Understanding how our genes are passed down And … · And how to calculate the probabilities of our ... • 2 copies of recessive allele = homozygous recessive genotype ... widow’s

Unit3:GeneticsCalculatingthe

probabilityofourgenetics

UnderstandinghowourgenesarepasseddownAndhowtocalculatetheprobabilitiesofourtraits.

Unit3:GeneticsCalculatingtheprobabilityofourgenetics

Leading questions:

1. What do Punnett Squares mean?

2. How do we calculate the probabilities of our traits?

3. What traits will pass down and by how much?


Do you remember these?


What do the y and Y represent and where are they from?


Recall from meiosis

• A pair of homologous chromosomes (chromosomes of similar traits) are crossed over and separated.

• The gene that results in the sex cells are the “letters” from Punnett Squares.


What does that mean?If the red chromosome containsthe genes for Brown eyes (B).

If yellow contains genes for Blueeyes (b)

After crossing over, the genesswitch chromosomes with theother.

bB

bb

B

B

Sisterchromatids


What does that mean?• At the end of

meiosis, all 4 ofthese sisterchromatids willseparate (segregate) into 4 sex cells.

• Because theprobability of gettingB and b is the same, that is why we useBb!

bb

B

B

B

bb

bB


When the genes ofthe homologouschromosomesseparate into separatesex cells, Mendelclassified this as the“Law of Segregation”.

Thiswillresultinasexcellwithtraitsthatareeitherfromyourmom(browneyes)ordad(blueeyes).

Weusesymbolstorepresentdominantandrecessivegenes.• Thedominantgene isalwaysrepresentedbyacapitalletter.

• E.g.A

• Therecessivegeneisalwaysrepresentedbythesameletterinthelowercase.

• E.g.a

Towritethecombinationofatrait,twolettersmustbeused.Ie.AA,aa,Aa


Example

Brownhairisadominantgene

Blondehairisarecessivegene

Brownhair’sgenewillbeexpressed becauseitisthedominant gene


Otherimportantterms• Alleles - Alternativeformsofageneforagiventrait(Bluevsbrowneyes)

• Genotype – theallelespossessedbyanindividualforaspecifictrait.

• Eg:BB,Bb,andbbarepossiblegenotypesforeyecolour.

• 2copiesofdominantallele=homozygousdominantgenotype(BB).

• 2copiesofrecessiveallele=homozygousrecessivegenotype (bb).

• 1copyofeach=heterozygous genotype(Bb).


• Phenotype – thephysicalappearanceofatrait(ie:whatthegenesactually“do”tous).

• Ex.Blueeyes,short,longtoenails,Extrahormones.


Pforphysicaltraits!


Howdowecalculatetheprobabilityofourgenes?

Monohybrid Crosses• Across(combinationofalleles) foronetrait =amonohybridcross.

• Eg:Ifaheterozygouswomanwithawidow’speakreproduceswithaman withthesame genotype,whatkindofhairlinewilltheirchildrenhave?

• (Note:widow’speakisdominant overstraighthairline).

Unit3:GeneticsMonohybridcrosses

• Youmayalsobeaskedtodeterminegenotypicratiosandphenotypicratios intheoffspring:

Genotypicratios:1WW:2Ww :1ww (or,moresimply,1:2:1)

Phenotypicratios:3widow’speak:1straighthairline(3:1)

**Phenotypicratioscanalsobewrittenaspercentages:25%chanceofstraighthairline75%chanceofwidow’speak

Unit3:GeneticsMonohybridcrosses

W w

W WW Ww

w Ww ww

Whatifsomeoneshowsthedominantphenotype,butwedonotknowiftheyarehomozygousdominant orheterozygous?• Wecandoatestcross:

• Crosstheindividualwiththeunknowngenotypewithonethatishomozygousrecessive.

• Whyhomozygousrecessive?

Unit3:GeneticsTestcross

Example:Akidhasbrowneyes(dominanttrait)butwedon’tknowifheisheterozygousorhomozygousdominant.Howcanwedeterminehisgenotype?


B bb Bb bbb Bb bb

Kid’sgenotype

B Bb Bb Bbb Bb Bb

Kid’sgenotype

UseatestcrossThekid’sgenotypecouldbe2possibilities.IfALLoftheoffspringinthetestcrossshowthedominanttrait,thenthekid’sgenotypeishomozygousdominant.

InBorderCollies,blackcoat(B)isdominant toredcoat(b). Abreederhasablackmale thathaswonnumerousawards. ThebreederwouldliketousethedogforbreedingifheispurebredorBB. Tolearnthisinformation,shetestcrosseshimwitharedfemale(bb).


1.Iftheblackmaleis BB,whatkindofgamete(sperm)canheproduce?

2.Iftheredfemaleis bb,whatkindofgamete(eggs)cansheproduce?

B

b

InBorderCollies,blackcoat(B)isdominant toredcoat(b). Abreederhasablackmale thathaswonnumerousawards. ThebreederwouldliketousethedogforbreedingifheispurebredorBB. Tolearnthisinformation,shetestcrosseshimwitharedfemale(bb).


2.Ifanyofthepuppiesarered,whatisthefather'sgenotype?

Blackmalecouldbe:

BBorBb

Testwithbb


Trysomequestionsnow!


PunnettSquareswithBloodtypes• Thereare4mainbloodtypesin

humans:A,B,AB,andO

• BloodtypeAandBaredominant,TypeOisrecessive.

• WeuseIA toindicateadominantAallelegeneandIBforalleleBblood.

• iirepresentsrecessiveforOblood.


PunnettSquareswithBloodtypes

• InordertogettypeABblood,bothdominant allelesofAandBmustbeexpressedtogether.Theyareco-dominant(seelater).

• Ex.IAIB


BloodtypeExample• CouldamanwithtypeBbloodandawomanwithtype

ABproduceachildwithtypeOblood?

• 2possibilitiesfortypeBblood:IBIB orIBi


BloodtypeExample• CouldamanwithtypeBbloodandawomanwithtype

ABproduceachildwithtypeOblood?

• Sincethere’snohomozygousrecessive(ii),TypeObloodisnotpossible.


BloodtypeExample1.WhatifthemotheristypeOandthefatherisA?Whatwouldtheoffspring'sbloodtypebe?


BloodtypeExample2. A woman with Type O blood and a man who is Type AB have are expecting a child. What are the possible blood types of the kid?


Sex-linkedproblemswithPunnettSquares• Traitscontrolledbygenesonthesexchromosomesare

calledsex-linkedtraits.• WeusuallyfocusontraitscontrolledbytheX

chromosome• ThisisbecauseithasmanymoregenesthantheY

chromosome!


Sex-linkedproblemswithPunnettSquares• Notationforsex-linkedgenes:• GenescarriedontheXchromosome:

• UsetheletterXtoindicatethatthegeneisontheXchromosome

• Useasuperscript letterforthetraititself.• Eg:XB =dominantgenefornormalcolourvision

Xb =recessivegeneforcolour-blindness


Sex-linkedproblemswithPunnettSquaresExamples:Red-greencolour-blindness(X-linked)

XB =normal;Xb =colour-blind

Genotype Phenotype

XBXB Female; normalvision

XBXb Female;normalvision(carrier)

XbXb Female; colourblind

XBY Male;normalvision

XbY Male;colourblind

Noticehowthegenotypesarewrittenandwhatthephenotypesare!


Sex-linkedproblemswithPunnettSquares• Note:Becausemalesonlyhaveone Xchromosome,

theycannotbecarriers ofsex-linkeddisordersorconditions.

• Theyeitherhavethegeneandshowthecondition,ordon’thaveitanddon’tshowthecondition.

Genotype Phenotype

XBXB Female; normalvision

XBXb Female;normalvision(carrier)

XbXb Female; colourblind

XBY Male;normalvision

XbY Male;colourblind

Cannotbeacarrier!!!


Sex-linkedproblemswithPunnettSquaresExamplescont’d:• Hemophilia(deficiency

inbloodclottingfactors)• Duchennemuscular

dystrophy• (Theseareallrecessive

conditions)

Sex-LinkedInheritancePracticeProblems• Boththemotherandthefatherofacolorblindmaleappeartobenormal.Colourblindness isrecessive.Fromwhomdidthesoninheritthealleleforcolorblindness?Whatarethegenotypesofthemother,father,andtheson?

Sex-LinkedInheritancePracticeProblems• Awomaniscolorblind.Whatarethechancesthathersonwillbecolorblind?Ifsheismarriedtoamanwithnormalvision,whatarethechancesthatherdaughterswillbecolorblind?Willbecarriers?

Sex-LinkedInheritancePracticeProblems• Boththehusbandandthewifehavenormalvision.Thewifegivesbirthtoacolorblinddaughter.Isitmorelikelythefatherhadnormalvisionorwascolorblind?Whatdoesthisleadyoutodeduceaboutthegirl’sparentage?

Sex-LinkedInheritancePracticeProblems• Whatisthegenotypeofacolorblindmalewithlongfingersiss=longfingers?Ifallhischildrenhavenormalvisionandshortfingers,whatisthelikelygenotypeofthemother?

Unit3:GeneticsTracingx-linkedwithpedigree

PedigreeCharts!

PedigreeChartstrackingx-linkeddisorders


PedigreeCharts!Mydog’sPurebredPedigree:http://www.k9data.com/pedigree.asp?ID=116679

Pedigreechartstracksthegenetictraitsovergenerationssimilartoafamilytree.

SquaresindicateMALE

CirclesindicateFEMALES


Example1:Leopold’sfamily(QueenVictoria’sson)

(a)Whatistheprobabilitythatherothersonwashemophilic?

(b)Whatistheprobabilitythatherdaughterwasacarrier?Hemophilic?

(c)Whatistheprobabilitythatbothchildrenwerenormal?


Example1:Victoria’syoungestdaughterBeatrice’sfamily

(a)Lookingatthepedigreeoftheroyalfamily,identifywhichofBeatrice’schildrenreceivedthehemophilicgene;whycanyoumakethisconclusion?

(b)NoticethatBeatrice’sdaughter,Eugenie,marriedKingAlfonsoXIIIofSpainandhadsixchildren,oneofwhomwasthefatherofJuanCarlos,thecurrentKingofSpain.WouldyoupredictthatJuanCarloswasnormal,acarrier,orahemophilic?


DrawingExample

Amanandwomanmarry.Theyhavefivechildren,2girlsand3boys.Themotherisacarrierofhemophilia,anX-linkeddisorder.Shepassesthegeneontotwooftheboyswhodiedinchildhoodandoneofthedaughtersisalsoacarrier.Bothdaughtersmarrymenwithouthemophiliaandhave3children(2boysandagirl).Thecarrierdaughterhasonesonwithhemophilia.Oneofthenon-carrierdaughter’ssonsmarriesawomanwhoisacarrierandtheyhavetwindaughters.

Whatisthepercentchancethateachdaughter(twins)willalsobeacarrier?

Unit3:GeneticsDihybridCrosses

DihybridCrosses(2alleles)!

Mendel’s2nd LawistheLawofIndependentAssortment:

• TheLawofIndependentAssortmentiswhenthereare2differenttraits onthehomologouschromosome(eyecolourandhaircolour)

• Eachofthesetraitswillindependently separate.

• Forexample:Bb(eyes)andHh (hair)willseparateintoBHBh Hb andhb.

• Theimportantpartisthatthetraitsdonot“sticktogether”



Thisleadsustodihybridcrosses

Trackingtheinheritanceof2alleles

Ex.ApersonwhoisWwSs (widow’speakandshortfingers)andapersonwhoisalsoWwSs havechildren:

1. Figureoutthegametesforeachparent• WS,Ws,wS,andws forboth• PlacethesecombinationsontheedgesofaPunnettsquare

2. Figureoutthegenotypesoftheoffspring(useaPunnettsquare)


FromthePunnettsquare:• Phenotypicratio - 9widow’speak,shortfingers;

• 3widow’speak,longfingers;

• 3straighthairline,shortfingers;

• 1straighthairline,longfingers(9:3:3:1)

Tocalculatethechancesofhavingtwotraits,youmustmultiply theprobabilitiesofhavingeachindividualtraitandreporttheresult.

• Eg:Fromthepreviousproblem,whatarethechancesofhavingachildwithshortfingers andawidow’speak?

• Chanceofshortfingers=¾• Chanceofwidow’speak=¾• Chanceofhavingboth=¾x¾=9/16



Trysomequestionsnow!

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Understanding how our genes are passed down And … · And how to calculate the probabilities of our ... • 2 copies of recessive allele = homozygous recessive genotype ... widow’s