AFDAA 2012 WINTER MEETING Population Statistics Refresher Course - Lecture 3: Statistics of Kinship Analysis

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1 AFDAA 2012 WINTER MEETING Population Statistics Refresher Course - Lecture 3: Statistics of Kinship Analysis Ranajit Chakraborty, PhD Center for Computational Genomics Institute of Applied Genetics Department of Forensic and Investigative Genetics University of North Texas Health Science Center Fort Worth, Texas 76107, USA Tel. (817) ; Fax (817) ranajit.chakraborty@unthsc.edu Lecture given as a part of the AFDAA Population Statistics Refresher Course Held at the AFDAA 2012 Winter Meeting at Auston, TX on February 2, 2012

2 Statistics of Kinship Analysis Learning Objectives Attendees of this lecture should be able to understand Objectives of kinship analysis from DNA evidence data Possible conclusions of kinship analysis Questions answered from kinship analysis Concept of exclusion probability and their limitations Likelihood ratio approach of kinship analysis Paternity analysis, reverse parentage analysis, and kinship analysis for missing person identification Advanced issues mutation, need of linage markers

3 Lecture 3: Statistics for Kinship Analysis from DNA Evidence Topics Covered What is kinship analysis and its special cases Possible conclusions from kinship analysis of DNA evidence Questions answered in kinship analysis Requirements of data for kinship analysis Likelihood ratio: Paternity Index, Kinship Index General formulation of statistics for kinship analysis Advanced issues (mutation, missing person identification with multiple remains and choice of informative reference samples)

4 Kinship Analysis Kinship Determination Objectives: - Evidence sample s DNA is compared with that of one or more reference profiles - The objective is to determine the validity of stated biological relatedness among individuals, generally in reference to the placement of a specific target individual in the pedigree of reference individuals profiles

5 Kinship Analysis Types of Kinship Analysis - Standard Paternity Analysis - Deficient Paternity Analysis (e.g., Mother-less cases) - Reverse Parentage Analysis - Familial Search (i.e., Pairwise relationship testing) - Missing person identification

6 Three Types of Conclusions Exclusion (Match), or Inclusion Inconclusive

7 What is an Exclusion? In all types of Kinship Analyses Allele sharing among evidence and reference samples disagrees with the Mendelian rules of transmission of alleles with the stated relationship being tested

8 When is the Observation at a Locus Inconclusive? Compromised nature of samples tested failed to definitely exclude or include reference individuals May occur for one or more loci, while other loci typed may lead to unequivocal definite inclusion/ exclusion conclusions Caused often by DNA degradation (resulting in allele drop out), and/or low concentration of DNA (resulting in alleles with low peak height and/or area) for the evidence sample

9 What is an Inclusion? In all types of Kinship Analyses Allele sharing among evidence and reference samples is consistent with Mendelian rules of transmission of alleles with the stated relationship being tested; i.e., the stated biological relationship cannot be rejected (Note: In the context of Kinship analyses, the terminology of match is not appropriate)

10 Exclusion Nope Nope

11 Inclusion

12 PATERNITY TESTING MOTHER ALLEGED FATHER CHILD Two alleles for each autosomal genetic marker

13 Language of Paternity Testing Maternal Contribution Obligate Paternal Allele Dual Obligate Paternal Alleles

14 Three Genetic Profiles are Determined Mother A B Child B m C p Alleged father C D

15 Typical Paternity Test Two possible outcomes of test: Inclusion The obligate paternal alleles in the child all have corresponding alleles in the Alleged Father Exclusion The obligate paternal alleles in the child DO NOT have corresponding alleles in the Alleged Father

16 Exclusion Nope Nope

17 Results The Tested Man is Excluded as the Biological Father of the Child in Question

18 Inclusion

19 Results The Tested Man Cannot be Excluded as the Biological Father of the Child in Question Several Statistical Values are Calculated to Assess the Strength of the Genetic Evidence

20 Language of Paternity Testing PI CPI W PE RMNE Paternity Index Combined Paternity Index Probability of Paternity Probability of Exclusion Random Man Not Excluded

21 summarizes information provided by Likelihood Ratio Paternity Index genetic testing Probability that some event will occur under a set of conditions or assumptions Divided by the probability that the same event will occur under a set of different mutually exclusive conditions or assumptions

22 Paternity Index Observe three types from a man, a woman, and a child Assume true trio the man and woman are the true biologic parents of child Assume false trio woman is the mother, man is not the father In the false trio the child s father is a man of unknown type, selected at random from population (unrelated to mother and tested man)

23 Standard Paternity Index Mother, Child, and Alleged Father PI is a likelihood ratio = X/Y Defined as the probability that an event will occur under a particular set of conditions (X). Divided by the probability that the event will occur under a different set of conditions (Y).

24 Standard Paternity Index In paternity testing, the event is observing three phenotypes, those of a woman, man and child. The assumptions made for calculating the numerator (X) is that these three persons are a true trio. For the denominator (Y) the assumptions is that the three persons are a false trio.

25 Paternity Biological Relationship Parents M F Child C

26 Paternity Analysis?

27 Paternity Analysis Hypothetical case DNA Analysis Results in Three Genotypes Mother Child Alleged Father () (BC) (CD)

28 Paternity Analysis CD BC An mother and a CD father can have four possible offspring: AC, AD, BC, BD

29 Standard Paternity Index PI determination in hypothetical DNA System PI = X / Y Numerator X = is the probability that (1) a woman randomly selected from a population is type, and (2) a man randomly selected from a population is type CD, and (3) their child is type BC.

30 Paternity Analysis CD BC

31 Standard Paternity Index PI determination in hypothetical DNA System PI = X / Y Denominator Y = is the probability that (1) a woman randomly selected from a population is type, (2) a man randomly selected and unrelated to either mother or child is type CD, and (3) the woman s child, unrelated to the randomly selected man is BC.

32 Paternity Analysis CD Tested Man BC CD Untested Random Man

33 Standard Paternity Index When mating is random, the probability that the untested alternative father will transmit a specific allele to his child is equal to the allele frequency in his race. We can no look into how to actually calculate a Paternity Index

34 Hypothetical DNA Example First Hypothesis Numerator Person Mother Child Alleged Father Type BC CD In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type b) Man randomly selected from population is type CD, and c) Their child is type BC

35 Paternity Analysis Paternity Index Numerator 2p A p B CD 2p C p D BC Probability = 2p A p B x 2p C p D x 0.5 x 0.5

36 Hypothetical DNA Example Second Hypothesis Denominator Person Mother Child Alleged Father Type BC CD In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type b) An alternative man randomly selected from population is type CD, and c) The woman s child, fathered by random man, is type BC

37 Paternity Analysis Paternity Index Denominator 2p A p B CD 2p C p D 0.5 BC p C Probability = 2p A p B x 2p C p D x 0.5 x p C

38 Paternity Analysis Paternity Index PI = PI = 2p A p B x 2p C p D x 0.5 x 0.5 2p A p B x 2p C p D x 0.5 x p C 0.5 p C

39 Hypothetical DNA Example Probability Statements Person Mother Child Alleged Father Type BC CD One might say (Incorrectly) a) Numerator is probability that tested man is the father, and b) Denominator is probability that he is not the father

40 Hypothetical DNA Example Probability Statement Person Mother Child Alleged Father Type BC CD A Correct statement is a) Numerator is probability of observed genotypes, given the tested man is the father, and b) Denominator is probability of observed genotypes, given a random man is the father.

41 Paternity M and C share one allele and AF is heterozygous for the other allele Parents M? CD AF Child BC C AF has a 1 in 2 chance of passing C allele Random Man has p chance of passing the C allele PI = 0.5/p

42 There are 15 possible combinations of genotypes for a paternity trio

43 Paternity Biological Relationship Parents M F Child C

44 Paternity Index M and C share one allele and AF is homozygous for the obligatory allele Parents? C M AF Child BC C AF can only pass C allele Random Man has p chance of passing the C allele PI = 1/p

45 Paternity Analysis Paternity Index Numerator 2 2p A p p C B C BC Probability = 2p A p B x p C 2 x 0.5 x 1

46 Paternity Analysis Paternity Index Denominator 2p A p B C p C BC p C Probability = 2p A p B x p C 2 x 0.5 x p C

47 Paternity Analysis Paternity Index PI = PI = 2p A p B x p 2 C x 0.5 x 1 2p A p B x p 2 C x 0.5 x p C 1 p C

48 Paternity Index M and C share both alleles and AF is heterozygous with one of the obligatory alleles Parents? BC M AF Child C M has a 1 in 2 chance of passing A or B allele AF has a 1 in 2 chance of passing B allele RM has (p + q) chance of passing the A or B alleles PI = 0.5/(p+q)

49 Paternity Analysis Paternity Index Numerator 2p A p B BC 2p B p C 0.5 A 0.5 B Probability = 2p A p B x 2p B p C x 0.5 (ma) x 0.5 (fb)

50 Paternity Analysis Paternity Index Denominator 2p A p B BC 2p B p C 0.5 probability = p A + p B 2p A p B x 2p B p C x (0.5 (ma) x p B (mb) x p A )

51 Paternity Analysis Paternity Index PI = 2p A p B x 2p B p C x 0.5 (ma) x 0.5 (fb) 2p A p B x 2p B p C x (0.5 (mb) x p A (ma) x p B ) PI = PI = p A + 0.5p B 0.5 p A + p B

52 Paternity Index M and C share both alleles and AF is heterozygous with both of the obligatory alleles Parents? M AF Child C M has a 1 in 2 chance of passing A or B allele AF has a 1 in 2 chance of passing A or B allele RM has (p + q) chance of passing the A or B alleles PI = 1/(p+q)

53 Paternity Analysis Paternity Index Numerator 2p A p B 2p A p B 0.5 A B 0.5 A B Probability = 2p A p B x 2p B p C x (0.5 (ma) x 0.5 (fb) (mb) x 0.5 (fa) )

54 Paternity Analysis Paternity Index Denominator 2p A p B 2p A p B 0.5 A B probability = p A + p B 2p A p B x 2p A p B x (0.5 (ma) x p B (mb) x p A )

55 Paternity Analysis Paternity Index PI = 2p A p B x 2p B p C x (0.5 (ma) x 0.5 (fb) (mb) x 0.5 (fa) ) 2p A p B x 2p B p C x (0.5 (mb) x p A (ma) x p B ) PI = PI = p A + 0.5p B 1 p A + p B

56 Paternity Biological Relationship Parents M F Child C

57 Paternity Index M and C share both alleles and AF is heterozygous with one of the obligatory alleles Parents? B M AF Child C M has a 1 in 2 chance of passing A or B allele AF can only pass the B allele RM has (p + q) chance of passing the A or B alleles PI = 1/(p+q)

58 Paternity Analysis Paternity Index Numerator 2p A p B p B B Probability = 2p A p B x p B 2 x 0.5 (ma) x 1 (fb)

59 Paternity Analysis Paternity Index Denominator 2p A p B B p B probability = p A + p B 2p A p B x p B2 x (0.5 (ma) x p B + 1 (mb) x p A )

60 Paternity Analysis Paternity Index PI = 2p A p B x p B 2 x 0.5 (ma) x 1 (fb) 2p A p B x p B2 x (0.5 (mb) x p A (ma) x p B ) PI = PI = p A + 0.5p B 1 p A + p B

61 PI Formulas Single locus, no null alleles, low mutation rate, codominance M A A A BC BC BD C A A A AF BC AC AC AC Numerator Denominator a b b 0.5a 0.5a 0.5a 0.5a 0.5a PI 0.5/a 0.5/b 0.5/b 0.5/a 0.5/a 0.5/a 0.5/a 0.5/a

62 PI Formulas Single locus, no null alleles, low mutation rate, codominance M A B BC C A A AF A A A A Numerator Denominator a 0.5a a 0.5a PI 1/a 1/a 1/a 1/a

63 PI Formulas Single locus, no null alleles, low mutation rate, codominance M C AF AC Numerator 0.25 Denominator 0.5(a+b) PI 0.5/(a+b)

64 PI Formulas Single locus, no null alleles, low mutation rate, codominance M C AF A Numerator Denominator 0.5(a+b) 0.5(a+b) PI 1/(a+b) 1/(a+b)

65 Effect of Population Substructure

66 SAMPLING THEORY OF ALLELE FREQUENCIES Under the mutation-drift balance, the probability of a sample in which n i copies of the allele A i is observed, for any set of i = 1, 2,... is given by n ( ) ( ) i i ni Pr( Ai ) ( n ) ( ) i i i Where n = Σn i, i = p i (1 - )/,. = i = (1 - )/, p i = frequency of allele A i in the population and ( ) is the Gamma function, in which is the coefficient of coancestry (equivalent to F st or G st, the coefficient of gene differentiation between subpopulations within the population)

67 GENOTYPE FREQUENCY Using the general theory, 2 i i i i i Pr( A A ) p p (1 p ) Pr( A A ) 2 p p (1 ) i j i j

68 CONDITIONAL MATCH PROBILITY [2 (1 ) pi][3 (1 ) pi] Pr( Ai Ai Ai Ai ) (1 )(1 2 ) Pr( A A A A ) i j i j 2[ (1 ) p ][ (1 ) p ] i (1 )(1 2 ) j Thus, for = 0, 2 i i i i i i i Pr( A A ) Pr( A A A A ) P Pr( A A ) Pr( A A A A ) 2PP i j i j i j i j

69 PI Formulas population substructure considered M A A A BC BC BD C A A A AF BC AC AC AC PI (1+3 ) / 2[3 +a(1- )] (1+3 ) / 2[ +b(1- )] (1+3 ) / 2[ +b(1- )] (1+3 ) / 2[2 +a(1- )] (1+3 ) / 2[2 +a(1- )] (1+3 ) / 2[ +a(1- )] (1+3 ) / 2[ +a(1- )] (1+3 ) / 2[ +a(1- )]

70 PI Formulas Single locus, no null alleles, low mutation rate, codominance M A B BC C A A AF A A A A PI (1+3 ) / [4 +a(1- )] (1+3 ) / [3 +a(1- )] (1+3 ) / [2 +a(1- )] (1+3 ) / [2 +a(1- )]

71 PI Formulas Single locus, no null alleles, low mutation rate, codominance M C AF AC PI (1+3 ) / 2[3 +(1- )(a+b)]

72 PI Formulas Single locus, no null alleles, low mutation rate, codominance M C AF A PI (1+3 ) / [4 +(1- ) (a+b)] (1+3 ) / [4 +(1- ) (a+b)]

73 Combined Paternity Index

74 Combined Paternity Index When multiple genetic systems are tested, a PI is calculated for each system. This value is referred to as a System PI. If the genetic systems are inherited independently, the Combined Paternity Index (CPI) is the product of the System PI s

75 Combined Paternity Index What is the CPI? The CPI is a measure of the strength of the genetic evidence. It indicates whether the evidence fits better with the hypothesis that the man is the father or with the hypothesis that someone else is the father. continued...

76 Combined Paternity Index The theoretical range for the CPI is from 0 to infinity A CPI of 1 means the genetic tests provides no information A CPI less than 1; the genetic evidence is more consistent with non-paternity than paternity. A CPI greater than 1; the genetic evidence supports the assertion that the tested man is the father.

77 Probability of Paternity

78 Probability of Paternity The probability of paternity is a measure of the strengths of one s belief in the hypothesis that the tested man is the father. The correct probability must be based on all of the evidence in the case. The non-genetic evidence comes from the testimony of the mother, tested man, and other witnesses. The genetic evidence comes from the DNA paternity test.

79 Probability of Paternity The probability of paternity (W) is based upon Baye s Theorem, which provides a method for determining a posterior probability based upon the genetic results of testing the mother, child, and alleged father. In order to determine the probability of paternity, an assumption must be made (before testing) as to the prior probability that the tested man is the true biological father.

80 Probability of Paternity The prior probability of paternity is the strength of one s belief that the tested man is the father based only on the non-genetic evidence.

81 Probability of Paternity Probability of Paternity (W) = CPI x P [CPI x P + (1 P)] P = Prior Probability; it is a number greater than 0 and less than or equal to 1. In many criminal proceedings the Probability of Paternity is not admissible. In criminal cases, the accused is presumed innocent until proven guilty. Therefore, the defense would argue that the Prior Probability should be 0. You cannot calculate a posterior Probability of Paternity with a Prior Probability of 0.

82 Probability of Paternity In the United States, the court system has made the assumption that the prior probability is equal to 0.5. The argument that is presented is that the tested man is either the true father or he is not. In the absence of any knowledge about which was the case, it is reasonable to give these two possibilities equal prior probabilities.

83 Probability of Paternity With a prior probability of 0.5, the Probability of Paternity (W) = = CPI x 0.5 [CPI x (1 0.5)] CPI x 0.5 CPI x = CPI CPI + 1

84 Posterior Odds in Favor of Paternity Posterior Odds = CPI x Prior Odds Prior Odds = P / (1 - P) Posterior Odds in Favor of Paternity = CPI x [P / (1 - P)] If the prior probability of paternity is 0.7, then the prior odds favoring paternity is 7 to 3. If a paternity test is done and the CPI is 10,000, then the Posterior Odds in Favor of Paternity = 10,000 x (0.7 / 0.3) = 23,333 Posterior Odds in Favor of Paternity = 23,333 to 1

85 Probability of Exclusion

86 Probability of Exclusion The probability of exclusion (PE) is defined as the probability of excluding a random individual from the population given the alleles of the child and the mother The genetic information of the tested man is not considered in the determination of the probability of exclusion

87 Probability of Exclusion The probability of exclusion (PE) is equal to the frequency of all men in the population who do not contain an allele that matches the obligate paternal allele of the child.

88 Probability of Exclusion PE = 1 - (a 2 + 2ab) a = frequency of the allele the child inherited from the biological father (obligate paternal allele). The frequency of the obligate allele is determined for each of the major racial groups, and the most common frequency is used in the calculation.

89 Probability of Exclusion PE = 1 - (a 2 + 2ab) b = sum of the frequency of all alleles other than the obligate paternal allele. b = (1 a) PE = 1 [a 2 + 2a(1 a)] PE = 1 [a 2 + 2a 2a 2 ] PE = 1 [2a a 2 ] PE = 1 2a + a 2 PE = (1 a) 2

90 Probability of Exclusion If the Mother and Child are both phenotype, men who cannot be excluded are those who could transmit either an A or B allele (or both). In this case the: PE = [1 - (a + b)] 2

91 Probability of Exclusion In addition, if population substructure is considered PE = (1-a b)(1- )[1-(1- )(a+b)] (1-2 ) (1-3 )

92 PE formulas M A C A OA A PE (1- )(1-a)[1-a(1- )] / (1+2 )(1+3 ) A A [2 +(1- )(1-b)] [3 +(1- )(1-b)] / (1+2 )(1+3 )

93 PE formulas M C AA OA A PE [1-a(1- )][1+ -a(1- )] / (1+2 )(1+3 ) BB B [1-b(1- )][1+ -b(1- )] / (1+2 )(1+3 ) A/B (1-a-b)(1- )[1-(a+b)(1- )] / (1+2 )(1+3 )

94 PE formulas M C AC OA C PE [(1- )(1-c)+2 ][(1- )(1-c)+3 ] / (1+2 )(1+3 ) BC C [(1- )(1-c)+2 ][(1- )(1-c)+3 ] / (1+2 )(1+3 )

95 Combined Probability of Exclusion The individual Probability of Exclusion is calculated for each of the genetic systems (loci) analyzed. The overall Probability of Excluding (CPE) a falsely accused man in a given case equals: 1 [(1 PE 1 ) x (1 PE 2 ) x (1 PE 3 ) x (1 PE N )]

96 Random Man Not Excluded PE = 1 - (a 2 + 2ab) RMNE This component of the equation represents the proportion of the male population that contain the obligate paternal allele and therefore would not be excluded as the father of the tested child for a given genetic test.

97 Random Man Not Excluded PE = 1 RMNE Or RMNE = 1 - PE For a given group of genetic loci tested, the proportion of the male population that would not be excluded is equal to: RMNE 1 x RMNE 2 x RMNE 3 x. RMNE N

98 PowerPlex TM 1.1 CSF1PO D16S539 TPOX D7820 TH01 D13S317 vwa D5S818 M C AF M C AF P P M C AF M C AF P P-41414

99 Typical Paternity Trio P M C AF Allele Frequency HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH01 7 9p 9 (11p15.5) 7m 6 HUMvWA m 16 (12p p13.2) 18 16p 15 8 = = = = = =

100 Typical Paternity Trio P M C AF PI Formula HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH01 7 9p 9 (11p15.5) 7m 6 HUMvWA m 16 (12p p13.2) 18 16p /(a+b)] 1/(a+b) 0.5/a 0.5/a

101 Typical Paternity Trio P M C AF Paternity Index HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH01 7 9p 9 (11p15.5) 7m 6 HUMvWA m 16 (12p p13.2) 18 16p

102 Typical Paternity Trio P M C AF PE Formula HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH01 7 9p 9 (11p15.5) 7m 6 HUMvWA m 16 (12p p13.2) 18 16p 15 [1 (a+b)] 2 [1 (a+b)] 2 (1 a) 2 (1 a) 2

103 Typical Paternity Trio P M C AF PE HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH01 7 9p 9 (11p15.5) 7 7m 6 HUMvWA m 16 (12p p13.2) 18 16p

104 Typical Paternity Trio P M C AF Allele Frequency D16S m 12 (16p24 - p25) 11p 11 D7S p 11 (7q) 9 9m 10 D13S m 11 (13q22 - q31) 10 8p 8 D5S (5q21 - q31) = = = =

105 Typical Paternity Trio P M C AF PI Formula D16S m 12 (16p24 - p25) 11p 11 D7S p 11 (7q) 9 9m 10 D13S m 11 (13q22 - q31) 10 8p 8 D5S (5q21 - q31) /a 0.5/a 0.5/a 1/a

106 Typical Paternity Trio P M C AF D16S m 12 (16p24 - p25) 11p 11 D7S p 11 (7q) 9 9m 10 D13S m 11 (13q22 - q31) 10 8p 8 D5S (5q21 - q31) Paternity Index

107 Typical Paternity Trio P M C AF D16S m 12 (16p24 - p25) 11p 11 D7S p 11 (7q) 9 9m 10 D13S m 11 (13q22 - q31) 10 8p 8 D5S (5q21 - q31) 11 PE Formula (1 a) 2 (1 a) 2 (1 a) 2 (1 a) 2

108 Typical Paternity Trio P M C AF D16S m 12 (16p24 - p25) 11p 11 D7S p 11 (7q) 9 9m 10 D13S m 11 (13q22 - q31) 10 8p 8 D5S (5q21 - q31) 11 PE

109 PowerPlex 2.1 PENTA E FGA D18S51 TPOX D8S1179 D21S11 TH01 vwa D3S1358 M C AF M C AF P P M C AF M C AF P P-41414

110 Typical Paternity Trio P M C AF Allele Frequency FGA 24 24m 23 (4q28) 23 21p 21 D18S (18q21.3) D21S m 29 (21q q21) 28 29p 28 D3S (3p) D8S p 15 (8) 10 14m = = = = = =

111 Typical Paternity Trio P M C AF PI Formula FGA 24 24m 23 (4q28) 23 21p 21 D18S (18q21.3) D21S m 29 (21q q21) 28 29p 28 D3S (3p) D8S p 15 (8) 10 14m /a 1/(a+b) 0.5/a 0.5/a 0.5/a

112 Typical Paternity Trio P M C AF Paternity Index FGA 24 24m 23 (4q28) 23 21p 21 D18S (18q21.3) D21S m 29 (21q q21) 28 29p 28 D3S (3p) D8S p 15 (8) 10 14m

113 Typical Paternity Trio P M C AF PE Formula FGA 24 24m 23 (4q28) 23 21p 21 D18S (18q21.3) D21S m 29 (21q q21) 28 29p 28 D3S (3p) D8S p 15 (8) 10 14m 13 (1 a) 2 [1 (a+b)] 2 (1 a) 2 (1 a) 2 (1 a) 2

114 Typical Paternity Trio P M C AF PE FGA 24 24m 23 (4q28) 23 21p 21 D18S (18q21.3) D21S m 29 (21q q21) 28 29p 28 D3S (3p) D8S p 15 (8) 10 14m

115 Typical Paternity Trio 13 Core CODIS Loci Combined Paternity Index 431,602 Probability of Paternity % Probability of Exclusion %

116

117 What if. we don t have the mother s genetic data? We can still develop a likelihood estimation for parentage. Lets examine the following logic:

118 Paternity Index Only Man and Child Tested Observe two types from a man and a child Assume true duo the man is the father of the child Assume false duo the man is not the father of the child (simply two individuals selected at random) In the false duo the child s father is a man of unknown type, selected at random from population (unrelated to tested man)

119 Paternity Index Only Man and Child Tested Hypothetical case DNA Analysis Results in Two Genotypes Mother Child Alleged Father Not Tested () (AC)

120 Motherless Paternity Index PI determination in hypothetical DNA System PI = X / Y Numerator X = is the probability that (1) a man randomly selected from a population is type AC, and (2) his child is type. X = Pr{AF passes A} x Pr {M passes B} + Pr{AF passes B} x Pr{M passes A}

121 Motherless Paternity Index PI determination in hypothetical DNA System PI = X / Y Denominator Y = is the probability that (1) a man randomly selected and unrelated to tested man is type AC, and (2) a child unrelated to the randomly selected man is. Y = Pr{RM passes A} x Pr {M passes B} + Pr{RM passes B} x Pr{M passes A}

122 Motherless Paternity Index When the mother s genetic data is present, Pr{M passes A} is 0, 0.5, or 1, and Pr{M passes B} is 0, 0.5, or 1 Without the mother s data, Pr {M passes A} becomes the frequency of the gametic allele, p and Pr {M passes B} becomes the frequency of the gametic allele, q.

123 Motherless Paternity Index So, if we have a heterozygous child, and a heterozygous Alleged Father AC then X = Pr{AF passes A} x Pr {M passes B} + Pr{AF passes B} x Pr{M passes A} X = Pr{AF passes A} x q + Pr{AF passes B} x p Pr{AF passes A} = 0.5 Pr{AF passes B} = 0 X = 0.5 x q + 0 x p X = 0.5q

124 Motherless Paternity Index So, if we have a heterozygous child, and a heterozygous Alleged Father AC then Y = Pr{RM passes A} x Pr {M passes B} + Pr{RM passes B} x Pr{M passes A} Y = p x q + q x p Y = 2pq

125 Motherless Paternity Index So, if we have a heterozygous child, and a heterozygous Alleged Father AC then PI = X / Y X = 0.5q Y = 2pq PI = 0.5q / 2pq PI = 0.25/p PI = 1/4p

126 Paternity Index Only Man and Child Tested Parents M? AC AF Child C The untested Mother could have passed either the A or B allele AF has a 1 in 2 chance of passing A allele RM has (p + q) chance of passing the A or B allele

127 Paternity Index Only Man and Child Tested AC

128 Paternity Index Only Man and Child Tested Numerator AC 2p A p C p B 0.5 A 2p A p B Probability = 2p A p C x 2p A p B x 0.5 (fa) x p B

129 Paternity Index Only Man and Child Tested Denominator AC 2p A p C p A + p B p A + p B 2p A p B probability = 2p A p C x 2p A p B x (p (ma) x p (fb) + p (mb) x p (fa) )

130 Paternity Index Only Man and Child Tested PI = 2p A p B x 2p A p C x 0.5 (ma) x p B 2p A p B x 2p A p C x (p (ma) x p (fb) + p (mb) x p (fa) ) PI = PI = 0.5p B 2p A p B 0.25 p A

131 Paternity Index Only Man and Child Tested Parents M? A AF Child C The untested Mother could have passed either the A or B allele AF can only pass A allele RM has (p + q) chance of passing the A or B allele

132 Paternity Index Only Man and Child Tested A

133 Paternity Index Only Man and Child Tested Numerator A p A 2 p B 1 2p A p B Probability = p A 2 x 2p A p B x 1 (fa) x p B

134 Paternity Index Only Man and Child Tested Denominator A p A 2 p A + p B p A + p B 2p A p B probability = p A2 x 2p A p B x (p (ma) x p (fb) + p (mb) x p (fa) )

135 Paternity Index Only Man and Child Tested PI = p A 2 x 2p A p C x 1 (ma) x p B p A 2 x 2p A p C x (p (ma) x p (fb) + p (mb) x p (fa) ) PI = PI = p B 2p A p B 0.5 p A

136 Paternity Index Only Man and Child Tested Parents M? AF Child C The untested Mother could have passed either the A or B allele AF can pass either A or B allele RM has (p + q) chance of passing the A or B allele

137 Paternity Index Only Man and Child Tested

138 Paternity Index Only Man and Child Tested Numerator 2p A p B p A + p B 0.5 A B 2p A p B Probability = 2p A p B x 2p A p B x (0.5 (fa) x p B (fb) x p A )

139 Paternity Index Only Man and Child Tested Denominator 2p A p B p A + p B p A + p B 2p A p B probability = 2p A p B x 2p A p B x (p (ma) x p (fb) + p (mb) x p (fa) )

140 Paternity Index Only Man and Child Tested PI = 2p A p B x 2p A p B x (0.5 (fa) x p B (fb) x p A ) 2p A p B x 2p A p B x (p (ma) x p (fb) + p (mb) x p (fa) ) PI = 0.5p B + 0.5p A 2p A p B PI = p A + p B 4p A p B

141 Paternity Index Only Man and Child Tested Parents? A M AF Child A C The untested Mother would have to pass an A allele AF can pass only the A allele RM has p chance of passing the A allele

142 Paternity Index Only Man and Child Tested A A

143 Paternity Index Only Man and Child Tested Numerator A p A 2 p A 1 p A 2 A Probability = p A 2 x p A 2 x 1 (fa) x p A

144 Paternity Index Only Man and Child Tested Denominator A p A 2 p A p A p A 2 A probability = p A 2 x p A 2 x p (ma) x p (fa)

145 Paternity Index Only Man and Child Tested PI = p A 2 x p A 2 x 1 (fa) x p A p A 2 x p A 2 x p (ma) x p (fa) PI = PI = p A p A x p A 1 p A

146 Paternity Index Only Man and Child Tested Parents? M AF Child A C The untested Mother would have to pass an A allele AF would have to pass the A allele RM has p chance of passing the A allele

147 Paternity Index Only Man and Child Tested A

148 Paternity Index Only Man and Child Tested Numerator 2p A p B p A 0.5 p A 2 A Probability = 2p A p B x p A 2 x 0.5 (fa) x p A

149 Paternity Index Only Man and Child Tested Denominator 2p A p B p A p A p A 2 A probability = 2p A p B x p A 2 x p (ma) x p (fa)

150 Paternity Index Only Man and Child Tested PI = 2p A p B x p A 2 x 0.5 (fa) x p A 2p A p B x p A 2 x p (ma) x p (fa) PI = PI = 0.5p A p A x p A 0.5 p A

151 Paternity Index Only Man and Child Tested Formulas Single locus, no null alleles, low mutation rate, codominance C A A AF AC A AC A Numerator 0.5b 0.5(a+b) b 0.5a a Denominator PI 2ab 0.25/a 2ab (a+b)/4ab 2ab 0.5/a a 2 0.5/a a 2 1/a PE [1-(a + b)] 2 [1-(a + b)] 2 [1-(a + b)] 2 (1-a) 2 (1-a) 2

152 C Paternity Index Only Man and Child Tested Formulas: with Population Substructure AF AC PI (1+2 ) / 4[(1- )a+ ] PE (1- )(1-a-b)[1-(1- )(a+b)] / (1+ )(1+2 ) A (1+2 )[(a+b)(1- )+2 ] / 4[(1- )a+ ] [(1- )b+ ] (1+2 ) / 2[(1- )a+2 ]

153 C A Paternity Index Only Man and Child Tested Formulas: with Population Substructure AF AC PI (1+2 ) / 2[(1- )a+2 ] PE (1- )(1-a)[(1- )(1-a)+ ] / (1+ )(1+2 ) A A (1+2 ) / [(1- )a+3 ]

154 41376 PowerPlex 2.1 FGA TPOX PENTA E D18S51 D8S1179 D21S11 TH01 vwa D3S1358 C AF C AF

155 MOTHERLESS PATERNITY CASE P C AF Allele Frequencies HUMCSF1PO = (5q q34) = HUMTPOX = (2p23-2pter) = HUMTH = (11p15.5) = HUMvWA = (12p p13.2) =

156 41376 TPOX 13 C AF C PI = X / Y AF X = 0.5b + 0.5a Y = 2ab PI = (a + b) / 4ab PI = (0.5b + 0.5a) / 2ab

157 41376 THO C AF C AF PI = X / Y X = 0.5b Y = 2ab PI = 0.25 / a PI = 0.5b / 2ab

158 41376 vwa C AF C AF PI = X / Y X = b Y = 2ab PI = 0.5 / a PI = b / 2ab

159 MOTHERLESS PATERNITY CASE P C AF PI Formula HUMCSF1PO /a (5q q34) HUMTPOX 8 8 (a+b)/4ab (2p23-2pter) HUMTH /a (11p15.5) HUMvWA /a (12p p13.2) 16

160 MOTHERLESS PATERNITY CASE P C AF PI HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH (11p15.5) HUMvWA (12p p13.2) 16

161 MOTHERLESS PATERNITY CASE P C AF PE Formulas HUMCSF1PO [1-(a+b)] 2 (5q q34) HUMTPOX 8 8 [1-(a+b)] 2 (2p23-2pter) HUMTH [1-(a+b)] 2 (11p15.5) HUMvWA [1-(a+b)] 2 (12p p13.2) 16

162 MOTHERLESS PATERNITY CASE P C AF PE HUMCSF1PO (5q q34) HUMTPOX (2p23-2pter) HUMTH (11p15.5) HUMvWA (12p p13.2) 16

163 MOTHERLESS PATERNITY CASE P C AF Allele Frequencies D16S = (16p24 - p25) = D7S = (7q) = D13S = (13q22 - q31) D5S = (5q21 - q31) =

164 MOTHERLESS PATERNITY CASE P C AF PI Formulas D16S /a (16p24 - p25) D7S /a (7q) D13S /a (13q22 - q31) D5S /a (5q21 - q31) 13 12

165 MOTHERLESS PATERNITY CASE P C AF PI D16S (16p24 - p25) D7S (7q) D13S (13q22 - q31) D5S (5q21 - q31) 13 12

166 MOTHERLESS PATERNITY CASE P C AF PE Formulas D16S [1-(a+b)] 2 (16p24 - p25) D7S [1-(a+b)] 2 (7q) D13S (1-a) 2 (13q22 - q31) D5S [1-(a+b)] 2 (5q21 - q31) 13 12

167 MOTHERLESS PATERNITY CASE P C AF PE D16S (16p24 - p25) D7S (7q) D13S (13q22 - q31) D5S (5q21 - q31) 13 12

168 MOTHERLESS PATERNITY CASE P C AF Allele Frequencies FGA = (4q28) = D18S = (18q21.3) 20 D21S = (21q q21) 29 D3S = (3p) = D8S = (8) =

169 41376 FGA C AF C AF PI = X / Y X = 0.5b Y = 2ab PI = 0.5b / 2ab PI = 0.25 / a

170 41376 D18S C AF C AF PI = X / Y X = 0.5a Y = a 2 PI = 0.5 / a PI = 0.5a / a 2

171 C PI = X / Y AF D21S11 C AF X = 0.5a Y = a 2 PI = 0.5a / a 2 PI = 0.5 / a

172 41376 D3S C PI = X / Y X = 0.5b Y = 2ab AF C AF PI = 0.25 / a PI = 0.5b / 2ab

173 D8S1179 C AF C PI = X / Y AF X = 0.5b + 0.5a Y = 2ab PI = (0.5b + 0.5a) / 2ab PI = (a + b) / 4ab

174 MOTHERLESS PATERNITY CASE P C AF PI Formulas FGA /a (4q28) D18S /a (18q21.3) 20 D21S /a (21q q21) 29 D3S /a (3p) D8S (a+b)/4ab (8) 13 13

175 MOTHERLESS PATERNITY CASE P C AF PI FGA (4q28) D18S (18q21.3) 20 D21S (21q q21) 29 D3S (3p) D8S (8) 13 13

176 MOTHERLESS PATERNITY CASE P C AF PE Formulas FGA [1-(a+b)] 2 (4q28) D18S (1-a) 2 (18q21.3) 20 D21S (1-a) 2 (21q q21) 29 D3S [1-(a+b)] 2 (3p) D8S [1-(a+b)] 2 (8) 13 13

177 MOTHERLESS PATERNITY CASE P C AF PE FGA (4q28) D18S (18q21.3) 20 D21S (21q q21) 29 D3S (3p) D8S (8) 13 13

178 Motherless Paternity 13 Core CODIS Loci ( =0) Combined Paternity Index 1,676 Probability of Paternity 99.94% (prior=0.5) Probability of Exclusion 99.94%

179 Motherless Paternity 13 Core CODIS Loci ( =0.01) Combined Paternity Index 867 Probability of Paternity 99.88% Probability of Exclusion 99.90%

180 Motherless Paternity 13 Core CODIS Loci ( =0.03) Combined Paternity Index 309 Probability of Paternity 99.68% Probability of Exclusion 99.80%

181 Reverse Paternity Testing Can we identify crime scene evidence to a deceased/missing individual? Applications: no-body homicides victims of mass disasters

182 REVERSE PATERNITY INDEX BODY IDENTIFICATION ALLEGED EVIDENCE ALLEGED MOTHER FATHER A B B C C D

183 Three genotypes: Alleged Mother Child (missing) Alleged Father

184 Missing child?

185 Reverse Paternity Index RPI = X / Y Numerator X = is the probability that (1) a woman randomly selected from a population is type, and (2) a man randomly selected from a population is type CD, and (3) their child is type BC.

186 Reverse Paternity Analysis CD BC

187 Reverse Paternity Index RPI = X / Y Denominator Y = is the probability that (1) a woman randomly selected from a population and unrelated to missing child is type, (2) a man randomly selected from a population and unrelated to missing child is type CD, and (3) a child, randomly selected from a population is BC.

188 Reverse Paternity Analysis Missing child scenario CD BC

189 Hypothetical DNA Example First Hypothesis Numerator Person Alleged Mother Child Alleged Father Type BC CD In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type b) Man randomly selected from population is type CD, and c) Their child is type BC

190 Reverse Paternity Analysis Missing child scenario Numerator 2p A p B CD 2p C p D BC Probability = 2p A p B x 2p C p D x 0.5 x 0.5

191 Hypothetical DNA Example Second Hypothesis Denominator Person Alleged Mother Child Alleged Father Type BC CD In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type b) Man randomly selected from population is type CD, and c) A child, randomly selected and unrelated to either woman or man, is type BC

192 Paternity Analysis Paternity Index Denominator 2p A p B CD 2p C p D BC 2p B p C Probability = 2p A p B x 2p C p D x 2p B p C

193 Reverse Paternity Analysis Missing child scenario CD BC

194 Reverse Paternity Analysis Missing child scenario LR = LR = 2p A p B x 2p C p D x 0.5 x 0.5 2p A p B x 2p C p D x 2p B p C p B p C

195 Reverse Paternity Analysis Missing child scenario CC BC

196 Reverse Paternity Analysis Missing child scenario Numerator 2 2p A p p C B C BC Probability = 2p A p B x p C 2 x 0.5 x 1

197 Paternity Analysis Paternity Index Denominator 2p A p B C p C 2 BC 2p B p C Probability = 2p A p B x p C 2 x 2p B p C

198 LR = LR = p B p B x 2p C p D x 0.5 x 1 p B p B x 2p C p D x 2p B p C 0.5 2p B p C

199 Example of Failure of Pairwise Comparison of profiles to detect True Relationship K3 K1 AA AA K4 Q1 K2? AC Q1 = Evidence Sample; K1 ~ K4 = Reference samples, with family relationships known; Q1 is investigated as a Full Sibling in the family Pairwise comparison of Q1 with others cannot exclude P- O, F-S, H-S relationships Five profiles together exclude (barring mutation) Q1 being the missing family member mtdna and Y-STR, along with autosomal STRs, can discriminate H-S vs. completely unrelated scenarios for Q1

200 A Real Case of Missing Person Identification Father (untyped) Mother (untyped) ? (bone) Data Autosomal STRs typed for: , , , and Mitochondrial DNA (mtdna) typed for: and Y-STRs typed for: , , , and Question: Does (bone) belong to the above pedigree?

201 Example (Contd.) Process of Identification comparison of likelihoods of two pedigrees Father (untyped) Mother (untyped) Father (untyped) Mother (untyped) (Bone) V S and (Bone)

202 Statistics for Example Case (Autosomal STRs) Likelihood ratio (LR), θ= 0 with mutation (STRBase) Population IQQ CCP SCL TMC PUQ LR 7.47E E E E E+09 Prior odds = 1/6 Probability of Positive Identification (PIP) = (LR*prior odds)/(lr*prior odds + 1) Population IQQ CCP SCL TMC PUQ LR (Bone)

203 Example Contd. (Mitochondrial DNA Match) mtdna Haplotype frequency and LR per population Population Sample Size Counts Exact frequency Mismatch = 0 CI(0.95) UpperBound LR Counts General frequency Mismatch<=2 CI(0.95) UpperBound LR Total E E E E+02 African E E E E E+02 Asian E E E+01 Caucasian E E E+02 Hispanic E E E+02 Native Ame E

204 Example Contd. (Y Chromosome STRs Match) Y STR Haplotype frequency and LR per population Population Counts Sample Size θ Conditional frequency CI(0.95) UpperBound LR Total E E E E+03 African Ame E E E E+02 Asian E E E E+02 Caucasian E E E E+02 Hispanic E E E E+02 Indian E E E E+02 Native Ame N/A N/A 3.42 E E+01

205 Thank you!