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福建农林大学期末考试试卷2008 2009学年第 学期课程名称: 分子生物学 考试时间 120 分钟 专业 年级 班 学号 姓名 题号一二三四总得分得分评卷人签字复核人签字得分一、填空题(每空0.5分,共10分)1、The enzymes needed in the bacterial DNA replication are topoisomerase, helicase, _, DNA polymerase and _.(引物酶;DNA连接酶)2、There are five families of histones:_, _, _ and _, known as the core histones, and H1.(H2A;H2B;H3;H4)3、The holoenzyme of E.coli RNA polymerase is_,the core enzyme is made up of _,while the_ factor has the ability to recognize specific binding sites.(2;2;)4、There are two kinds of mutagens, one is physical mutagens, such as_,the other is chemical mutagens, such as_.(各种射线;碱基类似物/烷化剂/亚硝酸/EB)5、The two-dimensional structure of tRNA is . And the three-dimensional structure of tRNA is . (三叶草型、L型)6、 was suggseted by Crick to explain the redundancy of the genetic code.(摆动假说)7、The start codon is . There are three stop codons: , , .(ATG或AUG、TAA或UAA、TAG或UAG、TGA或UGA)8、 site is where incoming aminoacyl-tRNA molecules bind, and site is where the growing polypeptide chain is usually found.(A、P)得分二、名词解释(每小题2分,共30分)1、semi-conservative replication:半保留复制,亲代双链DNA以每条链为模板,按碱基配对原则各合成一条互补链,这样一条亲代DNA双螺旋,形成两条完全相同的子代DNA螺旋,子代DNA分子中都有一条合成的“新”链和一条来自亲代的旧链。2、Tm:核酸热变性温度范围的中点。3、asymmetry of transcription:转录的不对称性,指双螺旋的DNA进行转录时只有一条链能被转录,另一条链可能起调节作用。4、transition:在基因突变中,一个嘌呤被另一个嘌呤所代替或一个嘧啶被一个另嘧啶所代替的现象。5、exon:外显子,指真核生物基因中,含有蛋白质编码信息区段、在成熟mRNA的加工过程中被保留下来的部分。6、transposon:转座子,基因组上一些可由一个位置移动到另一位置的核酸序列。7、operon:原核基因表达调控的单位,包括启动子、操纵基因和一系列功能相关的结构基因。8、Synonymous codon:编码同一氨基酸的密码子。9、Ribosome:核糖体,蛋白质合成的场所。10、Shine-Dalgarno sequence :SD序列,由Shine和Dalgarno发现原核生物起始密码子上游813个nt的保守多嘌呤序列,与核糖体16S小亚基的特定序列碱基配对,相对于起始密码子正确定位核糖体,保证翻译的准确。11、housekeeping gene:生物体不同发育时期所有类型细胞中均保持表达状态的基因。12、Ribozyme:核酶,能够催化特定生化反应的RNA分子。13、Chaperone:分子伴侣,帮助蛋白质分子形成正确空间构象的辅助分子。14、DNA clone:DNA克隆,将片段相对较短的目的核酸片段与能进行独立复制的载体连接形成重组DNA,重组DNA能够在寄主细胞内增殖,从而获得大量的目的核酸片段。15、phage:侵染细菌的病毒。得分三、单项选择题(每小题1分,共15分)( D )1、DNA replication in eukaryotes:A. usually starts at the same place every timeB. only involve 2 different DNA polymerasesC. is faster than in bacterial DNA replicationD. all of the above are false( D )2、Which of the following are true statements regarding telomeres:A. they stabilize linear chromosomesB. they consist of a short repeating sequenceC. they are located at the chromosome endsD. all of the above are true( C )3、Mismatches produced from replication errors are first recognized by _ for the purposes of repair.A. MutH B. MutM and MutLC. MutL and MutS D. Exonuclease I( B )4、Transcription in prokaryotes and the eukaryotic nucleus:A. is catalyzed by the same enzymeB. involves Watson-Crick type pairing of nucleotidesC. requires a primerD. proceeds from 3 to 5( C )5、Which of the following is an example of highly-repetive DNA?A. Alu element B. Histone gene clusterC. Minisatellite DNA D. KpnI family( C )6、Which of the following is not involved in homologous recombination in E. coli that we studied in class:A. RecA B. RecBCD C. RecQ D. RuvA( B )7、Regulation of the Lac operon is belong to:A. repression of negative control B. induction of negative controlC. repression of positive control D. induction of positive control( D )8、which of the following descriptions of subunits of DNA polymerase III holoenzyme is true?A. the subunits clamp the polymerase to the DNAB. the subunits link the two catalytic cores togetherC. the subunits contribute the DNA polymerase activityD. all of the above are ture( B )9、Nuclear mRNA splicing is performedA. in nucleosomes B. in spliceosomesC. in the nucleolus C. in ribosome( C )10、Cap functions not include A. enhancing splicing of the first intron B. enhancing translation of the mRNA C. enhancing degradation of the mRNA in the cytoplasm D. enhancing the stablitity of the mRNA( D )11、Which of the following is true of the genetic codeA. it is read 3 nucleotides at a timeB. it is not overlappingD. it is degenerate (more than one triplet codes for most amino acids)D. a and b are true( A )12、The region(s) of tRNA recognized by amino-acyl tRNA synthetases (so that they can charge the right tRNA) isA. the CCA B. the D loop C. the anticodon C. the T or pseudouridine loop( C )13、In nuclear mRNA splicing, the intron is released as a A. circular RNA B. linear RNA C. lariat RNA D. a “Y” RNA( B )14、The first and main polyadenylation signal is foundA. 20-30 bp from the ORFB. 20-30 bp from the polyadenylation siteC. 20-30 bp from the promoter siteD. 20-30 bp from the end of the RNA( A )15、During initiation step of the translation in prokaryotes, the components involved are A. ribosome, mRNA, initiator tRNA charged, initiator factors, GTP B. ribosome, mRNA, initiator tRNA charged, initiator factors, ATP C. ribosome, mRNA, initiator tRNA charged, initiator factors, CTP D. ribosome, mRNA, initiator tRNA charged, initiator factors, TTP得分四、问答题(45分)1、Please illuminate the main content of central dogma and draw the sketch map.(10 points)DNA是生物遗传的主要物质基础。遗传信息以特定的脱氧核苷酸序列储存在DNA分子中(1分)。DNA能准确地自我复制(1分),并把遗传信息转录到mRNA分子上(1分),然后通过mRNA翻译成为具有特定氨基酸顺序的蛋白质(1分),由蛋白质行使各种生物学功能。Crick把这种遗传信息的流动称为分子遗传的中心法则。后来人们又发现反转录病毒可以其RNA作为模板指导合成DNA(1分);某些病毒RNA可以进行自我复制,即以自身RNA为模板,合成子代RNA(1分)。这些发现是对中心法则的补充和发展。 DNA RNA 蛋白质 (4分)2、what is the promoter?Please describe the most conserved structure of E.coli promoters.(7 points)定义:RNA聚合酶特异性结合和起始转录所需的保守序列。(2分)结构:TTGACA1618bpTATAAT58bpG/A -35序列 -10序列 +1 1分 1分 1分 1分 1分3、Please briefly describe the structure features of the Watson-Crick double-helix.(7 points)(1)两条单链,右手螺旋,反向平行,大沟和小沟;(2分)(2)嘌呤和嘧啶位于螺旋的内侧,磷酸和脱氧核糖基位于螺旋外侧,彼此以3 ,5 -磷酸二酯键连接,形成DNA分子的骨架。(1.5分)(3)直径2 nm,10bp/螺旋,螺距3.4nm。(1.5分)(4)碱基互补配对存在。(2分)4、What are the characters of expression vectors? (5 points)(1)能够进行独立于基因组的自我复制 (1分)(2)容易从细胞中分离提取 (1分)(3)含有多克隆位点,允许外源片段的插入 (1分)(4)含有至少一个选择标记 (1分)(5)具有启动子、终止子等表达元件 (1分)5、What are the differences of the ribosome between eukaryotes and prokaryotes?(10 points)核糖体的差异 真核生物(80S)是60S(5S、28S、5.8S)、40S(18S)的大小亚基组成 (5分)原核生物(70S)是50S(5S、23S)、30S(16S)的大小亚基组成(5分)6、Write out 3 great events which are important in the development histroy of molecular biology and illustrate their significance.(6 points)(1)DNA双螺旋结构的提出:第一次说明了遗传信息是如何储存,并如何传递给后代的;(2)限制性内切酶的发现:成为基因工程的“手术刀”,使体外基因操作成为可能;(3)PCR技术的发明:体外快速扩增目的DNA片段(4)核酸测序技术的发明:快速准确地测定核酸序列,使对其产物的结构、功能的推测成为可能;(2)(4)技术的发明是基因工程时代的三个标志性技术。(5)人类基因组计划:人类第一次在基因水平上认识自己,(6)克隆羊多利的诞生:从实践上证实了动物细胞的全能性。福建农林大学半期考试试卷参考答案2009 2010学年第 学期课程名称: 分子生物学 考试时间 120分钟 专业 年级 班 学号 姓名 题号一二三四总得分得分评卷人签字复核人签字得分一、名词解释(每小题2分,共30分)1、Splitting gene:真核基因的编码序列被非编码序列所隔断的现象。2、Semi-discontinuous replication:DNA复制时,一条链的合成是连续的,另一条链的合成是不连续的。3、Restricition endonuclease:能识别特异碱基序列,并在特定位点切断核酸双链的核酸内切酶。4、Nucleosome:构成真核生物染色质的基本结构单位,是DNA和蛋白质构成的紧密结构形式。其核心为H2A、H2B、H3、H4各2分子组成八聚体,DNA在其表面上缠绕1.75圈,约146bp,H1结合在连接相邻核小体之间的DNA上。核小体DNA的长度平均为200bp。5、Operon:原核基因表达调控的单位,包括启动子、操纵基因和一系列功能相关的结构基因。6、Frameshift mutations:DNA分子中插入或缺失一个或少数几个核苷酸,使整个阅读框改变而引起的突变。7、Induction:培养基中添加了某些物质(常常是底物),导致某些酶的合成。8、Synonymous codon:编码同一氨基酸的密码子。9、cis-acting element:顺式作用元件,是同一DNA分子中具有转录调节功能的特异DNA序列。按功能特性,真核基因顺式作用元件包括启动子、增强子及沉默子等。10、Ribosome binding site:核糖体结合位点,由Shine和Dalgarno发现原核生物起始密码子上游813个nt的保守多嘌呤序列,与核糖体16S小亚基的特定序列碱基配对,相对于起始密码子正确定位核糖体,保证翻译的准确。11、Exon:外显子,在mRNA剪接后仍会被保存下来,并可在蛋白质生物合成过程中被表达为蛋白质。12、Leucine zipper:亮氨酸拉链,是含有4个或5个亮氨酸残基, 彼此之间精确的相距7个氨基酸残基。这样, 在螺旋的每一个侧面就出现一个Leu, 这些Leu排成一排, 两个蛋白质分子的螺旋之间靠Leu残基之间的疏水作用形成一条拉链。这类蛋白与DNA的特异性结合都是以二聚体形式起作用的, 但与DNA结合的结构域并不在拉链区。13、Vector:载体,能够携带外来DNA片段在寄主细胞中进行独立自我复制的DNA分子。14、Housekeeping gene:生物体不同发育时期所有类型细胞中均保持表达状态的基因。15、DNA clone:DNA克隆,将片段相对较短的目的核酸片段与能进行独立复制的载体连接形成重组DNA,重组DNA能够在寄主细胞内增殖,从而获得大量的目的核酸片段。得分二、填空题(每空0.5分,共10分)1、Transcription starts at the _,and stops at the _.(启动子;终止子)2、The enzymes needed in the bacterial DNA replication are topoisomerase, helicase, primase, _, DNA polymerase I and _.(DNA聚合酶III;连接酶)3、The nucleotide is composed of the _, _ and _. The linkage between the adjacent nucleotides is called _.(磷酸;戊糖;碱基;3,5-磷酸二酯键)4、DNA replication can be divided into three stages,including _, _ and _.(起始;延伸;终止)5、The start codon is . There are three stop codons: , , .(ATG或AUG;TAA或UAA;TAG或UAG;TGA或UGA)6、 site is where incoming aminoacyl-tRNA molecules bind, and site is where the growing polypeptide chain is usually found.(A;P)7、In the stage of translation termination, recognize the , make peptidyl transferase join the polypeptide chain to , thus releasing it.(释放因子;终止密码子;水分子)得分三、单项选择题(每小题1分,共15分)( A )1、Which of these DNAs will have the highest Tm :A. A+T=15% B. A+T=30%C. G+C=15% D. G+C=30%( A )2、What enzymatic activity is essential to avoid mutations during DNA replication?A. 3 5 exonuclease B. 5 3 exonucleaseC. 3 5 endonuclease D. 5 3 endonuclease( C )3、Watson & Crick won the noble prize in 1962 for their discoveries concerning:A. the interaction between tumour viruses and the genetic material of the cellB. the mobile genetic elementsC. the molecular structure of nucleic acidsD. the invention of the PCR method( D )4、Which of the following is correct regarding DNA structure:A. it is a left-handed helix B. it consists of two parallel strandsC. there are about 10 base per turn in the DNA double helixD. between the backbone strands run the major and minor grooves( D )5、The specificity of DNA binding by the E. coli RNA polymerase is provided mainly by the:A. core subunits B. subunitC. subunit D. subunit( D )6、Telomerase is a kind of:A. RNA polymerase B. DNA polymerase C. primerase D. reverse transcriptase( B )7、In the classic model of the Lac operon, the operatorA. is located downstream of the lacZB. binds the lactoseC. is located upstream of the promoterD. is able to translate the repressor( C )8、Which of the following is not one kind of ribozyme? A. RNase P B. Tetrahymena rRNA intron C. 18S rRNA D. Mitochondrial RNA intron( B )9、Which of the following is transcripted by RNA polymerase A. rRNA B. mRNA C. tRNA D. all snRNA( D )10、Addition of polyA to the 3 end of mRNA requiresA. polyA polymerase (PAP) B. RNA cleavageC. polyadenylation signal D. all of the above( B )11、Nuclear mRNA splicing is performedA. in nucleosomes B. in spliceosomesC. in the ribosome D. in the polysome( A )12、During initiation step of the translation in prokaryotes, the components involved are A. ribosome, mRNA, initiator tRNA charged, initiator factors, GTP B. ribosome, mRNA, initiator tRNA charged, initiator factors, ATP C. ribosome, mRNA, initiator tRNA charged, initiator factors, CTP D. ribosome, mRNA, initiator tRNA charged, initiator factors, TTP( A )13、The region(s) of tRNA linkage with amino acid isA. the CCA B. the D loop C. the anticodon D. the T & pseudouridine loop( B )14、The promoter for 5S RNA genes in X. laevis, A. is located upstream of the geneB. is located downstream of the gene near startpointC. is located toward the middle of the gene D. is located the 3end of the gene ( B )15、The wobble hypothesis was suggested byA. J Watson B. F CrickC. T. Cech D. P. Sharp得分四、问答题(共45分)1、Please summary the basic characters of DNA replication.(5 points)1)半保留:最基本特征,对生物遗传有重要意义;(1分)2)半不连续:前导链、滞后链、冈崎片段;(1分)3)方向:53;(1分)4)可以是单向,也可以是双向(更常见);(1分)5)需要RNA引物。(1分)2、Please illuminate the main content of central dogma.(6 points)DNA是生物遗传的主要物质基础。遗传信息以特定的脱氧核苷酸序列储存在DNA分子中(1分)。DNA能准确地自我复制(1分),并把遗传信息转录到mRNA分子上(1分),然后通过mRNA翻译成为具有特定氨基酸顺序的蛋白质(1分),由蛋白质行使各种生物学功能。Crick把这种遗传信息的流动称为分子遗传的中心法则。后来人们又发现反转录病毒可以其RNA作为模板指导合成DNA(1分);某些病毒RNA可以进行自我复制,即以自身RNA为模板,合成子代RNA(1分)。这些发现是对中心法则的补充和发展。3、Please draw the sketch map of the secondary structure of tRNA,and write out the function of each part.(7 points) (2分)氨基酸接受臂:携带活化氨基酸;(1分)TC区:结合核糖体中的rRNA;(1分)额外环:是不同tRNA的区别所在;(1分)反密码区:通过其中的反密码子与mRNA上密码子的配对,来决定Aa在肽链上的正确位置;(1分)D区:结合氨酰tRNA合成酶。(1分)4、what is the promoter?Please describe the most conserved structure of E.coli promoters.(6 points)定义:RNA聚合酶特异性结合和起始转录所需的保守序列。(1分)结构:TTGACA1618bpTATAAT58bpG/A -35序列 -10序列 +1 1分 1分 1分 1分 1分5、Please illuminate the difference between promoter and enhancer. (5 points) 不同点 启动子 增强子 (1分) 1 基本转录调控元件 调控诱导型表达基因 (1分) 2 位于基因附近 与基因间隔一段距离 (1分)3 单方向作用 双方向作用 (1分)4 基因特异 非基因特异 (1分) 6、What is the polymerase chain reaction? Please illuminate the steps of PCR. (6 points)原理:多聚酶链式反应(polymerase chain reaction , PCR)的原理类似于DNA的天然复制过程。在待扩增的DNA片段两侧和与其两侧互补的两个寡核苷酸引物,经变性、退火和延伸若干个循环后,DNA扩增2n倍。(3分) 典型的PCR反应体系由如下组分组成:DNA模板、反应缓冲液、dNTP、MgCl2、两个合成的DNA引物、耐热Taq聚合酶。步骤:预变性变性(1分)-退火(1分)-延伸(1分),循环 聚合7、What are the differences of the ribosome between eukaryotes and prokaryotes? Which is faster in the peptide bond formation?(10 points)核糖体的差异 真核生物(80S)是60S(5S、28S、5.8S)、40S(18S)的大小亚基组成 (4分)原核生物(70S)是50S(5S、23S)、30S(16S)的大小亚基组成(4分)真核生物核糖体合成肽键速度更快。(2分)福建农林大学期末考试试卷参考答案2009 2010学年第 学期课程名称: 分子生物学 考试时间 120 分钟 专业 年级 班 学号 姓名 题号一二三四总得分得分评卷人签字复核人签字得分一、填空题(每空0.5分,共10分)1、There are five types of DNA repair, include photo-reactivation repair, _, _,_ and error-prone repair.(切除修复;重组修复;错配修复)2、The holoenzyme of E.coli RNA polymerase is_,the core enzyme is made up of _,while the_ factor has the ability to recognize specific binding sites.(2;2;)3、The eukaryotes have a well-defined cell cycle.The phase are:G1,_,_ and M.(S;G2)4、Only one strand of a DNA duplex will be transcribed. So the RNA is identical in sequence with another strand of the DNA which is called _,and the template strand is _.(有义链;反义链)5、In eukaryotes, mRNA processing includes the following steps: addition of at 5end, addition of at 3end, removal of , joining together the ,And these take place in .(M7G帽子结构;poly(A)尾巴;内含子;外显子;细胞核)6、The structure is a common secondary structural reprensentation of tRNA molecules. The tertiary structure of tRNA molecules is . (三叶草;L)7、In the large 60S subunit of the eukaryote ribosome, there are rRNA, and rRNA, and rRNA. (5S;5.8S;28S) 得分二、名词解释(每小题2分,共30分)1、Promoter:RNA聚合酶结合并起始转录的一段DNA序列。2、Semi-conservative replication:半保留复制,亲代双链DNA以每条链为模板,按碱基配对原则各合成一条互补链,这样一条亲代DNA双螺旋,形成两条完全相同的子代DNA螺旋,子代DNA分子中都有一条合成的“新”链和一条来自亲代的旧链。3、Intron:真核生物DNA编码序列中间隔的非编码序列,它们也能被转录成RNA,但在mRNA成熟加工中会被切除。4、Transversion:在基因突变中,嘌呤被嘧啶所代替的现象,反之亦然。5、Gene recombination:指由于不同DNA链的断裂和连接而产生的DNA片段的交换和重新组合,形成新的DNA分子的过程。因此,新的DNA分子中含有原来的两个DNA 分子的片段。6、Hybridization:不同来源的核酸单链通过碱基互补配对原则结合成双链的过程。7、Transcription:以DNA为模板,在RNA聚合酶的催化下,合成RNA的过程。8、Proofreading activity:即DNA聚合酶的35外切酶活性,当复制过程中插入错配碱基时该活性被激活,进而将错配碱基切除。9、RNA processing:RNA加工,前提RNA转变为成熟RNA的过程,发生在细胞核内。10、Triplet code:三联体密码,3个连续的核苷酸编码一个氨基酸。11、Chaperone:分子伴侣,帮助蛋白质分子形成正确空间构象的辅助分子。12、Trans-acting element:反式作用元件,指能直接或间接地识别或结合在各类顺式作用元件核心序列上参与调控靶基因转录效率的蛋白质。13、Open reading frame:特定的核酸序列由一个起始密码子(AUG/ATG)和三个终止密码子组成,核酸序列能翻译为蛋白质序列。14、Ribozyme:核酶,具有催化作用的RNA分子。15、Zinc finger:锌指,一种常出现在DNA结合蛋白中的一种结构基元。是由一个含有大约30个氨基酸的环和一个与环上的4个Cys或2个Cys和2个His配位的Zn2+构成,形成的结构像手指状。得分三、单项选择题(每小题1分,共15分)( D )1、What is the percentage of A in a piece of RNA if the percentage of C is 30%?A. 30% B. 20%C. 60% D. Unable to calculate with the information provided.( D )2、Which of the following is NOT a function of RNA?A. It acts a template during translation.B. It is required for RNA splicing.C. It is essential for the structure of ribosomes.D. It attaches the correct amino acid to tRNA.( C )3、Histones are found in close association with DNA. Which of the following statements is incorrect?A. Histone H1 is less evolutionarily conserved than core histones.B. A nucleosome is composed of 146 bp of DNA and a histone octomerC. There are only four types of histonesD. Nucleosomes are evenly distribution within chromatin( B )4、The elongation of the lagging strand during DNA synthesis:A. In the 3-5 directionB. Produces Okazaki fragments C. Polymerizes
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