Bacteriology 303- Exam 4 practice questions- Spring 05

Bacteriology 303- Exam 4 practice questions- Spring 05 -...

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Unformatted text preview: (jenetics answers 7 , ' ' ' Page 1 of 6. ._' *' W ML/ Bact 303: Genetics Study Problems With Answers Multiple Choice Questions from Previous Bact 303~Examinations 1 His mutants (histidine auxotrophs) can be isolated by A direct plating on medium containing histidine B direct plating on medium without histidine C. replica plating from a medium without histidine to a medium with histidine - i} replica plating from a medium with histidine to a medium Without histidine E. none of the above - anew-A1 mm. wm WW“ my: Ann Mm“. .mmrmmfimvmfiumwfimnmnnmmmml:fivm‘fimflowmwmhmmxmwwnxwmnewmmrrmmow«Mmavanm‘wmwrmmymwmuon 2. His+ revertants (prototrophs) could be isolated f1 om a population of His mutants by A direct plating on medium containing histidine ' - ‘18:. direct plating on medium without histidine C replica plating from a medium Without histidine to a medium with histidine e. replica plating from a medium with histidine to a medium without histidine ~ ~E. two oi the above 3. An environmental scientist wants to determine if two chemicals X and Y are possible mutagens The Ames Test is employed using a Trp strain of Sallizonella typhimm mm. The following results are . obtained. - NUMBER OF REVERTANT ('1‘er COLONIBS Chemical Control plate Test plate X 38 36 Y 42 = 44 The conclusion from these results is: A. X is a- possible mutagen; Y is probably not a mutagen B. X is probably not a mutagen; Y probably is a mutagen C. both X and Y are probable mutagens ‘ it}? neither X nor Y are mutagens - E none of the above “—1 w- .4 «4me rmwwm-‘W WWW-"7'!lu-qmywn‘:mt-Wummmmqfimmvvrpawmfiwwmwuvmmrm”awn-rum"mmfiumavmmmvwnvw;v:rwwnwvnmrfiimmmwrwmuxmmmmmfiwwmz 4. A molecular biologist extracts the DNA from a lactose positive (Lac+ ) strain of E 6011' and a penicillin- resistant (PCr) strain of S. am eus Using restriction enzymes and DNA ligases she cuts and joins together fragments of the DNA derived f1 cm the two organisms hoping that the lac gene of E coli will recombine with the PC resistance gene of S. am ens She uses a process of artificial transformation to insert (transform) the DNA fragments into a E. coli strain which is penicillin-sensitive (PCS) and lactose-negative (Lac ) What medium should she plate the transformants on in order to select for a penicillin—resistant lactose positive recombinant? A. minimal medium plus glucose - 3 minimal mediumplnspenicillin ,and.lactose,- - , _ -- A B 4. , , , . , - . C. minimal medium plus penicillin and glucose ' D. minimal medium plus lactose E. none of the above http://www.bact.wisc.edu/Bact303/geneticsanswers _ . ,- 5/1/2005 Geneticsanswers . , , , , PageZ of-6 W:Hrrg—MWHWMW,IM~,»Wan—m—wwymmanW—anmm ,MflmmWMWWWmFWWrdwwwmme-W 5 A bacterial geneticist mixes together 108 cells which require'methionine and are streptomycin sensitive and 108 cells which require phenylalanine and tyrosine and are streptomycin resistant In order to detect whether genetic recombination had occurred between the two strains it would be best to grow the mixture of cells 1n A minimal glucose medium with no supplements B. minimal glucose medium plus methionine and phenylalanine _ C minimal glucose medium plus methionine, phenylalanine and tyrosine D minimal glucose medium plus phenylalanine, tyrosine and streptomycin . "~33. minimal glucose medium plus streptomycin fixsflwunuwmrwsfihw¢fiIn”?:1A:fivmmwpwm~mfimwfirvbmmmmwvmmwmm"aw-“5w:nmmvmmww}crui-$Wmmm2mmm, _ 1 A . wwfim'firrmvmweflw—wnuw _ 6. Before mixing these strains'together the geneticist (above) plates 108 cells of the auxotroph that 5 requires both phenylalanine and tyrosine onto unsupplemented minimal glucose medium. Approximately 100 colonies arise on the plate The geneticist concludes that A genetic recombination lS Occurring between cells of this strain at a rate of 10 6 B. the culture is contaminated C. the rate of mutation to auxotrophy is greater than the rate of reversion to prototrophy . D.the rate of reversion to prototrophy is greater than the rate of mutation to auxotrophy —E. the rate of reversion suggests that the double auxotroph results from mutation in a single gene mm,Nymmm—«wmmmmwmmmmwmummf$nwmymmwuwmw.mumw~wwnmmwmum—mmmmm»md—®munww«In-manna ~~~~~~ .—F r. ' 7 Which one of the following strains of E 6011' would be able to conjugate with E. coli F ' ,,Met‘ Arg‘, B1 to produce a recombinant that grows on minimal medium supplemented with vitamin B1? A F“ Met+ ,+,Arg Bl‘ ' 13.1? ,Met,Arg,Bl+ C. Hfr, Met-, Arg‘, Bl+ as Hfr Met+ mg, 131- E F“ Met+ ,+,Arg 131+ mmw.» mun-aw Nev-may ago-4M0 :Wulum xw wamaw ’mm W “Wang.” war—- Mmmm fawn—m HSMMMW «WI-VF .Wy-vro' -W.MX MN: m mvavwmwmw wvmarmmmmtm’aw‘xn- ”income!” 8.5 X 108 cells of each of two strains of Salmonella typhimurium, one which is Cys' His and TCr, and the other which is Arg Met and TCS, are mixed in a tube. After 30 minutes the mixture is plated onto vaiious media and the following results after growth are noted: Minimal medium (glucose plus inorganic salts): no cOlonies Minimal medium plus Arg210 colonies Minimal medium plus His: 20 colonies ' Minimal medium plus'iArg and TC: no Colonies Minimal medium plus'Arg, Met and TC: 50 colonies Arg=arginine Cys = cysteine His— “ histidine 777Me7t= meth1on1ne 777777 77 777.777 77 777777777 7 7 - ' TC= tetracycline (r= resistant s — sensitive) The results of the experiment suggest that http://wWw.bact.wisc.edu/Bact3 03/ge11eticsanswers ' , ‘ . 5/1/2005 “a Uenetrcs answers ' Page 3 of 6 1 A. Genetic recombination has taken place between the two strains of bacteria. B The Arg and TC genes are physically close to one another on the bacterial chromosome C DNA transfer was mediated by transduction as opposed to conjugation. BThe frequency of mutation from TCS to TCr 15 approximately 10 7. B All of the above w. A...“ «W ..Mmmwa W~~WWWVW.~~NWM*MNMMWH WWWMV WWWWMW MWWWWWWN. 133. Resistance transfer factors (RTFs) in E. coli are A. insertion elements on the F .factor x7137. transmissible plasmids that encode genes for drug resistance C. genes that move from place to place on the bacterial chromosome D. outer membrane proteins that exclude passage of antibiotics and hydrophobic agents E plasma membrane proteins that mediate the passage of solutes into and out . , of the cell M.“ ”It '3», WWW» w nymmwyWWfl mi... WW... :mwv . Mvrsmz—flwwmvm ”WwW:NWFM1VWWW-MWH wwrmwvm-‘W mnwm-agmsw—n—HWWWM. 134. In the Griffith experiment the substance present in the suspension of heat-killed virulent cells of Sireptococcuspnemnoniae that when mixed with living non virulent cells transformed them into living virulent cells was "7531. DNA B. capsular material from the virulent cells C. mouse phagocytes D. RNA ' E Spn toxin W.- “W rm.“ “my... , INAKV-IHVV‘fflnrz-LW mflwwy mm»... ”WW; mkmmfi. ~.-.,me_—_.«WP_— 1—”. V5... rmwmwfiranryal l‘I’.‘.J')_‘="Hmvm‘fifw‘fifiwvm'nmwmmvfi l3 5. In the Zinder and Lederberg U—tube experiment, DNAase was added to the U—tube to inhibit the process of A translation . , fit. transformation C lysogeny D conjugation E. transduction .. w::nu-mnfimf3F”=um¢¢15xevmm~=nnveumu>mesewfimajwmsvmw-asrfiwA:Ema“:lvrawnw.”Wmwas:mammmmukmmmawwnwsmfimwflmxmnwul:v’vflvnln'nwnamm“Waumwmmmmwmmrwurmmrmw 13 6 The penicillin enrichment technique rs useful when isolating auxotrophic mutants (for example Trp- ) because penicillin A prevents the growth of wild type bacteria but not mutants B. does not affect eukaryotic cells W715. kills only the growing bacterial cells, not the mutants in the population - D increases the rate of mutation toward auxotrophy B does not affect E coli under the conditions of the procedure WMHW- .— .rrrmm.vn1nv:~r .v’sw "W. .m..-m—:nv.... _ vm ..wn. . -.-NV...m...m—r.-.w. mm . .nfitwwwm ..n ...-,msrmm..«_. .7_.--<.. ._........».-v,-.«.-m._:—-. . 7 137 Trp7- mutants {tryptophan auxotrophs) can be isolated by 7i7v77777 7 77777777777 7777777 777 7 7 77 A. direct plating on medium containing tryptophan B. direct plating on medium without tryptophan 7C. replica plating from a medium without tryptophan to a medium with tryptophan http://www.bact.wisc.edu/Bact303/geneticsanswers ' A I ' » i . 5/1/2005 Genetics answers : . Page 4 of 6. 41’}. replica plating from a medium with tryptophan to a medium without tiyptophan E. none of the above mmmmwmmwmwmnamnmw.wnwmmm.wnwWmmmywq... ”NMFWW“NW.W MW! 138. Lys+ revertants (prototrophs) could be isolated from a population of lys- mutants by A. direct plating on medium containing lysine (~11. direct plating on medium without lysine . C. replica plating from a medium without lysine to a medium with lysine D- two of the above E. none of the above 139. A bacterial geneticist mixes together 108 cells which require methionine and are ampicillin sensitive and 108 cells which require phenylalainineand tyrosine and are ampicillin resistant In order to detect whether genetic recombination had Occurred between the two strains it would be best to grow the mixture of cells in A minimal glucose medium with no supplements B minimal glucose medium plus methionine and phenylalanine C. minimal glucose medium plus methionine, phenylalanine and tyrosine- D. minimal glucose medium plus phenylalanine tersine and ampicillin v 1?. minimal glucose medium plus ampicillin - mwrmrmnwmmwumw ....... mwmmmmmmmmpnun—Fur:r3-NWum-rmvmfimmmflinwmmrmwwwnwfi-'fi1W‘mm‘r'.‘V-an-$“HE 140. Before mixing these strains together the geneticist (above) plates 108 cells of the auxotroph that- requires methionine and is ampicillin—sensitive onto minimal glucose medium with no supplements. Approximately 100 colonies arise on the plate. The geneticist cOncludes that r A. the rate. of reversion to methionine prototrophy is 10'6 B. the rate of mutation to ampicillin resistance is 106. C. the combined rate of reversion methionine prototrophy and mutation to ampicillin resistance ile'12 D. all of the above E. none of the above anwW—me . . ran-mm , . . .. 4. «am—,9.- .. ,. .W. . WWWHWWWMW‘WWmthmummmm—Wmmmmmmmwmw ' 141. Two strains of Salmonella are mixed, one which is arg- his - cob- and CMr, and the other which Is cob- ,thi-, and CMs In order to determine if genetic recombination takes place between these organisms which medium would you plate the mixture of cells on in order to detect the recombinants? arg = arginine cob = cobalamin'(vit B12) his = histidine thi = thiamin (vit B 1) CM = chloramphenicol (r = resistant; s # sensitive) ' ' A. Glucose minimal medium plus cob, arg' B. glucose minimal medium plus cob, thi C; glucoseminimalmediumpluscob; 'hi'5"**"”"’ ., D. glucose minimal medium plus his, thi, CM ”133. glucose minimal-medium plus cob, CM Irfl—n-qneq”In".Is;'5'1awn‘lzmmflflvxrlmlmmgew;The-YL'FIWMU-fi'm‘sw’wlwwu-onwnuhm7:7!Infi¢r¢~ruMnewr‘ewx-mmrrW-Wl'wwmwxrlwn1mmafi’flmt'fll' .-. http://www.bact.wisc.edu/Bact3 03/geneticsanswers .- . 5/1/2005 I. . 1....-_..1 V q Uenetrcs answers 7 7 N Page 5 of 6 142. A molecular biologist extracts the DNA from a lactose positive (Lack) strain of E coli and a penicillin—resistant (PCr) strain of S. aureus. Using restriction enzymes and DNA ligases, she cuts and joins together fragments of the DNA derived from the two organisms hoping that the lac gene of E. coli ’ Will recombine With the PC resistance gene of S. aureus She uses a process of artificial transformation to insert (transform) the DNA fragments into a E. colistrain Which IS penicillin—sensitive (PCs) and lactose-negative (Lac-). What medium should she plate the transformants on in order to select for a penicillin—resistant, lactose positive recombinant? A minimal medium plus glucose ‘ v 3 minimal medium plus penicillin and lactose ' i 1 C minimal medium plus penicillin and glucose D minimal medium plus lactose E none of the above meml. ”.1 W»m~q~m mw 1M“...”.1~1~WW~11.-mwwWmMWWWWHWHWWuwmwm—MWMMMWWWWHW \« 143. During the process of generalized transduction A cell to cell contact is required '3. a bacteriophage capsid can enclose and transfer any part (gene) of the bacterial chromosome , C. the transfer of DNA 1s inhibited 1n the presence of DNAase 1—3. the recipient must be susceptible to infection by the same bacteriophage as the donor .351. 11111.1 11? the shore Study Questions (from Procaryotic Molecular Biolog x) I 166. Indicate by a + or — in the columns below whether growth would'or would not occur if Escherichia coli cultures with the indicated characteristics were inoculated onto each medium. E. coli Characteristics Medium contains ' ' ' , B,L,T M,L,T L,T,SM . B,M,L . B,M,L,T,SM B-M- T+ L+. SMR - m ,_ -. + 1- ' + B.+ M+ T- L— Sh/IR + . + + 1‘“ 13+ M+ T- L- SMS ' -+ + __ l ._ B+ M- T+ L- SMR . + + + B— means that the organism cannot make biotin and must be supplied With it to grow. B+ means that the organisms can make biotin and can grow if biotin is not in the medium. L=leucine T=threonine M=methionine SM=streptomycin SMR=streptomycin resistant SMS=streptomyci11 sensitive. -1.~ ”an. our-v .m. 14.1.... ”-1—. m wrwuvmuww :umivu—vwl .,mm.» n,-1-1.»,-w1.1,.Wm1mm.WM,11,..-q1.WWWNHM”.mwww‘.wqwmwwwwww..wwmw,,,,m.,,,m,.wrn.w.,My.9...”.,):.,n.,u...:.m.,...,,wflww_.1-W1m:m 178 The following experiment was done: 108 cells of strain 1 of a bacterium requiring the amino acid proline (P) and the vitamin thiamine (T) were mixed with 108 cells of strain 2 which requires the amino acids methionine (M) and leucine (L). After 30 minutes incubation the mixture was spread at various dilutions onto plates of minimal medium (containing glucose as the only organic substance) After overnight incubation 310 colonies were obtained on the plate containing a 10'3 dilution, and 34 colonies "on the plate containing a 10'4 dilution Each of the following statements refers to the above experiment; indicate whether eachb' 1s true or false: (9 l NU http://www.bact.wisc.edu/Bact3 03/geneticsanswers ' . _ ' 5/1/2005 ,1.._.A LW_VT...._.“. 1 _/‘ Genetics answers . Page 6 of 6 + If each requirement is considered due to a single gene,- the genotype of strain 1 for the four characteristics isP T M+L+(cor1e1:tf ~ This experiment provides evidence that gene transfer. in bacteria is unidirectionaltcan’t 1833 the if genes 1110116 11-11111 strain 3 to 81121311 2 111 it they move from strain 211.1 strain 3 111131111; tbrrnation cf recombinants) ' , ~ ' ' ' ' ' ' ~1- In this experiment the results could have been due to transformation transduction, or conjugation 113111111331 them-1131111133111 of transterit‘ can: were mixed 101313131131 in the absence 131" IWAase )1 m It is impossible to calculate the frequency of recombination from the date given {can caisuiate the rate at" iecntninnatinnt by dividing- 1.311311131133111 or 13.13335 'eiateci 31111:: the number of11313011133311111113111113 11331;} 2311111313) ' . ~ Strain 1 is an Hfr strain {no way at kntwving because we 1313111 31111511113112 13116171211311 of gene 1nove111e11t ' i or whether 11.111111511111111 was the meet 3111103111313) This experiment has shown that genetic recombination has occurred by gene transfer rafter cell-to cell" contact (true because the 13113331213113 able t0 0323113312330116 111111131131 31111. 111113111131 not 331111111 11011311131 1.131111} w “W1... mwmmwwmfl. .«- 1mm“- v—w—n—Mw _ , ,m—WVWW, —..,v. "mm”.wqu men—g—wwmfi F—Rw—F—fw-‘WWM M "Wmmqwmm . 179 Two cultures are mixed one which is B— M— T+ and SMR, and the other which is B+ M+ T— and SMS After being mixed they are plated on the media indicated. Show by + or- in the columns below whether any growth will be observed if genetic recombination between the two cultures does or (10155 not I occur. (31th ifno _ . -- Growth if - Medium contains recombination ’ recombination occurs occurs B, T, SM 1 t SM ~ + . B, M ’ .1- -1. T + ' . . ' ~1— T, SM _ (33.311313; muta' 113g1.'13111) ~3- No additions . _ . p + B- means that the organism cannot make biotin and must be supplied with it to grow. B+ means that the organism can make biotin and can grow if biotin 15 not in the medium T=threOnine, M=methionine, SM=streptomycin; SMR=strept0mycin resistant an mm" M v' wm-mxw'mm mmaui- .1»:me :rwsr, mm.- .\-- .» mmy- 3U)Mw9\wns-mmmp em" m 11,11 .5“ mas-“Wu..-.1»w..gm»11mmm-vmm-musm1mmwmspxsmmemamavnwwumwmuem“smug.“unumrmmwmmwmwwmwm http://Www.bact.wisc.edu/Bact303/geneticsanswers I - . .' 5/1/2005 ...
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