往届细胞考试题.docx

1、往届细胞考试题 一、填空题(每空0.5分，共10分)1. 分泌蛋白在内质网中通过加上寡聚糖进行翻译后修饰。蛋白质1. 在内质网和高尔基体的腔中被被修饰。分泌蛋白通过出芽并形成蛋白包被小泡进行转运。有两种类型的包被小泡，一种是笼形蛋白包被小泡，它介导 调节型 的运输；另一种是包被蛋白复合体小泡，它介导组成型的运输。低密度的脂蛋白通过其表面的 辅基蛋白B-100 与细胞质膜中的 受体 结合，然后通过受体介导的内吞作用 被运进细胞内。2. 控制芽殖酵母细胞周期有几个关卡，其中G1关卡主要受 START 基因的控制。3. 染色质由DNA包装成染色体压缩了8400倍，其中压缩率最高的是从 螺线管 压缩成

2、 超螺线管 ，有 40倍。4. 2002年的生理学/医学诺贝尔奖颁给了两位英国科学家和一位美国科学家，以表彰他们为研究器官发育和程序性细胞死亡过程中的 基因调节作用 所作出的重大贡献。5. 在线粒体内膜的呼吸链上有四种类型的电子载体，它们是 铁硫蛋白 ； 辅酶Q ； 黄素蛋白；细胞色素 。二、判断题(正确的标T，错误的标F,或写出必要的答案，共15分)1.Indicate whether each of the following statements is true of the G1 phase of the cell cycle, the S phase, the G2 phase, o

3、r the M phase. A given statement may be true of any, all, or none of the phases.（每题0.5分，共5分）(a) The amount of nuclear DNA in the cell doubles.(b) The nuclear envelope breaks into fragments.(c) Sister chromatids separate from each other. (d) Cells that will never divide again are likely to be arreste

4、d in this phase.(e) The primary cell wall of a plant cell forms.(f) Chromosomes are present as diffuse, extended chromatin.(g) This phase is part of interphase.(h) Mitotic cydin is at its lowest level.(i) A Cdk protein is present in the cell.(j) A cell cycle checkpoint has been identified in this ph

5、ase.Answer: (a)S ; (b)M ; (c)M; (d)G1 ; (e)M; (f)Gl;(g)Gl, G2, S;(h)M, G2, S; (i)Gl, S, G2, M;(j)Gl, G2, M2. 同一个体不同组织的细胞中, 核仁的大小和数目都有很大的变化, 这种变化和细胞中蛋白质合成的旺盛程度有关。( T, 正确。 )3. 将同步生长的M期细胞与同步生长的S期细胞融合,除了见到正常的染色体外,还可见到细线状的染色体。( 错误,粉末状的染色体 )4. 在有丝分裂后期，通过对周期蛋白的遍在蛋白多聚化，介导周期蛋白被蛋白酶体降解，从而退出M期。( T,正确 )5. 核纤层是由核

6、纤层蛋白A、核纤层蛋白B和核纤层蛋白C构成的，其中只有核纤层蛋白A与内核膜相连，核纤层蛋白B和C则与染色质相连。( F,错误，核纤层蛋白B与内核膜相连 )6. 在细胞周期中，如果纺锤体装配不正常，则被阻止G2期。（F,错误，被阻止在M期。）7. 结合有核糖体的内质网被称为粗面内质网，脱去核糖体的内质网则称为光面内质网。( 不正确，因为二者的膜蛋白不同 )8. 同源异型框是一类同源异型基因表达产物中60个氨基酸的保守序列, 它的突变可以改变发育的方向。( 答：正确。 )9. 叶绿体的核酮糖二磷酸羧化酶是由16个亚基组成的聚合体, 其中8个大亚基是核基因编码的。( 答：错误, 8个小亚基是核基因编

7、码; )10. 有丝分裂器中有三种类型的纺锤体微管，其中星微管的可能作用是给核分裂传递信号。（ 答：错误, 给胞质分裂传递信号 ）11. 在减数分裂过程中,染色体间发生的分子重组是随机发生的。( 答: 错误, 同源染色体间的分子重组是随机发生的。 )三、选择题(请将正确答案的代号填入括号，每题1分，共15分)1. Ethyl alcohol is detoxified in the liver. You would expect alcohol to have which of the following effects on liver cells? （ B ） a. Nuclear deg

8、eneration b. Growth of the smooth ER c. Increased lysosomes d. Growth of rough ER e. None of the above2. Which of the following proteins would not be found in the smooth endoplasmic reticulum?（ D ） a. Ca2+-pumping enzymes b. cytochrome P450 c. glucose 6-phosphatase d. signal peptidase3. Which of the

9、 following explains why microsomes cant be seen in cells viewed with the electron microscope?（ B ） a. They are far too small. b. They are artifacts of homogenization and centrifugation. c. They are transparent to electrons. d. They actually can be seen in electron micrographs of cells.4. If you comp

10、ared the proteins in a cis Golgi compartment with those in a trans Golgi compartment, you would find:（ C ） a. the proteins in the two compartments are identical. b. the proteins in the cis compartment are glycosylated and contain modified amino acids, whereas those in the trans compartment are not m

11、odified. c. the proteins in the cis compartment are glycosylated, whereas those in the trans compartment are glycosylated and contain modified amino acids. d. the proteins of the cis compartment are shorter than those of the trans compartment.5. Which type of vesicle of the trans Golgi network would

12、 be most likely to carry hormones destined for regulated secretion?（ B ） a. lysosomal vesicles b. clathrin-coated vesicles c. non-clathrin-coated vesicles d. all of the above6. If you treated cells with a drug that interferes with microtubules, such as colchicine, which of the following would result

13、?（ D ） a. Cell shape would be disrupted. b. Mitosis and meiosis would not occur. c. The intracellular location of organelles would be disrupted. d. All of the above would result.7. First you dissolve the membrane from an intact flagellum, using the detergent Triton X-100. Next you soak the axoneme i

14、n a solution containing EDTA, which removes the Mg2+. What remains of the axoneme after these treatments?（ A ）a. peripheral tubules onlyb. peripheral tubules and central tubules, but no side arms or ATPase activity c. peripheral tubules, central tubules, side arms, and ATPase activity d. peripheral

15、tubules, central tubules, side arms, ATPase activity, and a Membrane8. The sarcoplasmic reticulum must have integral membrane proteins that can: a. release and pump Ca2+.（ A ） b. bind to tropomyosin and troponin. c. undergo action potentials. d. contract.9. When chromatin is treated with nonspecific

16、 nucleases, what is the length of the reulting pieces of DNA? ( D) a. random numbers of base pairs b. about 60 base pairs c. about 8 base pairs d. about 200 base pairs10. What do telomeres do?（ D ） a. They protect the chromsomes from degradation by nucleases. b. They prevent the ends of chromosomes

17、from fusing with one another. c. They are required for complete chromosomal replication. d. all of the above11. Cyclin concentrations are highest during which periods of the cell cycle?（ C ） a. late G1 and early S b. late G2 and early M c. late G1 and late G2 d. late M and late S12. ARF是一种单体G蛋白, 它有一

18、个GTP/GDP结合位点, 当结合有GDP时, 没有活性。若ARF-GDP同( D )结合, 可引起GDP和GTP的交换。 a.GTPase; b.GTP酶激活蛋白; c.Ca2+-ATPase d.鸟嘌呤核苷释放蛋白。13. 用剧烈方法分离到的叶绿体是型叶绿体，不能( D ) 。 a. 产生O2 b.不能合成ATP c. 不能产生NADPH d.不能固定CO214. 细胞的生长和分化在本质上是不同的, 生长是细胞数量的增加、干重的增加；而细胞分化则是: ( D ) a. 形态结构发生变化; b. 生理功能发生变化; c. 生化特征发生变化; d. 以上都是正确的。15. 真核生物的基因表达调

19、控发生在四个水平上。通过对DNA的甲基化来关闭基因的调控则是属于( A )。 a. 染色质活性水平的调控；b. 转录水平调控； c. 转录后加工水平的调控；d. 翻译水平的调控。四、简答题(选做4题，每题5分，20分)1. How does regulated secretion differ from constitutive secretion?Answer. Regulated secretion occurs only in response to a signal. The proteins to be secreted are stored in special secretory

20、 vesicles. Sorting into the regulated secretory pathway is controlled by selective protein aggregation. Constitutive secretion appears to occur by default with secretory proteins, which do not selectively aggregate being included in transport vesicles.2. Dynamic instability causes microtubules eithe

21、r to grow or to shrink rapidly. Consider an individual micro-tubule that is in its shrinking phase. What would happen if the solution contained an analogue of GTP that cannot be hydrolyzed?Answer If GTP is present but cannot be hydrolyzed, microtubules will continue to grow until all free tubulin su

22、bunits have been used up.3. State the conclusion that can be drawn from the following finding: When an animal cell is treated with colchicine, its microtubules depolymerize and virtually disappear. If the colchicine is then washed away, the MTs appear again, beginning at the centrosome and elongatin

23、g outward at about the rate (1 gm/min) at which tubulin polymerizes in vitro. Answer: The centrosome serves as a microtubule-organizing center in vivo, andall of the microtubules radiating from the centrosome apparently have the same polarity.4. 什么是蛋白质N-连接糖基化和O-连接糖基化？发生在何种部位？答：加在于粗面内质网上合成的蛋白质上的糖基可由两

24、种途径连接：通过天冬氨酸残基的N原子或通过丝氨酸和苏氨酸残基的O原子。N-连结糖蛋白合成的第一步在粗面内质网上进行，糖链是从磷酸多萜醇转移至新生肽链上。这种糖基化在高尔基体中继续被修饰。O-连结的糖基化是在高尔基体中进行的。5. 过氧化物酶体是怎样进行氧浓度调节的?有什么意义?答: 过氧化物酶体中的氧化酶都是利用分子氧作为氧化剂, 催化下面的化学反应: RH2 + O2 - R + H2O2这一反应对细胞内氧的水平有很大的影响。例如在肝细胞中,有20%的氧是由过氧化物酶体消耗的,其余的在线粒体中消耗。在过氧化物酶体中氧化产生的能量以产热的方式消耗掉, 而在线粒体中氧化产生的能量贮存在ATP中。

25、线粒体与过氧化物酶体对氧的敏感性是不一样的,线粒体氧化所需的最佳氧浓度为2%左右,增加氧浓度,并不提高线粒体的氧化能力。过氧化物酶体与线粒体不同, 它的氧化率是随氧张力增强而成正比地提高(图7-44)。因此,在低浓度氧的条件下,线粒体利用氧的能力比过氧化物酶体强,但在高浓度氧的情况下,过氧化物酶体的氧化反应占主导地位,这种特性使过氧化物酶体具有使细胞免受高浓度氧的毒性作用。五、计算与推理(第1题必做，2、3选一题，每题5分，共10分)1. In an electron micrograph of a human chromosome spread, you observe a thick fi

26、ber with a length of about 900 nm and an apparent diameter of 30 nm, which is expected for the solenoid structure of condensed chromatin.What is the length in base pairs of the double-helical DNA present in this fiber? Assume, for simplicity, that there is one helical turn of the solenoid per 30 nm

27、along the fiber. Answer： There are six nucleosomes per helical turn of the solenoid structure, and one helical turn of the solenoid corresponds to slightly less than 30nm along the length of a chromatin thick fiber.Assuming, for simplicity of calculation, one helical turn per 30 nm, then there are 6

28、 nucleosomes per 30-nm stretch of thick fiber. A 900-nm-long, thick fiber thus has 30 solenoid turns (900 nm divided by 30 nm/turn) and contains 180 nucleosomes (6 nucleosomes/turn 30 turns).The DNA content of each human nucleosome plus the linker DNA connecting it to adjacent nucleosomes is about 2

29、00 bp. This thick fiber thus contains 36,000 bp of DNA: (200 bp/nucleosome) (180 nucleosomes/900-nm thick fiber).2. One of the functions of the mitotic Cdk (the MPF protein kinase) is to cause a precipitous drop in cyclin concentration halfway through M phase. Describe the consequences of this sudde

30、n decrease and suggest possible mechanisms by which it might occur. Answer：Loss of cyclin leads to inactivation of the mitotic Cdk. As the result, its target proteins become dephosphorylated by phosphatases, and the cells exit mitosis-they disassemble the mitotic spindle, reassemble the nuclear enve

31、lope, decondense their chromosomes, and so on. Cyclin is degraded by ubiq-uitin-dependent destruction in proteosomes, and the activation of the mitotic Cdk most likely causes the ubiquitination of the cyclin, but with a substantial delay. As discussed in Chapter 5, ubiquitination tags proteins for d

32、egradation in proteasomes.2. A protein that inhibits certain proteolytic enzymes (proteases) is normally secreted into the bloodstream by liver cells. This inhibitor protein, antitrypsin, is absent from the bloodstream of patients who carry a mutation that results in a single amino acid change in the protein. Antitrypsin deficiency lung tissue, because of the uncontrolled activity of proteases.