武汉大学分子生物学试题
分子生物学试题(附答案)
蛋白质的生物合成(一)名词解释1．翻译 2．密码子 3．密码的简并性 4．同义密码子 5．变偶假说 6．移码突变 7．同功受体 8．多核糖体(二)问答题1．参与蛋白质生物合成体系的组分有哪些?它们具有什么功能?2．遗传密码是如何破译的?3．遗传密码有什么特点?4．简述三种RNA在蛋白质生物合成中的作用。
5．简述核糖体的活性中心的二位点模型及三位点模型的内容。
6．氨基酸在蛋白质合成过程中是怎样被活化的?7．简述蛋白质生物合成过程。
8．蛋白质合成中如何保证其翻译的正确性?9．原核细胞和真核细胞在合成蛋白质的起始过程有什么区别。
10．蛋白质合成后的加工修饰有哪些内容?11．蛋白质的高级结构是怎样形成的?12．真核细胞与原核细胞核糖体组成有什么不同?如何证明核糖体是蛋白质的合成场所?13. 已知一种突变的噬菌体蛋白是由于单个核苷酸插入引起的移码突变的，将正常的蛋白质和突变体蛋白质用胰蛋白酶消化后，进行指纹图分析。
结果发现只有一个肽段的差异，测得其基酸顺序如下：正常肽段 Met-Val-Cys-Val-Arg突变体肽段 Met-Ala-Met-Arg（1）什么核苷酸插入到什么地方导致了氨基酸顺序的改变？（2）推导出编码正常肽段和突变体肽段的核苷酸序列.提示：有关氨基酸的简并密码分别为Val: GUU GUC GUA GUGArg: CGU CGC CGA CG AGA AGGCys: UGU UGCAla: GCU GCC GCA CGC14. 试列表比较核酸与蛋白质的结构。
15. 试比较原核生物与真核生物的翻译。
(三)填空题1．蛋白质的生物合成是以___________为模板，以___________为原料直接供体，以_________为合成杨所。
2．生物界共有______________个密码子，其中___________个为氨基酸编码，起始密码子为_________；终止密码子为_______、__________、____________。
分子生物学习题库与参考答案
分子生物学习题库与参考答案一、单选题（共49题，每题1分，共49分）1.可以实现体细胞克隆的技术是:A、基因编辑B、聚合酶链式反应C、诱导多能干细胞D、体细胞核移植正确答案：D2.在PCR反应中,引物与模板的结合温度约为:A、37°CB、55°CC、72°CD、95°C正确答案：B3.大肠杆菌对于基因克隆的主要作用是:A、提供连接酶B、合成引物C、表达目的蛋白D、作为宿主正确答案：D4.将单链DNA合成双链的酶是:A、连接酶B、DNA聚合酶C、裂解酶D、RNA聚合酶正确答案：B5.可以增加靶标DNA拷贝数的技术是:A、PCRB、电泳C、测序D、印迹杂交正确答案：A6.在Southern杂交中起探针作用的是:A、DNAB、RNAC、载体D、引物正确答案：A7.编码氨基酸顺序信息的DNA序列称为:A、启动子B、基因C、外显子D、启动密码子正确答案：B8.可以自我复制的核酸是:A、mRNAB、rRNAC、tRNAD、miRNA正确答案：B9.DNA聚合酶在PCR反应过程中不需要的元素是:A、铜离子B、镁离子C、锰离子D、钾离子正确答案：A10.启动子序列通常位于:A、转录起始点上游B、编码区C、基因内含子D、终止子上游正确答案：A11.可以直接导入植物细胞的方法是:A、阳离子脂质体法B、农杆菌法C、微粒射击法D、电穿孔法正确答案：B12.DNA的组成单位是:A、氨基酸B、核苷酸C、核糖D、脱氧核糖正确答案：B13.农杆菌可以将T-DNA转入植物细胞的原因是:A、具有连接酶B、具有限制性内切酶C、具有转座子D、可以与植物细胞膜融合正确答案：C14.DNA测序中的Sanger方法利用了:A、引物延伸终止B、核酸杂交C、蛋白质切割D、荧光标记正确答案：B15.可以用于克隆目的基因的载体是:A、慢病毒B、质粒C、线性DNA分子D、mRNA正确答案：B16.属于真核生物的模型生物是:A、小鼠B、酵母C、果蝇D、以上所有正确答案：D17.可以将质粒DNA转入宿主细胞的是:A、限制性内切酶B、DNA连接酶C、显微注射器D、电穿孔仪正确答案：C18.可以直接将外源基因导入植物细胞的是:A、电穿孔法B、微注射法C、生物炮法D、农杆菌法正确答案：D19.检测蛋白质的方法不是:A、Western印迹B、质谱C、Northern印迹D、免疫印迹正确答案：C20.对DNA进行切割的酶类包括:A、分裂酶B、连接酶C、制限酶D、聚合酶正确答案：C21.可以直接检测蛋白质的方法是:A、PCRB、Northern blotC、Western blotD、Southern blot正确答案：C22.提供能量驱动翻译反应的化学键是:A、糖苷键B、磷酸酯键C、硫磷键D、氢键正确答案：B23.编码线粒体蛋白质的基因主要位于:A、细胞质DNAB、线粒体DNAC、细胞核DNAD、质粒DNA正确答案：B24.下列DNA酶类能切断磷酸二酯键的是:A、连接酶B、DNA聚合酶C、替代酶D、制限酶正确答案：D25.在制备重组DNA时,使用琼脂糖的目的是:A、提供营养B、连接DNA段C、物理分离片段D、催化连接反应正确答案：C26.是mRNA而不是tRNA或rRNA的特征是:A、可翻译成蛋白质B、可与核糖体结合C、含有多肽链D、富含无义密码子正确答案：A27.对蛋白表达的后翻译调控方式是:A、剪接体B、RNA编辑C、蛋白质水解D、磷酸化正确答案：C28.编码tRNA的基因位于:A、线粒体B、核糖体C、细胞核D、细胞质正确答案：C29.单克隆抗体技术利用的真核细胞是:A、诱导的多能干细胞B、杆状病毒感染的淋巴细胞C、转化的肿瘤细胞D、融合瘤细胞正确答案：D30.原核生物基因组中不含有的序列是:A、终止子B、编码区C、启动子D、外显子正确答案：D31.制备cDNA文库常用的反转录酶来源于:A、大肠杆菌B、反刍动物C、艾滋病毒D、枯草杆菌正确答案：C32.编码氨基酸的三联密码所在的核酸是:A、mRNAB、rRNAC、tRNAD、盖帽RNA正确答案：C33.参与DNA复制的关键酶是:A、RNA聚合酶B、连接酶C、DNA聚合酶D、选择酶正确答案：C34.编码rRNA的基因通常组织成:A、基因簇B、可变剪接体C、假基因D、反义基因正确答案：A35.编码 rRNA 的基因位于:A、线粒体DNAB、质粒DNAC、细胞核DNAD、细胞质DNA正确答案：C36.将DNA上的遗传信息转录为RNA的过程称为:A、翻译B、转录C、复制D、修复正确答案：B37.PCR技术依赖的关键酶是:A、连接酶B、聚合酶C、裂解酶D、制限酶正确答案：B38.编码氨酰tRNA合成酶的RNA是:A、mRNAB、rRNAC、tRNAD、siRNA正确答案：B39.可以实现定点诱变的技术是:A、CRISPR/Cas9B、ZFNsC、TALENsD、以上均可正确答案：D40.利用生物信息学分析推测基因功能的方法是:A、突变分析B、蛋白质互作C、序列比对D、同源建模正确答案：C41.可以改变染色体DNA序列的技术是:A、基因敲除B、基因转染C、基因沉默D、基因编辑正确答案：D42.下列不属于PCR反应体系的组成部分是:A、DNA模板B、聚合酶C、dNTPD、琼脂糖正确答案：D43.检测目的蛋白表达的方法不是:A、Southern blottingB、Western blottingC、Eastern blottingD、Northern blotting正确答案：A44.从cDNA库中可以获得的是:A、所有DNA片段B、编码区序列C、非编码区序列D、全部基因组序列正确答案：B45.可以直接克隆cDNA的是:A、质粒B、YACC、λ噬菌体D、人工染色体正确答案：A46.病毒载体导入宿主细胞的方法是:A、共转化B、显微注射C、电穿孔D、吸附感染正确答案：D47.可以识别启动子序列的转录因子是:A、β因子B、α 因子C、σ 因子D、Rho因子正确答案：C48.可以直接提取基因组DNA的方法是:A、PCRB、Northern blotC、Southern blotD、盐析法正确答案：D49.基因敲除实验中所用对照组应为:A、目的基因缺失组B、野生型组C、质粒载体组D、siRNA处理组正确答案：B二、多选题（共35题，每题1分，共35分）1.PCR反应的原料不包括:A、引物B、载体酶C、聚合酶D、dNTPE、模板DNAF、琼脂糖G、无橡皮管封闭正确答案：BFG2.对DNA序列进行PCR扩增需要以下原料:A、模板DNAB、载体酶C、引物D、连接酶E、脱氧核苷三磷酸F、DNA聚合酶G、以上ACF正确答案：G3.质粒载体应具有下列哪些特征:A、大片段插入区B、可自主复制C、含有筛选位点D、与宿主互作E、含有克隆位点F、编码病毒蛋白G、低拷贝数正确答案：BCE4.在制备cDNA文库时需要用到的关键酶包括:A、连接酶B、PCR酶C、限制性内切酶D、RNA聚合酶E、反转录酶F、RNase HG、链化酶正确答案：EF5.编码氨基酸序列的核酸为:A、rRNAB、mRNAC、tRNAD、mRNA前体E、单链RNAF、双链RNAG、环状RNA正确答案：B6.制备重组质粒的主要步骤是:A、载体线性化B、消化插入片段C、连接反应D、感受态细胞制备E、转化宿主细胞F、克隆鉴定G、所有以上步骤正确答案：G7.对肿瘤基因组的检测可以应用:A、Southern印迹B、Northern印迹C、Western印迹D、Eastern印迹E、基因检测F、测序G、基因芯片正确答案：AEFG8.基因表达调控的机制包括:A、转录水平调控B、RNA水平调控C、翻译水平调控D、蛋白活性调控E、基因增幅F、肽链释放G、以上AD均可正确答案：G9.参与翻译过程的RNA包括:A、mRNAB、rRNAC、tRNAD、siRNAE、snRNAF、反义RNAG、以上ABC均参与正确答案：G10.编码氨基酸序列的核酸为:A、rRNAB、mRNAC、tRNAD、cDNAE、基因F、反义链G、互补链正确答案：B11.PCR反应的原料组成包含:A、模板 DNAB、上游引物C、下游引物D、DNA聚合酶E、脱氧核苷三磷酸F、缓冲液G、以上ABDEF正确答案：G12.基因敲入的方法可以包括:A、siRNAB、基因敲除C、ZFNsD、TALENsE、CRISPR/Cas9F、Cre-Lox重组系统G、反义RNA抑制正确答案：CDEF13.编码脱氧核糖的DNA单链为:A、编码链B、反义链C、互补链D、下游链E、正义链F、上游链G、载体链正确答案：B14.基因编辑技术包括:A、ZFNs技术B、TALENs技术C、CRISPR/Cas技术D、基因敲除E、RNAi技术F、慢病毒介导G、以上所有正确答案：ABC15.制备重组DNA的步骤包括:A、载体选择B、插入DNA获得C、双酶切D、连接反应E、转化F、筛选G、以上全部正确答案：G16.参与制备cDNA文库的关键酶类有:A、连接酶B、限制性内切酶C、聚合酶D、反转录酶E、核酸酶F、RNase HG、以上DE正确答案：G17.制备重组质粒的关键步骤不包括:A、载体的选择B、消化载体及插入片段C、连接反应D、PCR扩增插入片段E、转化感受态细胞F、筛选重组克隆G、测序验证正确答案：D18.启动子序列的特征包括:A、定位于基因转录起始点上游B、通常为AT富集区C、具有内含子D、与编码区互补E、与编码区反向验配F、与编码区部分重叠G、保守性非常低正确答案：AB19.直接参与蛋白质翻译的分子包括:A、DNAB、mRNAC、rRNAD、tRNAE、RNA聚合酶F、释放因子G、所有以上分子正确答案：BCDF20.制备重组质粒需要下列步骤:A、载体选择B、消化载体和插入DNAC、连接反应D、感受态细胞制备E、转化宿主细胞F、克隆筛选G、以上全部正确答案：G21.检测mRNA的方法包括:A、Northern杂交B、Western印迹C、Southern印迹D、荧光in situ杂交E、实时定量PCRF、RNA序列表达谱分析G、以上ABDF正确答案：G22.启动子通常位于:A、翻译起始点上游B、转录终止点下游C、翻译终止点下游D、基因内含子E、编码区F、转录起始点上游G、终止子下游正确答案：F23.编码氨基酸序列信息的核酸为:A、DNAB、RNAC、mRNAD、tRNAE、rRNAF、cDNAG、质粒正确答案：C24.基因敲入的技术可以包括:A、siRNAB、基因敲除C、ZFNsD、TALENsE、CRISPR/Cas9系统F、Cre-Lox重组系统G、反义DNA正确答案：CDEF25.制备重组 DNA 需要以下技术:A、载体准备B、PCR 扩增插入片段C、引物设计D、双酶切连接E、宿主细胞转化F、筛选重组克隆G、以上全部正确答案：G26.编码氨基酸序列的核酸为:A、DNAB、mRNAC、rRNAD、tRNAE、miRNAF、cDNAG、基因正确答案：B27.编码mRNA的DNA单链被称为:A、编码链B、上游链C、下游链D、正义链E、反义链F、互补链G、载体链正确答案：E28.Polymerase Chain Reaction (PCR) 的关键步骤不包括:A、模板变性B、引物与模板杂交C、新链延伸合成D、反向转录E、新链变性F、产物检测G、增幅后转化正确答案：D29.Polymerase Chain Reaction (PCR) 的关键步骤包括:A、模板预变性B、引物与模板退火C、新链延伸合成D、产物检测E、反义链合成F、连接酶反应G、以上全部正确答案：ABCD30.编码氨基酸的三联密码存在于:A、DNA双链上B、mRNA分子上C、tRNA分子上D、rRNA上E、盖帽RNA上F、启动子区域G、终止子区域正确答案：C31.抑制基因在体细胞中的表达方法有:A、siRNAB、基因敲除C、反义RNAD、CRISPR/Cas9E、基因激活F、启动子激活G、剪切反应激活正确答案：ABD32.检测mRNA表达水平的技术是:A、北方印迹B、西方印迹C、南方印迹D、东方印迹E、In situ杂交F、免疫组化G、芯片杂交正确答案：AEG33.编码氨基酸的三联密码存在于:A、DNA双链上B、mRNA分子上C、tRNA分子上D、rRNA 上E、盖帽RNA上F、启动子区域G、终止子区域正确答案：C34.可以提高基因在异源表达载体中的表达水平的方法不包括:A、扩增子克隆B、终止子序列调控C、引入变位信号D、改良启动子序列E、引入选择标记F、优化文库构建方法G、优化编码序列正确答案：F35.可以提取基因组DNA的方法有:A、PCR扩增B、Northern印迹C、Southern印迹D、过滤法E、质谱法F、限制性酶切G、盐析法正确答案：CG三、判断题（共38题，每题1分，共38分）1.基因敲入可以使用双链DNA分子实现。
2022武汉大学分子生物学真题
2022武汉大学分子生物学真题一、名词解释1、Ribozyme核酶2、MieneMutation错义突变3、Inulator绝缘子4、RNAtran-plicingRNA反式剪接5、Nucleotidee某ciionrepair 核苷酸切除修复6、Tranpoon转座子7、Dammethylae甲基化酶8、pliceoome剪接体9、corepromoter核心启动子10、SNP单核苷酸多态性二、简答题1、请简述真核生物mRNA的5’帽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怎样的影响？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型内含子剪切过程。
ORF中单个碱基改变对基因表达产物的影响5、简述Weternblot的原理和步骤。
三、论述题1、研究基因功能通常是降低或者升高基因表达水平，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。
请阐述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。
（二者只需选择一种进行阐述）3、某实验需要将1kb的cDNA片段插入某氨苄青霉素抗性的表达载体，在片段两端各有一个EcoRl位点，靠近5’端300bp的地方有一EcoRv位点，载体多克隆位点从5’端依次为BamHl，Hindllll，EcoRl，EcoRv和某hol位点。
实验设计步骤如下：（1）用EcoRl处理载体，然后用碱性磷酸酶处理。
（2）用EcoRl处理片段，然后与步骤（1）中的载体混合，加入DNA连接酶，在适当的条件下使cDNA片段与载体连接。
（3）连接产物转化到大肠杆菌，并在含有氨苄青霉素的LB平板上生长。
除此之外，还设了以下对照：对照1：在含有抗生素的平板上涂布未被转化的大肠杆菌感受态细胞。
对照2：用未被酶切处理的载体转化大肠杆菌感受态细胞，并在含有抗生素的平板上生长。
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。
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
武汉大学分子生物学_2007期末试卷A
武汉大学生命科学学院2007－2008学年第一学期期末考试《分子生物学》A 试卷Final exam of Molecular Biology Course (Fall 2007)年级(Grade) ______ 专业(Major) ________姓名Name _______ 学号(Student ID)_________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 could be included in your answer, as necessary.1.Trombone model2.TFIID3.Telomerase4.RNA editing5.Tandem mass spectrometry6.tRNA synthetase7.Viral-like retrotransposons8.Transcriptional silencing9.Suppression mutation10.Shine-Dalgarno sequencePART II: MULTIPLE/SINGLE CHOICES (2 points each)1.The following molecule is ____1)2’-deoxyadenosine 5’-phosphate2)2’-deoxyadenosine3)dAMP4)dATP2.The strictness of the rules for “Waston-Crick” pairing derives from thecomplementarities___ between adenine and thymine and between guanine andcytosine.1)of base stacking2)of shape3)of hydrogen bonding properties4)of size3.Which of the following statements correctly describe the difference between DNA andRNA? ____1)The major groove of the regular DNA double helical structure is rich in chemicalinformation.2)RNA contains deoxyribose and uracil.3)All RNAs are single-stranded while DNA is double-stranded.4)Some RNAs can fold up into complex tertiary structures, and function asenzymes.4.Nucleosomes are the building blocks of chromosomes, which of the followingstatements are CORRECT in describing the structure and function of nucleosome? ___1)The nucleosome is composed of a core containing eight histone proteins and theDNA (~147 bp) wrapped around them.2)The core histones specifically contact the major groove of the DNA.3)The interaction of DNA with the histone octamer is very stable and cannot bealtered unless during DNA replication.4)Nucleosome remodeling by the action of enzymes such as histone acetylase isvery important for gene expression.5.Which of the following statements regarding genomes of living organism is NOTcorrect? ___1)Genome size is roughly related to the complexity of the organism.2)The number of genes in a genome cannot explain the complexity of the organism.3)The average number of introns per gene increases with the organism complexity.4)The gene density (genes/Mb) increases with the organism complexity.6.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.7.Which of the following repair mechanisms is involved in repair of the damaged DNAwith a double-stranded break?1)Base excision repair2)Nucleotide excision repair3)Translesion repair4)RecBCD pathway repair5)Mismatch repair8.Which of the following factors/elements regarding translation are CORRECT?1)Ribosome binding site (RBS) is essential for the translation initiation in bacterialand eukaryotic cells.2)The p olyA tail in the 3’ end of an mRNA promotes the efficient recycling ofribosomes, and therefore the translational efficiency.3)Ribosome is able to discriminate between correctly or incorrectly charged tRNAs.4)The translocation factor EF-G mimics a tRNA molecule so as to displace thetRNA bound to the A site.5)The ribosome is a ribozyme because the large rRNA is responsible for thepeptidyl transferase activity.9.RNA polymerase III 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.10To obtain the sequence of a genome, which of the following steps are required? __1)Obtain a genomic library2)Obtain a cDNA library3)Shotgun sequencing on automated sequencers4)Sequence assembly on computers5)BLAST searchPART III: SHORT QUESTIONS (CHOOSE SIX QUESTIONS TO ANSWER) (5 points each, total of 30 points)1.Who is your favorite scientist among those introduced in this course? Please describehis/her research achievement and contribution to the molecular biology knowledge that you’ve learned and how his/her experience influences your research attitude.2.How the transcription of an mRNA is terminated in eukaryotic cells?3.Please describe the similarity and difference between group II intron andspliceosome-mediated pre-mRNA splicing.4.What are the general principles of transcription regulation in both prokaryotic andeukaryotic cells?5.Please give an example to demonstrate that the RNA secondary structure can regulategene expression in bacteria.Example 1: The attenuation regulation of the tryptophan operon.Example 2: Ribo-switch regulation6.How the activators and repressors regulate gene expression in eukaryotic cells?7.The following DNA sequence contains a small open reading frame (ORF) whichencodes only 5 amino acids. Please list the 5 genetic codons and the stop codon of the ORF. Which strand of the DNA (upper or lower strand) is the template for RNAtranscription? The promoter of the gene is in the right or left side of the sequence?5’ TCATGCTAGACACGTAATAGCATATGGGA –3’3’ AGTACGATCTGTGCATTATCGTATACCCT –5’PART IV: MAJOR QUESTIONS (10 points each, total of 30 points)1.Please discuss what you have learned from this course, including (1) the generalknowledge framework, (2) your most interested knowledge and why this knowledge has impressed you, (3) the value of teamwork, (4) any change of your learningattitude and/or the construction of your interest in science.2.Please discuss the similarity and difference between miRNA and siRNA, and describethe distinct contributions of these two small regulatory RNAs to the fundamentalbiology and application, respectively.3.It has been recently reported that a new protein X functions in repressing thetranscription of an oncogene gene Y. Could you design experiments to test if Xprotein binds to the promoter region of Y gene (1) in vitro and (2) in vivo? If it does bind, could you design an experiment to test if the binding is essential for thetranscriptional repression?[Notes: You have all DNA sequences, plasmid vectors, cloning enzymes and other reagents that he needs.]武汉大学生命科学学院2007－2008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 2008)PART I:1.Trombone modelThis model is proposed to explain the coordinated synthesis of the leading strand and the lagging strand to the direction of the repl ication folk movement at a replication folk (2’)2.TFⅡDA transcription factor composed of TBP and TAFs for RNA polymeraseⅡ;(1’)TBP recognizes TA TA box and TBP-DNA complex provides a platform for other transcription factors and polymerase to the promoter;(1’)Two of TAFs bind the core promoter elements such as Inr and DPE. Several of histone-like TAFs are also associated with some histone modification enzymes.(1’)3. Telomerase:Solve the End Replication Problem (1’) through adding the telomeric sequence to the 3’ end of the telomere. (1’) No extra primer nor template needed.4. RNA editing: a way of changing the sequence of RNA after transcription(1’) by site-specific deamination of insertion.(1’)5. Tandem mass spectrometryAnswer 1: A method that determines the protein sequence based on the accurate mass of protein fragments obtained by mass spectrometry (2’)Answer 2: (2’)6.Aminoacyl tRNA synthetaseAn enzyme that catalyzes the attachment of an amino acid to a cognate tRNA (2’) through two steps: adenylylation of amino acid and tRNA charging (+1’)7. Viral-like retrotransposons: also called long terminal reapeat (LTR) retrotransposons. The element includes two long terminal repeat sequences that flank a region encoding two enzymes: integrase and reverse transcriptase. It mediate transposition reaction through a RNA intermediate.8Transcriptional silencing is a specialized form of repression that can spread along chromatin,(1’) switching off multiple genes without the need for each to bear binding sites for specific repressors(1’). The mechanism of this repression is the propagation of certain repressing histone modifications over stretches of chromatin. (1’)Insolator elements can block this spreading, thus protect some inserted gene from si lencing.(0.5’) 9. Suppressor Mutation: 抑制突变，抑制基因突变，抑制因子突变Key points: (1) a second mutation(2) the GENOTYPE is mutationally altered but the wild type PHENOTYPE restores(1)maybe on a different gene (intergenetic) or on the same gene (intragenetic)10. Shine-Dalgarno sequence: 又称RBS, ribosome binding siteKey points: (1) in prokaryotic cells(2) a stretch of RNA 3 to 9 nucleosides upstream of the start codon in a mRNA(3) contains conservative sequence: 5’-AGGAGG-3’, which is complimented to certain region of 16s rRNA(2)the conservation and spacing decides the activeness of the following OFRPart III: Short questions2Each mRNA gene contains a poly-A signal sequence near the termination site(1’). Eukaryotic transcription termination is highly coupled with polyad enylation:(1’)(1)CstF/CPSF bound at the CTD tail is transferred to poly-A signal sequence after ti istranscripted, resulting in mRNA cleavage and recruitment of enzymes for polyadenylation .(2’)(2)Poly-A polynerase(PAP) adds about 200 As to RNA’s 3’ end.(2’)Two models for polymerase recycle:(3)Transfer of 3’-processing enzymes from CTD tail to RNA triggers comformational change inPol, reduce processivity, leading to spontaneous termination.(0.5’)(4)Absense of 5’-cap is sensed by the Pol, recognizes the transcript as improper andterminates.(0.5’)3.Please describe the similarity and difference between group II intron and spliceosome-mediatedpre-mRNA splicing.(答案有点长，大家可以在精简一下。
分子生物学练习题+答案
分子生物学练习题+答案一、单选题（共40题，每题1分，共40分）1、在DNA双链中连接碱基对的力是:A、共价键力B、离子键力C、氢键力D、范德瓦尔斯力正确答案：C2、可以切割磷酸二酯键的酶是:A、连接酶B、聚合酶C、DNA酶D、制限性内切酶正确答案：D3、可以识别特异序列的分子是:A、质粒B、酶C、探针D、引物正确答案：C4、识别到停止密码子后,释放肽链和RNA的复合物的是:A、rRNAB、mRNAC、tRNAD、释放因子正确答案：D5、可以识别启动子序列的转录因子是:A、Rho因子B、α 因子C、σ 因子D、β因子正确答案：C6、编码氨酰tRNA合成酶的RNA是:A、mRNAB、rRNAC、tRNAD、siRNA正确答案：B7、在转录过程中起模板作用的分子是:A、RNAB、rRNAC、tRNAD、DNA正确答案：D8、DNA聚合酶在PCR反应过程中不需要的元素是:A、铜离子B、镁离子C、锰离子D、钾离子正确答案：A9、原核生物基因组中不含有的序列是:A、启动子B、外显子C、终止子D、编码区正确答案：B10、可以直接将外源基因导入植物细胞的是:A、电穿孔法B、生物炮法C、微注射法D、农杆菌法正确答案：D11、识别启动子启动转录的是:A、Rho因子B、RNA聚合酶C、螺旋酶D、拓扑异构酶正确答案：B12、DNA的组成单位是:A、氨基酸B、核苷酸C、核糖D、脱氧核糖正确答案：B13、将单链DNA合成双链的酶是:A、连接酶B、DNA聚合酶C、裂解酶D、RNA聚合酶正确答案：B14、用于筛选重组克隆的抗性基因常来源于:A、乳酸菌B、大肠杆菌C、枯草杆菌D、链霉菌正确答案：D15、在PCR反应中DNA聚合酶的最适反应温度是:A、37°CB、55°CC、72°CD、95°C正确答案：C16、将DNA上的遗传信息转录为RNA的过程称为:A、翻译B、转录C、复制D、修复正确答案：B17、启动子序列具有下列哪个特征:A、富含GCB、富含ATC、具有内含子D、保守性低正确答案：B18、可以直接导入细胞质的方法是:A、生物枪B、显微注射C、化学转染D、电穿孔正确答案：B19、加入对照组的目的是:A、减少实验误差B、增加结果可重复性C、证明结果可靠性D、以上皆是正确答案：D20、编码tRNA的基因位于:A、线粒体B、核糖体C、细胞核D、细胞质正确答案：C21、可以永久存在于宿主细胞中的载体是:A、人工染色体B、黏粒C、病毒向量D、质粒正确答案：A22、可以直接将外源DNA导入细胞的是:A、琼脂糖凝胶电泳B、质粒载体C、PCRD、生物枪技术正确答案：D23、下列不属于RNA聚合酶的职能的是:A、识别启动子B、解离DNA双链C、催化磷酸二酯键形成D、终止转录正确答案：B24、用于初步筛选重组克隆的方法是:A、PCRB、测序C、杂交D、蓝白斑筛选正确答案：D25、下列不属于核酸杂交的技术是:A、Northern印迹B、In situ杂交C、Southern印迹D、Western印迹正确答案：D26、可以实现定点诱变的技术是:A、CRISPR/Cas9B、ZFNsC、TALENsD、以上均可正确答案：D27、在基因编辑技术中,靶向特定位点的核酸酶是:A、ZFNsB、TALENsC、CRISPR/CasD、Restriction enzymes正确答案：C28、大肠杆菌对于基因克隆的主要作用是:A、表达目的蛋白B、合成引物C、作为宿主D、提供连接酶正确答案：C29、可以自我复制的核酸是:A、mRNAB、rRNAC、tRNAD、miRNA正确答案：B30、在制备重组DNA时,使用琼脂糖的目的是:A、提供营养B、连接DNA段C、物理分离片段D、催化连接反应正确答案：C31、用于分离核酸片段的凝胶包括:A、琼脂糖凝胶B、纤维蛋白凝胶C、丙烯酰胺凝胶D、以上所有正确答案：D32、编码 rRNA 的基因位于:A、线粒体DNAB、质粒DNAC、细胞核DNAD、细胞质DNA正确答案：C33、在PCR反应中,引物与模板的结合温度约为:A、37°CB、55°CC、72°CD、95°C正确答案：B34、在Southern杂交中起探针作用的是:A、DNAB、RNAC、载体D、引物正确答案：A35、下列内切酶与其识别位点不正确匹配的是:A、EcoRI - GAATTCB、BamHI - GGATCCC、HindIII - AAGCTTD、PstI - CTGCAG正确答案：B36、基因敲除实验中所用对照组应为:A、目的基因缺失组B、野生型组C、质粒载体组D、siRNA处理组正确答案：B37、下列不属于PCR反应体系的组成部分是:A、DNA模板B、聚合酶C、dNTPD、琼脂糖正确答案：D38、在PCR反应中起引物作用的分子是:A、脱氧核糖B、Taq酶C、dNTPD、引物正确答案：D39、属于真核生物的模型生物是:A、小鼠B、酵母C、果蝇D、以上所有正确答案：D40、在基因芯片技术中,利用荧光探针可以检测:A、蛋白质表达B、DNA突变C、mRNA表达D、蛋白质结构正确答案：C二、多选题（共30题，每题1分，共30分）1、可以直接转入植物细胞的方法有:A、电穿孔B、农杆菌介导C、生物炮D、微注射E、病毒感染F、质粒转化G、以上ABC均可正确答案：G2、可以改变DNA序列的技术不包括:A、CRISPR/Cas9基因编辑B、ZFNs技术C、TALENs技术D、单链RNA技术E、慢病毒感染转导F、同源重组G、随机诱变正确答案：D3、用于快速扩增特定 DNA 序列的技术是:A、聚合酶链式反应B、印迹杂交C、克隆技术D、基因编辑E、免疫沉淀F、连接酶链式反应G、线性DNA合成正确答案：A4、编码氨基酸的三联密码存在于:A、DNA双链上B、mRNA分子上C、tRNA分子上D、rRNA上E、盖帽RNA上F、启动子区域G、终止子区域正确答案：C5、编码氨基酸序列信息的核酸为:A、DNAB、RNAC、mRNAD、tRNAE、rRNAF、cDNAG、质粒正确答案：C6、下列关于mRNA加工的说法正确的是:A、在细胞核中加帽B、在细胞质中加多聚A尾C、在细胞核中增加内含子D、在细胞质中切除内含子E、在细胞质中加帽F、在细胞核中加尾巴G、在细胞质中进行剪接正确答案：AB7、制备重组DNA的关键步骤包括:A、获取载体质粒B、载体和插入DNA消化C、两DNA段连接D、构建感受态细胞E、转化宿主细胞F、蓝白斑筛选G、所有以上步骤正确答案：G8、可以使细胞产生瘤变的 DNA 片段有:A、激活型原癌基因B、失活型抑癌基因C、缺失型原癌基因D、增强子区域激活E、缺失型抑癌基因F、重复型基因座G、重复型端粒DNA正确答案：ABE9、DNA测序中的Sanger方法基于:A、引物延伸终止B、引物延伸解离C、二代测序D、三代测序技术E、荧光定量PCRF、比较法G、质谱分析正确答案：AB10、参与制备cDNA文库的关键酶类有:A、连接酶B、限制性内切酶C、聚合酶D、反转录酶E、核酸酶F、RNase HG、以上DE正确答案：G11、可以提取基因组DNA的方法有:A、PCR扩增B、Northern印迹C、Southern印迹D、过滤法E、质谱法F、限制性酶切G、盐析法正确答案：CG12、基因表达调控的机制包括:A、转录水平调控B、RNA水平调控C、翻译水平调控D、蛋白活性调控E、基因增幅F、肽链释放G、以上AD均可正确答案：G13、制备重组质粒需要下列步骤:A、载体选择B、消化载体和插入DNAC、连接反应D、感受态细胞制备E、转化宿主细胞F、克隆筛选G、以上全部正确答案：G14、用于快速扩增特定DNA序列的技术是:A、聚合酶链式反应B、基因芯片C、印迹杂交D、连接酶反应E、线性引物延伸F、链置换扩增G、同源重组正确答案：A15、编码mRNA的DNA单链被称为:A、编码链B、上游链C、下游链D、正义链E、反义链F、互补链G、载体链正确答案：E16、启动子通常位于:A、编码区B、翻译终止点下游C、转录终止点下游D、翻译起始点上游E、终止子下游F、转录起始点上游G、基因内含子正确答案：F17、对肿瘤基因组的检测可以应用:A、Southern印迹B、Northern印迹C、Western印迹D、Eastern印迹E、基因检测F、测序G、芯片技术正确答案：AEFG18、制备重组DNA的步骤包括:A、载体选择B、插入DNA获得C、双酶切D、连接反应E、转化F、筛选G、以上全部正确答案：G19、编码氨基酸序列的核酸为:A、rRNAB、mRNAD、mRNA前体E、单链RNAF、双链RNAG、环状RNA正确答案：B20、可以提高基因在异源表达载体中的表达水平的方法不包括:A、引入选择标记B、改良启动子序列C、优化编码序列D、扩增子克隆E、引入变位信号F、优化文库构建方法G、终止子序列调控正确答案：F21、质粒载体应具有下列哪些特征:A、含有克隆位点B、编码病毒蛋白C、大片段插入区D、与宿主互作E、可自主复制F、含有筛选位点G、低拷贝数正确答案：AEF22、制备 cDNA 文库需要哪些关键技术:A、模板 RNA 提取B、反转录C、PCR 扩增D、连接酶反应E、内切酶反应F、克隆载体G、上述 AB正确答案：G23、编码蛋白质的核酸为:B、mRNAC、rRNAD、tRNAE、siRNAF、miRNAG、反转录病毒RNA正确答案：B24、检测mRNA的方法包括:A、Northern杂交B、Western印迹C、Southern印迹D、荧光in situ杂交E、实时定量PCRF、RNA序列表达谱分析G、以上ABDF正确答案：G25、模式生物的研究应用包括:A、遗传学研究B、药物筛选平台C、人类疾病模型D、功能基因组学E、发育生物学F、进化生物学G、所有以上正确答案：G26、基因编辑技术包括:A、ZFNs技术B、TALENs技术C、CRISPR/Cas技术D、基因敲除E、RNAi技术F、慢病毒介导G、以上所有正确答案：ABC27、下列关于DNA的描述错误的是:A、由脱氧核糖组成B、含有腺嘌呤和胞嘧啶C、双链螺旋结构D、含有关键遗传信息E、可以自主复制F、可以直接进行蛋白质转译G、可以经过转录形成RNA正确答案：F28、可以作为核酸探针的分子有:A、单链DNAB、双链DNAC、RNAD、寡核苷酸E、蛋白质F、生物素G、以上除E均可正确答案：ACDF29、基因敲入的技术可以包括:A、siRNAB、基因敲除C、ZFNsD、TALENsE、CRISPR/Cas9F、Cre-Lox重组系统G、反义DNA正确答案：CDEF30、制备重组质粒的主要步骤是:A、载体线性化B、消化插入片段C、连接反应D、感受态细胞制备E、转化宿主细胞F、克隆鉴定G、所有以上步骤正确答案：G三、判断题（共30题，每题1分，共30分）1、启动子与编码区距离越远,转录效率越高。
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）。
2017年武汉大学885分子生物学考研真题
一、名词解释(共8小题，每小题5分，共计40分）
1、碱基翻出；
2、RISC；
3、组蛋白折叠域；
4、ambermutation；
5、氨酰-tRNA合成酶；
6、EMSA；
7、DNAhelicase；
8、PCR；
二、简答题(共5小题，每小题10分，共计50分）
1、DNA分子的组成、结构特点。
2、假基因的形成机制及其特点。
3、原核生物内蛋白质的合成过程。
4、说明染色质免疫共沉淀技术的原理。
5、增强子、绝缘子的作用特点？
三、论述题(共3小题，每小题20分，共计60分）
1、一个基因在人体正常组织内编码出来的蛋白质大小为50kDa，但在人的乳腺癌组织内编码的两种蛋白质的大小分别为50kDa、40kDa，研究发现该基因含有五个外显子，其中四号外显子编码90个氨基酸（大小约为10kDa），因此猜想这种现象是由于可变剪切引起，设计实验证明你的猜想。
2、简述真核生物转录因子的特点，描述一种根据此特点而产生的一种分子生物学技术，并介绍其作用。
3、增强子和绝缘子在基因表达调控中的作用和特点有哪些？
885分子生物学初试参考书目
《Molecular biology of the gene》（原版）（第六版）PEARSON出版此书有中文翻译版，也可作为参考
《现代分子生物学》（第三版）（朱玉贤）。
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。
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。
武汉大学分子生物学真题2009年_真题-无答案
武汉大学分子生物学真题2009年(总分150,考试时间90分钟)一、名词翻译与解释1. Regulatory gene2. Real-time PCR3. Nick translation4. Generic promoter5. Histone code6. Proteomics7. Gene targeting8. Single nucleotide polymorphism9. Alternative mRNA processlng10. Replicon二、简答题1. 什么是DNA损伤?生物体DNA损伤与哪些因素有关?生物细胞有哪些机制来处理DNA 损伤，并进行DNA损伤修复?2. 表观遗传(epi-genetics)修饰包括哪些内容?并举一例说明其在基因的表述调控和染色体重塑等方面有哪些重要作用?3. 反式作用因子是直接或间接地识别或结合在各类顺式作用元件序列上而参与调控靶基因转录效率的蛋白质。
试问反式作用因子的DNA结合结构域有哪几种类型?特点如何?反式作用因子，即转录调控因子，是一类特殊的DNA结合蛋白。
不同的转录调控因子能与DNA上特异的顺式作用元件相互作用从而调控转录。
这些因子中的重要功能结构域有：螺旋一转角一螺旋，锌指结构，碱性亮氨酸拉链，螺旋一环一螺旋、同源域等，转录因子的磷酸化对转录调控起很大作用。
4. PCR的原理是什么?在PCR反应中是否循环次数越多，所合成产物数量也越多?为什么?5. 增强子和绝缘子在基因表达调控中的作用和特点有哪些?三、论述题1. 试论述转座子的类别、结构和转座机制及在基因表达与分子进化中的作用。
2. 在进行遗传重组和基因克隆与表达的研究中，对分子克隆的载体与宿主系统的选择有非常重要的作用。
请论述各主要载体系统的特点及宿主系统。
3. 从转录后水平上进行比较，论述原核生物和真核生物RNA加工的异同点。
转录后水平上的RNA加工即mRNA的加工。
4. 什么是HapMap计划、ENCODE计划、Jim计划、Proteome计划?这些计划与人类基因组计划有何联系?这些计划对分子生物学的发展和生命活动的分子机制有什么影响和意义?。
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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武汉大学2001 年攻读硕士学位研究生入学考试试题一、解释概念(20 分,每个4分)1. Satellite DNA-卫星DNA,又称串联重复序列，由2~6个核苷酸组成的重复单位组成的串联重复。
2. Replisome-复制体,是指由多种复制相关蛋白组成的复合物。
3.逆转座子，指另一类能够通过逆转酶和整合酶共同作用介导转座的转座子4. 反式激活因子，一类能结合到顺式作用元件上形式激活作用的蛋白质5. 衰减子，原核生物操纵子上一段能够显著减弱或终止转录作用的核苷酸序列衰减作用，指原核生物中根据底物浓度变化来调控基因转录进程的中断与否的一种调控机制。
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DNA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 年攻读硕士学位研究生入学考试试题三、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)武汉大学2003 年攻读硕士学位研究生入学考试试题一、名词翻译,并简要解释(共10小题,每题4分,共40分)1. Domains and motifs 结构域和结构基序，分别指的是反式作用因子中与蛋白质和DNA分子相互作用的两种蛋白质亚基。
2. Alternative splicing 选择性剪接，指一个基因在不同发育阶段、分化细胞或生理状态下，通过不同的剪接方式，来产生多个不同mRNA的过程。
3. Reporter genes 报告基因，指的是一类表达产物容易被检测或观察到的基因。
4. The PCR cycle PCR循环，完成一次PCR中变性-退火-延伸的过程为一次循环。
5. Restriction mapping 限制性酶切作图，一种利用限制酶切位点标定DNA分子相对位置的方法。
6. Multiple cloning sites 多克隆位点，指的是DNA中含有两个以上密集的能被多种限制酶识别并切割的位点。
7. DNA libraries DNA文库，是指某种生物的基因组DNA切割成适宜大小，与载体连接后导入宿主细胞进行克隆后所获得重组体内基因组片段的集合。
8. Proteomics 蛋白质组学，指的是一门在蛋白质组水平上研究蛋白质特征，包括表达、修饰、互作等，从而进一步了解基因功能的学科。
9. Replicon 复制子，指的是细胞中能独立复制的一个DNA单元。
10. Semi-conservative replication 半保留复制，指DNA复制中，每条单链分别作为模板合成新链，从而在每条子代DNA中保留一条亲代DNA链的过程。
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?武汉大学2004年攻读硕士研究生入学考试试题一,名词翻译与解释（共10小题，每题4分，共40分）1. Synonymous codons 同义密码子，指能编译同一种氨基酸的密码子。
2. RNA editing RNA编辑，一种通过改变碱基或插入核苷酸来对遗传信息进行修改的转录后加工方式。
3. Spliceosome 剪切体，指的是有多种snRNP组成的介导mRNA前体中内含子剪切的蛋白核酸复合物。
4. Microarray 微阵列，一种通过将基因组核酸固定在固相介质上，利用核酸杂交，结合电脑检测分析来高效检测基因表达的技术。
5. Plaque hybridization 噬菌斑杂交，指的是将噬菌斑拓印在硝酸纤维膜上变性处理后通过核酸杂交，来筛选含有特定基因的噬菌斑的方法。
6. Open reading frame 开放式阅读框，指基因组中一段连续的可编码一条整多肽的核苷酸序列。
7. Ribozyme 核酶，一种具有催化活性的RNA，能通过催化靶位RNA链中磷酸二酯键的断裂来特异性剪切底物分子。
8. RFLP 限制性片段长度多态性，指从不同个体制备的DNA经过同一种限制酶切割后产生不同长度DNA片段的现象。
9. Site specific recombination 位点特异性重组，指的是基因组中在重组酶作用下发生于具有特异性重组位点间的基因重排现象。
10.RNA interference RNA干扰，是指在进化过程中高度保守的，由双链RNA诱发的同源mRNA高效特异性降解的现象。
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（共4题，每题15分，共60分）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 聚合酶能识别不同的结构基因?3.在进行基因工程时,载体是携带靶DNA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乘10的9次方bp(而大肠杆菌的基因组约为4。
6 乘10的6次方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 年武汉大学硕士研究生入学考试分子生物学试卷一、名解（共10小题，每题4分，共40分）1. Ubiquitin 泛素，一种高度保守的由76个氨基酸构成的，可作用于介导蛋白质降解的蛋白质。
2. Ribosomal binding site 核糖体结合位点，指mRNA上一段可引导核糖体与mRNA结合的核苷酸序列。
3. siRNA 小干扰RNA，一种可由RNA依赖性RNA聚合酶扩增并经过Dicer切割形成的可介导同源mRNA降解或翻译抑制的小双链RNA。
miRNA 微RNA，一种由基因编码产生，可作用于同源mRNA降解或翻译抑制的单链小RNA。
4. Proteome 蛋白质组，指的是由一个基因组编码产生的所有蛋白质。
5. DNA foot printing DNA足迹，指的是一种利用蛋白质结合DNA可使结合部位不受DNase降解的特点来检测蛋白质结合位点的方法。
6. Single nucleotide polymorphism 单核苷酸多态性，主要指的是在基因组水平上由单个核苷酸变异而引起的DNA序列多样性的现象。
7. Chromosome walking 染色体步移，指通过连续将上一个已知克隆的末端序列作为引物来从文库中分离下一个克隆序列，从而获取已知克隆序列的旁侧序列的方法。
8. Haplotype 单倍型，指的是位于一条染色体上或某一区域内的一组相关联的SNP位点。
9. Zoo blot 动物园印迹（种间印迹）即用southern blot 来检测某一种属的DNA探针与多个种属基因组的DNA杂交活性，从而评估种间关系一段DNA序列在进化过程中的的保守程度。
10. Transcription unit 转录单位，指一段能够被RNA聚合酶转录成一条连续酶mRNA的DNA序列。
二、简答(共5小题，每题10分，共50分)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、论述题（共4题，每题15分，共60分）1.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?‘’武汉大学2006 年武汉大学硕士研究生入学考试分子生物学试卷一、名解（共10小题，每题4分，共40分）1、DNA Microarray DNA微阵列，一种将核酸固定在固相载体上，然后通过核酸杂交、电脑分析检测来高效检测基因组表达情况的技术。