1 / 14

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:.

enid
Download Presentation

Dominance Inheritance Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related