Solution Manual for Genetics Laboratory Investigations, 14th Edition

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’s ManualGENETICSLaboratory InvestigationsFourteenth EditionThomas R. MertensBall State UniversityRobert L. HammersmithBall State UniversitySOLUTION MANUAL

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iiiTo TheMany suggestions helpful to users ofGenetics Laboratory Investigationshavebeen incorporated directly into the fourteenth edition. Sources of genetic stocksand specific chemicals, supplies, and procedures are often included with theinvestigations. Additional information can be found in thisManual.We wish to call to the attention of those adoptingGenetics LaboratoryInvestigationsthe need for long-range planning on the part of the course instructoror coordinator. For example, if you were to do one investigation per week and waituntil the fourteenth week of the semester to begin Investigation 13 on "open-ended"Drosophilamatings, it would be impossible to complete the investigation in theremaining weeks of the semester. Since the manual has been designed to meetinstructional needs in a variety of academic settings, you can be selective inchoosing which investigations to do and in determining the sequence in which to dothem. We never use all of the 26 investigations in one semester, nor would we expectothers to do so.While theManualprovides answers to many of the questions inthe investigations, those questions dependant on data collected in the investigationsare generally not answered here. In at least two instances (Investigations 3 and 24),we have provided sets of data collected by students in our classes that instructorsmay wish to share with their students. Finally, we invite your comments concerning theusefulness of our manual; please call to our attention errors or unclear writing thatcan be corrected in future printings. We hope thatGenetics Laboratory Investigations,Fourteenth Edition will serve you and your students well.

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ivTableof ContentsAnswers to Queries and ProblemsPageTo the..................................................................iiiInvestigation 1:Drosophilaand Maize Experiments in Genetics:Monohybrid and Dihybrid Crosses ...................................1Investigation 2:Principles of Probability .........................................6Investigation 3:The Chi-Square Test ...............................................8Investigation 4:Cell Reproduction: Mitosis ........................................13Investigation 5:Meiosis in Animals: Oogenesis and Spermatogenesis .................15Investigation 6:Meiosis in Angiosperms: Microsporogenesis .........................17Investigation 7:Polytene Chromosomes fromDrosophilaSalivary Glands ..............19Investigation 8:Sex Chromosomes and Gene Transmission .............................20Investigation 9:The Sex Check: A Study of Sex Chromatin in Human Cells ............23Investigation 10: Human Chromosomes .................................................24Investigation 11: Linkage and Crossing Over .........................................26Investigation 12: Genetics of Ascospore Color inSordaria:An Investigation of Linkage and CrossingOver Using Tetrad Analysis ........................................28Investigation 13: Open-Ended Experiments UsingDrosophila:Locating a Mutant Gene in Its Chromosome ..........................30Investigation 14: The Genetic Material: Isolation of DNA ............................33Investigation 15: Restriction Endonuclease Digestion andGel Electrophoresis of DNA ........................................35Investigation 16: Amplification of DNA Polymorphisms by PolymeraseChain Reaction (PCR) and DNA Fingerprinting .......................38Investigation 17: Transformation ofEscherichia coli ................................40Investigation 18: Gene Action: Synthesis ofβ-GalactosidaseinEscherichia coli ...............................................41Investigation 19: Chromatographic Characterization ofDrosophila melanogasterMutants ...................................42Investigation 20: Bacterial Mutagenesis .............................................43Investigation 21: Gene Recombination in Phage .......................................44Investigation 22: Polygenic Inheritance: Fingerprint Ridge Count ....................45Investigation 23: Population Genetics: The Hardy-Weinberg Principle .................46

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vInvestigation 24: Population Genetics: The Effects of Selectionand Genetic Drift ................................................ 48Investigation 25: Applied Human Genetics ........................................... 50Investigation 26: NCBI and Genomic Data Mining ..................................... 53SUPPLEMENTAL LABORATORY TOPICSSupplemental 1 ..................................................................... 56Supplemental 2 ..................................................................... 59Supplemental 3 ..................................................................... 61

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1ANSWERS TO QUERIES AND PROBLEMSInvestigation 1I.MEDIUMDemerec and Kaufmann’sDrosophilaguide includes a number of recipes fordifferent media.1The ingredients of another useful medium, which has been attributedto C.B. Bridges of the California Institute of Technology is as follows:A.Ingredients of C. B. Bridge’sDrosophilaMedium20 g of agar200 g of cornmeal145 ml of Karo Syrup145 ml of Br’er Rabbit Molasses2400 ml of distilled water4.5 g of Dowicil-200 dissolved in 15 ml of distilled water2B.Preparing The MediumFirst,dissolvetheagarbyheatingitwith1000ml ofthedistilledwater,andthenaddanadditional1000ml ofwater. Carefully bringthemixturetoaboil.Beforeaddingthecornmealtothismixture,mixthecornmealwiththeremaining400 mlofunheateddistilledwater.(Mixingthecornmealwiththeunheatedwaterpreventsitfromforming lumpswhenyouaddittothehotagarsuspension.)Addthecornmealtothedissolvedagarmixture,stirringcontinuouslytopreventthemediumfromsticking.Nowaddthemolassesandsyrup, andboilthewholemixtureforabout15minutes.DissolvetheDowicil-200(moldinhibitor)in15mlofdistilledwater;addthistothemainmixtureandmixthoroughly.Thecornmealshouldnotsettleoutinthefinalproduct.Pourthemediumintochemically cleanbottles.Thesebottlesmaybeofanysize,but4-oz.wide-mouthedbottlesorhalf-pintmilkbottlesaresatisfactory.Transferthemediumtoabeakerforpouringand fillthebottlestoadepthofabout1in.Takecaretopreventthemediumfromcomingintocontactwiththeneckofthebottle.Placeapieceofnonabsorbentpaper,suchasbrownwrappingpaper,ineachbottlesothatitextendsdownintothemedium (adouble pieceofpaperismoresatisfactory).Usepaperthatisabout1in.wideandbecertainthatitextendsupwardtoapointabout 0.5in.belowtheneckofthe bottle.ThispaperprovidesadryplaceonwhichDrosophilalarvaecanpupate.Cottonplugsmaybeusedtostopperthebottles, butfoamdiSPoplugsaremoreconvenient.3____________________________1DrosophilaGuidemaybedownloadedfromCarnegieInstitutionforScience,BooksOnline,http://carnegiescience.Edu/publications/books_online.2Dowicil-200isavailablefromSigma-Aldrich,P.O.Box14508,StLouis,MO631783FoamdiSPoplugsareavailablefromBaxterScientificProductsDivisionofVWRScientificProducts(Sargent-Welch),P.O.Box5229,BuffaloGrove,IL60089,andCarolinaBiologicalSupplyCompany:www.carolina.com.____________________________

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2Oncethebottlesofmediumareplugged,sterilizetheminanautoclaveat20lbs.pressurefor20minutes.Other itemstobeusedin handlingthefliesmaybesterilizedatthesametime.Waterwillcondenseontheinsidesofthe bottlesastheycoolaftersterilization.Allow48hoursinawell-ventilatedroomforthiswatertoevaporate.Immediatelybeforeyouplacefliesintothebottlesofmedium,shakeasmallamountofdryyeastontothemedium.Theyeastwillgrowandserveasfoodforthedevelopingflylarvae.V.EXPERIMENTAL MATINGSC.Making Crosses1.F1females will be mated to F1males to produce the F2. This mating canoccur prior to or after placing flies in a fresh bottle.2.A test cross is a controlled mating to a recessive homozygote; therefore,the F1female must not have mated before she is placed with the recessivemale.3.Any cross designed to determine the genotype of the individual tested; across to an individual homozygous recessive for genes in question.VI.SUGGESTED MONOHYBRID CROSSESA.DrosophilaCrosses3 and 4.Sample answer: red eyes. The F1flies all had red (wild type) eyecolor, and the F2consisted of about 3 red to 1 sepia (brown).Genes that behave as dominants may be symbolized with capitalletters or with a+subscripted above the gene symbol.VII.INDEPENDENT ASSORTMENTA.Dihybrid Cross WithDrosophila2.F1phenotype: Wild type for both traits.F2phenotypic frequencies: 9 wild type (long wing, gray body) : 3 long,ebony : 3 vestigial, gray : 1 vestigial, ebony.5.Table 1.5PhenotypeObservedNumberExpectedNumberDeviationO-Elong, gray________________________________________________vestigial, gray________________________________________________long, ebony________________________________________________vestigial, ebony________________________________________________6.P1e/evg+/vg+(ebony)×e+/e+vg/vg(vestigial)F1e+/evg+/vg(wild type)F21e+/e+vg+/vg+, 2e+/evg+/vg+, 2e+/e+vg+/vg, 4e+/evg+/vg(all wildtype); 1e+/e+vg/vg, 2e+/evg/vg(both vestigial); 1e/evg+/vg+, 2e/evg+/vg(both ebony); 1e/evg/vg(ebony, vestigial)

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3B.Dihybrid Crosses in MaizeAnumberofalternativecornexperimentsareavailablecommerciallyandmaybeusedtodemonstratetheclassicaldihybridF2ratioof9:3:3:1.Eitherendospermtraitsalone,seedlingtraitsalone,oracombinationofoneendospermandoneseedlingtraitmaybeinvolvedinsuchcrosses.Consultthecurrentcatalogsof suchfirmsastheCarolinaBiologicalSupplyCompanytodeterminewhatmaterialsareavailable.Amongthepossibilitiesarethefollowing:oY/Y wx/wx×y/y Wx/WxF1Y/y Wx/wxF29:3:3:1yellow waxywhite starchyyellow starchyoR/R Su/Su×r/r su/suF1R/r Su/suF29:3:3:1purple starchyyellow sweetpurple starchyoAn F2segregating for two seedlings traits such as a gene for dwarf (e.g.,d5) and one for albinism (e.g.,lw3) would produce an F2ratio of 9 tallgreen : 3 tall albino : 3 dwarf green : 1 dwarf albino.oAn F2in which one trait is an endosperm trait (e.g.,Su, starchy vs.su,sugary) and the other trait a seedling condition (e.g.,Gl, normal vs.gl,glossy) will also assort into a 9:3:3:1 ratio.2.Gene SymbolPhenotypeSustarchy (smooth)susugary (wrinkled)Rcolored (purple)rcolorless (yellow -- no purple)(a)Dominant: starchy (smooth) and colored (purple).(b)Recessive: sugary (wrinkled) and colorless (yellow).3.Genotypes of the two possibilities: (1)R/R Su/Su×r/r su/suor (2)R/Rsu/su×r/r Su/Su.(a)Phenotypes: (1) purple starchy×yellow sugary or (2) purple sugary×yellow starchy.(b)More than one original cross could have been used as shown above.4.R/r Su/supurple, starchy.Table 1.6PhenotypeObservedNumberExpectedNumberDeviationO-Epurple, starchy___________________________________________________purple, sugary___________________________________________________yellow, starchy___________________________________________________yellow, sugary___________________________________________________

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4VIII. GENE INTERACTIONSA.Gene Interactions inDrosophila(This particular mating can be part of anopen-ended experiment in which different crosses involving eye color mutantscan be made. See Investigation 13.)Parental (scarlet)sc/sc Bw/Bw×(brown)Sc/Sc bw/bwF1Sc/sc Bw/bwWild-type (brick-red)F2SymbolPhenotypeExpected RatioSc/- Bw/-Wild-type9/16Sc/- bw/bwbrown3/16sc/sc Bw/-scarlet3/16sc/sc bw/bwwhite1/16B.Gene Interactions in MaizeAmongthepossiblekindsofearsofcornthatshowepistasisorgeneinteractionarethefollowing:Phenotypes of theTrue-Breeding ParentsPhenotype ofthe F1KernelsPhenotypes and Ratiosof the F2Kernels1. purple×whitepurple9 purple : 3 red : 4 white2. yellow×yellowyellow13 yellow : 3 purple3. white×whitepurple9 purple : 7 white4. purple×yellowpurple12 purple : 3 yellow : 1 whiteGivenbelowarethegenotypesforthefourcrossescitedabove.Notethatofficialmaizegenesymbolsareused.1.Pr/Pr R/Rxpr/pr r/rF1Pr/pr R/rF29Pr/- R/-(purple)purplewhitepurple3pr/pr R/- (red)4 --r/r(white)2.CI/CIR/R×C/C r/rF1CI/ C R/rF29CI/- R/-(yellow)yellowyellowyellow133CI/-r/r(yellow)1C/C r/r3C/C R/-(purple)3.C/C r/rxc/c R/RF1C/c R/rF29C/- R/-(purple)whitewhitepurple3C/- r/r73c/c R/-(white)1c/c r/r

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54.y/y R/R×Y/Y r/rF1Y/y R/rF29Y/- R/-purpleyellowpurple12(purple)3y/yR/-3Y/- r/r(yellow)1y/y r/r(white)Other F2ears of “genetic corn” showing epistasis can be obtained.All of the ears cited in this investigation involve two segregating geneloci.The epistasis ratios (9:3:4, 13:3, 9:7, and 12:3:1) are all modifications of theclassical 9:3:3:1 F2ratio. Ears of genetic corn in which three gene loci aresegregating may also be obtained. For example, the F2trihybrid epistasis ratioof 27:37 (a variation of the classical 27:9:9:9:3:3:3:1) could be used for thispurpose.The epistatic F2maize ears are becoming more difficult to obtain from vendors.For this reason, we have printed FIGURE 1.11 in a larger format so that studentscan count each of these different maize ears. Careful counting of kernels willprovide reasonable data for analysis. For the ear in (B), hue of the kernel ismore important than intensity of the color. For example, light red and dark redkernels are all counted as red, and the same is true for purple.

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6Investigation 2II.INDEPENDENT EVENTS OCCURRING SIMULTANEOUSLY1.Both heads: 1/2×1/2=1/4; one head, one tail: 1/2×1/2=1/4; head onone coin and tail on the other: 1/4+1/4=1/2; both coins tails: 1/2×1/2=1/4. Two coins fall heads, heads about 1/4 of the time; heads, tails(and vice versa) about 1/2 of the time; and tails, tails, about 1/4 of thetime. Stated as a ratio instead of a fraction, the expected result is1:2:1.Table 2.3ClassesCombinationsClass OccurringObservedExpected(O-E)3 headsHHH1/2×1/2×1/2 = 1/872 heads,1 TailHHT, HTH,THH3(1/2×1/2×1/2) = 3/8211 head,2 TailsHTT, THT,TTH3(1/2×1/2×1/2) = 3/8213 tailsTTT1/2×1/2×1/2 = 1/87Total8 possible8/8 = 156564.Table 2.4ClassesCombinationsProbability of Each ClassOccurring4 headsHHHH1/2×1/2×1/2×1/2 = 1/163 heads : 1 tailHHHT, HHTH,HTHH, THHH4(1/2×1/2×1/2×1/2) = 4/162 heads : 2 tailsHHTT, HTTH, THHT,TTHH, HTHT, THTH6(1/2×1/2×1/2×1/2) = 6/163 tails : 1 headHTTT, THTTTTHT, TTTH4(1/2×1/2×1/2×1/2) = 4/164 tailsTTTT1/2×1/2×1/2×1/2 = 1/165.a.(1/2)4= 1/16b.4(1/2)3(1/2) = 4/16 = 1/4c.6(1/2)2(1/2)2= 6/16 = 3/8d.Two boys and two girls. There are more ways (6) in which a family canconsist of 2 boys and 2 girls.e.A boy 1/2, a girl 1/2.

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7III.BINOMIAL EXPANSION1.a.(1/2)5= 1/32 =a5d.10(1/2)2(1 / 2)3= 10/32 = 5/16 = 10a2b3b.5(1/2)4(1/2) = 5/32 = 5a4be.5(1/2)(1/2)4= 5/32 = 5ab4c.10(1/2)3(1/2)2= 10/32 = 5/16 = 10a3b2f.(1/2)5= 1/32 =b52.a.1 boy and 5 girls: 6!/5!1! (1/2)(1/2)5= 6/64 = 3/32b.3 boys and 3 girls: 6!/3!3! (1/2)3(1/2)3= 20/64 = 5/16c.All 6 girls: 6!/0!6! (1/2)0(1/2)6= (1/2)6= 1/643.A normal child: 3/4; an albino: 1/4.a.All 4 normal: (3/4)4= 81/256b.3 normal and 1 albino: 4(3/4)3(1/4) = 108/256 = 27/64c.2 normal and 2 albino: 6(3/4)2(1/4)2= 54/256d.1 normal and 3 albinos: 4(3/4)(1/4)3= 12/256e.All 4 albinos: (1/4)4= 1/256IV.EITHER-OR SITUATIONS (MUTUALLY EXCLUSIVE EVENTS)1.EitherC orc gametes; 1/2 + 1/2 = 1 or 100%2.Either the genotypeAAor the genotype Aa: 1/4 + 2/4 = 3/4a.EitheraaB-oraabb: 3/16+1/16 = 4/16 = 1/4b.EitheraabborAaBb: 1/16 + 4/16 = 5/16c.EitherA-bborAAbb: 3/16 + 1/16 = 5/16d.EitherA-B-oraabb: 9/16 + 1/16 = 10/16 = 5/8V.PROBABILITY AND GENETIC COUNSELINGa.4×7:1(aa)×1(Aa)×1/2 = 1/2b.5×1:1(Aa)×2/3(Aa)×1/4 = 2/12 = 1/6c.6×13:1(Aa)×1/2(Aa)×1/4 = 1/8d.10×14:2/3(Aa)×1/2(Aa)×1/4 = 2/24 = 1/12e.3×17:2/3(Aa)×1/3(Aa)×1/4 = 2/36 = 1/18Note: #17 has a 1/3 probability because his overall is his mother’sprobability of being heterozygous (2/3) times his probability (1/2) if hismother was heterozygous.f.3×15:2/3(Aa)×1/2(Aa)×1/4 = 2/24 = 1/12g.16×17:1/2(Aa)×(2/3×1/2)(Aa) x 1/4 = 2/48 = 1/24

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8Investigation 3I.A. APPLICATION OF CHI-SQUARE. COLORED AND WHITE BEANSPurchase four four-pound bags of white beans from a local supermarket. Place thebeans from two of the bags into a large container, cover them with an aqueous solutionof Basic Fuchsin (approximately 0.05%), and allow the beans to stain for approximately30 minutes. Pour off the stain and rinse the beans several times with water; drain andspread the beans out on a sheet of plastic to dry. After the stained beans have dried,place them in a large container. Add the remaining two bags of white beans and mixthoroughly. Sixteen pounds of beans easily provide enough beans for a class of 30.As an alternative method of getting a one-to-one ratio of beans, we have usedequal volumes (not weight) of white navy beans to black beans. Make sure when you buythe beans that the navy beans are of the same size as the black beans. Be sure tocount multiple samples of the bean mix to see that they are indeed a 1:1 ratio.Sample Data from a Class of 24Sample #RedWhiteTotalSample #RedWhiteTotal132435467813322306628232732665314329352681329034163115340332672431932164016331318649533729863517334302636633830764518312311623733931765619328298626832933065920294291585930832563321294346640103173466632228429057411295316611233093026111234030464424316282598Use these or similar data for chi-square tests as described in Part I A of thisinvestigation. Complete Tables 3.3 and 3.4 as instructed.II.CALCULATION OF CHI-SQUAREA.Practice Problems1.OEO–E(O–E)2(O–E)2/ENormal272249235292.1245Dumpy6083–235596.3735Totals3323322χ=8.4980’s Calculation of2χ=8.4980

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9a.1 degree of freedom.c.Less than 1%.b.Reject.2.PhenotypeOEO–E(O–E)2(O–E)2/ER-Su-134149.5-15.5240.251.6070R-susu150149.50.5.250.0017rrSu-146149.5-3.512.250.0819rrsusu168149.518.5342.252.2893Totals5985982χ=3.9799’s Calculation of2χ=3.9799a.3 degrees of freedom.c.10% to 30%.b.Accept.B.Assigned Problems1.PhenotypeOEO-E(O-E)2(O-E)2/ER-Su-348364.5-16.5272.250.7469R-susu119121.5-2.56.250.0514rrSu-139121.517.5306.252.5206rrsusu4240.51.52.250.5556Total6486482χ=3.8745a.3 degrees of freedom.b.Accept.c.10% to 30%.d.Hypothesis2χvalueP-valueAccept or RejectHypothesis3 R- : 1 rr4671812.9765-10%Accept3 Su- : 1 susu4871610.008290-95%Accept

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102.PhenotypeOEO-E(O-E)2(O-E)2/EA-B-10484.2519.75390.06254.6298A-bb6384.25-21.25451.56255.3598aaB-6984.25-15.25232.56252.7604aabb10184.2516.75280.56253.3301Total337337.002χ=16.0801a.3 degrees of freedom.b.Reject.c.Less than 1%.d.Hypothesis2χvalueP-valueAccept or RejectHypothesis1 A- : 1 aa1671700.0267180-90%Accept1 B- : 1 bb1731640.2403650-70%Accepte.Genes A and B are linked -- lack of independent assortment preventsobtaining the expected 1:1:1:1 dihybrid test cross ratio. Each gene,however, behaves according to Mendel’s first law, yielding a 1:1ratio.3.PhenotypeOEO-E(O-E)2(O-E)2/EA601613.77-12.77163.07290.2657B170134.7335.271243.97299.2331AB3644.91-8.9179.38811.7677O690703.59-13.59184.6880.2625Totals14971497.002χ=11.529a.3 degrees of freedom.b.Reject.c.Less than .01%.d.A, 40.15%;B, 11.36%;AB, 2.40%;O, 46.09%.

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114.Family SexRatiosOEO-E(O-E)2(O-E)2/EAll boys170151.437518.5625344.56642.27533 B : 1 G642605.7536.251314.06252.16932 B : 2 G912908.6253.37511.39060.01251 B : 3 G572605.75-33.751139.061.8804All girls127151.4375-24.4375597.19143.9435Totals242324232χ=10.281a.4 degrees of freedom.b.Reject.c.1% to 5%.d.SexOEO-E(O-E)2(O-E)2/EMale50024846156243365.0219Female46904846–156243365.0219Totals969296922χ=10.0438df =1P =< 1%e.Reject.f.5002/4690 = 1.0665/1.0g.Selective advantage of Y-bearing spermatozoa in fertilizing eggs;possible production or survival of more Y-bearing spermatozoa so thatmore male children are conceived.5.FamilyRatiosOEO-E(O-E)2(O-E)2/EAll A-14401458–181210.22223 A- : 1 aa198619444217640.90742 A- : 2 aa970972–240.00411 A- : 3 aa197216–193611.6713All aa1518390.5000Totals460846082χ=3.3050
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