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Eric Jorgenson Epidemiology 217 2/21/12

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Presentation on theme: "Eric Jorgenson Epidemiology 217 2/21/12"— Presentation transcript:

1 Eric Jorgenson Epidemiology 217 2/21/12
Linkage Analysis Eric Jorgenson Epidemiology 217 2/21/12

2 Worldwide Distribution of Human Earwax SNP rs17822931
Yoshiura et al., Nature Genetics 2006

3 Geographic Distribution of PTC Phenotype
High Low Wooding Genetics 2006, adapted from Cavalli-Sforza 1994

4 Bimodal Distribution of PTC

5 Your Phenotypes and Genotypes
Taste SNP Ear wax SNP Sample Name taster ear wax rs rs rs BU10 Y D CT AG TT BU12 N None BU14 D? AA BU15 W CC BU17 BU19 GG BU20 BU21 mild BU22 BU23 BU24 W? BU25 BU26 BU27 BU28 BU29 Y mild BU30 From Joe Wiemels

6 Types of Genetic Studies
Family Studies Compare trait values across family members Linkage Analysis Compare trait values with inheritance patterns Association Compare trait values against genetic variants

7 Family Studies Familial Relationships Phenotype information
Twins Siblings Parents/offspring Phenotype information Affected/Unaffected (Prostate Cancer) Quantitative measure (Blood Pressure) No Genotype information required

8 Why do Family Studies? Is the trait genetic?
What is the mode of transmission? Dominant Recessive Additive Polygenic (Multiple genes involved)

9 Mutation and Meiosis

10 Recessive trait

11 First PTC Family Study L. H. Snyder Science 1931 Both: 15% One: 32%
Neither: 100% L. H. Snyder Science 1931

12 Linkage Analysis Narrow down position of disease gene
No biological knowledge needed Genetic markers (not disease gene) Recombination

13 Recombination a A a a b B b b A a a a A a a a b b b b B b B b
2 markers-A and B Know the orientation A a a a A a a a b b b b B b B b

14 Recombination a A a a b B b b R NR NR R A a a a A a a a b b b b B b B

15 Independent Assortment
b B b b 25% 25% 25% 25% A a a a A a a a b b b b B b B b

16 No recombination a A a a b B b b 0% 50% 50% 0% A a a a A a a a b b b b

17 Recombination Fraction
61 420 442 77 A a a a A a a a b b b b B b B b

18 Recombination Fraction
Recombination Fraction q = Recombinants / Total = / = 138 / 1000 = 13.8% 61 420 442 77 A a a a A a a a b b b b B b B b

19 Linkage Linkage Analysis Recombination fraction q < 50%
Two traits: PTC and KELL blood group Two genetic markers One trait and one genetic marker Linkage Analysis

20 Human Linkage Analysis
RFLP Markers for Linkage (1980) Huntington’s Disease Linkage (1983) Cystic Fibrosis Linkage (1985) Cystic Fibrosis Gene (1989) Huntington’s Disease Gene (1993)

21 Genomewide Linkage Analysis
Genetic Markers q = 10% on average Genes

22 Linkage Analysis LOD score based on recombination
LOD (q) = log (q)R (1 - q)NR ____________________ (q = 1/2) R + NR

23 Dominant Trait D d d d D d D d d d

24 Linkage Analysis R NR NR

25 LOD score LOD (q) = log (q)1 (1 - q)2 q ____________________
0.01 -1.11 0.05 -0.44 0.1 -0.19 0.2 0.3 0.07 0.4 0.06 0.5 0.00

26 IBD Identity by descent Allele Sharing methods
Often used for affected sib pairs

27 Identity By Descent a A a A 25% 25% 25% 25% A A a A A a a a

28 Identity By Descent a A A A a A A a a a Parent 1 Alleles shared IBD
A A a A A a a a

29 Identity By Descent a A A A a A A a a a Parent 1 1 1 1 1 2 0% 1 100%
Alleles IBD Frequency 2 0% 1 100% a A A A a A A a a a

30 Identity By Descent A A A A a A A a a a Sibling 1 Alleles shared IBD
A A a A A a a a

31 Identity By Descent A A A A a A A a a a Sibling 1 2 1 1 0 2 25% 1 50%
Alleles IBD Frequency 2 25% 1 50% A A A A a A A a a a

32 Identity By Descent IBD can be used for linkage analysis
Expect 50% alleles shared between siblings Look for IBD > 50% for concordant pairs Look for IBD < 50% for discordant pairs

33 PTC Linkage Analysis Utah Genetic Reference Project 27 large families
269 subjects

34 PTC Linkage Analysis

35 Human Chromosomes 23 pairs of chromosomes TAS2R38

36 Fine Mapping Linkage markers Genes Kim et al. Science 2003

37 Linkage Disequilibrium

38 Linkage Disequilibrium

39 Linkage Disequilibrium

40 Linkage Disequilibrium

41 Linkage Disequilibrium
Time

42 Linkage Disequilibrium Mapping
Genetic Markers Genes

43 PTC Linkage Disequilibrium Mapping
Kim et al. Science 2003

44 TAS2R38 Receptor Structure
Kim et al. J Dent Res 2004

45 3 SNPs in the TAS2R38 Gene P A V P A I P V V P V I A A V A A I A V V
Haplotype definition Each individual has two haplotypesdiplotype Haploytpeallele diplotypegenotype A A I A V V A V I

46 TAS2R38 Diplotype and PTC Score
2 haplotypes3 diplotypes AVI = 2 PAV = 10 Heterozygote = 9 A multivariate analysis explains most of the variance with a p-value of < 10-33 Kim et al. Science 2003

47 Confirm Mode of Inheritance
Both: 15% One: 32% Neither: 100% L. H. Snyder Science 1931 47

48 Explain Linkage Signal

49 Geographic Distribution of PTC Phenotype
Wooding Genetics 2006, adapted from Cavalli-Sforza 1994

50 Geographic Distribution of PTC Haplotypes
Kim et al. Science 2003

51 Diplotype and PTC Score
You may notice a 4th diplotype  due to a 3rd, rare haplotype. Kim et al. Science 2003

52 3 SNPs form 3 Haplotypes P A V A V I A A V Taster Non-taster Rare
3rd haplotype is the result of recombination. A of non-taster AV of taster Allows us to compare the effect of the 1st SNP vs. the 2nd and 3rd. Rare-not in all combinations Rare A A V

53 Comparing Diplotypes Mean for taster is 9
Mean for rare haplotype is 7. Difference is P vs. A Mean for non-taster is 2 Difference is AV vs. VI Both have an effect, but 2nd and 3rd have a greater effect.

54 Predicted Effect of the 3 SNPs
3 a.a. substitution matrices Sequence alignment Scale Chemical change 1st and 2nd most severe.

55 TAS2R38 Haplotype Function

56 PTC Diplotype and Taste
Sandell and Breslin Current Biology 2006

57 Next Week Next Generation Sequencing

58 Appendix: Phase Unknown Linkage

59 Phase Unknown ? ? ? ? ? ? ? ? ? ? D d d d 1 2 3 3 D d D d d d 1 3 2 3
? ? ? ? Phase Unknown ? ? ? ? ? ? D d d d D d D d d d

60 Phase Unknown ? ? ?

61 What if we don’t know phase?
We calculate the LOD score for each phase Divide by 2

62 Phase Unknown + = -0.02 for q = 0.44 LOD (q) = ½ log (q)1 (1 - q)2
____________________ (q = 1/2) 1 + 2 + ½ log (q)2 (1 – q)1 (q = 1/2) 2 + 1 = for q = 0.44


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