胞生物学期中考试试题及答案03.doc

细胞生物学期中试卷3一、填空题(每空0.5分，共10分)1. 细胞中水对于维持细胞温度的相对稳定具有重要作用，其原因是 。2. 纤粘连蛋白与细胞结合的结构域具有特征性的三肽结构，简称RGD序列，代表的三个氨基酸是 。3. _ _ is an intracellular protein that binds calcium and activates enzymes.4. A substance that fits into a specific binding site on the surface of a protein molecule is called a _ .5. Two products of phospholipase C activity that serve as second messengers are _ and .6. Cyclic AMP is produced by the enzyme _ _ and degraded by the enzyme _ _7. NO是一种气体信号分子,在体内主要由精氨酸产生, 它的半衰期很短, 故只作用于邻近细胞, 它作用的靶酶是 。8. junctions enable cells to get a hold on the extracellular matrix by connecting their actin filaments to the matrix.9. 动物细胞质膜靠 建立膜两侧的质子动力势，而植物细胞则是靠 。10. 蛋白质合成时多聚核糖体的形成对生命活动的意义在于: 。11. 限制哺乳动物细胞体积大小的主要因素有两个:一种是 的关系,另一种是 。12. Ca2+-ATPase是一个 次跨膜的膜蛋白，有 个细胞质环，工作时每次可以将 个Ca2+泵到细胞外。13. 选择蛋白介导的细胞粘着是 依赖性的, 而免疫球蛋白超家族蛋白介导的细胞粘着是 依赖性的。二、判断题(若是正确的标号， 错误的标号，每题1分，共20分)1. K+ 离子很难通过人工膜, 但加入短杆菌肽A后, 对K+ 的透性大大增加, 原因是短杆菌肽A可作为离子载体运输K+。( )2. One can determine if a membrane protein is exposed on the external side of the plasma membrane by covalent attachment of a labeling reagent or by protease digestion only if the membrane is intact. ( )3. 脂锚定蛋白是通过蛋白的氨基酸残基同膜脂的脂肪酸链形成共价键而锚定在质膜上。( )4. CsCl密度梯度离心法分离纯化样品时, 样品要和CsCl混匀后分装, 离心时, 样品中不同组分的重力不同, 停留在不同区带。( )5. 载体蛋白, 又称为通透酶, 它象酶一样, 不能改变反应平衡,只能增加达到反应平衡的速度; 但与酶不同的是, 载体蛋白不对被运的分子作任何修饰。( )6. 型内含子的剪接与核剪接机理相似, 都要形成剪接体和套索结构。( )7. 5SrRNA是由RNA聚合酶转录的, 它使用的是内部启动子。( )8. The surface area/volume ratio is generally greater for a prokaryotic cell than for a eukaryotic cell.（ ）9. The ribosomes found in the mitochondria of your musde cells are more like those of the bacteria in your intestine than they are like the ribosomes in the cytosol of your musde cells.（ ）10. Facilitated diffusion of a cation occurs only from a compartment of higher concentration to a compartment of lower concentration.( )11. Active transport is always driven by the hydrolysis of high-energy phosphate bonds.( )12. Treatment of an animal cell with an inhibitor that is specific for the Na+/K+ pump is not likely to affect the uptake of glucose by sodium cotransport.( )13. Tight junctions get their name from their property of holding cells together so tightly that they cannot be separated by mechanical forces.（ ）14. Gap junctions connect the cytoskeletal elements of one cell to a neighboring cell or to the extracellular matrix.（ ）15. NO acts only locally because it has a short half-life-about 5 to 10 seconds-in the extracellular space before it is converted to nitrates and nitrites by oxygen and water.( )16. Water-soluble signaling molecules, which have very short circulating life-times, usually mediate responses of short duration, whereas water-insoluble ones, which persist in the blood for hours to days, tend to mediate longer-lasting responses.( )17. Ribozyme(核酶)的化学本质是RNA, 但具有酶的活性, 专司切割RNA。( )18真核生物的28S、18S和5S的rRNA属于同一个转录单位，先转录成一个45S的前体， 然后边加工边装配核糖体的大、小两个亚基。( )19水是细胞的主要成分, 并且多以结合水的形式存在于细胞中。( )20. 为了使光学显微镜或电子显微镜标本的反差增大, 可用化学染料对它们进行染色。( )三、选择题(请将正确答案的代号填入括号，每题1分，共20分)1. 下列内容中除了( )以外, 都是细胞学说的要点。 a. 所有生物都是由一个或多个细胞构成 b. 细胞是生命的最简单形式 c. 细胞是生命的结构单元d. 细胞从初始细胞分化而来2. 在动物细胞培养过程中要用( ) 来进行观察。 a. 相差显微镜; b. 荧光显微镜; c. 倒置显微镜; d. 普通光学显微镜。3. 下述哪一种情况能够支持核糖体RNA具有催化活性?( ) a.rRNA碱基序列是高度保守的,而核糖体蛋白的氨基酸序列则不然b.具有对抗生素抗性的细菌在rRNA分子中有取代的碱基,而在核糖 体蛋白质中没有被取代的氨基酸 c.肽酰转移酶的反应对核糖核酸酶敏感 d.上述都是4. 用抗纤连蛋白的抗体注射胚体，发现在神经系统发育过程中神经嵴细 胞的移动受到抑制。这些实验说明：（ ） a. 神经嵴发育包括抗体基因的表达 b.发育中的神经无需要合成纤连蛋白 c.纤连蛋白-抗体复合物形成神经移动途径d.胚胎中的神经元在移动过程中必须与纤连蛋白暂时结合5. 心钠肽是心房肌细胞产生的肽激素,对血压具有调节作用。心钠肽作 为第一信使作用于受体, 并在细胞内产生第二信使, 下面四种第二信 使中哪一种对于心钠肽具有应答作用?( )a.cAMP b.cGMP c. Ca2+ d. DAG6. The distribution of K+ across an artificial membrane was measured, and the concentrations were found to be equal on both sides. Which of the following statements is true about the distribution of K+?( ) a.K+ must be at equilibrium across the cell membrane b.K+ cannot be at equilibrium across the membrane c.There cannot be a membrane potential under these conditionsd. More information is needed to determine whether K+ is at equilibrium7.Which mutant form of ras is likely to cause malignancy?( ) a.ras that cannot hydrolyze GTP b.ras that cannot bind to GTP c.ras that cannot bind to Grb2 or Sos d.ras that cannot bind to Raf8. The hormone glucagon stimulates the breakdown of stored glycogen in liver and muscle cells by the following enzymes. Which is the first enzyme that must be activated?( ) a. Protein kinase A b. Phosphorylase kinase c. Glycogen phosphorylase d. Protein phosphatase e. Protein phosphatase inhibitor-19. A cancer-causing gene makes G protein permanently bind to GTP, this results in( ). a. Continual production of cAMP. b. Continual production of GDP. c. Inhibition of cGMP production. d. Inhibition of adenylyl cyclase. e. None of the above10. Which one of the following is not part of a prokaryotic ribosome?( ) a. A 30S subunit consisting of 16S rRNA b. A 50S subunit consisting of 23S and 5S rRNA c. A 30S subunit consisting of 21 proteins d. A 40S subunit consisting of 18S rRNA e. None of the above11. Which one of the following was used to determine the structure of the DNA molecule?( ) a. Transmission electron microscope b. Scanning electron microscope c. Differential centrifugatioin d. X-ray crystallographye. None of the above12. You stain membrane proteins with a fluorescent-labeled antibody and notice the membrane is evenly stained. After a few hours, all of the fluorescence is at one end of a cell. You can conclude that( ). a. Lipids flip-flop in the membrane. b. Proteins move laterally in the membrane. c. The fluorescent dye was bleached. d. Proteins act as transporters. e. All of the above13. You have a preparation of ghosts and need to collect only the right-side out ghosts. You decide you can trap the right-side out ghosts in an affinity column packed with( ) a. SDS. b. A carbohydrate-binding molecule. c. A protein-binding antibody. d. A serine-binding molecule. e. None of the above14.All of the following individuals contributed to cell theory except( )a.Robert Hooke. b.Matthias Schleiden.c.Theodor Schwann. d.Rudolf Virchow.15.The addition of a peptide with an RGD sequence would probably ? the binding of cultured cells to a fibronectin coated dish.( )a.enhance b.inhibitc.have no effect on d.none of the above16. 适于观察细胞表面及断面超微结构三维图像的仪器是( ) a普通光镜 b荧光显微镜 c相差光镜 d扫描电镜 e透射电镜17. 红细胞中O2与CO2交换作用主要依靠( ) a血影蛋白 b血型糖蛋白 c. 带蛋白 d锚定蛋白 e. 肌动蛋白18. 能与胞外信号特异识别和结合，介导胞内信使生成，引起细胞产生效应的是( ) Acarrier protein Bchannel protein Creceptor D1iand Eenzyme19. 下列连接方式中属于与中间纤维相连的锚定连接的是( )。 a粘合带； b粘合斑； c桥粒； d紧密连接。20. 带电的离子之所以不能自由通过质膜,其原因是( )a.膜的孔径不够大;b.膜的电性与离子电性相斥; c.带电的离子聚集成团; d.带电离子通常与水形成水性外壳,以至于体积过大。四、简答题(每题4分， 选做5题，共20分)1. 基因敲除(knockout)实验中有哪些关键环节影响试验的成败？2. 真核生物与原核生物的rRNA基因在组织结构、转录和加工中有什么差异？3. G protein-linked receptors activate heterotrimeric G proteins through inter-actions between the receptor and the Gb subunit. Upon binding to the receptor, the Gb subunit then catalyzes GDP/GTP exchange by the Ga subunit. How is this similar to the activation of Ras by a receptor tyrosine kinase?4. 如何理解“被动运输是减少细胞与周围环境的差别,而主动运输则是努力创造差别,维持生命的活力”?5. 1981年，Thomas Cech和他的同事在世界上首次发现核酶，并获得了支持这一发现的实验证据。从他们的发现中你有何感想？6. 为什么说水是细胞中优良的热缓冲体系？五、实验设计与分析。(简要说明，不需详细步骤。每小题5分,共10分)。1. Some integral membrane enzymes depend upon the lipids in their microenvironment not only for scaffolding, but for enzymatic activity. The Na+-K+ ATPase is one example. Please design an experiment to test the influence of membrane fluidity on the velocity of the Na+-K+ ATPase.2. Describe the experimental approach used to identify functional domains of G-protein coupled receptors.六、分析、计算与思考(20分)1. 分析题（任选1题，5分）(1) Hypertension, or high blood pressure, is often seen in elderly people. A typical prescription to reduce a patients blood pressure includes compounds called beta-blockers, which block b-adrenergic receptors through-out the body. These receptors bind epinephrine, thereby activating a cellular response. Why do you think beta-blockers are effective in reducing blood pressure?(2)“解铃还需系铃人”这句谚语在细胞活动中能找到对应的故事吗？请说明。2. 计算与思考（5分）Assume you were given a mixture consisting of one molecule each of all possible sequences of a smallish protein of molecular weight 4800. How big a container would you need to hold this sample? Assume that the average molecular weight of an amino acid is 120. What does this calculation tell you a