分子生物学 翻译习题
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 UAA
B UAU
C UAC
D UAG
E UGA
6 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tRNA
B 识别相同密码子的多个tRNA
C 代表相同氨基酸的多个tRNA
D 由相同的氨酰tRNA合成酶识别的多个tRNA
8 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 CGA
B IG
C C CIG
D CGI
10 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GTP
D 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tRNA
16 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 )。
8 核酸复制时，DNA聚合酶沿模板链( )方向移动；转录时，RNA聚合酶沿模板链( )方向移动；翻译时，核糖体沿模板链( )方向移动｡n
9 肽链合成的终止因子又称为( 释放因子 )，能识别并结合到( 终止位点 )。
10 蛋白质合成后通过(n端的信号肽在内质网中)被定向输送到线粒体､叶绿体､细胞核内执行其特定的功能｡。
同济考博-分子生物学真题-答案
9、原癌基因细胞癌基因，正常情况下静止或极少表达，其表达产物对于细胞生长、分化、增殖有重要作用。一旦激活可以导致细胞恶变，诱发肿瘤的基因。激活方式有：点突变、获得启动子、基因扩增、染色体易位或重排。
10、多克隆位点：载体中有多个限制性内切酶的识别位点，能够插入多个外源基因并且能否复制的位点。
4.提取IL-2高表达的组织的mRNA，RT-PCR，T载体克隆或直接酶切克隆至表达载体中、测序。
2.真核细胞中基因表达的特异性转录调控因子是指什么？根据它们的结构特征可以分为哪些类型？它们和DNA相互识别的原理是什么？
特异性转录调控因子是指是除基本转录因子之外，与转录核心区以外的顺式作用元件相结合，影响基因转录的因子，分为DNA结合结构域和转录激活结构域。与GGGCGG结合的特异性转录调控因子叫Sp1。与CCAAT结合的转录因子叫CBF。大多数类固醇激素与受体结合之后，构成特异性转录因子，与顺式作用元件结合，激活基因转录。也有转录调控抑制因子，可以覆盖转录活化区，覆盖转录起始区。还有HSP。与DNA的识别模式有：(1)螺旋-转角-螺旋，helix3识别结合DNA，helix1,helix2结合其它蛋白质因子。(2)Zenic finger与DNA大沟相结合，可以与RNA-pol3及其它转录因子相互作用。(3)亮氨酸拉链：N-末端由碱性氨基酸构成的a-helix可以与DNA相结合.(4)helix-loop-helix兼性a-helix具有疏水侧面和亲水侧面，两个具有这种motif的反式因子可以形成dimer。紧靠helix N-末端的碱性氨基酸组成的序列可以结合DNA。
5.回文结构：两个序列相同的DNA单链反向碱基互补，串联重复，又叫反向重复序列，是限制性内切酶II的识别位点，也是转录终止的识别区。可以构成十字状结构。
分子生物学习题及答案（3,4,5章）汇总
分⼦⽣物学习题及答案（3,4,5章）汇总第3章⼀．名词解释（考试时，名词解释为英⽂，要写出中⽂并解释）1、复制(replication): 亲代双链DNA分⼦在DNA聚合酶的作⽤下，分别以每单链DNA分⼦为模板，聚合与⾃⾝碱基可以互补配对的游离的dNTP,合成出两条与亲代DNA分⼦完全相同的⼦代DNA分⼦的过程。
2、复制⼦(replicon)：也称复制单元，是基因组中具有⼀个复制起点（origin，ori）和⼀个复制终点（terminus，ter）并能在细胞中⾃主复制的基本单位。
3、半保留复制（Semi-Conservation Replication）：DNA复制过程中亲代DNA的双链分⼦彼此分离，作为模板，按碱基互补配对原则，合成两条新⽣⼦链，这种⽅式称为半保留复制。
4、冈崎⽚段（Okazaki fragment）冈崎⽚段是相对⽐较短的DNA链（⼤约1000核苷酸残基），是在DNA的后随链的不连续合成期间⽣成的⽚段，这是Reiji Okazaki在DNA合成实验中添加放射性的脱氧核苷酸前体观察到的，因此DNA的复制是半不连续复制。
5、DNA复制的转录激活(transcriptional activation）：RNA聚合酶使双链DNA分⼦局部开链，在合成10~12个核苷酸的RNA⽚段之后，再由DNA聚合酶完成前导链DNA的合成，在完成近1000~2000个核苷酸的DNA合成后，后随链才在引发酶的作⽤下开始启动冈崎⽚段的引物RNA的合成，将这⼀过程称为DNA复制的转录激活。
6、单链DNA结合蛋⽩(single strand DNA binding protein，SSB):在复制中维持模板处于单链状态并保护单链的完整性。
7、复制体(replisome)：DNA复制过程中的多酶复合体。
8、端粒（Telomere）：是真核⽣物染⾊体末端的⼀种特殊结构，是为了保证染⾊体稳定的⼀段⾼度重复序列，呈现四股螺旋。
分子生物学 双语习题及精解
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.。
分子生物学 翻译习题教学文稿
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 )。
第五章 基因表达2：蛋白质翻译 分子生物学习题
第五章基因表达2：蛋白质翻译名词解释：SD序列、无义突变和错义突变、EF-Tu、同义密码填空：1.可使每个氨基酸和它相对应的tRNA分子相偶联形成一个2.核糖体包括两个tRNA分子的结合位点：即P位点，紧密结合与多肽链延伸尾端相连接的tRNA分子；即A位点，结合带有一个氨基酸的tRNA 分子。
3.蛋白质合成的起始过程很复杂，包括一系列被催化的步骤4.tRNA的三叶草型结构中，其中氨基酸臂的功能是_________，反密码环的功能是___________。
5.核糖体沿着mRNA前进时，它需要另一个延伸因子，这一步需要的水解。
当核糖体遇到终止密码、、的时候，延伸作用结束，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，并且需要一套因子。
6.tRNA的反密码子为GGC，它可识别的密码子为和7.遗传密码中第个碱基常很少或不带有遗传信息8.真核细胞多肽合成的起始氨基酸均为，而原核细胞的起始氨基酸应为。
9.蛋白质生物合成是从_____端到______端10.原核生物中的释放因子有三种，其中RF-1识别终止密码子_____________、____________；RF-2识别__________、____________；真核中的释放因子只有___________一种。
判断：1、因为AUG是蛋白质合成的起始密码子，所以甲硫氨酸只存在于蛋白质的N末端（）2、原核生物蛋白质合成的起始氨基酸为甲硫氨酸，真核生物蛋白质合成起始氨基酸为甲酰甲硫氨酸（）3、核糖体小亚基最基本功能是连接mRNA和tRNA大亚基则催化肽键的形成。
（）选择题：1、反密码子中哪个碱基参与了密码子的简并性（）A、第一个B、第二个C、第三个D、第一个与第二个E、第二个与第三个2、“同工tRNA”是指（）A、识别同义mRNA密码子（具有第三个碱基简并性）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tRNA3、核糖体的E位点是（）A、真核mRNA加工位点B、tRNA离开原核生物核糖体的位点C、核糖体中受EcoR限制的位点D、电化学势驱动转运的位点4、蛋白质合成所需的能量来自（）A、ATPB、GTPC、ATP和GTPD、CTP5、mRNA的5’-ACG-3’密码子相应的反密码子是（）A、5′-UGC-3′B、5′-TGC-3′C、5′-CGU-3 ′D、5 ′ -CGT-3 ′6、在蛋白质合成过程中，下列哪些说法是正确的？（）A、氨基酸随机地连接到tRNA上去B、新生肽链从C一端开始合成C、通过核糖核蛋白体的收缩，mRNA不断移动D、合成的肽链通过一个tRNA与核糖核蛋白相连问答题：1、简述遗传密码的性质2、原核生物蛋白质合成的过程3、蛋白质前体加工包括哪些？有一个被认为是mRNA的核苷酸序列，长300个碱基，你怎样才能：1.证明此RNA是mRNA而不是tRNA或rRNA。
分子生物学英文试题
Multiple Choice(1) The attachment site for RNA polymerase in bacteria is called the:a. Initiatorb. Operatorc. Promoterd. Start codon(2) The specificity of bacterial RNA polymerase for their promoters is due to which subunit?a. αb. βc. γd. σ(3) The first protein complex to bind to the core promoter for a protein-coding gene in eukaryotes is;a. RNA polymerase IIb. General transcription factor TFIIBc. General transcription factor TFIIDd. General transcription factor TFIIE(4) Which modification must be made to RNA polymerase II in order to activate the preinitiation complex?a. Acetylationb. Methylationc. Phosphorylationd. Ubiquitination(5) What is the name of the DNA sequence that is located near the promoter of the lactose operon, and which regulates expression of the operon in E. coli?a. Activatorb. Inducerc. Operatord. Repressor(6) Which of the following types of sequence module enables transcription to respond to general signals from outside of the cell?a. Cell-specific modulesb. Developmental modulesc. Repression modulesd. Response modules(7) Which of the following is NOT a type of activation domain?a. Acidic domainsb. Glutamine-rich domainsc. Leucine-zipper domainsd. Proline-rich domains(8) Which of the following is NOT a experiment used to define the site on a DNA molecule to which a protein binds?a. Gel retardation assayb. DNA footprinting assayc. Modification interference assayd. Y east two hybrid assay(9) Which of the following DNA sequences can increase the rate of transcription initiation of more than one gene/promoter?a. Activatorsb. Enhancersc. Silencersd. T erminators(10) Approximately how many base pairs form the attachment between the DNA template and RNA transcript during transcription?a. 8b. 12-14c. 30d. The entire RNA molecule remains base-paired to the template until transcription is finished.(11) Which factor is thought to be most important in determining whether a bacterial RNA polymerase continues or terminates transcription?a. Nucleotide concentrationb. Structure of the polymerasec. Methylation of termination sequencesd. Thermodynamic events(12) What is the role of the Rho protein in termination of transcription?a. It is a helicase that actively breaks base pairs between the template and transcript.b. It id s DNA-binding protein that blocks the movement of RNA polymerase along the template.c. It is a subunit of RNA polymerase that binds to RNA hairpins and stalls transcription.d. It is a nuclease that degrades the 3’ ends of RNA transcripts.(13) Antitermination is involved in regulation of which of the following?a. Operons encoding enzymes involved in the biosynthesis of amino acids with regulation dependent on the concentration of the amino acids.b. Operons encoding enzymes involved in the degradation of metabolites, regulation dependent on the presence of the metabolitec. Genes present in the upstream region of the operond. Genes present in the downstream region of the operon.(14) What is the major transcriptional change that occurs during the Stringent Response in E. coli?a. Transcription rates are increased for most genes.b. Transcription rates are increased only for the amino acid biosynthesis operons.c. Transcription rates are decreased for most genes.d. Transcription rates are decreased only for the amino acid biosynthesis operons.(15) Which of the following is necessary for the RNA endonuclease activity of RNA polymerase that occurs when RNA polymerase is stalled during transcription?a. Rhob. RelAc. GreAd. RNAse H(16) How is the lariat structure formed during splicing of a GU-AG intron?a. After cleavage of the 5’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of the nucleotide at the 3’ splice site.b. After cleavage of the 5’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of an internal adenosine.c. After cleavage of the 5’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of the nucleotide at the 5’ splice site.d. After cleavage of the 3’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of an internal adenosine.(17) What are cryptic splice sites?a. These are splice sites that are used in some cells, but not in others.b. These are splice sites that are always used.c. These are splice sites that are involved in alternative splicing, resulting in the removal of exons from some mRNA molecules.d. These are sequences within exons or introns that resemble consensus splicing signals, but are not true splice sites.(18) What statement correctly describes trans-splicing?a. The order of exons within an mRNA transcript is rearranged to yield a different mRNA sequence.b. Exons are deleted from some mRNA transcripts but not others.c. Intron sequences are not removed from RNA transcripts and are translated into proteins.d. Exons from different RNA transcripts are joined together.(19) The chemical modification of eukaryotic rRNA molecules takes place in the:a. Cytoplasm.b. Endoplasmic reticulum.c. Nuclear envelope.d. Nucleolus.(20) Which of the following is an example of RNA editing?a. Removal of introns from an RNA transcript.b. Degradation of an RNA molecule by nucleases.c. Alteration of the nucleotide sequence of an RNA molecule.d. Capping of the 5’ end of an RNA transcript.(21) Nonsense-mediated RNA decay (NMD) is a system for the degradation of eukaryotic mRNA molecules with what features?a. NMD degrades mRNA molecules with stop codons at incorrect positions.b. NMD degrades mRNA molecules that encode nonfunctional proteins.c. NMD degrades mRNA molecules that lack a start codon.d. NMD degrades mRNA molecules that lack a stop codon.(22) Which of the following describes RNA interference?a. Antisense RNA molecules block translation of mRNA molecules.b. Double-stranded RNA molecules are bound by proteins that block their translation.c. Double-stranded RNA molecules are cleaved by a nuclease into short interfering RNA molecules.d. Short interfering RNA molecules bind to the ribosome to prevent the translation of viral mRNAs.(23) How are RNA molecules transported out of the nucleus?a. Passive diffusion through the membrane.b. Through the membrane pores in an energy-dependent process.c. Through membrane pores in an energy independent process.d. Through a channel in the membrane that leads to the endoplasmic reticulum.(24) Match protein/RNA with its function (answers can be used more than once or not at all)Spliceosome a. small nuclear ribonucleoproteins (snRNP)microRNAs b. guanylyl transferasemRNA capping c. ribozymeautocatalytic RNA splicing d. dicere. poly-A polymerase(25) Match protein with its function (answers can be used more than once or not at all)JAK a. G-proteinGlucocorticoid receptor b. EndonucleaseRAS c. DNA-binding proteinIF-2 d. RNA binding proteine. Kinase(26) Match lambda gene with its function (answers can be used more than once or not at all)cI a. Anti-terminationN b. Transcriptional repressorCRO c. Transcriptional activatorcII d. Transcriptional terminatore. Translation factorAnswers to practice exam #3How is it possible for microRNAs to regulate eukaryotic gene expression by binding to the 3’ untranslated end of an mRNA ?Binding to the 3’-UTR initiates an RNA cleavage event that removes the polyA tail and begins the mRNA degradation processWhy is attenuation absent in eukaryotic organisms ?Attenuation is the mechanism whereby amino acid biosynthesis operons are regulated by the cellular concentration of the amino acid that is the product of the genes in the operon by transcription termination. The attenuation mechanism requires that translation by ribosomes and transcription occur in the same subcellular compartment. In eukaryotes transcription and translation are carried out in different compartments, so attenuation would not be possible in eukaryo tes.What are the differences between activator and coactivator proteins ?An activator is a DNA binding protein that stabilizes construction of the RNA polymerase II transcription initiation complex. A coactivator is a protein that stimulates transcription initiation by binding nonspecifically to DNA or via protein-protein interactions.Explain what a “modification protection assay” is intended to discover and how it is carried out .Modification protection is a technique used to identify nucleotides i nvolved in interactions with a DNA-binding proteinHow are Caenorhabditis elegans and Drosophila melanogaster good model organisms for development in higher eukaryotes ?Developmental pathways in animals utilize similar regulators, therefore discovery of regulators in lower animals can reveal how development is controlled in higher animals.How does the anchor cell of C. elegans induce the vulva progenitor cells to differentiate into vulva cells? Why do the vulva progenitor cells follow different pathways upon receiving the signal from the anchor cell ?The anchor cell produces a diffusible signal that stimulates differentiation of vulva cells. Different vulva cells undergo different differentiation pathways because they are exposed to differing concentr ations of the signal molecule, and the vulva cells themselves produce secondary signaling molecules that control differentiation in nearby vulva cells.The process of excision of a GU-AG intron and splicing of exons is defined as requiring two transesterification reactions. What does this mean ?A transesterification reaction is the simultaneous cleavage and reformation of a phosphodiester bond. During intron splicing the donor site phosphodiester bond is cleaved and then reformed with the branchpoint nucleotide within the intron, forming a lariat structure. In the second transesterification, the branch point phospodiester bond I cleaved and simultaneously formed between the donor and acceptor sites. The net effect is that there is no change in the number of phosphodiester bonds. During sporulation in Bacillus σE and σF are present in both the prespore and mother cells. How is σF activated in the prespore?Sigma F is activated in the prespore by when it is released from protein-protein interaction with AB. Sigma F is inactive when it is bound to AB.Explain how the iron response protein (IRP) functions to activate expression of Ferritin and at the same time inhibit expression of Transferrin.The iron response protein can bind to iron response elements in RNA only when it is not bound to iron. In the case of ferritin, binding of IRP to the 5’-IRE blocks translation of the ferritin mRNA, so when it is not bound ferritin protein is produced. In the case of transferrin, binding to the 3’-IRE blocks degradation of the transferrin mRNA thereby increasing half life of the mRNA and stimulating transferrin protein production.。
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————————————————————————————————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。
分子生物学试题第四章
一、写出英文缩写的全称IF:起始因子EF:伸长因子RF:释放因子ORF:开放读码框架(开放阅读框)open reading frameNLS:核定位序列(Nudear Iocalization signal)二、选择题1.外源基因在大肠杆菌中的高效表达受到很多因素的影响,其中SD序列的作用是( B )A.提供一个mRNA转录终止位点B.提供一个mRNA转录起始子C.提供一个核糖体结合位点D.提供翻译的终点2.与tRNA中的反密码子为GCU相配对的mRNA中的密码子是( B )A. UGAB. CGAC. AGCD. AGI3. 稀有碱基常出现于(C )A. rRNAB. mRNAC. tRNAD. hnRNA4. 下列表述不正确的是(A )A. 共有20个不同的密码子代表遗传密码B. 每个核苷酸三联子编码一个氨基酸C. 不同的密码子可能编码同一个氨基酸D.密码子的第三位具有可变性5. ( B )的密码子可以作为起始密码子。
A. 酪氨酸B.甲硫氨酸C.色氨酸D. 苏氨酸6. 核糖体的E位点是(B )A.真核mRNA的加工位点B. tRNA离开原核生物核糖体的位点C. 核糖体中受EcoRI限制的位点D.真核mRNA起始结合位点7. 关于蛋白质合成描述正确的是( B )A. 转录转录起始位点+1处是蛋白质翻译的起始部位B. 所谓翻译就是把mRNA上携带的碱基序列转变成氨基酸的过程C. 翻译时mRNA上必须有SD序列D. 翻译起始后携带氨基酸的氨酰tRNA首先进入核糖体的A位点8. tRNA在发挥其功能时的两个重要部位是(D )A. 反密码子臂和氨基酸臂B.氨基酸臂和D环C. TψC环与可变环D. TψC环与反密码子臂9. 反密码子的化学性质属于(A )A. tRNAB. mRNAC. rRNAD. DNA三、填空题1. 核糖体上存在三个位点,A位点是新的氨酰tRNA进入的位点,是肽酰tRNA 的结合位点,E位点是空的tRNA释放位点。
分子生物学第五章翻译专选课件
IF1 9
15% 循 环 因 子 ？
二. 起始tRNA的特点
翻译时的第一个密码子是怎样被识别的呢？ (一)小亚基上16SrRNA3′端的六核苷酸(3′-
UCCUCC-5′ ） 和 Shine-Dalgarno 顺 序 （ 5′AAACAGGAGG-3′ ） 互 补 ， 相 互 结 合 ， 使 下游的AUG起始密码子定位在P位上。 (二) 核糖体结合位点也含有一个信号起始密 码子（initiation codon）－AUG。
PA 图15- 翻译起始时的进位反应
原核生物翻译起始因子
表 15-2 E.coli蛋 白 质 合 成 起 始 所 需 的 三 种 起 始 因 子
因 子质量因 子 /核功 能
（ KDa）糖 体
IF3 23
25% 亚 基 解 离 与 mRNA的 结 合
IF2 97.3 ?
起 始 tRNA的 结 合 与 GTP水 解
三 核糖体的作用位点
⑴A位点（或称 acceptor site）可以进入 氨基酰-tRNA(aminoacyl-tRNA)。
⑵ P位点(或称供位，donor site) 是被肽基 酰-tRNA(peptidyl-tRNA)所占据。
(3)E位点(Exit site) 脱酰tRNA(deacylatedtRNA)短暂地占据。
分子生物学第五章翻译
催化功能区：ATP tRNA受体双 tRNA反密码 寡聚物的
和氨基酸位点
螺旋结合区 子结合区
形成
核苷酸折叠或反向 平行的β-折叠
α-螺旋或 β-折叠桶
Ⅰ类酶
Ⅱ类酶
图15-2氨基酰tRNA合成酶含有3－4个不同的功能区 (仿B.Lewin:《GENES》Ⅵ,1997，Fig9.8)