武汉大学分子生物学真题2001-2014
分子生物学选择题试题及答案
分子生物学选择题试题及答案备注（爱写啥写哈） [填空题]_________________________________001 DNA二级结构模型是（） [单选题]A.α一螺旋B.走向相反的右手双螺旋(正确答案)C.三股螺旋D.走向相反的左手双螺旋E.走向相同的右手双螺旋002在核酸分子中核苷酸之间连接的方式是（） [单选题]A.2′－3′磷酸二酯键B.2′－5′磷酸二酯键C.3′－5′磷酸二酯键(正确答案)D.肽键E.糖苷键003自然界游离核苷酸中，磷酸最常见是位于（） [单选题]A.戊糖的C-5′上(正确答案)B.戊糖的C-2′上C.戊糖的C-3′上D.戊糖的C-2′和C-5′上E.戊糖的C-2′和C-3′上004可用于测量生物样品中核酸含量的元素是（） [单选题]A.碳B.氢C.氧D.磷(正确答案)005下列哪种碱基只存在于RNA而不存在于DNA（） [单选题]A.尿嘧啶(正确答案)B.腺嘌呤C.胞嘧啶D.鸟嘌呤E.胸腺嘧啶007核酸对紫外线的最大吸收峰在哪一波长附近？（） [单选题]A.280nmB.260nm(正确答案)C.200nmD.340nmE.220nm008 DNA分子中的碱基组成是（）A.A+C=G+T(正确答案)B.C=G(正确答案)C. A=T(正确答案)D. C+G=A+T009某DNA分子中腺嘌呤的含量为15%，则胞嘧啶的含量应为（） [单选题]A.15%B.30%C.40%D.35%(正确答案)E.7%010 DNA水解后可得到下列哪些最终产物（）A.磷酸(正确答案)C.腺嘌呤、鸟嘌呤(正确答案)D.胞嘧啶、尿嘧啶011关于DNA的碱基组成，正确的说法是（）A.腺嘌呤与鸟嘌呤分子数相等，胞嘧啶与胸嘧啶分子数相等B.不同种属DNA碱基组成比例不同(正确答案)C.同一生物的不同器官DNA碱基组成不同D.年龄增长但DNA碱基组成不变(正确答案)012 DNA二级结构特点有（）A.两条多核苷酸链反向平行围绕同一中心轴构成双螺旋(正确答案)B.以A-T，G-C方式形成碱基配对(正确答案)C.双链均为右手螺旋(正确答案)D.链状骨架由脱氧核糖和磷酸组成(正确答案)013真核生物DNA的高级结构包括有（）A.核小体(正确答案)B.环状DNAC.染色质纤维(正确答案)D.α-螺旋014下列是几种DNA分子的碱基组成比例。
2001-2013年武汉大学分子生物学研究生入学考试真题
武汉大学2001年攻读硕士学位研究生入学考试试题科目名称:分子生物学科目代码: 477一、解释概念(20分,每个4分)卫星DNA 复制体逆转座子反式激活因子衰减子与衰减作用二、填空(30分,每空1分,请将答案写在答卷上)1. 从病毒到高等生物的研究表明,遗传物质是。
2. 冈崎片段的发现证实了双链DNA的复制,在复制过程中,一条新生链的合成是的,称为链;而另一条链的合成是的,称为链。
3. 大肠菌中有三种DNA聚合酶,其中的pol I的作用是,而pol III的作用是。
pol I和pol III都有的三种活性是、、。
4.由于真核细胞染色体DNA的复制要有一段RNA为引物,因此线状的DN A复制后必须存在着5’端缩短的问题。
已发现有一种端粒蛋白称为,它由构成,可以使单链DNA的5’延长。
5. 对DNA损伤有几种修复系统,其中只有修复系统可以造成DNA变异,与这一系统有关的一套基因平时受到一称为的抑制蛋白所抑制,它发挥抑制作用是结合在一段约20bp长的称为的DNA序列上,当DNA损伤时,另一种蛋白质称为把这种抑制蛋白水解后,修复系统的基因才会被激活。
6.真核细胞中有三种依赖于DNA的RNA聚合酶分别合成不同的RNA,RNA pol I负责合成,RNA pol II负责合成,RNA pol III负责合成。
7.大分子互相作用是分子生物学的重要内容,包括蛋白质之间、蛋白质与DNA或RNA之间的互相作用,蛋白质有四种重要的结构花式与大分子互相作用有关,这些结构花式是, , 。
8. NO是气体小分子信号,它可由脱氨产生,它的作用方式是直接与酶作用使产生cGMP(环式GMP)。
9. 真核mRNA的5’ 端通常有帽子结构,3’ 端有polyA。
在polyA上游有一保守序列称为polyA信号,其序列为。
polyA能提高mRNA的翻译水平是由于:(1) (2) 。
10. G-蛋白关联受体是一类重要的细胞表面受体,它的结构特色是,它发挥信号传递作用的两条途径是途径和途径。
分子生物学试题及答案
分子生物学试题及答案一、单项选择题（每题2分，共20分）1. DNA分子的双螺旋结构是由谁提出的？A. 沃森和克里克B. 达尔文C. 孟德尔D. 摩尔根答案：A2. 以下哪个不是DNA聚合酶的功能？A. 合成DNA链B. 修复DNA损伤C. 催化RNA转录D. 校对新合成的DNA链答案：C3. 真核生物的mRNA帽子结构位于其5'端，其主要功能是什么？A. 促进翻译B. 保护mRNA不被降解C. 促进mRNA的剪接D. 促进mRNA的运输答案：B4. 以下哪种RNA分子在蛋白质合成中不直接参与？A. mRNAB. tRNAC. rRNAD. snRNA5. 基因表达调控中，转录因子的作用是什么？A. 提供转录所需的能量B. 识别并结合到特定的DNA序列上C. 催化DNA复制D. 促进DNA修复答案：B6. 以下哪种技术用于研究基因功能？A. PCRB. DNA测序C. 基因敲除D. DNA指纹分析答案：C7. 以下哪个不是DNA复制的特点？A. 半保留复制B. 需要引物C. 双向复制D. 需要逆转录酶答案：D8. 以下哪个不是RNA干扰（RNAi）的作用机制？A. 降解特定的mRNAB. 抑制基因表达C. 促进基因突变D. 沉默特定基因答案：C9. 以下哪个是真核生物中常见的非编码RNA？B. siRNAC. tRNAD. rRNA答案：A10. 以下哪个是原核生物和真核生物共有的基因表达调控机制？A. 转录后修饰B. 转录因子调控C. mRNA剪接D. 核糖体结合位点调控答案：B二、填空题（每空1分，共20分）1. DNA分子的基本组成单位是_______，而RNA分子的基本组成单位是_______。
答案：脱氧核苷酸；核糖核苷酸2. 在DNA复制过程中，_______酶负责解开双链DNA，而_______酶负责合成新的DNA链。
答案：解旋酶；DNA聚合酶3. 真核生物的基因表达调控主要发生在_______水平，而原核生物的基因表达调控主要发生在_______水平。
武汉大学分子生物学2004,2008,2013--2014年考博真题
四，简答题：16分 1.为什么rRNA分子和tRNA分子比mRNA分子更为稳定？ 2.试比较一下原核和真核热休克基因转录起始机制有何异同点？
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3．请叙述一下你的硕士论文的情况，具体方法，结果和意义等等 4.用限制性内切酶切割DNA后，经电泳检查，发现有脱尾现象，其可能的原因 是什么呢？ 五，问答题：33分 1.请叙述原核生物当中从核酸指导到蛋白？ 3.请叙述一种从接受信号到调控基因表达的信号转导途径
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6.分裂间期的早期，DNA处于（ ）状态 A.单体连续的线性双螺旋分子 B.半保留复制的双螺旋结构 C.保留复制的双螺旋结构 D.单链DNA E.以上都不正确 7、能够证明DNA是遗传物质的2个关键性实验是：肺炎链球菌在老鼠体内的毒 性和T2噬菌体感染大肠杆菌。这两个实验主要的论点证据是： (A)从被感染的生物体内重新分离得到DNA，作为疾病的致病剂 (B)DNA突变导致毒性丧失 (C)生物体吸收的外源DNA（而并非蛋白质）改变了其遗传潜能 (D)DNA是不能在生物体间转移的，因此它一定是一种非常保守的分子 8.沉默基因座 A.因为沉默子区域的存在与MAT基因座不同 B.在SIR基因产物的作用下，保持转录失活 C.存在几个DNase I超敏位点 D.与DNA复制起点结合在一起 E.因为染色质结构保持转录失活 9.基因组是 A．一个生物体内所有基因的分子总量 B.一个二倍体细胞中的染色体数 C.遗传单位 D.生物体的一个特定细胞内所有基因的分子总量 10、原核细胞mRNA含有几个功能所必需的特征区段，它们是 （A）启动子，SD序列，起始密码子，终止密码子，茎环结构 （B）启动子，转录起始位点，前导序列，由顺反子间区序列隔开的SD序列和OR F，尾部序列，茎环结构 （C）转录起始位点，尾部序列，由顺反子间区序列隔开的SD序列和ORF，茎环 结构
武汉大学历年生物化学真题
名词解释1、基因：能够表达和产生蛋白质和RNA的DNA序列，决定遗传性状的功能单位。
2、基因组：指含有一个生物体存在、发育、活动和繁殖所需要的全部遗传信的整套核酸，即一特定生物体的整套遗传物质的总会。
3、端粒：指以线性染色体的形式存在于真核基因组DNA末端的特殊结构。
4、操纵子：是指数个功能上相关的结构基因串联在一起，构成信息区，连同其上游的调控区（包括启动子和操纵基因）以及下游的转录终止信号所构成的基因表达单位，所转录的RNA为多顺反子5、顺式作用元件：是指那些与结构基因表达调控相关、能够被基因调控蛋白特异性识别和结合的特异DNA序列。
包括启动子、上游启动子元件、增强子、加尾信号和一些反应元件等6、反式作用因子：是指真核细胞内含有的大量可以通过直接或间接结合顺式作用元件而调节基因转录活性的蛋白质因子。
7、启动子：是RNA聚合酶特异性识别和结合的DNA序列。
8、增强子：位于真核基因中远离转录起始点，能明显增强启动子转录效率的特殊DNA序列。
它可位于被增强的转录基因的上游或下游，也可相距靶基因较远。
9、基因表达：是指生物基因组中结构基因所携带的遗传信息经过转录、翻译等一系列过程，合成特定的蛋白质，进而发挥其特定的生物学功能和生物学效应的全过程。
10、错义突变：DNA分子中碱基对的取代，使得mRNA的某一密码子发生变化，由它所编码的氨基酸就变成另一种的氨基酸，使得多肽链中的氨基酸顺序也相应的发生改变的突变。
11、无义突变：由于碱基对的取代，使原来可以翻译某种氨基酸的密码子变成了终止密码子的突变。
12、同义突变：碱基对的取代并不都是引起错义突变和翻译终止，有时虽然有碱基被取代，但在蛋白质水平上没有引起变化，氨基酸没有被取代，这是因为突变后的密码子和原来的密码子代表同一个氨基酸的突变。
13、移码突变：在编码序列中，单个碱基、数个碱基的缺失或插入以及片段的缺失或插入等均可以使突变位点之后的三联体密码阅读框发生改变，不能编码原来的蛋白质的突变。
(完整版)分子生物学习题与答案
第0 章绪论一、名词解释1 .分子生物学2 .单克隆抗体二、填空1．分子生物学的研究内容主要包含（）、（）、（）三部分。
三、是非题1、20 世纪60 年代，Nirenberg 建立了大肠杆菌无细胞蛋白合成体系。
研究结果发现poly（U）指导了多聚苯丙氨酸的合成，poly（G）指导甘氨酸的合成。
（×）四、简答题1. 分子生物学的概念是什么？2. 你对现代分子生物学的含义和包括的研究范围是怎么理解的？3. 分子生物学研究内容有哪些方面？4. 分子生物学发展前景如何？5. 人类基因组计划完成的社会意义和科学意义是什么？6．简述分子生物学发展史中的三大理论发现和三大技术发明。
7. 简述分子生物学的发展历程。
8. 二十一世纪生物学的新热点及领域是什么？9. 21 世纪是生命科学的世纪。
20 世纪后叶分子生物学的突破性成就，使生命科学在自然科学中的位置起了革命性的变化。
试阐述分子生物学研究领域的三大基本原则，三大支撑学科和研究的三大主要领域？答案：一、名词解释1 .分子生物学：分子生物学就是研究生物大分子之间相互关系和作用的一门学科，而生物大分子主要是指基因和蛋白质两大类；分子生物学以遗传学、生物化学、细胞生物学等学科为基础，从分子水平上对生物体的多种生命现象进行研究。
2 .单克隆抗体：只针对单一抗原决定簇起作用的抗体。
二、填空1. 结构分子生物学，基因表达与调控，DNA 重组技术三、是非题四、简答题1. 分子生物学的概念是什么？答案：有人把它定义得很广：从分子的形式来研究生物现象的学科。
但是这个定义使分子生物学难以和生物化学区分开来。
另一个定义要严格一些，因此更加有用：从分子水平来研究基因结构和功能。
从分子角度来解释基因的结构和活性是本书的主要内容。
2. 你对现代分子生物学的含义和包括的研究范围是怎么理解的？分子生物学是从分子水平研究生命本质的一门新兴边缘学科，它以核酸和蛋白质等生物大分子的结构及其在遗传信息和细胞信息传递中的作用为研究对象，是当前生命科学中发展最快并正在与其它学科广泛交叉与渗透的重要前沿领域。
武汉大学《分子生物学》复习题库及答案
武汉大学《分子生物学》复习题库及答案考试复习重点资料（最新版）资料见第二页封面第1页分子生物复习题及答案一、填空题1．病毒ΦX174及M13的遗传物质都是单链DN。
2．IDS病毒的遗传物质是单链RN。
3．X射线分析证明一个完整的DN螺旋延伸长度为3.4nm。
4．氢键负责维持-T间（或G-C间）的亲和力5．天然存在的DN分子形式为右手B型螺旋。
二、选择题（单选或多选）1．证明DN是遗传物质的两个关键性实验是：肺炎球菌在老鼠体内的毒性和T2噬菌体感染大肠杆菌。
这两个实验中主要的论点证据是（C）。
．从被感染的生物体内重新分离得到DN作为疾病的致病剂B．DN突变导致毒性丧失C．生物体汲取的外源DN（而并非蛋白质）改变了其遗传潜能D．DN是不能在生物体间转移的，因此它一定是一种非常保守的分子E．真核心生物、原核生物、病毒的DN能相互混合并彼此替代2．1953年Wtson和Crick提出（）。
．多核苷酸DN链通过氢键连接成一个双螺旋B．DN的复制是半保留的，常常形成亲本-子代双螺旋杂合链C．三个连续的核苷酸代表一个遗传密码D．遗传物质通常是DN而非RNE．分离到回复突变体证明这一突变并非是一个缺失突变3．DN双螺旋的解链或变性打断了互补碱基间的氢键，并因此改变了它们的光汲取特性。
以下哪些是对DN的解链温度的正确描述？（C、D）．哺乳动物DN约为45℃，因此发烧时体温高于42℃是十分危险的B．依赖于-T含量，因为-T含量越高则双链分开所需要的能量越少C．是双链DN中两条单链分开过程中温度变化范围的中间值D．可通过碱基在260nm的特征汲取峰的改变来确定E．就是单链发生断裂（磷酸二酯键断裂）时的温度4．DN的变性（、C、E）。
．包括双螺旋的解链B．可以由低温产生C．是可逆的D．是磷酸二酯键的断裂E．包括氢键的断裂5．在类似RN这样的单链核酸所表现出的“二级结构”中，发夹结构的形成（、D）。
．基于各个片段间的互补，形成反向平行双螺旋B．依赖于-U含量，因为形成的氢键越少则发生碱基配对所需的能量也越少C．仅仅当两配对区段中所有的碱基均互补时才会发生D．同样包括有像G-U这样的不规则碱基配对E．同意存在几个只有提供过量的自由能才能形成碱基对的碱基6．DN分子中的超螺旋（、C、E）。
武汉大学分子生物学真题2001-2014汇总.
一.解释概念(20分,每个4分)卫星DNA 复制体逆转座子反式激活因子衰减子与衰减作用三、问答题(50分)1. 说出双链DNA复制起始有关的五种重要的酶或蛋白并简述它们的功能。
(15分)2. 简述增强子的特点和性质及作用机制。
(10分)3. 简述真核RNA聚合酶II的转录起始复合物装配过程和转录起始(15分)4. DNA限制性内切酶EcoRI是人们熟悉的常用内切酶,它是在大肠杆菌(E.coli)R株中发现的,它被广泛用于分子克隆操作和DNA分析。
pUC质粒是常用克隆载体之一,它的多克隆位点上有EcoRI、BamHI、KpnI、HindIII等酶切点。
假如要你把一段由EcoRI切割产生的外源DN**段克隆到pUC质粒中,并把重组质粒转化大肠杆菌R株来扩增,已知条件是所用的R菌株中只有EcoRI一种限制性内切酶,你设计如何做才能确保成功?为什么?(10分)武汉大学2002分子生物学三.问答:1.简述(或绘图说明)真核细胞RNA聚合酶II转录的起始需要哪些基本转录因子及其装配过程(15分)2.简述(或绘图说明)色氨酸操纵子弱化的机制(15分)3.在讨论基因家庭时经常提到胚胎、胎儿和成体形成的蛋白质,这些述语是指什么现象?可用什么术语来描述这一类基因家族(5分)4. 你正在进行Southern blot分析,并刚刚完成凝胶电泳部分,下一步是将胶浸泡在NaOH溶液中使DNA变性为单链,为了节约时间,你跳过这一步,直接把DNA 从胶上转到硝酸纤维素膜上,你将标记好的探针与膜杂交,却发现放射自显影结果是一片空白,哪里错了呢?(5分)一、下列名词翻译成中文，并简要解释1、Domains and motifs2、Alternative splicing3、Reporter genes4、The PCR cycle5、Restriction mapping6、Multiple cloning sites7、DNA libraries8、Proteomics9、Replicon10、Semi-conservative replication二、简答题(共5题,每题8分,共40分)1、请列举三种以上蛋白质纯化技术,并说明不同纯化技术的简单原理。
武大分子生物学考研历年真题
武大分子生物学考研历年真题对于许多立志在生物学领域深造的学子来说，报考武汉大学分子生物学专业的研究生是一个极具吸引力的选择。
而深入研究历年真题，无疑是备考过程中的关键环节。
武大分子生物学考研真题涵盖了丰富多样的知识点和题型。
从细胞生物学基础，到分子遗传学的原理，再到基因表达调控等诸多方面，都有深入且细致的考查。
在细胞生物学基础部分，真题常常涉及细胞的结构与功能、细胞的信号转导通路等内容。
例如，会要求考生阐述细胞膜的组成成分和结构特点，以及它们如何影响细胞内外物质交换和信息传递。
又或者，详细描述细胞内各种细胞器的功能，以及它们之间是如何协同工作以维持细胞正常生理活动的。
分子遗传学原理方面，真题可能会聚焦于 DNA 复制、转录、翻译的过程和机制。
比如，让考生解释 DNA 聚合酶在复制过程中的作用和特点，或者描述 RNA 聚合酶如何识别启动子并启动转录。
基因表达调控更是重点中的重点。
这部分真题可能会让考生分析不同层次的基因表达调控机制，包括转录水平、转录后水平、翻译水平和翻译后水平的调控。
例如，要求阐述转录因子如何结合到 DNA 上调控基因的表达，或者解释 microRNA 是如何通过影响 mRNA 的稳定性来调节基因表达的。
真题的题型也多种多样。
常见的有选择题，考查考生对基本概念和知识点的理解和记忆。
判断题则侧重于考察考生对一些易混淆概念的辨析能力。
简答题要求考生能够简洁明了地回答问题，展现对知识点的准确把握和概括能力。
论述题通常是综合性较强的题目，需要考生将多个相关知识点串联起来，进行深入的分析和阐述，以展现其对知识的系统理解和运用能力。
通过对历年真题的分析，可以发现一些出题的规律和趋势。
一方面，重点知识点会反复出现，只是考查的角度和形式可能有所不同。
另一方面，随着学科的发展和研究的深入，真题也会逐渐融入一些新的研究成果和前沿知识。
例如，近年来，随着基因编辑技术的迅速发展，真题中可能会出现与CRISPRCas 系统相关的题目，要求考生了解其原理和应用。
武汉大学分子生物学真题2001-2014
一.解释概念(20分,每个4分)卫星DNA 复制体逆转座子反式激活因子衰减子与衰减作用三、问答题(50分)1. 说出双链DNA复制起始有关的五种重要的酶或蛋白并简述它们的功能。
(15分)2. 简述增强子的特点和性质及作用机制。
(10分)3. 简述真核RNA聚合酶II的转录起始复合物装配过程和转录起始(15分)4. DNA限制性内切酶EcoRI是人们熟悉的常用内切酶,它是在大肠杆菌(E.coli)R株中发现的,它被广泛用于分子克隆操作和DNA分析。
pUC质粒是常用克隆载体之一,它的多克隆位点上有EcoRI、BamHI、KpnI、HindIII等酶切点。
假如要你把一段由EcoRI切割产生的外源DN**段克隆到pUC质粒中,并把重组质粒转化大肠杆菌R株来扩增,已知条件是所用的R菌株中只有EcoRI一种限制性内切酶,你设计如何做才能确保成功?为什么?(10分)武汉大学2002分子生物学三.问答:1.简述(或绘图说明)真核细胞RNA聚合酶II转录的起始需要哪些基本转录因子及其装配过程(15分)2.简述(或绘图说明)色氨酸操纵子弱化的机制(15分)3.在讨论基因家庭时经常提到胚胎、胎儿和成体形成的蛋白质,这些述语是指什么现象?可用什么术语来描述这一类基因家族(5分)4. 你正在进行Southern blot分析,并刚刚完成凝胶电泳部分,下一步是将胶浸泡在NaOH溶液中使DNA变性为单链,为了节约时间,你跳过这一步,直接把DNA 从胶上转到硝酸纤维素膜上,你将标记好的探针与膜杂交,却发现放射自显影结果是一片空白,哪里错了呢?(5分)一、下列名词翻译成中文，并简要解释1、Domains and motifs2、Alternative splicing3、Reporter genes4、The PCR cycle5、Restriction mapping6、Multiple cloning sites7、DNA libraries8、Proteomics9、Replicon10、Semi-conservative replication二、简答题(共5题,每题8分,共40分)1、请列举三种以上蛋白质纯化技术,并说明不同纯化技术的简单原理。
武汉大学分子生物学2001—2015考研真题(1)
武汉大学2015年分子生物学武汉大学2014年分子生物学一.名词解释1、complementDNA2、RNAsplicing3、Gel filtration chromatography4、gene5、snRNP6、OKAZIKIN sequence7、operon8、enhancesome一.简答题1.从分子角度谈谈为什么DNA比RNA更广泛地用作遗传信息的携带者2.试说明中介蛋白在转录调控中的作用原理3.以下是DNA的电泳条带，说明DNA有哪三种拓扑异构构型？设计实验验证该结果。
4.翻译过程需要哪些组分？各自的功能分别是什么三.论述题1.CTD是RNAPOL2的末端结构，说明其结构特点。
它在转录各过程的作用是什么？2.试从产生过程，作用机制和生物学功能几个方面比较miRNA和siRNA的异同点3蛋白质A可以上调蛋白质B的表达水平，说明蛋白A可能在哪些水平发挥作用并设计实验证明。
武汉大学2013年分子生物学一、名词解释（共10小题，每题4分，共40分）1.Ribozyme2.Missense Mutation3.Insulator4.RNA trans-splicing5.Nucleotide excision repair6.Transposon7.Dam methylase8.Spliceosome9.Core promotor10. SNP二、简答题（共5小题，每题10分，共50分）1.真核生物mRNA的5’帽子结构有何生物学功能？其缺失会造成什么后果？2.某一基因开放阅读框中的一个碱基突变会对该基因编码产物产生什么影响？3.请简述真核生物的mRNA和原核生物的mRNA有何不同。
4.请简述Ⅰ型内含子剪切的过程.5.请简述Western blot 的原理和步骤。
三、论述题（共3小题，每题20分，共60分）1.研究基因功能通常是降低或升高基因表达水平，请简要说出三种相关研究方法及原理。
2. mRNA和蛋白质的降解是一个有序的过程，其中microRNA和泛素起了非常重要的作用，请简述microRNA的概念、产生过程、作用机制，或者泛素的概念和蛋白质泛素化的过程。
武汉大学分子生物学题库1
GLOSSARYAbundance of an mRNA is the average number of molecules per cell.Abundant mRNAs consist of a small number of individual species, each present in a large number of copies per cell.Acceptor splicing site—see right splicing junction.Acentric fragment of a chromosome (generated by breakage) lacks a centromere and is lost cell division.Acrocentric chromosome has the centromere located nearer one end than the other.Active site is the restricted part of a protein to which a substrate binds.Allele is one of several alternative forms of a gene occupying a given locus on a chromosome. Allelic exclusion describes the expression in any particular lymphocyte of only one allele coding for the expressed immunoglobulin.Allosteric control refers to the ability of an interaction at one site of a protein to influence the activity of another site.Alu family is a set of dispersed, related sequences, each～300 bp long, in the human genome. The individual members have Alu cleavage sites at each end (hence the name).Alu-equivalent family is the set of sequences in a mammalian genome that is related to the human Alu family.α-Amanitin is a bicyclic octapeptide derived from the poisonous mushroom Amanita phalloides; it inhibhits transcription by certain eukaryotic RNA polymerases, especially RNA polymerase II.Amber codon is the nucleotide triplet UAG, one of three codons that cause termination of protein synthesis.Amber mutaion describes any change in DNA that creates an amber codon at a site previously occupied by a codon representing an amino acid in a protein.Amber suppressors are mutant genes that code for tRNAs whose anticodons have been altered so that they can respond to UAG codons as well as or instead of to their previous codons.Aminoacyl-tRNA is transfer RNA carrying an amino acid; the covalent linkage is between the NH2group of the amino acid and either the 3’-or-2’-OH group of the terminal base of the tRNA.Aminoacyl-tRNA synthetases are enzymes responsible for covalently linking amino acids to the 2’ or 3’-OH position of tRNA.Amphipathic structures have two surfaces, one hydrophilic and one hydrophobic. Lipids are amphipathic; and some protein regions may form amphipathic; and some protein regions may form amphipathic helices, with one charged face and one neutral face.Amplification refers to the production of additional copies of a chromosomal sequence,1found as intrachromosomal or extrachromoxomal DNA.Anchorage dependence describes the need of normal eukaryotic cells for a surface to attach to in order to grow in culture.Aneuploid chromosome constitution differes from the usal diploid constitution by loss or duplication of chromosomes or chromosomal segments. Annealing is the pairing of complementary single strands of DNA to form a double helix.Antibody is a protein (immunoglobulin) produced by B lymphocyte cells that recognizes a particular foreign ‘antigen,’and thusw triggers the immune res ponse.Anticoding strand of duples DNA is used as a template to direct the synthesis of RNA that is complementary to it.Antigen is any molecule whose entry into an organism provokes synthesis of an antibody (immunoglobulin).Antiparallel strands of the d ouble helix are organized in opposite orientation, so that the 5’ end of one strand is aligened with the 3’ end of the other strand.Antitermination proteins allow RNA polymerase to transcribe through certain terminator sites.Ap endonucleases make incisio ns in DNA on the 5’ side of either apurinic or apyrimidinc sites.Apoinducer is a protein that binds to DNA to switch on transcription by RNA polymerase. Archebacteria comprise a minor line of prokaryotes, and may have introns in the genome. Ascus of a fungus contains a tetrad or octad of the (haploid) spores, representhing the products of a sihngle meiosis.att sties are the loci on a phage and the bacterial chromosome at which recombination integrates the phage into. or excises it from , the bacterial chromosome.Attenuation describes the regulation of termination of transcription that is involved in controlling the expression of some bacterial operons.Attenuato r is the terminator sequence at which attenuatioj occurs.Autogenous conhtrol describes the action of a gene product that either inhibits (negative autogenous control) or activates (positive autogenous control) expression of the gene coding for it.Autonomous controlling element in maize is an active transposon with the ability to transpose (cf nonautonomous controlling element).Autoradiography detects radioactively labeled molecules by their effect in creating an image on pholtographic film.Autosomes are all the chromosomes except the sex chromosomes; a diploid cell has two copies of each autosome.Blymphocytes (or B cells) are the cells responsible for synthesizing antibodies.2Backcross is another (earlier) term for a testcross.Back mutation reverses the effect of a mutation that hand inactivated a gene; thus it restores wild type.Bacteriophages are viruses that infect bacteria; often abbreviated as phages.Balbaini ring is an extremely large puff at a band of a polytene chromosome.Bands of polytene chromosomes are visible as dense regions that contain the majority of DNA;bands of normal chromosomes are relatively much larger and are generated in the form of regions that retain a stain on certain chemical treatments.Base pair (bp) is a partnership of A with T or of C with G in a DNA double helix; other pairs can be formed in RNA under certain circumstances.Bidirectional replication is accomplished when two replication forks move away from the same origin in different directions.Bivalent is the structure containing all four chromatids (two representing each homologue) at the start of meiosis.Blastoderm is a stage of insect embryogenesis in which a layer of nuclei or cells around the embryo surround an internal mass of yolk.Blocked reading frame cannot be translated into protein because it is interrupted by termination codons.Blunt-end ligation is a reaction that joins two DNA duplex molecules directly at their ends. bp is an abbreviation for base pairs; distance along DNA is measured in bp.Branch migration describes the ability of a DNA strand partially paired with its complement in a duplex to extend its pairing by displacing the resident strand with which it is homologous. Breackage and reunion describes the mode of genetic recombination, in which two DNA duplex molecules are broken at corresponding points and then rejoined crosswise (involving formation of a length of heteroduplex DNA around the site of joining).Buoyant desity measures the ability of a substance to float in some standard fluid, for example, CsCl.C banding is a technique for generating stained regions around centromeres.C genes code for the constant regions of immunoglobulin protein chains.C value is the total amount of DNA in a haploid genome.CAAT box is part of a conserved sequence located upstream of the startpoints of eukaryotic transcription units; it is recognized by a large group of transcription factors.Cap is the structure at the 5’ end of eukaryotic mRNA, introduced after transcripton by linking the terminal phosphate of 5’ GTP to the terminal base of the mRNA. The added G (and sometimes some other bases) are methylated, giving a structure of the form 7Me G5’ppp 5’Np…3CAP(CRP)is a positive regulator protein activated by cyclic AMP. It is needed for RNA polymerase to initiate transcription of certain (catabolitesensitive) operons of E. coli.Capsid is the external protein coat of a virus particle.Catabolite repression describes the decreased expression of many bacterial operons that results from addition of glucose. It is caused by a decrease in the level of cyclic AMP, which in turn inactivates the CAP regulator.cDNA is a single-stranded DNA complementary to an RNA, synthesized from it by reverse transcription in vitro.cDNA clone is a duplex DNA sequence representing an RNA, carried in a cloning vector. Cell cycle is the period from one division to the next.Cell hybrid is a somatic cell containing chromosomes derived from parental cells of different species (e.g. a man-mouse somatic cell hybrid), generating by fusing the cells to form a heterokaryon in which the nuclei subsequently fused.Centrioles are small hollow cylinders consisting of microtubules that become located near the poles during mitosis. They reside within the centrosomes.Centromere is a constricted region of a chromosome that includes the site of attachment to the mitotic or meiotic spindle (see also kinetochore).Centrosomes are the regions from which microtubules are organized at the poles of a mitotic cell. In animal cells, each centrosome contains a pair of centrioles surrounded by a dense amorphous region to which the microtubules attach. See also MTOC.Molecular chaperone is a protein that is needed for the assembly or proper folding of some other protein, but which is not itself a component of the target complex.Chemical complexity is the amount of a DNA component measured by chemical assay.Chi sequence is an octamer that provides a hotspot for RecA-mediated genetic recombination in E. coli.Chi structure is a joint between two duplex molecules of DNA revealed by cleaving an intermediate of two joined circles to generate linear ends in each circle. It resembles a Greek chi in outline, hence the name.Chiasma (pl. chiasmata)is a site at which two homologous chromosomes appear to have exchanged material during meiosis.Chromatids are the copies of a chromosome produced by replication. The name is usually used to describe them in the period before they separate at the subsequent cell division. Chromatin is the complex of DNA and protein in the nucleus of the interphase cell. Individual chromosomes cannot be distinguished in it .It was originally recognized by its reaction with stains specific for DNA.Chromocenter is an aggregate of heterochromatin from different chromosomes. Chromomeres are densely staining granules visible in chromosomes under certain conditions, especially early in meiosis, when a chromosome may appear to consist of a series of4chromomeres.Chromosome is a discrete unit of the genome carrying many genes. Each chromosome counsists of a very long molecule of duplex DNA and an approximately equal mass of proteins. It is visible as a morphological entity only during cell division.Chromosome walking describes the sequential isolation of clones carrying overlapping sequences of DNA, allowing large regions of the chromosome to be spanned. Walking is often performed in order to reach a particular locus of interest.cis-acting locus affects the activity only of DNA sequences on its own molecule of DNA; this property usually implies that the locus does not code for protein.cis-acting protein has the exceptional property of acting only on the molecule of DNA from which it was expressed.cis configuration describes two sites on the same molecule of DNA.cis/trans test assays the effect of relative configuration on expression of two mutations. In a double heterozygote, two mutations in the same gene show mutant phenotype in trans configuration, wild-type in cis configuration.Cistron is the geneti unit defined by the cis/trans test; equivalent to gene ib comprising a unit of DNA representing a protein.Class switching is a change in the expression of the c cregion of an immunoglobulin heavy chain during lymphocyte differentiation.Clone describes a large number of cells or molecules identical with a single ancestral cell or molecule.Cloning vector is a plasmid or phage that is used to ‘carry’ inserted foreign DNA for the purposes of producing more material or a protein product.Closed reading frame contains termination codons that prevent its translation into protein. Coated vesicles are vesicles whose membrane has on its surface a layer of the protein clathrin.Coconversion is the simultaneous correction of two sites during gene conversion.Coding strand of DNA has the same sequence as mRNA.Codominant alleles both contribute to the phenotype; neither is dominant over the other. Coevolution —see concerted evolution.Cognate tRNAs are those recognized by a particular aminoacy-tRNA synthetase. Coointegrate structure is produced by fusion of two replicons, one originally possessing a transposon, the other lacking it; the cointegrate has copies of the transposon present at both junctions of the replicons, oriented as directrepeats.Cold-sensitive mutant is defective at low temperature but functional at normal temlperature. Colony hybridization is a technique for using in situ hybridization to identify bacteria carrying chimeric vectors whose inserted DNA is homologous with some particular sequence. Compatibility group of plasmids contains members unable to coexist in the same bacterial5cell.Complementation refers to the ability of independent (nonallelic)genes to provide diffusible products that produce wild phenotype when two mutants are tested in trans configuration in a heterozygote.In vitro complementation assay consists of identifying a component of a wid-type cell that can confer activity on an extract prepared from a mutant cell. The assay identifies the component rendered inactive by the mutation.Complementation group is a series of mutations unable to complement when tested in pairwise combinations in trans; defines a genetic unit (the cistron) that might better be called a noncomplex mentation group.Complex locus (of D. melanogaster) has genetic properties inconsistent with the function of a gene representing a single protein. Complex loci are usually very large (>100kb) at the molecular level.Complexity is the total length of different sequences of DNA present in a given preparation. Compostie transposons have a central region flanked on each side by insertion sequences, either or both of which may enable the entire element to transpose.Concatemer of DNA consists of a series of unit genomes repeated in tandem. Concatenated circles of DNA are interlocked like rings one a chain.Concerted evolution describes the ability of two related genes to evolve together as though constituting a single locus.Condensation reaction is one in which a covalent bond is formed with loss of a water molecule, as in the addition of an amino acid to a polypeptide chain.Conditional lethal mlutations kill a cell or virus under certain (nonpermissive) conditions, but allow it to survive under other (permissive) conditions.Conjugation describes ‘mathing’ between two bacterial cells, when (part of ) the chromosome is transferred from one to the other.Consensus sequence is an idealized sequence in which each position represents the base most often found when many actual sequences are compared.Conservative recombination involves breakage and reunion of preexisting strands of DNA without any synthesis of new stretches of DNA.Conservative transposition refers to the movement of large elements, originally classified as transposons, but now considered to be episomes. The mechanism of movement resembles that of phage lambda.Constant regions of immunoglobulins are coded by C genes and are the parts of the chain that vary least. Those of heavy chains identify the type of immunoglobulin.Constitutive genes are expressed as a function of the interaction of RNA polymerase with the promoter, without additional regulation; sometimes also called household genes in the context of describing functions expressed in all cells at a low level.67Constitutive heterochromatin describes the inert state of permanently nonexpressed sequences, usually satellite DNA.Constitutive mutations cause genes that usually are regulated to be expressed without regulation.Contractile ring is a ring of actin filaments that forms around the equator at the end of mitosis and is responsible for pinching the daughter cells apart.Controlling elements of maize are transposable units originally identified solely by their genetic properties. They may be autonomous (able to transpose independently) or nonautonomous (able to element).Coordinate regulation refers to the common control of a group of genes.Cordycepin is 3’ deoxyadenosine, an inhibitor of poly adenylation of RNA.Core DNA is the 14.6 bp of DNA contained on a core particle.Core particle is a digestion product of the nucleosome that retains the histone octamer and has 14.6 bp of DNA; its structure appears similar to that of the nucleosome itself.Corepressor is a small molecule that triggers repression of transcription by binding to a regulator protein.Cosmids are plasmids into which phage lambda cos sites have been inserted; as a result, the plasmid DNA can be packaged in vitro in the phage coat.Cot is the product of DNA concentration and time of incubation in a reassociation reaction. Cot 21 is the Cot required to proceed to half completion of the reaction; it is directlyproportional to the unique length of reassociating DNA.Cotransfection is the simultaneous transfection of two markers.Crossing-over describes the reciprocal exchange of material between chromosomes that occurs during meiosis and is responsible for genetic recombination.Crossover fixation refers to a possible consequence of unequal crossing-over that allows a mutation in one member of a tandem cluster to spread through the whole cluster (or to be eliminated).Cruciform is the structure produced at inverted repeats of DNA if the repeated sequence pairs with its complement on the same strand (instead of with its regular partner in the other strand of the duplex).Cryptic satellite is a satellite DNA sequence not identified as such by a separate peak on a density gradient; that is, it remains present in main-band DNA.ctDNA is chloroplast DNA.Cyclic AMP (cAMP) is a molecule of AMP in which the phosphate group is joined to both the 3’ and 5’ positions of the ribose; its binding activates the CAP, a postive regulator of prokaryotic transcription.Cyclins are proteins that accumulate continuously throughout the cell cycle and are thendestroyed by proteolysis during mitosis. (see also MPF).Cytokinesis is the final process involved in separation and movement apart of daughter cells at the end of mitosis.Cytological hybridization—see in situ hybridization.Cytoplasm describes the material between the plasma membrane and the nucleus. Cytoplasmic inheritance is a property of genes located in mitochondria or chloroplasts (or possibly other extranuclear organelles).Cytoplasmic protein synthesis is the ranslation of mRNAs representing nuclear genes; it occurs via ribosomes attached to the cytoskeleton.Cytoskeleton consists of networks of fibers in the cytoplasm of the eukaryotic cell.Cytosol describes the general volume of cytoplasm in which organelles ( such as the mitochondria ) are located.D loop is a region within mitochondrial DNA in which a short stretch of RNA is paired with one strand of DNA, displacing the orignal partner DNA strand in this region. The same term is used also to describe the displacement of a region of one strand of duplex DNA byu a single-stranded invader in the reaction catalyzed by RecA protein.Degeneracy in the genetic code refers to the lack of an effect of many changes in the third base of the codon on the amino acid that is represented.Deletions are generated by removal of a sequence of DNA, the regions on either side being joined together.Denaturation of DNA or RNA describes its conversion from the double-stranded to the singlestranded state; separation of the strands is most often accomplished by heating. Denaturation of protein describes its conversion from the physiological conformation to some other (inactive) conformation.Derepressed state describes a gene that is turned on. It is synonymous with induced when describing the normal state of a gene; it has the same meaning as constitutive in describing the effect of mutation.Dicentric chromosome is the product of fusing two chromosome fragments, each of which has a centromere. It is unstable and may be broken when the two centromeres are pulled to opposite poles in mitosis.Diploid set of chromosomes contains two copies of each autosome and two sex chromosome. Direct repeats are identical (or related) sequences present in two or more copies in the same orientation in the same molecule of DNA; they aer not necessarily adjacent.Discontinuous replication refers to the synthesis of DNA in short (Okazaki) fragments that are later joined into a continuous strand.Disjunction describes the movement of members of a chromosome pair to opposite poles during cell division. At mitosis and the second meiotic division, disjunction applies to sister8chromatids; at first meiotic division it applies to sister chromatid pairs.Divergence is the percent difference in nucleotide sequence between two related DNA sequences or in amino acid sequences between two proteins.Divergent transcription refers to the intitiation of transcription at two promoters facing in the opposite direction, so that transcription proceeds away in both directions from a central region.dna mutants of bacteria are temperature-sensitive; they cannot synthesize DNA at 42℃, but can do so at 37℃.DNAase is an enzyme that attacks bonds in DNA.DNA-driven bybridization involves the reaction of an excess of DNA with RNA.DNA polymerase is an enzyme that synthesizes a daughter strand(s) of DNA (under direction from a DNA template). May be involved in repair or replication.DNA replicase is a DNA-synthesizing enzyme required specifically for replication.Domain of a chromosome may refer either to a discrete structural entity defined as a region within which supercoiling is independent of other domains; or to an extensive region including an expressed gene that has heightened sensitivity to degradation by the enzyme DNAase I.Domain of a protein is a discrete continuous part of the amino acid sequence that can be equated with a particular function.Dominant allele determines the phenotype displayed in a heterozygote with another (recessive) allele.Donor splicing site—see left splicing junction.Down promoter mutations decrease the frequency of initiaton of transcription. Downstream identifies sequences proceeding farther in the direction of expression, for example, the conding region is downstream of the initiation condon.Early development refers to the period of a phage infection before the start of DNA replication.Extopic expression describes the expression of a gene in a tissue in which it is not usually expressed; for example, in a transgenic animal.Elongation factors (EF in prokaryotes, eEF in eukaryotes) are proteins that associate with ribosomes cyclically, during addition of each amino acid to the polypeptide chain.End labeling describes the addition of a radioactively labeled group to one end (5’ or 3’) of a DNA strand.End-product inhibition describes the ability of a product of a metabolic pathway to inhibit the activity of an enzyme that catalyzes an early step in the pathway.Endocytosis is a process by which proteins at the surface of the cell are internalized, being transported into the cell within membranous vesicles.9Endocytic vesicles are membranous particles that transport proteins through endocytosis; also known as clathrin-coated vesicles.Endonucleases cleave bonds within a nucleic acid chain; they may be specific for RNA or for singlestranded of double-stranded DNA.Endoplasmic reticulum is a highly convoluted sheet of membranes, extending from the outer layer of the nuclear envelope into the cytoplasm.Enhancer element is a cis-acting sequence that increases the utilization of (some) eukaryotic promoters, and can function in either orientation and in any location (upstream or downstream) relative to the promoter.Envelopes surround some organelles (for example , nucleus or mitochondrion) and consist of concentric membranes, each membrane consisting of the usual lipid bilayer.Epigenetic changes influrence the phenotype without altering the genotype. They consist of changes in the properties of a cell that are inherited but that do not represent a change in genetic information.Episome is a plasmid able to integrate into bacterial DNA.Epistasis describes a situation in which expression of one gene wipes out the phenotypic effects of another gene.Essential gene is one whose deletions is lethal to the organism (see also lethal locus). Established cell lines consist of eukaryotic cells that have been adapted to indefinite growth in culture (they are said to be immortalized).Eubacteria comprise the major line of prokaryotes.Euchromatin comprises all of the genome in the interphase nucleus except for the heterochromatin.Evolutionary clock is defined by the rate at which mutations accumulate in a given gene. Excision-repair systems remove a single-stranded sequence of DNA containing damaged or mispaired bases and replace it in the duplex by synthesizing a sequence complementary to the remaining strand.Exocytosis is the process of secreting proteins from a cell into the medium, by transport in membranous vesicles from the endoplasmic reticulum, through the Golgi, to storage vesicles, and finally (upon a regulatory signal) through the plasma membrane.Exocytic vesicles (also secretory vesicles) are membranous particles that transport and store proteins during excytosis.Exon is any segment of an interrupted gene that is represented in the mature RNA product. Exonucleases cleave nucleotides one at a time from the end of a polynucleotide chain; they may be specific for either the 5’ or 3’ end of DNA or RNA.Expression vector is a cloning vector desined so that a coding sequence inserted at a particular site will be transcribed and translated into protein.Extranuclear genes reside outside the nucleus in organelles such as mitochondria and10chloroplasts.F factor is a bacterial sex or fertility plasmid.F1 generation is the first generation produced by crossing two parental (homozygous) lines. Facultative heterochromatin describes the inert state of sequences that also exist in active copies—for example, one mammalian X chromosome in females.Fast component of a reassociation reaction is the first to reature and contains highly repetitive DNA.Fate map is a map of an embryo showing the adult tissues that will develop from the descendants of cells that occupy particular regions of the embryo.Figure eight describes two circles of DNA linked together by a recombination event that has not yet been completed.Filter hybridization is performed by incubating a deatured DNA preparation immobilized on a nitrocellulose filter with a solution of radioactively labeled RNA or DNA.Fingerprint of DNA is a pattern of polymorphic restriction fragments that differ between individual genomes.Fingerprint of a protein is the pattern of fragments (usually resolved on a two dimensional electrophoretic gel) generated by cleavage with an enzyme such as trypsin.Fluidity is a property of membranes; it indicates the ability of lipids to move laterally within their particular monolayer.Focus formation describes the ability of transformed eukaryotic cells to grow in dense clusters, piled up on one another.Focus forming unit (ffu) is a quantitative measure of forcus formation.Foldback DNA consists of inverted repeats that have renatured by intrastrand reassociation of denatured DNA.Foot printing is a technique for identifying the site on DNA bound by some protein by virtue of the protection of bonds in this region against attack by nucleases.Forward mutations inatctivate a wild-type gene.Founder effect refers to the presence in a population of many individuals all with the same chromosome (or region of a chromosome) derived from a single ancestor.Frameshift mutations arise by deletions or insertions that are not a multiple of 3bp; they change the frame in which triplets are translated into protein.G banding is a technique that generates a striated pattern in metaphase chromosomes that distinguishes the members of a haploid set.G1 is the period of the eukaryotic cell cycle between the last mitosis and the start of DNA replication.G2 is the period of the eukaryotic cell cycle between the end of DNA replication and the start11。
分子生物武大考研试题
笔试题（满分100分）1. PCR的条件是什么？设置的条件有什么依据？2. 去离子水与蒸馏水是否相同，若相同，为什么？不相同，又是为什么？3. 关于菌株保存有以下两个方案（1）不加甘油逐渐降温后，然后保存在-80℃中，（2）加甘油直接保存在-80℃中。
分别分析两种方案是否正确，若正确，为什么？不正确，又是为什么？（3）菌株保存中，甘油的作用是什么？依据是什么？4. 高压灭菌的温度为多少？灭菌是为什么要排尽灭菌锅内的冷空气？5. 克隆所用的载体有哪些必要的结构？6. 电泳分离DNA所用的缓冲液pH为多少？此时DNA带什么电荷，为什么?7. 电泳分离DNA、RNA、蛋白质分别用什么染色？8. 现有100uLDNA水溶液（装在eppendorf管中），请设计一个实验沉淀DNAWuda武大---考试方法说明及各章课外思考题一、课程的学习与考试作为生命科学领域的专业基础课，本课程强调学生对微生物学基本理论、基本概念、基本实验原理及微生物基本实验操作的掌握与运用，鼓励学生在教学过程中对教学的积极参与和平时学习的积累，不鼓励期末突击复习、考试的投机学习行为。
为此，特制定期末总成绩的评定标准。
一）理论课期末考试卷面成绩占总成绩的55％，其余45％为平时成绩。
平时成绩包括：1、课后思考题（0～4分），每章均会根据教学内容布置思考题，要求按时上交，教师每次将随机抽查20％～30％的作业本进行评分，在学期末考试结束后，作业本全部上交作为评定平时成绩的依据。
2、学习笔记（0～4分），要求并鼓励课堂记录或课后整理学习内容，所有学习笔记在期末考试结束后上交作为评定平时成绩的依据。
3、参与教学活动（0～17分），主要是鼓励学生在学习过程中的主动学习能力，在课堂、网站论坛或与教师平时就学习问题的探讨将作为此项成绩评定的依据。
为了配合这项工作，论坛中的同步教学论坛只有在经过认证后方可发言，是对同学参与教学活动项目评定的主要依据。
(完整版)分子生物学试题及答案
三、简答 1．分别说出 5 种以上 RNA 的功能？ 转运 RNA tRNA 转运氨基酸 核蛋白体 RNA rRNA 核蛋白体组成成 信使 RNA mRNA 蛋白质合成模板 不均一核 RNA hnRNA 成熟 mRNA 的前体 小核 RNA snRNA 参加 hnRNA 的剪接 小胞浆 RNA scRNA/7SL-RNA 蛋白质内质网定位合成的信号识别体的组成成分 反义 RNA anRNA/micRNA 对基因的表达起调整作用 核酶 Ribozyme RNA 有酶活性的 RNA 2．原核生物与真核生物启动子的主要差别？ 原核生物 TTGACA --- TATAAT------起始位点-35 -10 真核生物
蝶呤对叶酸还原酶有抑制作用，因此不能生长。骨-脾融合细胞：在 HAT 中能生长，脾细胞可 以利用次黄嘌呤，骨细胞供应细胞分裂功能。
6、利用双脱氧末端终止法（Sanger 法）测定 DNA 一级结构的原理与方法？ 原理是采纳核苷酸链终止剂—2，，3，-双脱氧核苷酸终止 DNA 的延 长。由于它缺少形成 3/5/磷酸二脂键所需要的 3-OH，一旦参入到 DNA 链中，此 DNA 链就 不能进一步延长。依据碱基配对原则，每当 DNA 聚合酶需要 dNMP 参入到正常延长的 DNA 链中 时，就有两种可能性，一是参入 ddNTP,结果导致脱氧核苷酸链延长的终止；二是参入 dNTP,
7、激活蛋白（CAP）对转录的正调控作用？ 环腺苷酸（cAMP）受体蛋白 CRP（cAMP receptor protein），cAMP 与 CRP 结合后所形成 的复合物称激活蛋白 CAP （cAMPactivated protein ）。当大肠杆菌生长在缺乏葡萄糖的培 育基中时，CAP 合成量增加，CAP 具有激活乳糖（Lac）等启动子的功能。一些依靠于 CRP 的 启动子缺乏一般启动子所具有的典型的-35 区序列特征（TTGACA）。因此 RNA 聚合酶难以与其 结合。CAP 的存在（功能）：能显著提高酶与启动子结合常数。主要表现以下二方面： ①CAP 通过转变启动子的构象以及与酶的相互作用关心酶分子正确定向,以便与-10 区结 合，起到取代-35 区功能的作用。 ②CAP 还能抑制 RNA 聚合酶与 DNA 中其它位点的结合，从而提高与其特定启动子结合的概 率。
2001-2016年武汉大学885分子生物学考研真题及答案解析 汇编
2017版武汉大学《分子生物学》复习全书学长寄语2017版武汉大学《885分子生物学》考研复习全书是武汉大学高分已录取的学长收集整理的，全国独家真实、可靠，是真正针对武汉大学考研的资料。
我们将所有的资料全部WORD 化，高清打印。
真题编写了详细的答案解析，即使是小题也明确指出了考察的知识点，对于做题帮助更大。
同时，我们在分析历年考研真题的基础上，针对武汉大学考研，编写了详细的复习备考讲义，明确列出考研的重点、难点和考点，可在短时间内快速把握重点，提升成绩。
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2017版武汉大学考研资料《分子生物学》考研复习全书（备考经验、真题及答案解析、高分版笔记）配套教材：[1]《现代分子生物学》（第四版），朱玉贤李毅著，高等教育出版社，2012[2]《遗传学》（分子遗传学部分），王亚馥戴灼华，高等教育出版社，1999[3]《Instant Notes in Moecuar Biology》，Turner P C、Mcennan A D、Bates A D、White M R H，科学出版社。
目录第一部分：备考篇 (1)一、武汉大学介绍 (1)二、武汉大学历年复试分数线 (3)1、2016年复试分数线 (3)2、2015年复试分数线 (4)3、2014年复试分数线 (4)三、报考难度分析 (4)四、备考方法与策略 (4)1、考研政治 (4)2、考研英语 (6)3、专业课复习方法 (8)第二部分：真题篇 (10)1、2016年武汉大学《分子生物学》考研真题 (10)2、2016年武汉大学《分子生物学》考研真题答案解析 (11)3、2015年武汉大学《分子生物学》考研真题 (17)4、2015年武汉大学《分子生物学》考研真题答案解析 (19)5、2014年武汉大学《分子生物学》考研真题 (26)6、2014年武汉大学《分子生物学》考研真题答案解析 (28)7、2013年武汉大学《分子生物学》考研真题 (35)8、2013年武汉大学《分子生物学》考研真题答案解析 (37)9、2012年武汉大学《分子生物学》考研真题 (46)10、2012年武汉大学《分子生物学》考研真题答案解析 (47)11、2011年武汉大学《分子生物学》考研真题 (53)12、2011年武汉大学《分子生物学》考研真题答案解析 (55)13、2010年武汉大学《分子生物学》考研真题 (63)14、2010年武汉大学《分子生物学》考研真题答案解析 (64)15、2009年武汉大学《分子生物学》考研真题 (73)16、2009年武汉大学《分子生物学》考研真题答案解析 (75)17、2008年武汉大学《分子生物学》考研真题 (85)18、2008年武汉大学《分子生物学》考研真题答案解析 (87)19、2007年武汉大学《分子生物学》考研真题 (95)10、2007年武汉大学《分子生物学》考研真题答案解析 (97)21、2006年武汉大学《分子生物学》考研真题 (105)22、2006年武汉大学《分子生物学》考研真题答案解析 (107)23、2005年武汉大学《分子生物学》考研真题 (115)24、2005年武汉大学《分子生物学》考研真题答案解析 (117)25、2004年武汉大学《分子生物学》考研真题 (127)26、2004年武汉大学《分子生物学》考研真题答案解析 (129)27、2003年武汉大学《分子生物学》考研真题 (139)28、2003年武汉大学《分子生物学》考研真题答案解析 (141)29、2002年武汉大学《分子生物学》考研真题 (153)30、2002年武汉大学《分子生物学》考研真题答案解析 (157)31、2001年武汉大学《分子生物学》考研真题 (163)32、2001年武汉大学《分子生物学》考研真题答案解析 (165)第三部分：2017版精品复习笔记（高分版） (171)第一章染色体与DNA (171)知识要点 (171)考点综述 (171)真题分析 (172)复习建议 (172)核心内容 (172)第二章遗传信息传递 (185)知识要点 (185)考点综述 (185)真题分析 (186)复习建议 (188)核心内容 (189)第三章基因表达调控 (220)知识要点 (220)考点综述 (220)真题分析 (221)复习建议 (224)核心内容 (225)第四章基因工程与蛋白质工程 (234)知识要点 (234)考点综述 (234)真题分析 (235)复习建议 (240)核心内容 (241)第五章系统生物学 (247)知识要点 (247)考点综述 (247)真题分析 (248)复习建议 (249)核心内容 (249)第六章病毒分子生物学 (260)知识要点 (260)考点综述 (260)真题分析 (260)复习建议 (261)核心内容 (261)第七章病毒与人类健康 (283)知识要点 (283)考点综述 (283)真题分析 (283)复习建议 (284)核心内容 (284)第八章常用的分子生物学实验技术 (291)知识要点 (291)考点综述 (291)真题分析 (292)复习建议 (294)核心内容 (294)第四部分诺贝尔奖专题 (313)2015年诺贝尔生理学或医学奖 (313)2014年诺贝尔生理学或医学奖 (313)2013年诺贝尔生理学或医学奖 (314)2012年诺贝尔生理学或医学奖 (315)2011年诺贝尔生理学或医学奖 (315)2010年诺贝尔生理学或医学奖 (318)2009年诺贝尔生理学或医学奖 (319)2008年诺贝尔生理学或医学奖 (319)2007年诺贝尔生理学或医学奖 (320)2006年诺贝尔生理学或医学奖 (322)2005年诺贝尔生理学或医学奖 (322)第五部分武汉大学《分子生物学》习题集 (325)第一部分：备考篇一、武汉大学介绍江城多山，珞珈独秀；山上有黉，武汉大学。
武汉大学历年分子生物学考研真题
武汉大学分子生物学历年考研真题 (2001年-2013年）武汉大学2001 年攻读硕士学位研究生入学考试试题科目名称:分子生物学科目代码: 477注意：所有答题内容必须写在答题纸上，凡写在试题和草稿纸上的一律无效。
一、解释概念 (20 分,每个4分)1. 卫星 DNA2. 复制体3. 逆转座子4. 反式激活因子5. 衰减子与衰减作用二：填空(30 分,每空 1 分,请将答案写在答卷上)1.从病毒到高等生物的研究表明,遗传物质是。
2.冈崎片段的发现证实了双链 DNA的复制,在复制过程中,一条新生链的合成是的,称为链;而另一条链的合成是的,称为链。
3.大肠菌中有三种DNA聚合酶,其中的polI的作用是 ,而pol III的作用是。
pol I和pol III都有的三种活性是、、。
4.由于真核细胞染色体DNA的复制要有一段RNA为引物,因此线状的DNA复制后必须存在着5’端缩短的问题。
已发现有一种端粒蛋白称为 ,它由构成,可以使单链DNA的5’延长。
5.对DNA 损伤有几种修复系统 ,其中只有修复系统可以造成DNA变异,与这一系统有关的一套基因平时受到一称为的抑制蛋白所抑制,它发挥抑制作用是结合在一段约20bp长的称为的DNA序列上,当DNA损伤时,另一种蛋白质称为。
把这种抑制蛋白水解后,修复系统的基因才会被激活。
6.真核细胞中有三种依赖于DNA的RNA聚合酶分别合成不同的RNA,RNA pol I 负责合成,RNA pol II 负责合成,RNA pol III负责合成。
7.大分子互相作用是分子生物学的重要内容,包括蛋白质之间、蛋白质与DNA或RNA之间的互相作用,蛋白质有四种重要的结构花式与大分子互相作用有关, 这些结构花式是 , , 。
8.NO 是气体小分子信号,它可由脱氨产生,它的作用方式是直接与酶作用使产生cGMP(环式GMP)。
9.真核mRNA的5’端通常有帽子结构,3’端有 polyA。
在polyA上游有一保守序列称为polyA 信号,其序列为。
武汉大学微生物考研历年真题汇总2001-2004、2010-2015
武汉大学2001年攻读硕士学位研究生入学考试试题科目名称：微生物学科目代码：931注意：所有答题内容必须写在答题纸上，凡写在试题和草稿纸上的一律无效。
一：试写出5种细胞型微生物的名称〔包括拉丁文和相应的中文名称，每题1分，共5分〕二：翻译并解释名词〔每题5分，共25分〕1.Prion2.Archaea3.Ames test4.Anaerobic jar5.Nucleic acid probe三：简答题〔每题6分，共30分〕1.革兰氏染色的原理及步骤？2.微生物有哪些营养类型？3.在产能代谢中微生物可以通过哪几种方式形成ATP？4.试描绘并绘图表示细菌在液体培养过程中的生长规律？5.何谓共生？试以地衣为例说明微生物间的共生关系？四：问答题〔选做4题，每题10分，共40分〕1.细菌的接合、转导、转化〔包括人工转化和自然转化〕各有什么特点？假设二个不同营养缺陷标记〔A-B-C+D+和A+B+C-D-〕的菌株经混合后，能在根本培养基平板上形成原养型重组菌落，请设计实验来证明确定该遗传转移过程是转化、转导或接合？2.微生物的最显著特征就是个体微小，通常只能通过显微镜进展观察，试列举在显微观察中通过改变样品的反差以改善观察效果的技术方法，并简述其原理?3.为什么说Koch等建立的微生物纯培养技术是微生物建立与开展的基石？你认为该技术能否完全满足现今微生物学进一步开展的需要？试阐述这方面的最新研究进展并谈谈你的看法？4.描绘细菌染色体DNA和质粒DNA提取的一般方法及原理？试分析并绘图表示相应的电泳检测结果？5.试论病原菌的致病力？为什么说感染不是疾病的同义词？武汉大学2021年硕士研究生入学考试试题(学术型硕士〕科目名称：微生物学科目代码：887注意：所有答题内容必须写在答题纸上，凡写在试题和草稿纸上的一律无效。
一、名词解释〔共10小题，每题4分，共40分〕1. Archaea2. L form of bacteria3. Enrichment culture technology4. Chemotaxis5. Continuous culture6. Biofilm7. Antimicrobial Agents8. Type strain9. Prion10. Retrovirus二、简答题〔共5小题，每题10分，共50分〕1.试接合微生物在自然界中分布的特点，谈谈为什么无菌操作是微生物学研究和应用的根底，可以分别采用哪些方法对实验所用到的器具和实验材料进展消毒和灭菌？2.试根据蓝细菌的根本特性分析其在自然界分布广泛的原因？3.抗药性细菌采取哪些方法逃避抗生素的攻击？4.什么是巴斯德效应？其机理如何？5.阐述病毒感染细胞的完好生命周期？三：阐述题1.什么是细菌群体感应〔Quorum sensing〕系统？有哪几种根本类型？试举一例说明细菌如何通过群体感应系统调节其生物行为。
武汉大学2013至2014学年第一学期分子生物学期末考试试题
武汉大学2013至2014学年第一学期分子生物学期末考试试题《分子生物学》试卷Final exam of Molecular Biology Course (Spring 2014)年级______ 专业________ 学号_________ 姓名_______成绩______PART I: DESCRIPTION (2 points each)Your answer should describe what each item is and how it functions in the cell. Diagrams, structure and sequence information should be included in your answer, as necessary.1. Yeast artificial chromosome2. RNA interference3. Proteomics4. Shine-Dalgarno sequence5. Alternative splicing6. Ribozyme7. r-dependent termination8. RNA editing9. DNA lesions10. Protein targetingPART II: MULTIPLE CHOICES (1 points each)Select the one best answer for each question.1. The catalytic activity for peptide bond formation (the peptidyl transferase activity) is located in the:1) RNA of the large ribosomal subunit.2) leader sequence of the messenger RNA.3) RNA of the small ribosomal subunit.4) proteins of the small ribosomal subunit.5) proteins of the large ribosomal subunit.2. Bidirectional and semi-conservative are two terms that refer to:1) transcription.2) translation.3) replication.4) all of the above.5) none of the above.3. The fact that most amino acids are specified by multiple codons is known as:1) the “wobble” phenomenon.2) the universality of the genetic code.3) codon bias.4) the anticodon hypothesis.5) the redundancy of the genetic code.4. RNA polymerase I is the eukaryotic enzyme responsible for:1) transcription of ribosomal RNA.2) transcription of transfer RNA and other small RNA species.3) transcription of messenger RNA.4) initiation of Okazaki fragment synthesis in DNA replication.5. Restriction enzymes can cleave DNA that is either single-stranded or double-stranded, as longas it contains the appropriate recognition site.1) True 2) False6. Information about the sequence of the coding region of a gene is best obtained from:1) a YAC clone.2) a genomic clone.3) a cDNA clone.4) the protein.7. A chromatography method that can be used specifically to purify proteins based on their chargeis:1) gel filtration chromatography.2) ion-exchange chromatography.3) DNA affinity chromatography.4) antibody affinity chromatography.8. A nonsense mutation is a change in the DNA sequence that results in:1) a small deletion or insertion.2) an amino acid change in the protein encoded by the gene.3) a premature stop codon.4) all of the above.5) none of the above.9. A protein complex involved in degradation of proteins within the cell is known as the:1) ubiquitin/proteasome system.2) molecular chaperone.3) chaperonin.4) ribosome.5) Krebs/TCA cycle.10. ___binds to the repressor and turn on the transcription of the structural genes in the Lac operon.1) cAMP2) lactose3) allolactose4) CRP11. Which of the following RNA species is involved in degradation of the mRNA containing complementary sequence1) miRNA2) siRNA3) tRNA4) 5S RNA5) U3 snRNA12. The genome sequencing projects are confirming the theory that genome size is directlyproportional to the number of genes contained within that genome. In other words, a genome that is 10 times as big will contains approximately 10 times as many protein coding genes.1) True 2) False13. HeLa cells, derived from a human cervical carcinoma, are able to propagate indefinitely inculture and are therefore known as a(n):1) tissue culture.2) tumor.3) transgenic cell line.4) immortalized cell line.14. E. coli cells are smaller than yeast cells.1) True 2) False15. Which of the following domains is not a DNA binding domain1) Proline-rich domains2) Helix-turn helix domains3) Zinc finger domains4) Basic domains16. The aminoacyl-tRNA synthetases distinguish between about 40 different shaped tRNA molecules in the cells.1) True 2) FalsePART III: SHORT QUESTIONS (8 points each)1. How do bacterial replication start and accomplished. Remember to include theproteins/enzymes and important DNA sequence involved in this process.2.Design experiments to clone a yeast gene and express this gene in yeast.3. Below is the multiple cloning site (MCS) of the plasmid vector pUC18 and theN-terminal and C-terminal sequence of protein X. Note that the MCS constitutes a part of the LacZ open reading frame. Suppose that you are going to clone the protein X gene into pUC18, so that your target gene is transcribed under the control of LacZ promoter, and translated with the LacZ gene to produce a fusion protein. You are requested to use Bam HI and Pst I to the clone X gene, please add these restriction sites on the corresponding position of the X gene. Remember to maintain the reading frame of the X gene with the LacZ gene 4.(1) MCS of pUC18Eco RI Sac I Kpn I Sma I Bam HI Xba I Sal I Pst IACG AAT TCG AGC TCG GTA CCC GGG GAT CCT CTA GAG TCG ACC TGC AGG CAT GCAThr Asn Ser Ser Ser Val Pro Gly Asp Pro Leu Glu Ser T hr Cys Arg His Ala(2) N-terminal sequenceX gene. ATG ACC CCU CAU AAC…Met Thr Pro His Asn Gly Asp…(3) C-terminal sequence of X gene. …GAU AGU ACA GCU GCC AAG TAA…Asp Ser Thr A la Ala LysPART IV: MAJOR QUESTIONS (201:Please describe how an mRNA gene is transcribed, processed and translated in human cells. What are the possible mechanisms in regulating the expression of this gene?2 (20 points): A bacterium is found to metabolize a rare sugar produced by a plant that the bacteria grow on. However, the bacteria prefer glucose as the energy source. The problem is, if you want to finish this course with a satisfied score, you must figure out the regulatory mechanism that the bacteria used to determine the sugar choice.The gene involving in the rare sugar metabolism has been identified as fun3. You can use northern blot to analyze the expression of fun3and use DNA footprinting to analyze the binding of proteins to the control elements of fun3gene. The following table shows the experimental resultsQuestions:1. Please propose a mechanism to explain the above results. You should focus onthe question “How does the expression of fun3gene is tightly regulated so that it is only highly expressed when the rare sugar is the only carbon source”. You must answer what proteins A, B and C are. (8 points)2. How is protein A regulated? (2 points)(1) glucose turns the repressor on(2) glucose turns the repressor off(3) the rare sugar turns the repressor on(4) the rare sugar turns the repressor off3. How is protein C regulated? (2 points)(1) glucose turns the activator on(2) glucose turns the activator off(3) the rare sugar turns the activator on(4) the rare sugar turns the activator off4. How could you make the bacteria always use the rare sugar as the energy source even in the presence of glucose? (8 points)武汉大学生命科学学院2013－2014学年第一学期期末考试《分子生物学》试卷及参考答案Final exam of Molecular Biology Course (Spring 2014)写在参考答案前面的话：Ø该课程考试目的是考查学生对所学知识掌握的情况，除选择题外，其他题目的答案基本都不是唯一的。
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一.解释概念(20分,每个4分)卫星DNA 复制体逆转座子反式激活因子衰减子与衰减作用三、问答题(50分)1. 说出双链DNA复制起始有关的五种重要的酶或蛋白并简述它们的功能。
(15分)2. 简述增强子的特点和性质及作用机制。
(10分)3. 简述真核RNA聚合酶II的转录起始复合物装配过程和转录起始(15分)4. DNA限制性内切酶EcoRI是人们熟悉的常用内切酶,它是在大肠杆菌(E.coli)R株中发现的,它被广泛用于分子克隆操作和DNA分析。
pUC质粒是常用克隆载体之一,它的多克隆位点上有EcoRI、BamHI、KpnI、HindIII等酶切点。
假如要你把一段由EcoRI切割产生的外源DN**段克隆到pUC质粒中,并把重组质粒转化大肠杆菌R株来扩增,已知条件是所用的R菌株中只有EcoRI一种限制性内切酶,你设计如何做才能确保成功?为什么?(10分)武汉大学2002分子生物学三.问答:1.简述(或绘图说明)真核细胞RNA聚合酶II转录的起始需要哪些基本转录因子及其装配过程(15分)2.简述(或绘图说明)色氨酸操纵子弱化的机制(15分)3.在讨论基因家庭时经常提到胚胎、胎儿和成体形成的蛋白质,这些述语是指什么现象?可用什么术语来描述这一类基因家族(5分)4. 你正在进行Southern blot分析,并刚刚完成凝胶电泳部分,下一步是将胶浸泡在NaOH溶液中使DNA变性为单链,为了节约时间,你跳过这一步,直接把DNA 从胶上转到硝酸纤维素膜上,你将标记好的探针与膜杂交,却发现放射自显影结果是一片空白,哪里错了呢?(5分)一、下列名词翻译成中文，并简要解释1、Domains and motifs2、Alternative splicing3、Reporter genes4、The PCR cycle5、Restriction mapping6、Multiple cloning sites7、DNA libraries8、Proteomics9、Replicon10、Semi-conservative replication二、简答题(共5题,每题8分,共40分)1、请列举三种以上蛋白质纯化技术,并说明不同纯化技术的简单原理。
2、简述DNA损伤与DNA突变之间的区别与相互关系。
3、简述密码的简并性(degeneracy)和同义密码子(synonymous codon)及其在生物学上的重要性。
4、简述原核生物转录起始与转录终止过程中所涉及的主要蛋白质和核酸结构及其具体作用。
5、简述cDNA文库的构建过程。
三、论述题(共5题,1-4题每题15分,第5题10分,共70分)1、人类基因组计划完成的社会意义和科学意义是什么？2、什么是操纵子(operon)？试说明色氨酸操纵子(Trp operon)在原核基因表达调控中的调控机制和重要作用。
3、请简要解释顺式作用元件与反式作用因子,并举二例加以说明它们的相互作用方式。
4、试说明真核细胞与原核细胞在基因转录、翻译及DNA的空间结构方面存在的主要差异,表现在哪些方面？5、限制性核酸内切酶有哪几种类型？哪一种类型的限制酶最适合于基因工程,为什么？请简要说明其理由。
一、名词翻译与解释1、synonymous codons2、RNA editing3、Spliceosome4、Microarray5、Plaque hybridization6、Open reading frame7、Ribozyme8、RFLP9、Site specific recombination10、RNA interference二、简答题(共5小题,每小题10分,共50分)1、简述真核生物 rRNA基因,tRNA 基因和mRNA基因的转录机制。
2、原核生物和真核生物中存在哪些类型的转座子？其转座机制有那些？3、简述真核生物 DNA复制中,端粒复制与染色体其他部分DNA复制的异同和生物学意义。
4、原核生物的蛋白质合成可分为哪些阶段？简述各阶段的主要事件。
5、列举4种可用来检测,鉴定转基因动,植物的分子生物学技术和方法,简述选择这些技术的理由。
三、论述题1、试述环境因素对DNA的损伤以及生物体中存在的DNA损伤修复系统。
如果DNA损伤没有被修复会造成什么后果？2、试比较真核生物RNA聚合酶Ⅱ识别的启动子与原核生物RNA聚合酶所识别的启动子的结构特点,各结构单元的功能是什么？为什么原核生物一种RNA聚合酶能识别不同结构基因？3、在进行基因工程时,载体是携带靶DNA片段进入宿主细胞进行扩增和表达的工具,请问一个载体应具有哪些基本特性和结构特点？4、人类基因组计划基本完成表明：人类基因组约有3×109 bp（而大肠杆菌的基因组约为4.6 ×106 bp),其中仅有百分之一的基因组DNA直接编码蛋白质,约有百分之二十四的基因组DNA为内含子,而百分七十五的基因组DNA为其他非编码序列。
试从(1)传代中遗传信息的保持；(2)基因表达调空的角度来论述人类基因组如此排列的可能生物学意义。
武汉大学2005年分子生物学一、名解(4/10)1、Ubiquitin2、ribosomal binding site3、siRNA miRNA4、proteome5、DNA foot printing6、single nucleotide polymorphism7、chromosome walking8、haplotype9、zoo blot10、transcription unit二、简答(5/10)1、简述真核细胞rRNA 、tRNA和mRNA转录使用的聚合酶及各 RNA前体(precursor)加工的基本过程？2、研究表明蛋白质组比基因组更复杂，蛋白质组中存在的蛋白质数目比基因组中存在的基因数目要多许多，这是如何造成的，你如何解释这种现象？3、简述原核与真核细胞染色体结构的差别4、在分子生物学研究中经常用各种载体进行研究工作，试问有哪些不同大小类型的载体,各自的主要特点是什么？5、简述DNA损伤修复的机制,在什么情况下DNA损伤会引起DNA突变？三、论述4/151、蛋白质与核酸的相互作用存在于基因表达的各个水平上,请分别举例说明在基因复制RNA 转录和蛋白质翻译过程中二者的相互作用。
2、一个真核细胞的蛋白质编码基因有哪些结构元件(或组建)组成(包括各种调控元件)？请画图表示,并说名各元件的功能与作用？3、请设计一组试验来 (1)克隆一个你所感兴趣的人类基因 (2)并对基因产物大量表达与纯化 (3)然后研究该基因的生物学功能4、生命科学的研究已进入后基因组时代,试从“分子生物学”的角度来谈谈你对后基因组时代的认识,并预测后基因组时代里“分子生物学”发展的未来一、名解4*101、DNA Microarray2、Spliceosome3、Cosmid vector4、Proteasome5、Zinc finger domain6、DNase I hypersensitivity7、Insulator8、Post-transcriptional gene silencing9、Reverse transcription PCR10、Missense mutation二、简答题 10*51、简述microRNA 的基因调控功能与机制。
2、简述染色体端粒复制的问题,人正常细胞和癌细胞的端粒复制有何不同？3、真核生物基因组中含有几种类型的DNA序列？你如何设计实验来证明？4、简述RNA编辑(RNA editing)的机制及其对基因表达的影响。
5、下面的DNA序列中含有一个假象基因,编码仅5个氨基酸。
请写出这些氨基酸的密码子以及终止密码子,并指出RNA转录的模版是哪条链？启动子在该序列左侧还是右侧？5-TCATGCTAGACACGTAATAGCATATGGGA-3 3-AGTACGATCTGTGCATTATCGTATACCCT-5三、论述题15*41、论述一个真核基因产生有功能的蛋白质需要经过哪些分子生物学工程(event)？列出每个过程发生的细胞位置和参与的主要酶和蛋白因子。
2、有哪几种类型的遗传重组?其重组的特点如何？3、DNA的甲基化在遗传信息的传递、DNA复制和DNA损伤修复等过程中有什么重要意义？请举例说明。
4、假如你进入实验室开始研究一个小鼠DNA结合蛋白的生物学功能: (1)、请设计实验确定其编码基因在小鼠细胞内的表达水平？ (2)、请设计实验确定该蛋白的那个结构域具有DNA 结合功能？ (3)、如何确定该蛋白在小鼠体内(in vivo)的生物学功能？一、名词翻译与解释(共10小题 ,每小题4分 ,共40分)1、Transcriptome2、Translesion replication3、Riboswitch4、Synonymous mutation5、Tandem gene cluster6、Frameshift7、Nucleosome positioning8、Non-autonomous transposon9、Holliday junction 10、Polymerase switching二、简答题(共5小题 ,每小题10分 ,共50分)1、试从DNA和RNA结构的不同解释为什么DNA被广泛作为遗传信息的承载者？RNA在细胞内行使哪些功能？2、什么是蛋白质组?什么是蛋白质组学?如何理解高等生物细胞中一个基因组可以产生多少个蛋白质组？3、什么是解旋酶(helicase)？请设计实验并画设计图证明一个DNA解旋酶是具有3’→5’还是5’→3’解旋活性？4、论述原核生物基因表达调控的主要策略,并举例进行详细说明？5、核糖体是蛋白质合成的主要机器。
请问原核生物有哪些亚基和分子组成?核糖体在蛋白质翻译过程中有哪些功能位点？起什么作用？三、论述题(共4小题 ,每题15分 ,共60分)1、在真核生物中,成熟的mRNA序列往往与基因组DNA序列之间存在许多差异,有些序列甚至在基因组DNA上根本没有相应的互补序列。
另外，同一种基因的DNA常常会转录出多种mRNA。
请解释并说明产生这些现象的机制及其意义。
2、真核细胞mRNA的Ⅱ型RNA聚合酶转录,需要Ⅱ型启动子和转录因子TFⅡA ,TFⅡB ,TFⅡD ,TFⅡE 和TFⅡH等。
(1)请设计实验证明这些转录因子和RNA聚合酶Ⅱ结合到启动子的顺序；(2)请设计实验证明TFⅡD复合体可以单独准确结合到启动子TATA box上；(3)TFⅡD 复合体包括哪些蛋白质？如果你有其中一个蛋白质的抗体,并知道其他几个蛋白质的大小,请设计实验验证TFⅡD复合体的组分。
3、DNA复制具有很高的准确性(accuracy)或保真度(fidelity), 出错率低于1010 分之一。