Statistics of Linked Markers in Relationship Testing

Transcription

1 Statistics of Linked Markers in Relationship Testing Max P. Baur Rolf Fimmers Inst. f. Med. Biometry, University of Bonn Peter Schneider Inst. f. Legal Medicine, University of Cologne

2 Population Genetics Linkage Disequilibrium Max. P. Baur

3 Likelihood Statements in Relationship Testing How many individuals are involved in a simple Trio - Case child (tested) 2 mother (tested) 4 3 P-father (tested) 4 Random man (not tested) Hypotheses H: P - father is the biological father of the child H2: P - father is not the biological father of the child

4 Likelihood Statements in Relationship Testing How many individuals are involved in a simple Trio - Case child (tested) 2 mother (tested) 4 3 P-father (tested) 4 Random man (not tested) Hypotheses H: P - father is the biological father of the child H2: P - father is not the biological father of the child Wrong

5 Likelihood Statements in Relationship Testing How many individuals are involved in a simple Trio - Case child (tested) 2 mother (tested) 4 3 P-father (tested) 4 Random man (not tested) Hypotheses H: P - father is the biological father of the child H2: P - father is not related to the child H H2 Representation

6 Likelihood Statements in Relationship Testing All arguments are based on two (or more) mutually exclusive hypotheses Each hypothesis can be represented by a pedigree displaying the postulated relationships of the individuals involved Likelihood of a relationship hypothesis is the likelihood of a pedigree defining the relationships given the genetic information of marker typing

7 Likelihood Statements in Relationship Testing Example: Two persons, who know that they have the same mother want to find out, if they are full sibs or half sibs. Hypotheses: Persons H H2 How do you calculate the coresponding likelihoods given the marker information?

8 General Likelihood Algorithm Elston -Stewart -Algorithm see also: Ihm P., Hummel K. Z. Immun. Forsch. 49, (975) a b c d b c a d e f a d a c a e d f L(P) = Σ... Σ πpen (i k i ) πψ (k i ) πtrans (k i k i, k i ) k G k n G n i obs i unrel i, i 2, i 3 2 3

9 Independence Traits in individuals ) Sex - male, female P(M) = /2 P(F) = /2 2) Intelligence - intelligent, not so intelligent P(I) = /4 P(D) = 3/4 Question: What is the probability to find by chance an intelligent man P(M I) =?

10 Independence M F Individuals D /4 0 (M I) /2 I If sex and intelligence are independent P(M I) = /2 /4 = /8 Two stochastic events A and B are independent, if P(A B) = P(A) P(B)

11 Independence - Dependence L(P) = Σ... Σ πpen (i k i ) πψ (k i ) πtrans (k i k i, k i ) k G k n G n i obs i unrel i, i 2, i Independence - dependence of: Individuals Definition of the pedigree structure. Unrelated individuals: multiplication of population frequencies Descendents: multiplication of transmission probabilities conditional on the genotypes of both parents

12 Independence - Dependence Independence - dependence of: Marker loci 3 rd Law - Gregor Mendel Two traits segregate independently Only true for unlinked loci θ : proportion of recombinant gametes - θ : proportion of non-recombinant gametes θ only affects transmission within pedigree 0 θ ½

13 Independence - Dependence Independence - dependence of: Alleles from different loci

14 Independence - Dependence Two - Locus Linkage Disequilibrium A a L L 2 - P B = P b b What is the relationship between haplotypefrequencies and the corresponding genefrequencies of the two involved loci? P B 0 P A. P B +D AB B P A P a = -P A P(AB) = : P AB D AB = P AB = P A. P B - P A. P B + D AB

15 Independence - Dependence Independence - dependence of: Alleles from different loci Linkage disequlibrium δ only affects the calculation regarding population frequencies - unrelated individuals in the pedigree

16 Independence - Dependence Independence - dependence of: Alleles from different loci

17 Formal Genetics Penetrance - not important, marker loci Phase - very important for multilocus systems Individual typing results: Possible diplotypes: L : 2 / L2 : / L3 : 2 / L4 : 2 / 2 L5 : 2 / ) 2) 3) ) 6) 7) Closely linked SNPs 4) )

18 H Trio Case H A A Mother: Child: A P-father: 3 A 2 A 3 A 4 L =.. 2a a 2 (Mother) L 2 = 2a a 2 (Mother) 2a 3 a 4 ½ ½ (Father) (Child)... a 3 ½ 2a 3 a 4 (unknown biological father) (Child) (unrelated PF) = a a 2 a 3 a 4 = a a 2 a 3 2a 3 a 4 PI = 2a 3

19 Trio Case - Two Linked Markers A Mother: B A Child: B A P-father: 3 B 3 A 2 B 2 A 3 B 3 A 4 B 4 L = L 2 =. PI =

20 Trio Case Two independent markers Two (linked) markers θ = 0 0 < θ < ½ θ = ½ PI = 2a 3 δ = 0 PI 2 = 2b 3 independent markers independent markers independent markers PI t = PI PI 2 = 4a 3 b 3 δ 0 haplotype frequencies haplotype frequencies + recomb. fraction theoretical

21 Grandparent Case H H2 Mother: A B A 2 B 2 Child: A B A 3 B 3 P-GF: A 3 B 3 A 4 B 4 P-GM: A 5 B 5 A 6 B 6 L = L 2 = PI =

22 Grandparent Case Two independent markers Two (linked) markers θ = 0 0 < θ < ½ θ = ½ PI = 4a 3 δ = 0 = PI 2 = 4b 3 8a 3 b 3 haplotype haplotype + recomb. fraction independent markers PI t = PI PI 2 = 6 a 3 b 3 δ 0 haplotype frequencies haplotype frequencies + recomb. fraction theoretical

23 H Grandparent Case 2 H2 Mother: A B A 2 B 2 Child: A B A 3 B 5 P-GF: A 3 B 3 A 4 B 4 P-GM: A 5 B 5 A 6 B 6 L = L 2 = PI =

24 H Half - sibs: Full sibs H2 Mother: A B A 2 B 2 Child : A B A 3 B 3 Child 2: A B A 3 B 3 PI = Specialcases:

25 Forensic Genetics Linkage and Linkage Disequilibrium Problems to be solved: You do not want: Identification Paternity Testing Complex relationship testing Mixed stain analysis Mixed stain analysis involving relatives Polymorphisms, which are (closely) linked, because the phenotype-genotype relation is no longer unique. Possible recombination! Polymorphisms, with alleles in linkage disequilibrium because LD reduces information in comparison to the equilibrium state. Haplotypes! Multi - Locus - Situations ) Unknown phase - computational problems in extended haplotypes (remember HLA!!!) 2) Linkage disequilibria - reduced information; large samples to estimate 3) Recombination fractions - complex to incorporate, if not equal to 0 or ½

26 Examples 22 Trios typed with the GENPLEX panel (48 SNPs)

27 Examples 22 Trios typed with the GENPLEX panel (48 SNPs) Fam PI W Excl. when permuted V4,09E+06 99, V6 9,40E+03 99, PI Value 8 8 V7,24E+04 99,9993 V8,72E+04 99,99477 V9 2,0E+03 99,95239 V 2,88E+04 99, V6 7,03E+04 99, V8 7,93E+06 99, V23 8,46E+06 99, Frequency ,00E+04,00E+05,00E+06,00E+07 Class V25 8,60E+06 99, V27,69E+05 99, V29,02E+03 99,90207 W Value 7 V33 3,08E+05 99, V37 2,22E+05 99, V45 2,38E+04 99,9958 V52 6,83E+03 99, V56 9,00E+04 99, V58 3,85E+06 99, Frequency ,99 99,999 99, , V59 8,24E+06 99, Class 4 V60,60E+05 99, V6 3,34E+04 99, V62 2,74E+05 99,

28 References Linkage disequilibrium - Haplotype frequencies SNPs - Population Genetics Jennings Genetics 2 : 97 (96) Robbins Genetics 3 : 73 (97) Lewontin + Evolution 4 : 458 (960) Kojima Greiringer Ann. Math. Stat. 5 : 25 (944) Benett Ann Eugenics 8 : 3 (954) C.A.B. Smith (975) Cepellini (955, 967) Piazza (957) Weir (979) Baur (980). Ardlie et al Nature Review Genetics 3 : 299 (2002) Gabriel et al Science 296 : 2225 (2002) Weiss et al Trends in Genetics 8 : 9 (2002). SNPs - Paternity Testing Sanchez et al FSI Genetics 2 : 76 (2008) Borsting et al FSI Genetics 2 : 292 (2008) Philips et al FSI Genetics 2 : 98 (2008) Hepler + Weir FSI Genetics 2 : 66 (2008)

29 Forensic Genetics Linkage and Linkage Disequilibrium For every complex problem there is a simple solution - and it is wrong Mencken s Law