2005 Final

2005 Final - Name FINAL EXAM BICD100 GENETICS FALL 2005...

Info iconThis preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name FINAL EXAM BICD100 GENETICS FALL 2005 PROF. REINAGEL DECEMBER IS, 2005 3:00PM — 5:59PM g Check this box if you want your graded exam left in the hallway on the 3'“ F1001" of Pacific Hal! for you to pick up. QUESTION POINTS POSSIBLE EARNED 1 g c,— 2 .1 - 3 _ 5 3.5 -6 6 7 _a 8 _! '- 9 —I 10 ‘3 Fl" ‘ ' ‘ ‘15 giddy anal Qanéi hr +113 A*...if I e’r an 3 ‘3 Page lof 16 1. In an experiment like Mendel's, true-bree_ding plants with flat flower petals were crossed with true—breeding plants with ruffled flower petals.“T’he F1 generation plants had ruffled flower petals. You allow F1 plants to @- pgflrtate to tom 3 second generation, and analyzed the seeds of the resulting F2 generation. V a i/b . Assign yoursymbols Dominant allele: A phenotype: EHEHQ Newer Recessive allele:,a phenotype: flat £125.55." . What is the genotypgot the F1 plant? An What types of gametes does the F1 plant produce. in what ratio? [A t_ turmeric It: ‘ eme'la {adj} rs haploirfi What are the of the F2 generation plants. in what ratio? Awe—+1 Mil An: I am . What are the Met the F2 generation plants, In what ratio? ' ' 3 rial—Fred” Ha‘f Two genotypes of F2 plants have the same phenotype. To which of the pure breeding parental plants should you cross them to tell them apart? What ratio of phenotypes do you expect in the resulting progeny in each case? pimps oi the parent to which you will cross: fig For F2 plants with 3pm 53 expect Emmeggtypes in ratiom «II tamed For F2 plants with genotype £5 expect progeny phenotypes in ratlol lgEEtgil [|g+ Page 20f 16 g. In your actual experiment you at F2 plants. Use 51 Chi Square test to eerermrne wnemer me uuserveu data 51E C_ON§_I§TENT Willi Mendei's predicted ratio. with a confidence threshold ct. 9.115. Show your work and put a box 51.01.111.41 the 312 value. the P value. and your game answr 0111ch. 1'355 DLEGIVEJJ equr'l-EA 0411:9104 PM 3511101; I61! I52 1 1.11 16; '1111141156_ L1.51 12”; 17.11 107:.304lt 91;: ‘lfl‘ifiefi-J—E— :_ " NUI 1.141% 7101* Lenin-(«1811+ wH'L-i Mew-tails pref-ifc’rflél 1‘31‘Hd Circle one: YES the data are consistent with the predicted ratio ,fi‘ NO 1 the data are significantly different from the predicted ratio 1d Table 2-2 Critirul Values of the x4 Distribution r] 211' 0.005 05175 0.11 11.5 0.1 11.05 [0.023 P “(fig 1 0.005 111 1 .000 .000 0.015 0.455 2.705 3.541 IE1 05 5.535] 7.570 I 2 0.010 0.1151 0211 1.355 4.505 5.091 7.3711 T131111 10.5117 2 3 0.072 0.215 0.554 2.355 5.251 7.1315 9.3411 11.345 12.5311 3 4 11.207 0.4114 1.1154 3.357 7.7711 9.4115 1 1.143 13.277 14.550 4 5 0.412 0.031 1.5111 4.351 0235 11.070 12.1132 15.0115 15.7511 5 5 0.1175 1237 2204 5.3411 10.545 12.532 14.449 15.512 13.545 5 7 0.905 1.590 2.533 5.345 12.017 14.057 15.013 13.475 20.275 7 5 1.344 2.1130 3.41111 7344 13.352 15.507 17.535 20.090 21.355 13 9 1.735 2.700 4.155 5.343 14.554 15.913 10.023 21.555 23.589 0 10 2.155 3.247 4.1155 11.342 15.9117 113.307 20.453 23.2017 25.1313 10 11 2.503 3.1115 5.5714 10.341 17.275 151.575 21.11211 24.725 25.757 11 12 3.074 4.404 5.304 11.340 10.549 21.025 23.337 25.217 23.300 12 13 3.555 5.000 7.042 12340 19.1112 22.352 24.735 27.5130 23.5117 13 14 4.075 5.53 7.790 13.3351 21.054 23.555 25.1111 251.141 31.319 14 15 4.501 5.252 11.547 14.339 22.307 24.095 27.453 30.5711 32.801 15 ———__._—_._,__——_—___________ Pagc 30f 16 Nan 2. MEL-Mia fruit flies havegrgyvn bodies and Eagles. You have a pure- breeding ygllow-QDQLBQ strain of_ Hits (othenvise normal). and another pure- breeding strain oftlies with peach-colored "eyes (otherwise normal). The allele tor yellow-body is recessive to the allele ior brown-body. and the allele tor peach- eyes ls recessive to the allele for red-eyes. You cross males from the yellow- Lodleg strain to females irom the peach-eyed strain to obtain an F1 generation. m k Mn "3 .4qu I] a. You cross the F1 flies to one another and analyze the F2 progeny. If everything is working the way Mendel predicted fora simple dlhybrid cross with independently segregating traits. what phenotypic classes do you expect to find. and in what ratios? (Simply list the possible phenotypes and state the relative frequencies}. |2| kvoun horlyl'lrul syfl 1 'r-owfn keel f.en<l.g,5 'u yellow Lory r-e—rI eye I yell-w booty-truah aye. b. he yellow-body trait turns out to be caused by a recessive allele of a gene located on the X chromosomeof flies. Using this additional intormation. state the—genotyfi and phenotype of male and female flies in the parental and F1 generations for both eye color and body color. K" :— LI nwt- Lindy‘ X1“ vellfiw boilyl H 3 rte-"l {yell L: peach eye, Genoggpe phenoype Female parent Xfili‘l-l. 1-; mm Letty, Fem-J1 a7; Male parent VYEB Wllw tum-lyl rerl eye, Female F1 X" Willa lore-urn homily, IE3! 93,5 Male F1 Whit Lieer Lair“ mi 97.9, Page 40f 16 Ns c. Even though body-color is sex linked color and eye-color traits still Ja_sson inoepanogntjy lrorn one another. just as Mendel would have wipes—ta. Predict the onenogpic ratios in the F2 generation in this case. Give the phenotypic ratios for males, for females, and overall. Use Funnel sguaregsl, tree diagrams), or probaElllty equation(s) to justify your conclusions. You must SHOW W15 for full credit. \‘"t"'i~k x X‘YEa'l, iFI) >611 30% Yfi H __ W. X‘X'fili "5(ame wtnlzfxafib ‘ mam. xwlt Y‘Ysl. WLL KWth wxwtzr wins 'X‘YEL x“"‘\'"gt, mm. E'Yrul‘ Wm xx T733:— Phenotypio ratios among FEMALE F2: .1 J-ruw-n Luv-1y. rsri sysfil irnrsun Loclyl renoL‘ 1-.le Phenotyplc ratios among MALE F2: 3 LFUWH k"MM. :94 off] New“ Loris“ [55’1er eye: ’5‘. yaliow ind)” Haiti 9ch I yCwa’ bl'JlJYI 15in; | Ey‘.‘ {‘3'} 3| F‘il‘b) Phenotyplc ratios In F2 overall: ’I irran Emir/l r94 eye '1" Vans“: L‘Myl H93] flyt body. Fem-11 em: I ifs-“on! Lion's. F'r‘fflLl’t aye. Page 50f 16 QUESTION 3 ' 3. An individual came into a genetic clinic because of a harmless but very 1313 trait of having point! earg. She reported that she had an uncle with pointy ears too. but no other family history of the trait. She reported her family tree as follows: tat -.i i.--..t thug—H" a. Assuming the trait is hereditary. state the most liker mode of inheritance. and provide the MOST convincing reason for EACH conclusion you reached about the mode of inheritance. Mode of inheritance nwi-gggmfli regéeiv'e Reason] n gaggifi ha a I" N 5 Reason 2 LIV] " In, ' t5 ‘4." 1i“-r_t:i’t'!!.i oHiC-Pt b. All individuals are identified by numbers in the pedigree above. List the individuals that MUST be heterozygous for the trait: Page 60f 16 Name QUESTIONfl-I‘TOTAL POINTS: - I : I _ D]? _ _ ‘\ 4. You cross a mouse with genotype AABBEMII to a mouse with genotype aabbdd to make 5 C Tr an F1 mouse: AaBde. You know that the A gene is tightlyiinkefi to the‘gentmmere of Q: / chromosome 1. the B gene is somewhere on chromosome 1, and the D gene is on /' chromosome 4. You cross your F1 mouse with a mouse from the 3.8be parental strain. One of the resulting F2 progeny has the phenotype A_bbdd. ‘Diagrum the flrfl and _sec_or_Id divisions of a spegifiinleigsis in the Iii mouse that could have produced the necessary gamete to explain the F2 mouse‘s phenotype. Your diagram should clearly indicate lhe_di_t§_qlion of segregationgf the homologs and sister chromatids. any necessary gossover em. and the genogpjg-of all four meiotic products. Circle the gamete that explains—the phenotype of this" F2 mouse in question. fl '_ _ r.“ Hum] neat} ALA framed-e» From mghgg A F' T A El 1 ,l A A [I r _ ,: Z: M II:+_‘:._I+ ifh+ __5_..f " " N ———'a A b J or _ ___ _ .4 L t in _ A I __ ,l L _ J _ "La—a” ' I4L+ ~+rflwa it ‘ h l 49 4 l" T [I 44mg. 3, -————l—.— —L.._ I g l L} . I 1‘ "t L “E |-—r—l- ——+—ur “I I" F I —-—I—-—1- ‘ ___I [at Page 70f 16 Nan - UFSTI‘ION 5.:1‘0TALrom'rs: 5. Unicorns have typical diploid eukaryotic genetics. Normal Unicorns are mm. with blue eyes and ghiteggats. You are studying three genes in. unioorns whose W alleles cause the phenotypes of tatneneesjtarne). mottled coats {mottled}, and purple eyes [purple]. You cross unicorns from a Pure-Meg Dueteergie to unicums from were Melina mmeypflleimin til—6513i“ F1 nnicurns. SHOW YOUR WORK. ' a. State the phenotype andggngtmof the F1: m a fl —3 F1 :5 m (Wu, we; we“ wLne emu) ll Mp 'hnr -} MP You cross these Fl unicorns to a pure-breeding mottled, purple. lame strain and analyze 500 F2 progeny. You obtain the following data; mi; 3‘ 1:1 lll mottled purple lame L" - 'l, H P T’ HP NP 4m? twp l3 purple tame + p N? I” NF 14 tame ’r NP NP f' ‘_ 119 mottled tame + .. P F P 11 mottled purple to in HF N? P 114 wild type r NP If? [06 purple P I' l? l 12 mottled ' m I Elf. "_ E ' TV} I mi up P1 b. Determine the recombination frequency (map distance} between the mottled and tame enes: .-F:1F_ 'I00=fi?v+lll+llvl1)-lflfl:lg LM lnlul gm, 0. Determine the recombination frequency [map distance) betWeen the mottled and purple genes: I'l : IHI" “4' [UM "J unliluliéfll SOD d. Determine the recombination frequency (map distance) between the purple and lame genes: rt: Mli*ll‘-l*12_)-IDU'—SU.-,M-*m|;.qlred 30th Page 80f 16 Nam 6. You are interested in t_hr_e§.geu;§ in bacteriophage. The recessive mutant angles cause plaque phenotypes that were creatively namdlfltzzy. My. and Ml}. Another lab published the following map of the three genes: fuzzy shaky Eurple 2.9 .1?.4 I It To verify the published map. you cross a Mamie with a fuzzy ghage strain by ecu-infecting E. coli at a high multiplicity of infection (every bacterium infected with both types of phage). You plate the resulting iyaate and analyze me phenotypes of the plaques caused by the progeny phage. a. State the genotypes of the two parent phage strains: ‘l 9. J ‘5 ‘F ‘ + r‘ {Pu-11; phage sheila fl Evil Fl: finally Phat}: hm:n b. List all eight of the flssible phenotypg that could result fmm this cross. If 400 phage plaques were examined from this cross. how many plaques of each of the eight phenmypie classes would you expect? SHOW YOUR WORK. cm Wt at H00 Pest-"He e‘kem‘lyfdfi‘ 1 s P {5'1an Fare-16‘] mine IEO ‘F ‘I 1 (fuzzy) hone 1130 4 + -| {WAN—lyre) I 5 i' G Hui—r: . ehhky, nil" lo} I 5 t 5 f (a gag) P P It 34 'l’ ‘ f: {TRIBE Purple) 11 all + ‘i f1 [Par I i I F g 1 Midway, elfiltr] I "E l "4 I run Flt {dawn-Hr“? no u'nlehher'anee) I {IV-[171 $LfiLy m mam... It: (fiHimN‘Iml—l : Qfi-e at} aha-thy (.U‘I‘ fuzzy, (aw-Fla Lac; reason T- (Tfimcaoflqool- I: was “14-4”; 5 [IRE-Jar, Shekyl meple, M, £0 4m-10—e?—2= (MO—HM; that“ Put“er ' “30 {II-1117 Page 90f 15 : QUESI'Ions-irrtnAE-TGINTSe- - ' 8. You have a wild-type strain of E. coli with the genotype A B C D E F You introduce an F+ plasmid into your wild-type strain and isolated a few Hfr derivative strains that you call Hfrl. Hfr2. and Hfr3. You are studying several new genes in E. coli with interesting phenotypes. You obtain a multiply mutant strain with chromosomal genotype: fihsdef a) You mate each Hfr strain to your multiply-mutant strain in a separate experiment. At various times you interrupt the matings and plate the bacteria under conditions in which only the recipient strain can grow. You obtain the following earliest-time—of—entry data, in minutes: Draw a map of these genes that is consistent with the data, including all the genes and Hfr origins, the mm [in minutes). and the direction of transfer of each Hf!- " — . 2 __ ' *——‘—-————__ i‘I‘i'r I- <——4——|—Li—l—i— E C D l in 4 it 5 H Li 1 I; , (Ill-5 mag in min” .91, erg, +—1l——e “I 1 —l #5 db :1 F A E L D H . I: _ .A __ _E_ __.c’""fl_5i'flfi'infl n'i‘ Hint-13:0 _ _ Hes , B [(13 mo silt H‘Fr'l unknown H ienji'ln {knead mare E mite) iii-'r'l Page 120i" 16 N 13) Among the progeny from the Hfr3 mating abow notype: AbcdeF Draw out the gene transfer and crossovefls) that produced this outcome :J H Fr 3 H r 3 é—lfi—t—l—tfi (In-10+ {‘0 5541‘s,) E, .?_ q E o ______._______5 W 5 wt 61 a. C at b F A a a at .fi-erpmofiama the-omoSomg c) Among the progeny from the Hfr3 mating above, you find one that has the genotype: A B o d e f Draw out the gene transfer and crossover-(5) that produced this outcome: "H" '3 (again. «a rm ' {.o genie] H__I__y__ F- I: A E C. E} b F 4 g F;— L t I t I t- ——-—k————t————|---—t ‘1 + b ‘F 5-1 e e A b 'F A 9— n 4 01:1 Mmasamg n Lfirfimoé‘mmg Page 1301’ 16 [om TOTALPOINTS! -- ' | 10. The yeast Fina} exams-hello can live as a haploid or a diploid and has unordered tetrads. This yeast normally has the ability to survive on black coffee as its sole nutrition source. You get a big grant from Starbucks to determine how it achieves this feat. You decide to look for mutants with a cof— phenotm: mutants that cannot grow on coffee'alone. but can grow on a rich nutrient source. You plate out cof+ wild type haploid yeast onto n'ch nutrient plates, then replicate to coffee-only plates and identify colonies that fail to grow on the coffee plates. You obtain 8 inde ndent cof— mutants in this way. a. You cross each mutant to a wild type cof+ haploid of the opposite mating type and analyze the diploids. All of the resulting diploids are cof+. What does this tell you? ( > It” +he mmha'l‘ianfi are f-acefifilw, {cull all—plolél‘qw'l Pheno‘lypa.) cal '— I b. You sporulate the diploids from (a) far each of the mutants. For every mutant, you find that all the tetrads have 2 cof+ : 2 cof- spores. What does 1 this tell you? I’m.) malnhoos are coat/I") m a single firing (is? 3:“ ’ 1 flap] .2 mpg) c. You take cof~ spores of both mating types from the tetrads in (b) from each of the mutants, so that you can cross them tolone another. The table below gives the phenotype of the resulting diploids, where — means the diploid did not grow on coffee, and + means it did grow: 1 d.— What are the complementation groups defined by these mutants? [ll ‘ EL {11(5'Mr95 1: 2‘I Q: \I [5, 4:] What do you predict the result is for the cross not done (? in table)? ) 1 x 7 -—)-— (nu ,fiman, and 7 have mark-Hone in Hue? smote. gem?! ,' ,z' 50 no cm'nF- lemanl‘fi‘l'lbfl scours] Page 1501' 16 d. You take the diploids from each of .porulate them. The possible types of tetrads you could get are: I 4 cof+ : O cof- [I 3 cof+ : l cof- III 2 cof+ : 2 cof- V 1 cof+ : 3 cof- O cof+ : 4 cofi <H Assuming there are no unusual interactions between the genes, which type(s) of tetrads do you expect to get from the diploids from the table in (c) that have cof« phenotypes? Explain your answer. ' .— meiosis "— ——'9 Li Cei" {singlet-{IS} ya {‘YFQ lain“; a... l\l mmi‘nilgne tn saw-a aerial so no Ezfiragufi'l‘ien e‘F H+ Inn i-elmdlf, Assuming there are no unusual interactions between the genes, which type(s) of tetrads do you expect to get from the diploids from the table in (c) that have cof+ phenotypes? Explain your answer. “Fax - * — mast-lies: than (at 2 + —, :- Li M has n I‘ ‘ * News m.- + when \ 1co++12ral— {Hrn:q t+l2:_9f_xd, e. Using these and other mutants, you eventually identify many genes that are necessary for growth on coffee, including the genes COF 1, COFZ, and COPS. You conduct a cross—feeding experiment with the following results: - cofl- mutants incubated in coffee medium won‘t grow, but accumulate spmethin g that con- mutants can grow on .‘.- _ “F1. , cel- . In nevi lea ——3 halal-Aries le‘ growl» 0 cof3- mutants incubated in coffee medium won't grow, but accumulate something that either cofl - or con- mutants can grow on chit-i» coil'Lee. iE—h’ r-hei-uLeIi-i‘es 1C5“ answi‘iq Suggest a pathway for conversion of a precursor in coffee to an essential nutrient. showing the point of action of the products of the three COF genes consistent with the cross-feeding data: ch1 coFs eclipse E X ——3Y——-) melaboiiieg, Far amt-Jib Page Ifiof 16 ...
View Full Document

Page1 / 13

2005 Final - Name FINAL EXAM BICD100 GENETICS FALL 2005...

This preview shows document pages 1 - 13. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online