November 16 Remaining deadlines: Research paper Friday Dec 4

1. November 16 Remaining deadlines: Research paper Friday Dec 4 Three point cross lab report Dec 2 or 3 Exam Dec 14 at 12:00 noon Forensics lab results Genetic identification of the gene responsible for cystic fibrosis

2. 1857 bp 1058 bp 929 bp 383 bp 121 bp Tuesday’s lab class pMCT118 alleles VNTR polymorphisms 16 bp repeats 14-40 repeats 29 known alleles ranging in size from 369-801 bp) Common alleles? Primer-dimer

3. Wednesday’s lab class pMCT118 alleles

4. Forensics lab: Are there some especially common alleles? Rare alleles can be more useful for forensics. The type of alleles that are rare will be different in isolated populations. Question: The sizes of the PCR products of the pMTC118 locus in one family were 531 and 643 bp in the mother and 435 and 531 bp in the father. What are all possible fingerprints their children could have and what is the probability of any child getting each combination of alleles? 531 homozygous ½ X ½ 531 435 heterozygous ½ X ½ 643 435 heterozygous ½ X ½ 643 531 heterozygous ½ X ½

5. Mapping of the gene responsible for cystic fibrosis • Identification of polymorphic DNA markers linked to disease • Location of DNA on Chromosome • Determination of region in which polymorphic markers are tightly linked – no recombinants • Contig assembly and sequence analysis of region • Compare polymorphisms in candidate gene between normal and disease chromosomes to establish all affected family members have mutation • Test expression of gene, in expected tissues? • Identify potential function of protein and explain its role in disease

6. White et al 1985 Nature show that MET DNA is polymorphic when cut with Taq1 and polymorphism segregates as a Mendelian trait

7. White et al Nature 1985 Met polymorphism associated with cystic fibrosis Most families are single backcrosses with respect to the Met locus LOD score at recombination value 0 In most families, affected offspring have same genotype for Met locus and cf locus suggesting close linkage

8. LOD score is used to determine if two traits are linked in human pedigrees Odds of linkage is: (Probability gene and marker are linked at a certain map distance) divided by (Probability they are unlinked). Maximum likelihood odds of linkage; Change estimated linkage distance (θ) to get the best Odds of linkage score for the data. LOD is the log10 of the Odds of linkage score LOD is used so that information from separate families, in which parental allele combinations are distinct, can be combined.

9. Combine odds of linkage for many families: p1(L)/p1(NL) x p2(L)/p2(NL) xp3(L)/p3(NL) In practice we combine the log of odds: LOD1 + LOD2 + LOD3. Continue until LOD > 3.0 before linkage is accepted Linkage distance is based on the linkage distance that gives the maximum value for the data.

10. If genes and markers are unlinked, the odds of linkage will be <1.0 in some families and the final LOD score will be negative (<0). Therefore, as you add more families the LOD score will only increase if the data from the majority of families supports linkage.

11. Tsui et al 1985 Science identify another polymorphic DNA marker D0CRI-917 Partial restriction digest fragment 17.5 kb (polymorphic HindIII site)

12. Tsiu et al 1985 Science RFLP analysis shows affected children have common alleles for DNA marker and cystic fibrosis HincII digested DNA HindIII digested DNA Cystic fibrosis linked to D0CRI-917 lambda phage clone insert 18 kb

14. Knowlton et al 1985 Nature Marker linked to cf in family studies by Tsui et al shown to be on Chromosome 7 fragment by hybridization to somatic hybrid lines. Other markers potentially more closely linked can be identified using these hybrid lines.

15. Kerem et al 1989 Table of RFLP markers associated with cf by identification in family studies Specific alleles of these markers were found to be linked to disease in families with genetic recombination in chromosome 7, the chromosome that carried the disease. In this paper the linkage of these markers with disease is validated

16. How do we go from a list of linked markers to a map of the chromosome? Use end probes and fingerprinting to generate contigs

17. Fig. 10.8

18. Fig. 10.11 Combination of mapped polymorphic sequences and genomic DNA clones enables reconstruction of chromosome sequence STS are polymorphic DNA sequences BACS are cloning vectors with genomic DNA inserts

19. Kerem et al 1989 Science. Chromosome alignment of CF region. Gaps between contigs filled in with jumping technology.

20. DNA jumping Collins 1987 Science

21. Rommens et al Science 1989 put together DNA contig of cystic fibrosis region defined by DNA markers. Restriction Map of CF region Vertical bars represent mRNA identified as cDNA clones

22. Several transcribed regions were found in the 500 kb segment of chromosome 7 Located between polymorphic sequences associated with a recombination event Between the marker and cf in family studies Having the physical map of the entire region made it possible to identify many transcribed sequences. Riordan et al 1989 and Kerem et al 1989 found a transcript for an ion channel encoded in the region. That transcript was expressed in lung tissue. Many affected family members had a small deletion in the coding region of this transcript that would lead to deletion of a single amino acid. That polymorphism was the only candidate gene in the region that could be demonstrated to be homozygous in affected individuals and not in healthy individuals.

23. Riordan et al Science 1989Candidate gene expressed in tissues affected by CF

24. Riordan et al. Science 1989 All carriers in a family had ΔF deletion

25. Quinton 1983 CF defects in chloride permiability of sweat glands from CF patients Riordan et al 1989 Science predicted structure of CF protein - chloride ion channel transporter

26. Nature review 2009 Helen Pearson, editor

27. The following slides are for information only and will not be discussed in lecture. There will not be exam questions about these slides. LOD nomenclature Slides from Greg Copenhaver • θ = recombinant fraction • L(θ) = likelihood of linkage at 0 > θ < 0.5 • L(0.5) = likelihood of independent assortment • Log[L(θ)/L(0.5)] = log-of-odds ratio = LOD = Z • LOD scores > 3 indicate linkage • LOD scores < -2 indicate non-linkage

28. Example linkage of marker and trait Zmax = maximum likelihood score (MLS)

29. Inheritance Probabilities Assuming a doubly heterozygous parent: Aa•Bb The probability of progeny inheriting either non-recombinant chromosome from that parent is: ½ (1-θ) + ½ (1-θ) = 1- θ The probability of inheriting a recombinant chromosome is: ½(θ) + ½(θ) = θ

30. Sibship Probabilities Aa•Bb The probability of 5 progeny inheriting a non-recombinant chromosome is: (1-θ)5 The probability of 4 progeny inheriting a recombinant chromosome is: (θ)4 The probability of 9 progeny with 5 inheriting a non-recombinant chromosome and 4 inheriting a recombinant is: (1-θ)5(θ)4

31. Calculating lod (logarithm of odds) Scores lod = logarithm of odds = Z ( ) probability of pedigree if linked Z = log probability of pedigree if not linked • calculated for different linkage values, q • q = 0.1 means 10 cm (10% recombinants) • q = 0.25 means 25 cm (25% recombinants)

32. Linkage of ABO & NPS1 Is blood type (alleles O and B) linked to Nail-patella syndrome (alleles N and D where D is autosomal dominant mutant allele with full penetrance – shown as affected)?

33. Linkage of ABO & NPS1 What must the NPS1 genotype be? BO OO BO BO BO BO OO OO OO BO OO OO BO

34. Linkage of ABO & NPS1 What must the NPS1 genotype be? BO OO ND BO BO BO BO OO OO OO BO OO OO BO

35. Linkage of ABO & NPS1 BO OO NN ND BO ND BO ND BO NN BO ND OO ND OO NN OO NN BO ND OO NN OO NN BO NN Without knowing the phase we can’t identify recombinant (R) and non-recombinant (NR) progeny…. …so we calculate both.

36. Linkage of ABO & NPS1 Phase 1 BD/ON OO NN BO ND NR R BO ND NR R BO NN R NR BO ND NR R OO ND R NR OO NN NR R OO NN NR R BO ND NR R OO NN NR R OO NN NR R BO NN R NR 1

37. Linkage of ABO & NPS1 Phase 2 BN/OD OO NN BO ND NR R BO ND NR R BO NN R NR BO ND NR R OO ND R NR OO NN NR R OO NN NR R BO ND NR R OO NN NR R OO NN NR R BO NN R NR 1 2

38. Linkage of ABO & NPS1 Phase 2 BN/OD OO NN BO ND NR R BO ND NR R BO NN R NR BO ND NR R OO ND R NR OO NN NR R OO NN NR R BO ND NR R OO NN NR R OO NN NR R BO NN R NR 1 2 Phase-1 L(θ) = (1-θ)8(θ)3 Phase-2 L(θ) = (1-θ)3(θ)8 L(θ) = ½(1-θ)8(θ)3 + ½(1-θ)3(θ)8

39. Linkage of ABO & NPS1 Calculate likelihood for specific values of θ (e.g. 0.1) L(0.1) ½(1-0.1)8(0.1)3 + ½(1-0.1)3(0.1)8 =

40. Linkage of ABO & NPS1 Calculate likelihood ratio for specific values of θ (e.g. 0.1) L(0.1) ½(1-0.1)8(0.1)3 + ½(1-0.1)3(0.1)8 L(0.5) ½(1-0.5)8(0.5)3 + ½(1-0.5)3(0.5)8 0.441 = =

41. Linkage of ABO & NPS1 Calculate likelihood ratio for specific values of θ (e.g. 0.1) L(0.1) ½(1-0.1)8(0.1)3 + ½(1-0.1)3(0.1)8 L(0.5) ½(1-0.5)8(0.5)3 + ½(1-0.5)3(0.5)8 Calculate the LOD by taking the log of the likelihood ratio Log(0.441) = -0.356 Z(0.1) = -0.356 Do this for several θ values (0.001, 0.01, 0.05, 0.1, 0.2, 0.3, 0.4, 0.45) 0.441 = =

42. Linkage of ABO & NPS1 Zmax = maximum likelihood score (MLS)

43. Below Z = -2 so not linked at these θ Cumulative Z scores from multiple pedigrees Linkage of ABO & NPS1 θ 0.001 0.01 0.05 0.1 0.2 0.3 0.4 Z -6.0 -3.0 -1.1 -0.4 0.1 0.2 0.1 θ 0.001 0.01 0.05 0.1 0.2 0.3 0.4 Σ(Z) - - 28.2 31.2 27.8 19.4 9.0 LOD score are additive!

44. Linkage of ABO & NPS1 If the phase of the doubly heterozygous parent were known the calculation is siplified Calculate log of likelihood ratio for specific values of θ (e.g. 0.1) L(0.1) (1-0.1)8(0.1)3 L(0.5) (1-0.5)8(0.5)3 log -0.055 = =

45. Computer-Generated lod Scores 4 evidence for linkage, 11-19 cM, most likely 13 cM 3 2 1 lod score 0 q 0.1 0.2 0.3 0.4 0.5 -1 inconclusive for linkage between 3 and 11 cm or above 13 cm -2 -3 linkage excluded below -2 cm -4