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Dominance Inheritance Review

Dominance Inheritance Review. Incomplete Inheritance Co-dominance Inheritance SBI3U. • When one (1) allele is stronger (so dominant) than the other allele in. heterozygous genotype , resulting in phenotype of the strong (dominant). Examples:.

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Dominance Inheritance Review

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  1. DominanceInheritanceReview IncompleteInheritance Co-dominanceInheritance SBI3U

  2. •Whenone(1)alleleisstronger(sodominant)thantheotherallelein•Whenone(1)alleleisstronger(sodominant)thantheotherallelein heterozygousgenotype,resultinginphenotypeofthestrong(dominant) Examples: a)ForhairtexturetraitCurlyhairisdominant(C) Straighthairisrecessive(c) Therefore,CcgenotypetranslatestoCurlyhairphenotype b)FordimpletraitHavingdimpleisdominant(D) Nothavingdimpleisrecessive(d) Therefore,Ddgenotypetranslatestohavingdimplephenotype c)ForhitchhikerthumbtraitHavingthehitchhikerthumbisdominant(_____) Nothavingthehitchhikerthumbis_______(___) Therefore,______genotypetranslatesto_______

  3. Sampleproblem: Dwarfism,isadominanttrait. Ifbothparentswithdwarfismphenotypehavenormallookingoffspring, a)determinethegenotypeoftheparents b)thepercentageoftheirchildlookingnormal c)thepercentageoftheirgrandchildrenfromthenormaloffspringslookingdwarfish giventhattheirlifepartnersarealsonormal Answers: a)P:DdxDd(becauseifeitheroneof theparentshasDD,noneoftheir childrenwilllooknormal) 1 4 2 4 1 4 b)Dd DDDDd dDddd DD= Dd= dd= =25%(dwarfism) =50%(dwarfism) =25%(normal) c)Zeropercent(0%)F1=ddxdd

  4. Sampleproblem: BreastcancerhasbeentrackedtoassociatewithgenesBRCA1&BRCA2,whichare tumoursuppressorsandfunctiontoinhibitthegrowthoftumour(cellcyclegenes) Itisarecessivetrait.Thatmeans,ifthegeneisnegativelymutatedandthe offspringscarrybothmutatedalleles,sheismorelikelytogetbreastcancer. a)Determinethephenotypeofthegenotypevariantsforthistrait b)Ifthemotherisheterozygousforthetraitandthefatherhasnormalalleles, howmanyoftheir6childrenwill i)becarrierofthecancertrait(heterozygous) ii)behomozygousfornormalBRCA1 iii)developbreastcancer BRCA1brca1 BRCA1BRCA1 BRCA1brca1 BRCA1 BRCA1 BRCA1 brca1 Answers: BRCA1 BRCA1 a)BRCA1+BRCA1normal;leastriskforcancer BRCA1+brca1carrierofthecancertrait brca1+brca1developcancer 2 4 2 4 b)i)6 ii)6 =3children =3children iii)zero

  5. •AlsoknownasBlendingInheritance •Whentwo(2)allelesareequallydominant(samestrength),they interacttoproduceanew,third,phenotype Thismeans,theheterozygousgenotypehasitsownphenotype Three(3)phenotypesareproducedfromsuchcross •AlthoughMendeldidnotobservesuchpatternofinheritancewith peas,therearemanyexamplesfoundinnature. Whenthe heterozygous genotypeproduces adifferent/new phenotype

  6. Aredhibiscusiscrossedwithawhitehibiscus Whataretheirgenotypes? P:(HRHR)x(HWHW) F1:HRHR=¼=25%Red HRHW=½=50%Pink HWHW=¼=25%White F2:Predictthephenotyperatio forself-pollinatedhybrid hibiscus.

  7. YourTurn: Inguineapigs,colourofcoatisdeterminedbyatleastthreealleles. YellowcoatisdeterminedbythehomozygousgenotypeYY, WhitebythehomozygousgenotypeWW,and CreambytheheterozygousgenotypeYW. DeterminetheexpectedgenotypeandphenotyperatiooftheF1 generationwhichwouldresultfromacrossbetween: a)twocreamcolouredguineapigs; •b)ayellowcoatedandacreamcoatedanimal

  8. LetGYGYrepresentyellowcoat LetGWGWrepresentwhitecoat LetGYGWrepresentcreamcoat a)Pgenerationphenotypes: genotypes: …Solutions creamxcream GYGWxGYGW GY GW F1:genotypes GYGY=¼=25%(yellow) GY GYGYGYGW GYGW=½=50%(cream) GWGW=¼=25%(white) GWGYGWGWGW Therefore,theexpectedgenotypicratiois:1:2:1 theexpectedphenotypicratiois:1:2:1 b)Pgenerationphenotypes: genotypes: creamxyellow GYGWxGYGY GY GY F1:genotypes GYGY=½=50%(yellow) GYGW=½=50%(cream) GY GYGYGYGY Therefore,theexpectedgenotypicratiois:1:1 theexpectedphenotypicratiois:1:1 GWGYGWGYGW

  9. 1.Howdy!MynameisBobHoward,andIown20purebredredcows. Somethingstrangehappenedseveralmonthsago.Duringaviolentstorm,all ofthefencesthatseparatemycattlefrommyneighborscattleblewdown. Duringthetimethatthefencesweredown,threebulls,onefromeach neighbor,hadaccesstomycows.Forawhile,Ithoughtthatnoneofthebulls foundmycows,butoverthemonths,Ihavecometotheconclusionthatallof mycowsareexpectingcalves.Oneofthebullsisthefather.Whichbullisit? Alocalcollegeprofessortoldmetousealittlegeneticsdetectiveworkto figureoutwhothefatheris.Hetoldmetocollectinformationabouteachof thebulls,andtoreadarticlesaboutgeneticsandGregorMendel's experimentsingenetics.So,Ididexactlywhathesaid.Icompiledthe information.Now,Ineedyourhelptomakesenseofthedataandtofigure outwhothefatheris. Afterreadingthroughtheinformation,maybeyoucantellmewhymyred cowshad9roancalvesand11redcalves.Idon’treallyunderstandhowthis happened.Whenyouhavedeterminedwhichbullisthefather,pleasetell metheanswer.

  10. ThisisRocky.Heisa2,200poundRedbull.ThecolourofRocky&’s calves,ifmatedwitharedcow,canbedeterminedbyusingaPunnettsquare. Hisoffspringwillalsobeuniqueincolourcomparedtotheothertwobulls. ThisisRufus.Heisa1,920poundWhitebull.ThecolourofRufus’ calvescanalsobefiguredout,ifheismatedwitharedcow,byusinga Punnettsquare.Thecolourofhiscalveswillalsodifferfromtheothertwo bullsoffspring. ThisisFerdinand.Heis2,000poundRoanbull.Thelawsofgeneticstell usthattheoffspringheproduceswillprobablybedifferent,incolour,thanthe othertwobulls’offspring.UsingaPunnettsquare,youcanseethegene combinations,forcolour,thatFerdinand’soffspringcouldhaveifhemates witharedcow.ThecolourofFerdinand’scalveshastodowithprobability. Whodidit? …abonusmarkforfirstcorrectanswerpostedonWikidiscussion

  11. Co-dominance

  12. •Whenthedominancy(strength)oftheallelechangesunpredictably Thatis,sometimesonealleleisdominant;othertimesitisnot Thisresultsinthere-appearanceofthebothparentalphenotypes whichcausestheoffspringphenotypetolookpatchy! •ThegenotypesarenotblendedandtheystillobeyMendel’slawof segregation,resultinginamixtureofthephenotype. normalred bloodcell Haemoglobin ThegeneforhaemoglobinHb hastwoco-dominantalleles: i)HbA(thenormalgene) ii)HbS(themutatedgene) sickledred bloodcell HbA HbS Heterozygous: isgoodforprotectingfromMalaria

  13. Catcoat:orange(OO)xblack(oo)=Tortoiseshell paternal ‘o’allele maternal ‘O’allele mosaicadult

  14. PhenotypeGenotypePicture Antibodies Notes B–Antibodies A-Antibodies IAIAorIAi IBIBorIBi TypeA TypeB IAIB TypeAB Noantibodies Universal Recipient A–Antibodies B–Antibodies TypeO ii Universal Donor http://gslc.genetics.utah.edu/units/basics/blood/types.cfm

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