Exam2Fall2001

Exam2Fall2001 - EXAM #2 Prineiples of Genetics [Biol 2153}....

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Unformatted text preview: EXAM #2 Prineiples of Genetics [Biol 2153}. Lurkin Full Zflfll 9- Remember to fill in your name and ID number on the answer sheet! Choose the 11m answer for etielt question by filling.r in Lhe uppropriule spsee on your answer sheet. There are 33 questions. 1. Which of the following was one of the people who won the Nobel Prize in l962 for discovering the structure of DNA? at Maurice 'Wilkins. bl Erwin Churgol'l". L‘l {Isis-'on Avery. Ell Alfred Hershey. e} Rosulind Fruuklin. 2. lit in pttt'tieulur kind of Dmsopht'n. wild-type individuals have dull red eyes. Reeessive mutations in nit ]e[oj'1th genes. t-ermii'lt'ret tt-l or garnet tie}. result itt bright red eyes. Ft homozygous wild—type fly w‘ns crossed to it l'tontoeypous- recessive fly to generate F1 lteterozygotes. A female with wild—type eyes Iteteroeypnus [or both 1' uhtl i,- wtis test-crossed with it tnttle Iuwing bright retl eyes tltttt was homozygous [hr the reeess'iye ttlleles of both genes. 1t‘t’ltttt results wtntld he espeeted if these two genes t-ttte ttnlittltetl'.i it] “4 wild-type. 3M bright red. bl 3M wild—type. 1M bright red. el It? wild—type. HE bright red. til lflowilil—tyre. [fifltihnght red. e] lfitlh wild—type, It'io bright red 3. The hetqu phenotypes obtained from the cross described in Q. 3 IWere us follows: 1|l't’ild—type 45 Bright red tilt-{t totnl fire tltese results consistent with independent essonntent. or do the genes uppeitr to be linked? The Table of 5% eritiesl yulues is on the lust pnge of the exam. NOTE: (Inly use the two observed or expeeierl phenotypie classes in your 32 test. Lit the four genotypie classes. it} 32:1.D. so eeeept Independent Assortment. The genes ere not linked. h} x2=3 i .3. so reject Independent Assortment. The genes are linked. c} xflzflrflfli so rejeet Independent Assortment- The genes are linked. dI x2=4ilil so rejeet Independent Assortment. The genes ole linked. 4. “What is the recombination lrequeney between the genes? Hint: If you know the frequency of just one parental eless. you nun deduce Lhe t'racLion that one recombinunts. o} The genes are unlinked. so about 5092. Questions 5 through T refer to the results of the following test cross involving three genes: In Dresnphr'ln. the reeessit-‘e trait hlael: body is eontrolled hy the 1!: allele {h’ results in the normal grey body). The recessiVe trait “durnpy” short wings is controlled by the nip allele tdp’ results in the non’nel long wings]. The reeessiye trait hooked hrisdes is controlled by the hit allele thk“ results in norrrtal bristlesl. A phenotypieafly n-‘ild—type fly heterozygous for alleles of rd] three genes is erossed with it fly that is phenolypieally wild—type for all three eharaeters is crossed to a dumpy. hlnel-t. hooked male. Progeny with the phenotypes given below result from this cross. Assume that units not mentioned in :1 particular phenotypie class are wild—type. phenotypes oi" ni'ogeny # of flies eontpletely wild-type loll lilaek lEl' hluolt. hooked EDI dtiinpy. hooked it hooked it hooketi.du1'r1fiy.hi:1eit T2 tltnripy. liltielt t5 tltinipy ELIE Illlltl total 5. 11' these three genes assorted independently. we would have exneeted: it] Iiitit' ohenoiynie classes. 11] ton: lilienoiynie classes ot'ithotit Still l'lies eeeh. e] more than eight oltenotyn'tt' elasses. tl] shout 1’25 flies in eaeh oi'the eight phenotype: classes. ti. 1t‘l’hieli ol'the Following ohenotypie elttsses results from only a single eross—oyerheto'een.:1‘y11111dl1'£r it} completely wild—type. h] hliielt. e} dunipy. hooked. Ll} hooked. e} dumpy, hlaek. T. What is the correct gene order for these [time genes?I e} rtp—h—lii; a} lil’— e’p — h bl dp— ltk— t: {1} none of the shove. it. 1|ill-That is Lhe map distenee in old hetWeen rip and h? a} 5.4 em 11} 1D.2 eM e} 54.1eM d} 35.5 CM e} none of the shove. 9. Consider the following genetic map: If- CM 3 CM A—B—C From the testeross A B Cfn i: e 3: a i: efn h .5, what proportion of individUHln Would have the A B C phenotype? ai {13813 b} 3.38? ei H.412 Eli H.426 if}. The following eronn involvianr three genes was perfonneti: An Bi: Ce .1; mt hi;- ce, Hflli the phenotypes Lim‘l nutnher of progeny are Shown below. Each pair of alleles exhibits. snnple dominance. and rooennivenens. Biting-133' “ht-fly]ng Numbfl oi mggguv A E C BIG :1 h e Elli :1 B C l'ii‘ii A l“: L: liifi .4": B I: 5.3 n h L" 5| n B t: 47 A h c in ill-flit Which poi: oi’genen in clearly linked {no x2 teat neeessnry‘i'.’ n} A tuni B. h} B nntl C. e} A nntl C. d} All three genes are linked. o} None oflhe genes are liniteti. ll. Which of the following eonld enuee formation of n tiieentrio chromosome? :1; Single cross—over in n helerozvgote with one normal and one porioentrie invereion hontolog. hi Two snnnd tieo. in n heterozygote wiLh one normal and one perioentrio inversion homolog. oi Single eroits—over in n heletozygote with one normal nod one pornoontrie invention homolog. d} Two strand {Leo in a heterozygote with one normal nod one pornoenn'ie inversion homolog. e} None of the above. l2. A certain gene. A. has two alleles. A and n. lDine population consists of [El AA homozygotes. if} Aa heterozygotes. and EU as homozygotes. 1|What is the allele frequency of the A allele”:1 al 411% hi 3W2 c} 20% d} toss ci 5% i3. Human albinism is an autosomal recessive trait. .='t geneticist finds an isolated village tvhere li'4 ofthe population is albino in phenotype. If the village contains 1901} people. how many of these people would you expect to he heterozygous? ai 5D hi lfiii cl Eilti d] 5Uii e] ifiti iii. The disease phenyiketoneona {PKth is a recessive aotosoniai disease. About li’fitiliti children in Iltc IIS are horn with Hill. A pedigree is shown heiotv fora family. viith Frill-affect individuals indicated hy filled circles. Based on both the. pedigree and your khan-ledge of the allele frequency in the population, tvhat is the chance the child lahcied X. 1.t-iil he have PRU? Note that the child‘s father has married into the family I'roni the general population. c} nests [flaw] X e} None. of the above. a} Uiiijftfi lo] [H.133 l5. Cystic fibrosis is a recessive autosomal disease. Affected individuals rarely survive to reproductive age. The assumptions of Hardy-Weinberg equilibrium allow us to make a reasonable estimate of the frequency of heterozygous carriers in a population. based on the frequency of affected individuals. However. this estimate is not entirely accurate. Which of the following is the most signific'flt reason that estimates of cystic fibrosis allele frequencies using Hardy—Weinberg are not completely accurate?1 a} Small population size. bl Non—random mating. c} Mutation. :1} increased fitness of heteroaygotes in regions +with malaria. e} Selection againut recessive homozy'gows. 16. The following information eoneerns four lines in a particular population. George is a very large. health}: and strong lion. Sport};r is a very agile and adeniahie lion. aisle to quickly change his feeding habits in response to unfavorable conditions. Sandy was killed by an infection in his foot. Additional information is as follows: Ehvsieai Tm iLs jig George Sn ortv 5311511; Bod}i weight 15D REID 175 ll'fl {lhsi Tote] euhs fathered Ill i5 14 19 Age at denih {yrs} 115 13 1?. 9 Which lion had the highest relative fitness“? n} Ben h) George e) Spom' d'] Send}; I?. B-"s eon‘tplelely reoessiw allele 3 is lethal when homozygous. The dominant allele L routines in the lelhal i allele at: £1 title of lil"' per generation. Whtit will be it: frequeney oi' the l allele when il reaches rnuratiow seieelion equilibrium? ii} 0.5 h} {3.1 e} [LU] dinner it ellxlUfi u a} X! I! “II-5”" id. The moleeule in the right is ____T ' _ . r-. ea . a .H ainnTP - I “'i ' N hidGTP iv L e) ATP m ore -\¥_%/{F§x 19. Which etTow points to The 5' posiLion of the moleeeie referred to in Line previous question? it} #1 h} #2 e} #3 d} #4 20. Upon analyzmg the nucleotide ham cnmpositinn of Lhe: DNA fur saver-a] virus genomes, you Ubfifin tha following rasulta'. Basa Campoaifion {LE-I Virus A G T C I 21 29 3] 29 2 3?. 13 33 13 3 9 4] 9 ~11 4 19 3] SI 19 Which oflhe viruses. 1111151 haw a singIE—MTLm-EI DNA 3311011137 a} Virus 3 1:} Virus, 2 :2} firm 3 {I} Virus: 4 45'! [Em-1‘1 {all i'mm the Lima given. 2i. Imagine Ihm Maselfinn and Slah] had repealed 1hei1' famnus experiment. [his Lime lam-ling "I": will: MN {lighl N}. and 111911 nunsfcring In 15K [Ecum- '!. Which hf the t'uilnwing centrifuge {uhmz shows Lhr: I'EHUJL 3mm wnuld cxmct ui‘lurlfig gcnuralmns mfg;th in ESN‘.’ a] h} m d} H H H *3» H H H 22. Which of the: foilnwihg prntcins is M 13311 of Um core nuclenmmt? a} HJSICII'JE H1. 1:} Histnne HEA. c} Histcma 1-1213. :1} Hismne H3. 3} HiRlunE H4. 23. You have assembled all of the components to perform DNA replication in a test tube. The template molecule is a covalently closed circular plasmid contending an crimp of replication. In addition to dNTPa and other components needed you add radioactively-labeled fimntflceotide n'iphosphates {NTPai in which the carbon atoms in the rihose sugar have been labeled with “C. After the reaction has run to completion, in what form will you find the radioactivity? of] Free in solution as nucleotide uiphosphates tNl'Psi. h'} Free in solution as nucleotide monophoephates tNMPsl. c_‘I Incorporated into the plasmid as REA polymer. d} Incorporated into the plasmid. but converted to DNA polymer. 24. You perform the same experiment as in the previous question. again including radioactively-labeled IiilllIlUiEEutide triphosphates {EMTPS}. but this time you leave out the numeric enzyme. All other components needed for replication are added. After the reaction has run to completion. in what form will you find the radioactivity? a] Free in solution as nucleotide triphosphntcs {I‘ll—Pa}. b} Free in solution as nucleotide monophosphates i'leTsi. c} Incorporated into the plasmid as RNA polymer. d} incorporated into the plasmid. hut coni’crtcd to DNA polymer. 25. In the diagram at the right. Ute arrow labeled A points to: a} DNA polymerase 1. hi DNA polymerase Ll]. cl Primaae. d} Lipase. e} Helicase. Ed. .In the diagram at the right. the arrow la‘oelch points to: a) DNA polymerase I. b] DNA polymerase III. e] Primase. d} Ligase. e} Helicase. I'I LI] 2?. Consider the Sheri, partially base—paired DNA sequence beiew. 5 ' CTMCCTE ‘ 3 ’ CGCETGCCTCGATTGGPATCAGTE ' What is the full base that will he ineemerared ii dIXTPE and DNA pelg.-‘mera5e are added te the lube? a] A b} (3 e] C d] T e} U 28. A single strand ef a deubIe—atranded DEA Sequence is ShflW'fl heiew. Feur lfl-beee lung primer ee ueneea are alee shown. All fear primers ean baae~pair with ene ef' the twe strands ef the DN in the regime that are nei underiined. Sequence ef tep strand ef' deDNA: 5 ' CC'CTTGATIH'Tfifi—j‘flfl" -: - ~ .—'1rrI‘-II—|‘-r-_'1’r F‘uf-Ifi—lfl-l rr-r-I-In- -r| -'- I " ' "'-...'-...i-'-._1-"'--.'_.;J I? "I II In T-‘J: :i.r.'|-':1' f1 1 : E ‘ CCC’J‘TGAT‘A'I I: ' Fern 'F-fiEE-ifl'f CAGE}: ’ [-' r."i me-I.‘ 'r'l 3 : ’ ETATCMGSGE ' :r' r 231$: 1' ‘i ‘. E ' G‘IT'EA'I'TC'TT] ' Which ef Ihe feilewing primer pairs Wequ he able he :impiify [he underiined regieu ei’ DNA in a PCR rerueLieIi';J :1} primers ] and 1 hi primer}; 2 and 3 e} primers 1 and 4 d} primere 3 and 4 '29. A mutatien in a paru'euJar gene eliminates funeiien ef the TATA he); Which ef the feline-wing weuld eeeer'r1 a] Sigma faeterweuld fail Le bind. b] The 5' eap weeld her he added re The eanseripi. e} Aeljviadrig eanseripe'en faerers would [10! bind. d} Neeleeeemes weuld fail eff at the premeter. e} TBP weuld ner‘hied. 3th. The following. sequence shows a small fragment of DNA that is transcrihecl. The upper strand is the ending sn‘nna‘. E- “ -_31TT$C—El ' 3 ’ -TFLACGG- E“ The RNA that will he produced by n~anscription of this DEA is: a] 5 ’ -GGCP.AU-3 '. bl 5 ' -UAELCGG—3 ‘. El 3 ' -GGC‘3‘-_=t'f—5 '. a} s ' -aunucc-3 *. cl 3 ' ~auuecc-st. 31. The sequence ola hypothetical primary RNA transcript. prior to RN fit splicing is given helow: 5‘JACUGGQJACCAL‘UAQCLIL'AG-S' The sequences in the RNA that guide splicing tsplicejuactioa sequences} are underlined. What will he the length of the spliced niRl‘Cs’t'l a] 2i} nucleotides. [1} H nucleotides. e; 1?. nucleotides. d] H] nucleotides. 32. 1Which ofthc following plays an essential role in RNA seducing“.1 at HHRHA. hit pol}th tail. el {35 sequence. tl} promoter. c} none of the above. 33. Which tilthe following does not contain RNA” a} TBF. h} Riheseme. c} Spiieeosnn'te. ell CendensecL functional [2". cell chromosome. e] All 01' the fll'lt't‘t't- contain RNA. ...
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This note was uploaded on 05/05/2008 for the course BIOL 2153 taught by Professor Larkin during the Spring '03 term at LSU.

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Exam2Fall2001 - EXAM #2 Prineiples of Genetics [Biol 2153}....

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