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Wright BIO 2110 - BIO2110f13_exam2answers

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Name _______________________ BIO 2110 – Genetics Exam 2 1 November 2013 10 questions / 10 points per question calculators allowed print outs of alignments and ML phylogenies allowed see last page for potentially useless formulae.1. Draw a phylogeny of Caenorhabditis based on the ama-1 DNA sequence files that were sent to you earlier this week. P. pacificus should be the outgroup in this phylogeny.2. In the alignment shown below segregating sites are shaded in. Based on these sequence comparisons, determine θT and θW for these region of the genome. 1 GATAT CTCGA GATGC CCCAG TATAC GTGTA ATCCA GAATT CCGCG GATAT 2 GATAT CTCGA GATGC CCCAG TATAC GAGTA ATCCA GAATT CCTCG GATAT 3 GATAT CTCGA GATGC GCCAG TATAC GTGTA ATCCA GAATT CCTCG GATAT 4 GATAT CTCGA GATGC GCCAG TATAC GTGTA ATCCA GAATT CCGCG GATTT comparison # differences # diff. / 50 length of comparison = 50 nucleotides i.e. normalize by dividing # diff. by 50 θT = π = 0.0467 1-2 2 0.04 1-3 2 0.04 1-4 2 0.04 2-3 2 0.04 2-4 4 0.08 3-4 2 0.04 average = π = 0.0467 # segregating sites, K = 4 (note that this is UPPER CASE K, i.e. not normalized k = K/L = 4/50 = 0.08 (i.e. normalized by dividing K by length of comparison) θW = k/a a = 1 + ½ + … + 1/(n-1) n = 4 a = 1 + ½ + 1/3 = 1.833 θW = 0.08/1.8333 = 0.04363. Wallaby tails can be long or short. Their ears can be pointed or rounded. Variation of tail length and ear shape both result from allelic variation at a single gene. Based on the data from the dihybrid cross shown below, are these genes unlinked or linked? Justify you answer statistically (and show your work). P0 long tail, pointed ears x short tail, rounded ears ⤋ F1 long tail, rounded ears x long tail, rounded ears ⤋ obs exp F2 long tail, rounded ears 84 90 long tail, pointed ears 37 30 short tail, rounded ears 35 30 short tail, pointed ears 4 10 160 c2 = (84 – 90)2/90 + (37 – 30)2/30 + (35 – 30)2/30 + (4 – 10)2/10 = 0.400 + 1.633 + 0.833 + 0.091 = 6.467 df = 4 – 1 = 3 P > 0.05, ∴ accept null hypothesis of independent assortment (i.e. no linkage)4. When during meiosis do homologous chromosomes; i. pair with each other, prophase I ii. recombine with each other, and prophase I iii. disjoin from each other. anaphase I5. Wombats can have yellow or green eyes. This variation in eye color results from allelic variation in a single X-linked gene. The yellow eye color is dominant. What frequencies of yellow- and green-eyed female and male wombats would be observed in the F1 and F2 generations from a cross of P0 homozygous yellow-eyed females to hemizygous green-eyed males? P0 XyXy x XgY F1 XyXg x XyY F2 ½ yellow-eyed ♀♀ (XyXy and XyXg) ¼ yellow-eyed ♂♂ (XyY) ¼ green-eyed ♂♂ (XgY )6. In separate populations of dachsunds, you find two recessive mutations, long-leg1 and long-leg2, that result in dachsunds with exceptionally long legs. Describe genetic cross that would allow you to determine whether or not these mutations were in the same or different genes. From this experiment, describe what result you would expect if the mutations were in the same gene and what result you would expect if the mutations were in different genes. complementation test long-leg1 homozygotes x long-leg2 homozygotes i. if progeny have long legs, then long-leg1 and long-leg2 fail to complement and are in the same gene. ii. if progeny have short legs, then long-leg1 and long-leg2 complement and are in different genes.7. Opiliones normally have eight legs. Mutations in some genes result in opiliones with six legs. Mutations in other genes result in opiliones with ten legs. Based on the epistatic interactions described below, construct a genetic pathway for the regulation of leg number in opiliones. six legged mutant genes hexapodal two-gone ten legged mutant genes decapodal tenfooter plus-two doubly mutant animals hexapodal; decapodal = ten legs decapodal ---| hexapodal hexapodal; tenfooter = six legs hexapodal ---| tenfooter hexapodal; plus-two = ten legs plus two ---| hexapodal two-gone; decapodal = six legs two-gone ---| decapodal two-gone; tenfooter = six legs two-gone ---|tenfooter two-gone; plus-two = six legs two-gone ---|plus-two two-gone ---| decapodal ---| hexapodal ---| tenfooter plus-two8. The sxu, gnu and krk are located on the same chromosome. sxu is located 30 cM to the left of gnu and krk is located 10 cM to the right of gnu. i. Based on these map locations, how frequent would you expect double-crossover (DCO) chromosomes to be in this interval? ii. If 20 DCOs were observed out of 1,000 chromosomes scored, how strong interference be in this interval? sxu gnu krk ------|----------------30 cM----------------|----10 cM----|---------- i. expected DCO frequency = (0.30)(0.10) = 0.03 ii. DCOexp = (0.03)(1000) = 30 DCOobs = 20 C = 20/30 = 0.667 I = 1 – C = 0.3339. The eyes gone, stripe body and minute bristles genes all are located on chromosome III in Drosophila melanogaster. Dominance relationships for these genes are shown below. Also shown are data from a cross of a triple heterozygote to a tripe recessive homozygote. Based on the segregation data for this cross, which of these three genes is located between the other two. gene genotype phenotype EYG/EYG eyes present eyes gone, eyg EYG/eyg eyes present eyg/eyg eyes absent SR/SR solid color stripe body, sr SR/sr solid color sr/sr striped MB/MB large bristles minute bristles, MB MB/mb large bristles mb/mb small bristles eyg sr mb x eyg sr mb EYG SR MB eyg sr mb ⤋ eyes present, solid color, large bristles 676 eyes absent, stripes, small bristles 676 eyes present, solid color, small bristles 192 eyes absent, striped, large bristles 192 eyes present, striped, large bristles 29 \___ DCO phenotypes eyes absent, solid, small bristles 29 / ∴ sr is the middle gene eyes present, striped, small bristles 103 eyes absent, solid, large bristles 10310. Allelic variation at the fat gene determines whether yellow-spotted lizards are Fat or Thin. Allelic variation at the tailess gene


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