往届细胞考试题.doc

1、豆萁中山大学考研论坛 1一、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

2、中压缩率最高的是从 螺线管 压缩成 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 ph

3、ase, the G2 phase, or 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

4、are likely to be arrested 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 be

5、en identified in this phase.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, 这种变化和豆萁中山大学考研论坛 2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

6、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.

7、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

8、which of the following effects on liver cells? （ B ）a. Nuclear degenerationb. Growth of the smooth ERc. Increased lysosomesd. Growth of rough ERe. None of the above2. Which of the following proteins would not be found in the smooth endoplasmic reticulum?（ D ）a. Ca2+-pumping enzymesb. cytochrome P450

9、c. glucose 6-phosphatase豆萁中山大学考研论坛 3d. signal peptidase3. Which of the 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 ac

10、tually can be seen in electron micrographs of cells.4. If you compared 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 modifi

11、ed amino acids, whereas those in the trans compartment are not modified.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 compart

12、ment.5. Which type of vesicle of the trans Golgi network would be most likely to carry hormones destined for regulated secretion?（ B ）a. lysosomal vesiclesb. clathrin-coated vesiclesc. non-clathrin-coated vesiclesd. all of the above6. If you treated cells with a drug that interferes with microtubule

13、s, such as colchicine, which of the following would result?（ 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 t

14、he detergent Triton X-100. Next you soak the axoneme in 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豆萁中山大学考研论坛 4c. peripheral tubule

15、s, central tubules, side arms, and ATPase activityd. peripheral 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 potentia

16、ls.d. contract.9. When chromatin is treated with nonspecific nucleases, what is the length of the reulting pieces of DNA? ( D)a. random numbers of base pairsb. about 60 base pairsc. about 8 base pairsd. about 200 base pairs10. What do telomeres do?（ D ）a. They protect the chromsomes from degradation

17、 by nucleases.b. They prevent the ends of chromosomes 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 Sb. late G2 and early Mc. late G1 and

18、 late G2d. late M and late S12. ARF 是一种单体 G 蛋白, 它有一个 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 ATPc. 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 ) 豆

19、萁中山大学考研论坛 5a. 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 r

20、esponse to a signal. The proteins to be secreted are stored in special secretory 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

21、 included in transport vesicles.2. Dynamic instability causes microtubules either 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 ca

22、nnot be hydrolyzed, microtubules will continue to grow until all free tubulin subunits 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 th

23、en washed away, the MTs appear again, beginning at the centrosome and elongating 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 h

24、ave 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。豆萁中山大学考研论坛 65. 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 + H 2O2这一反应对细

25、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。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

26、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 fiber 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

27、 DNA present in this fiber? Assume, for simplicity, that there is one helical turn of the solenoid per 30 nm 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

28、 chromatin thick fiber.Assuming, for simplicity of calculation, one helical turn per 30 nm, then there are 6 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).T

29、he DNA content of each human nucleosome plus the linker DNA connecting it to adjacent nucleosomes is about 200 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

30、 a precipitous drop in cyclin concentration halfway through M phase. Describe the consequences of this sudden decrease and suggest possible mechanisms by which it might occur. Answer：豆萁中山大学考研论坛 7Loss of cyclin leads to inactivation of the mitotic Cdk. As the result, its target proteins become dephos

31、phorylated by phosphatases, and the cells exit mitosis-they disassemble the mitotic spindle, reassemble the nuclear envelope, 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 u

32、biquitination of the cyclin, but with a substantial delay. As discussed in Chapter 5, ubiquitination tags proteins for degradation in proteasomes.2. A protein that inhibits certain proteolytic enzymes (proteases) is normally secreted into the bloodstream by liver cells. This inhibitor protein, antit

33、rypsin, 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. Surprisingly, when the mutant antitrypsin is synthesized in the laboratory, it is

34、as active as the normal antitrypsin at inhibiting proteases. Why then does the mutation cause the disease? Think of more than one possibility and suggest ways in which you could distinguish between them. Answer：The actual explanation is that the single amino acid change causes the protein to misfold

35、 slightly so that, although it is still active as a pro-tease inhibitor, it is prevented by chaperone proteins in the ER from exiting this organelle. It therefore accumulates in the ER lumen and is eventually degraded. Alternative interpretations might have been: (1) the mutation affects the stabili

36、ty of the protein in the bloodstream so that it is degraded much faster in the blood than the normal protein, or (2) the mutation inactivates the ER signal sequence and prevents the protein from entering the ER. (3) Another explanation could have been that the mutation altered the sequence to create

37、 an ER retention signal, which would have retained the mutant protein in the ER. One could distinguish between these possibilities by using fluorescent-tagged antibodies against the protein to follow its transport in the cells (see Panel 5-3, pp. 158-159).六、比较题（每题 5 分，共 10 分）1. Compare and contrast

38、the following:cytoplasmic dynein vs. kinesinAnswer: 1. Both dynein and kinesin are large motor proteins that convert the chemical energy 豆萁中山大学考研论坛 8of ATP into movement. Both are found affiliated with microtubules, although only dynein occurs on the microtubules of cilia and flagella. Kinesin is a

39、plus-end directed microtubular motor, and dynein, among its other roles, is a minus-end directed microtubular motor. In spite of their similarities in function, they are not homologous proteins, and they assume quite different three-dimensional shapes. They are not members ora proteinfamily.2. 后期 A

40、与后期 B七、综合问答题(任选一题，20 分)1. 细胞内蛋白质有那些分选途径？各自的机理如何？答：翻译后转运与共翻译转运跨膜运输小泡运输核孔运输2. 比较裂殖酵母、芽殖酵母和哺乳动物细胞周期调控的异同。答：同：有关卡，有周期蛋白与周期蛋白激酶；异：CDC2,CDC28,哺乳动物不同的激酶与多种周期蛋白。八、附加题（每题 5 分，共 15 分）1. State the conclusion that can be drawn from the following finding:Extracts from nondividing frog eggs in the G2 phase of the

41、 cell cycle were found to contain structures that could induce the polymerization of tubulin into microtubules in vitro. When examined by immunostaining, these structures were shown to contain pericentrin（中心粒旁侧蛋白）. Answer: The extracts appear to contain structures that are functionally equivalent to

42、 centrosomes (as evidenced by the presence of pericentrin), which nucleate microtubule growth.2. The signal recognition particle (SKP) is involved in regulating the elongation of nascent secretory proteins and targeting them to the endoplasmic reticulum. Describe an experiment in which these functio

43、ns of SRP have been demonstrated. Answer:The functions of SRP were demonstrated in a series of experiments utilizing a cell-free protein-synthesizing system and mRNA encoding pre-prolactin, a typical secretory protein. When the mRNA was incubated in the cell-free translational system in the absence

44、of SRP and microsomes, the complete protein with its signal sequence was produced. The addition of SRP to the incubation mixtures caused protein elongation to cease after 70-100 amino acids had been incorporated. When 豆萁中山大学考研论坛 9microsomes containing the SRP receptor also were added to the incubati

45、ons, the block in protein synthesis was relieved and the complete protein minus the signal sequence was extruded into the lumen of the microsomes.3. Dephosphorylation is an important event that affects cellular structures during mitosis. Describe two of these events.Answer:Dephosphorylation events d

46、uring mitosis include protein phosphatases removing the regulatory phosphates from lamins A, B, and C, permitting reassembly of the nuclear laminae of the two daughter cell nuclei. When MPF activity falls during anaphase, a constitutive phosphatase dephosphory-lares inhibitory sites on myosin light

47、chain, allowing cytokinesis to proceed.一、填空题(每空 0.5 分，共 10 分)1. 定位于线粒体基质和内膜的蛋白都是由导肽引导的，但是参与运输的导肽的数量是不同的，前者需要一个导肽，后者需要两个导肽。2. 在动物细胞的运动中，关于运动力的产生有两种假说，一种是:微丝装配与去装配假说，另一种是肌动蛋白与肌球蛋白相互作用。3. The flow of electrons across the inner membrane generates a pH gradient and a membrane potential, which together ex

48、ert a Proton-motive force.4. 2001 年诺贝尔生理学与医学奖授予了美国科学家利兰哈特韦尔和英国科学家蒂莫西亨特、保罗纳西，以表彰他们发现了细胞周期的关键调节机制。其中，利兰哈特韦尔发现了“START ”基因；保罗纳西的贡献是发现了 CDK;蒂莫西亨特的贡献是:发现了调节 CDK 的功能物质 CYCLIN.5. 把有丝分裂期间出现的纺锤体、中心体、星体及染色体统称为有丝分裂器。6. 当核基因编码的线粒体蛋白进入线粒体时, 需要胞质溶胶中的 ATP和跨线粒体内膜的质子动力势提供能量来推动。7. 类胡萝卜素在光合作用中主要是帮助叶绿素提高光吸收效应，作用机理是除去叶绿素不能吸收的杂色光。8. 笼型蛋白包被小泡介导的是选择性分泌，而包被蛋白包被小泡介导的是非选择性分泌。9. 在细胞分裂中, 微管的作用是形成纺锤体, 将染色体拉向两极; 微丝的作用是 胞质分裂;.10. Bip 蛋白是一种分子伴侣，它在内质网中至少有两个作用：引导新生蛋白进入内质网； 帮助蛋白质正确装配。11. 染色体的着丝粒有两个基本功能连接两个姊妹染色单体；为动粒微管的装配提供位点。豆萁中