大二下學(xué)習(xí)分子試題2004exam-anwser

1、武漢大學(xué)生命科學(xué)學(xué)20032004 學(xué)年第一學(xué)期期末考武漢大學(xué)生命科學(xué)學(xué)20032004 學(xué)年第一學(xué)期期末考FinalexamofMolecularBiologyCourse(Spring寫(xiě)在參考答案前面的BestwishestoallofPARTI:DESCRIPTION(2pointsYouranswershoulddescribewhateachitemisandhowit functionsinthecell.Diagrams, structure and sequence information should be included in your answer, as necess

2、ary.Yeastartificial酵母人工染色體（0.5containscomponentsrequiredforreplicationandsegregationofthenaturalyeast chromosome, including two telomeric sequences (TEL), one centromere (CEN) and one autonomously replicating sequences (ARS) (0.75 point )containsgenesactasselectivemarkersinyeastandproperrestrictions

3、ites(0.75 point )can accommodategenomic DNAfragmentsofmore than1 Mb (0.5RNARNA（0.5 aconservedbiologicalresponsetodouble-strandedRNA(1regulatestheexpressionofprotein-codinggenesthroughsiRNAormiRNA(1 thedsRNAisrestrictedbyDICER, thenRISC mediatesthesiRNAandmiRNA related RNA degradation and translation

4、 inhibition, respectively. (1 point)（0.5point）Thestudyoftheproteomeusingtechniquesofresolutionproteinseparationandidentification(1.5The best separation method is two dimensional gel electrophoresis, the individual protein spotsarethencutfromthegelandtreatedwithproteasetoproduce aset of peptides char

5、acteristic of that protein. The precise masses of each peptide in the sample are then determined by MALDI mass spectrometry. The resulted peptide mass fingerprint of that protein is then compared to a database to deduce the function of that protein etc. (1.5 point)Shine-DalgarnoSD （0.5Aconservedsequ

6、ence8-13ntupstreamofthefirstcodontobetranslated(1 ThissequencewasdiscoveredbyShineandDalgarno(0.5Thesequenceispurine-richandcontainsallorpartofThesequenceispurine-richandcontainsallorpartof5-AGGAGGU-3(0.5 Canbase-pairwiththe3-endofthe16SrRNA(0.5point) Alternative splicing（0.5 Thegenerationofdifferen

7、tmature mRNAsfromaparticulartypeofgene transcript by choosing different 5- and 3-splice sites (2 point).（0.5 an RNAmoleculecapable of catalyzing achemical reaction (2-dependent-（0.5 During bacterialtranscription,some terminator sitesdo notform stronghairpins, thus termination of the transcription by

8、 bacterial RNA polymerase requirestheassistanceofanaccessoryfactorcalledrho(protein(2RNARNA（0.5 A form of RNA processing in which nucleotide sequence of the primary transcript isaltered byeitherchanging,insertingordeletingresiduesatthe specific points along the molecule (2 point).DNADNA（0.5Analterat

9、ion tothe normal chemicalor physicalstructureoftheDNA(2ThelesionscanleadtocelldeathorDNAmutation(1point). Protein targeting（0.5 Synthesis of eukaryotic proteins is usually occurred in cytoplasm (0.5 point). However,manyproteinsneedtobetransportedtospecificcellularlocations,such as nucleus, mitochond

10、rion or chloroplast, to exert their biological functions. This process is called protein targeting(1 point).Theultimatecellularlocationofproteinsisoftendeterminedbyspecific,relative short, aminoacid sequenceswithin the proteinsthemselves. The sequence inside of a protein determining the cellular loc

11、ation of the protein is called signal sequence (1 point).PARTII:MULTIPLECHOICES(1pointsSelecttheonebestanswerforeach1. Thecatalyticactivityforpeptidebondformation(thepeptidyltransferaseactivity)is located in the:RNA of the large ribosomal subunit. leadersequenceofthemessengerRNA. RNA of the small ri

12、bosomal subunit. proteinsofthesmallribosomalsubunit. proteins of the large ribosomal subunit.2. Bidirectionalandsemi-conservativearetwotermsthatreferall of the above. noneoftheabove.3. Thefact thatmost aminoacidsare specified bymultiplecodonsisknown all of the above. noneoftheabove.3. Thefact thatmo

13、st aminoacidsare specified bymultiplecodonsisknown the“wobble”theuniversalityofthegeneticcode. codon bias.the anticodon theredundancyof thegenetic 4.RNApolymeraseIistheeukaryoticenzymeresponsibletranscriptionofribosomaltranscriptionoftransferRNAandothersmallRNAspecies. transcription of messenger RNA

14、.initiationofOkazakifragmentsynthesisinDNA5.RestrictionenzymescancleaveDNAthatiseithersingle-strandedordouble-stranded, as long as it contains the appropriate recognition site.1) 2) 6. Informationabout thesequenceof thecodingregion of ageneisbest obtainedaYACagenomicclone. a cDNA clone. the protein.

15、7. Achromatographymethodthatcanbeusedspecificallytopurifyproteinsbasedontheir charge is:gel filtration chromatography. ion-exchangechromatography. DNAaffinityantibody affinity8. AnonsensemutationisachangeintheDNAsequencethat resultsasmall deletion oranamino acid changeinthe protein encoded bythe apr

16、ematurestopall of the above. noneoftheabove.9. Aprotein complexinvolved indegradationof proteins within thecellisknownasmolecular chaperone. Krebs/TCAcycle.10. bindstotherepressorandturnonthetranscriptionofthestructuralgenesinthe Lac operon.11. WhichofthefollowingRNAspeciesisinvolvedindegradationoft

17、he11. WhichofthefollowingRNAspeciesisinvolvedindegradationofthemRNA containing complementary sequence5SRNAU312. Thegenomesequencingprojectsareconfirmingthetheorythatgenomesizeisdirectly proportional to the number of genes contained within that genome. In other words, a genome that is 10 times as big

18、 will contains approximately 10 times as many protein coding genes.1) 2) 13. HeLacells, derivedfromahumancervicalcarcinoma, areabletopropagate indefinitely in culture and are therefore known as a(n):tissueculture. transgeniccellimmortalizedcell14. E. coli cellsaresmallerthanyeast1) 2) 15. Which of t

19、hefollowingdomainsisnota DNAbinding Proline-rich domains Helix-turnhelixdomains Zinc finger domains Basic domains16.Theaminoacyl-tRNAsynthetasesdistinguishbetweenabout40differentshapedtRNA molecules in the cells.1) 2) PARTIII:SHORTQUESTIONS(8points1. Howdobacterialreplicationstartandaccomplished. Re

20、membertoincludethe proteins/enzymes and important DNA sequence involved in this process. Initiation (3 points):replication of the bacterial chromosome is tightly coupled to the growth cycle. TheE. colioriginiswithinthegeneticlocus oriCthatcontainsfour9bpbinding sites for the initiator protein DnaA.S

21、ynthesisof DnaAiscoupledtogrowthratesothatthereplicationofthebacterial chromosome is coupled to the growth cycle.Once the cellular level of DnaA reaches a critical level, DnaA protein forms a complexof30-40moleculesattheoriCDNA,whichfacilitatesmeltingofthree13bpAT-richrepeatsequencesto13bpAT-richrep

22、eatsequencestoallowbindingofDnaBDnaBprotein isaDNAhelicasethatutilizestheenergyofATPhydrolysisto melt dsDNA.ThessDNAcreatedbyDnaBiscoatedwithsingle-strandedbindingprotein(Ssb) to protect it from breakage and to prevent the DNA renaturing.TheDNAprimasethenattachestotheDNAandsynthesizesashortRNA prime

23、r to initiate synthesis of the leading strand of the first replication fork.Unwinding(1Forreplicationtoproceedawayfromtheorigin,DNAhelicasesmusttravel along the template strands to open the double helix for copying, which is accomplished by the joint efforts of DnaB and Ssb.Unwindingcausespositivesu

24、percoilingoftheunwoundDNA;thepositive supercoilingisrelaxedcontinuouslybytheintroductionoffurthernegative supercoils by type II topoisomerase called DNA gyrase.Elongation(3Initiationofthereplicationofthelaggingstrand.Asthenewlyformedreplication fork displaces the parentallagging strand,a mobilecompl

25、ex called a primosome, which includes the DnaB helicase and DNA primase, synthesizes RNA primers every 1000-2000 nt on the lagging strand.BothleadingandlaggingstrandprimersareelongatedbyDNApolymerase III holoenzyme. This multisubunit complex is a dimmer, one half synthesizing the leading strand and

26、the other the lagging strand.Lagging strand synthesis. DNA polymerase III holoenzyme: subunit -the actual polymerase,-a35proofreadingexonuclease.DNApolymeraseIremovesthe lagging strand primers and fills in the resulted gaps. DNA ligase makes the final phosphodiester bond between fragments.Terminatio

27、nandsegregation(1Tworeplicationforksmeetat theterminatorsitestusgeneproductbindstotheterminusandactsasaninhibitorof DnaB Whenreplicationiscompleted,thetwointerlinkeddaughter circlesare unlinked by topoisomerase IV.2. Design experimentsto cloneayeastgeneandexpressthisgenein Clone(4 PCRamplificationof

28、thegeneofinterest,includingtwoproperrestrictionsitesat the ends of the PCR amplified DNA Choose a cloning vector: it could be a regular cloning vector that is capable of propagatinginE.coli,easytoextractfromE. coliandmanipulatedintesttubes. It could also be a shuttle and expression vector that also

29、contains the yeast 2 origintoallowtheplasmidreplicationinyeast,theselectivemarker toallowtheplasmidcontainingcellsgrow,thepromotertoallowthegeneofinteresttobe transcribed, a poly(A) site for addition of poly(A) tail on the transcript.Nomatter whattypeofplasmidcontainingcellsgrow,thepromotertoallowth

30、egeneofinteresttobe transcribed, a poly(A) site for addition of poly(A) tail on the transcript.Nomatter whattypeofthevectorischosen, plasmidshallbepreparedfrom theE. coli cells.Restrictiondigestionofbothplasmidandthegeneofinterestwiththesame restriction sites.Ligation and Selectionandidentificationo

31、fthetransformantstoobtaintheclonecontainingthe right recombinant plasmid.Expression(4If a shuttle and expression vector is chosen as described in step, the expression process will include 1preparation of the recombinant DNA from E. coli cells, 2transformation of yeast cells with the recombinant DNA,

32、 3selection of the transformants, 4growth of the transformants, 5expression of the gene by induction ofthetranscriptionfromthepromoterand6detectionoftheexpressedRNAand/or protein. Induction is not required if a constitutive promoter is used. (4 points)Ifaregularcloningvectorischosenasdescribedinstep

33、, asubcloneneedtoobtainedtoallowthegeneofinterestbeingclonedintoashuttleandexpression vector same as described in step. Then expression is performed the same as described in .Belowisthemultiplecloningsite(MCS)oftheplasmidvector pUC18andN-terminalandC-terminalsequenceofproteinX.NotethattheMCSconstitu

34、tesapart of the LacZ open reading frame. Suppose that you are going to clone the protein X gene into pUC18, so that your target gene is transcribed under the control of LacZ promoter, and translated with the LacZ gene to produce a fusion protein. You are requested to use BamHI and PstI to the clone

35、X gene, please add these restriction sites on the corresponding position of the X gene. Remember to maintain the reading frame of the X gene with the LacZ gene(1)MCSofKpnIACGAATTCGAGCTCGGTACCCGGGGATCCTCTA GAGACCTGCAGGCATThr Cys Arg His Thr Asn Ser Ser Ser Val Pro Gly Asp Pro Leu Glu LacN-terminalseq

36、uenceofXgene. ATGACCCCUCAUAACGGCMet Thr Pro His Asn Gly C-terminalsequenceofX gene.GAUAGUACAGCUGCCAAG Asp Ser Thr Ala Ala GGAT CCNATGACCCCUCAUAACGGCGAC(N-GAUAGUACAGCUGCCAAGTAACTGCAG(C-PARTIV:MAJORQUESTIONS(20points1:PleasedescribehowanmRNAgeneistranscribed,processedandtranslatedinhuman cells. What a

37、re the possible mechanisms in regulating the expression of this gene?Transcription:focusing1:PleasedescribehowanmRNAgeneistranscribed,processedandtranslatedinhuman cells. What are the possible mechanisms in regulating the expression of this gene?Transcription:focusingontranscriptioninitiationdescrib

38、edonp222ofthe text book. (5 points)Processing: 5 capping (addition of a m7G to the 5 end of the RNA pol II transcriptcatalyzedbymRNAguanyltransferase/cappingenzyme)(1.5point), intron splicing (removal of intron by splicesome that contains 5 snRNAs and many proteins. First step- cleavage at 5 splice

39、site and the conserved A being linked to the 5-end of the intron, second step-3cleavage and exon ligation)(2 points), 3cleavage and polyadenylation (First-assembly of the recognition complex on the polyadenylation site, second-poly(A) polymerase/PAPcleavesthetranscriptatthissite,third-PAPaddsupto250

40、 A residues to the 3end of the transcript) (1.5 point).Translation: Initiation, elongation and termination (p276, 278) (5 points) Regulation: Firstly, regulation can occur at the transcription level, for example, throughinteraction of thetranscriptionfactors with the promoter or URE to activate or r

41、epress the transcription (2.5 point). Secondly, regulation canoccuratthepost-transcriptionlevel,forexample,alternativetranscription or alternative poly(A) site (2.5 points).2 (20points):Abacteriumisfoundtometabolizeararesugar produced by aplant thatthe bacteria grow on. However, the bacteriaprefer g

42、lucose as the energy source. Theproblem is, if you want to finish this course with a satisfied score, you must figure out the regulatory mechanism that the bacteria used to determine the sugar choice.The gene involving in theraresugar metabolism has been identified as fun3. You can use northern blot

43、 to analyze the expression of fun3 and use DNA footprinting to analyze thebindingofproteinstothecontrolelementsof fun3gene.Thefollowingtableshowsthe experimental resultsPlease propose a mechanism to explain the above results. You should focus on the question“Howdoestheexpressionof fun3geneistightlyr

44、egulatedsothatitisonly highly expressed when the rare sugar is the only carbon source”. You must answer what proteins A, B and C are. (8 points)Raresugar Levelsfun3 Binding of -Protein A binds to the promoter region ProteinCbindsupstreamofthepromoter+-ProteinAbindsto thepromoter-+Protein B binds to the pro