清华大学生物化学试卷2005_final_exam_A

清华大学生物试题及答案一、选择题（每题2分，共20分）1. 细胞膜的基本骨架是：A. 糖蛋白B. 磷脂双分子层C. 蛋白质D. 核酸答案：B2. 叶绿体中光合作用的主要色素是：A. 叶绿素B. 胡萝卜素C. 叶黄素D. 藻蓝素答案：A3. DNA复制的起始点称为：A. 复制起点B. 复制终点C. 复制中心D. 复制终止点答案：A4. 细胞周期中，DNA复制发生在：A. G1期B. S期C. G2期D. M期答案：B5. 以下哪个不是真核细胞的细胞器？A. 核糖体B. 线粒体C. 高尔基体D. 质粒答案：D6. 基因表达调控的主要方式是：A. 转录后修饰B. 转录调控C. 翻译调控D. 翻译后修饰答案：B7. 以下哪个是原核生物？A. 酵母菌B. 大肠杆菌C. 草履虫D. 蘑菇答案：B8. 以下哪个是细胞凋亡的特征？A. 细胞膨胀B. 细胞膜破裂C. 细胞核浓缩D. 细胞质凝聚答案：C9. 蛋白质合成的主要场所是：A. 核糖体B. 内质网C. 高尔基体D. 线粒体答案：A10. 细胞呼吸的主要场所是：A. 细胞质B. 线粒体C. 内质网D. 高尔基体答案：B二、填空题（每空1分，共20分）1. 细胞膜的流动性主要取决于膜中________的含量。

答案：磷脂2. 在有丝分裂过程中，染色体的分离发生在________期。

答案：M3. 真核细胞中，mRNA的帽子结构位于________端。

答案：5'4. 细胞周期的调控主要依赖于________蛋白。

答案：周期依赖性5. 细胞凋亡是由________蛋白家族介导的程序性细胞死亡。

答案：Caspase6. 细胞内蛋白质合成的主要场所是________。

答案：核糖体7. 细胞内蛋白质的折叠和修饰主要发生在________。

答案：内质网8. 细胞膜上的________蛋白负责细胞间的识别和粘附。

答案：糖蛋白9. 细胞内能量的主要储存形式是________。

清华大学XX年生物化学1本科期末考试试题考试科目：生物化学考试时刻：考试类型：期末试题I.1: how many carbons does Arachidic acid have (20 carbons)2: how many double bonds does Arachidonic acid have (4 double bonds)3: list two advantages that fats have over sugars as stored fuels (more energy gram for gram; no hydration needed)4: where inside the cells are most of the phospholipids degraded (lysosomes)5: oligosaccharide head groups determine the blood type of an individual. How are they attached to the plasma membrane (glycosphingolipids or lipids and surface proteins)6: list at least one genetic disease that could result from abnormal accumulation of membrane lipids (Tay-Sachs, Sandhoff’s, Fabry’s, Gaucher’s, or Niemann-Pick diseases)7: list the three main eicosanoids that produced from arachidonic acid (prostaglandins; thromboxanes; and leukotrienes).8: list one NASID you know (aspirin, ibuprofen, or acetaminophen or meclofenamate)9: list two fat-soluble vitamins (A, D, E, K)10: which vitamin can be derived from beta-carotene (A).11: which year was the fluid mosaic model proposed (1972)12: why the thickness of most biological membranes is thicker than 3nm, the standard thickness of lipid bilayer ( due to association of proteins to the membrane and carbohydrates on the membrane)13: Please define the transition temperature of the lipid bilayer (the temperature above which the paracrystalline solid changes to fluid)14: If a membrane protein has its N-terminus exposed to the outside of the cell while itsC-terminus resides in the cytosolic compartment, is it a type I transmembrane protein (yes).15: How can you predict if a protein has a transmembrane domain (hydropathy or hydropathy index, or hydropathy plot).16: Name the two cell surface receptors that HIV use to enter cells (CCR5 and CD4).17: For the Na+ K+ ATPase, how many Na+ and K+ can it move across the membrane for the hydrolysis of one ATP (2 K+ in, 3 Na+ out).18: what drives F-type ATPases to synthesize ATP (proton or proton gradients)19: The acetylcholine receptor is a _____-gated channel (Ligand)20: The neuronal Na+ channel is a _____-gated channel (voltage)II.D and H(3 points) The antiparallel orientation of complementary strands in duplex DNA was elegantly determined in 1960 by Arthur Kornberg by nearest-neighbor analysis. In this technique, DNA is synthesized by DNA polymerase I from one (alpha-32P)-labelled and three unlabelled deoxynucleoside triphosphates. The resulting product is then hydrolyzed by a Dnase that cleaves phosphodiester bonds on the 3’ s ides of all deoxynucleotides. For example, in the labeled dATP reaction,ppp*A + pppC + pppG +pppT --ô€ƒ† …pCpTp*ApCpCp*ApGp*Ap*ApTp… -ô€ƒ†â€¦+Cp+Tp*+Ap+Cp+Cp*+Ap+Gp*+Ap*+Ap+TpT…If in the dATP labeled reaction the relative radioactivity of Tp*, Gp*, Cp* and Ap* is 0.2, 0.3, 0.4 and 0.1.In the labeled dGTP reaction, Tp* radioactivity will beA) 0.1B) 0.2C) 0.3D) 0.4E) 0In the labeled dCTP reaction, Tp* radioactivity will beF) 0.1G) 0.2H) 0.3J) 0(B)(1 point) A molecule of amylopectin consists of 1000 glucose residues and is branched every 25 residues. How many reducing ends does it haveA) 0B) 1C) 25D) 40E) 41F) 1000(1 point) (D)A mirror image ofB form DNA isA) Z-DNAB) A-DNAC) B-DNAD) does not exist in nature(1 point) (A)Which of the following statements is not trueA) all natural occurring DNA is in B formB) A-DNA probably does not exist in vivoC) Both major groove and minor grove of B-DNA are deepD) B-DNA is right handed, while Z-DNA is left handed.(1 point) (A)Lung fish has 100,000,000 kb (kb=1000 base pairs) for its haploid genome. The total length of this DNA isA) 34 mB) 3.4 mC) 1 mD) 17 mF) 34 mm(3 points) (1.5 point) A, D, EBoth tubes X and Y contain the same DNA in 0.1 M NaCl solution. In tube X more NaCl is added and in tube Y some ethanol is added. Which of the following is (are) correctA) tube X will have higher Tm than YB) tube Y will higher Tm than XC) tube X and Y will have the same TmD) tube X will have higher Tm than originalE) tube Y will have lower Tm than original(1 point) (D)Which of the following is NOT correctA) Glycosaminoglycan chains are linearB) Glycosaminoglycan chains are acidicC) Glycosaminoglycans are composed of repeating disaccharide unitsD) Different glycosaminoglycans have the same disaccharide units(1 point) (C)The so-called table sugar or cane sugar is a disaccharide ofA) galactose and glucoseB) 2 glucose unitsC) glucose and fructoseD) galactose and fructose(1 point) (A)Lactose is a disaccharide ofA) galactose and glucoseB) 2 glucose unitsC) glucose and fructoseD) galactose and fructose(1 point) (B)Choose the INCORRECT statement. In dideoxy sequencing or Sanger’s method,A) the newly synthesized DNA is labeled while the template is notB) the template has to be labeledC) a primer has to be usedD) ddNTPs are used to terminate elongation(1 point) (B)Choose the INCORRECT statement. In Maxam-Gilbert sequencing or chemical method,A) the template is labeledB) a primer is neededC) enzyme is not neededD) nucleotide analogues are not used(1 point) (D)uv radiation in the solar light directly cause which kind of the following DNA damageA) deaminationB) methylationC) DNA breakageD) pyrimidine dimer formationE) depurination(1 point) (C)RNA is sensitive to alkaline hydrolysis becauseA) RNA has uridine instead of thymidine in its sequenceB) RNA has hydroxyl group at 4th and 5th positions of its riboseC) RNA has hydroxyl group at 2nd and 3rd positions of its riboseD) RNA is mostly single strandedE) RNA forms three dimensional structure(1 point) (D)Which of the following is NOT a carbohydrateA) glucoseB) maltoseC) glycogenD) proteglycanE) glycosaminoglycan(3 points)(1.5 point) C, E, FWhich of the following is (are) NOT branchedA) glycogenB) starchC) celluloseD) amylopectinE) amyloseF) chitin(1 point) (A)Shown on the right isA) AB) GC) TD) CE) U(1 point) (C)Shown on the right isA) AB) GC) TD) CE) UIII.1) The follows are the important features of signal transduction systems EXCEPT :A. AmplificationB. Desensitization/turn-offC. SpecificityD. SimplicityE. Integration2) Relative to the outside of a cell, calcium concentration inside the cell is:A. HigherB. LowerC. Approximately equal3) Hydrolysis of GTP to GDP is essential for the normal function of all the GTP-binding proteinsA. TrueB. False4) cAMP-dept kinase (PKA) regulates only sugar metabolismA. TrueB. False5) The follows are among the intracellular second messengers EXCEPT:A. cAMPB. cGMPC. Ca2+D. IP3E. ATP6) How does IP3 work as a second messengerAï¼ژBy activating phaspholipase CBï¼ژBy activating protein kinase CC. By activating Ca2+ channelsD. By activating Gخ±S7) The insulin receptor has tyrosine kinase activity which phsophorylates several different proteins including itselfA. TrueB. False8) The insulin signal is amplified via the MAP kinase cascadeA. TrueB. False9) All signaling pathways are very unique and very specific, and they don’t talk to each otherA. True10) Which of the followings is UNIQUE to TGFخ²receptorsA. transmembrane proteinsB. Ser/The receptor kinasesC. Ligand binding activity present in the extracellular domainD. Become activated upon ligand binding11) The activity of CDK proteins is tightly regulated by the following events EXCEPT :A. PhosphorylationB. DephosphorylationC. Cyclin bindingD. UbiquitinationE. CDK inhibitors12) Caspases play essential roles in apoptosis. The following are among their effects EXCEPT:A. Activation of DNaseB. Activation of other caspasesC. Inducing cell shrinkageD. Stimulating gene expression13) Cancer can be a result of the following events EXCEPTA. Activation of mitogenic signalsB. Cell cycle arrestC. Inactivation of negative regulators for cell growthD. Blockage of cell apoptosis14) In 2002, the Nobel Price for Physiology or Medicine was awarded for the research onA. PhosphorylationB. Cell cycleC. ApoptosisD. G proteinE. OncogenesAnswers for Biosignaling:1) D3) A4) B5) E6) C7) A8) A9) B10) B11) D12) D13) B14) CIV.1. Which of the following statement about 2,3-bisphosphoglycerate (BPG) is wrong: (d)(a) BPG binds at a site distant from the oxygen-binding site of hemoglobin.(b) BPG regulates the oxygen-binding affinity of hemoglobin in relation to the pO2 in the lungs.(c) BPG concentration in normal human blood at sea level is lower than that at high altitudes.(d) BPG greatly increases the affinity of hemoglobin for oxygen.2. Antibodies of the IgG class : (d)(a) consist of four subunits.(b) have noncovalent bonds and disulfide crosslinks.(c) are abundant in the blood.(d) All three choices are correct.3. ELISA allows for rapid screening and quantification of the presense of an antigen in a sample. Which of the following steps of ELISA is wrong: (a)(a) Proteins in a sample are adsorbed to an inert surface, and the surface is washed with a solution of specific protein similar to the protein of interest, to block proteins in subsequence steps from also adsorbing to these surfaces.(b) The sample was treated with a solution containing antibodies against the protein of interest. Unbound antibody is washed away, and the sample is treated with with a solution containing antibodies against the primary antibody.(c) These secondary antibodies have been linked to an enzyme that catalyzes a reaction that forms a colored product.(d) After unbound secondary antibody is washed away, the substrate of the antibody-linked enzyme is added and the product formation is proportional to the concentration of the protein of interest in the sample.4. The major proteins of muscle are: (c)(a) Myosin and hemoglobin(b) Actin and troponin(c) Myosin and actin(d) Myoglobin and actin5. Positive cooperative binding can be identified by (c)(a) a hyperbolic binding curve.(b) a Hill plot with a slope less than one.(c) a Hill plot with a slope greater than one.(d) Choices a) and b) are both correct.6. Which pair of amino acids absorbs the most UV light at 280 nm (b)(a) Thr & His.(b) Trp & Tyr.(c) Phe & Pro.(d) Phe & Pro.7. The strong conclusion from Anfinsen's work on RNaseA was that: (b)(a) 100% enzyme activity corresponds to the native conformation.(b) the sequence of a protein determines its structure.(c) Cys-SH groups are not found in vivo.(d) disulfide bonds (S-S) in proteins can be reduced in vitro.(e) irreversible denaturation of proteins violates the "Thermodynamic Hypothesis".8. Which of the following statement about protein folding is wrong: (c)(a) Some proteins undergo assisted folding by chaperons.(b) Polypeptides fold rapidly by a stepwise process.(c) Misfolding may cause misfunctioning, but does not cause death.(d) A loss of 3-d protein structure sufficient to cause loss of function is called denaturation.9. Hydrogen bonds in a-helices are (d)(a) more numerous than Van der Waals interactions.(b) not present at Phe residues.(c) analogous to the steps in a spiral staircase.(d) roughly parallel to the helix axis.(e) about 5 أ… in length.10. What is the appoximate molecular weight of a protein with 200 amino acid residues in a single polypeptide chain: (b)(a) 11000(b) 22000(c) 44000(d) 66000V.1. What are the two most striking characteristics of enzymes(1 pt, 0.5 pt x 2)Answer:The two most striking characteristics of enzymes are their 1) catalytic power, 2) specificity. 2ï¼ژHow does an enzyme gain its catalytic power (2 pt, 1 pt x 2)Answer:1) Enzymes bring substrates together in an optimal orientation;2) They catalyze reactions by stabilizing transition states;3. Is the structure of the binding site of an enzyme complementary to its substrate or to transition state Why (1 pt, 0.5 pt x 2)Answer: An enzyme is in complementary in structure to the transition state of the substrate because the activation barrier is lowered during such a binding.4ï¼ژHow many general ways of regulations on enzyme Activity (2 pts, 0.5 pt x 4)Answer:1) Feed-back Inhibition2) Regulatory Proteins3) Covalent Modification4) Proteolytic Activation5. List the common features of enzymes (2.5 pts, 0.5 pt x 5)Answer:1) The active site take up small volume compare with the entire protein volume;2) The active sites are composed of amino acid residues located in different positions on a linear sequence;3) An enzyme binds to its substrate via multiple weak interactions;4) Active sites are clefts or crevices;5) The specificity of binding depends on precisely defined arrangement of atoms in an active site.6. How many types of weak interactions involved in the binding of a substrate by an enzyme (2 pts, 0.5 x 4)Answer:1) Electrostatic interactions; 2) Hydrogen bonds; 3) van der Waals forces; 4) Hydrophobic interactions.7. Write down the Michaelis-Menten equation (0.5 pt), what is the meaning of Km (0.5 pt) Under which condition the Km is a measure of the affinity of the enzyme for the substrate (1 pt), what is the meaning of Vmax (0.5 pt) What is the physical limit of the value of k3/km (1 pt)Answer:Michaelis-Menten equation: V = Vmax S/(S + Km)Km is equal to the substrate concentration at which the reaction rate is half its maximal rate value.Km is a measure of the enzyme’s affinity only when K2 >> K3 (i.e., the dissociation rate constant of ES is much greater than the catalytic rate constant)Vmax is reached when all the catalytic sites on an enzyme are saturated with substrate.The physical limit of the value of k3/km is that the enzyme-catalyzed reaction rate is rate of diffusion of both the substrate and the enzyme.VI.The following statements are true (T) or false (F)1. The biological system is an open system; living organisms are at equilibrium with their surroundings. 1.5False: The biological system is indeed an open system, able to absorb and exchange materials with the environment. However the living organisms use energy to, for example, keep their own concentrations of ions inside the cells, which are not necessarily of the same concentrations in the environment.2. Activation energy for a chemical reaction is the energy required for a chemical reaction to convert the reactant to transition state, but does not measure the free energy change between the reactants and products. 1.5True3. The mitochondrion is an organelle that functions as an energy regeneration powerhouse, but does not participate in the regulation of cell survival. 1.5False: Mitochondria indeed play the most important role in energy production; however, they are also a key organelle in the regulation of cell death. In fact, most critical apoptotic or antiapoptotic factors regulate cell death through functional interaction with mitochondria.4. The Buchners’ discovery of fermentation supported the view, as asserted by Pasteur, that fermentation can take place only in living cells. 1.55False: The Buchners’ finding that fermentation can take place in yeast cell extracts demonstrated that cell-free extracts possess all the elements for fermentation from sucrose to alcohol. Pasteur’s view that fermentation is inextricably tied to living cells was wrong. This finding has been regarded as one of the earliest biochemical experiments.5. Endoplasmic reticulum is a place for posttranscriptional RNA processing, translation, and protein modification.1.5False: The ER is a place for protein synthesis (translation), protein modification such as glycosylation and signal peptide cleavage, but not for RNA processing.6. Carbohydrates function as structural components in nucleic acids, amino acids, and protein modification. 1.5E. False: Everything but amino acids7. The present day biochemistry is the interweaving product of historical traditions of biochemistry, cell biology, molecular biology and genetics. 1.5True。

普生2011年选择题（2*40），判断题（3*10）简答（每题5分）1、为什么端粒和端粒酶对真核生物染色体复制是必需的，但是对细菌的环状染色体却不是必要的？【在真核生物中，端粒和端粒酶可以保证染色体的完整性，防止粘连和被水解，解决了5’-末端的合成问题（具体机制请查书）；而细菌是环状染色体，没有两端，双向复制，不存在染色体缩短的问题】2、如果动物细胞的细胞质中有反转录酶，会产生什么后果？【简单叙述反转录酶的作用机理，会导致基因组的稳定性无法维持】3、躯体神经系统和内脏神经系统在结构上有什么不同？【前者又称动物性（或随意性）神经系统，传出神经纤维直接到达效应器；后者又称植物性（或自主性）神经系统，它需要再中枢外的一个神经节中换一个神经元，分节前纤维和节后纤维（到肾上腺的交感神经例外）。

】4、计算重组率。

纯和亲本紫花、长花粉粒与红花、圆花粉粒杂交得F1，F1是紫花、长花粉粒。

F1自交，得到的表型和比例为：紫花、长花粉粒4797，紫花、圆花粉粒390，红花、长花粉粒398,，红花、圆花粉粒1170。

求基因重组频率。

【基因重组频率= 两种重组型个体之和/总个体= (390+398)/6755 ≈ 0.117】问答题（每题10分）1、说明糖皮质激素在人的生命活动中重要作用。

【主要为皮质醇（氢化可的松）等，由肾上腺皮质的束状带分泌，网状带也可分泌少量，ACTH（肾上腺皮质激素）可促进其分泌。

作用包括：（1）升糖作用：促进糖异生、升高血糖，同时促进糖原合成，促进肝外组织蛋白以及脂肪的分解；（2）调节水盐代谢：有弱的盐皮质激素作用；（3）抗感染、抗毒、抗敏】2、平衡系统中A和基因a，对隐性纯合的选择系数为1（即隐性纯合致死），设a起始频率为q0。

求经过n 代之后a的基因频率。

1q2 = q1/(1+q1) = q/(1+2q)……a的起始基因频率为q0，则q n = q0/(1+nq0)】（此题在普通生物学辅导与习题集p183）普生2012年今年普生只有四个题型：选择题（2*40），判断题（3*10），简答题（4*5），论述题（2*10）简答题：1、只有无性生殖的生物很简单，比如…几乎都是单细胞生物，为什么？【无性生殖中，子代遗传物质来自一个亲本，变异程度低，不利于适应环境、产生新性状。

清华大学生物学考研历年真题及参考资料普生03年一．名词解释（每个4分）1、朊粒2、端粒3、钠钾泵4、操纵子5、内稳态6、生态演替7、孤雌生殖8、共质体途径二．填空（每空1分）1、自然界最小的细胞是【支原体】2、细胞膜的脂类成分包括磷脂、糖脂和【胆固醇】3、神经元胞体中粗面内质网和游离核糖体组成的结构称为【尼氏体】4、细胞呼吸产生的二氧化碳和消耗的氧气的分子比称为【呼吸商】5、植物叶中光合作用的产物运输途径是【（韧皮部）筛管】6、植物细胞停止生长后所形成的细胞壁称为【次生细胞壁】7、植物对光照和黑暗时间长短的反应称为【光周期现象】8、C3途径中固定二氧化碳的受体是【RuBP，即核酮糖-1,5-二磷酸】9、骨骼肌纤维两条Z线之间的一段肌原纤维称为【肌小节/肌节】10、与调节血钙有关的一对拮抗激素分别是降钙素和【甲状旁腺素】11、响尾蛇探测温血动物所处方位的感受器是【红外探测器】12、鸟类体温调节中枢位于中枢神经系统的【下丘脑】13、胚珠中的大孢子母细胞来自【孢原细胞】14、哺乳动物相当于囊胚阶段的胚胎称为【胚泡】15、在生物数量性状表达上，每个基因只有较小的一部分表型效应，这类基因称为【微效基因】16、cDNA的中文表述是【互补DNA】（c：complementary）17、Mullis等科学家发明的PCR其中文表述是【聚合酶链式反应】18、限制性内切酶不能切开细菌本身DNA，是因为细菌DNA.的腺嘌呤和胞嘧啶【甲基化】19、细菌内形成的孢子称为【芽孢/内生孢子】20、生物种群在群落中的生活方式和在时间与空间上占有的地位称为【生态位】三．问答1、比较真核细胞和原核细胞的区别。

【这题能写满一张纸，主要从细胞结构、遗传结构及装置、基因表达这三方面回答，第三项属分子生物学范畴】2、简述两栖动物的循环系统。

【2房1室，心室无分隔，不完全双循环，每循环一次经过两次心脏】3、说明植物在登陆之前需要具备的条件。

【（1）维管组织（起支持和输导作用）；（2）表皮有蜡质和角质层保水，有气孔帮助呼吸；（3）出现孢子生殖，生殖完全摆脱了对水的依赖（后发展为种子生殖）。

中国科学院05年攻读硕士学位研究生入学试题《生物化学及分子生物学》一、判断题 20题，20题，每题1.5分，共30分.1、鞘磷脂的代谢过程主要与细胞质膜的流动有关与细胞生物活性分子的生成调节无关。

2、蛋白质的修饰与其运输和定位有关，而与其降解代谢无关。

3、蛋白质的豆蔻酰化是蛋白质脂肪酸化的一种形式。

4、可逆性膜锚定与蛋白激酶参与的信号转到有关，而与G蛋白（如Ras）参与的信号转导无关。

5、蛋白质溶液出现沉淀与蛋白质变性存在必然的关系。

6、 Km值是酶的特性常数之一，与酶的浓度、pH、离子强度等条件或因素无关。

7、一个酶的非竞争性抑制剂不可能与底物结合在同一个部位。

8、蛋白质泛素化（ubiquitination）过程需要三种蛋白质（酶）的参与，其中之一是泛素--蛋白连接酶。

9、往线粒体悬液中加入NADH可以还原线粒体的辅酶Q。

10、膜上有些七次跨膜受体在与配基结合时会形成二体。

11、低浓度不含钾离子的等渗缓冲液中悬浮着内含0.154M氯化钾的脂质体，此时往悬浮液中加入缬氨霉素，悬浮液的pH会下降12、内质网系膜结合的钙ATP酶在催化ATP水解时促进Ca2+/2H+交换。

13、辅酶I（NAD+ ）、辅酶II（NADP+）、辅酶A（CoA）、黄素单核苷酸（FMN）和黄素腺嘌呤二核苷酸（FAD）中都含有腺嘌呤基。

14、端粒酶（telomerase）是一种RNA蛋白质复合物，其作用机制是以RNA为模板，由蛋白质催化逆转录; 所以广义上说，端粒转录酶。

15、 Tm是DNA的一个重要特性，其定义为：使DNA双螺旋90％解开时所需的温度。

16、与DNA双螺旋相反方向缠绕而形成的超螺旋叫做“负超螺旋”。

17、细菌中的插入序列（IS）具有转座能力，能随机插入到任一DNA序列中，在靶点两侧形成一段短的正向重复序列。

18、细菌代谢酶的诱导和合成途径中酶的阻遏，调节蛋白都对操纵子起负调控作用。

19、真核RNA聚合酶II最大亚基C末端重复序列上的乙酰化导致RNA聚合酶II与其它转录的起始与延伸。

Name _______________ C lass _______________ Score ____________Biochemistry (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 (2.5x18 = 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 when thepH 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 Vai 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. Chymotrypsintreatment of the peptide produced a dipeptide that contained sulfur and has a UV absorbance, and a tripeptide. Exposure of the peptide to trypsinproduced 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 mechanisms11?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 maintain steadystateD)The concentration of the metabolic intermediate which is the product ofthe 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, AG'。

中国科学院2005年硕士研究生入学考试《生物化学与分子生物学》卷一、判断题1、鞘磷脂的代谢过程主要与细胞质膜的流动有关与细胞生物活性分子的生成调节无关。

2、蛋白质的修饰与其运输与定位有关，而与其降解代谢无关。

3、蛋白质的豆蔻酰化是蛋白质脂肪酸化的一种形式。

4、可逆性膜锚定与蛋白激酶参与的信号转到有关，而与G蛋白（如Ras）参与的信号转导无关。

5、蛋白质溶液出现沉淀与蛋白质变性存在必然的关系。

6、Km值是酶的特性常数之一，与酶的浓度、pH、离子强度等条件或因素无关。

7、一个酶的非竞争性抑制剂不可能与底物结合在同一个部位。

8、蛋白质泛素化（ubiquitination）过程需要三种蛋白质（酶）的参与，其中之一是泛素--蛋白连接酶。

9、往线粒体悬液中加入NADH可以还原线粒体的辅酶Q。

10、膜上有些七次跨膜受体在与配基结合时会形成二体。

11、低浓度不含钾离子的等渗缓冲液中悬浮着内含0.154M氯化钾的脂质体，此时往悬浮液中加入缬氨霉素，悬浮液的pH会下降。

12、内质网系膜结合的钙ATP酶在催化ATP水解时促进Ca+/2H +交换。

13、辅酶I（NAD+ ）、辅酶II（NADP+）、辅酶A（CoA）、黄素单核苷酸（FMN）与黄素腺嘌呤二核苷酸（FAD）中都含有腺嘌呤（AMP）残基。

14、端粒酶（telomerase）是一种RNA蛋白质复合物，其作用机制是以RNA为模板，由蛋白质催化逆转录; 所以广义上说，端粒酶是种逆转录酶。

15、Tm是DNA的一个重要特性，其定义为：使DNA双螺旋90％解开时所需的温度。

16、与DNA双螺旋相反方向缠绕而形成的超螺旋叫做“负超螺旋”。

17、细菌中的插入序列（IS）具有转座能力，能随机插入到任一D NA序列中，在靶点两侧形成一段短的正向重复序列。

18、细菌代谢酶的诱导与合成途径中酶的阻遏，调节蛋白都对操纵子起负调控作用。

19、真核RNA聚合酶II最大亚基C末端重复序列上的乙酰化导致RNA聚合酶II与其它转录的起始与延伸。

精心整理历年各高校生物化学考研真题中国农业大学1997--2008年生物化学考研真题1997年生物化学一、名词解释（每题3分，共30分）1操纵子2反馈抑制3密码子的简并性4蛋白质四级结构5盐析6碱性氨基酸7Z-DNA8ATP9核苷磷酸化酶10磷酸果糖激酶二、填空（每空1分，共28分）1DNA 损伤后的修复主要有共修复、______________和______________三种方式。

2DNA ，RNA 和肽链的合成方向分别是______________、_________________和______________。

3真核生物mRNA 前体的加工主要包括_______________________、___________________、4 5果糖17 8 9 10 111234在PH35为什么619981 234DNA 56在糖酵解过程中，___________________________是最重要的控制酶，另外_____________和___________________也参与糖酵解速度的调节。

7鱼藤酮能专一地阻断呼吸链上电子由______________流向____________________。

8线粒体的穿梭系统有________________和___________________两种类型。

9黄嘌呤核苷酸转变为__________核苷酸时需要氨基化，其氨基来自_________________。

10原核生物蛋白质合成中，蛋白因子IF-2与______________结合并协助其进入核糖体的________位。

11RNA 聚合酶全酶由______________和______________组成。

12密码子共______个，其中_________个为终止密码子，_____________个为编码氨基酸的密码子，起始密码子为_____________________。

二、是非题。

第二章氨基酸和蛋白质的一级结构习题2–1．图2—1的滴定曲线描述了谷氨酸的电离。

请回答下列问题：①指出三个pK’a的位置；②指出Glu－和Giu＝各一半时的pH；③指出谷氨酸总是带净正电荷的pH范围；④指出Glu±和Glu－能作为一种缓冲液的共轭酸碱对的pH范围.图2-1 谷氨酸的酸-碱滴定曲线2–2．为什么甘氨酸处在等电点时是以偶极离子的形式存在，而不是以完全不带电荷的形式存在？处在等电点时，其完全不带电荷的形式是多少?2–3．甘氨酸是乙酸甲基上的氢被氨基取代生成的，但是，甘氨酸羧基的p K’a 值比乙酸羧基p K’a低。

为什么?2–4．在pH9.0时，计算赖氨酸的两性离子、阳离子以及阴离子所占的比例。

已知赖氨酸三个可电离基团α-COOH，α–NH3+和ε- NH3+的p K’a值分别为2.18、8.95和10.53。

2–5．用强酸型阳离子交换树脂分离下述每对氨基酸，当用pH7.0的缓冲液洗脱时，下述每对中先从柱上洗脱下来的是哪种氨基酸?①天冬氨酸和赖氨酸；②精氨酸和甲硫氨酸；⑧谷氨酸和缬氨酸;④甘氨酸和亮氨酸；⑤丝氨酸和丙氨酸。

2–6．计算出由Ala、Gly、His、 Lys和Val所构成的可能的五肽数目。

2–7．在大多数氨基酸中，α–COOH的p K’a值都接近2.0，α–NH3＋的p K’a 值都接近9.0。

但是，在肽中，α–COOH的p K’a值为3.8，而α–NH3＋的p K’a 比值为7.8。

你能解释这种差别吗?2–8．某蛋白质用凝胶过滤法测定的表观分子量是90kD；用SDS-PAGE测定时，它的表观分子量是60kD，无论2-巯基乙醇是否存在。

哪种测定方法更准确？为什么？2–9．一种分子量为24,000、p I为5.5的酶被一种分子量类似、但p I为7.0的蛋白质和另外一种分子量为100,000、p I为5.4的蛋白质污染。

提出一种纯化该酶的方案。

2–10．下面的数据是从一个八肽降解和分析得到的，其组成是：Ala、Gly2、Lys、Met、Ser，Thr、Tyr。

) 1、蛋白质溶液稳定的主要因素是蛋白质分子表面形成水化膜,并在偏离等电点时带有相同电荷 2、糖类化合物都具有还原性 ( )3、动物脂肪的熔点高在室温时为固体，是因为它含有的不饱和脂肪酸比植物油多。

( )4、维持蛋白质二级结构的主要副键是二硫键。

( )5、ATP 含有3个高能磷酸键。

( )6、非竞争性抑制作用时，抑制剂与酶结合则影响底物与酶的结合。

( )7、儿童经常晒太阳可促进维生素D 的吸收，预防佝偻病。

( )8、氰化物对人体的毒害作用是由于它具有解偶联作用。

( )9、血糖基本来源靠食物提供。

( ) 10、脂肪酸氧化称β-氧化。

( ) 11、肝细胞中合成尿素的部位是线粒体。

( ) 12、构成RN Ａ的碱基有Ａ、Ｕ、Ｇ、T 。

( )13、胆红素经肝脏与葡萄糖醛酸结合后水溶性增强。

() 14、胆汁酸过多可反馈抑制7α-羟化酶。

( )15、脂溶性较强的一类激素是通过与胞液或胞核中受体的结合将激素信号传递发挥其生物()1、下列哪个化合物是糖单位间以α-1，4糖苷键相连： ( ) A 、麦芽糖 B 、蔗糖 C 、乳糖 D 、纤维素 E 、香菇多糖2、下列何物是体内贮能的主要形式 ( )3、蛋白质的基本结构单位是下列哪个：( )A、多肽B、二肽C、L-α氨基酸D、L-β-氨基酸E、以上都不是4、酶与一般催化剂相比所具有的特点是( )A、能加速化学反应速度B、能缩短反应达到平衡所需的时间C、具有高度的专一性D、反应前后质和量无改E、对正、逆反应都有催化作用5、通过翻译过程生成的产物是：( )Ａ、ｔRNA Ｂ、ｍRNA Ｃ、ｒRNA D、多肽链Ｅ、DNA6、物质脱下的氢经NADH呼吸链氧化为水时，每消耗1/2分子氧可生产ATP分子数量( )Ａ、１Ｂ、２C、3 Ｄ、４．E、57、糖原分子中由一个葡萄糖经糖酵解氧化分解可净生成多少分子ATP？( )A、1B、2C、3D、4E、58、下列哪个过程主要在线粒体进行( )A、脂肪酸合成B、胆固醇合成C、磷脂合成D、甘油分解E、脂肪酸β-氧化9、酮体生成的限速酶是( )A、HMG-CoA还原酶B、HMG-CoA裂解酶C、HMG-CoA合成酶D、磷解酶E、β-羟丁酸脱氢酶10、有关G-蛋白的概念错误的是( )A、能结合GDP和GTPB、由α、β、γ三亚基组成C、亚基聚合时具有活性D、可被激素受体复合物激活E、有潜在的GTP活性11、鸟氨酸循环中，合成尿素的第二个氮原子来自( )A、氨基甲酰磷酸B、NH3C、天冬氨酸D、天冬酰胺E、谷氨酰胺12、下列哪步反应障碍可致苯丙酮酸尿症( )D、色氨酸→5羟色胺E、酪氨酸→尿黑酸13、胆固醇合成限速酶是：( )A、HMG-CoA合成酶B、HMG-CoA还原酶C、HMG-CoA裂解酶D、甲基戊烯激酶E、鲨烯环氧酶14、关于糖、脂肪、蛋白质互变错误是：( )A、葡萄糖可转变为脂肪B、蛋白质可转变为糖C、脂肪中的甘油可转变为糖D、脂肪可转变为蛋白质E、葡萄糖可转变为非必需氨基酸的碳架部分15、竞争性抑制作用的强弱取决于：( )A、抑制剂与酶的结合部位B、抑制剂与酶结合的牢固程度C、抑制剂与酶结构的相似程度D、酶的结合基团E、底物与抑制剂浓度的相对比例16、红细胞中还原型谷胱苷肽不足，易引起溶血是缺乏( )A、果糖激酶B、６－磷酸葡萄糖脱氢酶C、葡萄糖激酶D、葡萄糖６－磷酸酶E、己糖二磷酸酶17、三酰甘油的碘价愈高表示下列何情况( )A、其分子中所含脂肪酸的不饱和程度愈高B、其分子中所含脂肪酸的不饱和程度愈C、其分子中所含脂肪酸的碳链愈长D、其分子中所含脂肪酸的饱和程度愈高E、三酰甘油的分子量愈大18、真核基因调控中最重要的环节是( )A、基因重排B、基因转录C、DNA的甲基化与去甲基化D、mRNA的衰减E、翻译速度19、关于酶原激活方式正确是：( )A、分子内肽键一处或多处断裂构象改变，形成活性中心D、分子内部次级键断裂所引起的构象改变E、酶蛋白与辅助因子结合20、呼吸链中氰化物抑制的部位是：( )A、Cytaa3→O2B、NADH→Ｏ2C、CoQ→CytbD、Cyt→CytC1E、Cytc→Cytaa31、基因诊断的特点是：( ) A、针对性强特异性高B、检测灵敏度和精确性高C、实用性强诊断范围广D、针对性强特异性低E、实用性差诊断范围窄2、下列哪些是维系DNA双螺旋的主要因素( )A、盐键B、磷酸二酯键C、疏水键D、氢键E、碱基堆砌3、核酸变性可观察到下列何现象( )A、粘度增加B、粘度降低C、紫外吸收值增加D、紫外吸收值降低E、磷酸二酯键断裂4、服用雷米封应适当补充哪种维生素( )A、维生素B2B、V—PPC、维生素B6D、维生素B12E、维生素C5、关于呼吸链的叙述下列何者正确？( )A、存在于线粒体B、参与呼吸链中氧化还原酶属不需氧脱氢酶C、NAD+是递氢体D、NAD+是递电子体E、细胞色素是递电子体6、糖异生途径的关键酶是( )A、丙酮酸羧化酶B、果糖二磷酸酶C、磷酸果糖激酶D、葡萄糖—6—磷酸酶E、已糖激酶7、甘油代谢有哪几条途径( )22D、合成脂肪的原料E、合成脂肪酸的原料8、未结合胆红素的其他名称是( )A、直接胆红素B、间接胆红素C、游离胆红素D、肝胆红素E、血胆红素9、在分子克隆中，目的基因可来自( )基因组文库B、cDNA文库C、PCR扩增D、人工合成E、DNA结合蛋白10关于DNA与RNA合成的说法哪项正确：( )A、在生物体内转录时只能以DNA有意义链为模板B、均需要DNA为模板C、复制时两条DNA链可做模板D、复制时需要引物参加转录时不需要引物参加E、复制与转录需要的酶不同四、填空题（每空0.5分，共15分）1、胞液中产生的N A DＨ经和穿梭作用进入线粒体。

大 连 理 工 大 学2005’限选课考试试卷A 答案课 程 名 称：《生物化学》 闭卷一、判断题（在括号中正确的写“√”，错误的写“×”。

每小题1分，共10分）（ × ）1. （ × ）2. （ × ）3. （ √ ）4. （ × ）5.（ × ）6. （ √ ）7. （ × ）8. （ √ ）9. （ √ ）10.二、选择题（在正确的选择上写“√”，每小题2分，共40分）1. A 大于82. D. 磷脂双层为骨架，蛋白质附着于表面或插入磷脂双层中。

3. C. 酶分子结合底物并发挥催化作用的关键性三维结构区4. D. AAGGTTCCGG5. A. 67%6. D. FMN7. A. V max 不变，K m 增加8. A. 3 CO 2 ，15 A TP9. C. 4，210. D. 糖酵解11. A. 凝胶色谱法12. C. 丙二酸单酰辅酶A13. C. 调节部位14. D. 6-磷酸葡萄糖酯酶15. B. 瓜氨酸+天冬氨酸------>精氨琥珀酸16. C. 乳酸17. C. cAMP18. C. 谷氨酸19. A. 柠檬酸20. D. 苯丙氨酸三、 填空题（每空1分，共10分）1. 脂质、蛋白质、多糖类2. 原核细胞、真核细胞3. 磷酸戊糖途径4. α-螺旋、β-折叠、β-转角5. 糖异生装 订 线四、简答题（6道题，每小题5分，总分30分）1.蛋白质是生物体的主要结构物质和功能物质，它在生命活动中具有重要的作用，可概括为催化作用、激素作用、运载作用、免疫作用、收缩作用、传递作用以及控制生长与分化作用等等。

牛奶、肌肉、蓖麻毒素、蜘蛛网、酶。

2. mRNA 信使RNA功能是将DNA的遗传信息传递到蛋白质合成基地—核糖核蛋白体，tRNA 转移RNA它在蛋白质生物合成中起着翻译氨基酸信息，并将相应的氨基酸转运到核糖核蛋白体的作用，每一个氨基酸至少有一个相应的tRNA ，rRNA核糖体RNA，是核糖核蛋白体的主要组成部分，与蛋白质生物合成相关。

清华大学生物工程试题,历年完整版生物工程前沿（吴庆余授课部分）考题与参考文献思考题：1.利用蓝细菌进行分子生物学研究和生物技术（生物工程）研究有哪些优点？2.根据课堂讲授内容，参考有关文献，利用已经构建的frxC-蓝细菌突变体，提出一个新的有意义的科学问题，并拟定一个解决该问题的实验技术路线。

3.根据课堂讲授内容，参考有关文献，提出一个蓝细菌生物技术产品的开发思路。

4.利用蓝细菌和重组DNA技术构建一个转基因突变体一般包括哪些实验步骤？5 举例说明，微藻细胞工程面临的问题和挑战，如何应对和解决。

1.Wu Qingyu and Wim Vermaas, 1995, Light-Dependent Chlorophyll aBiosynthesis upon ChlN Deletion in Wild-Type and Photosystem I-less Strains of the Cyanobacterium Synechocystis sp PCC 6803. Plant Molecular Biology，V ol. 29∶933-945.2.张大兵、吴庆余，1996，小球藻细胞的异养转化。

植物生理学通讯,32卷，第2期，140-144页。

3.徐红，吴庆余，1996，蓝藻光系统II的组成、相关基因和遗传修饰。

南京大学学报, 1996年第3期。

485-493。

4.吴庆余，Wim Vermaas, 1996, 蓝细菌Synechocystis sp PCC 6803中ORF469的分子克隆和缺失突变工程株的构建。

生物工程学报，1996，第3期。

311-317。

5.吴庆余，徐红，1997，蓝细菌Synechocystis sp PCC 6803中叶绿素的合成调控类囊体膜重建的研究。

植物学报，39(11): 1003-1009。

6.余九九、李宽钰、吴庆余，1997，螺旋藻的藻胆蛋白提取及稳定性研究，海洋通报，16（4）：24-267.Wu Qingyu, Junbiao Dai, Yoshihiro Shiraiwa, Guoying Sheng and Jiamo Fu,1999, A renewable energy source –hydrocarbon gases resulting from pyrolysis of the marine nanoplanktonic alga Emiliania huxleyi. Journal of Applied Phycology, 11(2): 147-152.8.Qingyu Wu, Yoshihiro Shiraiwa, Guoying Sheng and Jiamo Fu, 1999,Hydrocarbons resulting from pyrolysis of the marine coccolithophores Emiliania huxleyi and Gephyrocapsa oceanica. Marine Biotechnology, 1（4）：346-352. 9.戴俊彪、吴庆余，1999，培养海洋微藻Isocrysis galbana生产EPA和DHA。

生物化学 (清华大学).txt男人应该感谢20多岁陪在自己身边的女人。

因为20岁是男人人生的最低谷，没钱，没事业；而20岁，却是女人一生中最灿烂的季节。

只要锄头舞得好，哪有墙角挖不到？生物化学（清华大学）傅建富第一章1,氨基酸（amino acid）：是含有一个碱性氨基和一个酸性羧基的有机化合物，氨基一般连在α-碳上。

2，必需氨基酸（essential amino acid）：指人（或其它脊椎动物）（赖氨酸，苏氨酸等）自己不能合成，需要从食物中获得的氨基酸。

3，非必需氨基酸（nonessential amino acid）：指人（或其它脊椎动物）自己能由简单的前体合成不需要从食物中获得的氨基酸。

4，等电点（pI,isoelectric point）：使分子处于兼性分子状态，在电场中不迁移（分子的静电荷为零）的pH值。

5，茚三酮反应（ninhydrin reaction）：在加热条件下，氨基酸或肽与茚三酮反应生成紫色（与脯氨酸反应生成黄色）化合物的反应。

6，肽键（peptide bond）:一个氨基酸的羧基与另一个的氨基的氨基缩合，除去一分子水形成的酰氨键。

7，肽（peptide）:两个或两个以上氨基通过肽键共价连接形成的聚合物。

8，蛋白质一级结构（primary structure）：指蛋白质中共价连接的氨基酸残基的排列顺序。

9，层析（chromatography）:按照在移动相和固定相（可以是气体或液体）之间的分配比例将混合成分分开的技术。

10，离子交换层析（ion-exchange column）使用带有固定的带电基团的聚合树脂或凝胶层析柱11，透析（dialysis）：通过小分子经过半透膜扩散到水（或缓冲液）的原理，将小分子与生物大分子分开的一种分离纯化技术。

12，凝胶过滤层析（gel filtration chromatography）：也叫做分子排阻层析。

一种利用带孔凝胶珠作基质，按照分子大小分离蛋白质或其它分子混合物的层析技术。

写在前面的话本资料适用于清华大学生物学及海洋生物学考研复习，包括三门专业课（646生物学、844生物化学与分子生物学、871细胞生物学）的历年真题和部分期末考试题目。

普生和生化部分附有答案及注解，非选择题部分的用【】框出，方便查阅。

普生部分末尾有名词解释参考答案汇总，而生化部分的名词解释题型现已取消，故不提供答案。

如想进行考试模拟，可直接使用附带的“清华生物学历年真题（无答案版）”，或者到网上搜索并下载（仅个别年份缺失）。

由于考研时选考生化，不敢对细胞部分的题目妄作解释，所以只能把真题简单粘贴一下，仅供参考，希望选考细胞的有心人能够将这部分补充完整。

笔者考研时购买了200元的真题资料，后来震惊地发现里面的所有内容均可从百度文库、豆丁网等地方找到原版，还真是：只有不想找，没有找不到。

本来利用题目赚几个钱是无可厚非的，但那些资料贩子却下载前辈们整理并免费发到网上的真题，再加上几份同样可以随便下到的资料，就直接卖出，其行径令人发指。

更为重要的是，里面的所有资料均未经修改和排版，题目和答案错漏百出，还有好多缺失的年份。

浪费钱事小，误人子弟事大。

把TCA写成TAC、把σ因子安到真核生物中、而且还敢拿出去卖，笔者为有这种清华校友感到羞耻。

为了杜绝这种情况的出现，笔者耗费半个月将历年的真题和资料转为电子版，修改了错别字和病句，订正了知识上的错误，字体符号上下标等细节都得到了落实，质量远超市面上兜售的任何一份资料。

制作期间参考了课本、笔记、网络、购买的资料及参考答案，并加上了自己考研时的一些思考，保证真实性和准确度，可放心使用。

下面简单谈一下清华大学生物学考研的基本情况。

普通生物学复习时较为简单，但考起来却不容易。

知识面很广，考什么都不为过，因此重点很难把握。

与之相比，生物化学就俩字：靠谱。

复习时虽然比较艰难，但只要坚持下来，最终可以豁然开朗融会贯通，考试出不了那个范围。

至于细胞生物学，个人建议不要选，中英文出题，考的人又很少，题目出难了的话分数将无法和选考生化的人比肩，不过要是基础很好也可以考虑。

清华大学生物化学2科期末考试试题清华大学生物化学本科期末考试试题考试课程Biochemistry II考试时间：------考试类型：期末测试说明：1. 请选择正确答案，填在适当的横线处（1-30题）或写在答题纸上（31-33题）；2. 答案可能是一个或多个；3. 每题的分数在标在了题目后面；4. 本卷满分为100.1. If the ΔG` of the reaction A B is –12 kJ/mol, which of the following statements are correct? (Note the prime symbol means that a thermodynamic parameter is measured at pH.7.0) (2 points)A. The reaction will proceed spontaneously from left to right at the given conditions.B. The reaction will proceed spontaneously from right to left at standard conditions.C. The equilibrium constant favors the formation of B over the formation of A.D. The equilibrium constant could be calculated if the initial concentrations of A and B were known.E. The value of ΔG`o is also negative.Answer(s): A, D2. Which of the following statements about ATP and its roles in cells are true? (2 points)A. The ATP molecule is kinetically unstable and is thus consumed within about one minute following its formation in cells.B. ATP provides free energy to a thermodynamically unfavorable reactions by group transfer, always donating a Pi to form a covalent intermediate.C. ATP can be regenerated by coupling with a reaction that releases more free energy than does ATP hydrolysis.D. A transmembrane proton-motive force can drive ATP synthesis.E. The active form of ATP is usually in a complex with Mg2+.Answer(s): C, D, E3. A common moiety for NADP, NAD, FMN, FAD, and coenzyme A is: (2 points)A. A pyrimidine ring;B. A three ring structure;C. An ADP;D. A pyranose ring;E. A triphosphate group.Answer(s): C4. If the C-1 carbon of glucose were labeled with 14C, which of the carbon atoms in pyruvate would be labeled after glycolysis? (2 points)A. The carboxylate carbon;B. The carbonyl carbon;C. The methyl carbon.Answer(s): C5. Which of the following are metabolic products of pyruvate in higher organisms?(2 points)A. GlycerolB. Lactic acidC. AcetoneD. Acetyl-CoAE. EthanolAnswer(s): B, D6. Indicate whether each of the following statements about the pentose phosphate pathway is true (T) or false (F). (5 points)A. It generates NADH for reductive biosyntheses. FB. The reactions occur in the cytosol. TC. Transketolase and transaldolase link this pathway to gluconeogenesis. FD. It is more active in muscle cells than in fat-storage cells. FE. It interconverts trioses, tetroses, pentoses, hexoses, and heptoses. T7. Which of the following statements are correct? The citric acid cycle (2 points)A. does not exist as such in plants and bacteria, because its functions are performed by the glyoxylate cycle.B. oxidizes acetyl CoA derived from fatty acid degradation.C. produces most of the CO2 in anaerobic organisms.D. provides succinyl CoA for the synthesis of carbohydrates.E. provides precursors for the synthesis of glutamic and aspartic acids.Answer(s): B, E8. Match the cofactors of the pyruvate dehydrogenase complex with their corresponding enzyme components and with their roles in the enzymatic steps that are listed. (5 points)A. Coenzyme A: 3,7B. NAD+: 2, 9C. Thiamine pyrophosphate (TPP): 1, 5D. FAD: 2, 6E. Lipoamide: 3, 4,8(1). Pyruvate dehydrogenase component(2). Dihydrolipoyl dehydrogenase(3). Dihydrolipoyl transacetylase(4). Oxidizes the hydroxylethyl group(5). Decarboxylates pyruvate(6). Oxidizes dihydrolipoamide(7). Accepts the acetyl group from acetyllipoamide(8). Provides a long, flexible arm that coveys intermediates to different enzyme component.(9). Oxidizes FADH2.9. Matching the role in fatty acid oxidation and/or mobilization to the appropriate component listed below. (5 points)A. Bile salt: 2B. Serum albumin:5C. ApoC-II:4D. Apolipoprotein:3E. Carnitine:1(1). Acts as a “carrier” of fatty acids across the inner mitochondrial membrane.(2). Acts as a biological detergent, disrupting fat globules into small mixed micelles.(3). Binds and transports triacylglycerols, phospholipids, and cholesterol between organs.(4). Activates lipoprotein lipase, which cleaves triacylglycerols into their components.(5). Binds some fatty acids molibized from adipocytes and transports them in the blood to heart and skeletal muscle.10. Which of the following answers complete the sentence correctly? Surplus dietary amino acids may be converted into (2 points)A. proteins.B. Fats.C. ketone bodies.D. glucose.E. a variety of biomolecules for which they are precursors.Answer(s): A, B, C, D,E11. Which of the following compounds serves as an acceptor for the amino groups of many amino acids during metabolism? (2 points)A. GlutamineB. Asparagine.C. α-ketoglutarate.D. OxaloacetateE. GlutamateAnswer(s):C12. Match the functions for the coenzymes that are involved in amino acid metabolism. (4 points)A. Pyridoxal phosphate: (3)B. Coenzyme B12: (2)C. Tetrahydrobiopterin: (1)D. NAD+: (1)E. Biotin (4)(1). Carries electrons(2). Provides free radicals(3). Carries amino groups(4). Carries CO2.13. Which of the following experimental observations would not support the chemiosmotic model of oxidative phosphorylation? (3 points)A. If mitochondrial membranes are ruptured, oxidative phosphorylation cannot occur.B. Raising the pH of the fluid in the intermembrane space results in ATP synthesis in the matrix.C. Transfer of electrons through the respiratory chain results in formation of a proton gradient across the inner mitochondrial membrane.D. The orientation of the enzyme complexes of the electron transfer chain results in a unidirectional flow of H+.E. Radioactively labeled inorganic phosphate is incorporated into cytosolic ATP only in the presence of an H+ gradient across the inner mitochondrial membrane. Answer(s): B14. Some photosynthetic prokaryotes use H2S, hydrogen sulfide, instead of water as their photosynthetic hydrogen donor. How does this change the ultimate products of photosynthesis? (2 points)A. Carbohydrate (CH2O) is not produced.B. H2O is not produced.C. Oxygen is not produced.D. ATP is not produced.E. The products do not change.Answer(s): C15. Which of the following are constituents of chlorophylls? (2 points)A. Substituted tetrapyrrole.B. Plastoquinone.C. Mg2+.D. Fe2+.E. Phytol.F. Iron porphyrin.Answer(s): A, C, E16. The observation that the incubation of photosynthetic algae with 14CO2 in the light for a very brief time (5s) led to the formation of 14C-labeled3-phosphoglycerate suggested that the 14CO2 was condensing with sometwo-carbon acceptor. That acceptor was in fact which of the following? (2 points)A. AcetateB. Acetyl CoAC. Acetyl phosphateD. AcetaldehdydeE. Glycerol phosphateF. None of the aboveAnswer(s): F17. Which of the following are common features of the syntheses of mevalonate (an intermediate of cholesterol biosynthesis) and ketone bodies? (2 points)A. Both involve 3-hydroxyl-3-methylglutaryl CoA (HMG-CoA).B. Both require NADPH.C. Both require the HMG-CoA cleavage enzyme.D. Both occur in the mitochondria.E. Both occur in liver cells.Answer(s): A18. S-adenosylmethionine is involved directly in which of the following reactions. (3 points)A. Methyl transfer to phosphatidyl ethanolamine.B. Synthesis of glycine from serine.C. Synthesis of polyamines.D. Conversion of homocysteine to methionine.E. Generation of the 5` cap of the eukaryotic mRNAs.Answer(s): A, C, E19.Which of the following does not provide a carbon skeleton for the synthesis of amino acids? (2 points)A. succinate.B. α-ketoglutarateC. Pyruvate.D. Oxaloacetate.E. Ribose-5-phosphate.Answer(s): A20. Which of the following compounds directly provide atoms to form the purine ring? (3 points)A. Aspartate.B. Carbamoyl phosphate.C. Glutamine.D. Glycine.E. CO2.F. N5,N10-methylenetetrahydrofolate.G. N10-formyltetrahydrofolate.H. NH4+.Answer(s): A, C, D, E, G21. Which of the following statements about ribonucleotide reductase are true? (2 points)A. It converts ribonucleoside diphosphates into 2`-deoxyribonucleoside diphosphates in humans.B. It contains coenzyme B12, which generates free radicals needed for the catalysis.C. It accepts electrons directly from FADH2.D. It receives electrons directly from either thioredoxin or glutaredoxin.E. It contains two kinds of allosteric regulatory sites: one for controlling the overall activity and the other for controlling the substrate specificity.Answer(s): A, D, E22. Biosynthetic pathways that require NADPH include which of the following? (2 points)A. Gluconeogenesis.B. Fatty acid biosynthesis.C. Ketone body formation.D. Cholesterol biosynthesis.E. Tyrosine biosynthesis.Answer(s): B,D,E23. Homologous recombination in E. coli is likely to require which of the following?(3 points)A. DnaB protein.B. RecA protein.C. RecBCD complex.D. ATP.E. NAD+.F. Single-strand DNA binding protein.G. DNA-dependent RNA polymerase.H. DNA polymerase I.I. DNA ligase.J. dATP.Answer(s): B, C, D, E, F, H, I, J24. Which of the following statements about E. coli promoters are correct? (2 points)A. They may exhibit different transcription efficiencies.B. For most genes they include variants of consensus sequences.C. They specify the start sites for transcription on the DNA template.D. They have identical and defining sequences.E. They are activated when C or G residues are substituted into their –10 regions by mutation.F. Those that have sequences that correspond closely to the consensus sequences and are separated by 17 base pairs are very efficient.Answer(s): A, B, C, F25. The AAUAAA sequence on a RNA molecule marks (2 points)A. The site where ribosomes bind to initiate polypeptide synthesis.B. The site where transcription stops.C. The site near which the primary transcript is cleaved and a poly (A) sequence is added.D. The site where the release factor will bind to end polypeptide synthesis.E. The site where polyribonucleotide phosphorylase will add a stretch of random sequences.Answer(s): C26. The σ70 subunit of the E. coli RNA polymerase: (2 points)A. acts as the catalytic site for polymerization.B. Recognize promoters.C. Has a proofreading function.D. Increases the processivity of the enzyme.E. Recognizes termination signals.Answer(s): B27. The discontinuity of eukaryotic genes were first revealed by: (2 points)A. Using footprinting techniques.B. DNA sequence comparison.C. RNA sequence analysis.D. Electron microscopic analysis of RNA-DNA hybrid molecules.E. Plant genetic studies.Answer(s): D28. Which of the following mRNA codons can be recognized by the tRNA anticodon ICG. (2 points)A. UGC.B. CGA.C. UGA.D. CGU.E. CGC.Answer(s): B, D, E29. A new compound, vivekine, was recently discovered by a clever undergraduate student. It was isolated from bacteria found in deep sea-dwelling organisms. Vivekine inhibits protein synthesis in eukaryotes: Protein synthesis can initiate, but only dipeptides are formed and these remain bound to the ribosome. This toxin affects eukaryotic protein synthesis by blocking the: (2 points)A. binding of formylmethionyl-tRNA to ribosomes.B. activity of elongation factors.C. activation of amino acids.D. recognition of stop signals.E. formation of peptide bonds.Answer(s): B30. Indicate whether each of the following statements about prokaryotic translation is true (T) or false (F). (7 points)A. An aminoacyl-tRNA synthetase catalyzes formation of an ester bond. ( T )B. An mRNA molecule cannot be used to direct protein synthesis until it has been completely transcribed. ( F )C. The positioning of fMet-tRNA on the A site defines the reading frame. ( F )D. Incoming aminoacyl-tRNA are first bound to the A site. ( T )E. Formation of the 70S initiation complex requires an input of energy. ( T )F. The carboxyl group of the amino acid on the aminoacyl-tRNA is transferred to the amino group of a peptidyl-tRNA.( F )G. Release factors cause the peptidyl transferase activity of the ribosome to useH2O as a substrate. ( T )31. In an attempt to determine whether a given RNA was catalytically active in the cleavage of a synthetic oligonucleotide, the following experimental results were obtained. When the RNA and the oligonucleotide were incubated together, cleavage of the oligonucleotide occurred. When either the RNA or the oligonucleotide was incubated alone, there was no cleavage. When the RNA was incubated with higher concentrations of the oligonucleotide, saturation kinetics of the Michaelis-Menten type were observed. Do these results demonstrate that theRNA has catalytic activity? Explain. (8 points)Answer: These results alone do not establish that the RNA has catalytic activity. A catalyst must be regenerated. It is entirely possible that the results observed could be accounted for by a stoichiometric, as opposed to a catalytic, interaction between RNA and the oligonucleotide. In which the RNA may “commit suicide” as the oligonucleotide is cleaved. In such an interaction, a portion of the RNA would also be cleaved itself as a part of the reaction. Four reaction products would accumulate, two resulting from the cleavage of RNA and two from the cleavage of the oligonucleotide. To show that this particular RNA was catalytic, it would be necessary to demonstrate that it turns over and is regenerated in the course of the reaction.32.Translation involves conversion of the language of nucleotides to that of proteins. In the chain of events leading from a nucleotide sequence on DNA to the production of protein by ribosomes, where precisely does the process of translation occur? Explain. (6 points)Answer: Translation involves conversion of the language of nucleotides to that of proteins. The agent of translation is the appropriate aminoacyl-tRNA synthetase, which must recognize a particular amino acid and link it to a tRNA containing an anticodon for that amino acid.33.Suppose that a bacterial mutant is found to replicate its DNA at a very low rate. Upon analysis, it is found to have entirely normal activity of DNA polymerases I and III, DNA gyrase, and DNA ligase. It also makes normal amounts and kinds of dnaA,dnaB, dnaC, and SSB proteins. The oriC region of its chromosome is found to be entirely normal with respect to nucleotide sequence. What defect might account for the abnormally low rate of DNA replication in this mutant? Explain. (6 points) Answer: A decrease in the activity of primase would account for the low rate of DNA replication. DNA replication requires the prior synthesis of RNA primers. Decreased rates of dNTP synthesis would also slow replication.。