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Eric Jorgenson Epidemiology 217 2/21/12
Linkage Analysis Eric Jorgenson Epidemiology 217 2/21/12
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Worldwide Distribution of Human Earwax SNP rs17822931
Yoshiura et al., Nature Genetics 2006
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Geographic Distribution of PTC Phenotype
High Low Wooding Genetics 2006, adapted from Cavalli-Sforza 1994
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Bimodal Distribution of PTC
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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
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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
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Family Studies Familial Relationships Phenotype information
Twins Siblings Parents/offspring Phenotype information Affected/Unaffected (Prostate Cancer) Quantitative measure (Blood Pressure) No Genotype information required
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Why do Family Studies? Is the trait genetic?
What is the mode of transmission? Dominant Recessive Additive Polygenic (Multiple genes involved)
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Mutation and Meiosis
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Recessive trait
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First PTC Family Study L. H. Snyder Science 1931 Both: 15% One: 32%
Neither: 100% L. H. Snyder Science 1931
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Linkage Analysis Narrow down position of disease gene
No biological knowledge needed Genetic markers (not disease gene) Recombination
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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
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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
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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
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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
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Recombination Fraction
61 420 442 77 A a a a A a a a b b b b B b B b
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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
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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
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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)
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Genomewide Linkage Analysis
Genetic Markers q = 10% on average Genes
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Linkage Analysis LOD score based on recombination
LOD (q) = log (q)R (1 - q)NR ____________________ (q = 1/2) R + NR
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Dominant Trait D d d d D d D d d d
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Linkage Analysis R NR NR
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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
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IBD Identity by descent Allele Sharing methods
Often used for affected sib pairs
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Identity By Descent a A a A 25% 25% 25% 25% A A a A A a a a
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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
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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
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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
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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
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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
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PTC Linkage Analysis Utah Genetic Reference Project 27 large families
269 subjects
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PTC Linkage Analysis
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Human Chromosomes 23 pairs of chromosomes TAS2R38
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Fine Mapping Linkage markers Genes Kim et al. Science 2003
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Linkage Disequilibrium
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Linkage Disequilibrium
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Linkage Disequilibrium
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Linkage Disequilibrium
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Linkage Disequilibrium
Time
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Linkage Disequilibrium Mapping
Genetic Markers Genes
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PTC Linkage Disequilibrium Mapping
Kim et al. Science 2003
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TAS2R38 Receptor Structure
Kim et al. J Dent Res 2004
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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 haplotypesdiplotype Haploytpeallele diplotypegenotype A A I A V V A V I
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TAS2R38 Diplotype and PTC Score
2 haplotypes3 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
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Confirm Mode of Inheritance
Both: 15% One: 32% Neither: 100% L. H. Snyder Science 1931 47
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Explain Linkage Signal
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Geographic Distribution of PTC Phenotype
Wooding Genetics 2006, adapted from Cavalli-Sforza 1994
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Geographic Distribution of PTC Haplotypes
Kim et al. Science 2003
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Diplotype and PTC Score
You may notice a 4th diplotype due to a 3rd, rare haplotype. Kim et al. Science 2003
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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
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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.
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Predicted Effect of the 3 SNPs
3 a.a. substitution matrices Sequence alignment Scale Chemical change 1st and 2nd most severe.
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TAS2R38 Haplotype Function
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PTC Diplotype and Taste
Sandell and Breslin Current Biology 2006
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Next Week Next Generation Sequencing
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Appendix: Phase Unknown Linkage
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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
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Phase Unknown ? ? ?
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What if we don’t know phase?
We calculate the LOD score for each phase Divide by 2
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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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