21世紀是生命科學的世紀20世紀后葉分子生物學的突破性

PAGEPAGE54—簡答題

第一章 緒論2120主要領域？答案：同；生物大分子單體的排列（核苷酸，氨基酸）導致了生物的特異性。三大支撐學科：細胞學，遺傳學和生物化學。研究的三大主要領域：主要研究生物大分子結構與功能的相互關系，其中包括DNA之間的相互作用；激素和受體之間的相互作用；酶和底物之間的相互作用。答案：有人把它定義得很廣：從分子的形式來研究生物現象的學科。但是這個定義使分子生物學難以從分子角度來解釋基因的結構和活性是本書的主要內容。3二十一世紀生物學的新熱點及領域是什么？答案：結構生物學是當前分子生物學中的一個重要前沿學科，它是在分子層次上從結構角度特別是從理學、化學和計算數學等多學科交叉的，以結構（特別是三維結構）和異常病理現象的關系。分子發(fā)育生物學也是當前分子生物學中的一個重要前沿學科。人類基因組計劃，被稱21 世紀生命科學的敲門‖“人類基因組計劃”以及“后基因組計劃”的全面展開將進入從分子水平闡明生命活動本質的輝煌時代。目前正迅速發(fā)展的生物信息學，被稱為“21世紀生命科學迅速發(fā)展的推動力。尤應指出，建立在生物信息基礎上的生物工程制藥產業(yè)在21世紀將逐步成為最為重要的新興產業(yè)從單基因病和多基因病研究現狀可以看出這兩種疾病的診斷和治療在21世紀將取得不同程度的重大進展遺傳信息的進化將成為分子生物學的中心內‖的觀點認為，隨著人類基因組和許多模式生物基因組序列的測定通過比較研究，人類將在基因組上讀到生物進化的歷史，使人類對生物進化的認識從表面深入到本質；研究發(fā)育生物學的時機已經成熟。在21世紀，遺傳信息的進化研究成果，將成為解決發(fā)育問題的基礎，發(fā)育問題這一難題可望獲得突破性進展；21世紀，生物技術產業(yè)化的趨勢將不斷加劇基因工程技術轉基因技術和基因治療技術等將對21世紀的產業(yè)結構產生深遠的影響。當前，生命科學基礎研究中最活躍的前沿主要包括：分子生物學、細胞生物學、神經生物學、生態(tài)學，并由這些活躍的前沿引伸出諸如：基因組學、蛋白質組學、人類基因組計劃、后人類基因組計劃、克隆羊、克隆魚、腦的十年、生物的多樣性等時髦的名詞和熱門話題。相應的應用研究或技術研究也正趨成熟并逐漸普及，如生物工程，即基因工程、蛋白質工程、發(fā)酵工程、酶工程、細胞工程、胚胎工程等。由于生命科學與人類生存、人們健康、社會發(fā)展密切相關，必將成為21世紀全球關注的領域。4.簡述分子生物學的發(fā)展歷程。答案：從1847——識，而Morgan化學研究的進展，Watson和Crick又提出了脫氧核糖核酸的雙螺旋模型，為充分揭示遺傳信息的傳遞規(guī)律鋪平了道路。在蛋白質化學方面，繼Sumner1936利用紙1953Kendrew和Perutz利用X子氧過程中的特殊作用，成為研究生物大分子空間立體構型的先驅。20世紀401941年，曾在摩爾根實驗室工作過的美國遺傳的新概念（后來有所修改，40年代中期被普遍承認，從而建立了生物化學、遺傳學。M.德爾布呂克和其同事們在1946年，美國微生物學家J.萊德伯格同E.L.現象。這兩項突破以及他們對噬菌體和大腸桿菌的一些基本研究，對分子生物學的發(fā)展起了十分重要的作用。1944DNA證明DNADNA們對DNA化學組成和晶體結構的研究。1953425J.D.沃森和英國的──DNA雙螺旋結構的分子模型。這一成就后來被譽為20最偉大的發(fā)現，也被認為是分子生物學誕生的標志。50年代在蛋白質的結構分析方面也取得了重要成果。英國生物化學家F.桑格第一次分析出含有51個氨基酸的胰島素的氨基酸順序。這一成果對準確地研究蛋白質本身結構和功能之間的關系，以1973300多種蛋白質的氨基酸1977DNA堿基順序的分析方法并完成了分析φχ174噬菌體DNA的全部約5400的布口刺格父子及他們的學生創(chuàng)立并發(fā)展的X佩魯茨自30年代末開始，就系統地研究了血紅蛋白的結構。1969年完成了全部641961年法國細胞遺傳學家雅各布和J.美國分子生物學家H.M.特明和D.巴爾的摩長期從事腫瘤病毒研究的基礎上，于1970年分別獨立地發(fā)現雞肉瘤病毒和白血病病毒都是RNA病毒。在此基礎上他們發(fā)現了依賴于RNADNA聚合酶即反轉錄酶。反轉錄酶能使RNA鏈上的遺傳密碼反轉錄給DNA。這一發(fā)現不僅對某些腫瘤的病因作了分子生物學的闡明，而且動搖了中心法則的不可逆性，成為中心法則的重要補充。真核細胞內的調控機制要復雜得多，也是當前生物學家重點探索的問題之一。在此基礎之上，分子生物學發(fā)展的速度越來越快。一選擇題

第二章 基因的概念ADNAnucleotidemayconsistof DNA核苷酸可能有下列哪一項組成A?aribosesugar,aphosphategroup,andadenine.B?aphosphategroup,deoxyribose,andcytosine.C?uracil,deoxyribose,andaphosphategroup.D?deoxyribose,thymine,andahydroxylgroup.Whichisthemostaccuraterepresentationoftheorganizationlevelsofthegeneticinformationincells? 下列哪一項最準確地代表了細胞中遺傳信息的組織水平A?genesnucleotidechromosomesgenome.B?genomegenesnucleotideschromosomes.C?chromosomesgenesnucleotidesgenome.D?nucleotidesgeneschromosomesgenome.EukaryoticcellsdifferfromprokaryoticcellsinthatonlytheformercontainA?ribosomes.B?cytoplasm.C?DNA.D?anucleus.TheprokaryoticorganismthathasbeenthesubjectofmanygeneticstudiesisA?Saccharomycescerevisiae.B?Neurosporacrassa.C?E.coli.D?Drosophilamelanogaster.WhichelementisnotfoundinamoleculeofDNA?DNA分子中發(fā)現？A?carbon.B?sulfur.C?nitrogen.D?oxygen.WhichisfoundinRNA,butnotDNA?RNADNA分子中出現？A?Phosphate.B?Adenine.C?Ribose.D?Cytosine. ThetwopolynucleotidechainsinamoleculeofDNAareheldtogetherby whattype of DNA分子中的兩條多核苷酸鏈是依賴于哪一種類型的化學鍵結合在一起的？A?Phosphodiester.B?Phosphate.C?Peptide.D?hydrogen.TheDNAthatmakesupbacterialchromosomesA?single-stranded.B?circularandsupercoiled.C?complexedwithhistones.D?alltheabove.WhichofthefollowingisaDNAbasepair?A?A-T.B?T-C.C?A-U.D?G-T.WhichofthefollowingisaRNAbasepair?A?A-T.B?A-C.C?U-C.D?U-A.TheDNAandhistoneproteinsinaeukaryoticchromosomearecompactedintostructurescalled 和組蛋白組成的結構稱為A?proteosomes.蛋白體B?nucleosomes.C?telomeres.端粒D?centromeres.著絲粒Inhistransformationexperiments,FrederickGriffithobservedthatvirulentstrainsofStreptococcuspneumoniaeproducedcolonies.菌落A?smooth,shinyB?rough,dryC?exceptionallylargeD?unusuallycoloredViralgenomesmaybecomposedof病毒基因組的可能組成是A?RNAB?DNAC?eitherRNAorD?bothRNAandDNA.IntheHershey-Chaseblenderexperiments,theirmajorconclusionwasthatHershey 和Chase合實驗得出的主要結論是A?asinglegenedirectedthesynthesisofasinglepolypeptide.B?DNAwasthegeneticmaterial.C?DNAwasadoublehelix.D?thegeneticmaterialwaslocatedinthenucleusofcells.ThetypeofDNAfoundmostcommonlyinlivingcellsisthe form.DNA構型是A?A.B?B.C?D?ZGeneslocatedinwhichregionofaeukaryoticchromosomearemostlikelytobetranscribed? 基因于真核生物染色區(qū)域時最有可能被轉錄A?centromere著絲粒.B?telomere端粒.C?euchromatin常染色質D?heterochromatin異染色質Anucleosideconsistsofa 核苷的基本組成是A?pentosesugarandanitrogenousbase.B?phosphategroupandanitrogenousC?pentosesugarandaphosphategroup.D?pentosesugar,aphosphategroup,andanitrogenousbaseProkaryoticchromosomesconsistmostlyofA?uniquesequenceDNAonly.B?repetitivesequenceDNAC?eitheruniquesequenceorrepetitivesequenceD?bothuniquesequenceandrepetitivesequenceTocreateakaryotype染色體, chromosomesarespreadonaslideandstained.A?interphase.B?telophase.C?metaphase.D?anaphase20AchromosomefromanunknownmicroscopicorganismisexaminedandfoundtocontainonlyuniquesequenceDNA.ThisorganismismostlikelyA?virus.B?bacterium.C?fungus.D?protozoan.Ineukaryoticcells,thegeneticmaterialisfoundintheA?ribosomes.B?nucleus.C?endoplasmicreticulum.D?cytoplasm.Griffith’sexperimentinjectingamixtureofdeadandlivebacteriaintomicedemonstratedthat(choosethecorrectanswer):明（選擇正確的答案）A?DNAisdouble-strandedB?mRNAofeukaryotesdiffersfrommRNAofprokaryotesC?AfactorwascapableoftransformingonebacterialcelltypetoanotherD?BacteriacanrecoverfromheattreatmentiflivehelpercellsarepresentTheX-raydiffractiondataobtainedbyRosalindFranklinsuggested(choosethecorrect羅莎琳德?X射線衍射圖像的數據表明（選擇正確的答案）A?DNAisahelixwithapatternthatrepeatsevery3.4nanometersB?PurinesarehydrogenbondedtopyrimidinesC?DNAisaleft-handedhelixD?DNAisorganizedintonucleosomesGriffithinjectedmicewithdifferenttypesofbacteria.Foreachofthefollowingbacteriatypesinjected,indicatewhetherthemicelivedordied:的細菌類型，判斷老鼠是存活還是死亡？a. typeⅡR b. typeⅢS c. heat-killedⅢS d. typeⅡR+heat-killedⅢSA?lived lived died diedB?died lived died C?lived died lived diedD?died died lived lived二填空題1．基因敲除Geneknock-ou）即是（將特定基因失活的過程，它是研究（基因功能）遺傳學方法。2（pseudogen）因。3．在原核生物的基因表達調控中，因為沒有核膜，（轉錄）和（翻譯）是耦聯的。4．Anucleosomeiscomprisedoftwocopiesofhistones（H2A），（H2B），（H3），（H4），onecopyofhistone（H1），and（200）bpofDNA.三簡答題Fromachemical（bonding）view, whyisdouble-strandedDNAsostable?從化學（鍵）DNA比較穩(wěn)定的原因答案：Hydrogenbondsbetweenbasesandhydrophobicbondsduetobasestacking堿基之間的氫鍵以及疏水的堿基堆積力。Whatis―Chargaff’sRule‖規(guī)則的內容是什么？答案：A%=T%andG%=C%indsDNA.腺嘌呤和胸腺嘧啶的摩爾數相等，即A=T；鳥嘌呤和胞嘧啶的摩爾數相等，即G=C；含氨基的堿基（A和C）總數等于含酮基的堿基（G和T）總數，即A+C=G+T；嘌呤的總數等于嘧啶的總數，即A+G=C+T。SupercoilingDNArequiresenergyandallowsworktobedone.Whataretwobiochemicalfunctionsthatthispent-upenergyisusedfor?DNA中的能量的主要用于哪兩種生化作用？答案：Energycancausestrandseparationusedfor:（1）DNAreplication；（2）TranscriptionYouhaveisolatedaplasmidDNAthatisaclosedcircularmolecular1050bpinlengthwith5negativesupercoils. Whatarethelinkingnumber,helicalturns,writhe,andsuperhelicaldensity? 你分離出的質DNA1050bp5DNA螺旋數（纏繞數T）、扭曲數、以及超螺旋密度是多少？答案：Writhe=W=-5；helicalturns=T=1050/10.5=100；L=T+W=100+（-5）=95Superhelicaldensity=σ=W/T=-5/100=-0.05Whatisapseudogene,andhowdoyourecognizeone?什么是假基因？你如何識別出假基因？答案：Psuedogenesarearedefinedbytheirpossessionofsequencesthatarerelatedtothoseofthefunctionalgenes,butthatcannotbetranslatedintoafunctional假基因是指與正?；蚪Y構相似，DNA序列，即不能翻譯出有功能蛋白質的基因。canberecognizedbytheoccurrenceofoneormoremutationsthatobviouslyrenderthem可以利用補償其功能的突變體而區(qū)分出假基因。Ineukaryoticcells,DNAispackagedintochromatin.Therepeatingunitofchromatiniscallednucleosome.被包裝成染色質，其重復單元為核小體。Whatconstituteamononucleosome?每個核小體單元是由什么組成的？WhatistheconsequenceofDNApackagingontranscription?DNA怎樣的影響后果？答案1Amononucleosomeiscomposedof146bpofcoreDNAwrappedaroundahistoneoctamerof2copiesofeachofH2A,H2B,H3andH4.146bp核心DNA和各兩分子的組蛋H2AH2BH3、H4組成的八聚體構成。（2）TheconsequenceofDNApackagingistoinhibittranscriptionduetotheinabilityoftranscriptionmachineryandtranscriptionfactorstogainaccesstoDNA.DNA被包裝成染色質結構后，會抑制轉錄。因為這一結構能夠阻止轉錄機構及轉錄因子與DNA之間的相互接近。Whichcombinationsofhistonesform“histone-fold”dimerswitheachotherinthenucleosome?在核小體中，每個組蛋白二聚體分別由哪幾種組蛋白組成？H3andH4formonedimerpair,H2AandH2Bformtheother.組蛋白H3和H4形成一個二聚體；組蛋白H2A和H2B形成另一個二聚體。ThroughX-raydiffractionanalysisofcrystallizedDNAoligomers,differentformsofDNAhavebeenidentified.TheseformsincludeA-DNA,B-DNA,andZ-DNA,andeachhasuniquemolecularattributes.DNAX-射線衍射分析，鑒定出了DNADNA不同的構型即A-DNA、B-DNA、Z-DNA，每一種構型各有其獨特的分子特征。oftheseformsisthemostcommonforminlivingcells?在活的生物細胞中，普遍存在的分子構型是哪一種？（2）Z-DNAhasanunusualconformationresultinginmorebasepairsperhelicalturnthanB-DNA.Whatistheconformation?Doesthismoleculehaveanyfunctioninlivingcells?Z-DNA與B-DNA相比，每圈螺旋含有較多的堿基對數，請進一步說明Z-DNA具有怎樣的功能？（3）Whichoftheseformsisneverfoundinlivingcells? 現。答案：（1）B-DNAistheformmostcommontolivingcells.B-DNA是活細胞中DNA的普遍存在形式。（2）Z-DNAisalongandthin(about2nmwide)doublehelix,likeB-DNA.Unliketheright-handedB-DNAhowever,Z-DNAisleft-handed.Ithasanaxisthatrunsthroughtheminorgroove,andhas12basepairsperhelicalturnthatareinclined8.8°fromaplaneperpendiculartotheaxis.Incontrast,B-DNAhasanaxisthatrunsthroughthebasepairs,andhas10basepairsperhelicalturnthatareinclined2° fromaplaneperpendiculartotheaxis.ThemajorgrooveofZ-DNAisnotverydistinct,asitisthinandflattenedoutalongthehelixsurface,whilethemajorgrooveofB-DNAiswideandintermediateindepth(betweenthatofA-DNAandZ-DNA).TheminorgrooveofZ-DNAisextremelynarrowandverydeep,whilethatofB-DNAisnarrowandofintermediatedepth. IthasbeenproposedthatregionswithZ-DNAprovideastretchofleft-handedhelicalturnsthatareinvolvedinreplication,recombinationandtranscription.Stretchesofleft-handedturnsmayaidinunwindingright-handedhelicalturnsinB-DNAduringtheseprocesses.Z-DNAmayalsobemorestableunderextremeenvironmentalconditions. 與B-DNA相比，Z-DNA比較A-DNAisfoundonlywhentheDNAisdehydrated,soitisunlikelythatlengthysectionsofA-DNAwouldbefoundinlivingcells.Benzer用一般遺傳學方法測出T4rcistron中含有許多個突變子（或重組子）并且指出一個突變子的大小是1-3個核苷酸，后來證明這是科學上的一個驚人的預見。請回答：?你能說出在當時的條件下（沒有DNA序列分析技術，遺傳密碼還沒發(fā)現）1~3個核苷酸的結論。答案：基因是DNA基因內可以較低頻率發(fā)生基因內的重組，交換。通過大量的成對突變型的雜交，測得其最小的重組頻率為0.02%，即0.02個遺傳圖距。已知T415000.02少核苷酸：1.8×105÷1500×0.02=2.4bp因此指出一個突變子是1-3個核苷酸，暗示了三聯體密碼的存在。比較基因組的大小和基因組復雜性的不同：一個基因組有兩個序列，一個是A，另一個是B2000bp長。其中一個是由400bp的序550bp40樣？（2）這個基因組的復雜性如何？DNA（1）這個基因組的大小為4000bp；（2）這個基因組的復雜性為450bpWhatevidencedowehavethatinthehelicalformoftheDNAmoleculethebasepairsarecomposedofonepurineandonepyrimidine?我們有什么證據表明在DNA分子的螺旋構型中，在每一個堿基對中都含有一個嘌呤堿基和一個嘧啶堿基？答案：Twodifferentlinesofevidencesupporttheviewthatabasepairiscomposedofonepurineandonepyrimidine.有兩方面的證據支持在DNAWhenthechemicalcomponentsofdouble-strandedDNAfromawidevarietyoforganismswereanalyzedquantitativelybyChargaff,itwasfoundthattheamountofpurinesequaledtheamountofpyrimidines.Morespecifically,itwasfoundthattheamountofadenineequaledtheamountofthymine,andthattheamountofcytosineequaledtheamountofguanine.Thesimplesthypothesistoexplaintheseobservationswastheexistenceofcomplementarybasepairing,AononestrandpairedwithTontheotherstrand,andGpairedwithC.MoredirectphysicalevidencewasprovidedbyX-raydiffractionstudies. TheseestablishedthedimensionsoftheDNAdoublehelixandallowedforcomparisonwiththeknownsizesofthebases. diameterofthedoublehelixisconstantthroughoutitslengthat2nm. Thisistherightsizeaccommodateapurinepairedwithapyrimidine,buttoosmallforapurine-purinepair,andtoolargeforapyrimidine-pyrimidinepair.Thedouble-helixmodelofDNA,assuggestedbyWatsonandCrick,wasbasedondatagatheredonDNAbyotherresearchers.Thefactsfellintothefollowingtwogeneralcategories;givetwoexamplesofeach:(1)chemicalcomposition;(2)physicalstructure. WatsonCrickDNA雙螺旋模型，這些研究成果可以劃分為兩類即DNA的化學組成與DNA的物理結構。請詳細闡述這兩類研究成果的具體結論。答案：Thedouble-helixmodelofDNAsuggestedbyWatsonandCrickhadtoincorporateexistinginformationaboutitschemicalcompositionandphysicalstructure.Intermsofitschemicalcomposition,itwasknownthatDNAiscomposedofpolynucleotides,that(A)=(T)and(G)=(C)(Chargaff'srules),andthatwhilethepercentGCvariesbetweenorganisms,theA/TandG/Cratiosdonot.Intermsofitsphysicalstructure,thestructureandmoleculardimensionsofthecomponentmolecules(thebases,sugars,phosphates)wereknown. ItwasalsoknownfromstudiesofFranklinandWilkinsthatthemoleculeisorganizedinahighlyordered,helicalstructure,andthattherearetwodistinctiveregularitiesat0.34and3.4nmalongthemolecule'saxis.化學組成即DNA堿基組成的Chargaff規(guī)則：腺嘌呤和胸腺嘧啶的摩爾數相等，即A=T；鳥嘌呤和胞嘧啶的摩爾數相等，即G=C；含氨基的堿基C）總數等于含酮基的堿基T）總A+C=G+TA+G=C+TDNA異性。物理結構：DNA分子是由核苷酸組成，核苷酸有含氮的堿基、戊糖、磷酸構成。根據富蘭克林和威爾金斯的x射線衍射圖像表明：DNA分子是一個十分有序的雙螺旋結構，每兩個相鄰堿基平面的垂直距離是3.4?，每個螺旋包含10個堿基對。Hershey-Chase32P只標記在DNA35S只標記在35S35S標記的32P標記的噬菌體重復實驗，那么在子代病毒中是否可以找到帶32P標記的病毒？答案：因為DNA32P只標記在DNA35S只標記在蛋白質的外殼上。35S35S因此不帶標記。32P35P35P標記的病毒。Theco-crystalstructureofORC（originrecognitioncomplex）boundtoDNAshowsthat200bpofDNAiswrappedin3fullright-handedturnsaroundtheyeastORCcomplex.YouassembleORContoacircular8kbplasmidcontaininganARS（autonomouslyreplicatingBeforeassembly,theplasmidhadanaverageof10negativesupercoils.Afterassembly,youextractthereactionwithphenol-chloroform,recovertheDNA,andrunitonagel.(Hint:DNAiswrappedinaleft-handedturnaroundnucleosomes).PleaseWhatisthelinkingnumber,twist,andwritheoftheplasmidafterextraction(assume10.5bp/turnofrelaxedBDNA).答案：W=-10（tennegativesupercoils）T=8000bp/10.5bpperturn=761or762L=T+W=751or752WheredoestheextractedDNAmigrateonanagarosegellackingethidiumbromide?答案：AtthepositionofsupercoiledDNA.Whatwillhappentothemobilityoftheplasmidasyouaddincreasingamountsofethidiumbromidetothegelduringelectrophoresis?Explainbriefly.SinceETBrlocallyuntwistsDNA(twistgoesdown),theplasmidwillacquirepositivesupercoils(writhegoesup).Atfirst,thiswillresultinthelossofnegativesupercoilsandtheplasmidwillmigratelessrapidly.IfyouaddenoughETBr,theplasmidwillbecomepositively supercoiledandmigrateatthepositionofsupercoiledDNA.Nowyourepeattheexperiment,butyouincludeTopoisomeraseIintheassemblyreaction.Undertheseconditions,whatisthelinkingnumber,twist,andwritheoftheplasmidafterextraction?答案：W=+3T=8000bp/10.5bpperturn=761orL=764or765The10negativesupercoilspresentintheplasmidwillbeinstantlyremovedbytopoisomeraseI,butthesolenoidalsupercoils,constrainedbyORC,willnot.AssumingthateachORCwillgenerate3positiveplectonemicsupercoils,afterproteinextraction,W=+3,T=761(sinceDNAalwaysadoptsthemostfavorablevalue),L=T+W=764.Inadditiontojuststatingthenameofthestrain,wewerelookingforatleastsomeexplanationofhowitwouldwork.選擇題

第三章 DNA復制ThroughtheirexperimentswithDNAfromthebacteriumEscherichiac,eselsonandStahlshowedthatDNAreplicationDNA和StahlDNA復制是A?conservative.B?semi-conservative.C?DuplicativeD?dispersive.AttheconclusionofDNAreplication,thetworesultingDNAdoubleheliceseachDNA的復DNA雙螺旋的兩條鏈分別是A?oneparentalandoneprogenystrand.B?twoparentalortwoprogenystrands.C?stretchesofprogenyDNAinterspersedwithparentalDNAalongbothstrands.D?twonewlysynthesizedstrands.DNApolymerasesareenzymesthatcopyA?DNAintoB?DNAintoRNAC?RNAintoDNAD?RNAintoRNA.TobeginDNAreplication,ashort primermustfirstbeproduced.為了能夠起始DNA的復制，一段短的 引物必須預先合成A?DNAB?RNAC?polypeptideD?histoneWhichE.coliDNApolymerasehastheabilityto―proofread‖newlysynthesizedDNAandremoveerroneousbases?在大腸桿菌中，哪一種DNA聚合酶對新合成的DNA具有校正功能，能把錯配的堿基移去？A?DNApolymeraseIonlyB?DNApolymeraseIIIonlyC?DNApolymeraseIandIIID?allthreeDNApolymeraseshaveproofreadingabilityDuringsynthesis,allDNApolymerasesaddnucleotidesinwhichdirection?A?fromlefttorightB?from3’to5’C?from5’to3’D?inmorethanonedirectionatatimeIneukaryotes,DNAreplicationoccursduringwhichphaseofthecellcycle?A?B?G1C?G2D?MWhichofthefollowingisnotrequiredforDNAsynthesisreactions?A?dCTPsB?templateDNAC?DNApolymeraseD?calciumionsDNApolymerasecatalyzestheformationofaphosphodiesterbond betweenA?5’phosphateanda5’hydroxylgroup.B?3’phosphateanda5’hydroxylgroup.C?5’phosphateanda3’hydroxylgroup.D?3’phosphateanda3’hydroxylgroup.Thesequenceofnucleotides inonestrandofDNAis 5’-CCACTGG-3’,WhatisthesequenceofthecomplimentarystrandofA?5’-CCACTGB?3’-CCACTGC?5’-GGTCACC-3’D?3’-GGTGACInbacteriasuchasE.coli,replicationofthechromosomeisA?semidiscontinuousandbi-directional.B?discontinuousandunidirectional.C?continuousandbi-directional.D?semidiscontinuousandunidirectional.The3’5’exonucleaseactivityassociatedwithDNApolymerasereducesthefrequencyofreplicationerrorstoA?1/10.B?1/1,000.C?1/1,000,000.D?1/1,000,000,00.TheproofreadingactivityofDNApolymeraseremoveserrant nucleotidesfromthe ofastrand.A?3’endB?5’endC?3’and5’endD?middleOntheE.colichromosome,oriCA?encodesDNApolymeraseI.B?isabindingsiteforhistoneproteins.C?isthestartsiteforD?encodesanRNAprimer.TheenzymethatunwindsthedoublehelixtofacilitatereplicationisA?3’5’B?DNAhelicase.C?DNApolymerase.D?topoisomerase.WhentheDNAdoublehelixisreplicated,thenewlysynthesized5’ 3’strandisconsideredthe strand.A?B?laggingC?templateD?discontinuousSynthesisofthelaggingstrandA?continuously.B?conservatively.C?discontinuously.D?semidiscontinuously.WhichtypeofDNAisduplicatedbyrollingcirclereplication?A?bacteriophageλB?plasmidDNAC?bacteriophageΦX174D?alloftheaboveManytypesofmammaliancancercellsarenotablefortheirA?telomeraseactivity.B?lackoftelomeraseactivity.C?lackoftelomeres.D?increasednumberoftelomeres.TodeterminethenumberofreplicationsitesinE.coliandwhetherreplicationisunidirectionalorbidirectional,youexaminedtheresultsoftwodifferentexperiments.BothexperimentsinvolvedgrowingE.coliinamediumcontainingradioactivethymidine.Whatdidtheadditionofthymidinetothemediumallowyoutoobserveinbothexperiments? DNA什么？A?ThedifferencebetweentheleadingDNAstrandandthelaggingDNAstrand.B?ThedifferencebetweenthereplicatedandunreplicatedportionsofDNAC?ThedifferencebetweentheRNAprimerandthenewlysynthesizedDNA.ThedifferencebetweenthereplicationforkandthenewlysynthesizedDNA.D?Noneoftheabove.Basedontheexperimentsinthisactivity,whichofthefollowingistrueaboutE.colireplication?根據DNA復制的描述，下列那一項是正確的？A?Replicationbeginsatasinglesiteonthechromosome.B?Replicationisbidirectional.C?Synthesisbeginsatspecificsitesonthetemplatestrand.D?Alloftheabove.Duringreplication,proofreadingofthenewlysynthesizedDNAisperformedbyA?RNApolymeraseB?reversetranscriptaseC?topoisomeraseD?DNApolymerase尿嘧啶糖苷酶的功能是A?去除嘧啶二聚體B?切除RNA分子中的尿嘧啶C?切除DNAD?切除DNA分子中的尿苷酸E?切除RNA分子中的尿苷酸 DNADNA白質？A?DNA聚合酶Ⅰ、引發(fā)酶、SSB和連接酶B?SSB、解鏈酶、和拓撲異構酶C?連接酶、DNA聚合酶Ⅰ和ⅢD?DNASSBEDNA聚合酶ⅡDNA復制的幾種酶的作用次序是A?DNA解鏈酶→引發(fā)酶→DNA聚合酶→DNA連接酶→切除引物的酶B?DNAC?引發(fā)酶→連接酶→切除引物的酶D?DNA聚合酶E?DNA連接酶→切除引物的酶將兩段寡聚脫氧核苷酸片段和與DNA聚合酶一起加到含dGTPdCTPdTTPA2C∶1TB?1G∶1TC?3G∶2TD?E?5T∶4G∶3C∶1A噬菌體Φx174的基因組是一個由5386個堿基組成的單鏈DNA，該基因組編碼有種不同的蛋白質，這些蛋白質約有2380能力。對這種現象最好的解釋是A?它的基因組含有重疊基因B?氨基酸由二聯體密碼編碼C?細胞核糖體翻譯它的每一個密碼子不止一個D?蛋白質在使用后即被后加工二填空題DNA復制的方向是從端到端展開。維持DNA復制的高度忠實性的機制主要有聚合酶的高度選擇性（DNA校正功能）和（錯配修復。端聚酶由和（蛋白質）兩個部分組成，它的生理功能是（維持端粒的完整。染色體中參與復制的活性區(qū)呈Y型結構，稱為（復制叉三簡答題Explainthefollowingtermsandconcepts：請解釋下列術語或概念：leadingstrandandlaggingstrand前導鏈和滯后鏈；Sigmafactorandholoenzymeσ因子與全酶。strandistheDNAstrandthatsynthesizedinthe5′to3′directionastheparentalduplexisunwound.ItisprimedonetimeduringsynthesisandusestheDNAstrand3’to5’astemplate.前導鏈是指隨著親本雙螺旋的解開而按照5′3′方向連續(xù)合成的DNA子鏈，以方向的親本鏈為模板。LaggingstandusestheDNAstrandthatruns5’to3’astemplateandsynthesisDNAnon-continuouslyasOkazakifragment.It'ssynthesisrequiresmultipleprimingevents.Aseriesofthesefragmentsaresynthesized,each5′–3′;thentheyarejoinedtogethertocreateanintactlaggingstrand.滯后鏈是指以5′3′方向的親本鏈為模板，不連續(xù)合成一系列5′3′方向岡崎片斷，然后連接成一條完整的子鏈DNA。滯后鏈的合成需要若干次引發(fā)事件的發(fā)生。SigmafactorandholoenzymeSigmafactorisacomponentofthebacterialRNApolymeraseholoenzymewhichiscomposedofRNAcoreenzymeandSigmafactor.Sigmafactorisresponsibleforpromoterrecognitionandtranscriptioninitiation,whileRNAcoreenzymeisresponsibleforRNAsynthesis.σisaSpecificityFactor，whichdirectsthecoretotranscribespecific因子是RNA聚合酶全σσRNA核心酶負責RNA的合成。是一種特異性因子，能夠指導核心酶轉錄特異的基因。Whatisthetemplateusedbytelomerasetoaddtelomericrepeatsattheendsofchromosomes?端粒酶在向染色體末端添加寡聚重復單元時，以什么為模板？TelomerasecarriesitsownRNAtemplateforpolymerizationofthetelomereDNArepeats.Thatmeanstelomeraseisareversetranscriptase,i.e.anRNA-dependentDNApolymerase.端粒酶以自身攜帶RNADNARNADNA聚合酶。3．AspaceshiplandsontheEarthcarryingasampleofextraterrestrialbacteria.YouareassignedthetaskofdeterminingthemechanismofDNAreplicationinthisorganism.Yougrowthebacteriainunlabledmediumforseveralgenerations,thegrowitinpresenceof15Nforexactlyonegeneration.YouextracttheDNAandsubjectittoCsClcentrifugation.Thebandingpatternyoufindisasfollows:在利用飛船進行科DNA有標記的培養(yǎng)基上生長幾代15N標記的培養(yǎng)基中準確地繁殖一代DNA并進行CsClItappearstoyouthatthisevidencethatDNAreplicatesinthesemiconservativemanner,butyouarewrno.rohy?Whatotherexperimentcouldyouperform(usingthe15N15N14N14N ExperimentalsamplesamesampleandtechniqueofCsClcentrifugation)thatwouldfurtherdistinguishbetweensemiconservativeanddispersivemodesofreplication?根據這一結果，你認為DNA行復制，但你的觀點是不正確的，為什么？請你設計另一個實驗方案（手段即CsCl梯度離心）能夠有效地將半保留復制和彌散復制這兩種復制方式區(qū)分開來。TheCsClcentrifugationresulteliminatesthepossibilityoftheconservativemodelofreplication,butisstillconsistentwitheithersemiconservativeordispersivemodelsofDNAreplication.TodistinguishbetweenthesetwopossibilitiesusingthesamesampleandthetechniqueofCsClcentrifugation,onecoulddenaturetheDNA,andthensubjectthesingle-strandedsampletoCsClcentrifugation.ThiscouldbedoneinpracticebyusinganalkalineCsClgradient,asthetwoDNAstrandswilldenatureathighpH.Theexpectedresultsareshownbelow.CsCl梯度離心的結果雖然排除了全保留復制的可能性，但不能根據這一結果就可證明DNA相同的技術手段即CsCl梯度離心將半保留復制和彌散復制區(qū)分開來的方法是將DNA樣品變性，然后利用堿性CsCl梯度離心技術對單鏈DNA則出現一條帶，如下圖所示：

半保留復制變性后離心半保留復制變性后離心彌散復制變性后離心變性后離心彌散復制變性后離心KornbergisolatedDNApolymeraseⅠfromE.coli.DNApolymeraseⅠhasanessentialfunctioninDNAreplication.WhatarethefunctionsoftheenzymeinDNAreplication?科恩伯格從大腸桿菌中分離出了DNA聚合酶Ⅰ，在DNA聚合酶Ⅰ在DNA復制過程中具有什么功能？答案：DNApolymeraseⅠfunctionstoremovetheRNAprimersynthesizedduringtheinitiationofDNAreplicationandreplacethisRNAwithDNA.WhenDNApolymeraseⅢ,themainsyntheticenzymeforDNApolymerization,reachesanRNAprimer,itdissociatesfromtheDNA.DNApolymeraseⅠfunctionstocontinuesynthesisoftheDNAina5’-to-3’direction.ItsimultaneouslyremovestheRNAprimerusingits5’-to-3’exonucleaseactivity,andreplacestheRNAwithDNAnucleotides.DNA聚合酶ⅠDNARNA引物的功能，并合成一段DNARNADNA聚合功能的DNA聚合酶Ⅲ在遇到RNA引物時，就會從DNA5’，3’核酸外切酶的功能切除RNA5’3’方向合成DNA。DescribethemolecularactionoftheenzymeDNAligase.WhatpropertieswouldyouexpectanE.colicelltohaveifithadatemperature-sensitivemutationinthegeneforDNAligase?請描述DNA連接酶的DNA生怎樣的表型效應？答案：DNAligasecatalyzestheformationofaphosphodiesterbondbetweenthe3’-OHandthe5’-monophosphate groups on either side of a single-strand DNA gap, sealing the gap.Temperature-sensitiveligasemutantswouldbeunabletosealsuchgapsattherestrictive(high)temperature,leadingtofragmentedlaggingstrands,andpresumablycelldeath.IfabiochemicalanalysiswereperformedonDNAsynthesizedafterE.coliwereshiftedtoarestrictivetemperature,therewouldbeanaccumulationofDNAfragmentsthesizeofOkazakifragments. ThiswouldprovideadditionalevidencethatDNAreplicationmustbediscontinuousononestrand.什么是DNA聚合酶的進行性？如何測定一種DNA聚合酶的進行性？答案：DNA聚合酶的進行性是指某一種DNA解離的這段時間內，催化了多少脫氧核苷酸的摻入。測定某一DNA特異性DNA，當DNA聚合酶從模板上解離下來后，由于大量的非特異性DNA稀釋了原來的DNA模板，使得DNA聚合酶很難再與原來的模版結合，這樣就得到了新合成的DNA泳可大概知道片段的長度。盡管DNA聚合酶催化聚合反應既需要模板，又需要引物。但下面的單鏈DNA卻可以直接作為DNA聚合酶Ⅰ的有效的底物。試解釋其中的原因，并寫出由DNA結構。3’OH-TGGCTCATAGCCGGAGCCCTAACCGTAGACCACGAATAGCATTAGGp--’-5答案：由于此鏈DNA特殊的堿基組成，使得該DNA能夠形成如圖所示的莖環(huán)結構：T A OH3’G GAGCTAACCGTAGACCACGAATAGCATTAGG5’C C GDNA3’-OHDNA的外切酶活性將末端錯配的T水解掉而引入正確的G，然后再進行延伸反應，最終得到的產物是： T A CTCGGATTGGCATCTGGTGCTTTCGTAATCC-OH’GC C G

GAGCCCTAACCGTAGACCACGAATAGCATTAGGp5’簡述維持DNA復制的高度忠實性的機制。答案：維持DNA聚合酶所具有的3’5’RNADNA復制的忠實性，因為當DNARNA誤。在岡崎提出DNADNA是僅僅是后隨鏈才是不連續(xù)復制，存在著很大爭議。顯然前導鏈的合成不需要不連續(xù)復制，但是許多研究者發(fā)現經脈沖標記的岡崎片段可以和親代DNADNA的兩條鏈都可以作為岡崎片段的模板。試提出一種機制解釋前導鏈的復制也可能產生岡崎片段，并設計一個實驗證明你提出的機制。答案：前導鏈似乎也形成小的片斷，是因為在一個細胞中，含有微量的dUTP，它在DNA復制過程中有可能代替dTTP摻入到新合成的子鏈中，由于脫氧尿苷酸不是DNA分子中的正常核苷酸，所以體內修復系統會將它除去。去除的機制是先在尿嘧啶糖苷酶的作用下切除DNA然后AP核酸內切酶在AP位點將DNAAP位點在內的一段DNA鏈切除，再由DNAAP核酸內切酶的作用必然導致前導鏈形成小的DNA片斷。篩選AP核酸內切酶缺失的大腸桿菌突變株，觀察上述現象是否發(fā)生。Telomeresareuniquerepeatedsequences.WhereontheDNAstrandaretheyfound?Dotheyserveafunction?端粒是一類特殊的重復序列。端粒序列存在于DNA鏈的什么部位？具有怎樣的功能？答案：Telomeresarecharacteristicallyheterochromaticsequencesfoundattheendsofalinear,eukaryoticchromosome.Formostorganisms,thetelomericsequencesattheextremeendofachromosomearesimple,highlyrepeated,andspeciesspecific.Thesesequencesaresynthesizedbytheenzymetelomerase.Nearby,butnotattheveryendofachromosome,aretelomere-associatedsequences.Thesearerepeated,co