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) 735-2421; Fax (817) 735-2424 e-mail: [email protected]
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 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
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)
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
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
Three Types of Conclusions
Exclusion (Match), or Inclusion Inconclusive
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
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 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)
Exclusion
Nope Nope Inclusion PATERNITY TESTING MOTHER ALLEGED FATHER
CHILD
Two alleles for each autosomal genetic marker Language of Paternity Testing
Maternal Contribution Obligate Paternal Allele Dual Obligate Paternal Alleles Three Genetic Profiles are Determined
Mother A B
Child Bm Cp
Alleged father C D 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
Exclusion
Nope Nope Results
The Tested Man is Excluded as the Biological Father of the Child in Question Inclusion 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 Language of Paternity Testing
PI Paternity Index CPI Combined Paternity Index W Probability of Paternity PE Probability of Exclusion RMNE Random Man Not Excluded
Paternity Index summarizes information provided by genetic testing • Likelihood Ratio
• 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
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)
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).
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”. Paternity Biological Relationship
Parents M F
Child C Paternity Analysis
? Paternity Analysis Hypothetical case DNA Analysis Results in Three Genotypes
Mother (AB) Child (BC) Alleged Father (CD) Paternity Analysis
AB CD
BC
An AB mother and a CD father can have four possible offspring: AC, AD, BC, BD 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 AB, and (2) a man randomly selected from a population is type CD, and (3) their child is type BC. Paternity Analysis
AB CD
BC 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 AB, (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. Paternity Analysis
AB CD Tested Man
BC
CD Untested Random Man 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 Hypothetical DNA Example First Hypothesis Numerator Person Type Mother AB Child BC Alleged Father CD
In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type AB b) Man randomly selected from population is type CD, and c) Their child is type BC Paternity Analysis Paternity Index Numerator
CD 2pApB AB 2pCpD
0.5 0.5 BC
Probability = 2pApB x 2pCpD x 0.5 x 0.5 Hypothetical DNA Example Second Hypothesis Denominator Person Type Mother AB Child BC Alleged Father CD In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type AB b) An alternative man randomly selected from population is type CD, and c) The woman’s child, fathered by random man, is type BC Paternity Analysis Paternity Index Denominator
2p p 2pApB AB CD C D
pC 0.5 BC
Probability = 2pApB x 2pCpD x 0.5 x pC Paternity Analysis Paternity Index
2pApB x 2pCpD x 0.5 x 0.5 PI = 2pApB x 2pCpD x 0.5 x pC
0.5 PI = pC Hypothetical DNA Example Probability Statements
Person Type Mother AB Child BC Alleged Father 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 Hypothetical DNA Example Probability Statement
Person Type Mother AB Child BC Alleged Father 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. Paternity M and C share one allele and AF is heterozygous for the other allele
Parents AB ? CD M 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 There are 15 possible combinations of genotypes for a paternity trio Paternity Biological Relationship
Parents M F
Child C Paternity Index M and C share one allele and AF is homozygous for the obligatory allele
Parents AB ? 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 Paternity Analysis Paternity Index Numerator
2 C pC 2pApB AB
0.5 1 BC
2 Probability = 2pApB x pC x 0.5 x 1 Paternity Analysis Paternity Index Denominator
2 2pApB AB C pC
pC 0.5 BC
2 Probability = 2pApB x pC x 0.5 x pC Paternity Analysis Paternity Index
2 2pApB x pC x 0.5 x 1 PI = 2 2pApB x pC x 0.5 x pC
1 PI = pC Paternity Index M and C share both alleles and AF is heterozygous with one of the obligatory alleles
Parents AB ? BC M AF
Child AB 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) Paternity Analysis Paternity Index Numerator
2p p 2pApB AB BC B C
0.5A 0.5B AB
Probability = 2pApB x 2pBpC x 0.5(mA) x 0.5(fB) Paternity Analysis Paternity Index Denominator
2pApB AB BC 2pBpC
pA + pB 0.5 AB
probability =
2pApB x 2pBpC x (0.5(mA) x pB + 0.5(mB) x pA) Paternity Analysis Paternity Index
2pApB x 2pBpC x 0.5(mA) x 0.5(fB) PI = 2pApB x 2pBpC x (0.5(mB) x pA + 0.5(mA) x pB)
0.25 PI = 0.5pA + 0.5pB 0.5 PI = pA + pB Paternity Index M and C share both alleles and AF is heterozygous with both of the obligatory alleles
Parents AB ? AB M AF
Child AB 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) Paternity Analysis Paternity Index Numerator
2pApB AB AB 2pApB
0.5A + 0.5B 0.5A + 0.5B AB
Probability =
2pApB x 2pBpC x (0.5(mA) x 0.5(fB) + 0.5(mB) x 0.5(fA)) Paternity Analysis Paternity Index Denominator
2pApB AB AB 2pApB
pA + pB 0.5A + 0.5B AB
probability =
2pApB x 2pApB x (0.5(mA) x pB + 0.5(mB) x pA) Paternity Analysis Paternity Index
2pApB x 2pBpC x (0.5(mA) x 0.5(fB) + 0.5(mB) x 0.5(fA)) PI = 2pApB x 2pBpC x (0.5(mB) x pA + 0.5(mA) x pB)
0.5 PI = 0.5pA + 0.5pB 1 PI = pA + pB Paternity Biological Relationship
Parents M F
Child C Paternity Index M and C share both alleles and AF is heterozygous with one of the obligatory alleles
Parents AB ? B M AF
Child AB 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) Paternity Analysis Paternity Index Numerator
p 2 2pApB AB B B
0.5 1 AB
2 Probability = 2pApB x pB x 0.5(mA) x 1(fB) Paternity Analysis Paternity Index Denominator
2 2pApB AB B pB
pA + pB 0.5 AB
probability =
2 2pApB x pB x (0.5(mA) x pB + 1(mB) x pA) Paternity Analysis Paternity Index
2 2pApB x pB x 0.5(mA) x 1(fB) PI = 2 2pApB x pB x (0.5(mB) x pA + 0.5(mA) x pB)
0.5 PI = 0.5pA + 0.5pB 1 PI = pA + pB PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF Numerator Denominator PI A A AB 0.5 a 0.5/a A AB AB 0.5 b 0.5/b A AB BC 0.5 b 0.5/b AB A AB 0.25 0.5a 0.5/a AB A AC 0.25 0.5a 0.5/a BC AB AB 0.25 0.5a 0.5/a BC AB AC 0.25 0.5a 0.5/a BD AB AC 0.25 0.5a 0.5/a PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF Numerator Denominator PI A A A 1 a 1/a AB A A 0.5 0.5a 1/a B AB A 1 a 1/a BC AB A 0.5 0.5a 1/a PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF Numerator Denominator PI AB AB AC 0.25 0.5(a+b) 0.5/(a+b) PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF Numerator Denominator PI AB AB A 0.5 0.5(a+b) 1/(a+b) AB AB AB 0.5 0.5(a+b) 1/(a+b)
Effect of Population Substructure SAMPLING THEORY OF ALLELE FREQUENCIES
Under the mutation-drift balance, the probability of a sample in which ni copies of the allele Ai is observed, for any set of i = 1, 2, ... is given by
ni () () iin Pr(Ai ) ii()() n i
Where n• = Σni , i = pi(1 - )/, •.= i = (1 - )/, pi = frequency of allele Ai in the population and () is the Gamma function, in which is the coefficient of coancestry (equivalent to Fst or Gst, the coefficient of gene differentiation between subpopulations within the population) GENOTYPE FREQUENCY
Using the general theory,
Pr(A A ) p2 p (1 p ) i i i i i Pr(Ai A j ) 2 p i p j (1 ) CONDITIONAL MATCH PROBABILITY
[2 (1 )pp ][3 (1 ) ] Pr(AAAA | ) ii i i i i (1 )(1 2 )
2[ (1 )ppij ][ (1 ) ] Pr(AAAAi j | i j ) (1 )(1 2 )
Thus, for = 0 ,
2 Pr(AAAAAAPi i ) Pr( i i | i i ) i
Pr(AAAAAAPPi j ) Pr( i j | i j ) 2 i j PI Formulas population substructure considered
M C AF PI A A AB (1+3) / 2[3 +a(1- )] A AB AB (1+3) / 2[ +b(1- )] A AB BC (1+3) / 2[ +b(1- )] AB A AB (1+3) / 2[2 +a(1- )] AB A AC (1+3) / 2[2 +a(1- )] BC AB AB (1+3) / 2[ +a(1- )] BC AB AC (1+3) / 2[ +a(1- )] BD AB AC (1+3) / 2[ +a(1- )] PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF PI A A A (1+3) / [4 +a(1- )] AB A A (1+3) / [3 +a(1- )] B AB A (1+3) / [2 +a(1- )] BC AB A (1+3) / [2 +a(1- )] PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF PI AB AB AC (1+3) / 2[3 +(1- )(a+b)]
PI Formulas Single locus, no null alleles, low mutation rate, codominance
M C AF PI AB AB A (1+3) / [4 +(1- ) (a+b)] AB AB AB (1+3) / [4 +(1- ) (a+b)]
Combined Paternity Index 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
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...
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. Probability of Paternity 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. 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. 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. Probability of Paternity CPI x P Probability of Paternity (W) = [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. 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. Probability of Paternity With a prior probability of 0.5, the Probability of Paternity (W) = CPI x 0.5 [CPI x 0.5 + (1 – 0.5)] CPI x 0.5 = CPI x 0.5 + 0.5 CPI = CPI + 1 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 Probability of Exclusion 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 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.
Probability of Exclusion
PE = 1 - (a2 + 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. Probability of Exclusion PE = 1 - (a2 + 2ab) b = sum of the frequency of all alleles other than the obligate paternal allele. b = (1 – a) PE = 1 – [a2 + 2a(1 – a)] PE = 1 – [a2 + 2a – 2a2] PE = 1 – [2a –a2] PE = 1 – 2a + a2 PE = (1 – a)2 Probability of Exclusion
If the Mother and Child are both phenotype AB, 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 Probability of Exclusion
In addition, if population substructure is considered
(1-a–b)(1-)[1-(1-)(a+b)] PE = (1-2) (1-3)
PE formulas
M C OA PE A A A (1-)(1-a)[1-a(1-)] / (1+2)(1+3 )
A AB A [2+(1-)(1-b)] [3+(1-)(1-b)] / (1+2)(1+3 ) PE formulas
M C OA PE AB AA A [1-a(1-)][1+ -a(1-)] / (1+2)(1+3 )
AB BB B [1-b(1-)][1+ -b(1-)] / (1+2)(1+3 )
AB AB A/B (1-a-b)(1-)[1-(a+b)(1-)] / (1+2)(1+3 )
PE formulas
M C OA PE AB AC C [(1-)(1-c)+2][(1-)(1-c)+3] / (1+2)(1+3 )
AB BC C [(1-)(1-c)+2][(1-)(1-c)+3] / (1+2)(1+3 )
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 – PE1) x (1 – PE2) x (1 – PE3)… x (1 – PEN)] Random Man Not Excluded PE = 1 - (a2 + 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. 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:
RMNE1 x RMNE2 x RMNE3 x …. RMNEN PowerPlexTM 1.1
CSF1PO D16S539
D7820 TPOX
D13S317 TH01
vWA D5S818
M C AF M C AF M C AF M C AF P-41411 P-41414 P-41411 P-41414 Typical Paternity Trio P-41411 M C AF Allele Frequency
HUMCSF1PO 12 12 12 8 = 0.00493 (5q33.3 - q34) 8 8 10 12 = 0.32512
HUMTPOX 11 11 11 8 = 0.54433 (2p23 - 2pter) 8 8 8 11 = 0.03695
HUMTH01 7 9p 9 (11p15.5) 7m 6 9 = 0.16502
HUMvWA31 19 19m 16 16 = 0.20153 (12p13.3 - p13.2) 18 16p 15
Typical Paternity Trio P-41411 M C AF PI Formula
HUMCSF1PO 12 12 12 (5q33.3 - q34) 8 8 10 0.5/(a+b)]
HUMTPOX 11 11 11 (2p23 - 2pter) 8 8 8 1/(a+b)
HUMTH01 7 9p 9 0.5/a (11p15.5) 7m 6
HUMvWA31 19 19m 16 0.5/a (12p13.3 - p13.2) 18 16p 15 Typical Paternity Trio P-41411 M C AF Paternity Index
HUMCSF1PO 12 12 12 1.52 (5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11 (2p23 - 2pter) 8 8 8 1.72
HUMTH01 7 9p 9 (11p15.5) 7m 6 3.03
HUMvWA31 19 19m 16 (12p13.3 - p13.2) 18 16p 15 2.48 Typical Paternity Trio P-41411 M C AF PE Formula
HUMCSF1PO 12 12 12 [1 –(a+b)]2 (5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11 [1 –(a+b)]2 (2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9 (1 – a)2 (11p15.5) 7m 6
HUMvWA31 19 19m 16 (1 – a)2 (12p13.3 - p13.2) 18 16p 15 Typical Paternity Trio P-41411 M C AF PE
HUMCSF1PO 12 12 12 0.4488 (5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11 0.1753 (2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9 0.6972 (11p15.5) 7 7m 6
HUMvWA31 19 19m 16 0.6376 (12p13.3 - p13.2) 18 16p 15 Typical Paternity Trio P-41411 M C AF Allele Frequency
D16S539 12 12m 12 (16p24 - p25) 11p 11 11 = 0.27228
D7S820 10 11p 11 11 = 0.20197 (7q) 9 9m 10
D13S317 12 12m 11 8 = 0.09949 (13q22 - q31) 10 8p 8
D5S818 13 11 11 11 = 0.41026 (5q21 - q31) 11 Typical Paternity Trio P-41411 M C AF PI Formula
D16S539 12 12m 12 (16p24 - p25) 11p 11 0.5/a
D7S820 10 11p 11 0.5/a (7q) 9 9m 10
D13S317 12 12m 11 0.5/a (13q22 - q31) 10 8p 8
D5S818 13 11 11 1/a (5q21 - q31) 11 Typical Paternity Trio P-41411 M C AF Paternity Index
D16S539 12 12m 12 1.84 (16p24 - p25) 11p 11
D7S820 10 11p 11 (7q) 9 9m 10 2.48
D13S317 12 12m 11 (13q22 - q31) 10 8p 8 5.03
D5S818 13 11 11 (5q21 - q31) 11 11 2.44 Typical Paternity Trio P-41411 M C AF PE Formula
D16S539 12 12m 12 (1 – a)2 (16p24 - p25) 11p 11
D7S820 10 11p 11 (1 – a)2 (7q) 9 9m 10
D13S317 12 12m 11 (1 – a)2 (13q22 - q31) 10 8p 8
D5S818 13 11 11 (1 – a)2 (5q21 - q31) 11 Typical Paternity Trio P-41411 M C AF PE
D16S539 12 12m 12 0.5296 (16p24 - p25) 11p 11
D7S820 10 11p 11 0.6369 (7q) 9 9m 10
D13S317 12 12m 11 0.8109 (13q22 - q31) 10 8p 8
D5S818 13 11 11 0.3478 (5q21 - q31) 11 PowerPlex 2.1
PENTA E
FGA D18S51
TPOX
D21S11 D8S1179
TH01
vWA
D3S1358
M C AF M C AF M C AF M C AF P-41411 P-41414 P-41411 P-41414 Typical Paternity Trio P-41411 M C AF Allele Frequency
FGA 24 24m 23 (4q28) 23 21p 21 21 = 0.17347
D18S51 17 17 14 14 = 0.17347 (18q21.3) 14 14 17 = 0.15561
D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 29 = 0.18112
D3S1358 15 14 15 (3p) 14 14 14 = 0.14039
D8S1179 14 15p 15 (8) 10 14m 13 15 = 0.10969 Typical Paternity Trio P-41411 M C AF PI Formula
FGA 24 24m 23 (4q28) 23 21p 21 0.5/a
D18S51 17 17 14 (18q21.3) 14 14 1/(a+b)
D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 0.5/a
D3S1358 15 14 15 (3p) 14 14 0.5/a
D8S1179 14 15p 15 (8) 10 14m 13 0.5/a Typical Paternity Trio P-41411 M C AF Paternity Index
FGA 24 24m 23 (4q28) 23 21p 21 2.88
D18S51 17 17 14 (18q21.3) 14 14 3.04
D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 2.76
D3S1358 15 14 15 (3p) 14 14 3.56
D8S1179 14 15p 15 (8) 10 14m 13 4.56 Typical Paternity Trio P-41411 M C AF PE Formula
FGA 24 24m 23 2 (4q28) 23 21p 21 (1 – a)
D18S51 17 17 14 2 (18q21.3) 14 14 [1 –(a+b)]
D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 (1 – a)2
D3S1358 15 14 15 2 (3p) 14 14 (1 – a)
D8S1179 14 15p 15 2 (8) 10 14m 13 (1 – a) Typical Paternity Trio P-41411 M C AF PE
FGA 24 24m 23 (4q28) 23 21p 21 0.6831
D18S51 17 17 14 (18q21.3) 14 14 0.4501
D21S11 30 30m 29 0.6706 (21q11.2 - q21) 28 29p 28
D3S1358 15 14 15 (3p) 14 14 0.7389
D8S1179 14 15p 15 (8) 10 14m 13 0.7926 Typical Paternity Trio 13 Core CODIS Loci
Combined Paternity Index 431,602
Probability of Paternity 99.9997%
Probability of Exclusion 99.9997%
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: 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)
Paternity Index Only Man and Child Tested Hypothetical case DNA Analysis Results in Two Genotypes
Mother Not Tested Child (AB) Alleged Father (AC) 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 AB. X = Pr{AF passes A} x Pr {M passes B} + Pr{AF passes B} x Pr{M passes A} 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 AB. Y = Pr{RM passes A} x Pr {M passes B} + Pr{RM passes B} x Pr{M passes A} 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 . Motherless Paternity Index So, if we have a heterozygous child AB, 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
Motherless Paternity Index
So, if we have a heterozygous child AB, 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 Motherless Paternity Index
So, if we have a heterozygous child AB, 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 Paternity Index Only Man and Child Tested
Parents ? AC M AF
Child AB 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 Paternity Index Only Man and Child Tested
AC
AB Paternity Index Only Man and Child Tested Numerator
AC 2pApC
pB 0.5A AB 2pApB
Probability = 2pApC x 2pApB x 0.5(fA) x pB Paternity Index Only Man and Child Tested Denominator
AC 2pApC
pA + pB pA + pB
2pApB AB
probability =
2pApC x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA)) Paternity Index Only Man and Child Tested
2pApB x 2pApC x 0.5(mA) x pB PI = 2pApB x 2pApC x (p(mA) x p(fB) + p(mB) x p(fA))
0.5pB PI = 2pApB
0.25 PI = pA Paternity Index Only Man and Child Tested
Parents ? A M AF
Child AB 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 Paternity Index Only Man and Child Tested
A
AB Paternity Index Only Man and Child Tested Numerator
2 A pA
pB 1
2pApB AB
2 Probability = pA x 2pApB x 1(fA) x pB Paternity Index Only Man and Child Tested Denominator
2 A pA
pA + pB pA + pB
2pApB AB
probability =
2 pA x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA)) Paternity Index Only Man and Child Tested
2 pA x 2pApC x 1(mA) x pB PI = 2 pA x 2pApC x (p(mA) x p(fB) + p(mB) x p(fA))
pB PI = 2pApB
0.5 PI = pA Paternity Index Only Man and Child Tested
Parents ? AB M AF
Child AB 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 Paternity Index Only Man and Child Tested
AB
AB Paternity Index Only Man and Child Tested Numerator
AB 2pApB
pA + pB 0.5A + 0.5B
2pApB AB
Probability =
2pApB x 2pApB x (0.5(fA) x pB + 0.5(fB) x pA) Paternity Index Only Man and Child Tested Denominator
AB 2pApB
pA + pB pA + pB
2pApB AB probability =
2pApB x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA)) Paternity Index Only Man and Child Tested
2pApB x 2pApB x (0.5(fA) x pB + 0.5(fB) x pA) PI = 2pApB x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))
0.5p + 0.5p PI = B A 2pApB p + p PI = A B 4pApB 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 Paternity Index Only Man and Child Tested
A
A Paternity Index Only Man and Child Tested Numerator
2 A pA
pA 1 2 pA A
2 2 Probability = pA x pA x 1(fA) x pA Paternity Index Only Man and Child Tested Denominator
2 A pA
pA pA
2 pA A
2 2 probability = pA x pA x p(mA) x p(fA) Paternity Index Only Man and Child Tested
2 2 pA x pA x 1(fA) x pA PI = 2 2 pA x pA x p(mA) x p(fA)
p PI = A pA x pA 1 PI = pA Paternity Index Only Man and Child Tested
Parents ? AB 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 Paternity Index Only Man and Child Tested
AB
A Paternity Index Only Man and Child Tested Numerator
AB 2pApB
pA 0.5 2 pA A
2 Probability = 2pApB x pA x 0.5(fA) x pA Paternity Index Only Man and Child Tested Denominator
AB 2pApB
pA pA 2 pA A
2 probability = 2pApB x pA x p(mA) x p(fA) Paternity Index Only Man and Child Tested
2 2pApB x pA x 0.5(fA) x pA PI = 2 2pApB x pA x p(mA) x p(fA)
0.5p PI = A pA x pA 0.5 PI = pA Paternity Index Only Man and Child Tested Formulas
Single locus, no null alleles, low mutation rate, codominance
C AF Numerator Denominator PI PE AB AC 0.5b 2ab 0.25/a [1-(a + b)]2 AB AB 0.5(a+b) 2ab (a+b)/4ab [1-(a + b)]2 AB A b 2ab 0.5/a [1-(a + b)]2 A AC 0.5a a2 0.5/a (1-a) 2 A A a a2 1/a (1-a) 2
Paternity Index Only Man and Child Tested Formulas: with Population Substructure C AF PI PE AB AC (1+2) (1-)(1-a-b)[1-(1-)(a+b)] / (1+)(1+2) / 4[(1-)a+]
AB AB (1+2)[(a+b)(1-)+2] / 4[(1-)a+] [(1-)b+]
AB A (1+2) / 2[(1-)a+2] Paternity Index Only Man and Child Tested Formulas: with Population Substructure C AF PI PE A AC (1+2) (1-)(1-a)[(1-)(1-a)+] / 2[(1-)a+2] / (1+)(1+2)
A A (1+2) / [(1-)a+3]
41376 PowerPlex 2.1
PENTA E FGA D18S51 TPOX
D8S1179 D21S11
TH01
vWA
D3S1358
C AF C AF MOTHERLESS PATERNITY CASE P-41376 C AF Allele Frequencies HUMCSF1PO 10 11 10 = 0.25269 (5q33.3 - q34) 11 12 11 = 0.30049
HUMTPOX 8 8 8 = 0.54433 (2p23 - 2pter) 11 11 11 = 0.25369
HUMTH01 6 6 6 = 0.22660 (11p15.5) 9.3 7 9.3 = 0.30542
HUMvWA31 15 16 15 = 0.11224 (12p13.3 - p13.2) 16 16 = 0.20153 41376 TPOX
13 C AF 11 11
8 8 6
C AF
PI = X / Y
X = 0.5b + 0.5a PI = (a + b) / 4ab Y = 2ab
PI = (0.5b + 0.5a) / 2ab 41376 THO1 11 10 9.3 C AF 9.3 7
6 6 5 C AF PI = X / Y
X = 0.5b PI = 0.25 / a Y = 2ab
PI = 0.5b / 2ab 41376 vWA 21 C AF 16 16 15 13
11 C AF
PI = X / Y
X = b PI = 0.5 / a Y = 2ab
PI = b / 2ab MOTHERLESS PATERNITY CASE P-41376 C AF PI Formula
HUMCSF1PO 10 11 0.25/a (5q33.3 - q34) 11 12
HUMTPOX 8 8 (a+b)/4ab (2p23 - 2pter) 11 11
HUMTH01 6 6 0.25/a (11p15.5) 9.3 7
HUMvWA31 15 16 0.5/a (12p13.3 - p13.2) 16 MOTHERLESS PATERNITY CASE P-41376 C AF PI
HUMCSF1PO 10 11 0.83 (5q33.3 - q34) 11 12
HUMTPOX 8 8 1.44 (2p23 - 2pter) 11 11
HUMTH01 6 6 1.10 (11p15.5) 9.3 7
HUMvWA31 15 16 2.48 (12p13.3 - p13.2) 16 MOTHERLESS PATERNITY CASE P-41376 C AF PE Formulas
HUMCSF1PO 10 11 [1-(a+b)]2 (5q33.3 - q34) 11 12
HUMTPOX 8 8 [1-(a+b)]2 (2p23 - 2pter) 11 11
HUMTH01 6 6 [1-(a+b)]2 (11p15.5) 9.3 7
HUMvWA31 15 16 [1-(a+b)]2 (12p13.3 - p13.2) 16 MOTHERLESS PATERNITY CASE P-41376 C AF PE
HUMCSF1PO 10 11 0.1988 (5q33.3 - q34) 11 12
HUMTPOX 8 8 0.0408 (2p23 - 2pter) 11 11
HUMTH01 6 6 0.2190 (11p15.5) 9.3 7
HUMvWA31 15 16 0.4709 (12p13.3 - p13.2) 16 MOTHERLESS PATERNITY CASE P-41376 C AF Allele Frequencies D16S539 12 11 12 = 0.33911 (16p24 - p25) 13 12 13 = 0.16337
D7S820 11 11 11 = 0.20197 (7q) 12 14 12 = 0.14030
D13S317 11 11 11 = 0.31888 (13q22 - q31)
D5S818 11 11 11 = 0.41026 (5q21 - q31) 13 12 13 = 0.14615 MOTHERLESS PATERNITY CASE P-41376 C AF PI Formulas
D16S539 12 11 0.25/a (16p24 - p25) 13 12
D7S820 11 11 0.25/a (7q) 12 14
D13S317 11 11 1/a (13q22 - q31)
D5S818 11 11 0.25/a (5q21 - q31) 13 12 MOTHERLESS PATERNITY CASE P-41376 C AF PI
D16S539 12 11 0.74 (16p24 - p25) 13 12
D7S820 11 11 1.24 (7q) 12 14
D13S317 11 11 3.14 (13q22 - q31)
D5S818 11 11 0.61 (5q21 - q31) 13 12 MOTHERLESS PATERNITY CASE P-41376 C AF PE Formulas
D16S539 12 11 [1-(a+b)]2 (16p24 - p25) 13 12
D7S820 11 11 [1-(a+b)]2 (7q) 12 14
D13S317 11 11 (1-a)2 (13q22 - q31)
D5S818 11 11 [1-(a+b)]2 (5q21 - q31) 13 12 MOTHERLESS PATERNITY CASE P-41376 C AF PE
D16S539 12 11 0.2475 (16p24 - p25) 13 12
D7S820 11 11 0.4325 (7q) 12 14
D13S317 11 11 0.4639 (13q22 - q31)
D5S818 11 11 0.1968 (5q21 - q31) 13 12 MOTHERLESS PATERNITY CASE P-41376 C AF Allele Frequencies FGA 19 19 19 = 0.05612 (4q28) 21 25 21 = 0.17347
D18S51 16 16 16 = 0.10714 (18q21.3) 20
D21S11 29 28 29 = 0.18112 (21q11.2 - q21) 29
D3S1358 15 15 15 = 0.24631 (3p) 18 17 18 = 0.16256
D8S1179 11 11 11 = 0.05867 (8) 13 13 13 = 0.33929
41376 FGA 46.2 44.2 C AF 30 21 25
19 19 17 C AF
PI = X / Y
X = 0.5b PI = 0.25 / a Y = 2ab
PI = 0.5b / 2ab 41376 D18S51 25 C AF 16 20 15 16 12
9 C AF
PI = X / Y
X = 0.5a PI = 0.5 / a Y = a2
PI = 0.5a / a2 41376 D21S11 38 C AF 29 29
28 28 26 24.2 C AF
PI = X / Y
X = 0.5a PI = 0.5 / a Y = a2
PI = 0.5a / a2 41376 D3S1358 20 C AF 18 17
15 15 12 C AF PI = X / Y
X = 0.5b PI = 0.25 / a Y = 2ab
PI = 0.5b / 2ab 41376 D8S1179 18 C AF 13 13 11 11
8 C AF
PI = X / Y
X = 0.5b + 0.5a PI = (a + b) / 4ab Y = 2ab
PI = (0.5b + 0.5a) / 2ab MOTHERLESS PATERNITY CASE P-41376 C AF PI Formulas
FGA 19 19 0.25/a (4q28) 21 25
D18S51 16 16 0.5/a (18q21.3) 20
D21S11 29 28 0.5/a (21q11.2 - q21) 29
D3S1358 15 15 0.25/a (3p) 18 17
D8S1179 11 11 (a+b)/4ab (8) 13 13 MOTHERLESS PATERNITY CASE P-41376 C AF PI
FGA 19 19 4.45 (4q28) 21 25
D18S51 16 16 4.67 (18q21.3) 20
D21S11 29 28 2.76 (21q11.2 - q21) 29
D3S1358 15 15 1.02 (3p) 18 17
D8S1179 11 11 5.00 (8) 13 13 MOTHERLESS PATERNITY CASE P-41376 C AF PE Formulas
FGA 19 19 [1-(a+b)]2 (4q28) 21 25
D18S51 16 16 (1-a)2 (18q21.3) 20
D21S11 29 28 (1-a)2 (21q11.2 - q21) 29
D3S1358 15 15 [1-(a+b)]2 (3p) 18 17
D8S1179 11 11 [1-(a+b)]2 (8) 13 13 MOTHERLESS PATERNITY CASE P-41376 C AF PE
FGA 19 19 0.5935 (4q28) 21 25
D18S51 16 16 0.7972 (18q21.3) 20
D21S11 29 28 0.6706 (21q11.2 - q21) 29
D3S1358 15 15 0.3944 (3p) 18 17
D8S1179 11 11 0.3625 (8) 13 13 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% Motherless Paternity 13 Core CODIS Loci (=0.01)
Combined Paternity Index 867
Probability of Paternity 99.88%
Probability of Exclusion 99.90% Motherless Paternity 13 Core CODIS Loci (=0.03)
Combined Paternity Index 309
Probability of Paternity 99.68%
Probability of Exclusion 99.80% Reverse Paternity Testing
Can we identify crime scene evidence to a deceased/missing individual?
Applications:
no-body homicides victims of mass disasters REVERSE PATERNITY INDEX BODY IDENTIFICATION ALLEGED EVIDENCE ALLEGED MOTHER FATHER
A B B C C D Three genotypes:
• Alleged Mother • Child (missing) • Alleged Father Missing child
? Reverse Paternity Index RPI = X / Y Numerator
X = is the probability that (1) a woman randomly selected from a population is type AB, and (2) a man randomly selected from a population is type CD, and (3) their child is type BC. Reverse Paternity Analysis
AB CD
BC 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 AB, (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. Reverse Paternity Analysis Missing child scenario
AB CD
BC Hypothetical DNA Example First Hypothesis Numerator Person Type Alleged Mother AB Child BC Alleged Father CD
In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type AB b) Man randomly selected from population is type CD, and c) Their child is type BC Reverse Paternity Analysis Missing child scenario Numerator
CD 2pApB AB 2pCpD
0.5 0.5 BC
Probability = 2pApB x 2pCpD x 0.5 x 0.5 Hypothetical DNA Example Second Hypothesis Denominator Person Type Alleged Mother AB Child BC Alleged Father CD In order to explain this evidence Calculate Probability that a) Woman randomly selected from population is type AB 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 Paternity Analysis Paternity Index Denominator
2p p 2pApB AB CD C D
BC 2pBpC
Probability = 2pApB x 2pCpD x 2pBpC
Reverse Paternity Analysis Missing child scenario
AB CD
BC Reverse Paternity Analysis Missing child scenario
2pApB x 2pCpD x 0.5 x 0.5 LR = 2pApB x 2pCpD x 2pBpC
0.25 LR = 2pBpC Reverse Paternity Analysis
Missing child scenario
AB CC
BC Reverse Paternity Analysis Missing child scenario Numerator
2 C pC 2pApB AB
0.5 1 BC
2 Probability = 2pApB x pC x 0.5 x 1 Paternity Analysis Paternity Index Denominator
2 2pApB AB C pC
BC 2pBpC
2 Probability = 2pApB x pC x 2pBpC pBpB x 2pCpD x 0.5 x 1 LR = pBpB x 2pCpD x 2pBpC
0.5 LR = 2pBpC Example of Failure of Pairwise Comparison of profiles to detect True Relationship Q1 = Evidence Sample; K1 AA AB K2 K1 ~ K4 = Reference samples, with family relationships known; ? Q1 is investigated as a Full Sibling in the family K3 AA K4 AB Q1 AC
• 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 A Real Case of Missing Person Identification
Father Mother (untyped) (untyped)
6909.1 6997.1 6909.5 ? (bone)
6909.2 6909.3 6909.4 • Data – Autosomal STRs typed for: 6997.1, 6909.1, 6909.2, and 6909.3 – Mitochondrial DNA (mtDNA) typed for: 6997.1 and 6909.1 – Y-STRs typed for: 6997.1, 6909.1, 6909.2, and 6909.3 • Question: Does 6997.1 (bone) belong to the above pedigree? Example (Contd.)
• Process of Identification – comparison of likelihoods of two pedigrees
Father Mother Father Mother (untyped) (untyped) (untyped) (untyped)
6909.1 6997.1 6909.5 (Bone) V 6909.1 6909.5
6909.2 6909.3 S 6909.2 6909.3
6997.1 and (Bone) Statistics for Example Case (Autosomal STRs)
• Likelihood ratio (LR), θ= 0 with mutation (STRBase)
Population IQQ CCP SCL TMC PUQ LR 7.47E+09 1.62E+09 7.57E+09 2.71E+09 2.88E+09
• Prior odds = 1/6 • Probability of Positive Identification (PIP) = (LR*prior odds)/(LR*prior odds + 1)
Population IQQ CCP SCL TMC PUQ
0.999999999 0.999999996 0.999999999 0.999999997 0.999999997 LR (Bone) 1964352 2901903 2072461 7871493 9199534 Example – Contd. (Mitochondrial DNA Match) • mtDNA Haplotype frequency and LR per population
Mismatch = 0 Mismatch<=2 Sample Population Size Exact CI(0.95) General CI(0.95) Counts LR Counts LR frequency UpperBound frequency UpperBound
Total 5982 0 0 6.17E-04 1.62E+03 28 0.0047 6.76E-03 1.48E+02
African 1653 0 0 2.23E-01 4.49E+02 1 6.05E-04 3.37E-01 2.97E+02
Asian 937 0 0 3.93E-03 2.54E+02 10 0.0011 0.0195 5.12E+01
Caucasian 2116 0 0 1.74E-03 5.74E+02 6 0.0028 0.0062 1.62E+02
Hispanic 924 0 0 3.98E-01 2.51E+02 1 0.0011 0.006 1.66E+02
Native Ame. 352 0 0 0.0104 9.59E+01 10 0.0028 0.0516 19.37 Example – Contd. (Y Chromosome STRs Match) • Y STR Haplotype frequency and LR per population
Sample Conditional CI(0.95) Population Counts θ LR Size frequency UpperBound
Total 0 7812 9.34E-05 9.34E-05 4.73E-04 2.12E+03
African Ame. 0 1439 4.77E-05 4.77E-05 2.56E-03 3.91E+02
Asian 0 3018 3.72E-04 3.72E-04 1.22E-03 8.19E+02
Caucasian 0 1711 1.077E-04 1.077E-04 2.15E-03 4.64E+02
Hispanic 0 730 9.23E-05 9.23E-05 5.04E-03 1.98E+02
Indian 0 808 3.11E-04 3.11E-04 4.56 E-03 2.20E+02
Native Ame. 0 106 N/A N/A 3.42 E-02 2.92E+01 Thank you!