then AlaC) Asp, Ala then ArgD) Asp, Arg then AspE) Ala, Asp then Arg3. Proteins can be chromatographically separated by their differentA) Charge.B) Molecular weight.C) Hydrophobicity.D) Affinity for other molecules.E) All of above.4. A peptide was found to have a molecular mass of about 650 and uponhydrolysis produced Ala, Cys, Lys, Phe, and Val in a 1:1:1:1:1 ratio. The peptide upon treatment with Sanger's reagent (FDNB) produced NP-Cys and exposure to carboxypeptidase produced valine. Chymotrypsin treatment of the peptide produced a dipeptide that contained sulfur andhas a UV absorbance, and a tripeptide. Exposure of the peptide to trypsin produced a dipeptide and a tripeptide. The sequence of the peptide isA) val-ala-lys-phe-cysB) cys-lys-phe-ala-valC) cys-ala-lys-phe-valD) cys-phe-lys-ala-valE) val-phe-lys-ala-cys5. With regard to protein structure, which statement is false?A) The dominant force that drives a water-soluble protein to fold ishydrophobic interaction.B) The number of hydrogen bonds within a protein intends to beminimized.C) The conformations of a native protein are possibly the lowest energystate.D) The conformations of a native protein are countless.E) Disulfde bridges can increase the stability of a protein.6. Which structure is unique to collagen?A) The alpha helix.B) The double helix.C) The triple helix.D) The beta structure.E) The beta barrel.7. Which protein has quaternary structure?A) Insulin.B) A natural antibody.C) Chymotrypsin.D) Aspartate transcarbamoylase (ATCase).E) Myoglobin.8. Which of the following are "broad themes used in discussing enzymereaction mechanisms"?A) Proximity stabilizationB) Transition state stabilizationC) Acid-base catalysisD) Covalent catalysisE) All of the above9. Under physiological conditions, which of the following processes is not animportant method for regulating the activity of enzymes?A) PhosphorylationB) Temperature changesC) AdenylationD) Allosteric regulationE) Protein processing10. The conversion of glucose to pyruvate is a multistep process requiring tenenzymes. If a mutation occurs resulting in a lack of activity for one of these enzymes, which of the following happens?A) The concentration of the metabolic intermediate which is the substrateof the missing enzyme is likely to increase and accumulateB) The concentration of pyruvate will increaseC) The cell will produce more of the other nine enzymes to maintainsteady stateD) The concentration of the metabolic intermediate which is the productof the missing enzyme will decreaseE) A and D11. Indicate which is true about enzymes.A) Enzymes are permanently changed during the conversion of substrateinto product.B) Enzymes interact irreversibly with their substrates.C) Enzymes change the energy difference between substrates andproducts.D) Enzymes reduce the energy of activation for the conversion of reactantinto product.E) Enzymes increase the energy content of the products.12. Consider a reaction as follows:A +B <====>C + D, ΔG'0 = +10.0 Kj/molHow can this reaction go forward?A) Coupling of this reaction to a highly exergonic reaction through acommon intermediate.B) Immediate removal of the products.C) Addition of the appropriate enzyme.D) A and B.E) All of the above.13. Which statement is not true of enzyme inhibitors?A) A competitive inhibitor binds the active site of the enzyme.B) Allosteric enzymes exhibit Michaelis-Menton kinetics.C) A noncompetitive enzyme binds elsewhere than the active site.D) Noncompetitive inhibitors cannot be completely overcome by addingmore substrate.E) Competitive inhibition can be completely overcome by adding moresubstrate.14. You could most dramatically destabilize a duplex DNA molecule in anaqueous solution byA) Decreasing the temperature.B) Decreasing the concentration of salt.C) Decreasing the concentration of organic solvents.D) Adjusting the pH to a value near 7.E) Adding a detergent.15. With regard to DNA, which statement is false?A) Triple helix and tetraplex tend to appear at sites where replication,recombination or transcription is initiated or regulated.B) Hydrogen bonds are the only force to stabilize DNA structure.C) The absorbance of DNA at 260 nm will rise when it is denatured.D) Z-DNA tracts may play an role in the regulation of the expression ofsome genes or in genetic recombination.E) B-DNA is the most stable structure in physiological condition.16. With regard to biological membrane, which statement is false?A) The ratio of proteins to lipids is various in different membranes.B) The membrane with more proteins has more complicated function thanthat with fewer proteins.C) Each membrane has a characteristic composition of lipids.D) Lipids symmetrically distribute in the two leaflets of a membrane.E) The flip-flop movement of lipids is done by a kind of enzyme.17. Which of the following statements are true about the interactions betweensubstrate and active site? You can assume that the catalysis goes to completion.① Catalytic residues can be affected by pH② The majority of bonds in E-S interactions are covalent bonds such asdisulfide and peptide bonds③The components of a mixture of the two mirror-image isomers of acompound, called a racemic mixture, will be acted upon equally byan enzyme④ The active site can alter itself to better fit the substrate⑤ The active site is permanently altered by its interaction with substrate⑥ The substrate is permanently altered by its interaction with active site⑦ Competitive inhibitors have no affect on the active siteA) ①, ②, ③B) ②, ④, ⑤C) ③, ④, ⑤D) ②, ④, ⑦E) ①, ④, ⑥18. Which of the following statements refer to glycogen ("G") and whichrefer to cellulose ("C")?① Contains beta-1,4 glycosidic bonds② Cannot be digested by mammalian enzymes③ Straight-chain molecule of high tensile strength④ Contains alpha 1,6-glycosidic bonds⑤ Branched molecule⑥ Used as a structural component⑦ Used as a nutritional storage moleculeA) G= ①, ④, ⑦; C= ②, ③, ⑤, ⑥B) G= ④, ⑤, ⑦; C= ①, ②, ③, ⑥C) G= ②, ③, ④, ⑥; C= ①, ⑤, ⑦D) G= ⑤, ⑥; C= ①, ②, ③, ④, ⑦E) G= ①, ②, ④; C= ③, ⑤, ⑥,⑦Part II:According to the figures and conditions provided, fill the blanks (Fewer have two or more answers, 55 points).19. Protein secondary structures and their organization (5 points).A)Parallel β-pleated sheet;B)Antiparallel β-pleated sheet;C)Residues of Pro and Gly;D)Residues of Pro and Val;E)Residues of hydrophilic amino acids;F)Residues of hydrophobic amino acids;G)Residue of Arg, Lys, or His;H)Residue of Asp, or Glu.a)One domain of the protein is composed ofb)What amino acids are frequently found in 5c)If the residues at 6 is positively charged, then the residue at 7 is bestd)What residues occur mostly at 1 and 2?e)What residues occur mostly at 3 and 4?20. Several oxygen dissociation curves are shown in the following figure. (5points).Assume that curve C corresponds to purified hemoglobin placed in a solution containing physiological concentrations of CO2 and BPG at a pH of 7.0. Below, indicate which of the curves reflects the changes noted in physiological conditions by placing the curve # on the lines below (Use only one curve choice/physiological change):a. Decreased CO2 concentration______________b. Increase BPG concentration_______________c. Dissociation of hemoglobin into subunitsd. A mutation that eliminates ion pairs that stabilize the T ordeoxygenated statee. Fetal hemoglobin or a mutation which increase binding affinity foroxygen21. Supersecondary structure of proteins (5 points).A) β-α-β Loop;B) Greek Key;C) α-α Corner;D) Right-handed connection between β strands;E) Not observed;F) Very rare.a. b. c. Dd. e. f.22. Protein classification (5 points).A) All α; B) All β: C) α/β; D) α+β; E) Superfamily; F) Speciesa. Ad. e. f.23. Kinetics of enzyme inhibition or double-substrate reaction (5 points).a. Competitive inhibition Ab. Non-competitive inhibitionc. Uncompetitive inhibitiond. Mixed competitive inhibitionIf [S2] and ↓ substitute for [I] and ↑ respectively, then:e. Double-substrate enzyme reaction involving a ternary complexf. Double-substrate enzyme reaction with a ping-pong mechanism24. Structure about Nucleotides and DNA (5 points)a.Exo-2´ / Endo-2´configuration of riboseb.Exo-3´ / Endo-3´configuration of ribosec.Hoogsteen Pairsd.Can form cruciform structuree.Can form H-DNA25. Biomembrane lipids (5 points)a.Sphingolipidsb.Phospholipase A1c.Phospholipase A2d.Phospholipase Ce.Sphingosine26. Classes of membrane transports (5 points)a.Passive-transporters Cb.Primary Active transportc.Secondary Active transportd.Channelse.Pumpsf.Active co-transporters27. Mechanism of Ach-gated ion channel (left, a cross section at the gate) orvoltage-gated Na+ channel (right, vertical views).A)Inactive / activateable state of Na+ channel;B)Inactive / inactivateable state of Na+ channel;C)Inactivateable state of Na+ channel;D)Active state of Na+ channel;E)The 2nd tansmembrane helix of α, β, γ, or δ subunit;F)The 3rd tansmembrane helix of α, β, γ, or δ subunit;G)Small polar amino acid residue;H)Large non-polar amino acid residue;a)b)c)d) De)f)g) B28. Biosignalinga)Ligand-gated channel pathway Ab)Receptor enzyme pathwayc)G protein coupled pathway Cd)Nuclear receptor pathwaye)The pathway(s) can act in synapse transmission Af)The pathway(s) can regulate gene expressiong)The pathway(s) can regulate the concentration of glucose in bloodh)It can phosphorylate Tyr, Ser, or Thr of proteinsi)Molecules including cAMP, cGMP, IP3 etc.j)This signal molecule may be Ach Ek)This signal molecule is usually a steroid hormonel)This signal molecule may be insulinm)This signal molecule may be epinephrine Gn)The subunit(s) of G protein is(are) very conservativeo)The subunit(s) of G protein is (are) variousp)The subunit(s) of G protein can bind GDP &GTP LAnswer Sheet for Biochemistry (I) Final Exam (fall, 2004) Name Class ScorePart I1. 2. 3. 4. 5.6. 7. 8. 9. 10.11. 12. 13. 14. 15.16. 17. 18.Part II19: a) b) c) d) e)20: a) b) c) d) e)21: a) b) c) D d) e)f)22: a) a b) c) d) e)f)23: a) a b) c) d) e)f)24: a) b) c) d) e)25: a) b) c) d) e)26: a) c b) c) d) e)f)27: a) b) c) d) D e)f) g) B28: a) A b) c) C d) e) Af) g) h) i) j) Ek) l) m) G n) o)p) L生物化学I （2008年秋季学期, A 卷）1. Lysozyme destroys the cell wall of bacteria by catalyzing the hydrolysis of 1,4-β-linkages between N-acetylmuramic acid (NAM) and N-acetyl-D-glucosamine (NAG) residues in a peptidoglycan.. Figure A shows the hydrolytic process of 1,4-β-linkages between NAM and NAG, and figure B shows the pH-activity profile of lysozyme. (1) Describe figure A related to the catalytic mechanism of enzyme; (2) Calculate the ratio of the enzyme molecules which really work at pH 5.2 to all enzyme molecules, and say some thing possible for why it is not of 100% at the optimal pH (pH = pKa + lg([A -]/[HA])； pKa Glu = 5.9, pKa Asp = 4.5；100.7≈5; 12 points).（1）图1显示溶菌酶水解peptidoglycan 中的糖苷键时的酸碱催化和共价催化机理【1分】，其中Glu-35起酸催化和共价催化作用【2分】，Asp-52起碱催化作用【2分】（2）分子比例：【方法正确但计算错误得3分，方法和计算均正确得5分】lg([Asp-]/[HAsp]) ＝ pH – pKa ＝ 5.2 – 4.5 = 0.7，[Asp-]/[HAsp] ＝ 5起碱作用的Asp 所占比例: [Asp-] /（[HAsp]＋[Asp-]）＝ [Asp-]/（ [Asp-]/5＋[Asp-]）＝ 5/6 lg([Glu-]/[HGlu]) ＝ pH – pKa = 5.2–5.9 ＝ -0.7，[Glu-]/[HGlu] ＝ 1/5起酸作用的Glu 所占比例: [HGlu]/（[HGlu]＋[Glu-]）＝ [HGlu]/（[HGlu]＋[HGlu] /5）＝ 5/6 真正能工作的酶分子比例＝（5/6）×（5/6）≈ 69.4%（3） pH5.2是该酶的最佳pH 值，但工作的酶分子比例仅为69.4％而非100％。