BIO325 exam2 2006

BIO325 exam2 2006 - / Jarepk mach GENETICS (BIO 5325)...

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Unformatted text preview: / Jarepk mach GENETICS (BIO 5325) SUMMER 2006 0\\ 11 Exam 7/31/06 Total points: 120 There are 24 questions in this exam. Each question carries 5 points. Answer all 24 questions for a maximum of 120 points. Select ONE answer in questions with multiple choices. 1 A particular Hfr strain (that is pro+) normally transmits the pro+ marker as the last one in conjugation. In a cross of this Hfr strain with a F" strain (pro'), some pro+ R recombinants are recovered early in the mating process. When these pro+ cells are K // mixed with F'y/cells, the majority of the F" cells are converted into pro+ cells that i ' ’/ also carry th F factor. Explain these results. Tlnf/ Yret—\ (3W (C)q’\d‘ \c \Op\C-\g mk— (M(m~e(rtb\(3/m 'Tho3'/\ k (04% \/\0e\/{ WEWOJL Y— I cextg smoothnzowmt‘b W M“ 9'7)“ (C\\§, X {ii/V) I W3 0 N w [’7 g, M ¥ 5"}? {lb/ll" if" if"! 'i C VW 6M ‘ {Hi i I I ‘ ' . v ' I, J A Luv 1 g} j r ._ ; .5 r f t/ W 7 5 i W'WW’WVVI‘N . a i » (Vi i i w/ 915 ' v v imbue ‘1 "t w " J - w/(i‘V 1; w I 1“ ,,r. ,1 z 3 WW; ~ I. u V WI; in; (I! h. awn/mosww’ / t?" 014 )7' z ' ’ w '1 Jr ,-'i' .I 1 ' (wt -=‘e“’ , mew v w - w “r t x ' 2. A cross is made between two E. coli strains: Hfr arg+bio+leu+ x F" arg'bio'leu'. Interrupted-mating studies show that arg+ enters the recipient last, so arg+ recombinants are selected on a medium containing bio and leu only. These I] recombinants are tested for the presence of bio+ and leu+. The following numbers i (MS V1“ 4' of individuals are found for each genotype: arg+bio+leu+ 320 Mm“? Wu 4 HIV-47’: + . + . arg bio leu 8 — «ax- argJ'bio'leu+ 0 Q XXX?! J/— .. arg+bio'leu' 48 I m V WU- [at ‘ E f > _ (a) What is the gene order? is» u H \ 0mm?) /\>\o law (b) What are the map distances iii/recombination percentages? . / - 4 I ', avg-7* \p‘ro7‘-*\€U 0 \9 _ Lite/3w . 88% fl) l7—i 870 ’l .\ ’70 5,; “i ' if 1)) In a generalized transduction experiment. the Pl phage particles growing on E. coli with the genotypei'u/Tu/aW/‘p l are collected and allowed to infect coli cells with the genotype \‘ul' u/u‘ lrp‘. ’l'he cotransduction frequencies are calculated for the different gene combinations: ulu — 11']; = 0.2 (1/11 — \‘u/ I 0.8 ml — lrp = 0.6 “l’llCh ot‘the following is consistent with the data‘? (a) The gene order is a/u-i’ul-Irp; m/ is closer to [27) than it is to u/a. (b) The gene order is \‘(Il-a/CI-Il'p.‘ (1/(1 is closer to [177 than it is to ml. The gene order is ula-rul-Irp; ml is closer to alu than it is to Irp. (d) The gene order is a/a-HW—va/s 127) is closer to u/u than it is to m]. In bacterial matings. prophage can be transferred from lll‘r to F'. If the F' is nonlysogenic, the prophage is automatically induced when it enters the F‘ cell and the cell is lysed. This is because there is no phage repressor in the F' cell. which normally preVents lambda from entering the lytic cycle. Several new Ht‘r strains of E. coli were independently isolated. All were wild type. but Hfr l was lysogenic for phage lambda. All Hfrs were then mated to a nonlysogenic F" strain carrying mutations in the following genes: am, gal, [(115, pm. pi): r/za. The times oftirst appearance of individual lll‘r genes (wild-type alleles) among the recombinants were as follows (in minutes): Minutes ll t‘r marker H fr 1 Hfr 2 Hfr 3 am 8 60 73 gal 24 44 80 his No recombinants 21 12 [is No recombinants 4 29 pm 20 48 85 pvl'l‘ No recombinants 40 93 rha No recombinants 84 49 ’ Draw a complete map ofthe coli chromosome. showing the distance (in terms ot‘time) between each of the markers and the approximate location ol‘the lambda prophage. Assume a lOO-minute map. H‘lq‘r \ J4" . 8“ ‘~\« <1 i in I / \1 y/ / / \ 6. A temperate bacteriophage has the gene order a b e (I c fg /1 whereas the prophage ofthe same phage has the gene order g /7 u h c d of: What information does this permutation gix e you about the location ofthe phage attachment site? '1’ ‘ t: c (,l 'll V‘Q‘ H/k /.rl_ A ‘T‘XL tub/p13,- “\ 51-.th “tape; . c. I t t; , t In two isolates (one is resistant to ampicillin and the other is sensitiVe to ampicillin) ofa new bacterium. you find that genes encoding ampicillin resistance are being transferred into the ampieillin sensitive strain. To determine if the transfer ofthe ampicillin resistance gene to the sensiti\e strain is b_\ transduction or transformation. you treat the mixed culture ofcells with DNase (an enzyme that degrades DNA). How would addition of DNase into the mixed culture allow you to distinguish between these two modes of gene transfer? "Rm 6/ Dim «\s e- w v t t: l». (Ll t 1:3 v’emAV. W a D NA *t‘w? \ ‘\ \q( ll’\(j fbtfvefi (or. L )4 C.\ flit“ \,!\) I \\r\ ‘le (e v\ ~ t r‘Wwae“ *‘VV'V‘ S T“*v” t. /\'L 1 v ‘ \1 Ln (“W/\g :JC ‘\/ 6' - Lrvabv‘c wile-(’5‘. \\/\ Cn -'~_\ 1 ciflkxi C Lk k1, Ljy k," I V I‘LL < ( VN ,, {’\ i Suppose that a bacterium with the genotype met" his" lys' arg’ picked up from the em ironment a piece of DNA with the markers met' his' l_\'s‘ arg'. ti) What is the name ofthis mechanism (conjugation/ transduction" transformation) of gene transfer between bacteria? » m tow-«w (ii) The following diagram shows the position ofthe recombination events that occurred shortly after the bacterium picked up the piece of DNA. The inner circle represents the bacterial chromosome. Which ofthe possible answers listed below would be the genotype ofthe bacteria following these recombination events? @met’ his+ lys‘ arg+ tb) met; his lysr arg’ t’c) met- his" lys' arg~ (d) mef his' l_\‘s‘ arg' (e) inet' his" 135‘ arg’ 8. In a certain breed of dog. the alleles Band h determine black and brown coats respectiyely lloweyer. the allele Q oi'a gene on a separate chromosome is epistatic to the Band [7 color alleles resulting in a gray coat ((1 has no effect on color). li‘ animals oi‘ genotype 8 b s Q q are intercrossed. \\ hat phenotypic ratio is expected in the progeny? '1' r—\ r / / .' l ‘ / . W " l‘\,r‘ ‘ Li: 3 . . 4/1": t i)// H. (A I ~ l >wi l)‘t~c‘€ l) ” 'l i) rib) 4 i ate ‘ tr) / in it a? l i r /" Q. A true—breeding plant with red flowers was crossed to a true~breeding plant with white flowers. The FI progeny produced red flowers. When l7I x F] matings were made. the following F: progeny \y ere obsery ed: I b I h v Vfi < v; 1850 red , Lvln vie ) t7 W“ " "” J V (370 pink ,1 §3_l white 4 ,. i i lotal: 3351 ’ 'l ” l (a) [fixing your own clearly defined genetic symbols. give the appropriate parental. F]. and F} genotypes in order to explain the above results. fl / 7‘ \E /’ to " * f» ‘ ’l I I . ,) l"! Nv /\/0\2 '01 b 7 /“\/‘*\ / bx" ‘7 )‘r y J “it ' ‘7 a w (7! r A (r w i « . /\ /' ’ bl b .2) ~ \r i l L"\ M‘ (7/ x g: 2/ t (\/0\ \r)/ 5 l (b) What phenotypic ratio would be expected from a testcross oi'zm Fl plant? (l i 7) ‘ L’l t-‘C'K‘Klrr rt; (7,1 ,‘4 t\r VIV\/\ 10. Briefly explain the l‘ollowing ternis: (i) Penetrance ‘ I l , .g . . i l“, ‘r\ \7 va/x ‘vw/‘l g ‘7 V(.V\t,»{flg{ 7 “rm. IVA,\ l A\v\[}l \ V \ (itKN L «\r \V\c)\ (.‘l/ \"\C’» \ OK \r“ a \{A, \a live, wwe- l_ 7 \3( K“ til“ (til i;‘ —‘i‘L\\ (lie'g.h_‘vk,v\»\ '7‘]; ‘sl \iv\i~/l.iv,.(<‘rl‘~t>\ z-.\ ()\>m-._v“(~’\ \fi‘ b.‘ (.177 arc/l i-w guilt”, trail. .LL-\ g \ l4 3‘ \\C(1(/7c. \Lanr tn’("v\’\/> (it) lixprcssmty \{ |C’\\r)il\ ‘i\" L" V'V‘c \‘_l L‘» {Wig — J :j 7‘» "(1 ' \ff \lf Cl ll. liill in the following blanks: (a) ’lhe actiye region of a chromosome inyolyed in replication is a Y— shaped structure called a my. , 5.; (b) During DNA replication. the daughter strand that is synthesized continuously is called the lt'wgl M e) 3M an __. and the strand that is synthesized discontinuously is known as the l CM") the) i,"L‘=,,/\(1l\ (c) If DNA polymerase adds an incorrect nucleotide to the 3‘ terminus. a separate catalytic domain containing a 3‘ to 5' actiyity removes the mismatched base. (’5 “c motv use (d) Initiation 01‘ DNA synthesis on the lagging strand requires short _ up mic r g _ made by an enzyme called ‘9 w 'i ynag . which uses ribonucleotide triphosphates as substrates. (e) The unwinding ol‘the DNA helix at the replication fork is catalyzed by a hat, ( < 5c“ _ which uses the energy from ATP hydrolysis to moye unidirectionally along DNA. 12. The mechanism of DNA replication gives rise to the ‘end-rcplication problem" for linear chromosomes. Over time. this problem leads to loss 01‘ DNA from the ends ot‘chromosomes. Consider one round of replication in a human somatic cell. \Vhich ONE ot'the following statements correctly describes the status ol‘the two daughter chromosomes relatiye to the parent chromosome'.’ (Hint: Diagram the replication ot‘a linear DNA molecule using a single replication bubble) (a) ()ne daughter chromosome will be shorter at one end: the other daughter chromosome will be normal at both ends. (b) ()ne daughter chromosome will be shorter at both ends: the other daughter chromosome will be normal at both ends. (c) ()ne daughter chromosome will be shorter at both ends: the other daughter >_ chromosome will be shorter at only one end. Both daughter chromosomes will be shorter at one end; which is the v opposite end in the two chromosomes. (e) Both daughter chromosomes will be shorter at one end; which is the same end in the two chromosomes. (1‘) Both daughter chromosomes will be shorter at both ends. l4. _._. 'J] (i) \Vhy is a primer required for DNA replication? .- y iv to tJL'TfiC/J eta _,( {‘3 Vg‘rzah ,..r,/ A» i.) is "y; ._ (. N? (A :C’b W-l‘lx-tv: \(S, 8 t t. \k CJJ'\}‘ v {‘06 .0”. ‘re ‘> ” VET .KL " '/ /( ‘ v \ g r \ I, ‘ tf z‘-~ 1 ii’ M“ S 5‘ ‘C L41”; A? ll. [:ACC in L) ’i i“) MD :12» CI! .. : A 33 vhc‘ie._t>t\* I} ‘vk’t 5‘1“ . rants 't'r/ ‘ i . . * (it) \\ hat IS the function ol tlic lollomng enzymes during DNA r'ephcation? DNA Ligase: ._ [U l / 3k 9"} é) l’hn.:‘;+ are \w.’ iw‘ cf'l eel-w fr trvvcv‘ > '1 s A. L "< v \ ./»’)*\ CN ’1“ c 2k 'l‘opoisomerase: r") _ l v ., tvevt'tn’h (we: twin» u-~~‘ erg DNA 60 .r' (1411‘s “it. . ) l); : fink l’fc‘ \CIinqurp’i (In if \N'm (pew..- rc. o-rt xv" ‘ _ i All known DNA polymerases extend the polynucleotide chain 5' to 3' from a "primer." In yiy'o the primer. \xhich is RNA. is later remoyed. To amid loss of terminal sequences different strategies are used by E. calf and eukaryotic genomes. (i) What strategy does E. coli use? { i i e,r \vh,{r_l, \ S (17 AvVevf (A ( A Lg. D‘K} /\ [War ta. ti (7’ >¥‘s\ 'm‘ Peat ape-l (thisth yet“ (“V-Ck 0< H if ( ,cw A524" (ii) What strategy do eukaryotes use in overcoming this problem? 1‘. y ‘ > it \o, He .( org: 5 “(lick x E \t—zwcfc't M, "r <7) \ (46.3 C r.\ ‘p‘C La (Kwk (/igr‘q67i S ' l l A culture of bacteria is grown for many generations in a medium in which the only ayailable nitrogen is the heayy isotope (ISN). The culture is then switched to a medium containing only HN for one generation of growth. and then retumed to a lsN-containing medium for one more generation ofgrowth. lt‘the DNA from these bacteria is isolated and centrifuged to equilibrium in a CsCl density gradient. \y hat would be the ratio ofheay'y. light. and intermediate DNA molecules in this sample? ‘/ 2 \ i€)i\ NT \ i (i\'\’\ CM y'v-uC’C-l‘ I c2~k~<~f 6 (i) \Vhat are L'liargai‘l‘s rules? /\’ l /Xi c r cwi T Ci fijvlil’v\l1f1l»\§ ’ V‘Vm‘c 8 (ii) \Vhen Chargal‘t‘s rules are considered. which of‘tlie dimensions deriVed [‘rom the X-ray crystallograph} of DNA supports a double-stranded structure“? i.‘ dint v\{$ \grgr—c- 1' tL‘t") Left Lx \‘VV\N"( ‘ 6 , xii 17. (i) By differentially labeling the coat protein and the DNA of phage T2. Hershey and Chase demonstrated that (a) only the protein enters the infected cell. (b) the entire V'lI‘LlS enters the infected cell. (c) a metaphase chromosome is composed ot‘2 chromatids each containing a single DNA molecule. (d), the phage genetic material is most probably DNA. (e) the phage coat protein directs synthesis of new progeny phage. (ii) E. m/i cells are put into a medium containing tritiated thymidine for one generation only. The cells are then transferred back to normal medium with nonradioactiVe th_\midine. After one generation in the normal medium. what propotion ofthe cells will be labeled‘? (a) 116 (h) 18 (' ;) 1/4 1/2 (e) All 18. Describe the “GU—AG rule" ol‘splicing. /‘ \ t L A b kcvak A L.'\ \»C C '.: ‘y\1’vx ;>.r\ \sf fi»( \0 L C). l/\\/\ “~43 (it/\glylt x/hnc 3* {Wm 4 wk Mrm .les, the: wry-.94; (A In, x” [\i (A iQ L» U; (A . A f,’ “ a ('r' - Ark; vr - )\ C’p\ \‘f-K; I "‘ (Til \ "\ ‘3 J: \l‘» ' '~ ' ‘i 77 i ' 0 2pm I (1’ LV/ ’lfifi r 3””; \ ("v"~ 'i ‘ iii“ \‘ Act 19. The sigma subunit is a permanent component ofthe RNA polvmerase from E. c w... , t'o/i. allowing it to initiate transcription at appropriate promoters in the bacterial chromosome. Answer if this statement is true or false. Explain your answer. :. A ‘ 3 (Selig \ ,7: It 3( \3 mil"? "Cthi‘wwi'vjAl - ‘\ ’V l I ,t . A .-. /‘ r r\ . .’ ,t ,i c w \\,c c l Vv' VV‘ KN‘A Y‘ ‘VI l/thr'w-il—(V (qu’pk \p-st L(~-‘\\{‘V\ \: » rs. - r “ ‘fiest art is lvni,%qfl«Can .vttb«k' 44 i wirim) i'z’xvxS L « i\"’l .WK 0‘ ’l l7 i I)” W” l (1 i ’3 D’“ fi'\< l7» «r l ‘1'" i \ "I i a“, s; «t «, \' " Lvly' Cf‘jfi'vsi. Ile‘i i\)'\( (Z.(\lf\§" yytflmwfl Zn. l'—'l“-\/:—\ts ; ea. c ltl‘l 3_ WV“ pal luw’a. L‘Q/ efL L L/vk List three ways that the process olieukaryotic transcription dil‘l‘ers from the process ol‘ bacterial transcription. §\'i}:S|t'."\L("\i(y—t ‘i)t¢\((:"'\‘[:‘k("> 1x“ 7*; ,Cvkéy L\. 37 mer M \ arr-V19 ‘v mi W? W V V \3 (MW I 4 "K a.) ( lié l additional protein factors. r (b) In any one region of the DNA double helix. only one DNA strand is i usually used as a template. T (c) Bacterial mRi lAs‘ are processed bel‘ore they are translated. \: (dt lntrons are the coding regions within eukaryotic genes. F (e) RNA polymerases require both a template and a primer before they can initiate RNA synthesis. F The lollow ing table contains a list ol‘statements that apply to replication. transcription. both. or neither. In each empty box. put a check mark il‘that statement applies to replication or transcription. L“l1\t\/ ( at 1 3 c-i l ' V i' . . Indicate whether the l‘ollow ing statements are true or fail (a) Binding to the promoter orients RNA polymerase so that it transcribes the adjacent gene: however. the chpice oftemplate strand is dictated by 6. ... L..\7IC‘\'(’) Bk . £16”, Ltv’w c.) rt) C-5 \,) ‘ t; (y (‘ ‘VL»’\('1) ‘ A \V \ \r V. C l/fi y/t > - The new strand is made 5‘ to 3‘. . The new strand is made 3‘ to 5‘. I l‘he new strand is complementary to the template strand. ':l he template strand is RNA. . The product is DNA, The product is RNA. An RNAJirimer is required to initiate synthesis. Synthesis ol‘the new strand is initiated at a promoter. . ln eukaryotes. the newly synthesized strand is processed while it is being synthesized. + ln prokaryotes. translation is initiated on the newly synthesized t strand ey en before its synthesis is completed t H m. CTN 77 (it: i/ ‘ e, f: Mfiv M93? . V/jxwf: ' figs kj’l-‘Zc :‘K’{;@ .41 Y5 "Ciel Ell,sz no «a \-\’\‘ se. 77 Replication _ Transcription 24. The length ol‘a particular gene in human DNA. measured from the start site for transcription to the end ofthe protein-coding region. is 10.000 nucleotides. \xhereas the length ofthe mRNA produced from this gene is 4000 nucleotides. \Vhat is the most likel_\ reason for this discrepanc) '7 -/-. l: IA has {AA pl by L )~ vv (1" v c; < :7 ‘T0 (by pal/\C «"1 “it/{7 lvfi'tmts ‘\ \(\\.1 x” L w “(Vt (l . \L K \DNA (AU ty\ (“"9" (“’“(r‘ "H ( ‘l Elm li (Melt) ‘swclcc'd WW9 Ad“) 5" kl“ cw \\\ \ee (t mfg-re" ‘ 1 AV gm 3 The following is the sequence ol‘a piece ol‘mRNA. Write the sequence ofthe double—stranded piece of DNA from which it was derived. Make sure to label the strands "template" and "non—template" where applicable. as well as designate the 5‘ and 3‘ ends ol‘cach strand. 5‘ UCC UGA CGA UGC‘ me CGA 3‘ %1 ACWACTECT HLCwATbCLT 5‘ i 67 *LVWVK‘Z’V"? q 1c.Chit-JACAA'W “m: JAA it / ww‘a H’- 0 l1 ...
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This note was uploaded on 03/20/2008 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas.

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BIO325 exam2 2006 - / Jarepk mach GENETICS (BIO 5325)...

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