Question

Purple eyes (or), black body (b) and vestigial wings (vg) are recessive alleles in Drosophila melanogaster. Apr prb bvg" vg trihybrid is crossed to a fly homozygous for the recessive allele at each locus. The following numbers of offspring are observed. Phenotype Number of progeny Unordered genotypes Ordered genotypes Wild type 92 188 1241 vg 7 pr, b 8 pr, vg 1243 b, vg 190 pr, b, vg Total: 2990 1. Write the full, unordered genotypes of each category Comit the genotype of the tester, it is the same for everything), to the right of the number of progeny column 2. Which are double crossovers? Which are non-recombinants? 3. The non-recombinants represent the two chromosomes present in the trihybrid. Using linkage notation, write three possible orderings of the alleles on the trihybrid parent; each ordering must have a different locus in the middle. 4. Which ordering produces double crossover progeny consistent with the actually observed double crossover classes? 5. Rewrite all genotypes in the correct order, to the right of the unordered genotypes. Compare each ordered genotype to the non-recombinant genotypes to deduce the location of crossover, and indicate those crossovers with a "/" on the ordered genotypes. 6. Calculate recombination frequency, and draw a map. 7. Calculate the coefficient of coincidence and the interference

Answer 1

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ANSWER Phenotyper Number of P wild type 92 188 ри 1291 rogeny / onordered genetypes ordered genotypes ри*, * *9* bt/pet ugt b pet ugt b / prt/vgt bt pe vg b pert/vg | ь/ ря. v br/prtlug pe care le per, ug 1243 190 pot b vg brug po bug TOTAL 88 2990 firstly, we have tabulated the infamation already available Answer 1 unordered genotypes are written in the third collunan above (in black), arbetarity since we dont know Which one is the middle gene, so the order of the genes is arbitary for now! Note: wild type are always designated with symbol't'in genoty- Answer 2. Arranging the genotype frequencies in the descending order. prbrug = 12437 Parentals Note: Parentals non - recombinants 190 75001 11 ] single crossover 1) potb* vg+ 92 Sco - 88 (single crossover I) O S. మతతత్వతంత మత - (Double crossover) 2990

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Coco) - The one with the maximum frequencies are always parentali The one with the least frequencies are always Double crossovers and 361 and SW I refers to single crossovers, - Each of the pair of frequencies complement each other. ANSWER 3 (26) To know which is the middle gene we cross, Parentals X DCO - (pre bt ng OR P (pret b vgt so the middle gone is per so the ordered genotypes will be keeping pee as the middle gene and the order can be b.pre, vg or vg, peb ordered genotypes are written in the table only on Page 1 last collumn) and also below it descending order : bt Pret : 24 ] Parentals a 190 scor - cross between genes pr and 67 pr ugt - 188 92 7 SCOTT - cross bet ween genes baud శతవత తన వంతు PA b Toco 1990 As been asked specifically ordered notation of genes for tringbrid parent in answer 3, it is bt pr ug and b pertugt

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Answer 4. The ordering which produces Double crossovers Coco's) is bt pot ug and b prugt Now to know single cross over I (sco I) we cross Parentals x Sco I p. 6+ pr ug to or p b pr * Vgt) so bt pre vg si b prt vg so, sco I is a cross between genes pee and vg Now to know single crossover II (sco I we cross Parentals X so I P- (bt pr ug OR P. (b) pet vgt 33-6 pre ug SI - by prt vgh so scoT is a cross between genes b and pr. ANSWER 5 - All the correctly ordered genotypes have been written abready in the collumn on Page 1 and explained how' as a part of ANSWER 3. Location of each cross over has been deduced above already as sco I – single crossover between gunes pe and vg scos - single crossover between genes sco Double crossover between b b and per middle locus pr and both the side genes b' and vg (also indicated with symbel / in the collumn) sco Douaulo locus P and vg

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(502) U ANSWER 6 Recombination frequency O pou genres pre and ng = 190 + 188 + 8 + 7 x 100 2990 - 393 x 100 = 13.14% 2996 OR 13.14 cm gerres b and pu (Map distance) - = 92+ 88 + 8 +7 xin (10) 2990 175 x 100 = 6.52% 2११० OR 6.52 CM ( Map distance) Map will look like: +6.52.6M1 13.14.M b pr ANSWER 7. I coefficient of coincidence = Actual double crossovers Expected double crossovers Here actual double crossovers are = 8 +7=lisi Expected double crossovers can be calculated as :- - No. of progenies slowing recombination : • Between genes pu aud vg = 1907 188+8+7 • Bet ween genes band pe : S 92788 +8+7 195

Expected rate of double crossovers = ( 195 ) 2990 2990 -0.1314 X 0.0652 -0.00856 per 1 individual -0.008 56 X 2990 = 25.59 per 2990 individuals so, coefficient of coincidence = 15 (oc) 25.59 = 0.58 Interferance = 1-cool so Interferance = 1-0.58 = 0.42 - coefficient of coincidence calculated here is a measure of interferance in the - formation of chromosomal crossovers during meiosis. - It depicts a fact that there is a decreased lik lihood of occurence of a crossover adjacent to a crossover spot.