Midterm 2 - Name ID # Chem 130C Midterm 2 Spring 2009 l. (5...

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Unformatted text preview: Name ID # Chem 130C Midterm 2 Spring 2009 l. (5 points each) Choose the correct statement from the choices listed, and briefly state why each of the other choices is incorrect. a) RNA‘ i) incorporates both modified and unmodified bases during transcription ii) does not exhibit any double helical structure exhibits base stacking interactions and hydrogen-bonded base pairing iv) usually contains 65 to 100 nucleotides v) does not exhibit Watson-Crick base pairing H; . 1.25.1 (AF; ‘U M C) m MQLYIMWL (ERMA - K I: mxwxiug, ,iLea—mv» Wows» per-we _ WJ— Crg e. M KIM-LIA} chm—m. _’ . )§ _-_ “Ig\r _1’ V ‘ I. ‘ RNA L‘bRN/‘k‘ wag;qu (151; \bbflt Qi W/C P .UV‘Mfi m flyw‘mui‘vfin b) In eukaryotic transcription i) RNA polymerase does not require a template ii) all RNA is synthesized in the nucleolus iii) consensus sequences are the only known promoter elements phosphodiester bond formation is favored as a consequence of pyrophosphate hydrolysis v) RNA polymerase requires a primer RNA DNA wage mlmumlt i‘ MELer M 4‘3 RN A Cunt Ln. m MM“ '4 Vitamin/L} MM" \ 952M443» Myth 0;} (a, r K“ :L “A i . r. (tng ' ___ L. ELSE, *. EL L-ypfivw-kwrbb Sitk’Vm-‘tw'l R N A (Jo-WV» Ml flea. c) An enhancer i) is a consensus sequence in DNA located 70 to 90 nts upstream of the start site may be located either upstream or downstream of the gene it regulates iii) may be located on a different chromosome than the gene it regulates iv) functions by binding RNA polymerase v) stimulates transcription in both prokaryotes and eukaryotes E liar.» Wti-O. jivwrim Magi/DLL‘L 132%- 5 ‘xxgamw aw" film.” Lw’iutasvcva we; ML; - «fines—41.]. (.3 S - W LTF'a [21—Mme twat ce ‘ «WLE Chem 130C Midterm 2 Spring 2009 Page 2 d) The 0' subunit of prokaryotic RNA polymerase i) is part of the core enzyme ii) binds the antibiotic rifampicin iii) is inhibited by u—amanitin iv) must be present on the polymerase for transcription to occur @ specifically recognizes promoter sites w” “at” ‘W“% ,‘ m retract) £me - Mi D" W We;me (not; at» ‘K- Ta; A_&.(‘r~l:U‘LLtE-(\- 41m, attic; IN tTtAthv-t fiwL ‘ L m ml— twrquMet-iw. a”) 4 e) Termination of a prokaryotic transcript i) is a random process ii requires the rho subunit of the RNAP holoenzyme w may not require rho participation ifthe end of the gene contains a G-C rich palindrome iv) is most efficient if there is an T rich segment on the template strand at the end of the gene v) requires an ATPase in addition to rho factor RNA" mwcme > ‘ *mnrq ‘ rtgvajnjgwlt M A a Mir/c Lr’xl't— me W Mwofi. M WV;- w—Qtlk fink. 0‘4. enabler. WmaRNA mud .2 4 la. mimrmwh ’- as; Alexe- ML AT P we; (LC z 2. (10 points) The mob operon contains four DNA sequences (A through D) that include two structural genes encoding enzymes 1 and 2, a gene that encodes a regulatory protein, and a regulatory element (promoterloperator). The results listed below were obtained in a study examining the effects of mutations at various loci in the operon. (The plus and minus signs indicate the production, or lack of, of active enzyme; mob is a small, water soluble coregulator.) mob absent mob present Sequence Mutation Enzyme 1 Enzyme 2 Enzyme 1 Enzyme 2 None + + ~ k _ + .E m + + + ++ U0w> Chem 130C Midterm 2 Spring 2009 Page 3 Designate whether the mob operon is inducible or repressible, and assign roles to each of the DNA sequences in the operon. (Provide the reasoning that led to your conclusions.) 5% flv’L Whom-vol Wh— Alm, (luv... WWL W. use. ' - ' >E MWW-SMK} ac ONE © RAMA-'3’: Crfl'CALLA... Aweia- “>9me W 1W 'fzd‘fl-Jl 10. Weak LL cuanWL whit .erz. — JWmefL A =V fam‘va-‘QLA. / 01mg E- '7 p-L r1, gim”.—~_.~' - ‘Wx; X—l’lq". “mole Wm Lea amazes-we»: (mm WW mop)- W A W-rik/l— cam @| I WWW C “Mtg; @1) B Diome . 3. (3 points each) Superrepressed mutants (is) encode [ac repressors that bind operator but do not respond to the presence of inducer. Indicate the phenotypes of the following E. (:01! mutants with respect to inducibility and {S-galactosidase production, providing the reasoning behind your choices. a) ISO+Z+ UNiNDUClBLE; N0 IQ~WLWAM4 b) ISOCZ+ 1 coNe'TrTUTNE P1. H sot—4:“- ‘e—W fiA-fi-M-‘r r‘ I- WW4?» CANNOT i, i M. ed, 01% wees-«o- c) 150‘:z‘/1+o+z+ I I 5 UN-NVULtVbLEi No : 762nm 6:3ch I W“ l“— 01: L trams—(M) Iii—"hm «fin-ewflj :‘L W QAMXU‘ 0L! MW). NOT FAA;de accwa d) 1+0 2 /ISO+Z+ UNrUDUCIfiLE; No f‘yu‘rC-ML-CML'I met». If (ark—«M ,mo-dkkL-t‘i- . BOTH Q“... Ma. MVW/ '2 infill» (,wa 4. (12 points) Indicate the locations of the six base consensus sequence in the —10 and *35 regions as well as the location of the initiating nucleotide on the sense strand of the E. coli tRNATyr promoter shown below. CAACGTAACA TACA CGGCGCGTCATTTGA ATGA CGCCCTTCCCGATA ._ 13,0 -j_O ‘1- :L Chem 130C Midterm 2 Spring 2009 Page 4 (8 points) What is the probability that the 4026-nucle0tide sequence coding for the fi—subunit of E. coli RNA polymerase will be transcribed with the correct base sequence if the probability of inserting an incorrect base is 1/104? m .JJQJQ—‘L (“a iMLOfLQEc-T jt€vll=mmsbm M 01000.” ‘ x wk .1634, a? (3x. CORRECTflmMLN (\fi 0.05OI)’ c7“ o‘c‘qc‘q’ - r ‘tC CORRECT (Q,qqqq) :9 b) 5. a) bm}:€ n; L e .(‘Ji —(EI\-LI tor.- use: (8 points) The lac operator is 35 bp long, with 28 of these base pairs (14 on either side of the central point) related by palindromic symmetry. What is the probability that this symmetry arose by pure chance? a) (8 points) Describe the transcription of E. coli trp operon in the absence of both ribosomes and tryptophan. IM- X-‘Qu. KEV; [-1 1'?) @M/‘d‘mfl. 3"‘4- Aunt/301’s a. _ibwww¢La—x_ ALWAYS 3:? TM Wm agents by m aim/Lem CA N my: W ‘ F‘ a ‘ b) (10 points) Explain why inserting a 5 bp segment of DNA at the —50 position of a eukaryotic gene decreases the rate of RNAP II transcriptional initiation to a greater extent than inserting a 10 bp segment of DNA at the same sight. —— 50 CM —v toot 03a io'bp flit/(61W tux-.mch 4L~M 3-H?) ~ M 12.45% ' 43ng DNA va (imam—weeliiocl/fi MDT Wfiwufiflgufx C's/MW R! .‘Esa' ( 0.2 hf-"w*4.\.¢% (W SAM E flaw/L0 . A 5 lap Wimvtm menu; £044 AL—Lrw/L but 1“" m OPPDSiTE m a, "52;: ,t a; Mu‘Ch—t QRERT‘ER MW cm? i, , , . ( sq, _ c) (7 points) The introns in eukaryotic protein—coding genes are often quite lar ms in ge, but almost none are smaller than about 65 bp. Explain the reasons for this size minimum. WWW W WKrurggx 7QJ w Sn RN P RA. £5— www— fiym,mth A Z‘Woxtjfl, )CAL" 5" q iinvV-MLSSL r ...
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Midterm 2 - Name ID # Chem 130C Midterm 2 Spring 2009 l. (5...

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