Wright BIO 2110 - BIO2110_exam2_fall2015 (14 pages)

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BIO2110_exam2_fall2015



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BIO2110_exam2_fall2015

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Pages:
14
School:
Wright State University
Course:
Bio 2110 - Principles of Molecular and Classical Genetics
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Name BIO 2110 Genetics Exam 2 6 November 2015 calculators permitted see last page for potentially useless formulae Name Name 1 In a field of red flowers you find three different patches of white flowers You name the mutations that give these white flower m1 m2 and m3 These mutations all are recessive Explain the results of crosses 1 and 2 shown below Predict the results for cross 3 Explain the reason for your prediction Cross 1 P0 white m1 m1 x white m2 m2 F1 all red Cross 2 P0 white m1 m1 x white m3 m3 F1 all white Cross 3 P0 white m2 m2 x F1 white m3 m3 Name 2 In C elegans mutations in the him 8 gene result in defects in the pairing synapsis of X chromosomes As a result of these pairing defects him 8 mutants exhibit high rates of X chromosome nondisjunction i In what stage of meiosis would pairing defects first be observed ii What impact if any would these pairing defects have on meiotic recombination iii In which meiotic division would nondisjunction be observed Provide a concise brief explanation for your answer Name 3 Most loss of functions mutations mutations that disrupt gene function are recessive Provide a brief explanation for this observation Name 4 Below is shown a cis dyhybrid cross involving the dumpy wing and short legs genes of Drosophila melanogaster Both of these genes are on chromosome II Derive a map function to express the expected frequency of the Long Wing Short Leg phenotype in the F2 generation in terms of the recombination frequency p between these genes Note that in Drosophila melanogaster males there is no meiotic recombination P0 Dumpy Wings Short Legs x Long Wings Long Legs F1 Long Wings Long Legs x F2 Longs Wings Long Legs Long Wings Short Legs Dumpy Wings Long Legs Dumpy Wings Short Legs Long Wings Long Legs Name 5 In pitcher plants allelic variation in a single gene regulates whether the plant has sticky or slippery leaves and allelic variation in a different gene regulates whether the plant secretes acid or base Are the data presented below consistent with the null hypothesis of independent assortment for these genes Justify your answer statistically P0s are homozygous P0 sticky acidic F1 sticky acidic F2 sticky acidic slippery acidic sticky basic slippery basic x x slippery basic sticky acidic 1842 558 558 242 Name 6 The formation of a double stranded break DSB is the first step in meiotic recombination What happens to the DNA after this DSB is formed How does this second step lead to a meiotic crossover i e what is the second step after DSB formation pictures welcome but not required Name 7 Wild type eye color in Drosophila melanogaster is dark red Recessive mutations in the X linked white eye gene result in flies with white eyes From the cross shown below i what phenotypes would be observed in F1 females and males ii What frequencies would be expected for the indicated F2 phenotypes P0 White eyed x Red eyed F1 x F2 phenotype frequency Red eyed White eyed Red eyed White eyed Name 8 In cave populations of stickleback fish allelic variation in the spiny gene regulated phenotypic variation in spine number Allelic variation in the linked eyeless gene regulates the presence or absence of eyes Based on the segregation data shown below what is the recombination frequency between the spiny and eyeless genes P0s are homozygous SHOW YOUR WORK P0 three spined eyeless x five spined eyes present F1 three spined eyes present x F2 three spined eyes present five spined eyes present three spined eyeless five spined eyeles 816 384 384 16 three spined eyes present Name 9 The larry curly and mo genes are on the same chromosome see figure larry and curly are separated by 20 cM curly and mo are separated by 10 cM Given the data presented below for a three factor test cross determine how much if any interference is present in the region between larry and mo la 20 cM cu 10 cM mo phenotype gene recessive dominant la la LA cu cu CU mo mo MO LA CU MO la cu mo x la cu mo la cu mo LA CU MO la cu mo 349 357 LA CU mo la cu MO 49 45 LA cu MO la CU mo 5 1 LA cu mo la CU MO 98 96 Name 10 Coat color variation in Labrador Retrievers includes three colors black yellow and chocolate Based on the results shown below explain the genetic regulation of coat color in these dogs i e i how many genes regulate coat color ii what are the dominant phenotypes for each of these genes and iii what interactions are there between these genes P0 yellow x chocolate F1 black x F2 black yellow chocolate 162 72 54 black Name Name probability df 0 90 0 50 0 20 0 05 0 01 0 001 1 0 02 0 46 1 64 3 84 6 64 10 83 2 0 21 1 39 3 22 5 99 9 21 13 82 3 0 58 2 37 4 64 7 82 11 35 16 27 4 1 06 3 36 5 99 9 49 13 28 18 47 5 1 61 4 35 7 29 11 07 15 09 20 52 2 values 2 obs exp 2 exp Lod Log10 odds p2 q2 r2 2pq 2pr 2qr 1 p q 1 recombinant gamete frequency p T prob12 prob1 x prob2 pn 1 pn2WAA pnqnWAa W 1 nd 2 div 2 p N 2 x p q r 1 r ln N t2 N t1 t2 t1 1 TT 3 NPD 2 p N p2 classes 2n 1 W p2WAA 2pqWAa q2Waa df n 1 parental gamete frequency 1 p Nt N0 x e rt p2 2pq q2 1 b b 4ac 2a C DCOobs DCOexp par 1 p 1 cM equals a recombination frequency of 0 01 a 1 1 n 1 qn 1 qn2Waa pnq nWAa W 4Ne classes log10 freqext log10 1 W k a I 1 C odds probx prob0 5 rec p pn p0e n 1 1 3 1 2 3 Eco RI GAATTC 1 1 4 6 4 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 heterozygote frequency 2pq n 1 1 1 2Ne 1 2Ne 1 2Ne SCO DCO k fraction of segregating sites average frequency of nucleotide differences in pairwise sequence comparisons


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