中山大学生科院：《细胞生物学》2000级期中试卷_03（带答案）.doc_大学文库

1 中山大学生科院细胞生物学期中试卷参考答案 (2000 级生物科学、生物技术、药学专业,164 人) 任课教师∶ 王金发 2002 年 11 月 11 日姓名∶ 专业∶ 一、填空题(每空 0.5 分，共 10 分) 1. 水分间的氢键能够吸收较多的热能。 2. Arg-Gly-Asp。 3. _Calmodulin_ i 4. __Ligand. 5. _IP3_ ， DAG . 6. _Adenylyl cyclase_ ， _phosphodiesterase_ 7.鸟苷酸环化酶。 8. 粘着斑 9. K + -Na+泵， H +泵。 10. 节省了遗传信息量，减轻了核的负担。 11. 答：体积与表面积; 重要分子在细胞内的浓度 12. 10， 2 13.非钙二、判断题(若是正确的标√ 号，错误的标× 号，并做简要说明。每题 1分，共20分) 1.答：错误，短杆菌肽 A 形成离子通道。 2. (√) 3. ×,有两种情况,另一种情况是通过糖与脂的共价连接进行锚定 4. ( × ) 5．答：正确。 6.答: 错误, Ⅱ型内含子的剪接不形成剪接体。 7.答: 错误, 是 RNA 聚合酶Ⅲ转录的, 使用的是内部启动子。 8.（√） 9.（√） 10. Answer:F; the direction of movement depends on both the concentration gradient and the membrane potential, so facilitated diffusion can occur from a compartment of lower concentration to a compartment of higher concentration if the membrane potential is in the right direction and of sufficient magnitude. 11.Answer:F; hydrolysis of high-energy phosphate bonds is a common means of driving active transport, but so is an ion gradient, as in sodium-driven cotransport of sugars and amino acids into cells. 12.Answer:F; if the sodium-potassium pump is inhibited, the sodium gradient necessary for sodium-driven glucose uptake cannot be maintained and cotransport will decrease or even cease as the sodium gradient collapses. 13. Answer:False. Tight junctions provide molecule-tight seals between cells. 14.Answer.False. Gap junctions are communicating junctions. 15. Answer.True 16. Answer.True 17. 答：错误，除了水解酶的活性外，还有其他酶的活性，如磷酸酶活性。 18．答：错误，应是 5.8S。 19．答：错误，结合水只占 4～5%。 20. 答：错误，电镜要用电子染色。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. 答：b. 正确 2. 答：c. 3. 答: c 4. 答: d

5答:b 意义方面来理解这一说法。主动运输涉及物质 6.答d 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 8.答:a 9.答胞器的正常功能来说起三个重要作用①保证 10.答d 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答:d 物质如分泌物、代谢废物以及一些离子排到细 12.答b 13.答b 胞外即使这些物质在细胞外的浓度比细胞内 14.答:a 的浓度高得多;③能够维持一些无机离子在细 15.答:C 胞内恒定和最适的浓度特别是K+、Ca2和H 16.答:d 的浓度。概括地说,主动运输主要是维持细胞答:c 18.答:c 内环境的稳定以及在各种不同生理条件下细 19.答:C 胞内环境的快速调整这对细胞的生命活动来 20.答:d 说是非常重要的四、简笞题(每題4分,选做5题,共20分)5答:①利用现有的科学成果;②对新的发现 1答:①基因突变;②受体细胞一胚胎干细胞;要大胆提出假设并加以证实。选择标记 2.答:①真核生物的18 SrRNA585 6答:使一克水的温度上升一摄氏度所需要的 rRNA28S『RNA串联在一起;原核生物能量是1卡。这与其它液体相比是很高的。水的16 SrRNA-5 SrRNA-23 SrRNA串联吸收的大部分能量被用来破坏分子间氢键,这在一起;②真核生物的18 S rRNA-585些氢键是由于水分子的极性和不对称性造成 rRNA28 SrRNA物中不含tRNA基因的。因为吸收的能量要被用于断裂弱键,水的原核生物的16 S TRNA5 SrrNA23S 温度不象其它液体那样容易升高。在此种意义 rRNA转录物中有tRNA基因的转录产上,环境温度的变化可以在细胞中被缓冲 Answer: The activation of Ras requires五实验设计与分析。(简要说明,不需详细步 the assistance of another protein called骤。每小题5分共10分) SOS. Upon binding to the tyrosine kinase receptor, sos catalyzes the Ras GDP/GTP 1. Answer: ge reaction. In this respe Identify a tissue that would serve as a good are s heterotrimeric G protein. Here, sos acts source of the enzyme of interest, e.g, the plasma like the GBy subunits, whereas Ras is membranes of nerve tissue. Solubilize the similar to a go subunit Na+-K+ ATPase from its native membrane using 4.答:主要是从创造差异对细胞生命活动的 detergents. Reconstitute the enzyme into

2 5.答: b 6. 答:d 7. 答:a 8. 答:a 9. 答:a 10. 答:d 11. 答:d 12. 答:b 13. 答:b 14. 答:a 15. 答:c 16．答:d 17．答:c 18. 答:c 19. 答:c 20. 答:d 四、简答题(每题 4 分，选做 5 题，共 20 分) 1.答：①基因突变；②受体细胞—胚胎干细胞；选择标记 2. 答：①真核生物的 18S rRNA-5.8S rRNA-28S rRNA 串联在一起；原核生物的 16S rRNA-5SrRNA-23S rRNA 串联在一起；②真核生物的 18S rRNA-5.8S rRNA-28S rRNA 物中不含 tRNA 基因；原核生物的 16S rRNA-5SrRNA-23S rRNA 转录物中有 tRNA 基因的转录产物。 3. Answer:The activation of Ras requires the assistance of another protein called SOS. Upon binding 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 GB subunits, whereas Ras is similar to a G 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 (2)答:IP3的钙的释放和对钙浓度的消减。 By synthesizing different amounts of saturated and unsaturated 2.答案: 个分子量4800的蛋白质由约40个氨 fatty acids, the fluidity of the liposome can be 酸组成,因此有1.1X102(=2040) altered, and the enzyme can be tested under 个不同方式去组成这个蛋白,其中每种蛋白每个分子重8x10-g(=4800/(6X10)。因此由 conditions of different fluidity. (To measure the 每一种各一个分子组成的一个混合物重9X fluidity of the liposomes directly, without having10g(=8x10gX1.1x10),它是地球总重量6X10kg的15000倍。你真需要一个非常 infer it from the fatty acid composition, either 巨大的容器。 FRAP or SPT techniques can be applied. 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 2. Answer 可能的氨基酸序列中的很小一部分 The functional domains of G-protein coupled receptors were determined by2.综合思考(10分,按答题最多者计算每题得 experiments using recombinant chimeric receptor proteins containing parts of the 分) B2 and a2 adrenergic receptors. These Answe chimeric receptors were tested for their (a) Both the ECM of animal cells and the ligand binding specificity and the walls around plant cells ability to activate or inhibit adenylate consist of long id fibers cyclase. The results of these studies embedded in an amorphous demonstrated that a helix 7 and the hydrated matrix of branched- C-terminal domain of the receptor play a hain lecules ei ther role in determining ligand binding glycoproteins (ECM) specificity and the cyto-solic le polysaccharides (cell wall) between a helices 5 and 6 interacts with (b) For ECm, the fibers consist of G proteins ollagen and the matrix is a network of proteoglycans. For cell walls. the fibe 六、分析、计算与思考(20分) of cellulose and the 1.分析题(任选1题,5分) a network of polysaccharide and proteins ANSWER: (c) Both ECM and cell walls are The binding of epinephrine to B-adrenerg important in maintaining cell receptors causes a stimulation of shape and in retaining water heart function, both in terms of thereby resisting compression heart rate and with respect to the (d)Roles unique to the ECM include regulation mount of work done in pumping blood luI processes such as adhesion, The effect appears to be mediated by cyclic AMP. When an antagonist such motility, and differentiation embryonic development as the beta blocker propranolol is Role cell walls given to patients with hypertension he cellular response caused by the include protection of the cell binding of epinephrine to beta from mechanical in jury and eceptors is partially inhibited. microbial invasion, as well as Heart function is gradually provision of the mechanical restored over a period of time, with support necessary to withstand the turgor pressure that gives a corresponding decrease in blood plant tissues their rigidity. pressure

3 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 2 and 2 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  helix 7 and the C-terminal domain of the receptor play a role in determining ligand binding specificity and the cyto-solic loop between  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 -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-21 g 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 branchedchain 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