分子生物学双语练习

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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.。

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。

生物科技-中科院分子生物学试题库题库4 精品

GLOSSARYAbundance of an mRNA is the average number of molecules per cell.Abundant mRNAs consist of a small number of individual species, each present in a large number of copies per cell.Acceptor splicing site—see right splicing junction.Acentric fragment of a chromosome (generated by breakage) lacks a centromere and is lost cell division.Acrocentric chromosome has the centromere located nearer one end than the other.Active site is the restricted part of a protein to which a substrate binds.Allele is one of several alternative forms of a gene occupying a given locus on a chromosome. Allelic exclusion describes the expression in any particular lymphocyte of only one allele coding for the expressed immunoglobulin.Allosteric control refers to the ability of an interaction at one site of a protein to influence the activity of another site.Alu family is a set of dispersed, related sequences, each～300 bp long, in the human genome. The individual members have Alu cleavage sites at each end (hence the name).Alu-equivalent family is the set of sequences in a mammalian genome that is related to the human Alu family.α-Amanitin is a bicyclic octapeptide derived from the poisonous mushroom Amanita phalloides; it inhibhits transcription by certain eukaryotic RNA polymerases, especially RNA polymerase II.Amber codon is the nucleotide triplet UAG, one of three codons that cause termination of protein synthesis.Amber mutaion describes any change in DNA that creates an amber codon at a site previously occupied by a codon representing an amino acid in a protein.Amber suppressors are mutant genes that code for tRNAs whose anticodons have been altered so that they can respond to UAG codons as well as or instead of to their previous codons.Aminoacyl-tRNA is transfer RNA carrying an amino acid; the covalent linkage is between the NH2group of the amino acid and either the 3’-or-2’-OH group of the terminal base of the tRNA.Aminoacyl-tRNA synthetases are enzymes responsible for covalently linking amino acids to the 2’ or 3’-OH position of tRNA.Amphipathic structures have two surfaces, one hydrophilic and one hydrophobic. Lipids are amphipathic; and some protein regions may form amphipathic; and some protein regions may form amphipathic helices, with one charged face and one neutral face.Amplification refers to the production of additional copies of a chromosomal sequence,found as intrachromosomal or extrachromoxomal DNA.Anchorage dependence describes the need of normal eukaryotic cells for a surface to attach to in order to grow in culture.Aneuploid chromosome constitution differes from the usal diploid constitution by loss or duplication of chromosomes or chromosomal segments. Annealing is the pairing of plementary single strands of DNA to form a double helix.Antibody is a protein (immunoglobulin) produced by B lymphocyte cells that recognizes a particular foreign ‘antigen,’and thusw triggers the immune response.Anticoding strand of duples DNA is used as a template to direct the synthesis of RNA that is plementary to it.Antigen is any molecule whose entry into an organism provokes synthesis of an antibody (immunoglobulin).Antiparallel strands of the double helix are organized in opposite orientation, so that the 5’ end of one strand is aligened with the 3’ end of th e other strand.Antitermination proteins allow RNA polymerase to transcribe through certain terminator sites.Ap endonucleases make incisions in DNA on the 5’ side of either apurinic or apyrimidinc sites.Apoinducer is a protein that binds to DNA to switch on transcription by RNA polymerase. Archebacteria prise a minor line of prokaryotes, and may have introns in the genome. Ascus of a fungus contains a tetrad or octad of the (haploid) spores, representhing the products of a sihngle meiosis.att sties are the loci on a phage and the bacterial chromosome at which rebination integrates the phage into. or excises it from , the bacterial chromosome.Attenuation describes the regulation of termination of transcription that is involved in controlling the expression of some bacterial operons.Attenuato r is the terminator sequence at which attenuatioj occurs.Autogenous conhtrol describes the action of a gene product that either inhibits (negative autogenous control) or activates (positive autogenous control) expression of the gene coding for it.Autonomous controlling element in maize is an active transposon with the ability to transpose (cf nonautonomous controlling element).Autoradiography detects radioactively labeled molecules by their effect in creating an image on pholtographic film.Autosomes are all the chromosomes except the sex chromosomes; a diploid cell has two copies of each autosome.Blymphocytes (or B cells) are the cells responsible for synthesizing antibodies.Backcross is another (earlier) term for a testcross.Back mutation reverses the effect of a mutation that hand inactivated a gene; thus it restores wild type.Bacteriophages are viruses that infect bacteria; often abbreviated as phages.Balbaini ring is an extremely large puff at a band of a polytene chromosome.Bands of polytene chromosomes are visible as dense regions that contain the majority of DNA;bands of normal chromosomes are relatively much larger and are generated in the form of regions that retain a stain on certain chemical treatments.Base pair (bp) is a partnership of A with T or of C with G in a DNA double helix; other pairs can be formed in RNA under certain circumstances.Bidirectional replication is acplished when two replication forks move away from the same origin in different directions.Bivalent is the structure containing all four chromatids (two representing each homologue) at the start of meiosis.Blastoderm is a stage of insect embryogenesis in which a layer of nuclei or cells around the embryo surround an internal mass of yolk.Blocked reading frame cannot be translated into protein because it is interrupted by termination codons.Blunt-end ligation is a reaction that joins two DNA duplex molecules directly at their ends. bp is an abbreviation for base pairs; distance along DNA is measured in bp.Branch migration describes the ability of a DNA strand partially paired with its plement in a duplex to extend its pairing by displacing the resident strand with which it is homologous. Breackage and reunion describes the mode of genetic rebination, in which two DNA duplex molecules are broken at corresponding points and then rejoined crosswise (involving formation of a length of heteroduplex DNA around the site of joining).Buoyant desity measures the ability of a substance to float in some standard fluid, for example, CsCl.C banding is a technique for generating stained regions around centromeres.C genes code for the constant regions of immunoglobulin protein chains.C value is the total amount of DNA in a haploid genome.CAAT box is part of a conserved sequence located upstream of the startpoints of eukaryotic transcription units; it is recognized by a large group of transcription factors.Cap is the structure at the 5’ end of eukaryotic mRNA, introduced after transcripton b y linking the terminal phosphate of 5’ GTP to the terminal base of the mRNA. The added G (and sometimes some other bases) are methylated, giving a structure of the form 7Me G5’ppp 5’Np…CAP(CRP)is a positive regulator protein activated by cyclic AMP. It is needed for RNA polymerase to initiate transcription of certain (catabolitesensitive) operons of E. coli.Capsid is the external protein coat of a virus particle.Catabolite repression describes the decreased expression of many bacterial operons that results from addition of glucose. It is caused by a decrease in the level of cyclic AMP, which in turn inactivates the CAP regulator.cDNA is a single-stranded DNA plementary to an RNA, synthesized from it by reverse transcription in vitro.cDNA clone is a duplex DNA sequence representing an RNA, carried in a cloning vector.Cell cycle is the period from one division to the next.Cell hybrid is a somatic cell containing chromosomes derived from parental cells of different species (e.g. a man-mouse somatic cell hybrid), generating by fusing the cells to form a heterokaryon in which the nuclei subsequently fused.Centrioles are small hollow cylinders consisting of microtubules that bee located near the poles during mitosis. They reside within the centrosomes.Centromere is a constricted region of a chromosome that includes the site of attachment to the mitotic or meiotic spindle (see also kinetochore).Centrosomes are the regions from which microtubules are organized at the poles of a mitotic cell. In animal cells, each centrosome contains a pair of centrioles surrounded by a dense amorphous region to which the microtubules attach. See also MTOC.Molecular chaperone is a protein that is needed for the assembly or proper folding of some other protein, but which is not itself a ponent of the target plex.Chemical plexity is the amount of a DNA ponent measured by chemical assay.Chi sequence is an octamer that provides a hotspot for RecA-mediated genetic rebination in E. coli.Chi structure is a joint between two duplex molecules of DNA revealed by cleaving an intermediate of two joined circles to generate linear ends in each circle. It resembles a Greek chi in outline, hence the name.Chiasma (pl. chiasmata)is a site at which two homologous chromosomes appear to have exchanged material during meiosis.Chromatids are the copies of a chromosome produced by replication. The name is usually used to describe them in the period before they separate at the subsequent cell division. Chromatin is the plex of DNA and protein in the nucleus of the interphase cell. Individual chromosomes cannot be distinguished in it .It was originally recognized by its reaction with stains specific for DNA.Chromocenter is an aggregate of heterochromatin from different chromosomes. Chromomeres are densely staining granules visible in chromosomes under certain conditions, especially early in meiosis, when a chromosome may appear to consist of a series ofchromomeres.Chromosome is a discrete unit of the genome carrying many genes. Each chromosome counsists of a very long molecule of duplex DNA and an approximately equal mass of proteins. It is visible as a morphological entity only during cell division.Chromosome walking describes the sequential isolation of clones carrying overlapping sequences of DNA, allowing large regions of the chromosome to be spanned. Walking is often performed in order to reach a particular locus of interest.cis-acting locus affects the activity only of DNA sequences on its own molecule of DNA; this property usually implies that the locus does not code for protein.cis-acting protein has the exceptional property of acting only on the molecule of DNA from which it was expressed.cis configuration describes two sites on the same molecule of DNA.cis/trans test assays the effect of relative configuration on expression of two mutations. In a double heterozygote, two mutations in the same gene show mutant phenotype in trans configuration, wild-type in cis configuration.Cistron is the geneti unit defined by the cis/trans test; equivalent to gene ib prising a unit of DNA representing a protein.Class switching is a change in the expression of the c cregion of an immunoglobulin heavy chain during lymphocyte differentiation.Clone describes a large number of cells or molecules identical with a single ancestral cell or molecule.Cloning vector is a plasmid or phage that is used to ‘carry’ inserted foreign DNA for the purposes of producing more material or a protein product.Closed reading frame contains termination codons that prevent its translation into protein. Coated vesicles are vesicles whose membrane has on its surface a layer of the protein clathrin.Coconversion is the simultaneous correction of two sites during gene conversion.Coding strand of DNA has the same sequence as mRNA.Codominant alleles both contribute to the phenotype; neither is dominant over the other. Coevolution —see concerted evolution.Cognate tRNAs are those recognized by a particular aminoacy-tRNA synthetase. Coointegrate structure is produced by fusion of two replicons, one originally possessing a transposon, the other lacking it; the cointegrate has copies of the transposon present at both junctions of the replicons, oriented as directrepeats.Cold-sensitive mutant is defective at low temperature but functional at normal temlperature. Colony hybridization is a technique for using in situ hybridization to identify bacteria carrying chimeric vectors whose inserted DNA is homologous with some particular sequence. patibility group of plasmids contains members unable to coexist in the same bacterial cell.plementation refers to the ability of independent (nonallelic)genes to provide diffusible products that produce wild phenotype when two mutants are tested in trans configuration in a heterozygote.In vitro plementation assay consists of identifying a ponent of a wid-type cell that can confer activity on an extract prepared from a mutant cell. The assay identifies the ponent rendered inactive by the mutation.plementation group is a series of mutations unable to plement when tested in pairwise binations in trans; defines a genetic unit (the cistron) that might better be called a nonplex mentation group.plex locus (of D. melanogaster)has genetic properties inconsistent with the function of a gene representing a single protein. plex loci are usually very large (>100kb) at the molecular level.plexity is the total length of different sequences of DNA present in a given preparation. postie transposons have a central region flanked on each side by insertion sequences, either or both of which may enable the entire element to transpose.Concatemer of DNA consists of a series of unit genomes repeated in tandem. Concatenated circles of DNA are interlocked like rings one a chain.Concerted evolution describes the ability of two related genes to evolve together as though constituting a single locus.Condensation reaction is one in which a covalent bond is formed with loss of a water molecule, as in the addition of an amino acid to a polypeptide chain.Conditional lethal mlutations kill a cell or virus under certain (nonpermissive) conditions, but allow it to survive under other (permissive) conditions.Conjugation describes ‘mathing’ between two bacterial cells, when (part of ) the chromosome is transferred from one to the other.Consensus sequence is an idealized sequence in which each position represents the base most often found when many actual sequences are pared.Conservative rebination involves breakage and reunion of preexisting strands of DNA without any synthesis of new stretches of DNA.Conservative transposition refers to the movement of large elements, originally classified as transposons, but now considered to be episomes. The mechanism of movement resembles that of phage lambda.Constant regions of immunoglobulins are coded by C genes and are the parts of the chain that vary least. Those of heavy chains identify the type of immunoglobulin.Constitutive genes are expressed as a function of the interaction of RNA polymerase with the promoter, without additional regulation; sometimes also called household genes in the context of describing functions expressed in all cells at a low level.Constitutive heterochromatin describes the inert state of permanently nonexpressedsequences, usually satellite DNA.Constitutive mutations cause genes that usually are regulated to be expressed without regulation.Contractile ring is a ring of actin filaments that forms around the equator at the end of mitosis and is responsible for pinching the daughter cells apart.Controlling elements of maize are transposable units originally identified solely by their genetic properties. They may be autonomous (able to transpose independently) or nonautonomous (able to element).Coordinate regulation refers to the mon control of a group of genes.Cordycepin is 3’ deoxyadenosine, an inhibitor of polyadenylation of RNA.Core DNA is the 14.6 bp of DNA contained on a core particle.Core particle is a digestion product of the nucleosome that retains the histone octamer and has 14.6 bp of DNA; its structure appears similar to that of the nucleosome itself.Corepressor is a small molecule that triggers repression of transcription by binding to a regulator protein.Cosmids are plasmids into which phage lambda cos sites have been inserted; as a result, the plasmid DNA can be packaged in vitro in the phage coat.Cot is the product of DNA concentration and time of incubation in a reassociation reaction. Cot 21 is the Cot required to proceed to half pletion of the reaction; it is directly proportionalto the unique length of reassociating DNA.Cotransfection is the simultaneous transfection of two markers.Crossing-over describes the reciprocal exchange of material between chromosomes that occurs during meiosis and is responsible for genetic rebination.Crossover fixation refers to a possible consequence of unequal crossing-over that allows a mutation in one member of a tandem cluster to spread through the whole cluster (or to be eliminated).Cruciform is the structure produced at inverted repeats of DNA if the repeated sequence pairs with its plement on the same strand (instead of with its regular partner in the other strand of the duplex).Cryptic satellite is a satellite DNA sequence not identified as such by a separate peak on a density gradient; that is, it remains present in main-band DNA.ctDNA is chloroplast DNA.Cyclic AMP (cAMP) is a molecule of AMP in which the phosphate group is joined to both the 3’ and 5’ positions of the ribose; its binding activates the CAP, a postive regulator of prokaryotic transcription.Cyclins are proteins that accumulate continuously throughout the cell cycle and are then destroyed by proteolysis during mitosis. (see also MPF).Cytokinesis is the final process involved in separation and movement apart of daughter cells at the end of mitosis.Cytological hybridization—see in situ hybridization.Cytoplasm describes the material between the plasma membrane and the nucleus. Cytoplasmic inheritance is a property of genes located in mitochondria or chloroplasts (or possibly other extranuclear organelles).Cytoplasmic protein synthesis is the ranslation of mRNAs representing nuclear genes; it occurs via ribosomes attached to the cytoskeleton.Cytoskeleton consists of networks of fibers in the cytoplasm of the eukaryotic cell.Cytosol describes the general volume of cytoplasm in which organelles ( such as the mitochondria ) are located.D loop is a region within mitochondrial DNA in which a short stretch of RNA is paired with one strand of DNA, displacing the orignal partner DNA strand in this region. The same term is used also to describe the displacement of a region of one strand of duplex DNA byu a single-stranded invader in the reaction catalyzed by RecA protein.Degeneracy in the genetic code refers to the lack of an effect of many changes in the third base of the codon on the amino acid that is represented.Deletions are generated by removal of a sequence of DNA, the regions on either side being joined together.Denaturation of DNA or RNA describes its conversion from the double-stranded to the singlestranded state; separation of the strands is most often acplished by heating. Denaturation of protein describes its conversion from the physiological conformation to some other (inactive) conformation.Derepressed state describes a gene that is turned on. It is synonymous with induced when describing the normal state of a gene; it has the same meaning as constitutive in describing the effect of mutation.Dicentric chromosome is the product of fusing two chromosome fragments, each of which has a centromere. It is unstable and may be broken when the two centromeres are pulled to opposite poles in mitosis.Diploid set of chromosomes contains two copies of each autosome and two sex chromosome. Direct repeats are identical (or related) sequences present in two or more copies in the same orientation in the same molecule of DNA; they aer not necessarily adjacent.Discontinuous replication refers to the synthesis of DNA in short (Okazaki) fragments that are later joined into a continuous strand.Disjunction describes the movement of members of a chromosome pair to opposite poles during cell division. At mitosis and the second meiotic division, disjunction applies to sister chromatids; at first meiotic division it applies to sister chromatid pairs.Divergence is the percent difference in nucleotide sequence between two related DNA sequences or in amino acid sequences between two proteins.Divergent transcription refers to the intitiation of transcription at two promoters facing in the opposite direction, so that transcription proceeds away in both directions from a central region.dna mutants of bacteria are temperature-sensitive; they cannot synthesize DNA at 42℃, but can do so at 37℃.DNAase is an enzyme that attacks bonds in DNA.DNA-driven bybridization involves the reaction of an excess of DNA with RNA.DNA polymerase is an enzyme that synthesizes a daughter strand(s) of DNA (under direction from a DNA template). May be involved in repair or replication.DNA replicase is a DNA-synthesizing enzyme required specifically for replication.Domain of a chromosome may refer either to a discrete structural entity defined as a region within which supercoiling is independent of other domains; or to an extensive region including an expressed gene that has heightened sensitivity to degradation by the enzyme DNAase I.Domain of a protein is a discrete continuous part of the amino acid sequence that can be equated with a particular function.Dominant allele determines the phenotype displayed in a heterozygote with another (recessive) allele.Donor splicing site—see left splicing junction.Down promoter mutations decrease the frequency of initiaton of transcription. Downstream identifies sequences proceeding farther in the direction of expression, for example, the conding region is downstream of the initiation condon.Early development refers to the period of a phage infection before the start of DNA replication.Extopic expression describes the expression of a gene in a tissue in which it is not usually expressed; for example, in a transgenic animal.Elongation factors (EF in prokaryotes, eEF in eukaryotes) are proteins that associate with ribosomes cyclically, during addition of each amino acid to the polypeptide chain.End labeling describes the addi tion of a radioactively labeled group to one end (5’ or 3’) of a DNA strand.End-product inhibition describes the ability of a product of a metabolic pathway to inhibit the activity of an enzyme that catalyzes an early step in the pathway.Endocytosis is a process by which proteins at the surface of the cell are internalized, being transported into the cell within membranous vesicles.Endocytic vesicles are membranous particles that transport proteins through endocytosis; alsoknown as clathrin-coated vesicles.Endonucleases cleave bonds within a nucleic acid chain; they may be specific for RNA or for singlestranded of double-stranded DNA.Endoplasmic reticulum is a highly convoluted sheet of membranes, extending from the outer layer of the nuclear envelope into the cytoplasm.Enhancer element is a cis-acting sequence that increases the utilization of (some) eukaryotic promoters, and can function in either orientation and in any location (upstream or downstream) relative to the promoter.Envelopes surround some organelles (for example , nucleus or mitochondrion) and consist of concentric membranes, each membrane consisting of the usual lipid bilayer.Epigenetic changes influrence the phenotype without altering the genotype. They consist of changes in the properties of a cell that are inherited but that do not represent a change in genetic information.Episome is a plasmid able to integrate into bacterial DNA.Epistasis describes a situation in which expression of one gene wipes out the phenotypic effects of another gene.Essential gene is one whose deletions is lethal to the organism (see also lethal locus). Established cell lines consist of eukaryotic cells that have been adapted to indefinite growth in culture (they are said to be immortalized).Eubacteria prise the major line of prokaryotes.Euchromatin prises all of the genome in the interphase nucleus except for the heterochromatin.Evolutionary clock is defined by the rate at which mutations accumulate in a given gene. Excision-repair systems remove a single-stranded sequence of DNA containing damaged or mispaired bases and replace it in the duplex by synthesizing a sequence plementary to the remaining strand.Exocytosis is the process of secreting proteins from a cell into the medium, by transport in membranous vesicles from the endoplasmic reticulum, through the Golgi, to storage vesicles, and finally (upon a regulatory signal) through the plasma membrane.Exocytic vesicles (also secretory vesicles) are membranous particles that transport and store proteins during excytosis.Exon is any segment of an interrupted gene that is represented in the mature RNA product. Exonucleases cleave nucleotides one at a time from the end of a polynucleotide chain; they may be specific for either the 5’ or 3’ end of DNA or RN A.Expression vector is a cloning vector desined so that a coding sequence inserted at a particular site will be transcribed and translated into protein.Extranuclear genes reside outside the nucleus in organelles such as mitochondria and chloroplasts.F factor is a bacterial sex or fertility plasmid.F1 generation is the first generation produced by crossing two parental (homozygous) lines. Facultative heterochromatin describes the inert state of sequences that also exist in active copies—for example, one mammalian X chromosome in females.Fast ponent of a reassociation reaction is the first to reature and contains highly repetitive DNA.Fate map is a map of an embryo showing the adult tissues that will develop from the descendants of cells that occupy particular regions of the embryo.Figure eight describes two circles of DNA linked together by a rebination event that has not yet been pleted.Filter hybridization is performed by incubating a deatured DNA preparation immobilized on a nitrocellulose filter with a solution of radioactively labeled RNA or DNA.Fingerprint of DNA is a pattern of polymorphic restriction fragments that differ between individual genomes.Fingerprint of a protein is the pattern of fragments (usually resolved on a two dimensional electrophoretic gel) generated by cleavage with an enzyme such as trypsin.Fluidity is a property of membranes; it indicates the ability of lipids to move laterally within their particular monolayer.Focus formation describes the ability of transformed eukaryotic cells to grow in dense clusters, piled up on one another.Focus forming unit (ffu) is a quantitative measure of forcus formation.Foldback DNA consists of inverted repeats that have renatured by intrastrand reassociation of denatured DNA.Foot printing is a technique for identifying the site on DNA bound by some protein by virtue of the protection of bonds in this region against attack by nucleases.Forward mutations inatctivate a wild-type gene.Founder effect refers to the presence in a population of many individuals all with the same chromosome (or region of a chromosome) derived from a single ancestor.Frameshift mutations arise by deletions or insertions that are not a multiple of 3bp; they change the frame in which triplets are translated into protein.G banding is a technique that generates a striated pattern in metaphase chromosomes that distinguishes the members of a haploid set.G1 is the period of the eukaryotic cell cycle between the last mitosis and the start of DNA replication.G2 is the period of the eukaryotic cell cycle between the end of DNA replication and the start of the next mitosis.。

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 )。

大学分子生物学经典双语课件

2.1.2.2 Conformation polymorphism of the double helix
Alternative doublehelical structures of DNA
Base Obliquity
helix rise per base pair
bp number per turn
biological activity changed (even lost)； viscosity decreased，粘度 solubility decreased,溶解度 Hyperchromicity: the absorbance of ssDNA is greater than that dsDNA.增色 concentration = 50μg/ml： dNTPs A260 = 1.60 S.S DNA A260 = 1.37 D.S DNA A260 = 1.0
2.1.3
Triplex DNA
1953, Watson & Crick proposed D.S DNA model and found many redundant hydrogen bonding donor and receptors along big grooves. 1957, Felsenfeld proposed T.S DNA concept
transferring the other dsDNA through the break.
Type I topoisomerase
Type II topoisomerase
Contents
1
2 3 4 5 6 7
Structure of DNA Denaturation, renaturation and hybridization

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.。

分子生物学双语

A protein protects a series of bonds against nuclease attack
FIGURE 05: Micrococcal nuclease initially cleaves between nucleosomes
The nucleosome is a cylinder with DNA organized into ~1 2/3 turns around the surface
Photo courtesy of Barbara Hamkalo, University of California, Irvine.
The 10 nm fiber is a continuous string of nucleosomes
Basic Structures
• 30 nm fiber – A coil of nucleosomes.
Chapter 10 Chromatin
Individual nucleosomes are released by digestion of chromatin with micrococcal nuclease. The bar is 100 nm
Reprinted from Cell, vol. 4, P. Oudet, M. Gross-Bellard, and P. Chambon, Electron microscopic and biochemical evidence..., pp. 281-300. Copyright 1975, with permission from Elsevier [/science/journal/00928674]. Photo courtesy of Pierre Chambon, College of France.

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。

04级生物技术和生物科学本科分子生物学试题(A)

中英翻译（10分）1.upstream activating sequence2.linkage map3.terminator4.attenuator5.guide RNA6.色氨酸操纵子7. 帽子结蛋白Ⅰ(全称)8. RNA聚合酶9.聚合酶链式反应10. T4 DNA连接酶一、名词解释（20分）操纵子校正tRNA弱化作用增强子二、是非判断（20分）(请将答案写在题号前)1．DNA的一级结构包含了基因的信息，是基因表达的直接模板。

2．真核生物的TATA box 的作用是与TFⅡD结合并使转录在转录起点精确起始。

3．原生动物细胞中的大核是基因扩增的结果，小核没有量的变化。

4．T4 DNA连接酶的连接底物之一具有5’-磷酸基，另一底物应具有3’-OH。

5．在大肠杆菌中，ompF和omPc的蛋白总量保持恒定，这是外界渗透压影响下由mRNA的poly（A）的长度调节的。

6．Lac operon中，阻遏蛋白单体结合到操纵基因处，阻止了RNA聚合酶的有效结合。

7．所有的启动子都含有-10区、-35区和转录起点，且彼此之间总是保持适当的距离。

8．真核生物中，mRNA poly（A）—是组蛋白的翻译模板。

9．线粒体DNA的复制起点和终点的位置相同，所以其复制是单向的D-环复制。

10. 基因跳跃仅出现在原核生物中。

11．DNA的复杂性是指DNA中单拷贝序列的总长度。

12．环型DNA的复制起点和复制终点是唯一的。

13．嘧啶二聚体可以通过切除修复和重组修复从细胞中彻底清除。

14．随着细胞分裂的进行，线形越来DNA越短。

15．真核生物的核糖体上具有可与mRNA结合的部位，并且可与起始tRNA等形成起始复合物。

16．信号转导参与真核基因表达调节。

17．不对称PCR是特异性地扩增模板中的一条链，而另一条链不扩增。

18．通过DNA Walking ，可以将具有重叠序列的短片段拼接为长链。

19．含亮氨酸拉练的转录因子可以通过亮氨酸拉练的回折形成有活性的DNA结合域。

分子生物学习题[1]

PPT文档演模板
分子生物学习题[1]
大肠杆菌在含有葡萄糖的培养基中长势良好，当将其接种到只含有乳糖的培养基上时，起初大肠杆菌的长势很弱，但接种几代之后该菌株又恢复其良好长势。请运用相关知识解释这一现象。
这是基因的可诱导调节。所谓的可诱导调节是指一些基因在特殊的代谢化合物的作用下，由原来的关闭状态转变为工作状态，即在某些物质的诱导下使基因活化。例如，在只有乳糖培养基中，E. coli开始生长不好，直到合成了利用乳糖的一系列酶，具备了利用乳糖作为碳源的能力，E. coli才能在这一培养基中生存下来，E. coli获得这一能力的原因就是因为在诱导物乳糖的诱导下，开动了乳糖操纵子基因，表达了它编码的酶：β-半乳糖苷酶，这一类基因称为可诱导基因，这类酶称为诱导酶
分子生物学习题
PPT文档演模板
2020/11/10
分子生物学习题[1]
一段从E.coli中分离出来的DNA单链序列为：5'-
GTAGCCTACCCATAGG-3' (1)DNA复制时，另一条单链的
序列；(2)假设mRNA是由这段DNA单链的互补链作为模板转
录而来，mRNA的序列怎样？(3)如果转译从mRNA的5'端开
PPT文档演模板
分子生物学习题[1]
演讲完毕，谢谢听讲!
再见，see you again
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2020/11/10
分子生物学习题[1]
的功能。一些依赖于CRP的启动子缺乏一般启动子所具有的典型的
-35区序列特征（TTGACA）。因此RNA聚合酶难以与其结合。
CAP的存在（功能）：能显著提高酶与启动子结合常数。主要
表现以下二方面：CAP通过改变启动子的构象以及与酶的相互作用

(完整版)-朱玉贤分子生物学习题题库

i m e th e6.DNA 复制之初参与从超螺旋结构解开双股链的酶或因子是A.解链酶 B.拓朴异构酶Ⅰ C.单链结合蛋白D.引发前体E.拓朴异构酶Ⅱ7.关于真核生物DNA 复制与原核生物相比，下列说法错误的是A.引物长度较短B.冈崎片段长度较短C.复制速度较慢D.复制起始点只有一个E.由DNA 聚合酶α及δ催化核内DNA 合成8.端粒酶是一种A.DNA 聚合酶B.RNA 聚合酶C.DNA 水解酶D.反转录酶E.连接酶9.关于大肠杆菌DNA 聚合酶Ⅲ的说法错误的是A.催化dNTP 连接到DNA 片段的5′羟基末端B.催化dNTP 连接到引物链上C.需要4种不同的dNTP 为作用物D.是由多种亚基组成的不对称二聚体E.在DNA 复制链的延长中起主要作用10.在一个复制叉中，以下哪一种蛋白质数量最多A.DNA 聚合酶 B.引发酶C.SSBD.DNA 解链酶E.DNA 拓朴异构酶11.着色性干皮病（XP ）是一种人类遗传性皮肤病，是因什么缺陷而引起的A.DNA 复制B.转录C.转录后加工D.DNA 修复E.翻译12.在紫外线照射引起DNA 分子的损伤中最常见形成的二聚体是A.C-CB.C-TC.T-TD.T-UE.U-C13.下列对大肠杆菌DNA 聚合酶的叙述不正确的是A.DNA-Pol Ⅰ可分为大小两个片段B.DNA-Pol Ⅱ具有3′→5′外切酶活性C.DNA-Pol Ⅲ在复制链延长中起主要作用D.DNA-Pol Ⅲ由4个亚基组成E.以4种脱氧核苷三磷酸作为底物14.下列关于大肠杆菌DNA 聚合酶的叙述哪一项是正确的A.具有3′→5′核酸外切酶活性B.不需要引物C.需要4种不同的三磷酸核苷D.dUTP 是它的一种作用物E.可以将二个DNA 片段连起来15.DNA 复制需要（1）解链酶，（2）引发酶，（3）DNA 聚合酶，（4）切除引物的酶，（5）DNA 连接酶，其作用顺序是A.1,2,4,3,5B.1,2,3,4,5C.1,4,3,2,5D.1,4,2,3,5E.4,3,2,5,116.冈崎片段产生的原因是A.DNA 复制速度太快B.双向复制C.有RNA 引物D.复制与解链方向不同E.复制中DNA 有缠绕打结现象17.滚环复制A.是低等生物的一种复制形式B.不属于半保留复制C.内环链5′→3′延长，外环链3′→5′延长D.不需要DNA 连接酶的作用E.需要NTP 而不是dNTP 作原料18.比较真核生物与原核生物的DNA 复制，二者的相同之处是A.引物长度较短B.合成方向是5′→3′C.冈崎片段长度短D.有多个复制起点E.DNA 复制的速度较慢【单项选择题参考答案】1.E2.C3.D4.B5.C6.A7.D8.D9.A10.C 二、【解释名词】（1）DNA 半保留复制（2）半不连续复制（3）DDDP （4）冈崎片段（5）复制叉（6）反转录（7）基因突变（8）核酶（9）限制性核酸内切酶（10）基因重组（11）DNA 克隆（12）聚合酶链式反应参考答案略三、【简答题】1.比较原核生物和真核生物的DNA 复制有哪些异同点？2.DNA 半保留复制是如何被证实的？3.简述维持DNA 复制高度忠实性的机制。

山东大学分子生物学章节习题及参考答案05蛋白质的生物合成(翻译)

第五章蛋白质的生物合成（翻译）一、选择题1．仅有一个密码子的氨基酸是A．色氨酸、赖氨酸B．苏氨酸、甘氨酸C．甲硫氨酸、甘氨酸D．亮氨酸、丙氨酸E．色氨酸、甲硫氨酸2．密码与反密码配对时，不遵从碱基配对规律，称为A．密码的简并性B．密码的偏爱性C．密码的连续性D．密码的摆动性E．密码的通用性3．真核生物核蛋白体中没有的rRNA是A．18SB．23SC．5SD．28SE．5.8S4．反密码存在于A．DNAB．tRNAC．mRNAD．rRNAE．cDNA5．不符合密码的通用性的细胞器是A．细胞核B．微粒体C．线粒体D．内质网E．高尔基体6．氨基酰-tRNA合成酶的校正活性是A．水解酯键B．水解3’，5’磷酸二酯键C．水解磷酸酯键D．形成酸酐键E．形成磷酸酯键7．关于核蛋白体，错误的是A．由rRNA和多种蛋白质组成B．分为大小亚基C．是翻译的场所D．在细胞核内起作用E．一个mRNA上可附着多个核蛋白体8．能促使大小亚基解离的因子是A．IF1B．IF2C．IF3D．EF-TsE．IF1与IF39．EF-Tu的功能是A．协助氨基酰-tRNA进入A位B．促进核糖体亚基聚合C．促进核糖体解聚D．促进mRNA与核糖体分离E．促进肽酰-tRNA移位10．延长因子EFG具有哪种酶的活性A．转肽酶B．酯酶C．转位酶D．转甲酰酶E．转氨酶11．肽链延长过程的叙述，错误的是A．又称为核蛋白体循环B．每循环一次延长一个氨基酸C．分为进位，成肽和转位三步D．需要EFT、EFGE．需要ATP供能12. 翻译终止时激活转肽酶为酯酶活性的是A．RF-1B．RF-2C．RF-3D．RF-4E．RR13. 蛋白质合成中不消耗能量的阶段是A．氨基酸活化B．翻译起始C．进位D．成肽E．转位14. 关于多肽链一级结构的翻译后修饰，描述错误的是A．蛋白质合成过程中N端总是甲酰甲硫氨酸B．天然蛋白质N端多数不是甲酰甲硫氨酸C．脱甲酰基酶可除去N端甲酰基D．氨基肽酶可除去N端氨基酸E．翻译终止才能除去N端甲酰基15. 鸦片促黑皮质素原水解加工生成的是A．胰岛素B．糖蛋白C．脂蛋白D．ACTHE．TSH16. 可被信号肽酶裂解的部位是A．加工区B．疏水核心区C．碱性氨基末端区D．酸性羧基末端区E．亲水区17.关于信号肽识别粒子（SRP）的描述，错误的是A．由蛋白质与RNA组成的复合体B．能特异识别结合信号肽C．具有暂停蛋白质合成的作用D．可将正在合成蛋白质的核蛋白体带至膜外E．SRP需与对接蛋白结合18. 白喉毒素可共价修饰的因子是A．EF3B．eEF1C．EF1D．eEF2E．EF219. 干扰素通过何种方式使eIF2失活A．甲基化B．ADP核糖基化C．羧化D．磷酸化E．乙酰化20. 可辨认结合分泌蛋白新生肽链N端的是A．转肽酶B．信号肽识别颗粒C．GTP酶D．RNA酶E．对接蛋白二、名词解释1. 多聚核蛋白体(polyribosome)2. 信号肽(signal peptide)3．开放阅读框架（open reading frame, ORF）三、问答题1．三种RNA在蛋白质合成中各起何作用?2．原核与真核生物翻译起始阶段各有何异同？3．细胞核蛋白合成后如何靶向输送到细胞核？4．举例说明抗生素在翻译水平抑菌的作用机理。

分子生物学双语

Molecular Biology 分子生物学
分子生物学双语
Chapter 1 Welcome to Molecular Biology
➢ What is Molecular Biology? ➢ Why should we learn? ➢ How should we learn?
分子生物学双语
分子生物学双语
普通生物学
分类单位: 界门纲目科属种
分子生物学双语
微生物学
分子生物学双语
细胞生物学
分子生物学双语
生物化学
分子生物学双语
分子生物学的延伸
分子生物学
分子结构生物学分子发育生物学分子神经生物学分子育种学分子肿瘤学
分子细胞生物学分子免疫学分子病毒学分子生理学分子考古学
分子生物学双语
3)分子生物学的主要任务：
– 阐明这些生物大分子复杂的结构及结构与功能的
关系。 – 揭示生命的本质的分子机理
• Examples: Flash1
– Human immunodeficiency virus (HIV) is a lentivirus慢病毒(a member of the retrovirus 逆转录病毒 family) that causes acquired immunodeficiency syndrome (AIDS)
分子生物学双语
4. History of Molecular Biology 分子生物学发展简史
分为三个主要阶段:
• 对DNA和遗传信息传递的认识阶段
– 确立DNA为生物遗传物质的地位
• 重组DNA技术的建立和发展阶段 • 重组DNA技术的应用和分子生物学的迅猛发展
阶段

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 )。

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分子生物学习题[1]

(完整版)-朱玉贤分子生物学习题题库

山东大学分子生物学章节习题及参考答案05蛋白质的生物合成(翻译)

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