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Genotypes  X cannotbedeterminedwithcertaintyandcanonlybeaccomplishedthroughfamilystudies.4$$ B.0  DominantandRecessiveTraits$$  (#33"3"    392#  1  .3  0    Traits aretheobservedexpressionsofgenes.39r݌<$$ Ќ  "3"    3B2#  2  .3  0    Atraitthatismanifestedwhenthedeterminingalleleispresentinasingledoseiscalled dominant ; H thepersonmaybeheterozygousatthatlocusandstillrevealthetrait(i.e.,abrowneyedpersonmay $  alsohaveablueeyegeneandcantransmitthattooffspring).3B{݌  $$ Ќ  "3"    3R2#  3  .3  0    A recessive traitisrevealedonlywhenthealleleispresentinthehomozygousstate(musthave2blue  T  eyegenestoexpressblueeyecolor).3R݌ 0 $$ Ќ  "3"    32#  4  .3  0    Bloodgroupantigens,asarule,are _codominant_ traits;_heterozygotes_Ԁmanifesttheproductsofboth 8  allelespresent(ifapersonhasinheritedanAbloodgroupgeneandaBbloodgroupgenebothare   expressedandtheindividualisAB).3݌ $$ Ќ  "3"    32#  5  .3  0    Todeterminepotentialbloodgroupsinheritedbyoffspringtheparentsgenotypemustbeknown.37݌D$$ Ќ    *!ddd Xdd Xdd XH$H$,dd ,zdd ,dd +  LL  Z Genotypes  X @[['A  X   O  X @  O   @@&_AO_ '" 'OO " @  O  0 _AO_ '0" 'OO"0"  "";"    ;0  2#  a  .3  0P $$  ThisrepresentsoneparentbeinggroupAandtheothergroupO.;A݌hP $P $ Ќ  ";"    ;00  2#  b  .3  0P $$  Inthisexample50%ofthechildrenwillbegroupAand50%willbegroupO.;0i݌@P $P $ Ќ  "3"    3f 2#  6  .3  0    Attimesthegenotypecanbedeterminedbythephenotypesinheritedbythechildren.3f ݌!$$ Ќ  ";"    ;!0  2#  a  .3  0P $$  ChildisgroupO,oneparentisgroupA,oneparentisgroupB.;!!݌p"P $P $ Ќ    *"#dddd zdd zdd !H$H$,dd ,zdd ,dd +  x $x   Z Genotypes  % @[['A  &   O  ' @  B  H( @@&AB 'H)" 'BO H*" @  O   \+ _AO_ ' \," 'OO" \-"  "";"    ;n%0  2#  b  .3  0P $$  Inthisexamplethecouplehasa25%chanceofhavinggroupO,A,BorAB.;n%%݌"< .P $P $ Ќ  C.0  ParentagetestingD$!0$$ 0  1.0$$Manyofthebloodgroupantigensareexpressionsof_codominant_Ԁtraitswithastraightforwardmode %#2 ofinheritanceandareusefulindeterminingexclusionofpaternityandprobabilityofpaternity.&t$3$$ 0  2.0$$Ifoneassumesthatthemotheristrulythemotherandthatthetestingwasdoneproperly,therearetwo |($&5 typesofexclusions.T)&6$$  d. ,< 0  0$$a.0P $$ Directexclusion isestablishedwhenageneticmarkerispresentinthechild,butisabsent X fromthemotherandtheallegedfather.Example:4P $P $ 0  0$$0P $$0 P $P $Child0 $ $0X$$0X$X$Mother0$$0`$$Father `$`$ 0  0$$0P $$0 P $P $A0 $ $0X$$0X$X$O0$$0`$$O`$`$ 0  0$$0P $$ P $P $ 0  0$$0P $$ProvidedthatneitherthemothernorthefatherareoftherareOhphenotype,thechildhas < inheritedtheAgene,whichcouldnotcomefromeitherthemotherortheallegedfather.lP $P $ 0  0$$b.0P $$Inan indirectexclusion ,geneticmarkersareabsentfromthechildthatshouldbetransmitted   bytheallegedfather,givenhisobservedphenotype.Example:  P $P $ 0  0$$0P $$0 P $P $Child0 $ $0X$$0X$X$Mother0$$0`$$Father x `$`$ 0  0$$0P $$0 P $P $_Fy_(a+b)0X $ $0X$X$_Fy_(a+b)0`$$_Fy_(ab+) P `$`$ 0  0$$0P $$Inthiscasetheallegedfatherispresumablyhomozygousforthe_Fyb_andshouldhave X  transmitted_Fyb_tothechild.Sincethechildis_Fy_(b),thereisanindirectexclusion.0 P $P $ 0  3.0$$Directexclusionsprovidemoreconvincingevidencethattheallegedfatherisnotthebiologicfather   thanindirectexclusionsbecauseonlyrarelycanthetestresultsbeexplainedbyestablished ` mechanisms(_eg_,suppressorgenes).8$$ 0  4.0$$Apparentindirectexclusions,however,cansometimesresultfromthepresenceofasilentallele @ (_Fyb_/_Fy_).$$ 0  5.0$$InadditiontobloodgroupantigensDNAtesting,enzymetestingandtypingfor_HLA_Ԁantigens(tissue p antigens)arealsoutilizedindeterminingthepaternityofachild.H$$ D.0  PopulationGeneticsP$$ 0  1.0$$Basicunderstandingofpopulationgeneticsisimportantnotonlyinparentagetestingbutalsoinsuch  clinicalsituationsaspredictingthelikelihoodoffindingcompatiblebloodforapatientwithmultiple  antibodies.X$$ 0  2.0$$Phenotypefrequenciesaredeterminedbytestingredcellsfromalargenumberofrandompeopleof `  thesamerace,thencalculatingthepercentageofpositiveornegativereactionswithagivenantiserum.8!$$ 0  3.0$$Phenotypefrequenciesforagivenbloodgroupsystemshouldequal100%.Example:77%_Jk_(a+)and # 23%_Jk_(a).Ifbloodisneededforapatientwithanti_Jka_,23unitsoutof100(orapproximately1out  h$ offour)shouldbecompatible.!@%$$ 0  4.0$$Ifthepatienthasmultiplebloodgroupantibodiesitispossibletocalculatethefrequencyofthe H# ' combinedphenotypebymultiplyingtheindividualfrequencies.* Mustbeabletocalculateforexam . $!($$ 0  0$$0P $$0 P $P $0 $ $0X$$0X$X$PhenotypeFrequency%%|#*$$ 0  0$$Littlec0 $$0 $ $0X$$0X$X$0$$20&T$+$$ 0  0$$K0P $$0 P $P $0 $ $0X$$0X$X$0$$91',%,$$ 0  0$$_Jk_(a)0P $$0 P $P $0 $ $0X$$0X$X$0$$23\(&-$$ 0  0$$0.2x0.91x0.23=0.04 *'/$$ 0  0$$Fourunitsoutof100wouldpossiblybenegativeforall3antigenslistedabove.Insomecases,due +d)1 tothehighfrequencyofsomeoftheantigensinvolved,itmaybenecessarytocontacttheratedonor ,<*2 supplier.l-+3$$  Z| %&e  EXAM2ONLINE#&e %Z|c?#