1. 1 Balanced Translocation detected by FISH

2. 2 Red- Chrom. 5 probe Green- Chrom. 8 probe

3. 3 2D Protein Gels

4. 4

5. 5 MS-peptide size signature: match to all predicted proteins

6. 6 1.Follow the mutation 2. Follow which regions of DNA are co-inherited (linked) Positional Cloning by Recombination Mapping

7. 7 1.Follow the mutation To determine disease gene presence or absence (genotype) from phenotype you must first establish Dominant / recessive Aurosomal / sex-linked Positional Cloning by Recombination Mapping

8. 8 SINGLE GENE DEFECTS Modes of Inheritance To deduce who (likely) has one or two copies of mutant gene Affected FemaleUnaffected Male

9. 9 AUTOSOMAL DOMINANT +/+ D/+ +/+

10. 10 RECESSIVE AUTOSOMAL X-LINKED RECESSIVE a/+ a/a a/+ x/++/Y x/+ x/Y +/Y

11. 11

12. 12 2. Follow which DNAs are co-inherited (linked) Use DNA sequences that differ among individuals within a family- Polymorphisms. Positional Cloning by Recombination Mapping TG CA

13. 13 VNTR / STRP DETECTION

14. 14 A1 A3 A1 A3 A4 A2 A1 A2 A4 A3

15. 15 X X X 23 Parent Gamete Child A1 A2 B1 B2 C1 C2 B1 C1 A2

16. 16 Recombination Mapping Measures distance between 2 sites on a chromosome according to frequency of recombination Distance between 2 DNA markers or Distance between a “disease gene” and a DNA marker

17. 17 No fixed proportional Conversion between Genetic distance (cM) and Physical distance (kb, Mb)

18. 18 FAMILY A A1D A2+ NR R DDDD++

19. 19 FAMILY B A1D A2+ NR R A1+ A2D RRRRRNR

20. 20 INFORMATIVE MEIOSIS Ideally:- unambiguous inheritance of mutation and markers (requires heterozygosity for each in parent) knowledge of which alleles linked in parent (phase)

21. 21 Assign numbers to results of linkage analysis to deal with non-ideal meioses to sum data from many meioses in a family to sum data from several families

22. 22 Likelihood of R Likelihood of NR If linked and RF = If unlinked:-  1 -   1/2 Family A has 1 recombinant and 5 Non-Recombinants Likelihood, given linkage of L ( )   =. (1- )   5 L (1/2) = (1/2) 6 Z = Lod = log { L ( ) / L (1/2)}  Or given unlinked:-  Z0.580.620.510.30 0.1 0.2 0.3 0.40.5

23. 23 Z = 3  Lod

24. 24 FAMILY B A1D A2+ NR R A1+ A2D RRRRRNR

25. 25 Family B:- Disease gene may be linked to A1 or A2 Consider equally likely 50% chance Family B has 1 R and 5 NR 50% chance Family B has 5 R and 1 NR L ( )  L (1/2) = (1/2) 6 Z = Lod = log { L ( ) / L (1/2)}   Z0.280.320.220.080 0.1 0.2 0.3 0.40.5 =. (1- )  5 1/2 { } +. (1- )  5 1/2 { }

26. 26  Z0.280.320.220.080 0.1 0.2 0.3 0.40.5 Phase unknown  Z0.580.620.510.30 0.1 0.2 0.3 0.40.5 Phase known

27. 27 Z = Z1 + Z2 + Z3 + Z4 +….. Z = Z(A) + Z(B) + Z(C) + Z(D) + …. For family “A” with meioses 1, 2, 3, 4 ….. For multiple families, “A”, “B”, “C”, “D”….. Assumption: same gene responsible for disease in all families Problem: locus heterogeneity

28. 28 Z = 3  Lod

29. 29

30. 30 LINKAGE DISEQUILIBRIUM Many generations

31. 31 PCR test DNA segments

32. 32

33. 33 Testing for specific mutations

34. 34 ARMS 3’ mis-match of primer

35. 35

36. 36 OLA

37. 37

38. 38

39. 39

40. 40

41. Aa BB CC DD Ee FF Gg HH II JJAA BB CC Dd Ee FF GG HH II Jj AA BB CC Dd Ee FF Gg HH II JJ a B C D e/E F G H I J A B C d E/e F G H I J A B C D E/e F g H I J A B C D e/E F G H I j MotherFather Son/Daughter Family Trio SNP genotypes reveal haplotypes Deduced haplotypes- ignoring recombination

42. Creation of variant sequences Rearrangement of sequence variants by recombination First, consider just the creation of variant sequences within a short stretch of DNA where there is no significant rearrangement due to recombination (an assumption that turns out to be valid)

43. ABCDEFGHIJKLMNOPQRST AbCDEFGHIJKLMNOPQRST ABCDEFgHIJKLMNOPQRST AbCDEFGHIJKLMNOPqRST aBCDEFgHIJKLMNOPQRST AbCDEFGHIJkLMNOPqRST AbCDEFGhIJkLMNOPqRST ABCDEfGHIJKLMNOPQRST aBCDEFgHIJKLMNOPQrST aBCDEFgHIJKLMnOPQrST b bq bqk bqkh g ga gar garn f History

44. ABCDEFGHIJKLMNOPQRST AbCDEFGHIJKLMNOPQRST ABCDEFgHIJKLMNOPQRST AbCDEFGHIJKLMNOPqRST aBCDEFgHIJKLMNOPQRST AbCDEFGHIJkLMNOPqRST AbCDEFGhIJkLMNOPqRST ABCDEfGHIJKLMNOPQRST aBCDEFgHIJKLMNOPQrST aBCDEFgHIJKLMnOPQrST b bq bqk bqkh g ga gar garn f Retention & amplification of only a few haplotypes

46. For any short region of DNA typically only 4-6 haplotypes are found in a sampling of present day humans (of the many millions that must have existed in at least one copy en route). These local haplotypes provide some information about ancestry. Now consider how the major haplotypes of each short region of DNA are associated with neighboring haplotypes to see where recombination events took place.

47. aBCDEFgHIJKLMnOPQrSTUVwXyZ  High LD regions?

48. aBCDEFgHIJKLMnOPQrSTUVwXyZ  High LD segment Recombination hot-spot

50. 85% of genome made up of 5-20kb high LD blocks Only 4-5 different major haplotypes per block in the world!

51. Haplotype blocks

52. 100 kb 12345 6 7 8

53. DiseaseNo disease /2,000/3,000 Minor allele frequency SNP-2a93130 SNP-2b2127 SNP-3a14062 SNP-3b2435 SNP-3c140260 SNP-3d87120 …….….…. …….…...