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

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

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

Bimodal Distribution of PTC

Your Phenotypes and Genotypes Taste SNP Ear wax SNP Sample Name taster ear wax rs10246939 rs1726866 rs17822931 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

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

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

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

Mutation and Meiosis

Recessive trait

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

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

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

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

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

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

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

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

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

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)

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

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

Dominant Trait D d d d 1 2 3 3 D d D d d d 1 3 2 3 2 3

Linkage Analysis 1 2 3 3 R NR NR 1 3 2 3 2 3

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

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

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

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

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 1 1 1 1 A A a A A a a a

Identity By Descent A A A A a A A a a a Sibling 1 Alleles shared IBD 2 1 1 0 A A a A A a a a

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 2 1 1 0 A A a A A a a a

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

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

PTC Linkage Analysis

Human Chromosomes 23 pairs of chromosomes TAS2R38

Fine Mapping Linkage markers Genes Kim et al. Science 2003

Linkage Disequilibrium

Linkage Disequilibrium

Linkage Disequilibrium

Linkage Disequilibrium

Linkage Disequilibrium Time

Linkage Disequilibrium Mapping Genetic Markers Genes

PTC Linkage Disequilibrium Mapping Kim et al. Science 2003

TAS2R38 Receptor Structure Kim et al. J Dent Res 2004

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

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

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

Explain Linkage Signal

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

Geographic Distribution of PTC Haplotypes Kim et al. Science 2003

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

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

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.

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

TAS2R38 Haplotype Function

PTC Diplotype and Taste Sandell and Breslin Current Biology 2006

Next Week Next Generation Sequencing

Appendix: Phase Unknown Linkage

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 1 2 3 3 D d D d d d 1 3 2 3 2 3

Phase Unknown 1 2 3 3 ? ? ? 1 3 2 3 2 3

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

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 = -0.02 for q = 0.44