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4.1Linkage: basic diploid eukaryotic chromosome mapping

4.1Linkage: basic diploid eukaryotic chromosome mapping. Bateson and Punnet (early XX c.): Purple flowers and long pollen PPLL Red flowers and round pollen ppll F 2 Purple long ( P_L_ ) 4831 Purple round ( P_ll ) 390 Red long ( ppL_ ) 393 Red round ( ppll ) 1338 TOTAL 6952.

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4.1Linkage: basic diploid eukaryotic chromosome mapping

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  1. 4.1Linkage: basic diploid eukaryotic chromosome mapping

  2. Bateson and Punnet (early XX c.): • Purple flowers and long pollen PPLL • Red flowers and round pollen ppll • F2 • Purple long (P_L_) 4831 • Purple round (P_ll ) 390 • Red long (ppL_ ) 393 • Red round (ppll ) 1338 • TOTAL 6952 Linkage discovery Expected 9 3991 3 1303 3 1303 1 435

  3. T. H. Morgan • pr/pr+: purple/red • vg/vg+: vestigial winds/normal • pr pr vg vg X pr+pr+vg+vg+ • Test cross pr pr vg vg X pr+prvg+vg Linkage inDrosophila Expected porportion1:1:1:1 pr + vg151 pr vg+154 pr vg1195 pr + vg +1339 TOTAL 2839 Repulsion Coupling

  4. Crossing with alternated alleles: • pr+pr+vg vg X pr pr vg+vg+ • F1pr+ pr vg+ vg X prprvgvg Linkage in Drosophila (II) pr + vg 965 pr vg +1067 pr vg 146 pr + vg +157 TOTAL 2335

  5. Linked alleles tend to be inherited together

  6. Crossing over produces new allelic combinations

  7. Linkage detection

  8. For linked genes, recombinant frequencies are less than 50 percent

  9. Genotype and allele location in chromosomes can be displayed • Double heterozigote Aa Bb (linked) • AB / ab (coupling) • Ab / aB (repulsion) Linked genes notation Coupling Repulsion A B A b a b a B

  10. Morgan y Sturtevant (1911): recombination frequencies can be used for making maps, since they reflect the genes separation distances. • pr pr ct ct X pr+pr+ct+ct+ • Test cross pr ct/ pr ct X pr+ct+ /ctpr Recombination frequency pr + ct/ctpr21 pr ct + /ctpr23 pr ct /ctpr165 pr + ct + /ctpr191 TOTAL 400 Recombinants Parentals RF=recombinants/total

  11. 11 cM pr ct • FR= (21+23)/400=0.11 • X100: 11 maps units or centimorgans (cM) pr + ct/ctpr21 pr ct + /ctpr23 pr ct /ctpr165 pr + ct + /ctpr191 TOTAL 400 Recombinants Parentals Linkage maps RF=recombinants/total

  12. Rec. cv-v Rec. cv-ct Rec. cv-v Rec. cv-ct Rec. cv-v Rec. v-ct Rec. cv-v Rec. v-ct Rec. v-ct Rec. cv-ct Rec. v-ct Rec. cv-ct • vermillion: vermillion eyes, v • crossveinless: no veins in wings, cv • cut: cut end wings, ct • P: v+ cv ct / v+ cv ct X v cv+ ct+ / v cv+ ct+ • Test: v cv ct / v cv ct X v+ cv ct / v cv+ ct+ Three point cross v cv+ ct+ 580 v+ cv ct592 v cv ct+ 45 v+ cv+ ct40 v cv ct89 v+ cv+ ct+ 94 v cv+ ct3 v+ cv ct+ 5 TOTAL 1448 Parentals

  13. 13.2 cM 6.4 cM v ct cv 18.5 cM • FR v-cv=(45+40+89+94)/1448=18.5 cM • FR v-ct=(89+94+3+5)/1448=13.2 cM • FR ct-cv=(40+45+3+5)/1448= 6.4 cM • Linked genes (FR<50 cM) Recombination frequencies

  14. 13.2 cM 6.4 cM v ct cv • cv-ct = 0.064 • ct-v = 0.132 Coefficient of coincidence and interference • Positive interference: when the occurrence of one crossover reduces the probability that a second one will occur in the same region.

  15. Determine phenotypes and numbers of progeny • Determine parental genotypes • Determine the no recombinant phenotypic classes no recombinants and double ones (more and less frequent) • Determine the central locus • Determine the order • Determine the location of crossovers that led to phenotypic classes • Determine the recombination frequencies • Construct the map • Determine the coefficient of coincidence and the interference Steps to perform a three point maps

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