1. Faccurate calculation of large map distances Fmapping function Fanalysis of single meioses Fordered: gene  centromere Funordered: gene  gene LECTURE 07: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION II

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

3. ACCURATE MAPPING Fmapping large distances is less accurate Fbest estimate of map distance obtained by adding distances calculated for shorter intervals Fif possible, include more genes in the map

4. ACCURATE MAPPING Fproblems occur when you have... Fno intervening genes Fgenes very close together

5. ACCURATE MAPPING Faccount for multiple crossovers ? Fneed mapping function to correct for multiple events and accurately relate recombination to map distance

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

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

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

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

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

11. POISSON DISTRIBUTION

12. 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 Fproportion of class with i items Fdifferent m values

14. MAPPING FUNCTION Fuse Poisson to describe distribution of crossovers along chromosome Fif... Fcrossovers are random, Fwe know mean # / region on chromosome Fthen we can calculate distribution of meioses with 0, 1, 2... n multiple crossovers

15. Frecombination frequency (RF) = % recombinants Fmeiosis with 0 crossovers  RF of 0% MAPPING FUNCTION Fmeiosis with 1 crossover  RF of 50%

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

17. MAPPING FUNCTION Frecombinants make up half of the products of meioses with 1 or more crossovers F0 crossover class is the only critical one Fproportion of meioses with at least one crossover is 1 – 0 class; the 0 class is... f(0) = ———— = e -m e -m m 0 0 !

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

19. MAPPING FUNCTION Ffor 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 RF = ½ (1 – e -m ) or use the equation... ~40

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

21. MAPPING FUNCTION Ffor 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 RF = ½ (1 – e -m ) = 40 cM or use the equation... ~40

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

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

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

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

26. ANALYSIS OF SINGLE MEIOSES F2 kinds of mapping with tetrads / octads: Fordered analysis to map gene  centromere Funordered analysis to map gene  gene

27. ORDERED ANALYSIS Fno crossing over between gene A and centromere A & a segregate to different poles  “M I ” segregation

28. ORDERED ANALYSIS Fcrossing over between gene A and centromere A & a segregate to different poles  “M II ” segregation

29. ORDERED ANALYSIS F4 types of M II patterns Fequal frequencies

30. ORDERED ANALYSIS MIMI

31. # M I = 126 + 132 = 258 = 86 % # M II = 9 + 11 + 10 + 12 = 42 = 14 % MIMI

32. ORDERED ANALYSIS  divide by 2

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

34. ORDERED ANALYSIS F3 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 a b Fnow consider 2 genes......independent...???

35. ORDERED ANALYSIS Fcrossover between centromere and both genes...

36. ANALYSIS OF SINGLE MEIOSES F2 kinds of mapping with tetrads / octads: Fordered analysis to map gene  centromere Funordered analysis to map gene  gene

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

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

39. ORDERED ANALYSIS Fsecond allele determines pattern Fmaximum M II = 2/3 = 33.3 % = 33.3 cM Fmultiple crossovers !

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

41. UNORDERED ANALYSIS Fparental ditypes Fnon-parenal ditypes Ftetratypes = NCO + 1/4 DCO = 1/4 DCO = SCO + 1/2 DCO

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

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

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

45. UNORDERED ANALYSIS F 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 F2 kinds of mapping with tetrads / octads: Fordered analysis to map gene  centromere Funordered analysis to map gene  gene

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

48. e.g., ORDERED & UNORDERED ANALYSIS

50. ORDERED & UNORDERED ANALYSIS

51. Fordered analysis to map gene  centromere FM I... no recombination FM II... recombination ORDERED & UNORDERED ANALYSIS

52. Fordered analysis to map gene  centromere FM I... no recombination FM II... recombination RF = ½ (M II / TOTAL) ORDERED & UNORDERED ANALYSIS

53. Fordered analysis to map gene  centromere FM I... no recombination FM II... recombination RF = ½ (M II / TOTAL) RF x 100 = map distance (cM) ORDERED & UNORDERED ANALYSIS

54. RF = ½ (M II / TOTAL) RF x 100 = map distance (cM) r  cent. = ½ [(1+2+1+2+1)/200] x 100 = 1.75 cM t  cent. = ½ [(1+2+1+15+13+17+1)/200] x 100 = 16.75 cM ORDERED & UNORDERED ANALYSIS

56. Funordered analysis to map gene  gene Fconsider all possible gene pairs (here only 1) FPD  NPD?, unlinked or PD >> NPD, linked FPD = NCO + 2-strand DCO FTT = SCO + 3-strand DCO (x2) FNPD = 4-strand DC) (¼ of all DC0) ORDERED & UNORDERED ANALYSIS

57. Funordered analysis to map gene  gene RF = ½ [ TT + 6NPD ] / TOTAL ORDERED & UNORDERED ANALYSIS

58. Funordered analysis to map gene  gene RF = ½ [ TT + 6NPD ] / TOTAL RF x 100 = map distance (cM) ORDERED & UNORDERED ANALYSIS

59. Funordered analysis to map gene  gene RF = ½ [ TT + 6NPD ] / TOTAL RF x 100 = map distance (cM) PD (133) >> NPD (2)  linked r  t = ½ [ 65 + 6(2) / 200 ] x 100 = 19.25 cM ORDERED & UNORDERED ANALYSIS

61. but... 1.75 + 16.75 = 18.5 ??

62. ORDERED & UNORDERED ANALYSIS more accurate... calculation includes DCOs

63. EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION: PROBLEMS F in Griffiths chapter 4, beginning on page 141, you should be able to do questions #1-30 F begin with the solved problems on page 138 if you are having difficulty F look at the way Schaum’s Outline discusses linkage and mapping for alternative explanations - especially tetrad analyses F try Schaum’s Outline questions in chapter 4, beginning on page 208