Name ____________________________ BIO 2110 – Genetics Exam 2 6 November 2015 calculators permitted see last page for potentially useless formulaeName ____________________________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 white (m3/m3) ↓ F1 ?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 ♂ Long Wings, Long Legs* ⇓ F2 Longs Wings, Long Legs Long Wings, Short Legs Dumpy Wings, Long Legs Dumpy Wings, Short LegsName ____________________________ 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 x slippery, basic ↓ F1 sticky, acidic x sticky, acidic ↓ F2 sticky, acidic 1842 slippery, acidic 558 sticky, basic 558 slippery, basic 242Name ____________________________ 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 three-spined, eyes present ↓ F2 three-spined, eyes present 816 five-spined, eyes present 384 three-spined, eyeless 384 five-spined, eyeles 16Name ____________________________ 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 x la cu mo la cu mo la cu mo ⇓ LA CU MO 349 la cu mo 357 LA CU mo 49 la cu MO 45 LA cu MO 5 la CU mo 1 LA cu mo 98 la CU MO 96Name ____________________________ 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 black ↓ F2 black 162 yellow 72 chocolate 54Name ____________________________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 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 +