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LECTURE 07: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION II

LECTURE 07: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION II. accurate calculation of large map distances mapping function analysis of single meioses ordered: gene  centromere unordered: gene  gene. CHAPTER 4: FURTHER IDEAS.

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LECTURE 07: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION II

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  1. LECTURE 07: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION II • accurate calculation of large map distances • mapping function • analysis of single meioses • ordered: gene  centromere • unordered: gene  gene

  2. CHAPTER 4: FURTHER IDEAS • double & higher multiple crossovers  underestimates of map distances calculated from recombination • specialized mapping formulae  accurate map distance corrected for multiple crossovers • analysis of single meioses (in Fungi) can... • position centromeres on genetic maps (~ genes) •  mechanisms of segregation and recombination • crossovers occur occasionally in mitotic diploid cells

  3. ACCURATE MAPPING • mapping large distances is less accurate • best estimate of map distance obtained by adding distances calculated for shorter intervals • if possible, include more genes in the map

  4. ACCURATE MAPPING • problems occur when you have ... • no intervening genes • genes very close together

  5. ACCURATE MAPPING • account for multiple crossovers ? • need mapping function to correct for multiple events and accurately relate recombination to map distance

  6. POISSON DISTRIBUTION • low #s sampled from large population • possible numbers obtainable are large, but most samples are small • e.g. random distribution of 100 x 1$ in class of 100 students... few students receive many bills...

  7. POISSON DISTRIBUTION • e.g. random distribution of 100 x 1$ in class of 100 students

  8. POISSON DISTRIBUTION • here, average is 1 bill/student... m = 1 • # for a particular class... i = 0, 1, 2... 100

  9. POISSON DISTRIBUTION • here, mean is 1 bill/student... m = 1 because 100 bills & 100 students • # for a particular class... i = 0, 1, 2... 100 • how many students are predicted to capture 3 bills?

  10. POISSON DISTRIBUTION • here, mean is 1 bill/student... m = 1 because 100 bills & 100 students • # for a particular class... i = 0, 1, 2... 100 • how many students are predicted to capture 3 bills? f(i) = ———— e-m mi i !

  11. POISSON DISTRIBUTION

  12. POISSON DISTRIBUTION f(0) = 0.368 f(1) = 0.368 f(2) = 0.184 f(3) = 0.061 f(4) = 0.015 f(5) = . . .

  13. POISSON DISTRIBUTION • proportion of class with i items • different m values

  14. MAPPING FUNCTION • use Poisson to describe distribution of crossovers along chromosome • if... • crossovers are random, • we know mean # / region on chromosome • then we can calculate distribution of meioses with 0, 1, 2... n multiple crossovers

  15. MAPPING FUNCTION • recombination frequency (RF) = % recombinants • meiosis with 0 crossovers  RF of 0% • meiosis with 1 crossover  RF of 50%

  16. MAPPING FUNCTION • meioses with 0 crossovers  RF of 0% • meioses with > 0 crossovers  RF of 50% compare the non-recombinant chromatids recombinants shown darker

  17. MAPPING FUNCTION • recombinants make up half of the products of meioses with 1 or more crossovers • 0 crossover class is the only critical one • proportion of meioses with at least one crossover is 1 – 0 class; the 0 class is... f(0) = ———— = e-m e-m m0 0 !

  18. MAPPING FUNCTION • proportion of meioses with at least one crossover is 1 – 0 class, which is... f(0) = ———— = e-m • so the mapping function can be stated as... RF = ½ (1 – e-m) e-m m0 0 !

  19. ~40 MAPPING FUNCTION RF = ½ (1 – e-m) • for low m... m = 0.05, RF = ½ m m = 0.1, RF = ½ m m = 1, RF = 50 ... RF = m / 2 at the dashed line or use the equation...

  20. MAPPING FUNCTION RF = ½ (1 – e-m) • RF = 27.5 cM ? 0.275 = ½ (1 – e-m) 0.55 = 1 – e-m e-m = 1 – 0.55 = 0.45 m 0.8 (mean # of crossovers / meiosis) corrected RF = m/2 = 0.4 = 40 % or 40 cM

  21. ~40 MAPPING FUNCTION RF = ½ (1 – e-m) = 40 cM • for low m... m = 0.05, RF = ½ m m = 0.1, RF = ½ m m = 1, RF = 50 ... RF = m / 2 at the dashed line or use the equation...

  22. MAPPING FUNCTION • mapping large distances is less accurate • best estimate of map distance obtained by adding distances calculated for shorter intervals • if possible, include more genes in the map • put RF values through a mapping function

  23. ANALYSIS OF SINGLE MEIOSES • products of meiosis remain together • groups of haploid cells... either 4 (tetrads) or 8 (octads)

  24. ANALYSIS OF SINGLE MEIOSES • Neurospora crassa (we use Sordaria fimicola in Lab 5) • note pigment phenotypes of the ascospores

  25. ANALYSIS OF SINGLE MEIOSES • meiosis & post-meiotic mitosis in linear tetrad / octad

  26. ANALYSIS OF SINGLE MEIOSES • 2 kinds of mapping with tetrads / octads: • ordered analysis to map gene  centromere • unordered analysis to map gene  gene

  27. A & asegregate to different poles  ORDERED ANALYSIS • no crossing over between gene A and centromere “MI” segregation

  28. A & asegregate to different poles  ORDERED ANALYSIS • crossing over between gene A and centromere “MII” segregation

  29. ORDERED ANALYSIS • 4 types of MII patterns • equal frequencies

  30. MI ORDERED ANALYSIS

  31. MI ORDERED ANALYSIS # MI = 126 + 132 = 258 = 86 % # MII = 9 + 11 + 10 + 12 = 42 = 14 %

  32. ORDERED ANALYSIS  divide by 2

  33. MI ORDERED ANALYSIS # MII = 9 + 11 + 10 + 12 = 42 = 14 %  A  centromere = 14 / 2 = 7 cM

  34. a b ...independent a b ...independent a b ...??? ORDERED ANALYSIS • now consider 2 genes... • 3 possibilities: 1. the genes are on separate chromosomes 2. the genes are linked but on opposite sides of the centromere 3. the genes are linked and on the same side of the centromere

  35. ORDERED ANALYSIS • crossover between centromere and both genes...

  36. ANALYSIS OF SINGLE MEIOSES • 2 kinds of mapping with tetrads / octads: • ordered analysis to map gene  centromere • unordered analysis to map gene  gene

  37. ORDERED ANALYSIS • lots of crossovers between gene & centromere... appear to be unlinked

  38. ORDERED ANALYSIS • MII frequency never reaches 100% • theoretical maximum RF = 2/3 or 66.7% • theoretical maximum calculated map distance = 33.3% • >1 crossovers  with distance... • especially, DCO look like SCO

  39. ORDERED ANALYSIS • second allele determines pattern • maximum MII = 2/3 = 33.3 % = 33.3 cM • multiple crossovers !

  40. UNORDERED ANALYSIS • meioses with 0 crossovers  RF of 0% • meioses with > 0 crossovers  RF of 50%

  41. UNORDERED ANALYSIS = NCO + 1/4 DCO = 1/4 DCO = SCO + 1/2 DCO • parental ditypes • non-parenal ditypes • tetratypes

  42. UNORDERED ANALYSIS • NCO = PD – NPD PD but not NCO  = P  score 1x 

  43. UNORDERED ANALYSIS • SCO = TT – 2NPD T but not SCO  = 2P  score 2x 

  44. UNORDERED ANALYSIS • DCO = 4NPD DCO but not NPD  = 4P  score 4x 

  45. UNORDERED ANALYSIS • corrected map distance (cM) between genes = RF × 100 cM = [ ½ single events + double events ] / total × 100 cM = [ ½ ( TT – 2NPD ) + 4NPD ] / total × 100 cM = ½ [ TT + 6NPD ] / total × 100 cM

  46. ANALYSIS OF SINGLE MEIOSES • 2 kinds of mapping with tetrads / octads: • ordered analysis to map gene  centromere • unordered analysis to map gene  gene

  47. e.g., ORDERED & UNORDERED ANALYSIS or

  48. e.g., ORDERED & UNORDERED ANALYSIS

  49. e.g., ORDERED & UNORDERED ANALYSIS

  50. ORDERED & UNORDERED ANALYSIS

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