2002Exam

2002Exam - BI206/282 — Exam 1 Page2 Februar 14 2002 FormO...

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Unformatted text preview: BI206/282 — Exam 1 Page2 Februar 14 2002 FormO 1. For an organism with the genotype aBb, in which ‘ D. 3/4 AWE flQ/qxvtrycvw A... E. 1 " ‘ ' I . MMHlll-M‘fios‘lw / ' u WE- .1: .1 V y ‘ i if e,_1 if fig; "’33 '17“? 4. M“ J . ' 2. Among the crosses below, which wrll giv a ratio of genotypes? A. W A... .t 5:: ' B. AaBb X AaBb Q {3? C. AaBb X aabb '6‘” r ' is 5:2 C? AaBB X aaBB pm i, A n C) . V I I 4 Q a w: “I :L “A I > ‘_ l“ 23; . 3. What aspeotw‘of chromosome bweflhayior most olearly aooognts for Mendel'slaw of. segregation?“ . t: .--,_I.-.w_.-.-,., ' I g I ' . 3 Movement ofisrser chromatids to opposrte poles at anaphasev llvo‘f meioqsrsugM—A Moyement orwosofles *0 Opposite poles at anaphase l of eiosis. " C. Crossing over between homologous chromosomes during prophase | of meiosis." D. " Replication of chromosomes prior to meiosis. R ‘ Independent alignment of different homologous pairs in metaphase l. W \,,_.m *demwwm. x 4. If a woman and a man already have;"§_i)_<w_girls,, what is the probability that their next two childrenwwill be QQ s? warms.” ~. A . 1 (mt v») 9M) B. 7/8 C. 3/4 D. 1/2 1 / 4 ‘ l ' I” E B|206I282 — Exam 1 Page3 February 14, 2002 ‘ Form 0 as 5. Two mice with yélglowjg‘ pr du ' s. The F1 consists of 98 yellow and 50 black mice. Which of“ the following statements is (are) consistent with this result? ' Zil l. Yellow fur is causedflby a dominan all.“ _ Yellow'fur isfd'u'e to the expressionof a rgé’ewssyivfeqlethal fine. / lll. Crossing gf'any blaggfj rfiofis‘emto any yellow ‘E‘t'moUSe shOuld yield 50% black mice and Woe. of V " MM“ . 5;. Q R ' || m ‘— l and Ill all of the above 6. In humans, individuals can have the blood type of A, B, AB, and 0. Which of the following statements explains the relationship between thefidf anle alleles? @ IA and l3 are co-dominant to each other and both are dominant to i. W" B. All three alleles are co-dominant C. All three alleles are recessive D. All three alleles are dominant E. IA and l5 are epistatic to i 7. You are working with arelative of the fruitfly Drosophila melanogaster, called Drosophi/a viri/is, and you have discovered,a D. virilis mutant that has a bar-shaped eye similar to the bar mutant of D. melanogaster. You cross thetLbarZsha » .1“ ' ' a true-breeding wild'type Stock of vjri/is. (Wild type D. viri/is hampherical eyes similar to D. _melanggaster.) The fishowwofwild type and the other 50% show a novel got-eye‘phenotypme. You follow up thiS‘tirst cross by crossing two dot-eyed flies to“eachtothuer.mifiwt’h'evptogeny you find tom; 9 Mnotypes: dot-eyed, WETQ'jt’y‘Wp‘E‘f'bjal- gagged, and stamshapednat a ratio of 6:3:2:1” respectively. 420 1 Which of the following statements is/are consistent with your data from your crosses? W} 6) Two genes appear to be involved in the control of eye shape I“ 7 ll. One- gene appears to be involved in the control of eye shape Ill. The original was hwyggus for the allele that causes the bar- shxaped phenotype land I” L‘ _ All of the above Bl206/282 — Exam 1 7 Page4 February 14, 2002 Form 0 8. In tobacco, either of the, genes, Fl or J, are required to make purple flowers. Plants homozygous recessive for Fl or Jaré purple. in addition, if [rJJ is crossé‘kao RRjj, the resulting F1 IS purple. .R' and II encode similar enzymes in the biochemical pathway needed to synthesize purple pigment. What is the expected phenotypic ratio resUlting from self-fertilization of the F1.dihybrid? Q... 3 A. 9 purple: 3V‘o'range: 3 pink: 1 white B. 9 purple: 6 pink: 1 white C. 9 purple: 7 white Yiijp>wwm . D. 1 purple: 1 white . CE? 15 purple: 1 white Mia Wm M 953 \ M 9. You are a commercial berry farmer with two truebreeding strains, one with red berries and ogewwitbmwbite. You cross these strains and obtain all red berries in t‘heFt. in the F2, you observe the ratio of 12. red???) golden: 1 white berriesfwli you cross random individuals bearing white Berries with individuals Bearing golden berries, what mm”. phenotypic ratios do you expect to see in ing‘iy‘igua‘l’licfg’ssmes? - .» Wm M was...ng WWWWM r.» ire.“ my) fl @1111” A. all white ' E X ‘3 » 1 golden: 1 white ‘ all golden m... WWW LX rrWW 1 W WWW D. B a C y W E. A & B 1 (WW )1 Yr ’ WAT-a bylaw WWW-«‘A‘AW 10. In corn three dominant genes are necessary for kernel color. Therefore, the genotype B_D_Fi_ is colored. Any homozygous, recessive for one gene iszcolorless. Predict the phenotypic ratio of the offspring of the crossESB/BD BrXQQngFir. Q .— \" V 114.11“ # a. l a .0 g coloredzcolorless 7 Pl I Li "’ rt A- 927 ‘7." .1 :- 3-17.:- f f q: f; H; b l . 23:41 I " " ‘ ‘ ‘1 E. 1323 n, 27/“f ' r “ ' f‘: 7' ' pmerng ( 11. At which stage ofgmeiomsis do pairs offchvromosomes‘.align in theflc‘e’nte‘r‘not the Eu. gummmflmmg ,.._,._ < a. _ f\‘ g?” ‘prophase | W 2’ metaphase! C. anaphase l D. metaphase H z E anaphasell Bl206/282 — Exam 1 Page5 Februar 14 2002 ‘ FormO ' 12. The following diagram shows a somatic cell and a germline cell Ier the same diploid organism. ' ( ) ‘ ’ > somaticce” Z“: q germline cell .1 1, lg! What is the chromosome number of Vs‘o‘rnatic cell inthiswgigflfim ' j 2n=2 Xx Adm,“ I . 2n = 4 Hg,“ “‘“C. 2n = 8 i D. 2n = 23 E. 2n = 46 13. The following peflgLee is concerned with an inherited dentalabflmmality, amelogenesis imperfecta. KAXA w Which mode of inheritance céEfiot account for the transmission of this trait? A. AUtO‘somal recessive w B. Autosomal dominant C"). X-linked recessii/"e‘\ )g—linlged domi anti.me *2 Nate ‘arth’a’ébo “e1” Bl206/282 — Exam 1 Page6 February 14I 2002 " Form 0 14. Which of the following sex chromosome genotypes will be the same sex in both fruitflies and humans? (Assume a 2n complement of autosomes in all cases.) A. XO " @ C. XXY D. XXXY E. XXYY (15. and 16.) A mouse with normal gait was crossed to a mouse displaying an odd gait called "dancing." E1 animals all showed a ngrmflalgajt. When F1 animals were crossed to each—otherZZ normal and 5 dancers were obtained. Use chi square analysis to test if the data fitm'el in which "dancing" is caused by being homgzmous for the reggsive allele of ongflgeaggf ' 4225 v degrees of 0.05 37%;!“- A chi square table is given at left: .. I [Ff/’1- nmw " (DD V ‘l r, a"; a" d”. Mar f}, ), MN ‘1’ a e ;, g o '2»: 15. What do you decide about the proposed hypothesis? A. Reject the hypothesis. Fail to reject the hypothesis. M" C. cannot be determined. There are no answers D and E for this question 16. Regardless of your answer to the previous question, assume for thisquestion that the hypothesis is correct and "dancing" is caused by beinghomo; gougjor the recessiye’lallele"of oneg‘e‘ne. Which phenomenon could helpdexplain the rata. interference reduced penetrance unequal crossing over incomplete dominance epistasis .m .U Bl206/282 — Exam 1 ' Page7 Februar 14 2002 FormO 17. The following drawing represents the A chromosomes of a particularjemale frgitijh/Myou are working with. One chromosome contains a recwggsige mgtgnt allele of the camagQQ/Vgfiemnfle (can and dominant‘mutant all'eTe‘ST'Bar-eyeslBan. This chromosome also has a constriction that is microscopically visible. The other X chromosome has wild-type alleles of both carnation (calf) and Bar-eyes (Bah) and the chromosome has piece of the Y chromosome attached to one'end. After mating this female to wiigg-ypgurmgigs, a portion of the male offspring have, wild- pe carnation eyes and mwgtgrltfiaggygs. What is the structure of the X chromosome -« if these‘male flies? Bl206/282 — Exam 1 Page8 February 14, 2002 . Form 0 (18. and 19.) Three rmmutgtions that affect the singingvoicesflgfucanefles are: ginger, posh and scary. Among therogeny from a trihybr’idNFt female that was testcresse’d; the folIcWi'ng—‘phenotypes were obseryed: M ' " '” " P 5 r” --posh, scary 380““7*~~._ re s :1.» res» \ .gfnger ‘3 V may; ,4: emgen’sCary‘ ~ ' " p. posh _ ginger, posh 6L! ginger, posh, scary ‘ C9 e $435 § 25* ‘ wild type 18. Which of the following is the correct manic:.theebovedata? 9mg] er 13.5 cM sci” 8.5 cM poslh cary 13.5 CM flsca o h in er h "(V 13.5 cM {'0 ‘ 8.50M g g e U. . posh 12.7 cM £80 W 7.7 cM gmger A. ' ginger posh. 8.5 CM D 'n er \ . s r osh 9' g 22‘gM /08 y 8.5 CM p it c in 19. Approximately what is the yalue for interference in this cross? A. V 0.0 \ B" Q i' { M3 , 3;“; I K 0.7 If r { j I > *’Q 3 ’ \J l D. 1.0 \ E. 1.4 j} \ A /\ ~/\ , Q, . Bl206l282 — Exam 1 Page10 Februar 14 2002 Form 0 20 The yeast Saccharornyces cerevisiae has unordered tetrads. in a cross made to study the linkage relationships among threegenes, the following tetrads were obtained. The cross was between a strain of genotypé + b sand-one of genotypeiéflhy}; "i aw; Batté'rh .2? '2. , r " i‘r - I P 1» ‘ _ L I, a (I) I ‘ .. .‘ ’ . r . AMA % Which, if any~ of V I “2 w____e.,f£l.lowin ~>g_e,n§§‘§[§wljpjgedfltg‘ge h other? (Hi ? :eparately for each gene pair you Wish to test.) "*1: aandba- X m nt: Analyze B. b and c G ED a and c at. « D. a, b, and c 0 H j E. none. = } \.._. 4r l%5{/>lt4 éOL /. 21. A haploid Neurospora strair! requiring methirognjwngflmillsv ..iOSsedrto,thenrwild— ' A r%nissected to give the sp re orders shownin M. . ' ,;+ 4 am Em t “h '3m + + 8 5+ $+,l+ ‘ mzm m hat is the mggjifiiance (in map units) of thewgys to itsEe’nygmgrg 25 M 09w)“ -_ 30 A 35 w I E- 50 H I l0 '\/ 0 §ng mkwkm B. c, Bl206/282 — Exam 1 Page 1 1 February ‘14, 2002 Form 0 22. stranded DNA, then guanine would make up What percent of thflefl bases? lf adenine makes up 31% of the nucleotides in a given samgfilewofwgouble- .. Mmuflyu wrwafiwmmwmsw. A. 22 % “M” m A 7 C75 B. 30% if is», A: c. . 4o % ,, A Va 7~ 27 % L. - s . 15 % 23. Which of the following nucleic acid moleculesngguld»servegjrectlyas a simus‘trate forvaNA replication EQJvmerase? H ~ “0 ,nmw,w gflA. j 5'—guEa accccu—3' 3M; - B. 24. by making the second set of niclgs on different “DNA strands than the firstpair of nicks 3’—cagtatggggaccgtta—5? 5'—cagtatggggéccgtta—3' 3'—gtcata ~cct—5' 5'—gtcatacccCt—3' 3'—cagtaggggga—5‘ 5'—gtdata / ccct—3' 5'—gucauaccccu—3' 3'—caguauggggaccggpa—5' "@qu xzaaffl/V4 ’ When resolution of a Holliday structure (a recombination intermediate) occurs 3 will result in: recombination between two outside markers no observable recombination between outside markers no heteroduplex formation no gene conversion no meiosis Which of the following statements is/are true regarding heteroduplex DNA? H eroduplex DNA is formed when non-sister chromatids exchange strands Mismatches in heteroduplex DNA can leadto gene conversion Hereroduplex forms on, both chromatids involved in the. crossover I II III l and Ill all of the above Bl206/282 - Exam 1 Page12 Februar 14 2002 Form 0 26. A researcher set out to repeat the Luria-Delbrggk fluctuation experiment which described the spontaneous nature of mutation. The, researcher inoculated twenty small flasks each with a single bacterium derived from an antibiotic sensitive strain and let them grow overnight. The next morning the researcher noticed that all but one of the flasks had come open and thus was ruined. Not wishing to redo the experiment the researcher used the one remaining intact flask and plated aliquots from the culture onto twenty antjbjgtic-cogtaining plays. What results would you expect torsreeon These twentyplates and hOW wouldrthese- ffiflliiQQEflBfiEfi..I9.1.D.§.,.Q.rtgiflal Luria- Delbriick experiment? A. Each plate would hag/fife approximate same number of antibiotic resistant colonies thus replicating L ria and Delbrtick's results. B. Each plate would have extremely different numbers of antibiotic resistant colonies thus replicating Luria and DelbriLick's results. kw. Each plate would have the approximate same number of antibiotic resistant colonies which is different from Luria and Delbriick's results. D. Each plate would have extremely different numbers of antibigticnresistant colonies which is different from Luna and DelbriJck's results. ” E. Like Luria and Delbrtick's results, no antibiotic resistantcolonies.wjll be recovered because no antibiotic was present when the culture was originally grown UP: ' H 27. The following table is a complementation table for six”. recessive mutations that cause white flowers in peas. - indicates mutant phenotype, + indicates wild—type phenotype. ' If ’5 2: \fl (0 . 5 Using the table, how many different genetic Igngere identified? ’2 l A. 6 ,1 5 ~~~~ W ,2, 4/ L ,I' .1 . mpdw> Bl206/282 — Exam 1 Februar 14 2002 Page 13 Form 0 28. Which of the following statements is/are true regarding a oomplementation test? ’W‘mmwmm’lahu ...;..».‘.-»<.;: ,u I. Only reeessive alleles can betested ii. if two mutatioggjyein’lhe samevgene, they can complgm if the mutations are at different ends of gene v H H ‘ ' lll. A large deletion that removes two adjacentgenes should fail to complement single base mutations in either gene I ll III | and ill all of the above 29. Below is shown a map of a series of deletion mutants of the rllA gene (or cistron) of T4 phage. The region deleted is indicated by the size of the box. For example mutant 1272 deletes all 6 regions (A1 - A6) of the rl/A gene, while mutant PT1 deletes only region A4 - A6. }—————————————————rHAdamn___________________l deletion mutants :m;2t,m_w “ ;ffifi;fi jfiéff7fi Q; You have isolated a single base mutation‘in r/IA and you cross it to the above a - etion mutants. You find that yowuwoannotgenerate-wildjypée recombinants with any of the strains. In which regiqugfjflelllAgenedoesyour mutation lie? A. m ‘ m“ ' - B. A2 C. A4 D. A5 “MET; A6 m; Bl206/282 — Exam 1 Page14 February 14, 2002 Form 0 30. In Neurospora, compounds A, B, C, D, and E are chemical components of a biosynthetic pathway. Four mutations were found that affectedTh‘i'spathway and they were tested for their ability to be rescued by the various compounds. The table below shows the results of the experiment. "+" indicates growth; "—" indicates no growth. media supplement Which strain carries a‘muktationwifl the final step of the biosynthetic pathway? A. 1 ' ' B. 2 3 “D. 4 E. none of the above ‘31.) The followingxompares an RNAtr‘awnsgrflitled from a wild—type coding region to .QQ’RNAcontaining: a "+1", frameshift mutation "induced by the”intercalatingagent?" pro‘fla‘VinT'KA "+1" fra'me’shift mutation is an insertion of one base.) As in theurl/Bygene of bacteriophage T4, intragenic suppressors can be isolated by a deletion amm‘base. The underlined base is the original +1 mutation. The reading frame is shown’ras codons. wt 5’ aug ccc gau cgu gga ggc gcc ccg +1 5’ aug ccc gga:ucg ugg agg cgc ccc g Below are shown five of these potential imragem'c suppressors, Which one will ‘ obyigusly not functionflafis suppressor? ‘ W 33% +1, —1 5’ aug ccc gga cgu gga ggc gcc ccg +1, —1 5’ aug ccc gga ucg ggg ggc gcc ccg +1, —1 5/ aug ccc jgggwycg +1, —1 5’ aug cccfiggujegu qga ggc gcc ccg ,a. x ,r "a. ,. » 1x» +1, —1 5’ aug ccc ggagucgfiugg agg;cgclccg B|206l282 — Exam 1 " Page 15 February 14, 2002 Form 0 32. in a certain species of flowering plants with a diploid genome, four enzymesare W involved in flower color. The genes that encode these four enzymes are on different chromosomes. The biochemical pathway involved is as follows: whiteL reeni> blue» fl Tu le 9 —»‘/ l” (Note that either of \twogdifferent egz mes is sufficient to convert a blue pigment into a purp‘lfiepiflgwment.) A trUe-breeaingggggg‘hfijgwseargd plant is mated with a true-breeding blueiflvowered plant. All of the res'ultin’g'mF1 plants havepur lie towers. The F1 plants are allowedib‘iff‘s; If fertilize yielding an Eggrgfie‘nerfiation. What'flratig do you expect for 7% . “Oweer coler in. the F2 gehrergtioo" m :4 atawhiteadugreeoegoblueswlspurple WWWfiEégree-m 9 purple : 3 blue : 4 green . 15 purple : 1 blue . E. 9 purpleswaabluea3ugreene~1 white . {I I‘ \0 A_lb£9l0{ Gill W10? 6?“? ML ” 5L. 4% i g N I act at; mi 7 A" WC“ V P . 3. {a pariah/i») : 4% fl ' w W 3 , P llo'. 3&8“ ftp. 01 n FF ( F a P; K , a Q) w jig; C: {:9 BI206/282 — Exam 1 Page 16 February 14, 2002 Form 0 (33. and 34.) red-eyed, Iongfliflged tiies are bred together and. produce the offsprin in in thejollowi “table: ' phenotype Males mj red-eyedm red-eye vestigial-winged m— ,, H white-eyed, Ion-winged - “ white-eyedgvesxiiaI-winged - 33. Inthecross aboye which traitsisawwm' - A. ‘h ' ' WW B ,C- 7; vi all few—phenotypes-Me co-dominant Which statements is/are consistent with the data presented? ~Wi r e-,is“an;X-link_e_§i«trait Both eyecolormandmwingwsflgpe are autosomai traits /iH. ‘VEye coiowrwirswruan’. X-iinked trait} egg mg A. l = il - iii D. l and III E. all of the above END ...
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2002Exam - BI206/282 — Exam 1 Page2 Februar 14 2002 FormO...

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