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)
Nope Nope
Exclusion
Inclusion
Two alleles for each
autosomal genetic marker
PATERNITY TESTING
MOTHER
CHILD
ALLEGED FATHER
Language of Paternity Testing
Maternal Contribution
Obligate Paternal Allele
Dual Obligate Paternal Alleles
Three Genetic Profiles are
Determined
Child Bm Cp
Alleged father C D
Mother A B
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
Nope Nope
Exclusion
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
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)
Paternity Index
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
F M
Child
C
Paternity Analysis
?
DNA Analysis Results in Three
Genotypes
Mother (AB)
Child (BC)
Alleged Father (CD)
Paternity Analysis
Hypothetical case
Paternity Analysis
AB
BC
An AB mother and a CD father can
have four possible offspring:
AC, AD, BC, BD
CD
Standard Paternity Index PI determination in hypothetical DNA System
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.
PI = X / Y
Numerator
AB
BC
CD
Paternity Analysis
Standard Paternity Index PI determination in hypothetical DNA System
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.
PI = X / Y
Denominator
AB
BC
CD
CD
Tested Man
Untested Random Man
Paternity Analysis
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
In order to explain this evidence Calculate Probability that
b) Man randomly selected from population is type CD, and
c) Their child is type BC
Hypothetical DNA Example First Hypothesis
Numerator
Person
Mother
Child
Alleged Father
Type
AB
BC
CD
a) Woman randomly selected from population is type AB
Paternity Analysis
Paternity Index Numerator
AB
BC 0.5 0.5
2pApB 2pCpD
Probability = 2pApB x 2pCpD x 0.5 x 0.5
CD
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
Hypothetical DNA Example Second Hypothesis
Denominator
Person
Mother
Child
Alleged Father
Type
AB
BC
CD
In order to explain this evidence Calculate Probability that
AB
BC
2pApB 2pCpD
Probability = 2pApB x 2pCpD x 0.5 x pC
pC
CD
0.5
Paternity Analysis
Paternity Index Denominator
2pApB x 2pCpD x 0.5 x 0.5
2pApB x 2pCpD x 0.5 x pC
PI =
PI = 0.5
pC
Paternity Analysis
Paternity Index
Hypothetical DNA Example Probability Statements
a) Numerator is probability that tested man is the
father, and
b) Denominator is probability that he is not the father
One might say (Incorrectly)
Person
Mother
Child
Alleged Father
Type
AB
BC
CD
Hypothetical DNA Example Probability Statement
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.
A Correct statement is
Person
Mother
Child
Alleged Father
Type
AB
BC
CD
Paternity M and C share one allele
and AF is heterozygous for the other allele
Parents
M
Child
AB
AF
CD
C
BC
?
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 PI =
There are 15 possible
combinations of genotypes for
a paternity trio
Paternity Biological Relationship
Parents
F M
Child
C
Paternity Index M and C share one allele
and AF is homozygous for the obligatory allele
Parents
M
Child
AB
AF
C
C
BC
?
AF can only pass C allele
Random Man has p chance of passing the C allele
PI = 1/p PI =
Paternity Analysis
Paternity Index Numerator
0.5 1
2pApB pC
2
Probability = 2pApB x pC2 x 0.5 x 1
AB
BC
C
2pApB
Probability = 2pApB x pC2 x 0.5 x pC
pC
AB
BC
C
0.5
Paternity Analysis
Paternity Index Denominator
pC2
2pApB x pC2 x 0.5 x 1
2pApB x pC2 x 0.5 x pC
PI =
PI = 1
pC
Paternity Analysis
Paternity Index
Paternity Index M and C share both alleles and
AF is heterozygous with one of the obligatory alleles
Parents
M
AB
Child
C
AB
?
AF
BC
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)
M has a 1 in 2 chance of passing A or B allele
Paternity Analysis
Paternity Index Numerator
0.5A 0.5B
2pApB 2pBpC
Probability = 2pApB x 2pBpC x 0.5(mA) x 0.5(fB)
AB
AB
BC
2pApB
probability =
2pApB x 2pBpC x (0.5(mA) x pB + 0.5(mB) x pA)
pA + pB
AB
AB
BC
0.5
Paternity Analysis
Paternity Index Denominator
2pBpC
2pApB x 2pBpC x 0.5(mA) x 0.5(fB)
2pApB x 2pBpC x (0.5(mB) x pA + 0.5(mA) x pB)
PI = 0.25
0.5pA + 0.5pB
PI =
Paternity Analysis
Paternity Index
PI = 0.5
pA + pB
Paternity Index M and C share both alleles and AF is
heterozygous with both of the obligatory alleles
Parents
M
Child
AB
AF
AB
C
AB
?
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)
M has a 1 in 2 chance of passing A or B allele
Paternity Analysis
Paternity Index Numerator
AB
AB 0.5A + 0.5B 0.5A + 0.5B
2pApB 2pApB
Probability =
2pApB x 2pBpC x (0.5(mA) x 0.5(fB) + 0.5(mB) x 0.5(fA))
AB
AB
AB
2pApB
probability =
2pApB x 2pApB x (0.5(mA) x pB + 0.5(mB) x pA)
pA + pB
AB
Paternity Analysis
Paternity Index Denominator
2pApB
0.5A + 0.5B
2pApB x 2pBpC x (0.5(mA) x 0.5(fB) + 0.5(mB) x 0.5(fA))
2pApB x 2pBpC x (0.5(mB) x pA + 0.5(mA) x pB)
PI = 0.5
0.5pA + 0.5pB
PI = 1
pA + pB
PI =
Paternity Analysis
Paternity Index
Paternity Biological Relationship
Parents
F M
Child
C
Paternity Index M and C share both alleles and
AF is heterozygous with one of the obligatory alleles
Parents
M
AB
Child
C
AB
?
AF
B
AF can only pass the B allele
RM has (p + q) chance of passing the A or B alleles
PI = 1/(p+q)
M has a 1 in 2 chance of passing A or B allele
Paternity Analysis
Paternity Index Numerator
0.5 1
2pApB pB
2
Probability = 2pApB x pB2 x 0.5(mA) x 1(fB)
AB
AB
B
AB
AB
2pApB
probability =
2pApB x pB2 x (0.5(mA) x pB + 1(mB) x pA)
pA + pB
B
0.5
Paternity Analysis
Paternity Index Denominator
pB2
2pApB x pB2 x 0.5(mA) x 1(fB)
2pApB x pB2 x (0.5(mB) x pA + 0.5(mA) x pB)
PI = 0.5
0.5pA + 0.5pB
PI =
PI = 1
pA + pB
Paternity Analysis
Paternity Index
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
A
A
A
AB
AB
BC
BC
BD
C
A
AB
AB
A
A
AB
AB
AB
AF
AB
AB
BC
AB
AC
AB
AC
AC
Numerator
0.5
0.5
0.5
0.25
0.25
0.25
0.25
0.25
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
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
A
AB
B
BC
C
A
A
AB
AB
AF
A
A
A
A
Numerator
1
0.5
1
0.5
Denominator
a
0.5a
a
0.5a
PI
1/a
1/a
1/a
1/a
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
AB
C
AB
AF
AC
Numerator
0.25
Denominator
0.5(a+b)
PI
0.5/(a+b)
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
AB
AB
C
AB
AB
AF
A
AB
Numerator
0.5
0.5
Denominator
0.5(a+b)
0.5(a+b)
PI
1/(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
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)
( )( )Pr( )
( ) ( )
in i ii
i i i
nA
n
GENOTYPE FREQUENCY
Using the general theory,
2Pr( ) (1 )
Pr( ) 2 (1 )
i i i i i
i j i j
A A p p p
A A p p
CONDITIONAL MATCH PROBABILITY
Thus, for = 0 ,
[2 (1 ) ][3 (1 ) ]Pr( | )
(1 )(1 2 )
2[ (1 ) ][ (1 ) ]Pr( | )
(1 )(1 2 )
i ii i i i
i j
i j i j
p pA A A A
p pA A A A
2Pr( ) Pr( | )
Pr( ) Pr( | ) 2
i i i i i i i
i j i j i j i j
A A A A A A P
A A A A A A PP
PI Formulas population substructure considered
M
A
A
A
AB
AB
BC
BC
BD
C
A
AB
AB
A
A
AB
AB
AB
AF
AB
AB
BC
AB
AC
AB
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- )]
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
A
AB
B
BC
C
A
A
AB
AB
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- )]
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
AB
C
AB
AF
AC
PI
(1+3) / 2[3 +(1- )(a+b)]
PI Formulas Single locus, no null alleles, low mutation rate,
codominance
M
AB
AB
C
AB
AB
AF
A
AB
PI
(1+3) / [4 +(1- ) (a+b)]
(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
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.
Combined Paternity Index
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.
• 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
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.
• 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
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
• 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
b = sum of the frequency of all alleles
other than the obligate paternal allele.
Probability of Exclusion
PE = 1 – [a2 + 2a(1 – a)]
b = (1 – a)
PE = 1 - (a2 + 2ab)
PE = 1 – [a2 + 2a – 2a2]
PE = 1 – [2a –a2]
PE = 1 – 2a + a2
PE = (1 – a)2
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:
Probability of Exclusion
PE = [1 - (a + b)] 2
(1-a–b)(1-)[1-(1-)(a+b)]
(1-2) (1-3)
In addition, if population substructure is
considered
Probability of Exclusion
PE =
M
A
A
C
A
AB
OA
A
A
PE
(1-)(1-a)[1-a(1-)]
/ (1+2)(1+3 )
[2+(1-)(1-b)] [3+(1-)(1-b)]
/ (1+2)(1+3 )
PE formulas
M
AB
AB
AB
C
AA
BB
AB
OA
A
B
A/B
PE
[1-a(1-)][1+ -a(1-)]
/ (1+2)(1+3 )
[1-b(1-)][1+ -b(1-)]
/ (1+2)(1+3 )
(1-a-b)(1-)[1-(a+b)(1-)]
/ (1+2)(1+3 )
PE formulas
M
AB
AB
C
AC
BC
OA
C
C
PE
[(1-)(1-c)+2][(1-)(1-c)+3]
/ (1+2)(1+3 )
[(1-)(1-c)+2][(1-)(1-c)+3]
/ (1+2)(1+3 )
PE formulas
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:
Combined Probability of Exclusion
1 – [(1 – PE1) x (1 – PE2) x (1 – PE3)… x (1 – PEN)]
PE = 1 - (a2 + 2ab)
Random Man Not Excluded
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.
RMNE
PE = 1 – RMNE Or
RMNE = 1 - PE
Random Man Not Excluded
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
P-41411 P-41414
CSF1PO
TPOX
TH01
vWA
D16S539
D7820
D13S317
D5S818
P-41411 P-41414 M C AF M C AF M C AF M C AF
Typical Paternity Trio
P-41411
HUMCSF1PO 12 12 12
(5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11
(2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9
(11p15.5) 7m 6
HUMvWA31 19 19m 16
(12p13.3 - p13.2) 18 16p 15
M C AF Allele Frequency
8 = 0.00493 12 = 0.32512
8 = 0.54433 11 = 0.03695
9 = 0.16502
16 = 0.20153
Typical Paternity Trio
P-41411
HUMCSF1PO 12 12 12
(5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11
(2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9
(11p15.5) 7m 6
HUMvWA31 19 19m 16
(12p13.3 - p13.2) 18 16p 15
M C AF PI Formula
0.5/(a+b)]
1/(a+b)
0.5/a
0.5/a
Typical Paternity Trio
P-41411
HUMCSF1PO 12 12 12
(5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11
(2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9
(11p15.5) 7m 6
HUMvWA31 19 19m 16
(12p13.3 - p13.2) 18 16p 15
M C AF
1.52
Paternity Index
1.72
3.03
2.48
Typical Paternity Trio
P-41411
HUMCSF1PO 12 12 12
(5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11
(2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9
(11p15.5) 7m 6
HUMvWA31 19 19m 16
(12p13.3 - p13.2) 18 16p 15
M C AF
[1 –(a+b)]2
PE Formula
(1 – a)2
[1 –(a+b)]2
(1 – a)2
Typical Paternity Trio
P-41411
HUMCSF1PO 12 12 12
(5q33.3 - q34) 8 8 10
HUMTPOX 11 11 11
(2p23 - 2pter) 8 8 8
HUMTH01 7 9p 9
(11p15.5) 7 7m 6
HUMvWA31 19 19m 16
(12p13.3 - p13.2) 18 16p 15
M C AF
0.4488
PE
0.6972
0.1753
0.6376
Typical Paternity Trio
P-41411
D16S539 12 12m 12
(16p24 - p25) 11p 11
D7S820 10 11p 11
(7q) 9 9m 10
D13S317 12 12m 11
(13q22 - q31) 10 8p 8
D5S818 13 11 11
(5q21 - q31) 11
M C AF Allele Frequency
11 = 0.27228
11 = 0.20197
8 = 0.09949
11 = 0.41026
Typical Paternity Trio
P-41411
D16S539 12 12m 12
(16p24 - p25) 11p 11
D7S820 10 11p 11
(7q) 9 9m 10
D13S317 12 12m 11
(13q22 - q31) 10 8p 8
D5S818 13 11 11
(5q21 - q31) 11
M C AF PI Formula
0.5/a
1/a
0.5/a
0.5/a
Typical Paternity Trio
P-41411
D16S539 12 12m 12
(16p24 - p25) 11p 11
D7S820 10 11p 11
(7q) 9 9m 10
D13S317 12 12m 11
(13q22 - q31) 10 8p 8
D5S818 13 11 11
(5q21 - q31) 11 11
M C AF
1.84
Paternity Index
2.48
5.03
2.44
Typical Paternity Trio
P-41411
D16S539 12 12m 12
(16p24 - p25) 11p 11
D7S820 10 11p 11
(7q) 9 9m 10
D13S317 12 12m 11
(13q22 - q31) 10 8p 8
D5S818 13 11 11
(5q21 - q31) 11
M C AF
(1 – a)2
(1 – a)2
(1 – a)2
(1 – a)2
PE Formula
Typical Paternity Trio
P-41411
D16S539 12 12m 12
(16p24 - p25) 11p 11
D7S820 10 11p 11
(7q) 9 9m 10
D13S317 12 12m 11
(13q22 - q31) 10 8p 8
D5S818 13 11 11
(5q21 - q31) 11
M C AF
0.5296
PE
0.8109
0.6369
0.3478
FGA
TPOX
D8S1179
vWA
PENTA E
D18S51
D21S11
TH01
D3S1358
M C AF M C AF
P-41411 P-41414
PowerPlex 2.1
M C AF M C AF
P-41411 P-41414
Typical Paternity Trio
P-41411
FGA 24 24m 23 (4q28) 23 21p 21 D18S51 17 17 14 (18q21.3) 14 14 D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 D3S1358 15 14 15 (3p) 14 14 D8S1179 14 15p 15 (8) 10 14m 13
M C AF Allele Frequency
21 = 0.17347
14 = 0.17347 17 = 0.15561
29 = 0.18112
15 = 0.10969
14 = 0.14039
Typical Paternity Trio
P-41411
FGA 24 24m 23 (4q28) 23 21p 21 D18S51 17 17 14 (18q21.3) 14 14 D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 D3S1358 15 14 15 (3p) 14 14 D8S1179 14 15p 15 (8) 10 14m 13
M C AF PI Formula
1/(a+b)
0.5/a
0.5/a
0.5/a
0.5/a
Typical Paternity Trio
P-41411
FGA 24 24m 23 (4q28) 23 21p 21 D18S51 17 17 14 (18q21.3) 14 14 D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 D3S1358 15 14 15 (3p) 14 14 D8S1179 14 15p 15 (8) 10 14m 13
M C AF
2.88
Paternity Index
3.04
3.56
4.56
2.76
Typical Paternity Trio
P-41411
FGA 24 24m 23 (4q28) 23 21p 21 D18S51 17 17 14 (18q21.3) 14 14 D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 D3S1358 15 14 15 (3p) 14 14 D8S1179 14 15p 15 (8) 10 14m 13
M C AF
(1 – a)2
(1 – a)2
(1 – a)2
(1 – a)2
[1 –(a+b)]2
PE Formula
Typical Paternity Trio
P-41411
FGA 24 24m 23 (4q28) 23 21p 21 D18S51 17 17 14 (18q21.3) 14 14 D21S11 30 30m 29 (21q11.2 - q21) 28 29p 28 D3S1358 15 14 15 (3p) 14 14 D8S1179 14 15p 15 (8) 10 14m 13
M C AF
0.6831
PE
0.7389
0.4501
0.7926
0.6706
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:
• 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
DNA Analysis Results in Two Genotypes
Mother Not Tested
Child (AB)
Alleged Father (AC)
Paternity Index Only Man and Child Tested
Hypothetical case
Motherless Paternity Index PI determination in hypothetical DNA System
X = is the probability that (1) a man
randomly selected from a population
is type AC, and (2) his child is type
AB.
PI = X / Y
Numerator
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
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.
PI = X / Y
Denominator
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 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
X = Pr{AF passes A} x Pr {M passes B} +
Pr{AF passes B} x Pr{M passes A}
Motherless Paternity Index
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
So, if we have a heterozygous child AB, and a
heterozygous Alleged Father AC then
Motherless Paternity Index
Y = 2pq
So, if we have a heterozygous child AB, and a
heterozygous Alleged Father AC then
PI = X / Y
X = 0.5q
PI = 0.5q / 2pq
PI = 0.25/p
PI = 1/4p
Parents
M
Child
C
AB
?
AF
AC
AF has a 1 in 2 chance of passing A allele
RM has (p + q) chance of passing the A or B allele
The untested Mother could have passed either
the A or B allele
Paternity Index Only Man and Child Tested
AB
AC
Paternity Index Only Man and Child Tested
Numerator
pB 0.5A
2pApC
Probability = 2pApC x 2pApB x 0.5(fA) x pB
AB
AC
2pApB
Paternity Index Only Man and Child Tested
probability =
2pApC x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))
pA + pB
AB
AC
pA + pB
Denominator
2pApC
2pApB
Paternity Index Only Man and Child Tested
PI = 0.5pB
2pApB
2pApB x 2pApC x 0.5(mA) x pB
2pApB x 2pApC x (p(mA) x p(fB) + p(mB) x p(fA))
PI =
PI = 0.25
pA
Paternity Index Only Man and Child Tested
Parents
M
Child
C
AB
?
AF
A
AF can only pass A allele
RM has (p + q) chance of passing the A or B allele
The untested Mother could have passed either
the A or B allele
Paternity Index Only Man and Child Tested
AB
A
Paternity Index Only Man and Child Tested
Numerator
pB 1
pA2
Probability = pA2 x 2pApB x 1(fA) x pB
AB
A
2pApB
Paternity Index Only Man and Child Tested
probability =
pA2 x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))
pA + pB
AB
A
pA + pB
Denominator
pA2
2pApB
Paternity Index Only Man and Child Tested
PI = pB
2pApB
pA2 x 2pApC x 1(mA) x pB
pA2 x 2pApC x (p(mA) x p(fB) + p(mB) x p(fA))
PI =
PI = 0.5 pA
Paternity Index Only Man and Child Tested
Parents
M
Child
C
AB
?
AF
AB
AF can pass either A or B allele
RM has (p + q) chance of passing the A or B allele
The untested Mother could have passed either
the A or B allele
Paternity Index Only Man and Child Tested
AB
AB
Paternity Index Only Man and Child Tested
Numerator
pA + pB 0.5A + 0.5B
2pApB
Probability =
2pApB x 2pApB x (0.5(fA) x pB + 0.5(fB) x pA)
AB
AB
2pApB
Paternity Index Only Man and Child Tested
probability =
2pApB x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))
pA + pB
AB
AB
pA + pB
Denominator
2pApB
2pApB
Paternity Index Only Man and Child Tested
PI = 0.5pB + 0.5pA
2pApB
2pApB x 2pApB x (0.5(fA) x pB + 0.5(fB) x pA)
2pApB x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA)) PI =
PI = pA + pB
4pApB
Paternity Index Only Man and Child Tested
Parents
M
Child
C
A
?
AF
A
AF can pass only the A allele
RM has p chance of passing the A allele
The untested Mother would have to pass an A allele
Paternity Index Only Man and Child Tested
A
A
Paternity Index Only Man and Child Tested
Numerator
pA 1
pA2
Probability = pA2 x pA
2 x 1(fA) x pA
A
A
pA2
Paternity Index Only Man and Child Tested
probability = pA2 x pA
2 x p(mA) x p(fA)
pA
A
A
pA
Denominator
pA2
pA2
Paternity Index Only Man and Child Tested
PI = pA
pA x pA
PI = 1
pA
Paternity Index Only Man and Child Tested
pA2 x pA
2 x 1(fA) x pA
pA2 x pA
2 x p(mA) x p(fA)
PI =
Parents
M
Child
C
A
?
AF
AB
AF would have to pass the A allele
RM has p chance of passing the A allele
The untested Mother would have to pass an A allele
Paternity Index Only Man and Child Tested
A
AB
Paternity Index Only Man and Child Tested
Numerator
pA 0.5
2pApB
Probability = 2pApB x pA2 x 0.5(fA) x pA
A
AB
pA2
Paternity Index Only Man and Child Tested
probability = 2pApB x pA2 x p(mA) x p(fA)
pA
A
AB
pA
Denominator
2pApB
pA2
Paternity Index Only Man and Child Tested
PI = 0.5pA
pA x pA
PI = 0.5
pA
Paternity Index Only Man and Child Tested
2pApB x pA2 x 0.5(fA) x pA
2pApB x pA2 x p(mA) x p(fA)
PI =
Paternity Index
Only Man and Child Tested
Formulas
Single locus, no null alleles, low mutation rate,
codominance
C
AB
AB
AB
A
A
AF
AC
AB
A
AC
A
Numerator
0.5b
0.5(a+b)
b
0.5a
a
Denominator
2ab
2ab
2ab
a2
a2
PI
0.25/a
(a+b)/4ab
0.5/a
0.5/a
1/a
PE
[1-(a + b)]2
[1-(a + b)]2
[1-(a + b)]2
(1-a) 2
(1-a) 2
Paternity Index
Only Man and Child Tested
Formulas: with Population Substructure
C
AB
AB
AB
AF
AC
AB
A
PI
(1+2)
/ 4[(1-)a+]
(1+2)[(a+b)(1-)+2]
/ 4[(1-)a+] [(1-)b+]
(1+2)
/ 2[(1-)a+2]
PE
(1-)(1-a-b)[1-(1-)(a+b)]
/ (1+)(1+2)
Paternity Index
Only Man and Child Tested
Formulas: with Population Substructure
C
A
A
AF
AC
A
PI
(1+2)
/ 2[(1-)a+2]
(1+2)
/ [(1-)a+3]
PE
(1-)(1-a)[(1-)(1-a)+]
/ (1+)(1+2)
41376
FGA
TPOX
D8S1179
vWA
PENTA E
D18S51
D21S11
TH01
D3S1358
PowerPlex 2.1
C AF C AF
MOTHERLESS PATERNITY
CASE P-41376
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
C AF Allele
Frequencies
TPOX
C AF
41376
6
13
11 11
8 8
PI = X / Y
X = 0.5b + 0.5a
Y = 2ab
PI = (0.5b + 0.5a) / 2ab
PI = (a + b) / 4ab
C AF
THO1
C AF
41376
C AF
9.3 7 6 6
PI = X / Y
X = 0.5b
Y = 2ab
PI = 0.5b / 2ab
PI = 0.25 / a
5
11
9.3 10
vWA
C AF
41376
11
13
21
C AF
16 16
15
PI = X / Y
X = b
Y = 2ab
PI = b / 2ab
PI = 0.5 / a
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PI Formula
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PI
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PE Formulas
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PE
MOTHERLESS PATERNITY
CASE P-41376
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
C AF Allele
Frequencies
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PI Formulas
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PI
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PE Formulas
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PE
MOTHERLESS PATERNITY
CASE P-41376
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
C AF Allele
Frequencies
FGA
C AF
41376
C AF
21 25 19 19
PI = X / Y
X = 0.5b
Y = 2ab
PI = 0.5b / 2ab
PI = 0.25 / a
17
30
46.2 44.2
D18S51
C AF
41376
C AF
16 20
16
9
12
15
25
PI = X / Y
X = 0.5a
Y = a2
PI = 0.5a / a2
PI = 0.5 / a
D21S11
C AF
41376
C AF
29 29
28
PI = X / Y
X = 0.5a
Y = a2
PI = 0.5a / a2
PI = 0.5 / a
24.2
26
28
38
D3S1358
C AF
41376
C AF
18 17 15 15
PI = X / Y
X = 0.5b
Y = 2ab
PI = 0.5b / 2ab
PI = 0.25 / a
12
20
D8S1179
C AF
41376
C AF
8
18
13 13
11 11
PI = X / Y
X = 0.5b + 0.5a
Y = 2ab
PI = (0.5b + 0.5a) / 2ab
PI = (a + b) / 4ab
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PI Formulas
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PI
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PE Formulas
MOTHERLESS PATERNITY
CASE P-41376
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
C AF PE
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)
Combined Paternity Index 867
Probability of Paternity 99.88%
Probability of Exclusion 99.90%
Motherless Paternity 13 Core CODIS Loci (=0.01)
Combined Paternity Index 309
Probability of Paternity 99.68%
Probability of Exclusion 99.80%
Motherless Paternity 13 Core CODIS Loci (=0.03)
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
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.
RPI = X / Y
Numerator
AB
BC
CD
Reverse Paternity
Analysis
Reverse Paternity Index
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.
RPI = X / Y
Denominator
Missing child scenario
AB
BC
Reverse Paternity
Analysis
CD
In order to explain this evidence Calculate Probability that
b) Man randomly selected from population is type CD, and
c) Their child is type BC
Hypothetical DNA Example First Hypothesis
Numerator
Person
Alleged Mother
Child
Alleged Father
Type
AB
BC
CD
a) Woman randomly selected from population is type AB
Numerator
AB
BC 0.5 0.5
2pApB 2pCpD
Probability = 2pApB x 2pCpD x 0.5 x 0.5
CD
Missing child scenario
Reverse Paternity
Analysis
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
Hypothetical DNA Example Second Hypothesis
Denominator
Person
Alleged Mother
Child
Alleged Father
Type
AB
BC
CD
In order to explain this evidence Calculate Probability that
AB
BC
2pApB 2pCpD
Probability = 2pApB x 2pCpD x 2pBpC
CD
Paternity Analysis
Paternity Index Denominator
2pBpC
Missing child scenario
AB
BC
Reverse Paternity
Analysis
CD
2pApB x 2pCpD x 0.5 x 0.5
2pApB x 2pCpD x 2pBpC
LR =
LR = 0.25
2pBpC
Missing child scenario
Reverse Paternity
Analysis
Missing child scenario
AB
BC
CC
Reverse Paternity
Analysis
0.5 1
2pApB pC
2
Probability = 2pApB x pC2 x 0.5 x 1
AB
BC
C
Numerator
Missing child scenario
Reverse Paternity
Analysis
2pApB
Probability = 2pApB x pC2 x 2pBpC
AB
BC
C
Paternity Analysis
Paternity Index Denominator
pC2
2pBpC
pBpB x 2pCpD x 0.5 x 1
pBpB x 2pCpD x 2pBpC
LR =
LR = 0.5
2pBpC
Example of Failure of Pairwise Comparison of
profiles to detect True Relationship
• 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
K1 AA AB
AA AB
K2
K3 AC K4 Q1
Q1 = Evidence Sample;
K1 ~ K4 = Reference samples, with family relationships known;
Q1 is investigated as a Full Sibling in the family
?
A Real Case of Missing Person Identification
• 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?
Father (untyped)
Mother (untyped)
6909.1
6909.2
6997.1 (bone) ? 6909.5
6909.3 6909.4
Example (Contd.)
• Process of Identification – comparison of
likelihoods of two pedigrees
Father (untyped)
Mother (untyped)
6909.1
6909.2
6997.1 (Bone)
6909.5
6909.3
V
S
Father (untyped)
Mother (untyped)
6909.1
6909.2
6909.5
6909.3
6997.1 (Bone) and
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
Population IQQ CCP SCL TMC PUQ
LR (Bone) 0.999999999
1964352
0.999999996
2901903
0.999999999
2072461
0.999999997
7871493
0.999999997
9199534
• Prior odds = 1/6
• Probability of Positive Identification (PIP) =
(LR*prior odds)/(LR*prior odds + 1)
Example – Contd.
(Mitochondrial DNA Match)
• mtDNA Haplotype frequency and LR per population
Population Sample
Size
Mismatch = 0 Mismatch<=2
Counts Exact
frequency
CI(0.95)
UpperBound LR Counts
General
frequency
CI(0.95)
UpperBound LR
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
Population Counts Sample
Size θ
Conditional
frequency
CI(0.95)
UpperBound LR
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!
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