c. bind to different receptors and activate different second messengers. d. bind to the same receptors, one being intracellular and the other, extracellular. 答:b
17. Which mutant form of ras is likely to cause malignancy? a. ras that cannot hydrolyze GTP b. ras that cannot bind to GTP c. ras that cannot bind to Grb2 or Sos d. ras that cannot bind to Raf 答:a
The sequence of events that occurs after insulin binds to its receptor tyrosine kinase is:
a. binding of IRS -> phosphorylation of proteins with SH2 domains -> effect b. binding of proteins with SH2 domains -> phosphorylation of IRS -> effect c. autophosphorylation and phosphorylation of IRS -> binding of proteins with SH2 domains -> effect
autophosphorylation and binding of IRS -> phosphorylation of proteins with SH2 domains -> effect
19. 下列连接方式中属于与中间纤维相连的锚定连接的是( c )。 a．粘合带； b．粘合斑； c．桥粒； d．紧密连接。
20. 缬氨霉素是一种可形成通道的离子载体, 能够使( a )的运输速度提 高100,000倍。
a.K+ ; b.Na+ ; c.Ca2+ ; d.H+
四、简答题(每题4分， 选做5题，共20分)
1．Describe the similarities and differences in the cycling of Gs and Ras between the active and inactive forms.
Answer 1:. Binding of hormone to its receptor causes a con-formational change in the receptor. The receptor then binds to Gm protein causing a conformational change in the Gsa protein and release of the bound GDP. GTP then binds to Gsa which causes release of the GTP-Gsa complex from the b and ¡ subunits. The GTP-Gsa complex then activates adenylyl cyclase. Activation is short-lived because the intrinisic GTPase activity in Gsa hydrolyzes GTP to GDP and returns G~ to its inactive form.
2. 如何理解“被动运输是减少细胞与周围环境的差别,而主动运输则是努 力创造差别,维持生命的活力”?
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。
3. 紧密连接除了连接细胞外还有什么作用?意义何在?
答: 紧密连接除了连接细胞之外，还有两个作用：防止物质双向渗漏，并限制了膜蛋白在脂分子层的流动，维持细胞的极性。
4. 胰高血糖素和肾上腺素是如何使靶细胞中的cAMP的浓度升高的?
答: 胰高血糖素和肾上腺素作为第一信使作用于靶细胞的膜受体, 通过G蛋白偶联细激活腺苷酸环化酶,将ATP生成cAMP。
5. 核酶是如何被发现及证实的? 这一发现有什么意义?
答: 1981年，Thomas Cech和他的同事在研究四膜虫的26S rRNA前体加工去除基因内含子时获得一个惊奇的发现∶内含子的切除反应发生在仅含有核苷酸和纯化的26S rRNA前体而不含有任何蛋白质催化剂的溶液中,可能的解释只能是:内含子切除是由26S rRNA前体自身催化的,而不是蛋白质。
6. 为什么说水是细胞中优良的热缓冲体系？
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、实验证明题。(简要说明，不需详细步骤。每小题5分,任选2题， 共10分)。
1. Some integral membrane enzymes depend upon the lipids in their microenvironment not only for scaffolding, but for enzymatic activity. The Na+-K+ ATPase is one example. Please design an experiment to test the influence of membrane fluidity on the velocity of the Na+-K+ ATPase. Answers:
1. 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. 如何证明RNA聚合酶Ⅲ进行5S rRNA基因转录时, 使用的是内部启动子?
可通过基因操作。如将5S rRNA基因的5＇侧的上游序列完全除去，检测转录情况, 然后再切去5S rRNA基因的部分内部序列,再检测转录情况。实验结果表明: 将5S rRNA基因的5＇侧的上游序列完全除去并不影响5S rRNA基因的转录。但是，如果将5S rRNA基因内部缺失一部分序列(从50位到80位缺失)，RNA聚合酶Ⅲ不仅不能转录这段DNA，甚至不能结合上去(图6E-1)。如果将5S rRNA基因的内部启动子序列插入到基因组的其它部位，会使插入的部位形成一个新的转录起点。
图E6-1 RNA聚合酶Ⅲ转录5S rRNA基因启动子的鉴定
将图中5S rRNA基因的A片段或D片段缺失都不影响RNA聚合酶Ⅲ的转录,但是将B片段或C片段缺失,都会影响RNA聚合酶Ⅲ的转录。表明5S rRNA基因的启动子位于5S基因内部的控制区。
3. 请设计一个实验研究受体与配体结合的特异性
克采用非放射性标记的底物同放射性标记的配体竞争受体的结合位点的方法，原理是: 如果结合是特异性的话,只有信号分子能够同受体结合,而与信号分子无关的分子则不能同受体结合。例如,放射性标记的胰岛素与受体的结合不会受葡萄糖或ACTH(促肾上腺皮质激素)的抑制,但是能够被非放射性标记的胰岛素或胰岛素衍生物所抑制(图E5-2)。
图5-17 证明激素与受体结合特异性的实验
实验中,将放射性标记的胰岛素与分离的膜一起温育,同时加入各种不同浓度的葡萄糖或ACTH。与胰岛素无关的激素不会与放射性标记的胰岛素竞争质膜受体。通过检测放射性即可证明。
六、分析、计算与思考(20分)
1. “解铃还需系铃人”这句谚语在细胞活动中能找到对应的故事吗？请 说明。（5分）
答：IP3的钙的释放和对钙浓度的消减。
2. 计算题（5分）
① The average molecular weight of proteins in the cell is about 30,000 daltons. A few proteins, however, are very much larger. The largest known polypeptide chain made by any cell is a protein called titin (made by muscle cells), and it has a molecular weight of 3,000,000 daltons. Estimate how long it will take a muscle cell to translate an mRNA coding for titin (assume the average molecular weight of an amino acid to be 120, and a translation rate of two amino acids per second for eucaryotic cells). ② Transcription occurs at a rate of about 30 nucleotides per second. Is it possible to calculate the time required to synthesize a titin mRNA from the information given here? 2. Answer:
① A titin molecule is made of 25,000 amino acids. It therefore takes about 3.5 hours to synthesize a single molecule of titin in muscle cells.
② To calculate the time it takes to transcribe a titin mRNA, you would need to know the size of its gene, which is likely to contain many introns. Transcription of the exons alone requires about 42 minutes. Because
introns can be quite large, the time required to transcribe the entire gene is likely to be considerably longer.
2. 综合思考(10分, 按答题最多者计算每题得分)
① 物质运输是一个耗能过程，请举例说明有哪些能量来源？
答：①自由能；②ATP；③光能；④磷酸烯醇式丙酮酸；⑤动力势（协同）
②． What is the advantage of having an extracellular signal transmitted by a cascade of sequential events?
Answer②:Transduction of an extracellular signal via a cascade of sequential events is advantageous because of the amplification of the signal. The binding of a few molecules to a receptor can result in the synthesis of a large number of effector molecules because at each step of the cascade the signal is enzymatically amplified.
③． Heating of calf type I collagen fibers to 450C denatures the triple helices and separates the three chains from each other. Collagen that has been treated in this way does not renature to form a normal collagen triple helix. Why?
Answer③The N-terminal and C-terminal propeptides pre-sent 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.
试题二参考答案
一、填空题(每空0.5分，共10分) 1. 一个导肽，两个导肽。
2. 微丝装配与去装配假说，肌动蛋白与肌球蛋白相互作用。 3. Proton-motive
4. “START”基因；发现了CDK。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. 答：胞质溶胶中的ATP; 跨线粒体内膜的质子动力势; 7. 除去叶绿素不能吸收的杂色光。 8. 笼型蛋白，包被蛋白
9. 答： ①形成纺锤体, 将染色体拉向两极; ②胞质分裂; 10. B①引导新生蛋白进入内质网； ②帮助蛋白质正确装配。 11.①连接两个姊妹染色单体； ②为动力微管的装配提供位点。
二、判断题(若是正确的标√号， 错误的标×号，每题1分，共15分) 1. ∨ 2. ×
3. (×，Not only do plants need mitochondria to make ATP in cells that
do not have chloroplasts, such as root cells, but mitochondria make most of the cytosolic ATP in all plant cells.） 4. ×
5. 答: T,错误, 同源染色体间的分子重组是随机发生的。 6. 答：T, 正确。 7. 答: T, 正确;
8. 答：T, 错误, 8个小亚基是核基因编码;
9.答: T, 错误, 不一样,α亚基结合GTP后不被水解, 也不与GDP交换, 而β亚基上 的GTP结合位点是可交换位点;
10. 答: T, 错误, 中心粒中的微管是三联管。 11. 答: T, 正确;
12.答:T, 错误, M6P受体也位于细胞质膜上。 13. 答：T, 正确; 14. 答：T, 正确; 15. 错误,粉末状的染色体
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.（d） 2. (d) 3. (b) 4. (c) 5. (b) 6. (d) 7. t( b ) 8. (b) 9. ( a ) 10. ( d )。 11. ( b )。 12. ( b ) 13. ( d ) 14. ( c ) 15. ( a )。
四、简答题(选做4题，每题5分，20分) 1. Answer1:
(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.
Answer2. Regulated secretion occurs only in response to a signal. The proteins to be secreted