分子生物学 双语习题及精解

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双语分子生物学复习资料

双语分子生物学复习资料

1. (核酸性质—减色性,双链DNA,单链DNA,寡核苷酸)Which has the highest absorption per unit mass at a wavelength of 260 nm? ( B )A RNA.B mononucleotides.C double-stranded DNA.D protein.减色性跟紫外最大吸收峰Nucleotides(260)〉RNA〉ssDNA>double-stranded DNA4.(RNA的转录时,DNA的模板链和编码链(有义链和反义链))mRNA互补()的链是模板链(或称反义链,与反义链互补的是有义链)。

The sequence 5’-AGTCTGACT-3’ in DNA is equivalent to which sequence in RNA?( C )——给你DNA序列,找到对应的RNA序列,给你RNA序列,找到对应的DNA序列。

A5’-AGUCUGACU-3’B5’-UGTCTGUCT-3’C 5’-UCAGUCUGA-3’D 5’-AGUCAGACU-3’DNA双链维持的结构是什么?A H-bonds and disulfide bonds(二硫键)B base stacking interaction(碱基堆积力)and peptide bondsC hydrophobic interaction(疏水相互作用)and H-bondsD base stacking interaction and hydrophobic interaction 碱基堆积和疏水相互作用10.Which one of following statements about telomere(端粒)and teromerase(端粒酶)is false?( B )A Telomere contains up to hundreds copies of a short repeated sequence.B Teromerase is a typical DNA polymerase. (端粒酶不是经典的DNA聚合酶)C The activities of telomerase are related to the development stages and tissue types.D The telomere is synthesized by the enzyme telomerase and DNA polymerase.11. Which one of the following statements about telomere(端粒)and telomerase (端粒酶)is true? ( C)(概念)A The activity of telomerase is very low in cancer and germ cell.B Telomere is long stretches at the ends of the chromosome and rich in G-C.C Telomerase is the reverse transcriptase responsible for the extension of telomeric repeatD Telomerase is complex of DNA, protein and RNA.8种必须氨基酸14.Which is the major forces between the amino acids in the polypeptides(多肽链)? ( )A H-bonds and disulfide bonds(二硫键)B base stacking interaction(碱基堆积力)C hydrophobic interaction(疏水相互作用)D peptide bonds (肽键)核酸是3,5磷酸二酯键21. Which one of the following statements about DNA secondary structure is not true? ( D )A The basic features of DNA double helix were deduced by James D. Watson and Francis Crick in 1953l.B The structure identified by James D. Watson and Francis Crick is known as B-form DNA which is right-handed form(右手螺旋).C Under conditions of low humidity, DNA can be induced to form A-form DNA .D Z-form DNA is right-handed like the B-form DNA .(是左手)For example: telomere replication 端粒复制1.bp base pairs碱基对2.GMO Genetically modified organism 基因修饰生物3.The approximate purity of dsDNA preparations may be estimated by determination of the ratio of absorbance at 260 and 280 nm (A260/A280), pure dsDNA has an A260/A280 of 1.8, and pure RNA one of around 2.0.4. mononucleotide 单核苷酸5. 核酸Nucleic acid6. 染色体chromosome7.Gly and pro(脯氨酸) often occur in the beta-turn of protein secondary structure8. 核苷酸Nucleotide, ,核苷Nucleoside ,核小体Nucleosome2. Which one of the following statements about transcription in E.coli is true? ( D )A the–10 sequence is always exactly 10 bp upstream from the transcription start site.B the initiating nucleotide is always a G.C the intervening sequence between the –35 and –10 sequences is conserved.(保守的).D the distance between the –35 and –10 sequences is critical for transcription efficiency.3. E.coli release factor1 (RF1) recognizes which codons?( D )RF2识别UGA and UAAA UAA only.B UAG only.C UGA and UAA.D UAG and UAA.5. Which topoisomerase is to segregate daughter DNA after replication in prokaryote? ( D )原核生物DNA复制好后怎么把酯链的DNA分开?A topoisomerase (拓扑异构酶) IB topoisomerase IIC topoisomerase IIID topoisomerase IV8.In eukaryotes, the leading and lagging strands are synthesized by DNA polymerase(B ), respectively.A DNA pol a and DNA pol δB DNApolδand DNA pol aC DNA polεand DNA pol δD DNA pol a and DNA pol βδ70因子非常重要10.In eukaryotes, mtDNA(线粒体DNA) is replicated by the ( C ).A DNA pol aB DNA polδC DNA polγ(DNA聚合酶γ)D DNA pol β14. The following replication factors(复制因子)bind DNA in the initiation of bacterial (细菌)DNA replication except ( A ).A DnaAB DnaBC DnaGD RPABCD在起始的时候发挥作用15.The yeast(酵母)replication origin is called autonomously replicating sequence (自主复制序列ARS), also called ( D ).A Ori AB Ori BC Ori C (细菌的起始序列)D Ori R改下题目:在真核生物里面酵母菌的复制起始点Ori C和Ori R 又叫什么ARS 1.RF replicative form 复制叉2.In lactose operon, the lacZ gene expression is induced by rather than(而不是)lactose itself, however, in our experiment, is used to add to themedium(培养基)for induction of lacZ gene expression.异乳糖比乳糖诱导作用更强3.primer 引物4. replication复制5. tanscription转录6.The operon model consists of 启动子、操纵子、structrure gene and regulation genes(调节基因).1. Which one of the following statements about PCR is false? ( C )A PCR uses thermostable(热稳定的)DNA polymerases.B ideally, PCR primers should be of similar length and G+C content.C PCR optimization usually includes varying the magnesium(镁离子)concentration and thepolymerization temperature(聚合温度).镁离子浓度非常重要,不能随意调整浓度,一般恒定。

分子生物学L1 L6 问题及答案

分子生物学L1 L6 问题及答案

分子生物学L1 L6 问题及答案分子生物学l1-l6问题及答案l11.nucleicacidisthegeneticmaterial(toexplainviafourexamples)(1) DNA是细菌的遗传物质:细菌转化实验提供了DNA是遗传物质的初步证据。

从第一个菌株中提取并添加到第二个菌株的DNA可以将遗传特征从一个菌株转移到另一个菌株。

肺炎球菌可引起肺炎并导致小鼠死亡。

荚膜多糖有两种:S型和R型。

S型肺炎球菌能在杀死小鼠的同时杀死活的S型细菌,这表明存在一种转化物质。

纯化后,这种转化物质被发现是DNA,因此DNA是细菌的遗传物质。

(2)dna是病毒的遗传物质:噬菌体感染大肠杆菌的实验证明dna是病毒的遗传物质。

当细菌的dna和蛋白质组分被标记上不同的放射性同位素32p及35s时,实验后发现仅有dna被传递到感染细菌所产生的子代噬菌体中,这就很好的证明了dna是病毒的遗传物质。

(3) DNA也是动物细胞的遗传物质:当DNA被添加到培养基中培养的某个真核单细胞生物群落中时,核酸会进入细胞,其中一些会合成一些新的蛋白质。

例如,在胸苷激酶(TK)的合成实验中,DNA被引入受体细胞后,它就成为受体细胞的一部分。

它以与其他部分相同的方式继承。

导入DNA的表达将使细胞产生一些新的特性。

起初,这些实验仅在那些培养基中培养的单个细胞中成功。

现在,人们已经通过显微注射技术成功地将DNA导入小鼠受精卵,并使其成为其遗传物质的稳定部分。

这些实验直接表明,DNA不仅是真核生物的遗传物质,而且可以在不同物种之间转移并保持功能活性。

(4)有一些病毒如烟草花叶病毒(tmv)等就使用另一种核酸――核糖核酸(rna)作为遗传物质,其化学组成结构与dna只是略有不同。

烟草tmv重建实验很好的说明了rna在生命体中起着相同的作用。

由此可见,遗传物质的本质就是核酸。

实际上,除了一些rna病毒外,其余生物的遗传物质都是dna。

2.(1)3?― 也就是说,由核苷酸中五碳糖的第三个C原子连接的羟基末端,可以与另一个核苷酸的5'-磷酸基形成3',5'-磷酸二酯键。

分子生物学习题集精解2

分子生物学习题集精解2

分子生物学习题集精解2第5章DNA重组一、填空题1.自然情况下,在同一基因两个稍微不同拷贝(等位基因)间发生重组的过程中,一个等位基因经过基因转变过程会被另一等位基因代替。

2.通过位点专一的基因重组,游动DNA序列和一些病毒可进入或离开一条目的染色体。

3.一般性重组(同源重组)中,基因交换发生的同源DNA序列间,最常见是发生在同一染色体的两个拷贝。

4.在交换区域,一个DNA分子的一条链与另一个DNA分子的一条链相互配对,在两个双螺旋间形成一个异源双链连接。

5.同过DNA复性,两个单链的互补DNA分子一起形成一个完全双链螺旋,人们认为这个反应从一个慢的螺旋成核作用步骤开始。

6.大肠杆菌的染色体配对需要RecA蛋白,它与单链DNA结合并使同源双链DNA与之配对。

7.一般性重组(同源重组)的主要中间体是交叉链互换,也用它的发现者名字命名为Holliday连接。

8.重组通常从DNA 缺口处开始。

9.负责把RNA转录成互补DNA分子的反转录酶可以解释由反转录病毒引起的永久性基因转变。

10.利用自己的位点专一重组酶把自己从寄主基因组中的一个地方移到另一地方的遗传元件叫转座元件,也叫作转座子。

11.酵母的Tyl元件是一种反转录转座子,它的转座需一段完整RNA转录物的合成,这个转录物又被复制成一个双螺旋DNA,随后被整合到一个新的染色体位置。

12.F质粒的IS元件使F质粒与大肠杆菌染色体之间发生同源重组,产生一个Hfr 细菌。

13.转座元件是指能移动到基因组其他位的DNA序列。

转座元件在以下方面影响基因组:能够引起基因的重排,通过插入能够灭活基因,转座元件的启动子能够影响邻近基因的表达。

14.最简单的转座元件是IS元件。

IS元件由两段短的反向重复序列和一段夹在重复序列之间的负责转座的转座酶基因组成。

当整合到新位点后,转座元件总是在靶位点产生一段同向重复序列。

15.复合转座元件由两个IS元件与夹在中间的抗生素抗性基因组成。

分子生物学双语版本复习资料

分子生物学双语版本复习资料

细胞分类原核生物(prokaryote):最简单的活细胞直径1-10um 细胞膜、细胞壁(most)、环状染色体、质粒、RNA、核糖体、鞭毛、纤毛包括真细菌和古细菌真细菌(eubacteria):可为单细胞或多细胞 E.coli古细菌(archaea):膜脂由醚键连接而不是酯键真核生物(eukaryote):动物、植物、真菌(fungi)、原生生物(protists)直径10-100um分化(differentiation):拥有相同DNA但转录不同基因亚细胞器细胞核(nuclei):DNA的转录和RNA的加工场所核仁:rRNA合成和核糖体进行部分组装的场所线粒体:直径1-2um 内含一个小的环状DNA分子、线粒体特异RNA、合成线粒体蛋白的核糖体叶绿体:内膜腔内有第三膜体系---类囊体内质网:与核膜相连光面内质网:脂类物质合成和生物异源物质代谢,含解毒的酶粗面内质网:膜蛋白和分泌蛋白的合成微体:溶酶体—含降解性水解酶过氧化物酶体:高活性自由基和氢过氧化物的代谢乙醛酸酶体:植物的过氧化物酶体进行乙醛酸循环细胞器的分离:渗透压冲击、可控制的机械剪切和某些非离子去污作用可以破坏质膜差速离心法:根据沉降系数的不同分离细胞器密度梯度离心法:分离密度相似的细胞器生物大分子蛋白质:起结构和功能双重作用氨基酸聚合体核酸:核苷酸的聚合体多糖:单糖以糖苷键共价连接几丁质是N-乙酰胺基葡糖的聚合体(真菌细胞壁和节肢动物外骨骼中)黏多糖:结缔组织的重要组成部分脂类:饱和与不饱和脂肪酸的三酰甘油磷脂:两分子脂肪酸和一分子磷酸以酯键与甘油相结合鞘磷脂:磷脂胆碱+脑胺复杂大分子:核蛋白—核酸+蛋白质(端粒酶、核糖核酸酶P)糖蛋白、蛋白多糖(蛋白质+黏多糖)—糖类与蛋白质共价相连脂连接蛋白:共价相连脂蛋白:非共价相连大分子的组装蛋白质复合体:微管(微管蛋白构成)、微丝(肌动蛋白和肌球蛋白构成)、中间纤维(多种蛋白质构成)细胞骨架(一系列蛋白质微丝)微管蛋白:110kDa的球形蛋白是细胞骨架、鞭毛、纤毛的主要组分核蛋白:细菌70s核糖体由一个50s大亚基(23sRNA、5sRNA、31种蛋白质)和一个30s小亚基(16sRNA和21种蛋白质)组成真核生物的80s核糖体含有60s(28SRNA、5.8sRNA、多种5sRNA)和40s(18sRNA)两个亚基膜:膜磷脂和鞘磷脂形成了极性基团在外部、烃链在内部的双分子层膜蛋白的功能:1、信号分子的受体2、酶3、转运时的孔或通道4、细胞间相互作用的介质非共价相互作用:弱相互作用电荷与电荷、电荷与偶极、偶极与偶极之间的相互作用氢键疏水作用力氨基酸的分类20种常见氨基酸都有一个与质子、氨基、羟基相连的手性α-碳原子和侧链酸性氨基酸:Glu、Asp 带负电碱性氨基酸:Lys、His、Arg 带正电中性氨基酸:Ser、Thr、Asn、Gln、Cys 不带电荷非极性氨基酸:Gly、Pro、Ala、Val、Leu、Ile、Met芳香族氨基酸:Phe、Tyr、Typ(可吸收紫外光280nm处最大)蛋白质的结构与功能球蛋白:可溶性蛋白多数酶纤维蛋白:重要的结构蛋白如丝蛋白、角蛋白一级结构:α-氨基与α-羧基以肽键相连的氨基酸顺序二级结构:α-螺旋(每圈3.6个氨基酸,右手螺旋,链内氢键)β-折叠(平行与反向平行)β-转角无规则卷曲三级结构:不同二级结构区域和连接区的组合非共价相互作用亲水性氨基酸在外面,疏水性氨基酸在内部二硫键氢键范德华力疏水作用力伴娘蛋白:保证肽链的正确折叠四级结构:多个肽链亚基的组合别构效应(亚基间的相互作用)辅基:提供额外化学功能的非蛋白质分子NAD+、血红素、金属离子蛋白质的功能:1、酶2、信号传递3、转运与储存(血红蛋白转铁蛋白脂蛋白铁蛋白)4、结构与运动(胶原蛋白角蛋白肌动蛋白肌球蛋白)5、营养(酪蛋白卵清蛋白)6、免疫(抗体)7、调节结构域:同一多肽中有限的高度有序结构片段相连(由外显子编码)结构基序motif(超二级结构):蛋白质家族中从共同祖先进化过程中保留下来的保守的结合位点或催化位点的必要部分相似的结构基序可以在没有序列相似性的蛋白质中发现直向同源:不同物种的具有相同功能、承担相同生化角色的蛋白质家族成员共生同源:进化不同但功能相似的蛋白蛋白质分析法蛋白质纯化依据:凝胶过滤层析---蛋白质大小等电聚焦(形成PH梯度)、电泳、离子交换层析---所带离子电荷疏水作用层析---疏水性亲和层析---酶或受体与配体的特殊亲和性Edman降解法:从N端对多肽进行测序抗体:脊椎动物的免疫系统为了应对外来物质(抗原)入侵而产生的蛋白质,对抗原有很高的结合亲和力与特异性X射线衍射确定蛋白质结晶的三级结构蛋白质功能分析方法:分离纯化、研究突变体的表现核酸结构碱基:嘌呤为双环结构,嘧啶为单环结构核苷:碱基共价结合于戊糖分子的1位,DNA中的为2-脱氧核糖碱基+糖分子=核苷(糖苷键相连)核苷酸:一个或多个磷酸基团结合到核苷的3位、5位、2位上碱基+糖分子+磷酸分子=核苷酸NTP 5-三磷酸核糖核苷磷酸二酯键:前一个核糖的5-羟基与下一个核糖的3-羟基通过磷酸基团共价相连核酸序列:DNA或RNA链中的碱基A、C、G、T(或U)排列顺序由5端写至3端DNA双螺旋:两条独立的反向平行的单链DNA以右手螺旋缠绕,糖-磷酸骨架在外,氢键、碱基堆积力每螺旋10个碱基对标准的DNA双螺旋---B型,A型---右手螺旋(RNA链)每圈11个碱基对Z型---左手螺旋每圈12个碱基对RNA的二级结构:局部分子内碱基配对和其他氢键相互作用而维持的局部互补的螺旋结构DNA修饰:A和C的甲基化核酸的理化性质核酸螺旋的稳定性由疏水作用和堆积在碱基对间的偶极矩作用决定酸效应:强酸---核酸水解为碱基、糖和磷酸中度酸---脱嘌呤核酸碱效应:变性----碱基的互变异构态改变,特异碱基被破坏某些化学试剂破坏碱基间的疏水作用力使核酸变性DNA的水溶液具有高黏性平衡密度梯度离心(等密度梯度离心):分离DNA(RNA)与蛋白质DNA的密度梯度为1.7g/cm3。

《分子生物学》习题及答案解析.doc

《分子生物学》习题及答案解析.doc

分子生物学习题及答案第1章序言1.简述孟德尔、摩尔根和Waston等人对分子生物学发展的首要奉献。

孟德尔是遗传学的奠基人,被誉为现代遗传学之父。

他经过豌豆试验,发现了遗传学三大根本规律中的两个,别离为别离规律及自在组合规律。

摩尔根发现了染色体的遗传机制,创建染色体遗传理论,是现代试验生物学奠基人。

于1933年因为发现染色体在遗传中的效果,赢得了诺贝尔生理学或医学奖。

Watson于1953年和克里克发现DNA双螺旋结构一(包含中心法则),取得诺贝尔生理学或医学奖,被誉为''DNA之父”。

2.写出DNA、RNA、mRNA和siRNA的英文全名。

DNA: deoxyribonucleic acid 脱氧核糖核酸RNA: ribonucleic acid 核糖核酸mRNA: messenger RNA 信使RNAtRNA: transfer RNA 转运RNArRNA: ribosomal RNA 核糖体RNAsiRNA: small interfering RNA 搅扰小RNA3.试述''有其父必有其子”的生物学实质。

其生物学实质是基因遗传。

子代的性状由基因决议,而基因因为遗传的效果,其基因的一半来自于父方,一般来自于母方。

4.早期首要有哪些试验证明DNA是遗传物质?写出这些试验的首要进程。

1)肺炎链球菌转化试验:表面光滑的S型肺炎链球菌(有荚膜多糖一致病性);表面粗糙R型肺炎链球菌(无荚膜多糖)。

%1活的S型一打针一试验小鼠一小鼠死亡%1死的S型(经烧煮灭火)一打针一试验小鼠一小鼠存活%1活的R型一打针一试验小鼠一小鼠存活%1死的S型+活的R型一试验打针一小鼠死亡%1别离被杀死的S型菌体的各种组分+活的R型菌体一打针一试验小鼠一小鼠死亡(内只要死的S型菌体的DNA转化R型菌体导致致病菌)*DNA是遗传物质的载体2)噬菌体侵染细菌试验%1细菌培育基35S符号的氨基酸+无符号噬菌体一培育1-2代一子代噬菌体简直不含带有35S符号的蛋白质%1细菌培育基32N符号的核昔酸+无符号噬菌体一培育1-2代一子代噬菌体含有30% 以上32N符号的核昔酸*噬菌体传代进程中发挥效果的或许是DNA而不是蛋白质。

分子生物学习题集精解3

分子生物学习题集精解3

第8章翻译一、填空题1.氨酰tRNA合成酶可使每个氨基酸和它相对应tRNA分子相耦联形成一个氨酰tRNA分子。

2.核糖体包括两个tRNA分子的结合位点:肽酰tRNA结合区,即P位点,紧密结合与多肽链延伸属端连接的tRNA分子;氨酰tRNA结合区,即A位点,结合带有一个氨基酸的tRNA分子。

3.肽酰转移酶催化肽键的形成,一般认为这个催化反应是由核糖体大亚基上的rRNA分子介导的。

4.释放因子蛋白与核糖体上A位点的终止密码结合,导致肽基转移酶水解连接新生多肽与tRNA分子的化学键。

5.任何mRNA序列能以三种可读框的形式被翻译,而且一种都对应一种完全不同的多肽链。

6.蛋白质合成的起始过程很复杂,包括一系列被起始因子催化的步骤。

7.在所有细胞中,都有一种特别的起始tRNA识别起始密码子AUG,它携带一种特别的氨基酸,即甲硫氨酸,作为蛋白质合成的起始氨基酸。

8.核糖体沿着mRNA前进,它需要另一个延伸因子EF-G,这一步需要GTP的水解。

当核糖体遇到终止密码(UAG 、UGA、UAA)的时候,延长作用结束,核糖体和新合成的多肽被释放出来。

翻译的最后一步被称为终止,并且需要一套释放因子。

9.氨酰tRNA合成酶“补充”tRNA分子,而肽酰转移酶催化肽链的合成。

10.假定摆动假说是正确的,那么最少需要32种tRNA来翻译61种氨基酸密码子。

11.胶原蛋白通过在不同的脯氨酸残基上添加羟基基团而被化学修饰。

这个反应是由两种酶催化的,它们是脯氨酰-3-羟基化酶和脯氨酰-4-羟基化酶。

其他的一些蛋白质被则叫作蛋白激酶磷酸化。

蛋白质添加寡聚糖的过程叫作糖基化作用,而添加脂肪酸链则叫作酰基化作用。

O-寡聚糖是一种同丝氨酸或苏氨酸残基上氧连接的寡聚糖,而N-寡聚糖是通过与天冬酰胺上的氮原子连接而成。

N-寡聚糖又是从同一种叫作常醇脂的前体寡聚糖衍生而来。

12.阿黑皮素原是多聚蛋白被切割以后可以产生很多活性蛋白质的一个例子。

如骨髓灰质炎病毒之类的RNA病毒也能合成类似的结构物。

分子生物学考题及答案2

分子生物学考题及答案2

;14.Through their experiments with DNA from the bacterium Escherichia coli ,Meselson and Stahl showed that DNA replication is(A) conservative.;(B)dispersive ;(C) duplicative.;(D)semi-conservative15.A mutation changes a CG base pair to an AT base pair. This is a ___ mutation.(A )transversion ;(B )transition ;(C )transpositional ;(D )translocation16、Which of the following is an example of a nonsense mutation?(A )ACG to ACC ;(B )AUG to UUG ;(C )UAC to UAG ;(D )AAA to UUU17.A mutation occurs in which an AUU codon is changed to an AUC codon. Both of these codons signify the amino acid leucine. This is a ___ mutation.(A)Nonsense ;(B )missense ;(C )silent ;(D )neutral18.In a eukaryotic cell, when a positive regulatory protein interacts with a promoter element(A )transcription is activated ;(B )transcription is inhibited ;(C )translation is inhibited ;(D )replication is activated19.In eukaryotes, a protein is synthesized in the ___ and modified in the ___ .(A )nucleus; endoplasmic reticulum ;(B )endoplasmic reticulum; Golgi plex(C )Golgi plex; nucleus ; (D )nucleus; Golgi plexframeshift ;20.Utraviolet light usually causes mutations by a mechanism involving(A )one-strand breakage in DNA ;(B )light-induced change of thymine to alkylated guanine ;(C )inversion of DNA segments ;(D )induction of thymine dimmers ; (E )deletion of DNA segments21.氨酰tRNA 的作用由______决定(A )氨基酸;(B )反密码子;(C )固定的碱基区;(D )氨酰tRNA 合成酶的活性22.一个复制子是______(A )细胞分裂期间复制产物被分离之后的 DNA 片段;(B )复制的 DNA 片段和在此过程中所需的酶和蛋白;(C )任何自发复制的 DNA 序列(它与复制起始点相连);;(D )复制起点和复制叉之间的 DNA 片段23.下列哪些转录因子是装配因子(A )SP1;(B )TF ⅡB ;(C )TF ⅡH ;(D )都不是24.在原核生物复制子中以下哪种酶除去RNA 引发体并加入脱氧核糖核苷酸?(A )DNA 聚合酶Ⅲ;(B )DNA 聚合酶Ⅱ;(C )DNA 聚合酶Ⅰ;(D )DNA 连接酶25.DNA 依赖的RNA 聚合酶的通读可以靠_____(A )ρ因子蛋白与核心酶的结合;(B )抗终止蛋白与一个内在的ρ因子终止位点结合,因而封闭了终止信号;;(C )抗终止蛋白以它的作用位点与核心酶结合,因而改变其构象,使终止信号不能被核心酶识别;(D )NusA 蛋白与核心酶的结合只有在乳糖存在的条件下才能表达;在乳糖存在的条件下不能表达在乳糖不存在的条件下表达;不管乳糖存不存在都能表达三、 填空题(本大题共10小题,每空0.5分,共计10分) 1.转录因子可分为两类,即( )和( )。

分子生物学复习部分资料中英文题目

分子生物学复习部分资料中英文题目

分子生物学复习部分资料中英文题目————————————————————————————————作者: ————————————————————————————————日期:名词解释(probe)探针:分子杂交中和待测核苷酸链碱基互补的具有特定序列的被标记的核苷酸链,可用于检测核酸样品中存在的特定基因。

ﻫ(molecular hybridization)分子杂交:是利用DNA变性与复性这一基本性质来进行DNA或RNA定性或定量分析的一项技术。

(gene chip)基因芯片:指单位面积有规律地紧密排列的特定的DNA片段的支持物。

(gene library)基因文库:是指一个包含了某一生物体全部DNA序列的克隆群体。

ﻫ()cDNA文库:是包含某一组织细胞在一定条件下所表达的全部mRNA经逆转录而合成的cDNA序列的克隆群体,它以cDN A片段的形式贮存着该组织细胞的基因表达信息。

ﻫ(genomic DNAlibrary)基因组DNA文库:是指生物的基因组DNA的信息(包括所有的编码区和非编码区)以DNA片段形式贮存的克隆群体。

ﻫ(transgenic technology)转基因技术:采用基因转移技术使目的基因整合入受精卵细胞或胚胎干细胞,然后将细胞导入动物子宫,使之发育成个体的技术。

(transgenosis)转基因: 转基因技术中被导入的目的基因ﻫ(transgenic animal)转基因动物:转基因技术中目的基因的受体动物(Somatic cell nuclear transfer)核转移技术:将动物的一个体细胞核全部导入另一个体的去胞核的的激活的卵细胞内,使之发育成个体,即克隆(clone)。

gene knockout基因剔除:建立在同源重组基础上的有目的去除动物体内某种基因的技术。

(functional cloning)功能克隆:通过对一种致病基因功能的了解来克隆该致病基因。

ﻫ(positiona l cloning)定位克隆:从一种致病基因的染色体定位出发逐步缩小范围,最后克隆该基因。

双语分子生物学

双语分子生物学

老阚班双语分子生物学题型:填空、名解、选择、判断改错、简答、论述一、绪论1.molecular biology(分子生物学):分子生物学是研究核酸、蛋白质等所有生物大分子的形态、结构特征及其重要性、规律性和相互关系的科学。

2.DNA发现者:James Watson and Francis CrickDNA发现的实验:图1-1(P7)、图1-2(P8)3.DNA全称——Deoxyribonucleic AcidRNA全称——Ribonucleic Acid4.分子生物学主要研究内容:◆DNA重组技术◆基因表达调控研究◆生物大分子的结构功能研究—结构分子生物学◆基因组、功能基因组与生物信息学研究5.DNA recombinant technique(DNA重组技术):是20世纪70年代初兴起的技术科学,目的是将不同DNA片段按照人们的设计定向连接起来,在特定的受体细胞中与载体同时复制并得到表达,产生影响受体细胞的新的遗传性状二、染色体与DNA1.Nucleosome(核小体):核小体是染色体结构的基本单位、由H2A、H2B、H3、H4各两个分子生成的八聚体和由大约200bp的DNA组成的。

八聚体在中间,DNA分子盘绕在外,而H1则在核小体的外面。

每个核小体只有一个H1。

2.C值反常现象(C-value paradox):C值往往与种系进化的复杂程度不一致,某些低等生物却具有较大的C值。

3.关于DNA结构的词汇:double–helix(双螺旋)、base(碱基)sugar-phosphate backbonts(糖-磷酸骨架)、base pairs(碱基对)right/left-handed double-helix(右/左手双螺旋)4.DNA类型:右手螺旋:A-DNA,B-DNA左手螺旋:Z-DNA5.组蛋白:Octameric core (H2A、H2B、H3、H4)+外围(H1)6.真核生物重复序列类型:单拷贝、轻度(或低度)重复、中度重复、高度重复(四个空)不重复序列、中度重复序列、高度重复序列(三个空)7.真核生物基因组的特征:1)真核基因组庞大2)存在大量的重复序列3)大部分为非编码序列4)转录产物为单顺反子5)断裂基因,有内含子结构6)存在大量的顺式作用原件7)具有端粒结构8)细胞器基因, 转录和翻译在时间和空间上是不偶联的;9)有假基因化现象8.Z-DNA的特点:1)left-handed double helix with a zig-zag conformation of thebackbone (less smooth than B-DNA)2)Narrower, more elongated helix than A or B.3)Only one groove is observed, resembling the minor groove4) A high G-C content favours Z conformation.5)Z-DNA formation occurs during transcription of genes, attranscription start sites near promoters of actively transcribed.6)Base pairs nearly perpendicular to helix axis9.DNA复制所需要的元素(Requirements for DNA replication):1.原料(Precursor):四种脱氧核苷三磷酸(dATP、dGTP、dCTP、dTTP)2.模板(Template):以DNA的两条链为模板链,合成子代DNA3.引物(Primer):DNA的合成需要一段RNA链作为引物4.酶:a)引物合成酶(引发酶)(Primases)、b)DNA聚合酶(DNA polymerase)c)DNA连接酶(Ligases)d)DNA 拓扑异构酶(DNA Topisomerase)e)DNA 解螺旋酶/解链酶(DNA helicase)10.复制的几种方式(Modes of DNA replication):a)线性DNA双链的复制b)环状DNA双链的复制:1)Theta replication2)Rolling Circle Replication3) D loop11.The steps of BERa)Damaged baseb)DNA glycosylase recognize the damaged basec)Remove the base by hydrolysis of the N-glycosidic bondd)The AP endonuclease removes the AP site and neighboringnucleotidese)The gap is filled by DNA polymerase I and DNA ligase12.转座子(transposon, Tn):是存在于染色体DNA上可自主复制和位移的基因单位13.转坐(transposition):DNA转座是由可移位因子介导的遗传物质重排现象, 又称移位。

分子生物学习题及问题详解

分子生物学习题及问题详解

核酸一、填空题1. 碱基互变异构是指碱基的酮式与烯醇式或氨基式与亚氨基式异构体发生互变的现象,如果这种现象在DNA 复制的时候发生,如此可以导致碱基错配。

然而,在生理pH 下,碱基主要以酮式或氨基式形式存在。

2.DNA 双螺旋中只存在 2 种不同碱基对。

A 总是与_T___配对,G 总是与 C 配对,此规如此称为Chargaff 法如此。

但在RNA 双螺旋中,还含有第三种碱基配对,它是GU 。

3.X 线衍射证明,核苷中碱基与核糖环平面相互垂直。

4.一个双链RNA 分子与一个双链DNA 分子的差异有分别含U 与T 、核糖与脱氧核糖和A 型与 B 型双螺旋。

5.核酸在260 nm 附近有强吸收,这是由于碱基环上的共轭双键。

6.给动物食用3H 标记的胸苷,可使DNA 带有放射性,而RNA 不带放射性。

如果要让RNA 带有放射性,应该给动物食用3H 标记的尿苷。

7.Tm 是指DNA 热变性时候的熔链温度,双链DNA 中假设GC 含量多,如此其Tm 值高。

DNA 样品的均一性愈高,其熔解过程的温度X围愈窄。

DNA 所处溶液的离子强度越低,其熔解过程的温度X围越宽,熔解温度越低,所以DNA 应保存在较高浓度的盐溶液中。

8.双链DNA 热变性曲线通常呈S 形,这种曲线说明DNA 的变性具有协同效应;在DNA 发生热变性后,在pH 2以下,或pH 12 以上时,其A260 增加,同样条件下,单链DNA 的A260 不变。

9.核酸分子中的糖苷键均为β-N-糖苷键型,但假尿苷中的糖苷键为β-C-糖苷键。

核苷酸之间通过3,5-磷酸二酯键连接形成多聚物。

10. 细胞内总含有T 的RNA 是tRNA ,它是通过U 的后加工形成的。

11. DNA 双螺旋的构型可以有三种,它们分别为 B 型,A 型,Z 型。

B-DNA 的螺距为3.4 nm,每圈螺旋的碱基对数为10 ,细胞里的B-DNA 每圈螺旋的实际碱基对数为10.4 。

分子生物学习题答案

分子生物学习题答案

分子生物学习题答案第一章绪论Chapter 1 Introduction一名词解释1.人类基因组计划:与曼哈顿原子弹计划和阿波罗登月计划相媲美的美国人类基因组计划(human genome project, HGP),解读人基因组上的所有基因、24个染色体DNA分子中的碱基序列。

在―人类基因组计划‖中,分为两个阶段:DNA序列图以前的计划和DNA序列图计划。

序列图前计划包括遗传图、物理图、转录图。

2. RFLP (restrict fragment length polymorphism ):A variation from one individual to the next in the number of cutting sites for a given restriction endonuclease in a given genetic locus.3. DNA指纹:基因组中存在着多种重复序列,拷贝数从几个到数十万个,可分为串联重复序列和分散重复序列。

根据个体重复序列拷贝的位置和数目的差异,使用限制性内切酶,获得具有个体特异性的DNA片段。

可以作为亲缘关系或个人身份的鉴定。

4. SNP(single nucleotide polymorphism, 单核苷酸多态性):在一个群体中,基因组内某一特定核苷酸位置上出现2种或2种以上不同核苷酸的现象,在群体中相应频率为1-2%。

如果低于这个频率,可视为点突变。

二简答1. What is molecular biology?Molecular biology is the subject of gene structure and function at the molecular level.To explain the principle of development, metabolism, heredity and variation, aging at the molecular level. It grew out of the disciplines of genetics and biochemistry.2. Major events in the genetics century第二章核酸、蛋白质结构一选择题:B, E, D, A, A二名词解释1.Transfection:describes the introduction of foreign material into eukaryotic cells using a virus vector or other means of transfer. The term transfection for non-viral methods is most often used in reference to mammalian cells, while the term transformation is preferred to describe non-viral DNA transfer in bacteria and non-animal eukaryotic cells such as fungi, algae and plants.2.Configuration:The configuration of a molecule is the permanent geometry that results from the spatial arrangement of its bonds. The ability of the same set of atoms to form two or more molecules with different configurations is stereoisomerism.Configuration is distinct from chemical conformation, a shape attainable by bond rotations.3.构象:(Conformation, generally means structural arrangement),指一个分子中不改变共价键结构,仅是单键周围的原子旋转所产生的原子空间排列。

分子生物学习题集精解

分子生物学习题集精解

分子生物学习题集精解————————————————————————————————作者: ————————————————————————————————日期:第二章 DNA与染色体一、填空题1.病毒ΦX174及M13的遗传物质都是单链DNA。

2.AIDS病毒的遗传物质是单链RNA 。

3.X射线分析证明一个完整的DNA螺旋延伸长度为3.4nm 。

4.氢键负责维持A-T间(或G-C间)的亲和力。

5.天然存在的DNA分子形式为右手B型螺旋。

二、选择题(单选或多选)1.证明DNA 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 )。

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E.真核心生物、原核生物、病毒的DNA 能相互混合并彼此替代2.1953年Watson 和 Crick 提出(A )。

A.多核苷酸 DNA 链通过氢键连接成一个双螺旋B.DNA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 DNA 而非 RNAE.分离到回复突变体证明这一突变并非是一个缺失突变3.DNA 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吸收特性。

以下哪些是对 DNA 的解链温度的正确描述?(C、D )A.哺乳动物 DNA 约为45℃,因此发烧时体温高于42℃是十分危险的B.依赖于 A-T 含量,因为A-T 含量越高则双链分开所需要的能量越少C.是双链 DNA 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 的特征吸收峰的改变来确定E.就是单链发生断裂(磷酸二酯键断裂)时的温度4.DNA的变性( A、C、E )。

分子生物学 翻译习题

分子生物学 翻译习题

分子生物学翻译习题work Information Technology Company.2020YEAR翻译习题一选择题1 多数氨基酸都有两个以上密码子,下列哪组氨基酸只有一个密码子()A 苏氨酸、甘氨酸B 脯氨酸、精氨酸C 丝氨酸、亮氨酸D 色氨酸、甲硫氨酸E 天冬氨酸和天冬酰胺2 tRNA分子上结合氨基酸的序列是()A CAA-3′B CCA-3′C AAC-3′D ACA-3′E AAC-3′3 关于遗传密码的叙述不正确的是()A 20种氨基酸共有64个密码子B 碱基缺失、插入可致框移突变C AUG是起始密码D UUU是终止密码E一个氨基酸可有多达6个密码子4 tRNA能够成为氨基酸的转运体、是因为其分子上有()A -CCA-OH 3′末端B 3个核苷酸为一组的结构C 稀有碱基D 反密码环 E假腺嘌吟环5 蛋白质生物合成中的终止密码是( )。A UAAB UAUC UACD UAGE UGA6 Shine-Dalgarno顺序(SD-顺序)是指()A 在mRNA分子的起始码上游8-13个核苷酸处的顺序B 在DNA分子上转录起始点前8-13个核苷酸处的顺序C 16srRNA3'端富含嘧啶的互补顺序D 启动基因的顺序特征7 “同工tRNA”是( )A 识别同义mRNA密码子(具有第三碱基简并性)的多个tRNAB 识别相同密码子的多个tRNAC 代表相同氨基酸的多个tRNAD 由相同的氨酰tRNA合成酶识别的多个tRNA8 反密码子中哪个碱基对参与了密码子的简并性(摇摆)( )A 第—个 B第二个 C第二个 D 第一个与第二个9 与mRNA的GCU密码子对应的tRNA的反密码子是( )A CGAB IGC C CIGD CGI10 真核与原核细胞蛋白质合成的相同点是( )A 翻译与转录偶联进行B 模板都是多顺反子C 都需要GTPD 甲酰蛋氨酸是第一个氨基酸11 下列选项中翻译延长所必需的是( b );氨基酸与tRNA连接需要( d );遗传密码的摆动性是指(a)A mRNA上的密码子与tRNA上的反密码子不一定严格配对B 转肽酶C 氨酰-tRNA合成酶D 磷酸化酶E N-C糖甘键12 蛋白质生物合成时( )A mRNA与核糖体的大亚基结合B mRNA与核糖体的小亚基结合C tRNA与核糖体的大亚基结合D tRNA与核糖体的小亚基结合13 外源基因在大肠杆菌中高效表达受很多因素影响,其中SD序列的作用是( )A 提供一个mRNA转录终止子B 提供一个mRNA转录起始子C 提供一个核糖体结合位点D 提供了翻译的终点14 真核生物的翻译起始复合物在何处形成( )A 起始密码子AUG处B 5'端的帽子结构C TATA框D CAAT框15 氨酰-tRNA合成酶( )A 活化氨基酸的氨基B 利用GTP作为活化氨基酸的能量来源C 催化在tRNA的5'磷酸与相应氨基酸间形成酯键D 每一种酶特异的作用于一种氨基酸及相应的tRNA16 在研究蛋白质合成中,可利用嘌呤霉素,这是因为它( )A 使大小亚基解聚B 使肽链提前释放C 抑制氨酰-tRNA合成酶的活性D 防止多核糖体形成二填空题1 核糖体上可以区分出五个功能活性位点,其中A位主要在( )上,而P位点主要在( )。2 tRNA的二级结构为( 三叶草 )形,三级结构为(L形 )。3 tRNA的3'末端为( ),5'末端为( )。4 蛋白质合成时,起始密码子通常是( ),起始tRNA上的反密码子是( )5 tRNA反密码子的第一位碱基可出现I,它可与( )之间形成氢键而结合,这是最常见的摆动现象。6 参与蛋白质折叠的两个重要酶为(热休克蛋白)和 ( 伴侣素)。7 细胞内存在一种称为泛素的蛋白质,它的主要作用是(蛋白质的降解 )。

分子生物学(双语)2012A

分子生物学(双语)2012A

分子生物学(双语)2012A安徽大学2012 —20 13 学年第 1 学期《分子生物学(双语)》考试试卷(A 卷)(闭卷时间120分钟)考场登记表序号一、名词解释题(每小题3分,共36分)1. semiconservative replication2. Base Excision Repair3. rho dependent terminator4. polycistronic mRNA5. DNase Ⅰhypersensitivity6. C value paradox7. IS element8. primase9. elongation factor10. topoisomerase11. RNA interference12. enhancer题号一二三四五六七总分得分阅卷人得分院/系年级专业姓名学号答题勿超装订线 ------------------------------装---------------------------------------------订----------------------------------------线----------------------------------------二、填空题(每空1.5分,共21分)1. During DNA replication, unwinds double-stranded DNA,and removes RNA primers in the Okazaki fragments with its 5' to 3' exonuclease activity.2. A typical prokaryotic promoter has two consensus sequences, which are termed ________ and _______. The ideal spacing between the two consensus sequences is________ .3. Transposition occurs via two types of mechanisms. One is called ; the other is called .4. In negative control ,the regulator protein acts as a to block transcription of the operon.5. In SOS repair system, the two key regulatory proteins are and Lex A. Lex A is a (positive/ negative) regulator .6. , which locates upstream of the start codon of the mRNA, binds 3' end of the 16S rRNA in the ribosome to initiate protein synthesis7. In eukaryotes the tRNA genes are transcribed by RNA polymerase .8. When a double-strand DNA is denatured, the absorbance at A260 will (decrease/increase). 9. The generation of antibody diversity is mainly due to .三、单项选择题(每小题1.5分,共21分.请填入下表相应题号栏中)1 2 3 4 5 6 7 8 9 10 11 12 13 141. DNA polymerase can add nucleotides toA. a free 3′end of a grow ing DNA chainB. a free 5′end of a growing DNA chainC. either a 3′or 5′end of a DN A chainD. a 5′end of RNA primerE. all of the above2. What is the first protein to bind to a eukaryotic promoter, and at which site does it bind? A. TFIID binds to a TATA box. B.RNA polymerase II binds to a TATA box. C. A regulator protein binds to a regulator region. D. An activator protein binds to an enhancer.3. Splicing is the process that does which of the following?A. Add multiple adenosine bases to the end of a primary RNA transcriptB. Remove exons and join intronsC. Remove introns and join exonsD. Remove mutated regions of primary transcript RNA4. Which subunit of the E. coli RNA polymerase holoenzyme catalyzes the synthesis of RNA ?A. α-subunitB. β-subunit得分得分得分C. rho factorD. sigma factor5. The DNA in transcriptionally active chromatin tends to beA. highly methylatedB. fully-methylatedC. Under-methylated or non--methylatedD. Semi-methylated6. The most efficient control of eukaryotic gene expression is achieved at the level ofA. Replication.B. translation initiationC. post-transcriptionD. post-translation.E. .transcription initiation.7. Translation is terminated when a stop codon is presented at the ________.A. E siteB. A siteC. P siteD. either A or BE. either A or C8. In ara operon, C protein can serve asA. an activatorB. a repressorC. either a repressor or an activator depending on presence/absence of arabinoseD. both a repressor and an activator irrespective of presence/absence of arabinose9. In a standard recombinant DNA procedure, the step which follows the construction of recombinantDNA molecule should beA re-ligation of the vectorsB. ligating the vector with the gene of interestC. digesting the vector with a restriction endonucleaseD. transformation of the molecule into host cells10. Which of the following is NOT a step in a PCR reaction:A. AnnealingB. ExtensionC. DenaturationD. dehydration11. One important difference between DNA replication in prokaryotes and eukaryotes is thatA. prokaryotes do not use enzymes in the replication processB. there is only one replication origin in prokaryotesC. there are no Okazaki fragments in prokaryotesD. replication is conservative, not semiconservative, in prokaryotes12. A frameshift mutation could result fromA. a base insertion only.B. a base deletion only.C. a base substitution only.D. either an insertion or a deletion of a base.E. deletion of three consecutive bases.13. Which of the enzymes below is involved in repair of both thymine dimer and deaminate cytosine in E.coliA. DNA polymerase IB. PhotolyaseC. Uracil N-glycosylaseD. AP endonuclease14.An IRE (iron response element) regulates the expression of iron metabolism related genes byA. by binding to IRE-BPB. by forming a hair-pin structureC. at post-transcriptional or translational levelD. all of the above四、简答题(第1小题12分,第2小题10分,共22分)1. Describe how tryptophan concentration regulates the expression of the structural genes in the trp operon.2. How are eukaryotic RNA precursors processed ?。

分子生物学大实验习题及解答(精)

分子生物学大实验习题及解答(精)

分子生物学大实验习题及解答一、名词解释1、genome;2、基因芯片;3、持家基因;4、Operon;5、感受态细胞;6、逆转录酶;7、PCR技术;8、转化;9、重组DNA技术;10、基因沉默;11、hnRNA;12、复制子;13、反义RNA;14、衰减子;15、拟基因;16、RNA编辑;17、颠换;18、拓扑异构酶;19、变性;20、转座子二、基础理论单项选择题1、DNA连接酶的作用为()A、合成RNA引物B、将双螺旋解链C、去除引物、填补空隙D、使双螺旋DNA链缺口的两个末端连接2、DNA复制中RNA引物的主要作用是( )。

A、引导合成冈奇片段B、作为合成冈奇片段的模板C、为DNA合成原料dNTP 提供附着点D、激活DNA聚合酶3、下列哪一种蛋白不是组蛋白的成分()A、 H1B、H2A 、H2BC、 H3、H4D、 H54、DNA的变性:()A、包括双螺旋的解旋B、可以由低温产生C、是可逆的D、是磷酸二酯键的断裂5、转录需要的原料是:()A、 dNTPB、 dNDPC、 dNMPD、 NTP6、在原核生物复制子中以下哪种酶除去RNA引发体并加入脱氧核糖核甘酸:A、DNA聚合酶ⅢB、 DNA聚合酶ⅡC、DNA聚合酶ⅠD、外切核酸酶MFl7、下真核生物复制起始点的特征包括()A、富含GC区B、富含AT区C、 Z DNAD、无明显特征8、SDS凝胶电泳测定蛋白质的相对分子量是根据各种蛋白质()A、在一定pH值条件下所带的净电荷的不同B、分子大小不同C、分子极性不同D、溶解度不同9、DNA复制时不需要以下哪种酶?()A、 DNA指导的DNA聚合酶B、RNA指导的DNA聚合酶C、拓扑异构酶D、连接酶10、1、证明DNA是遗传物质的两个关键性实验是:肺炎链球菌在老鼠体内的毒性和T2噬菌体感染大肠杆菌。

这两个实验中主要的论点证据是:()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11、利用自己的位点专一重组酶把自己从寄主基因组中的一个地方移到另一个地方的遗传元件叫()A、启动子B、转座子C、T-DNAD、顺反子12、原核DNA合成酶中()的主要功能是合成前导链和冈崎片段A、DNA聚合酶ⅠB、DNA聚合酶ⅡC、DNA聚合酶ⅢD、引物酶13、在基因工程中,需使用特定的限制酶切割目的基因和质粒以便于重组和筛选。

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Section C - properties of nucleic acids1.The sequence 5'-AGTCTGACT-3' in DNA is equivalent to which sequence in RNA?A 5'-AGUCUGUGACU -3'B 5' -UGTCTGUTC -3'C 5' -UCAGUCUGA-3'D 5'- AGUCAGACU-3'2. Which of the following correctly describes A-DNA?A a right-handed antiparallel doublehelix with 10 bp/turn and bases lyingperpendicular to the helixaxis.B a left-handed antiparalleldouble-helix with 12 bp/turn formedfrom alternating pyrimidine-purinesequences.C a right-handed antiparallel doublehelix with 11 bp/turn and bases tiltedwith respect to the helix axis.D a globular structure formed by shortintramolecular helices formed in asingle-strand nucleic acid.3. Denaturation of double stranded DNA involves.A breakage into short double-stranded fragments.B separation into single strands.C hydrolysis of the DNA backbone.D cleavage of the bases from the sugar-phosphate backbone.4. Which has the highest absorption per unit mass at a wavelength of 260 nm?A double-stranded DNA.B mononucleotides.C RNA.D protein.5. Type I DNA topoisomeraes ...A change linking number by士2B require ATP.C break one strand of a DNA double helix.D are the target of antibacterial drugs. Section D - Prokaryotic and eukaryotic chromatin structure1.Which of the following is common to both E. coli and eukaryotic chromosomes?A the DNA is circular.B the DNA is packaged into nucleosomes.C the DNA is contained in the nucleus.D the DNA is negatively supercoiled.2.A complex of 166 bp of DNA with the histone octamer plus histone HI is known as a . . .A nucleosome core.B solenoid.C 30 nm fiber.D chromatosome.3.In what region of the interphase chromosome does transcription take place?A the telomere.B the centromere.C euchromatin.D heterochromatin.4.Which statement about CpG islands and methylation is not true?A CpG islands are particularly resistant to DNase I.B CpG methylation is responsible for the mutation of CpG to TpG in eukaryotes.C CpG islands occur around the promoters of active genes.D CpG methylation is associated with inactive chromatin.5.Which of the following is an example of highly-repetitive DNA?A Alu element.B histone gene cluster.C DNA minisatellites.D dispersed repetitive DNA.Section E - DNA replication1.The number of replicons in a typical mammalian cell is . . .A 40-200.B 400.C 1000-2000.D 50000-100000.2. In prokaryotes,the lagging strand primers are removed by . . .A 3' to 5' exonuclease.B DNA ligase.C DNA polymerase I.D DNA polymerase III.3. The essential initiator protein at theE. coli origin of replication is . . .A DnaA.B DnaB.C DnaC.D DnaE.4. Prokaryotic plasmids can replicate in yeast cells if they contain a cloned yeast. . .A ORC.B CDK.C ARS.D RNA.Section F - DNA damage, repair and recombination(此章不考)1. Per nucleotide incorporated, the spontaneous mutation frequency in E. coli is . . .A 1 in 106.B 1 in 108.C 1 in 109.D 1 in 1010.2. The action of hydroxyl radicals on DNA generates a significant amount of . . .A pyrimidine dimmers.B 8-oxoguanine.C O6- methylguanine.D 7-hydroxymethylguanine.3. In methyl-directed mismatchrepair in E. coli, the daughterstrand containing the mismatchedbase is nicked by . . .A M utH endonuclease.B U vrABC endonuclease.C A P endonuclease.D3' to 5' exonuclease.4. The excision repair of UV-inducedDNA damage is defective inindividuals suffering from ...A hereditary nonpolyposis colon cancer.B Crohn's disease.C classical xeroderma pigmentosum.D xeroderma pigmentosum variant. Section K - Transcription in prokaryotes1. Which two of the following statements about transcription are correct?A RNA synthesis occurs in the 3' to 5' direction.B the RNA polymerase enzyme moves along the sense strand of the DNA in a 5' to 3' direction.C the RNA polymerase enzyme movesalong the template strand of the DNA in a 5' to 3' direction.D the transcribed RNA is complementary to the template strand.E the RNA polymerase adds ribonucleotides to the 5' end of the growing RNA chain.F the RNA polymerase adds deoxyribonucleotides to the 3' end of the growing RNA chain.2. Which one of the followingstatements about E. coli RNA polymerase is false?A the holoenzyme includes the sigma factor.B the core enzyme includes the sigma factor.C it requires Mg2+ for its activity.D it requires Zn2+ for its activity.3. Which one of the following statements is incorrect?A there are two αsubunits in the E. coli RNA polymerase.B there is one β subunit in the E. coli RNA polymerase.C E. coli has one sigma factor.D the β subunit of E. coli RNA polymerase is inhibited by rifampicin.E the streptolydigins inhibit transcription elongation.F heparin is a polyanion, which binds to the β’ subunit.4. Which one of the following statements about transcription in E. coli is true?A the -10 sequence is always exactly10 bp upstream from the transcription start site.B the initiating nucleotide is always a G.C the intervening sequence between the -35 and -10 sequences is conserved.D the sequence of the DNA after thesite of transcription initiation is notimportant for transcriptionefficiency.E the distance between the -35 and -10 sequences is critical for transcription efficiency.5. Which one of the following statements about transcription in E. coli is true?A loose binding of the RNA polymerase core enzyme to DNA is non-specific and unstable.B sigma factor dramatically increasesthe relative affinity of the enzymefor correct promoter sites.C almost all RNA start sites consist of a purine residue, with A being more common than G.D all promoters are inhibited by negative supercoiling.E terminators are often A-U hairpin structures.Section L - Regulation of transcription in prokaryotes1. Which two of the following statements are correct?A the double stranded DNA sequencethat has the upper strand sequence5'-GGATCGATCC-3' is apalindrome.B the double stranded DNA sequencethat has the upper strand sequence5'-GGATCCTAGG-3' isapalindrome.C the Lac repressor inhibits binding of the polymerase to the lac promoter.D the lac operon is directly induced by lactose.E binding of Lac repressor to allolactose reduces its affinity for the lac operator.F IPTG is a natural inducer of the lac promoter.2. Which one of the following statements about catabolite-regulated operons is false?A cAMP receptor protein (CRP) andcatabolite activator protein (CAP)are different names for the sameprotein.B when glucose is present in the cell cAMP levels fall.C CRP binds to cAMP and as a result activates transcription.D CRP binds to DNA in the absence of cAMP.E CRP can bend DNA, resulting in activation of transcription.3. Which one of the following statements about the trp operon is true?A the RNA product of the trp operon is very stable.B the Trp repressor is a product of the trp operon.C the Trp repressor,like the Lac repressor, is a tetramer of identical subunits.D the Trp repressor binds totryptophan.E tryptophan activates expression from the trp operon.F the trp operon is only regulated by the Trp represso4. Which two of the following statements about attenuation at the trp operon are true?A attenuation is rho-dependent.B deletion of the attenuator sequenceresults in an increase in both basaland activated levels of tran- scriptionfrom th~ trp promoter.C the attenuator lies upstream of the trp operator sequence.D attenuation does not require tight coupling between transcription and translation.E pausing of a ribosome at twotryptophan codons in the leaderpeptide when tryptophan is in shortsupply causes attenuation.F a hairpin structure called thepnti-terminator stops formation ofthe terminator hairpin, resulting intranscriptional read-through into thetrpE gene, when tryptophan isscarce.Section M - Transcription in eukaryotes1. Which one of the followingstatements about eukaryotic RNApolymerases I, II and III is false?A RNA Pol II is very sensitive to α-amanitin.B RNA Pol II is located in th~ nucleoplasm.C RNA Pol III transcribes th~ genes for tRNA.D eukaryotic cells contain other RNApolymerases in addition to RNA PolI, RNA Pol II and RNA Pol III.E each RNA polymerase containssubunits with homology to subunitsof the E. coli RNA polymerase aswell as additional subunits,whichare unique to each polymerase.F the carboxyl end of RNA Pol IIcontains a short sequence of onlyseven amino acids which is calledthe carboxyl-terminal domain (CTD)and which may be phosphorylated.2. Which two of the following statements about RNA Pol I genes are true?A RNA Pol I transcribes the genes for ribosomal RNAs.B human cells contain 40 clusters of five copies of the rRNA gene.C the 185, 5.85 and 285 rRNAs aresynthesized as separate transcripts.D RNA Pol I transcription occurs in the nucleoplasm.E RNA Pol I transcription occurs in the cytoplasm.F rRNA gene clusters are known as nucleolar organizer regions.3. Which one of the following statements about RNA Pol I transcription is false?A in RNA Pol I promoters the coreelement is 1000 bases downstreamfrom the upstream control element(UCE).B upstream binding factor (UBF)binds to both the UCE and theupstream part of the core element ofthe RNA Pol I promoter.C selectivity factor SLl stabilizes the UBF-DNA complex.D SL1 contains several subunitsincluding the TATA-binding protein TBP.E in Acanthamoeba there is a single control element in rRNA gene promoters.4. Which two of the following statements about RNA Pol III genes are true?A the transcriptional control regions of tRNA genes lie upstream of the start of transcription.B highly conserved sequences in tRNA gene coding regions are also promoter sequences.C TFIIIC contains TBP as one of its subunits.D TFIIIB is a sequence specific transcription factor on its own.E in humans 5S rRNA genes are arranged in a single cluster of 2000 copies.Section 0 - RNA processing and RNPs 1. Which ribonucleases are involved in producing mature tRNA in E. coli?A RNases A, D, E and F.B RNases D, E, F and H.C RNases D, E, F and P.D RNases A, D, H and P.2. Most eukaryotic pre-mRNAs arematured by which of the followingmodifications to their ends?A capping at the 3’-end cleavage and polyadenylation at the 5'-end.B addition of a GMP to the 5'-end,cleavage and polyadenylation to create the 3'-end.C addition of a guanine residue to the5'-end cleavage and polyadenylationto create the 3'-end.D addition of a GMP to the 5'-end,polyadenylation,then cleavage to create the 3'-end.3. Which one of the followingstatements correctly describes thesplicing process undergone bymost eukaryotic pre-mRNAs?A in a two-step reaction, thespliceosome removes the exon as alariat and joins the two intronstogether.B splicing requires conservedsequences which are the 5ιsplicesite,the 3' -splice site thebranch-point and the polypurinetract.C the U1 snRNP initially binds to the5'-splice site,U2 to the branchpointsequence and then the tri-snRNP, U4,US and U6 can bind.D in the first step of splicing the G atthe 3'-end of the intron is joined tothe 2’-hydroxyl group of the Aresidue of the branchpoint sequenceto create a lariat.Section P - The genetic code and tRNA 1. Which of the following list of features correctly apply to the genetic code?A triplet degenerate nearly universal, comma-less, nonoverlapping.B triplet universal, comma-less, degenerate, nonoverlapping.C overlapping, triplet, comma-less, degenerate nearly universal.D overlapping, comma-less nondegenerate nearly universal triplet. 2. Which of the following statementsabout tRNAs is false?A most tRNAs are about 76 residues long and have CCA as residues 74, 75 and 76.B many tRNAs contain the modifiednucleosides pseudouridinedihydrouridine ribothymidine andmosme.C tRNAs have a common L-shapedtertiary structure with threenucleotides at one end able to basepair with an anticodon on amessenger RNA molecule.D tRNAs have a common cloverleafsecondary structure containing threesingle stranded loops called the D-,T- and anticodon loops.3.Which three statements are true? The aminoacyl tRNA synthetase reaction...A joins AMP to the 3’-end of the tRNA.B is a two step reaction.C joins any amino acid to the 2'- or 3' -hydroxyl of the ribose of residue A76.D is highly specific because thesynthetases use identity elements inthe tRNAs to distinguish betweenthem.E joins AMP to the amino acid to produce an intermediate.F releases PPi in the second step. Section Q - Protein synthesis1. Which statement about the codon-anticodon interaction is false?A it is antiparallel and can include nonstandard base pairs.B inosine in the 5' -anticodon position can pair with A,C or U in the 3'-codonpositionC inosine in the 3’-anticodon position can pair with A, C or U in the 5’-codon position.D A is never found in the 5'-anticodon position as it is modified by anticodon deaminase.2.Which one of the following statements correctly describes initiation of protein synthesis in E.coli?A the initiator tRNA binds to the Shine-Dalgarno sequence.B three initiation factors are involved and IF2 binds to GTP.C the intermediate containing IF1, IF2,IF3, initiator tRNA and mRNA is called the 30S initiation complex.D binding of the 50S subunit releases IF1, IF2, GMP and PPi.E the initiation process is completewhen the 70S initiation complex is formed which contains the initiator tRNA in the A site of the ribosome and an empty P site.3.Which statement about elongation of protein synthesis in prokaryotes is false?A elongation can be divided into threesteps: peptidyl-tRNA deliverypeptide bond formation andtranslocation.B the peptidyl transferase center of thelarge ribosomal subunit isresponsible for peptide bond for-mation.C in the EF-Tu-Ts exchange cycleEF-Tu-GTP is regenerated by EF-Tsdisplacing GDP.D EF-G is also known as translocaseand uses GTP in its reaction.4.Which two of the following statements about initiation ofeukaryotic protein synthesis aretrue?A eukaryotes use a mRNA scanning method to locate the correct start codon.B there are at least nine eukaryotic initiation factors (eIFs).C eukaryotic initiation uses N-formylmethionine.D the 80S initiation complexcompletes the initiation process andcontains the initiator tRNA base-paired to the start codon in the Asite.E ATP is hydrolysed to AMP and PPi during the scanning process.F the initiator tRNA binds after the mRNA has bound to the small subunit.。

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