to the tyrosine kinase receptor, SOS catalyzes the Ras GDP/GTP exchange reaction. In this respect, SOS and Ras together are similar to a heterotrimeric G protein. Here, SOS acts like the GBg subunits, whereas Ras is similar to a Ga subunit.
4. 答: 主要是从创造差异对细胞生命活动的意义方面来理解这一说法。主动运输涉及物质输入和输出细胞和细胞器,并且能够逆浓度梯度或电化学梯度。这种运输对于维持细胞和细胞器的正常功能来说起三个重要作用:① 保证了细胞或细胞器从周围环境中或表面摄取必需的营养物质,即使这些营养物质在周围环境中或表面的浓度很低;② 能够将细胞内的各种物质,如分泌物、代谢废物以及一些离子排到细胞外,即使这些物质在细胞外的浓度比细胞内的浓度高得多; ③能够维持一些无机离子在细胞内恒定和最适的浓度,特别是K+、Ca2+和H+的浓度。概括地说，主动运输主要是维持细胞内环境的稳定,以及在各种不同生理条件下细胞内环境的快速调整, 这对细胞的生命活动来说是非常重要的。
5.答：①利用现有的科学成果；②对新的发现要大胆提出假设并加以证实。
6.答：使一克水的温度上升一摄氏度所需要的能量是1卡。这与其它液体相比是很高的。水吸收的大部分能量被用来破坏分子间氢键，这些氢键是由于水分子的极性和不对称性造成的。因为吸收的能量要被用于断裂弱键，水的温度不象其它液体那样容易升高。在此种意义上，环境温度的变化可以在细胞中被缓冲。
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. Answer:
Identify a tissue that would serve as a good source of the enzyme of interest, e.g., the plasma membranes of nerve tissue. Solubilize the Na+-K+ ATPase from its native membrane using detergents. Reconstitute the enzyme into liposomes of simple and well-defined composition, then assay the enzyme for activity. By synthesizing a variety of liposomes with different amounts of saturated and unsaturated fatty acids, the fiuidity of the liposome can be altered, and the enzyme can be tested under conditions of different fiuidity. (To measure the fiuidity of the liposomes directly, without having to infer it from the fatty acid composition, either FRAP or SPT techniques can be applied.)
2. Answer:
The functional domains of G-protein coupled receptors were determined by experiments using recombinant chimeric receptor proteins containing parts of the b2 and a2 adrenergic receptors. These chimeric receptors were tested for their ligand binding specificity and their ability to activate or inhibit adenylate cyclase. The results of these studies demonstrated that a helix 7 and the C-terminal domain of the receptor play a role in determining ligand binding specificity and the cyto-solic loop between a helices 5 and 6 interacts with G proteins.
六、分析、计算与思考(20分) 1. 分析题（任选1题，5分） （1） ANSWER:
The binding of epinephrine to b-adrenergic receptors causes a stimulation of heart function, both in terms of heart rate and with respect to the amount of work done in pumping blood. The effect appears to be mediated by cyclic AMP. When an antagonist such as the beta blocker propranolol
is given to patients with hypertension, the cellular response caused by the binding of epinephrine to beta receptors is partially inhibited. Heart function is gradually restored over a period of time, with a corresponding decrease in blood pressure.
（2）答：IP3的钙的释放和对钙浓度的消减。
2. 答案：
一个分子量4800的蛋白质由约40个氨基酸组成，因此有1．1 X 1052(=2040)
个不同方式去组成这个蛋白，其中每种蛋白每个分子重8x10-21g(=4800／(6X1023))。因此由每一种各一个分子组成的一个混合物重9 X 1031g(=8x lO-21g X 1．1 X 1052)，它是地球总重量6X1024kg的15000倍。你真需要一个非常巨大的容器。
鉴于大部分细胞蛋白质甚至比这个例子中的大，据此显然在活细胞中只用全部 可能的氨基酸序列中的很小一部分。
2. 综合思考(10分, 按答题最多者计算每题得分) Answer:
(a) Both the ECM of animal cells and the walls around plant cells consist of long, rigid fibers embedded in an amorphous, hydrated matrix of branched- chain molecules, either glycoproteins (ECM) or polysaccharides (cell wall).
(b) For ECM, the fibers consist of collagen and the matrix is a network of proteoglycans. For cell walls, the fibers consist of cellulose and the matrix is a network of polysaccharides and proteins.
(c) Both ECM and cell walls are important in maintaining cell shape and in retaining water, thereby resisting compression.
(d) Roles unique to the ECM include regulation of cellular processes such as adhesion, motility, and differentiation during embryonic development. Roles unique to cell walls include protection of the cell from mechanical injury and microbial invasion, as well as provision of the mechanical support necessary to withstand the turgor pressure that gives plant tissues their rigidity.
（2） 答：①细胞培养技术 ②离心分离技术 ③流式细胞分离技术 ④基因敲除技术 ⑤干细胞培养技术
（3）． Answer③The N-terminal and C-terminal propeptides present in newly synthesized collagen monomers as-ist in alignment of the peptides to form the triple helix. These propeptides are removed after the trimers are transported to the extracellular matrix, and thus are not available to perform the same function in denatured calf type I collagen. In addition, inappropriate disulfide bridges can be generated during renaturation; these will also inhibit the generation of a normal triple helix.
（4）ANSWER:
The binding of epinephrine to b-adrenergic receptors causes a stimulation of heart function, both
in terms of heart rate and with respect to the amount of work done in pumping blood. The effect appears to be mediated by cyclic AMP. When an antagonist such as the beta blocker propranolol is given to patients with hypertension, the cellular response caused by the binding of epinephrine to beta receptors is partially inhibited. Heart function is gradually restored over a period of time, with a corresponding decrease in blood pressure. 试题四
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 进行翻译后修饰。蛋白质
在 和 的腔中被被修饰。分泌蛋白通过 并形成 进行转运。有两种类型的包被小泡，一种是 ，它介导 的运输；另一种是 小泡，它介导 的运输。低密度的脂蛋白通过其表面的 与细胞质膜中的 结合，然后通过 被运进细胞内。
控制芽殖酵母细胞周期有几个关卡，其中G1关卡主要受 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 压缩成 ，有 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 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∶ ① ；② ；③ ；④ 。
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, 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 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 been identified in this phase.
2. 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。( )
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。( )
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。( )
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。( )
6. 在细胞周期中，如果纺锤体装配不正常，则被阻止G2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。( )
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。( )
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。( ) 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。（ ）
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。( )
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? （ ） a. Nuclear degeneration b. Growth of the smooth ER c. Increased lysosomes d. Growth of rough ER e. None of the above
2. Which of the following proteins would not be found in the smooth endoplasmic reticulum?（ ） a. Ca2+-pumping enzymes b. cytochrome P450 c. glucose 6-phosphatase d. signal peptidase
3. Which of the following explains why microsomes can't be seen in cells viewed with the electron microscope?（ ）
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 compared the proteins in a cis Golgi compartment with those in a trans Golgi compartment, you would find:（ ）
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 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 compartment. 5. Which type of vesicle of the trans Golgi network would be most likely to carry hormones destined for regulated secretion?（ ） a. lysosomal vesicles b. clathrin-coated vesicles c. non-clathrin-coated vesicles d. all of the above
6. If you treated cells with a drug that interferes with microtubules, such as colchicine, which of the following would result?（ ） 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 in a solution containing EDTA, which removes the Mg2+. What remains of the axoneme after these treatments?（ ） peripheral tubules only
b. peripheral tubules and central tubules, but no side arms or ATPase activity c. peripheral tubules, central tubules, side arms, and ATPase activity
d. peripheral tubules, central tubules, side arms, ATPase activity, and a Membrane 8. The sarcoplasmic reticulum must have integral membrane proteins that can: a. release and pump Ca2+.（ ） b. bind to tropomyosin and troponin. c. undergo action potentials. d. contract.
When chromatin is treated with nonspecific nucleases, what is the length of the reulting pieces of DNA? a. random numbers of base pairs b. about 60 base pairs c. about 8 base pairs d. about 200 base pairs 10. What do telomeres do?（ ）
a. They protect the chromsomes from degradation 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 above
11. Cyclin concentrations are highest during which periods of the cell cycle?（ ） a. late G1 and early S b. late G2 and early M c. late G1 and late G2 d. late M and late S
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( )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Ⅱ型叶绿体，不能( ) 。 a. 产生O2 b.不能合成ATP c. 不能产生NADPH d.不能固定CO2
14. 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: ( )
a. 形态结构发生变化; b. 生理功能发生变化;