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 GATTT3. 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 ⤋ F2 long tail, rounded ears 84 long tail, pointed ears 37 short tail, rounded ears 35 short tail, pointed ears 44. When during meiosis do homologous chromosomes; i. pair with each other, ii. recombine with each other, and iii. disjoin from each other.5. 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?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.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 hexapodal; tenfooter = six legs hexapodal; plus-two = ten legs two-gone; decapodal = six legs two-gone; tenfooter = six legs two-gone; plus-two = six legs8. 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----|----------9. 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 eyes absent, solid, small bristles 29 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 determines whether yellow-spotted lizards are Tailed or Tailess. These genes are linked. Based on the data presented below, what is the recombination frequency between these genes? [P0s are homozygous.] P0 Fat, Tailess x Thin, Tailed ⤋ F1 Fat, Tailed x Fat, Tailed ⤋ F2 Fat, Tailed 816 Fat, Tailess 384 Thin, Tailed 384 Thin, Tailess 16probability 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 χ2 values Lod = Log10(odds) χ2 = Σ [(obs-exp)2/exp] p2 + 2pq + q2 = 1 p2 + q2 + r2 + 2pq + 2pr + 2qr = 1 df = n – 1 !!!!!! ¼ p2 !!!!!! p + q = 1 #classes = 2n + 1 p + q + r = 1 parental gamete frequency = ½ (1 – p) recombinant gamete frequency = ½ p Nt = N0 x e-rt r = ln[N(t2)/N(t1)]/(t2 - t1) θT = # prob12 = (prob1 x prob2) __ __ W = p2WAA + 2pqWAa + q2Waa pn+1 = (pn2WAA + pnqnWAa)/W C = DCOobs/DCOexp € p =12(TT) + 3(NPD)N € x =−b ± b2− 4 ac2a € p =12(2nddiv)N par = ½ (1-p) 1 cM equals a recombination frequency of 0.01 a = 1 + ½ + … + 1/(n – 1) __ qn+1 = (qn2Waa + pnqnWAa)/W θ = 4Neµ #classes = log10(freqext)/log10(¼) 1 θW = k/a I = 1 - C 1 1 Eco RI = GAATTC 1 2 1 odds = probx/prob0.5 1 3 3 1 (2Ne – 1)/2Ne