Final Exam 2008 - answer guide [fix_56]-1

Final Exam 2008- - Bio5Afinal(2008)Drs.ZidovetzkiandHaimo ExplanationpreparedbyDr.Maduro[3 ­5 ­2011] 1

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Unformatted text preview: Bio5Afinal(2008)Drs.ZidovetzkiandHaimo ExplanationpreparedbyDr.Maduro[3 ­5 ­2011] 1) LossofasingleaminoacidwouldhavetomeanthatanentirecodonismissingfromtheDNA andhencethemRNA.Theonlyanswerconsistentwiththisis(D). 2) Ifthehaploidnumberisn=5,therearefivepairsofhomologouschromosomes(2n=10).Foreach pairofhomologouschromosomes,twodistincttypesofgametesaremade onewiththe chromosomefromthemaleparent,andtheotherwiththechromosomefromthefemale parent.Ifweaddanotherpairofhomologouschromosomes,sincetheseassortindependently (alignatthemetaphaseIplateindependently),itaddsanother2xasmanytypesofgametes.In general,thenumberofgeneticallydifferentgameteswillbe2tothepowerofthenumberof chromosomepairs,inthiscase25,whichis2x2x2x2x2=32(C). 3) Ifacellishaploid,theneachchromosomeisrepresentedonlyonce.Foragivenallele,itisthe onlyoneinthecell.InG1,eachchromosomeisunreplicatedandconsistsofasingleduplex (double ­stranded)moleculeofDNA,sothereis1copyoftheallele.InG2,afterS(synthesis) phase,eachchromosomeconsistsoftwosisterchromatidsjoinedatacentromere,sothereare 2copies.1,2(C). 4) SisterchromatidsaretheproductsofDNAreplicationofasinglechromosome.SinceDNA replicationproducesidenticalcopiesofaparentmolecule,sisterchromatidsareidentical.(A) 5) Apromoterisfound5 tothestartoftranscription,anddetermineswhereRNApolymerasewill bind.AssuchitwillalsodeterminewhichDNAstrandisusedasatemplatetomakemRNA(i.e., istranscribed),sotheansweris(C). 6) Adipeptidewithtwoaminoacidshasonepeptidebondbetweenthem.Atripeptidehastwo peptidebonds.Ingeneral,apolypeptidewithnaminoacidswillhaven ­1peptidebonds.100 ­ 1=99(D). 7) Entropy(disorder)alwaysincreasesinaclosedsystem,(E). 8) TranscriptionisproductionofanRNAmoleculefromaDNAtemplate,butitisfromonlyone gene(chromosomesegment).ReplicationisproductionoftwonewDNAmoleculesfromone DNAtemplate.Hence,IIIiscorrect.Polymerasesworkonlyinthe5 to3 direction,sopointIis correct.DNApolymerasecannotsynthesisnewbaseswithoutapre ­existing3 OH,andthatis suppliedbyPrimase,whichmakesRNA.PrimaseandRNApolymerasebothmakeRNAfroma DNAtemplate,soIIiscorrect.Theanswerhastobe(E). 9) PointIiscorrect,sincemitosisgeneratesgeneticallyidenticaldaughtercells,whilemeiosis generatesfourproductsthatcouldbeverydifferentfromoneanotherduetoindependent assortmentofchromosomesandcrossingover.PointIIiscorrect,sincethebasicnumberof chromosomes(N)willnotchange thatis,thechromosomenumberappropriatetothespecies staysthesame.PointIIIisnotcorrect,asmitosisgives2n 2n(orn n)butmeiosisalways gives2n n,anddoesnotpreserveploidy.Theansweris(D). 10) Allthreepointsarecorrectaboutcheckpoints.(E). 11) Areducingagentwilldonateelectrons,soitloseselectrons.Insodoing,itwillalsoloseenergy. (D). 12) Icefloatsbecauseofthelowerdensityofwatermoleculesinsolidwater.(B). 13) 15to6isapproximatelya3:1ratio,suggestingE=earlyfloweringandisdominanttoe=late flowering.TheparentsmustbeEexEe(D). 14) Tetracyclineblocksproteinsynthesisinprokaryotesonly.(B). 15) NADHwillhavehigherenergythanNAD+.(B). 16) Chemiosmosisinchloroplastsandmitochondriainvolvesaprotongradient(D). 17) PointIisnotcorrect,becausetheDNAstrandshowncopiedintomRNAreads5  ­UUA ­UGA ­UAC ­ UAA ­GUA ­CAU ­3 ,whichdoesnotcontainanopenreadingframe(startcodon sensecodons  stopcodon).PointIIiscorrect,becausethemRNAmadewouldreadas5  ­AUG ­UAC ­UUA ­GUA ­ UCA ­UAA ­3 ,whichcontainsanopenreadingframe.PointIIIisnotcorrect,becausethe5 end ofanyRNAhastostartwitha5 phosphate(soitcannotbe3 UUA).(B). 18) ThecodonsforvalineareGUA,GUC,GUG,GUU.ThismeansthattheanticodonloopinthetRNA forvalinehastobeUAC,CAC,GACorAAC,alsoreading5 to3 .IftheDNAstrandshownisa template,thenthetRNAwouldreadas5  ­UUA ­UGA ­UAC ­UAA ­GUA ­CAU ­3 wherethetriplets arecomplementarytothecodonsindicatedinthefigureas6 ­5 ­4 ­3 ­2 ­1.Theonlypossiblecodon is#4,whichreadsUAC.(D). 19) TheonlyopenreadingframeisobtainedbytranscribingtheDNAstrandcomplementarytothe oneshown,andwillread5  ­AUG ­UAC ­UUA ­GUA ­UCA ­UAA ­3 .ThelastsensecodonisUCA whichspecifiesserine,whichwillbetheaminoacidatthecarboxy(carboxyl)end.(C). 20) Hydroxylis OHwhichhaspolarcharacter.Theonlyansweris(E). 21) Eachchildrepresentstherandomfusionofnewgametesandisnotdependentontheother children.Sotheansweristhesameasasking,whatistheproportionofhomozygousrecessive childrenborntoheterozygousparents ¼(B).[Note:Correctspellingofdisorderis Tay ­Sachs .] 22) UseH=wild ­type,dominantnormalalleleofthehemophiliagene,andh=recessiveallelecausing thecondition.TheyoungmanisXhY.ThewifemustbeXHXH.Iftheyhaveachild,itwouldhave genotypeXHY(aboy)orXHXh(agirl),bothnormal,sotheprobabilityofhemophiliainthechildis 0%.AllmalesnormallyinherittheirXchromosomefromtheirmother,sothemaninheritedhis Xhchromosomefromhismother.ShewouldhaveinheritedthisXhallelefromeitherherfather ormotherwithaprobabilityof50%.Ifherfather(theman smaternalgrandfather)hadthisXh chromosome,hewouldhavehadhemophilia.Theansweris0%,50%(B). 23) Aminoacidsaremadedifferentbythesidechainsattachedtothealphacarbon.(B). 24) Chlorineneedsonemoreelectrontofillitsouterelectronshell.Sodiumcandonatethiselectron toemptyitsouterelectronshell.(B). 25) InitiationinvolvesachargedtRNAmetbindingtoastartcodon,whichendsupinthePsitewhen theribosomeisassembled.(A). 26) TheonlycorrectorientationofnewDNAsynthesis,andtheshorterOkazakifragments,isshown in(A). 27) Activetransportmustinvolveproteinsintegraltothemembrane.(A). 28) Electrontransportchainproteinsareintheinnermembraneofthemitochondria.(E). 29) Independentassortmentofallelescanoccurbecausegenesondifferent(non ­homologous) chromosomesalignindependentlyatmetaphaseIofmeiosis.Theonlycompatibleansweris(C). 30) Useb=blackbody,B=yellowbody,c=cinnabareyes,C=redeyes.Thegenesarelinked,hence onthesamechromosome,andclosetogether.ThecrossisthenBC/BCxbc/bc.(Thisis equivalentinsymbolstoBBCCxbbcc,butweusetheothernotationtoshowthegenesareon thesamechromosome.)TheF1flieshavegenotypeBC/bc.WearetoldthattwoF1individuals arecrossedtogether.Ifthegeneswereunlinked,wewouldjustpredicta9:3:3:1ratioofyellow ­ red:yellow ­cinnabar:black ­red:black ­cinnabar.However,wearetoldthatthegenesare linked.Sowecanruleoutanswer(C)whichwouldbetrueifthegenesassortedindependently.  WenexthavetoconsidertheproportionofthegametetypesmadebytheF1s.Letusfirst assumethatthemostfrequentgametetypesaregoingtobeparental,thatisBCorbc. Assumingthegenesarecompletelylinkedandnorecombinantgametesaremade,thetypesof progenywouldbe¾yellow ­redand¼blackcinnabar.Sowecansummarizethetwoextremes asfollows:    Betweenthetwoextremesofcompletelylinkedandassortingindependently,therearealways black ­cinnabarflies,sowecanruleout(A).Notethatwithcompletelinkage,thereareno yellow ­cinnabarorblack ­redflies,whichtellsusthatcrossingovertoproducerecombinant gametesarenecessaryforthistypeofflytobeseenintheF2.Sosomewherebetween3/16 (0.1875)and0liestheproportionofyellow ­cinnabarandblack ­redflies.Thismeanswecanrule outanswer(E),because0.1875islessthan1/5(=0.2).  Thatleavesanswers(B)and(D).Answer(B)specifies1/16oftheflieshavetheblackbody,red eyephenotype.Thisispossibleifthegeneticdistancebetweenbandcwasjustright.Butwedo notknowthisfromtheinformationinthequestion.However,wecanseethat(D)hastobe true.Ifthereisindependentassortment,theproportionofblack ­redflieswillbe0.1875,andif thereislinkage,itwouldhavetobemuchlessthanthis.Thebestansweristherefore(D),aswe canbecertainthattheproportionofblack ­redfliesis<0.2.  [Note:InDrosophila,crossingoverdoesnotoccurinmales.Thequestiondoesnotindicatethis. However,onecanstillarriveatthecorrectanswer,becausethiswillmaketherecombinant classes(yellow ­cinnabarandblack ­red)evenrarer.] 31) Freeribosomesareusedtomakeprimarilycytoplasmicproteins,whichcouldbeenzymes,so thebestansweris(C). 32) Thefunctionofp53isnotrelevant.Thequestionisreallyaskingwhatmutationalchangewill resultinthemostdetrimentaleffectonaprotein.Answer(E)istheonlyconditionexpectedto produceadrasticchangeinthecodingregionandhencetheprotein. 33) TheF1generationtellsusthatthepurpleflowers(P)andtallsize(T)aredominant.Giventheir phenotypes,theparentsmusthavegenotypesPPttxppTT.TheF1sarethenPpTt.ThePunnett squareisnotnecessaryhere.Onehastorememberonlywhatphenotypeclassesoccurinthe 9:3:3:1ratio.The9isthePT(doubledominant)class,andthe1isthept(doublerecessive) class.The3sarethetwoclasseswhereoneisdominantandtheotherrecessive.The phenotypesoftheoriginalparentsarethese:Pt(purple,short)andpT(white,tall).Hence,3/16 +3/16oftheF2plantshavethephenotypeoftheoriginalparents.[Note:Thewordingofthe questionsuggeststhattheanswershouldbe1(readitcarefully) butthepossibleanswers clearlyindicatethatthequestionmeanstoask, Whatistheprobabilitythattheoffspringofa crossbetweentwoF1swillhaveaphenotypesimilartooneoftheoriginalparents? ] 34) Achemicalreactionratecanbeincreasedbyaddinganenzyme(D). 35) Aphosphodiesterbondformsbetweenthehydroxylgrouponthe3 carbonandthephosphate onthe5 carbon(B).[Note:Technically,thebondreferredtointhequestionismoreproperly referredtoas thecovalentbondthatformsduringRNAorDNAsynthesis. Thewhole phosphatebridgebetweenthe3 carbonofone(deoxy ­)riboseandthe5 carbonofthenextis itselfcalledaphosphodiesterbond,becauseitincludesthephosphorandtwooftheoxygens.] 36) Ingeneral,intronsarenotfoundinprokaryoticgenes,onlyeukaryotic ­(B). 37) Inmitosishomologouschromosomesdonotnormallypair (D). 38) Eachpositionisindependentoftheothers,so5x5x5=125(E).[Note:Itwouldbestrangetohave anoddnumberofpossiblenitrogenousbasesindouble ­strandedDNA,atleast,giventhatDNA isusuallyapairedduplex.Theoddoneoutwouldhavetoformabasepairwithitself.] 39) Conservativereplicationmeanstheoriginalheavy/heavyparentalDNAisregeneratedafter replication,whiletheseconddaughtermoleculeconsistsofalllight/lightDNA.Therewillbetwo bandsafteronegeneration.Aftermoregenerations,thelighterbandwouldaccumulatemore DNA,buttherewouldstillbetwobands (B). 40) Atequilibrium,thereisnonetchangeinfreeenergy.(E). 41) PointIisfalsebecausehomologouschromosomessegregateinMeiosisI.PointIIiscorrect. PointIIIisalsocorrect,becauseatG2(afterSphase)allchromosomesaretwosisterchromatids joinedatacentromere sotheDNAcontentis4x.AftermeiosisI,itis2xpercell,thenafter meiosisII,itis1xpercell(or¼thestart). 42) (E)istheoddoneout. 43) (E)isthecorrectanswer. 44) Thenumberofcentromeresisthesame,andthereisoneforeachchromosome.10,10is(B). 45) Thenewpeptidebondhastoformintheshadedregioninthecenter,joiningtheC=Oofthe growingpeptidetotheHNoftheincomingaminoacidonthechargedtRNA.(C). 46) Thecellhasthreereplicatedchromosomesconsistingoftwosisterchromatidsjoinedata centromere.Theoddnumberanddifferentsizesmeanthatthecellishaploid(n=3).PointI 47) 48) 49) 50) 51) 52) 53) 54) 55) 56) 57) 58) 59) 60) cannotbecorrectifthecellishaploid.InMeiosisIweshouldseebivalents.InmeiosisIIthis couldoccur,soIIIiscorrect (C). (D)isthecorrectanswer.IngeneralE.coliwillnotspliceoutintrons. TheprimarystructureofproteinsistheorderofaminoacidsalongtheN ­C ­C ­N ­C ­C backbone. Secondarystructureinvolveshydrogenbonds,butnotprimary.(C). (C)isclearlywrong.ExonsareneededbythemRNAfortranslation. (A)istheonlycorrectanswerhere. Wearecomparing[H+]to[H+]fromahigherpHunittothenextlowerone,sothesolutionis moreacidic(higherH+)byafactorof10.(E). Correctanswerhastobe(B). (A)forobviousreasons. Watermustbemovingintotheredbloodcells,whicharesemi ­permeable,toequalizethe waterconcentrations.Hencethebloodcellshavealowerwaterconcentration.(A). Oxygenistheterminalelectronacceptorincellularrespiration.(B) ThelightreactionsgenerateATP,NADPHandO2.Only(D)iscorrect. Theonlycorrectconfigurationis(D).DNAsynthesisoccursinSphaseandingeneral,genesare notexpressedinM(mitosis). (C)bestexplainstheincreasedprevalenceofsex ­linkedphenotypesinmales. mRNAissingle ­stranded,sonothingcanbesaiddefinitivelyabouttheotherbasesindividually, exceptthattheyalladduptotherest(E). Allofthesepointsarepropertiesofthegeneticmaterial(E).  ...
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This note was uploaded on 03/27/2011 for the course BILD 1 taught by Professor Chen during the Winter '11 term at UC Riverside.

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