Genetics Jeopardy Data & Hypoth Segregati on Indepen Assort. Pedigrees Epistasis Quant. Traits Pop. Gen. X-linked 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 Final Jeopardy
1 pt: You observe that these two corn rootworm larvae have different sizes even though they are the same age. These are the two broad categories of factors that control the observed size difference.
Parents: Fast X Slow F1: all Medium 2 pts: This is the simplest explanation for the genetic control of growth rate in corn rootworm larvae. F2: 270 fast 531 Medium 258 slow
3 pts: Match the ratio with the data with the Genetic Hypothesis Ratio 9:3:4 12:3:1 15:1 Data a. 147 white faced 13 solid faced b. 117 pink 36 red 7 white c. 36 fuzzy 89 hairy 35 smooth Hypothesis a. A_B_ phenotype X aab_ phenotype Y A_bb phenotype Z aabb phenotype Y b. A_B_ phenotype X A_bb phenotype X aab_ phenotype X aabb phenotype Y c. A_ B_ phenotype X A_bb phenotype X aab_ phenotype Y aabb phenotype Z
1 pt:. Manx cats come in many colors / patterns Manx X Manx mating data 344 Manx 175 tailed X 2 analysis shows: Significant deviations from 3:1 Non significant deviations from 2:1 Can you explain with one Punnet square? If yes, HOW? T t T t TT Tt Tt tt
2 pts: Cotton fibers can be tough (T_) or weak (tt). A tough F1 (Tt) is self-pollinated and 30 tough F2 offspring selected. This is the expected number of homozygous tough plants among the selected F2s. T cotton t T TT Tt t Tt tt
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 F3 offspring with weak fibers. cotton T t T TT Tt t Tt tt
1 pt: Two parent cockroaches, with genotypes AAbbCCdd and AABBccdd, are crossed. This is the number of different kinds of gametes an F1 cockroach from this cross can produce.
R is Roundup resistant; rr is susceptible to the herbicide. T has tough cotton fibers; tt has weak fibers. F1 is RrTt 2 pts: This is the chance that a F2 will be true-breeding for Roundup resistance and tough fibers. cotton
3 pts: Pishu s father is ttee, her mother is TtEe TT or Tt : Tabby tt solid EE or Ee: black ee brown Pishu: the Obaidi s black, tabby cat Pishu has the same phenotype as her mother. What is the probability that Pishu has the same genotype as her mother?
1 pt. The is the genotype of Dog F & the genotype of Dog G. brown bb E_ black B_ E_ yellow ee A B C D E F G H I?J?
2 pt. This is the chance that the next pup born from the Dog F X Dog G will be black? brown bb E_ black B_ E_ yellow ee A B C D E F G H I?J?
3 pt. What is the chance that B will have the N-spot trait? N spot : X n Y N spot: X n X n Non N: X N Y Non N: X N X N X N X n?? A? B C D
1 pt: These are the phenotype ratios that suggest epistasis involving two independently assorting gene pairs is controlling the differences observed in a trait. 9:7 9:3:4 2:1 3:1 1:2:1 15:1 None of these! Epistasis gives frequencies Instead of ratios
2 pts: This would be the genotype of Henry. bbe_ is brown B_E_ is black ee is yellow Ramona X Henry Their puppies DAILY DOUBLE
Resistant Inbred line #1 X Resistant Inbred line #2 3 pts: Two inbred corn lines resistant to European corn borer are crossed to produce F1plants and those are selfed to produce F2s. F1 all Resistant F2 148 Resistant 12 Susceptible How many gene pairs Control this ECB resistant trait? How many resistant genotypes are in the F2?
number 1 pt: The graph shows the phenotypic distribution of wool production in a herd of sheep. 20 10 This is the reason the distribution is continuous. 0 2 5 Wool yield 8 10
2 pts: In pounds of production, each capital letter adds 10 pounds, whereas each lower case letter subtracts 5. The pounds of wool expected in a F1 produced by 2 individuals with the same wool production levels: aabbccddeeff x AABBccDDeeff
3 pts: Parents with The highest Value are always Selected Each generation Match the selection progress line with the trait High Low Trait Heritability Wool production 0.42 Milk fat level 0.10 Length of 0.33 reproductive cycle B A C 2 4 6 8 10 12 14 Generations
Birds eat bugs as the bugs search for a mate. 1 pt: RR and Rr stink bugs are more attractive to birds than rr stink bugs. This would be the force imposed on our Stink bug population by birds that might change the frequency of the R allele.
2 pts: This is the frequency of the R allele in this stink bug population: RR : 14% Rr : 4% rr: 82%
3 pts: This is expected numbers of RR,Rr and rr Individuals in the 1000 member stink bug population if the population was randomly mating with respect to this trait. #obs #exp RR : 140 Rr : rr: 40 820
1 pts: Yellow bodies females are crossed with normal colored males. The following results are obtained. Yellow Male 346 Normal Male 0 Yellow Female 0 Normal Female 314 Provide a genetic explanation for these results including The genotypes of the yellow female and normal male Parents.
2 pt: How many females in this pedigree can diagnosed as carriers of the ALD disease allele (X d )? FIG A. ADRENOLEUKODYSTROPHY (ALD) PEDIGREE X D X D, X D X d X D Y A? B? D C? X d Y FILLED SYMBOLS HAVE THE DISEASE? - DISEASE DIAGNOSIS NOT YET MADE
3 pts: Chance that B will be have ALD? FIG A. ADRENOLEUKODYSTROPHY (ALD) PEDIGREE X D X D, X D X d X D Y A? B? D C? X d Y FILLED SYMBOLS HAVE THE DISEASE? - DISEASE DIAGNOSIS NOT YET MADE
FINAL Jeopardy: Chance that C will have fast twitch muscle fibers? F_ slow Muscle twitch phenotype unknown, from population in which fast twitch people are 4% of the population and the population is randomly mating for this trait ff fast A? B? D C? FIG F. Fast twitch muscle phenotype PEDIGREE