Bacteriology 303- Exam 3- Fall 03

Bacteriology 303- Exam 3- Fall 03 - u r 1...

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Unformatted text preview: u r - 1 Enemy -* , ""7“ 7 “ ’ V v Bactericiogy 303' J -/ Third Examination ‘9 N member 25, 2003 1. The DNA ccden ACT. is transcribed into an mRNA cedcn A. TGA . '. B. ACU C. TCA D, UGA , E. none of the above . 2.. The tPn‘JA enticeden fer the RNA ccdcn TJCA is A. AGT - A n. AGU : . ‘ A? 9 wt D. UGA ~ . ’ . 4/ ‘5] A .173. ncne ct‘theabc‘ve ‘ ‘ M - .. b , 3. The RNA codOns-GGA and GGC enccde for the amine acid glycine. This isnnexampteofln) 'f' ‘2 ' - "' - ‘ ' A. degeneracy of the genetic code _,___fi,”siient_mntaticn. ‘ ’ Cg pens-[mutation , , _ . D, transcription of the genetic cede. ‘ E. translation of the genetic code a . _ 4. What is the sequence ct” DNA nnctectide "bases that is centplementnn/ tn the ._ ’ ' first five bases (it‘ll-=5) en the strand shcwn abeve? (Read, left to tight) ' A. (anew , . ‘ . A- ' http://vmvwbacttniécnedtt/bacfim/Exanfi net 3 12/11 4/03» ’ ' ' " '7' "UN ”7 ‘ ' I A_A_Mfi,__n #771—445-3~T UL“;"F ,Amo'flilfli “ ' - AMA _ ;7_ w,” , A~7 ,, we ' ”A“ ’ """ ' D. phe ' E. ; none (termination) e , .. it. The situation described in questions 9 and it? is an example Of) (a) A. base substitution , ‘ ' 5 B. silent mutation ' . (C? frame shift mutation - ' D. degeneracy of the genetic code E. genetic recombination , 12. "What is the most iiieeiy phenotypie consequence of a point mutation_(e.;g.w +hasesuhstitutioni—in-a-haetenai gene that encodes for a particuiai enzyme? , D. The enZyine produced has an aitei‘ed primarysti'ucttneand may or may . not show any. alteration in activity. ‘ ' ' E0 The enzyme produced exhibits no change in ptiini‘aty structure but has . altered activity. - ' ' ‘ . I A, Transcriptionot‘ the entne gene is blocked because RNA poiynietase I ' cannot hind to its promoter in the gene. . ' ' ' B. Transiation is bioched because tihosoines cannot bind to the transcript _ (mRNA) of the gene , . _ http:/fuofimi.hact.u/ise.edu/[email protected] ~ I '. - , . ' I HIM/03 "fi ’2'}! -; ,,_ .L, L774 ,, fl . A i; , _, A _ A?” 7“” A lat-=5" 7 vs. L' a C... Transcription is halted at the point of the mutation. D. Transcription is halted downstream of the point of the mutation; E A complete transcript (mRNA) of the gene is produced, but an error (terminator or Stop codon) encountered during translation truncates (shortenS) the protein it in the Lederberg and Tatum experiment which provided the first evidence or“ genetic exchange in bacteria, the pumose in using triple auxotrophs of the mating strains of E Celt was to A minimize the development of rerertants which would he indistinguishable from recombinants . ‘ , _ B distinguish conjugation from transformation C. guarantee that one strain was a donor and one strain was recipient D prove that conjugation required cell-to=cell contact ‘ E. rationalize the develOpment of the replica plating technique for isolation of nutritional mutants ' ' q m an the Zinder and Lederberg U=tuhe experiment DNAase was added to the Untuhe to inhibit the bacterial process of ' ‘ t . A translation , B transformation . C lysogeny D conjugation E transduction b , v ' to What was the outcome or "the Griffith experiment when killed encapsulated cells of Streptococcus pneumoniae were mixed with living nonencapsulated cells then injected into mice? - - A. Mice were not killed and living nonencapsulated cells were recovered B Mice were ltilled and living nonencapsulated cells were recovered C Mice were killed and living encapsulated cells were recovered D Mice were not killed and no bacteria could he reCovered. _- E none of the above ' ' ' ' c . l7 The genetic acquisition ofmultiple drug resistance in Shigell’o as a result of conjugation with E. coir” is an example of A lysogeny , . , . g g /\ W”? .18 natural selection ' " . 06>? O i ’r‘ . .3? ”(3. genetic reversion, ' " , CD ‘ ../’ \/ "UK/f". J" / http://vvww.hectwiscedu/hacti«(B/Examihtml nus/o3 vertical evolution horizontal evolution 2, . l8. in matings by conjugation bet-ween l7+ mil F-' strains ofE. colt . A. recipients convert to an F+ genotype ._ . , - B. recipients remain F— unless they are able to recombine the F Factor , C° recipients remain F= because the F factor is spontaneously lost by the R- genotype . ' ’ - D. donors become it by donating the .F factor E° Any of the shove is possible ' ' OK , > V , _ . l9. The strain oi‘E colt Which donates chromosomal genes (such as _ polymerases anti hiosynthetic enzymes} daring conjngation with other bacteria is A. l3+ B. F“ Cu Fe- ‘ D. R'l‘ll E. l—lt‘r ' Q- P L 20 Which one "of the following strains ofE. Cali would be able to conjugate with E. coli he met =- arg =- Bl =— to produce a recombinant that grows on AA MMA~7 _ H HEfagUAJAVuznn-A“ 7w W, M, -wininirnal-nteclininsnoplena‘e‘ntetl‘with Vitamin Bi ‘2‘ ~ A. F+ met+ arg-l— Bl =. B, F+ met+ arg-l- Bi + .1 " C. Hfrmet-arg-B1 + D. Hfr inet+ arg+ B1 -- E. none of the ahoVe 2i, Which onée ot" the following strains ot‘E, colt" would he ahle to conjugate with E. coil" like trot lea a, Lac - to produce a recombinant that ‘ grows on minimal meditnn with lactose as a sole source of carbon? _, A. F+9hp+9len=9Lac+ " ' "' ' a ’ B. lit trp i=3 len-l-g Lac+ 3 -, (1. lift; trp =, len+9 Lac = D, l—lfig tho =9 lent Lac+ .. hf‘pflvvwlcactheeuWoacwlin/Eaamjhnm - 1 - , : tan/o3 I Augvvvi .5: ,' Hr: ndfigtéfimu_flgu ,,.;,,__i ;.,,A, a i,u._r,ni,r _ All. ..., M, ,_i,i.. “A _,._,,,,A “aw“ , 7.4 ,;AD,-A._A,, _ 4.7-- _-_m. a E Hfr, trp+, leu ,Lac+ 22 The acquisition ()1 °beta phage DNA in Corynebacterium diphtheriae that converts the bacterium to a virulent phenotype 18 an example of - . . _ . A. lysogenic conversion I B transformation ‘ C. transduction . D, enzyme induction . E enzyme repression , , ‘ ’ ox £23 During the process of generalized nanSduction ' A naked DNA is taken up Why a recipient bacterium B a phage accidentally encloses and transfers bacterial genes hem one bacterium to another - C bacteriophage DNA becomes integrated into host cell DNA D. DNA is transferred across a bridge that holds mating bacteria together E two ofthe above ’- ' . . _ _, _-._,_-.. 6» 2A Daring the process of restricted (Specialized) nansduction ‘ A lysogenic bacteriophages are required . - B. only genes linked to the prophage DNA are transferred htom a donor to recipient . . . C the prophage makes an imperfect excision from the bacterial chromdsome before replicating and lysing the host cell D two or fthe above ' . E. all of the above . 5b ' A 25 Which of the follbvving statements is true of the transformation systems of - both Haemophilus and StreptocOccus? A The transforination system is geneticallyaencoded by bacterial ‘ » chromosOmal DNA. B Highly artificial treatment of cells, such as exposure to high concentrations ' of divalent cations, is an absolute requirement for transformation to occur ' C Competence involves formation of membranous ”blobs" ”in the IGram= ' negative outer membrane 01. fthe recipient . 'D Only homologous DNA with a unique l l base pair sequence is taken up by the. recipient cells E A protein called competence factor must be produced in a critical concentration to trigger cellular ability to take upi transforming DNA naprat/embarrass.earl/"Danits/Eminent, . , rail/o3 , ___ Aéfhgfifacifif mayfoeeonsiriererifiansepisemeeecanse“ * M "W “ “ ” ” ' A. it can exist free in tire cflogsiasrn or it may integrate into tire iracteriai chromosome ~ , ' ~ ‘ B. it may be substituted. with genes which encode drug resistance C. it encodes for the genetic information invoiveti in its own transfer D. it isa seii—repiicating genetic factor . . ' B. it encodes for dispensabie genetic information ,a' a) . fl '7. Resistance transfer factors (RTFS) in bacteria are A. insertion eiernents on the F factor (3. 'transposons that more from piace to piace on the bacteriai chromosome D. chromosomai genes fortire synthesis of antiiiiotics , - Ea genetic factors that prevent tire transfer of drug resistance aniong mating strains ' ' ' ".128, 'Simiiai‘ities between bacteriophage lambda DNA anti the]? factor in E. coli include ~ ' - " , ' ‘ ' A? They are genetic eiernents that are repiicated in tire bacteriai host. B. They may contain genes removed. fromtire bacteriai chromosome. ‘ - C. They may integrate into the bacteriai chromosome, . _ . v D. They maybe transferred horizontaiiy flora a donor to recipient during the processes of genetic exchange. ‘ ‘ ' t ' ' E. ail ofthe above - V 2-9:“ During repiication of a iytic phage in a susceptible bacteriai ceii? tire ,eariy ,‘ proteins transiated flora eariy phagenrRNA are concerned with ‘ . wean A. attachmentof the Virus to its host ceii , ' B. entry of tire phage" nucieic acid into the host ceii C. .repiication of the virai nucleic acid ' D. viral assemniy " ' E. iysis oftire host ceii ' if, i ' 30. During the replication of a iys‘ogenic phage in its specific irost ceiig the * late proteins arcinvoived in a ' ' ' ’ A. transcription of virus DNA B. integration into host DNA C. replication oftne virus nucieic aciri . _- D. Virai assembiy and escape ficnr its host ceii A but:://Www.bact.wisc.edu/bact3fl3/Exam3Jami 12/14/03 E. two of the alscve 31,1:‘Tlse penicillin citriclsmeat (penicillin selecticn) technique is useful when . isclating auxctrcplsic mutants (for example trp «) because penicillin A. acts as a mutagen and thereby increases the rate of mutaticis thartl tsp auxctrcplsy - ‘ ‘ . B. dces not affect eukaryctic cells mutants to grew 32 15mg - mutants (arfgiaiae auxctrcplss) cats lse isclatetl trees air arg+ bacterial population by ' . - - ‘ - ' A; direct plating cf >lll8 bacteria cm a medium containing arginltre B direct plating of >108 bacteria on a medium lacking arginirie . _ replica plating of cclcnies grewirsg a medium containingargiaiiie tc a ‘ medium lacking argiisirie ' ‘ * I ' D. twc'cftlie above E. metre of the ache -~ hwy/WWW.lsact.Wiscedu/lsacfim/ExamB lsml 12/14/03 l \f.) f . #33 t 13) (It) i X 247 ' 230 Y 32 208' The conehaszon ° aoiy IS a mntagen both X and Y are probable nantngens Do neither X not 14’ are ntngens E none of the above * G 35. Two strains ofSanonefla whzmzmnm are maxed, one which is org - Ens =3 cob _ = and C1 3 and the other which ts cob ,. tnt a, and CMS In order to determine if £613.6th recombination tones ptaee between these organisms which of the medtn below would you plate the maxnne of Cells On In order to detect the re.eomtnnnnts9 erg ~ arginnze 12/14/03 ‘ Page’m DEM I 7 7 bacteria ‘ _ - ' g B. 10 2 / 14p? C. 10‘5 * W ’ 9:17 y . -7 [1,. [U ./ D. 10 L; ------ v, " E. Genetic mwmbmafimm has Gecmad but 145% ram: cannot be cafl'cuflamd fmm "thcdat'cagiven. u .‘ ' "I ' " U ' C. ‘ ' . 39. 5 X108 cells of 63.611 GfWO strafing 0fS’aZm0mefla kaimurmm 01:36 which as cys - has = and TCR‘ and the (flaw Whnch RS a V Minimal medium @1me phns margamc salts): yflfiamg httflj/Wmv.bactwiscfidu’iaacfi03/”Exam13.himil v ' 12/14/03 - h r.- x ‘I ‘ 1i ; if . ‘ r; 1 ‘ .‘J u. r V} ‘ 1:; 4 x‘ V L . f“ .,. mi . 1' w ‘ ,s; :c. , ;x :3, ’f’, 'Dxafian’mi M Pagell 0214 " Minnnai medium plus atg: 50 colonies Minimal medium plus his: 5 Colonies Minimal medium. pins erg and TC: no colonies . Minimal medium plus aig, met and TC: 50 colonies arg = aiginine cys = cysteine his— - histidine met: methionine TC: tetracycline (i: resistant; s— = sensitiye) The results of the eXpethnent suggest that A Genetic tecetnbination did not take place between the two strains of bacteria - B. The rate of reversien to cysteine protonophy (cysfl is lit-8 C The rate of reversion to methionine pmtottophy (mew) is 107 D The heqhency of mutation hem TCs to TCr is apptoXiinately TO7 B all of the above . . . Questions 40—42 refer to the folloWing pathway foil“ hyptophan bioSynthesis in. __“_E _0012 Enzymes a ~g—b e d and— e— are eneeded by the strnehnal genes of the ' ' hyptophan (tip) opeien in E colt . . , - a “ 'b » e . d e . ' - * ' a .. .' ' - Glutamine + Chorisniate ~~~~~~~~ > Anthtanilate m. ===== > B- ======== > C ===== > D ----4‘---> Tryptophan ' ' v ' 40 This pathway" is regulated in E colt“ by the ptoeesstes) of A feedback inhibition \/ E end product repression / C. enzyme induction. :3: D catabolite repression E two of the above _ .c _ ,- . All The activity of which enzymets) is inhibited by an"eXcess ot‘ hyptophan? A a only . ' , ' - .i B b only ‘ ,3- _ _ ,, ‘ ....-::—7 47%“ http://imyymbaetwisc.edu/[email protected]/Exasn3.hhnl _ ' ' . , i2/l4/03 A. a only B. h only C. e only D. h, cs da and e only E. a, h, c, d, antic 43. in the pathway of lactose utilization hy E. colt; when the lac opeion is induced ' ~ ' ' 'B. the lac repressot is inactive. and transcription of the lac ' . C, lactose Cannot he transported by the cells _ .' Dan lactose cannot he cleaved to entet glycolysis' E. none ofthe above a _ ,_ ' . ' « opeton does‘occur ' 44. in the positive control of ti'anSCtiption of the la.- _____ tepteSsionlmcycli-cw fiisteqniiett-for A, activating the lac repressoifl . Bo hiacti‘vating'the lac repressor C. activating the CAP protein D. inactivating the CAP protein . E. two cf the ahoVe ' ‘45. in a repreSSihle op‘ei'on (egg, the tip opeion), in the presenceot‘ the . effector (signal) molecule, the repressorptotehi is A. active and can hind to the operator site B; active and cannht hind to the operator site - . _ C; inactive and cannot hind to the operator site ~ - D. inactive and can hind to the promoter site _ A . _ ' E active theiehy allowing RNA polymerase to transctihe the genes for that Questions 466i) tei’et to the plot helow which illustrates. growth (as optical http://WW.hachwiscedn/haot?»OBZEXannB.hhnl - l2/l4/03 "Wit” 35538313533231 ‘ ' — ,4 ; Q , 4;“ _ "A; 7‘ .. A. A 7 _-, h 7___ .__VA.W. 77.— _~——~ A —7A A ~ ~‘rr-~?ag€’13‘0f‘i‘4 " “ 7 '7 7 U mam; as g ital-L904 3.9 g ugscu $12-13, lflfiélzifl, 3-0 9 Elm 3.0 g m 5.0 g Esau-ares a Sizer . 2‘ ma! density ‘3 a 4 a runs as éu“_zer zu ' Timelheurs} 46. The pattern of grewrh illustrated is referred to as A. typical bacterial growth curve ' ' I B. diauxic growth ‘ ~ C. continuous culture ' , _ D. synchronous growth ' o E. bieuulal grth 47,: Ar Whar'u'ure during the growth cruve ure'the‘icells uuuzéegglucuse' use source of curhou'uud energy? ' V * . ' . ' - A, 0-4 hours . ' B. ”478 hours . C. 8-10 hours B. lO-ZG hours B. none of the above 48. Ar where um “ , e does cyclic AMP begin-lo accumulate in the cells? "A. 0 hours " ' ’ - l ‘ B. 2 hours ‘ C. 8 hours . D. ll) hours . ' E. 20 hours C 149. Ar What rhue does the uuuscrlpuou of’rhe lac operou begin? A. ll hours ' ' - ' hupfli’W‘mv.hacr,u/isc.edu/hacfi03/Exuuu3himl .lZ/M/llZ-l , ' Page “13’, 0171747 ‘ density) of Escherichia coli in the following minimal medium: mazes; 2.0 g Reflex- 7-99 4:93:14 , £12 9 46 The pattern of grth illustrated rs referred to as A typical bacterial growth curve B diauxic growth - - C continuous culture _ D. synChrouous growth ' ' E bleuhlal groWlh ‘ 47; At What time d‘urlhg the growth curve are the cells uuhzmg glucose as a scurCe of carbon and energy? ' . , A 0-4 hours B. 44:- 8 hours C. 8-10 hours D. 10-20 hours B. none 'or the above D 48 At What rune does cyclic AMP begin to accumulate m the cells? . . A 0 hours _ ' ' . 7 , 1 B. 2 hours C. 8 hours D. ll) hours ‘ E‘ZOhours ' C 49 Ar Whar time does the urauScrlp‘llou of’lhe lac operou heghr? A 0 hours - - ' j ' ' hmm/IWW/WhactwiscedufbacfiO3/Exam3 ,hmrl , 12/14/03 ‘ ,i‘l 44L“ I)“ z :3 ‘ . 1 203362 mm. no? .26 038 9" 48:3 25 m<m_:m.nmo: mowing . . gov 3 mm: W 3qu Ummméhmm ”003 3? 3mm << nor—awe: w... Emgmmoz. S: mmwom, Amomvnmmémmw Ocmnoamq «h 3» _ZU_<_UC>ENmU Ema—20 Ow 010mmz <m. DOWN—m0... Wmvazwm _.o . , mm 989:? moamfimfi... $83588” <ocq4onm_moo$u oubo zmxmaca 18.0.65 wooqm” 398 . _ , _._.m_<_u Xm<mn <Ocmm“ _._.m_<_n X m<mu <Ocmm” AM 5 t a .5 3 g. $ 8 ~~t m > m m 0 > m w 0 .O m ._1 03 M} r- N O rn’orn>mcnooDUomw>>mm>Uom>mom>moomb>womo>wo » _ZU_D>._.mm.> _<_Cr._.=urm WmmUOme X _ZU_O>A.mm fiIm _._.m_<_ .<<>w O._<_._._1_.mU ...
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