Wright BIO 2110 - BIO2110_f2016_exam2 (14 pages)

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BIO2110_f2016_exam2



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BIO2110_f2016_exam2

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14
School:
Wright State University
Course:
Bio 2110 - Principles of Molecular and Classical Genetics
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Name BIO 2110 Genetics Exam 2 4 November 2016 calculators are permitted see last two pages for potentially useless information 1 In cuttlefish allelic variation in a single gene results in variation in swimming speed Allelic variation in a linked gene results in variation in the ability of cuttlefish to change their skin color Based on the segregation data presented below determine the recombination frequency between these genes P0s are homozygous SHOW YOUR WORK P0 Fast Polychrome F1 Fast Monochrome F2 Fast Monochrome Fast Polychrome Slow Monochrome Slow Polychrome x x Slow Monochrome Fast Monochrome 4 125 1 875 1 875 125 2 Below are shown segregation from reciprocal monohybrid crosses for two genes the fat head gene in grasshoppers and the tetrodotoxin gene pufferfish For each of these genes explain the segregation patterns of F1 and F2 phenotypes All P0s are from populations that are fixed for either the dominant or recessive allele reciprocal crosses for the fat head gene cross 1 fat head thin head P0 fat head fat head F1 thin head F2 156 fat head 0 thin head 153 fat head 0 cross 2 thin head fat head thin head thin head thin head 129 fat head 0 thin head 134 fat head 0 reciprocal crosses for the tetrodotoxin cross 1 toxic nontoxic P0 toxic toxic F1 toxic F2 137 nontoxic 0 toxic 68 nontoxic 69 cross 2 nontoxic toxic toxic nontoxic toxic 53 nontoxic 47 toxic 49 nontoxic 51 3 In octopi allelic variation in the tententacle gene results in variation in tentacle number Allelic variation in the fuscluster gene results in frequent fusions of clusters of suckers on the tentacles Based on the segregation data presented below are these two genes linked P0s are homozygous SHOW YOUR WORK P0 ten unfused x eight fused F1 eight unfused X eight unfused F2 eight unfused ten unfused eight fused ten fused 435 163 173 29 4 Meiotic recombination crossover is precise to the base pair Explain based on the molecular mechanism of recombination how such precision is possible 5 Based on the test cross data presented below place the evrs to tnkr and chnc genes on a recombination map in their proper order IT IS NOT NECESSARY to calculate recombination frequencies between these genes Progeny genotypes are shown only for the chromosomes derived from the heterozygous parent Among progeny reciprocal genotypes are grouped together cross 1 evrs tnkr chnc EVRS TNKR CHNC x evrs tnkr chnc evrs tnkr chnc EVRS TNKR CHNC evrs tnkr chnc 315 315 EVRS TNKR chnc evrs tnkr CHNC 35 35 EVRS tnkr chnc evrs TNKR CHNC 15 15 EVRS tnkr CHNC evrs TNKR chnc 135 135 cross 2 evrs to tnkr EVRS TO TNKR x evrs evrs EVRS TO TNKR evrs to tnkr 360 360 EVRS TO tnkr evrs to TNKR 40 40 EVRS to tnkr evrs TO TNKR 90 90 EVRS to TNKR evrs TO tnkr 10 10 to tnkr to tnkr 6 Based on the epistasis data presented below construct a genetic pathway for the regulation of eye number in Brachypelma genotype single mutants sxorb hexocular decocular xyes teneye double mutants sxorb decocular sxorb xyes sxorb teneye hexocular decocular hexocular xyes hexocular teneye phenotype six eyes six eyes ten eyes ten eyes ten eyes six eyes ten eyes six eyes ten eyes ten eyes ten eyes 7 Allelic variation in a single gene results in flower color variation in Laqueum diaboli Based on the segregation data presented below describe the dominance relationships for the alleles of this gene P0s are homozygous P0 purple x white F1 pink x F2 purple pink white 315 627 312 pink 8 In Gavialis gangeticus recessive mutations in the pink gene result in pink teeth the wild type tooth color is white Recessive mutations in the unlinked toothless gene result in animals without teeth From the dihybrid cross shown below i what phenotypes would be observed in the F2 generation ii out of 800 F2s how many of each phenotype would be expected iii briefly explain how you determined the number of expected animals in each phenotypic class P0s are homozygous P0 toothless x pink teeth F1 white teeth white teeth phenotypes F2 number expected hu ey du ey lou ie 9 The huey duey and louie genes all are on the same chromosome see below huey and duey are separated by 10 cM duey and louie are separated by 30 cM Based on the segregation data presented how much if any crossover interference is present in this region of this chromosome Progeny genotypes are shown only for chromosomes derived from the heterozygous parent Chromosomes are shown in reciprocal pairs SHOW YOUR WORK 10 cM huey duey louie HUEY DUEY LOUIE 30 cM x huey huey HUEY DUEY LOUIE huey duey louie 315 309 HUEY DUEY louie huey duey LOUIE 37 39 HUEY duey louie huey DUEY LOUIE 141 135 HUEY duey LOUIE huey DUEY louie 11 13 duey louie duey louie 10 Null mutations are mutations that completely eliminate gene function In some genes null mutations are dominant Provide a biochemical explanation for the dominance of loss of function mutations 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 Lod Log10 odds 2 obs exp 2 exp p2 q2 r2 2pq 2pr 2qr 1 p q 1 p q r 1 recombinant gamete frequency p r ln N t2 N t1 t2 t1 1 TT 3 NPD p 2 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 T prob12 prob1 x prob2 pn 1 pn2WAA pnqnWAa W 1 nd 2 div p 2 N b b 2 4ac x 2a C DCOobs DCOexp par 1 p 1 cM equals a recombination frequency of 0 01 a 1 1 n 1 qn 1 qn2Waa pnqnWAa W 4Ne classes log10 freqext log10 1 W k a odds probx prob0 5 rec p pn p0e n I 1 C 1 1 1 2 1 1 3 3 1 1 4 6 4 1 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 SCO DCO Eco RI GAATTC 2Ne 1 2Ne 1 2Ne k fraction of segregating sites p average frequency of nucleotide differences in pairwise sequence comparisons


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