清华大学生物化学试卷2005_final_exam_A

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清华大学生物试题及答案

清华大学生物试题及答案

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

清华大学XX年生物化学1本科期末考试试题

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。

清华大学生物学考研历年真题及参考资料

清华大学生物学考研历年真题及参考资料

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

清华大学生物学考研历年真题及参考资料

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2005年攻读硕士学位研究生入学试题《生物化学及分子生物学》试题及答案

中国科学院2005年攻读硕士学位研究生入学试题《生物化学及分子生物学》试题及答案

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。

生物化学期末考试试题12套答案在最后(大学期末复习资料).doc

生物化学期末考试试题12套答案在最后(大学期末复习资料).doc

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

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

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。

历年各高校生物化学考研真题

历年各高校生物化学考研真题

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。

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A.5
B.6
C.6.5
D.7.5
E.No ATP would form under these conditions.
Answer(s): D
3.Indicate which of the following events can occur during the processing of eukaryotic mRNA transcripts.
清华大学本科生考试试题专用纸(A)
考试课程Biochemistry II 2005年1月15日
说明:1.第1-40题为选择题,每题2分,答案可能是一个或多个,请将
正确答案填在适当的横线处;
2.第41-44题为问答题,每题的分数标在了题目后面,请将答案写
在答题纸上;
3.本卷满分为100。
1. Which compound links glycolysis, nucleotide synthesis, and glycogen synthesis?
Answer(s):D
8.Consider the mRNA sequence: (5’) AAUGCAGCUUUAGCA(3’). The sequence ofthe coding strand of DNA is:
A.(5’) ACGATTTCGACGTAA(3’)
B.(3’) TTACGTCGAAATCGT(5’)
E.The value ofG’ois also negative.
Answer(s): AD
13.Which of the following isnota physiological role of nucleotides?
A.allosteric regulators
B.intermediates for biosynthetic processes
A.hexokinase; ATP
B.phosphofructokinase-1; AMP
C.pyruvate kinase; ATP
D.pyruvate dehydrogenase: NADH
E.pyruvate carboxylase; ADP
Answer(s):B
16.Which of the following describes fatty acid transportinto the mitochondrial matrix?
A.substitution
B.insertion
C.deletion
D.silent
E.B and C
Answer(s): E
5.Which of the followingcorrectlydescribes the nitrogen cycle?
A.Fixation ofatmospheric nitrogen by nitrogen-fixing bacteria yields bioavailable nitrate.
A.All seven genes are transcribed into mRNA, which is then translated into five different proteins.
B.The repressor binds to the structural genes, preventing their transcription.
C.(5’) AATGCAGCTTTAGCA(3’)
D.(5’) AAUGCAGCUUUAGCA(3’)
E.(3’) AATGCAGCTTTAGCA(5’)
Answer(s):C
9.The synthesis of palmitate requires:
A.8acetyl-CoA.
B.14 NADH.
B.odd-number fatty acids.
C.saturated fatty acids.
D.unsaturated fatty acids with trans double bonds.
Answer(s):A
11.Which of the following is a characteristic of many aminotransferase reactions?
B.are important in the repair of damaged DNA.
C.involve RecA protein.
D.generategenetic diversity.
E.The two types of recombination have nothing in common.
A.They have a large, negativeG’o.
B.The amino group is transferred to an-keto acid (such as-ketoglutarate) to form the corresponding amino acid.
C.The amino group is transferred from an ammonia molecule.
A.UGC
B.CGA
C.UGA
D.CGU
E.CGC
Answer(s):BDE
7.Both general recombination and transposon-type recombination:
A.occurbetween homologous regions on chromosomes.
E.Splicing together of exons.
F.Differential cutting and splicing to produce two different proteins.
Answer(s):CE
4. AlthoughDNA replication has very high fidelity, mutations do occur. Which of the following types of single base-pair mutations would be most likely to be a lethal mutation?
D.Once fatty acyl groups have entered the matrix, they are committed to oxidation to acetyl-CoA.
E.All of the above are true.
Answer(s):E
17.In nucleotide metabolism, all of the following are trueexcept:
A.acetyl-CoA
B.oxaloacetate
C.citrate
D.glucose 6-phosphate
E.glycerol 3-phosphate
Answer(s): D
2.You have discovered acompound that inhibits fumarase. How many moles of ATP would you expect to be generated from one mole of acetyl-CoA in the presence of this inhibitor?
C.Efficient binding of RNA polymerase to DNA occurs only in the presence of CRP and cAMP.
D.cAMP acts as an inducer by binding to the Lac repressor.
E.In the absence of lactose, the repressor binds the operator.
C.7 ATP.
D.A and C.
E.A, B, and C.
Answer(s):D
10.Enoyl-CoA isomerase is needed for the complete-oxidation of
A.unsaturated fatty acids with cis double bonds.
A.It is the rate-limiting step in fatty acid oxidation.
B.It is regulated by malonyl-CoA.
C.The cytosolic and matrix pools of CoA are distinct and separate.
D.A and B
E.A, B, and C
Answer(s):C
6. Assuming the 5’3’ connection of writing nucleotide sequence, indicate which of the following mRNA codons can be recognized by the tRNA anticodon ICG.
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.
B.Nitrate is reduced to ammonia in a process known as denitrification.
C.Biological nitrogen fixation is carried out by a complex of proteins call the nitrogenase complex.
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