1
TheGeneTicsRevoluTion
PROBLEMS
Ineachchapter,asetofproblemsteststhereader’scomprehensionoftheconceptsinthechapterandtheirrelationtoconceptsinpreviouschapters.Eachproblemsetbeginswithsomeproblemsbasedonthefiguresinthechapter,whichembodyimportantconcepts. Thesearefollowedbyproblemsofamoregeneralnature.
WORKINGWITHTHEFIGURES
1.Ifthewhite-floweredparentalvarietyinFigure1-3werecrossedtothefirst-generationhybridplant inthatfigure,whattypesofprogenywouldyouexpecttoseeandinwhatproportions?
Answer:Youwouldgeta1:1ratioofpurpletowhite.Thisisbecausethefirst-generationhybridplanthasonecopyofthepurplealleleandonecopyofthewhiteallele,andasaresult,50percentof thegameteswouldcarrythepurplealleleand50percentofthegameteswouldcarrythewhiteallele. Thewhite-floweredparentalvarietyhastwocopiesofthewhiteallele,andallthegametesproduced fromthewhiteplantwillcarrythewhiteallele.Hence,acrossbetweenthetwowouldproducea1:1ratioofpurpletowhite.
Hybridplant P/p ¥ white plantp/p
Gametes 50%P 50%p ¥ 100%p
50%P/p:50% p/p
Purple: white
2.InMendel’s1866publicationasshowninFigure1-4,hereports705purple(violet)floweredoffspringand224white-floweredoffspring.Theratioheobtainedis3.15:1forpurple:white.Howdo you think he explained the fact that the ratio is not exactly 3:1?
Answer:Thisdependsonthesamplesize.Whenthesamplesizewaslarge,theproportionswerecloseto3:1(e.g.,forroundandwrinkledseedstheratiowas2.95:1andthetotalpopulationsize
1
2CHAPTER1TheGeneticsRevolution
was7324), whereasforasmall sample size suchasthe purpleandwhite petal floweredplants (929 plants), the ratio was not as close to 3:1.
3.InFigure1-6,thestudentshave1of15differentheightsplustherearetwoheightclasses(4ft11inand5ft0in)forwhichtherearenoobservedstudents.Thatisatotalof17heightclasses.Ifasingle Mendeliangenecanonlyaccountfortwoclassesofatrait(suchaspurpleorwhiteflowers),how manyMendeliangeneswouldbeminimallyrequiredtoexplaintheobservationof17heightclasses?
Answer:Ifasinglegenecanonlyaccountfortwoclassesofatrait,minimumof9genesarerequired to explain the 17 height classes.
4.Figure1-7showsasimplifiedpathwayforargininesynthesisinNeurospora.SupposeyouhaveaspecialstrainofNeurosporathatmakescitrullinebutnotarginine.Whichgene(s)arelikelymutantormissinginyourspecialstrain?YouhaveasecondstrainofNeurosporathatmakesneithercitrullinenor arginine but does make ornithine.Which gene(s) are mutant or missing in this strain?
Answer:Ifthemutantstrainmakescitrulline,thatmeansgenesAandBmustbefunctional.Therefore,theonlygenethatismissingormutantinthefirstNeurosporastrainmustbegeneC.
Inthesecondstrain,geneAmustbefunctionalsinceitisabletomakeornithine.GeneBmustbe missingormutantsinceitisunabletomakecitrulline.However,geneCmayormaynotbemissing/mutant.EnzymeCconvertscitrullineintoarginine(theyareinthesamesequentialpathway),andenzyme C is dependent on the availability of citrulline for its function.
5.Consider Figure 1-8a.
a.What do the small blue spheres represent?
b.What do the brown slabs represent?
c.Do you agree with the analogy that DNAis structured like a ladder?
Answer:
a.Theblueribbonrepresentssugarphosphatebackbone(deoxyriboseandaphosphategroup),while the blue spheres signify atoms.
b.Brown slabs show complementary bases (A,T, G, and C).
c.Yes,itisahelicalstructure.
6.InFigure1-8b,canyoutellifthenumberofhydrogenbondsbetweenadenineandthymineisthe sameasthatbetweencytosineandguanine?DoyouthinkthataDNAmoleculewithahighcontent ofA+Twould be more stable than one with a high content of G + C?
Answer:Therearetwohydrogenbondsbetweenadenineandthymine;threebetweenguanineand cytosine. No, the molecule with a high content of G-C would be more stable.
CHAPTER1TheGeneticsRevolution3
7.Whichofthreemajorgroups(domains)oflifeinFigure1-11isnotrepresentedbyamodelorganism?Answer:Archaea
8.Figure 1-13b shows the human chromosomes in a single cell.The green dots show the locationof agenecalledBAPX1.Isthecellinthisfigureasexcell(gamete)?Explainyouranswer.
Answer:Itisnotasexcell.ClonedBAPXIgenehashybridizedtotwochromosomesinthecell,indicatingtherearetwocopiesoftheBAPXIgene.Ifitwereagamete,itwouldhaveonlyonecopyofeachchromosomeandoftheBAPXIgene.
9.Figure1-15shows thefamilytreeorpedigreeforLouiseBenge(IndividualVI-1) who suffers fromthediseaseACDC because she has twomutantcopiesoftheCD73gene.Shehas foursiblings(VI-2,VI-3,VI-4,andVI-5)whohavethisdiseaseforthesamereason.Doallthe10childrenofLouiseandhersiblingshavethesamenumberofmutantcopiesoftheCD73gene,ormightthisnumberbedifferent for some of the 10 children?
Answer:All10childrenhaveonemutantcopyoftheCD73gene.ChildrengetoneCD73copyfromtheirmomandonefromtheirdad.SinceLouiseandherfoursiblingseachcarrytwodefectivegenes,alltheirchildrenwillgetonemutantCD73copy.
BASICQUESTIONS
10.BelowisthesequenceofasinglestrandofashortDNAmolecule.Onapieceofpaper,rewritethissequence and then write the sequence of the complementary strand below it.
GTTCGCGGCCGCGAAC
Comparing the top and bottom strands, what do you notice about the relationship between them?
Answer:
GTTCGCGGCCGCGAAC CAAGCGCCGGCGCTTG
Theyarecomplementarytoeachotherandrunintheoppositedirection(antiparallel).Thesequences are also palindromic; they read the same in either direction.
11.Mendelstudiedatallvarietyofpeaplantswithstemsthatare20cmlongandadwarfvarietywithstems that are only 12 cm long.
a.Underblendingtheory,howlongwouldyouexpectthestemsoffirstandsecondhybridstobe?
b.UnderMendelianrules,whatwouldyouexpecttoobserveinthesecond-generationhybridsifall thefirst-generationhybridsweretall?
4CHAPTER1TheGeneticsRevolution
Answer:
a.First-generationhybridswouldhavestems16cmlongbecausethatistheaveragebetween20cmand12cm.Thesecondgenerationwouldbeaproductoftwo16-cmstemmedplantsmating,so they would also have stems 16 cm long.
b.Ifthefirst-generationhybridsarealltall,thentall must bedominant,andyouwould expecta3:1ratiooftall:dwarfinthesecondgeneration.
12.IfaDNAdoublehelixthatis100basepairsinlengthhas32adenines,howmanycytosines,guanines, and thymines must it have?
Answer:Thymines = 32, Cytosines = 68, Guanines = 68
Numberofthyminesis32becausethymineandadeninearepaired,sothenumberofadeninesequals thenumberofthymines.Theremaining136basepairsmustbecytosinesandguanines.Thenumber ofcytosinesequalsthenumberofguaninesbecausethosebasesarepaired. Therefore,thereare136
÷ 2 = 68 base pairs of each.
13.ThecomplementarystrandsofDNAinthedoublehelixareheldtogetherbyhydrogenbonds:G≡CorA=T.Thesebondscanbebroken(denatured)inaqueoussolutionsbyheatingtoyieldtwosinglestrandsofDNA (seeFigure1-13a).HowwouldyouexpecttherelativeamountsofGCversus ATbasepairsinaDNAdoublehelixtoaffecttheamountofheatrequiredtodenatureit?HowwouldyouexpectthelengthofaDNAdoublehelixinbasepairstoaffecttheamountofheatrequiredtodenatureit?
Answer:Double-strandedDNAwithahighGCcontentwouldbemorestablebecauseGCpairs have three hydrogen bonds and hence would require more heat to denature compared with a double- stranded DNAwith high ATcontent.
Asthelengthofthedoublehelixincreases,theheatrequiredtodenaturewouldalsoincrease.Thisis because there would be more hydrogen bonds to break.
14.ThefigurethatfollowsshowstheDNAsequenceofaportionofoneofthechromosomesfromatrio(mother,father,andchild).Canyouspotanynewpointmutationsinthechildthatarenotineitherparent? In which parent did the mutation arise?
CHAPTER1TheGeneticsRevolution5
Answer:ThereisanewpointmutationfromanAtoaC.ThechildinheritedcopyF1fromthefather withoutanydenovomutations,asshownonthebottom.Theothercopy(asshownonthetop)must thereforebefromthemother.ThemotherhasanSNP(A/T)atthefifthnucleotideshown,soweknowthechildinheritedcopyM1(A).TheA-to-CmutationincopyM1isnotobservedineitheroftheparents,soitmustbeanewmutation.
CHALLENGINGPROBLEMS
15.a. Therearethreenucleotidesineachcodon,andeachofthesenucleotidescanhaveoneoffourdifferent bases. How many possible unique codons are there?
b.IfDNAhadonlytwotypesofbasesinsteadoffour,howlongwouldcodonsneedtobetospecifyall 20 amino acids?
Answer:
a.Thereare64uniquecodons(4¥4¥4=64).
b.Inordertospecify20aminoacidsusingonlytwobases,acodonmustbefivebaseslong(2¥2¥
2¥2¥2=32).Thiswouldgive32uniquecodonsthatcouldspecify20aminoacids.
16.Fatherscontributemorenewpointmutationstotheirchildrenthanmothers.Youmayknowfromgeneralbiologythatpeoplehavesexchromosomes—twoXchromosomesinfemalesandanXplus aYchromosome in males. Both sexes have the autosomes (As).
a.Onwhichtypeofchromosome(A,X,orY)wouldyouexpectthegenestohavethegreatest number of new mutations per base pair over many generations in a population?Why?
b.Onwhichtypeofchromosomewouldyouexpecttheleastnumberofnewmutationsperbase pair?Why?
c.CanyoucalculatetheexpectednumberofnewmutationsperbasepairforageneontheXandY chromosomeforevery1newmutationinageneonanautosomeifthemutationrateinmalesis twice that in females?
Answer:
a.Y chromosome.All,or100percentof, Y chromosomesarepassedthroughthemalelineagevia sperm.Theproductionofsperm(spermatogenesis)continuesthroughoutaman’slife,undergoingmanyroundsofcelldivision.EachroundofcelldivisionincludeDNAreplication,whichprovides opportunityfornewmutations,sothemalelineagehasahighermutationratethanthefemale lineage, as explained in the text.
b.Xchromosome.TwooutofeverythreeXchromosomes(66.7percent)aretransmittedthroughthefemalelineage,whichhasalowermutationratethanthemalelineage.Bycomparison,50percentofautosomesareinheritedthroughfemaleswiththelowermutationrateand50percentthroughmaleswiththehighermutationrate.
c.Setthemalemutationrate(mM)as2.0perunittimeandthefemalemutationrate(mF)as1.0perunittime.Sinceautosomesaretransmittedequallythroughmalesandfemales,thentheautosomalrateistheaverageofthemaleandfemalerates:
(mA)=(2+1)/2=1.5
6CHAPTER1TheGeneticsRevolution
Since genes on theYare transmitted only through males, theYrate equals the male rate:
mY=mM=2
Since 1/3 of X chromosomes are inherited through males and 2/3 through females:
mX=(1/3mM)+(2/3mF)=(1/3¥2)+(2/3¥1)=1.33
TheexpectednumberofnewmutationontheYforevery1onanautosomeis2/1.5=1.33.The expected number of new mutation on the X for every 1 on an autosome is 1.33/1.5 = 0.889.
17.For young men of age 20, there have been 150 rounds of DNAreplication during sperm production ascomparedwithonly23roundsforawomanofage20.Thatisa6.5-foldgreaternumberofcell divisionsandproportionatelygreateropportunityfornewpointmutations.Yet, onaverage,20-year-oldmencontributeonlyabouttwiceasmanynewpointmutationstotheiroffspringasdowomen.How can you explain this discrepancy?
Answer:Whileexperimentalevidencetoexplainthisobservationarenotavailable,onehypothesisis thatspermcellsarephysiologicallyweakandtheirnormalfunctioniseasilydisruptedbymutations. Thus,spermthatcarrydeleteriousnewmutationsarelesslikelytosurviveandformazygotewith an egg cell.Therefore, many sperm with new mutations are eliminated prezygotically.
18.Incomputerscience,abitstoresoneoftwostates,0or1.Abyteisagroupof8bits,whichhas28
=256possiblestates.Moderncomputerfilesareoftenmega-bytes(106bytes),orevengiga-bytes(109bytes),insize.Thehumangenomeisapproximately3billionbasepairsinsize.Howmanynucleotidesareneededtoencodeasinglebyte?Howlargeofacomputerfilewouldittaketostore the same amount of information as a single human genome?
Answer:Fournucleotides,withfourpossiblestates(A,T,C,G)canencode4¥4¥4¥4=44=256
=1byte.Ifonebytecanencodefournucleotides,then3¥109 nucleotidescanbeencodedin(3¥
109)/4=7.5¥108bytes=750¥106,or750megabytes.
19.The human genome is approximately 3 billion base pairs in size.
a.Usingstandard8.5-in¥11-inpaperwithone-inchmargins,a12-pointfontsizeandsingle-spacedlines,howmanysheetsofpaperprintedononesidewouldberequiredtoprintoutthehuman genome?
b.Areamof500sheetsofpaperisabout5cmthick.Howtallwouldthestackofpaperwiththe entire human genome be?
c.Would you want a backpack, shopping cart, or a semi trailer truck to haul around this stack?
Answer:
a.Assumingasingleone-sidedpagecanfit23lines,with56lettersineachline,23¥56=1288 letterspersheet.Thehumangenomeis3¥109basepairs,sothenumberofsheetsrequiredtoprint out theentirehumangenome =3 ¥109/1288= 2,329,192sheetsofpaper.
CHAPTER1TheGeneticsRevolution7
b.Ifareamof500sheets=5cmthick,thethicknessof2,329,192sheets=5cm/500sheets¥
2,329,192 sheets = 23,291 cm = 233 meters.
c.Semitrailer truck