exam1jeopardy sp 12.pdf

Genetics JeopardyData &HypothSegregationIndepenAssort.PedigreesEpistasisQuant.TraitsPop.Gen.X-linked1 11 1 1 1 1 122 2 2 2 2 2 233 3 3 3 3 33Final Jeopardy

1 pt:You observe that these two corn rootworm larvae havedifferent sizes even though they are the same age.These are the two broad categories of factors thatcontrol the observed size difference.Return to board

Parents: Fast X SlowF1: all Medium2 pts:This is the simplest explanation for the genetic control ofgrowth rate in corn rootworm larvae.Return to boardF2: 270 fast 531 Medium 258 slow

3 pts: Match the ratio with the data with theGenetic HypothesisRatio9:3:412:3:115:1Return to boardDataa. 147 white faced13 solid facedb. 117 pink36 red7 whitec. 36 fuzzy89 hairy35 smoothHypothesisa. A_B_ phenotype XaaB_ phenotype YA_bb phenotype Zaabb phenotype Yb. A_B_ phenotype XA_bb phenotype XaaB_ phenotype Xaabb phenotype Yc. A_ B_ phenotype XA_bb phenotype XaaB_ phenotype Yaabb phenotype Z

1 pt:.Manx cats come in many colors / patternsManx X Manx mating data 344 Manx175 tailedX 2 analysis shows:Significant deviations from 3:1Non significant deviations from 2:1Can you explain with one Punnet square?If yes, HOW?TtTtTTTtTtttReturn to board

2 pts:Cotton fibers can be tough (T_)or weak (tt). A tough F1 (Tt) isself-pollinated and 30 tough F2offspring selected.This is the expected number ofhomozygous tough plants among theselected F2s.TcottontTTTTtReturn to boardtTttt

3 pts:Many F2 cotton plants are self-pollinated.The 2 pt. question told us the F1 were Tt,with the genotypes T_ being tough tt weak.This is the expected fraction of the F3offspring with weak fibers.cottonTtTTTTtReturn to boardtTttt

1 pt:Two parent cockroaches, with genotypesAAbbCCdd and AABBccdd, are crossed.This is the number of different kinds of gametesan F1 cockroach from this cross can produce.Return to board

R__ is Roundup resistant;rr is susceptible to the herbicide.T__ has tough cotton fibers;tt has weak fibers.F1 is RrTt2 pts:This is the chance that a F2 will betrue-breeding for Roundup resistanceand tough fibers.cottonReturn to board

3 pts: Pishu’s father is ttEe , her mother is TtEeTT or Tt : Tabby tt solidEE or Ee: black ee brownPishu: the Obaidi’s black , tabby catPishu has the samephenotype as her mother.What is the probabilitythat Pishu has the samegenotype as her mother?Return to board

1 pt. The is the genotype of Dog F& the genotype of Dog G.brown bb E_black B_ E_yellow __ eeA B C D EF G HReturn to boardI?J?

2 pt. This is the chance thatthe next pup bornfrom the Dog F X Dog Gwill be black?brown bb E_black B_ E_yellow __ eeA B C D EF G HReturn to boardI?J?

3 pt. What is the chance that B will have the N-spot trait?N spot : X n YN spot: X n X nNon N: X N YNon N: X N X NX N X n??A?BCDReturn to board

1 pt:These are the phenotype ratios that suggestepistasis involving two independentlyassorting gene pairs is controlling thedifferences observed in a trait.9:7 9:3:4 2:1 3:1 1:2:1 15:1None of these! Epistasis gives frequenciesInstead of ratiosReturn to board

2 pts:This would be thegenotype of Henry.bbE_ is brownB_E_ is black__ ee is yellowRamona XHenryTheirpuppiesReturn to boardDAILY DOUBLE

ResistantInbred line #1XResistantInbred line #23 pts:Two inbred corn linesresistant to Europeancorn borer are crossed toproduce F1plants andthose are selfed toproduce F2s.F1 all ResistantF2 148 Resistant12 SusceptibleReturn to boardHow many gene pairsControl this ECBresistant trait?How many resistantgenotypes are in theF2?

number1 pt:The graph showsthe phenotypicdistribution ofwool productionin a herd ofsheep.2010This is thereason thedistribution iscontinuous.025Wool yield810Return to board

2 pts:In pounds of production, each capital letter adds 10pounds, whereas each lower case letter subtracts 5.The pounds of wool expected in a F1 produced by 2individuals with the same wool production levels:aaBBCCddEEff x AABBccDDeeffReturn to board

3 pts:Parents withThe highestValue are alwaysSelectedEach generationMatch theselectionprogressline with thetraitReturn to boardHighLowTraitHeritabilityWool production 0.42Milk fat level 0.10Length of 0.33reproductive cycleBAC2 4 6 8 10 12 14Generations

Birds eat bugs as the bugssearch for a mate.1 pt:RR and Rr stink bugs aremore attractive to birdsthan rr stink bugs.This would be the forceimposed on our Stink bugpopulation by birds thatmight change thefrequency of the R allele.Return to board

2 pts:This is the frequency of the R allelein this stink bug population:RR : 14%Rr : 4%rr: 82%Return to board

3 pts:This is expected numbers of RR,Rr and rrIndividuals in the 1000 member stink bugpopulation if the population was randomlymating with respect to this trait.#obs #expRR : 140Rr :rr:40820Return to board

1 pts: Yellow bodies females are crossed withnormal colored males. The following results areobtained.Yellow Male 346Normal Male 0Yellow Female 0Normal Female 314Provide a genetic explanation for these results includingThe genotypes of the yellow female and normal maleParents.Return to board

2 pt: How many females in this pedigreecan diagnosed as carriers of theALD disease allele (X d )?Return to boardFIG A. ADRENOLEUKODYSTROPHY (ALD) PEDIGREEX D X D , X D X dX D YA ?B?DC?X d YFILLED SYMBOLS HAVE THE DISEASE? - DISEASE DIAGNOSIS NOT YET MADE

3 pts: Chance that ‘B’ will behave ALD ?Return to boardFIG A. ADRENOLEUKODYSTROPHY (ALD) PEDIGREEX D X D , X D X dX D YA ?B?DC?X d YFILLED SYMBOLS HAVE THE DISEASE? - DISEASE DIAGNOSIS NOT YET MADE

FINAL Jeopardy:Chance that ‘C’ willhave fast twitch muscle fibers ?F_ slowReturn to boardMuscle twitch phenotypeunknown, frompopulation in whichfast twitch peopleare 4% of the populationand the population israndomly mating for thistraitff fastA ?B?DC?FIG F. Fast twitch muscle phenotype PEDIGREE