细胞生物学课前思考题99问题&答案

99级 四问答 1. Dynamic instability causes microtubules either to grow or to shrink rapidly. Consider an individual microtubule that is in its shrinking phase. (1). What would happen if only GDP, but no GTP, were present in the solution? (2). What would happen if the solution contained an analogue of GTP that cannot be hydrolyzed? （1）If only GDP were present, microtubules would continue to shrink and eventually disappear, because tubulin dimers with GDP have very low affinity for each other and will not add stably to microtubules. (2). If GTP is present but cannot be hydrolyzed, microtubules will continue to grow until all free tubulin subunits have been used up. 2. How does regulated secretion differ from constitutive secretion? Regulated secretion occurs only in response 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 included in transport vesicles. 3. Consider a protein that contains an ER signal sequence at its amino terminus and a nuclear localization sequence in its middle. What do you think the fate of this protein would be? Explain your answer. The protein is translocated into the ER. Its ER signal sequence is recognized as soon as it emerges from the ribosome. The ribosome then becomes bound to the ER membrane, and the growing polypeptide chain is transferred through the ER translocation channel. The nuclear localization sequence is therefore never exposed to the cytosol. It will never encounter nuclear import receptors, and the protein will not enter the nucleus. 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。 5. 过氧化物酶体是怎样进行氧浓度调节的?有什么意义? 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。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、看图描述(每题10分，共20分) 1．根据对核蛋白运输机制的研究及相关蛋白的发现, 提出了核蛋白的输入模型(图1：含有NLS信号的核蛋白从细胞质输入细胞核的推测模型), 请对这一模型作出文字说明(PCC1:Ran nucleotide-exchange factor1)。 答: 按照这一推测的模型，在细胞质中的核运输蛋白α、核运输蛋白β和货物蛋白（cargo protein）相互作用形成一个运输复合物，其中运输蛋白α亚基识别并与NLS结合。而运输蛋白β亚基与核孔复合物作用，将复合物转运到细胞核中。在此过程中需要消耗ATP。在细胞核中，Ran·GTP（一种小GTP结合蛋白）与输入蛋白β亚基相互作用，导致货物蛋白与复合物脱离，成为细胞核中的游离蛋白。为了进行下一个运输循环，输入蛋白α亚基和输入蛋白β亚基—Ran·GTP复合物重新回到细胞质。细胞质中的Ran GTP—激活蛋白（RanGAP）将Ran·GTP转变成Ran·GDP, 并使Ran·GDP与输入蛋白β亚基脱离，游离的输入蛋白β亚基和α亚基一起参与新的具有NLS信号的入核蛋白的运输。而Ran·GDP可通过核孔复合物回到细胞核中，在Ran核苷交换因子1(Ran nucleotide-exchange factor1,RCC1)的作用下，释放GDP, 重新结合GTP。 2. 图2（酵母MPF的活性调） 是关于MPF的活性调节，请作出文字说明： 答： 开始p34cdc2 蛋白激酶是无活性的, 同周期蛋白B结合后,仍然没有活性, 但此时的复合物成为两种激酶的底物, 一种是Weel激酶, 它使p34cdc2的靠近催化位点的酪氨酸残基(Tyr15)磷酸化, 阻止了它本身的激活, 防止了Cdc2作为成熟的MPF起作用。第二种蛋白激酶是MO15, 可以使Cdc2的另外一个位点的苏氨酸(Thr167)磷酸化, 这种磷酸化最大限度地激活了MPF的活性, 但是, 只要酪氨酸残基也是磷酸化的,Cdc2-周期蛋白复合物就没有活性, 这种无活性的MPF称为前MPF(Pre-MPF)。要使MPF具有活性，需要要靠第二种蛋白, 在酵母中是Cdc25蛋白将磷酸化的酪氨酸残基去磷酸化从而被激活, 诱导细胞从G2进入M期。不过，Wee1和Cdc25是相互竞争的，如果细胞生长得不够大，Wee1的活性就强，有利于MPF的磷酸化，若细胞生长得够大，就有利于脱磷酸，促进细胞进入M期。 六、综合题(20分): 生物膜是怎样合成的?可能的机理是什么? 答：关于膜的合成，曾提出两个模型:一个自装配模型(spontaneous self-assembly), 即膜是理由蛋白、脂和糖自动组装的, 但与体外实验结果不符。因为用纯化的脂和蛋白在体外装配时总是形成脂质体，这种脂质体与活细胞膜的一个根本区别是:脂质体的结构总是对称的, 而活细胞中膜结构则是不对称的。 第二个是不断更新模型, 该模型认为膜的合成通过不断地将脂和蛋白插入已有的膜，即由已有膜的生长而来。这一模型比较符合细胞膜结构的动态性质, 由于细胞的胞吞和胞吐作用以及小泡运输,使膜处于动态平衡状态, 这样膜也就不必重新合成,而是在原有的基础上不断更新。 膜的合成涉及脂、蛋白和糖的来源问题。 膜脂有两种来源:①通过磷脂转运蛋白,②通过出芽和膜融合, 关于膜脂的不对称性分布，有几种可能的方式∶一种是磷脂交换蛋白对磷脂的运输和插入是选择性的；第二种解释是热动力学驱使磷脂的不对称分布；另外在ER膜中有翻转酶(flippase),在新的磷脂合成之后,通过翻转酶的作用也会造成磷脂的不对称分布。 膜蛋白有整合蛋白和外周蛋白。发现膜整合蛋白是通过内膜系统经小泡转运到质膜上的, 而外周蛋白则是在游离核糖体上合成,并以可溶的形式释放到胞质溶胶中。然后再与细胞质膜的胞质溶胶面结合,成为外周蛋白。糖则是在内质网和高尔基体腔中通过对蛋白的修饰添加的。最后在与质膜融合时,通过外翻,糖的部分位于细胞质膜的外侧。这就是为何几乎所有质膜上的糖蛋白的糖都是朝向细胞外的原因。 脂锚定蛋白的形成有几种可能的机制: 糖脂锚定的膜蛋白是在粗面内质网上合成,然后在ER腔中被连接到ER膜的GPI上，随后通过小泡运输,经高尔基体出芽形成小泡,最后与质膜融合,含糖的一面外翻朝向细胞外侧。 脂肪酸锚定膜蛋白是水溶性的,在游离核糖体合成后释放到胞质溶胶中,然后与包埋在质膜中的脂肪酸共价结合。连接的脂肪酸包括豆蔻酸(myristic acid, 一种14碳的饱和脂肪酸)和棕榈酸(palmitic acid,一种16碳的饱和脂肪酸)。 七，附加题 1. 分析思考（5分） 基因的转录加工和蛋白质的合成是遗传信息流的两个关键环节，对此细胞核与细胞质有明确的分工：细胞核负责基因的转录加工，而细胞质则负责完成蛋白质的合成。核糖体是合成蛋白质的工厂， 由大小两个亚基组成，它们都是在细胞核着中装配的。从细胞的生命活动看，在长期的进化过程中选择了经济合理的路线，但是在核糖体的装配中却选择了一种违背经济规律的做法：组成核糖体大小亚基的80余种蛋白质在胞质溶胶中合成之后被运送到细胞核中；在核仁中装配成的核糖体大、小亚基又要被运送到胞质溶胶中，它们是核输出中体积最大的物质。由于细胞中核糖体的数量巨大，组成核糖体的rRNA的种类和数量都远比组成核糖体的蛋白质少，如果细胞核将合成好的rRNA运送到胞质溶胶中，在那里进行核糖体的装配和蛋白质的合成岂不更经济合理？假如是这种情况的话，有哪些不利？谈谈您的看法。 答：看起来，细胞核并不直接参与蛋白质的合成，但是它却控制着蛋白质合成的两个关键元件，即mRNA的合成和核糖体的装配。控制中着核糖体的合成就等于控制中蛋白质合成的装配线，有利于控制生命的节奏。因此，为求得安定，必要的花费是值得的。 ①如果核糖体亚基的装配在胞质溶胶中，不仅极大地削弱了细胞核对蛋白质合成的控制力，还会发生一些细胞核不良后果。②由于核糖体的空间构型复杂，装配过程长。如果mRNA同正在装配中的未成熟的核糖体亚基结合的话，会导致蛋白质合成的不正常。③由于组成核糖体的蛋白质种类多，数量大，并且胞质溶胶中代谢途径多，会影响核糖体的装配，从而影响其他的生化过程，最终影响细胞的正常声明活动。 2.（5分） Consider the v-SNAREs that direct transport vesicles from the trans Golgi network to the plasma membrane. They, like all other v-SNAREs, are membrane proteins that are integrated into the membrane of the ER during their biosynthesis and are then transported by transport vesicles to their destination. Thus, transport vesicles budding from the ER contain at least two kinds of v-SNAREs--those that target the vesicles to the cis Golgi cisternae and those that are in transit to the trans Golgi network to be packaged in different transport vesicles destined for the plasma membrane. (A) Why might this be a problem? (B) Suggest possible ways in which the cell might solve it. A. The problem is that vesicles having two different kinds of v-SNAREs in their membrane could dock on either of two different membranes. B. The answer to this puzzle is presently not known, but we can predict that cells must have ways of turning the docking ability of SNAREs on and off. This may be achieved through other proteins that are, for example, copackaged in the ER with SNAREs into transport vesicles and facilitate the interactions of the correct v-SNARE with the t-SNARE in the cis Golgi network. 3 （10分）。 Specialization of membranes is a characteristic of mammalian cells. What are the mechanisms by which the apical and basolateral surfaces of certain cells become differentiated? 答：Membrane proteins are sorted to either the apical or basolateral domains by several different mechanisms. One mechanism for targeting proteins to the appropriate domain of the plasma membrane involves sorting in the trans-Golgi network. Except for the GPI anchor, which acts as an apical or basolateral targeting signal, no unique sequences have been identified that target proteins to the apical or basolateral domain. Another mechanism operates in bepatocytes where all newly made apical and basolateral proteins are first delivered together from the trans-Golgi network to the basolateral membrane. From there, both apical and basolateral proteins are endocyto-sed to the same vesicles. Within endosomes, there is then sorting and transport to the appropriate domain. The attachment of integral membrane proteins to the cytoskeleton serves as a retention signal and may assist in the apicai-basolateral sorting of some proteins. Hence, depending on cell type, at least three different mechanisms for api-cal-basolateral sorting are possible. 4.(5分) Many mutant yeasts have been isolated that are defective in the control of their cell cycle. They proliferate normally at low temperatures (300C)but show abnormal patterns of cell growth and division when grown at a higher temperature (370C). Two mutant strains (called "gee" and “wee") with defects at different sites in the same gene have very different responses to elevated temperature. Gee strain cells grow until they become enormous but no longer divide. Wee strain cells have very short cell cycles and divide when they are very much smaller than usual. Suggest a possible model to explain these observations, and suggest what the normal protein encoded by this gene might do. 答：Normally, cells divide only when they have grown to a certain size. This size control is clearly defective in the two mutant strains. In the case of gee cells, cell size increases without ever triggering cell division, suggesting that the mutant cell-cycle control protein has lost its ability to monitor cell size. It might, for example, now permanently inhibit MPF. In wee cells, on the other hand, the mutant control protein triggers cell division prematurely, before cells have grown to the appropriate size. This could be a control protein, for example, that no longer inhibits MPF, so that MPF becomes active prematurely. In fact, there is a yeast cell-cycle control protein called Wee1, which is a kinase that phosphorylates MPF on a site that causes its inactivation; yeast cells with a mutation in the wee-1 gene have a short cell cycle and are small. 00级 四、简答题 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 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 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 cannot 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 then washed away, the MTs appear again, beginning at thecentrosome 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, and all 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。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 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 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 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).The 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 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： 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 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 ubiquitination of the cyclin, but with a substantial delay. As discussed in Chapter 5, ubiquitination tags proteins for degradation in proteasomes. 3. 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. Surprisingly, when the mutant antitrypsin is synthesized in the laboratory, it is 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 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 stability 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 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 the following: cytoplasmic dynein vs. kinesin Answer: Both dynein and kinesin are large motor proteins that convert the chemical energy of ATP into movement. Both are found affiliated with microtubules, although only dynein occurs on the microtubules of cilia and flagella. Kinesin is a 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 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 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 functions 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 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 microsomes containing the SRP receptor also were added to the incubations, 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 during 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 chain, allowing cytokinesis to proceed. 01级： 四、简答题(任选5题, 每题4分，共20分) 1. 肝细胞中除线粒体合成少量蛋白质外， 绝大多数的蛋白质都是在细胞质的游离核糖体和膜结合核糖体上合成的。请您推测在肝细胞那种核糖体上合成的蛋白质占多数，是游离核糖体还是膜结合核糖体（假定细胞内所有区室的蛋白质的平均密度和寿命都是相同的）？说明您推断的依据。 1.答：游离核糖体合成的蛋白质的分配去向包括胞质溶胶、线粒体、过氧化物酶体、细胞核等， 约占细胞体积的80%以上。而膜结合核糖体上合成的蛋白质的去向包括ER、高尔基体、溶酶体、质膜、细胞外等， 只占细胞体积的20%， 所以游离核糖体上合成的蛋白质起主导作用。据此，可以肯定地说，肝细胞中游离核糖体上合成的蛋白质占游离多数。 2. 线粒体内膜中的电子传递链的最主要的贡献是什么？ 答：线粒体内膜中的电子传递链的最主要的贡献是建立了质子动势。 3. 从不同的环境中分离到两种细菌:一种是从平均温度为～40℃的温泉中分离的, 另一种是从平均温度为～4℃的冷水湖中分离的。问: a. 请推测两种细菌的细胞质膜中, 哪一种具有较多的不饱和脂肪酸? b. 那一种细菌质膜中的脂肪酸链较长? c. 在27℃哪一种细菌质膜的流动性高? 答: a. 从冷水湖中分离的细菌的细胞质膜具有较多的不饱和脂肪酸, b. 来自温泉细胞的质膜中含有较多长链脂肪酸。 c. 在27℃, 来自冷水湖细菌的膜具有较大的流动性。 4. 简要说明在动物细胞的有丝分裂和胞质分裂中细胞质骨架起什么作用？如何 起作用？ 答：有丝分裂需要微管装配成钫锤体，然后通过微管线性分子发动机的作用将染色体拉向两极。胞质分裂需要肌动蛋白在质膜的下方装配成收缩环， 然后在肌球蛋白Ⅱ的作用下，通过收缩环的收缩将细胞质动力分开形成两个子细胞。 5. 紫杉醇与秋水仙碱的作用相反。紫杉醇与微管紧密结合并使微管稳定。若将紫杉醇添加到细胞中, 可促进游离微管蛋白亚基装配成微管。与之相反, 秋水仙碱则阻止微管的装配。紫杉醇与秋水仙碱都是细胞分裂的毒素, 都可用作抗癌剂。根据您对微管动力学的了解, 说明为什么这两种药物的作用相反但都是细胞分裂的致毒剂。 答: 细胞分裂取决与微管聚合与去聚合的能力。在有丝分裂期间, 细胞首先将大多数微管去聚合, 然后装配成纺锤体。用紫杉醇处理细胞则防止了微管的去聚合从而阻止了有丝分裂纺锤体的形成。用秋水仙碱处理细胞则阻止了新微管的聚合, 因此同样不能形成有丝分裂纺锤体。换个角度, 这两种药物都破坏了微管的动态不稳定性, 因此会干扰有丝分裂纺锤体正常工作, 既使能够形成纺锤体也是如此。 6. 假定您从线虫中分离到一些纯的蛋白质, 经分析, 该蛋白含有二硫键, 并且其疏水区不长于5个氨基酸。根据这些特性, 推测该蛋白位于线虫细胞的哪一区室? 依据是什么? 答: 由于该蛋白含有二硫键, 它必然通过易位从胞质溶胶进入ER, 并在ER腔内进行二硫键的形成。由于该蛋白不含典型的跨膜序列, 所以该蛋白不会成为膜蛋白。如果是GPI锚定蛋白, 很可能在细胞表面。另外, 由于该蛋白是可溶性蛋白, 该蛋白存在与细胞器的腔内(ER、高尔基体等), 也有可能分泌到细胞外。 五、计算题(10分)。 1. 一个核小体的长度是11nm, 由146bp的DNA组成(0.34nm/bp), 请问DNA压缩 成核小体的比值是多少? 如果加上两个核小体间的54bp的连接DNA, 请问“珠 -绳”的压缩比是多少?有丝分裂时DNA压缩了10000倍, 相当于原长度的百 分之多少? 1. 答: 核小体的包装比例是4.5[(146 bp x 0.34 nm/bp)/ (11 nm) = 4.5]. “珠-绳”的压缩比值是2.3 [(200 bp x 0.34 nm/bp)/ (11 nm + {54 bp x 0.34 nm/bp}) = 2.3]. 有丝分裂DNA压缩染色体的比值是0.023% (2.3/10,000)。 七、附加题(5分) 1. 从进化的角度，说明细胞骨架对动物细胞进化的作用。 答：① 动物细胞的体积很大，且形态多样、又没有细胞壁， 因此细胞骨架的形成对于维持细胞的形态起重要作用。 ② 动物细胞及所有真核细胞都具有细胞核，而细胞核的形态主要是由细胞骨架支持的。内核膜下的核纤层对于维持细胞核的形态具有至关重要的作用。 3 动物细胞通过形态的变化进行移动， 在动物细胞的移动中， 肌动蛋白纤 维是必须的。 4 动物细胞的基因组比细菌的基因组大得多， 并分成多个染色体。在细胞 分裂中， 染色体必须正确地分配到两个子细胞， 在此过程中， 微管功不可没。 5 动物细胞具有很多细胞内有很多细胞器，它们在细胞内的定位主要靠发动 机分子沿着微管运输的。特别是神经轴中的物质运输， 没有微管是不可能的。