Ranajit Chakraborty, PhD

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: 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

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

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

Paternity Analysis

DNA Analysis Results in Three

Genotypes

Mother (AB)

Child (BC)

Alleged Father (CD)

Paternity Analysis

Hypothetical case

Paternity Analysis

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

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

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

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

Alleged Father

a) Woman randomly selected from population is type AB

Paternity Analysis

Paternity Index Numerator

BC 0.5 0.5

2pApB 2pCpD

Probability = 2pApB x 2pCpD x 0.5 x 0.5

b) An alternative man randomly selected from population

is type CD, and c) The woman’s child, fathered by random man, is type

Hypothetical DNA Example Second Hypothesis

Denominator

Person

Mother

Alleged Father

2pApB 2pCpD

Probability = 2pApB x 2pCpD x 0.5 x pC

Paternity Analysis

Paternity Index Denominator

2pApB x 2pCpD x 0.5 x 0.5

2pApB x 2pCpD x 0.5 x pC

PI = 0.5

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

Alleged Father

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

Alleged Father

Paternity M and C share one allele

and AF is heterozygous for the other allele

Parents

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

Parents

Paternity Index M and C share one allele

and AF is homozygous for the obligatory allele

Parents

AF can only pass C allele

Random Man has p chance of passing the C allele

PI = 1/p PI =

Paternity Analysis

2pApB pC

Probability = 2pApB x pC2 x 0.5 x 1

Probability = 2pApB x pC2 x 0.5 x pC

Paternity Analysis

2pApB x pC2 x 0.5 x 1

2pApB x pC2 x 0.5 x pC

PI = 1

Paternity Analysis

Paternity Index

Paternity Index M and C share both alleles and

AF is heterozygous with one of the obligatory alleles

Parents

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

0.5A 0.5B

2pApB 2pBpC

Probability = 2pApB x 2pBpC x 0.5(mA) x 0.5(fB)

probability =

2pApB x 2pBpC x (0.5(mA) x pB + 0.5(mB) x pA)

pA + pB

Paternity Analysis

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

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

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

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))

probability =

2pApB x 2pApB x (0.5(mA) x pB + 0.5(mB) x pA)

pA + pB

Paternity Analysis

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

Paternity Analysis

Paternity Index

Parents

Paternity Index M and C share both alleles and

AF is heterozygous with one of the obligatory alleles

Parents

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

2pApB pB

Probability = 2pApB x pB2 x 0.5(mA) x 1(fB)

probability =

2pApB x pB2 x (0.5(mA) x pB + 1(mB) x pA)

pA + pB

Paternity Analysis

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 = 1

pA + pB

Paternity Analysis

Paternity Index

PI Formulas Single locus, no null alleles, low mutation rate,

codominance

Numerator

Denominator

codominance

Numerator

Denominator

codominance

Numerator

Denominator

0.5(a+b)

0.5/(a+b)

codominance

Numerator

Denominator

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

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

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

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

(1+3) / 2[3 +a(1- )]

(1+3) / 2[ +b(1- )]

(1+3) / 2[2 +a(1- )]

(1+3) / 2[ +a(1- )]

codominance

(1+3) / [4 +a(1- )]

(1+3) / [3 +a(1- )]

(1+3) / [2 +a(1- )]

codominance

(1+3) / 2[3 +(1- )(a+b)]

codominance

(1+3) / [4 +(1- ) (a+b)]

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...

• 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.

• 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.

• 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 (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.

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 + 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

• 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.

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.

b = sum of the frequency of all alleles

other than the obligate paternal allele.

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:

PE = [1 - (a + b)] 2

(1-a–b)(1-)[1-(1-)(a+b)]

(1-2) (1-3)

In addition, if population substructure is

considered

(1-)(1-a)[1-a(1-)]

/ (1+2)(1+3 )

[2+(1-)(1-b)] [3+(1-)(1-b)]

/ (1+2)(1+3 )

PE formulas

[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

[(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.

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

D16S539

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

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)

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

Paternity Index

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

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

0.6972

0.1753

0.6376

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 = 0.27228

11 = 0.20197

8 = 0.09949

11 = 0.41026

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

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

Paternity Index

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

PE Formula

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

0.8109

0.6369

0.3478

D8S1179

PENTA E

D18S51

D21S11

D3S1358

M C AF M C AF

P-41411 P-41414

PowerPlex 2.1

M C AF M C AF

P-41411 P-41414

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

21 = 0.17347

14 = 0.17347 17 = 0.15561

29 = 0.18112

15 = 0.10969

14 = 0.14039

P-41411

M C AF PI Formula

1/(a+b)

P-41411

M C AF

Paternity Index

P-41411

M C AF

(1 – a)2

[1 –(a+b)]2

PE Formula

P-41411

M C AF

0.6831

0.7389

0.4501

0.7926

0.6706

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)

Hypothetical case

Motherless Paternity Index PI determination in hypothetical DNA System

X = is the probability that (1) a man

is type AC, and (2) his child is type

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}

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

Y = 2pq

So, if we have a heterozygous child AB, and a

PI = X / Y

X = 0.5q

PI = 0.5q / 2pq

PI = 0.25/p

PI = 1/4p

Parents

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

Numerator

pB 0.5A

Probability = 2pApC x 2pApB x 0.5(fA) x pB

probability =

2pApC x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))

pA + pB

Denominator

PI = 0.5pB

2pApB x 2pApC x 0.5(mA) x pB

2pApB x 2pApC x (p(mA) x p(fB) + p(mB) x p(fA))

PI = 0.25

Parents

AF can only pass A allele

the A or B allele

Numerator

Probability = pA2 x 2pApB x 1(fA) x pB

probability =

pA2 x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))

pA + pB

Denominator

PI = pB

pA2 x 2pApC x 1(mA) x pB

pA2 x 2pApC x (p(mA) x p(fB) + p(mB) x p(fA))

PI = 0.5 pA

Parents

AF can pass either A or B allele

the A or B allele

Numerator

pA + pB 0.5A + 0.5B

Probability =

2pApB x 2pApB x (0.5(fA) x pB + 0.5(fB) x pA)

probability =

2pApB x 2pApB x (p(mA) x p(fB) + p(mB) x p(fA))

pA + pB

Denominator

PI = 0.5pB + 0.5pA

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

Parents

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

Numerator

Probability = pA2 x pA

2 x 1(fA) x pA

probability = pA2 x pA

2 x p(mA) x p(fA)

Denominator

PI = pA

pA x pA

PI = 1

pA2 x pA

2 x 1(fA) x pA

pA2 x pA

2 x p(mA) x p(fA)

Parents

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

Numerator

pA 0.5

Probability = 2pApB x pA2 x 0.5(fA) x pA

probability = 2pApB x pA2 x p(mA) x p(fA)

Denominator

PI = 0.5pA

pA x pA

PI = 0.5

2pApB x pA2 x 0.5(fA) x pA

2pApB x pA2 x p(mA) x p(fA)

Paternity Index

Only Man and Child Tested

Formulas

Single locus, no null alleles, low mutation rate,

codominance

Numerator

0.5(a+b)

Denominator

0.25/a

(a+b)/4ab

[1-(a + b)]2

(1-a) 2

Paternity Index

Formulas: with Population Substructure

/ 4[(1-)a+]

(1+2)[(a+b)(1-)+2]

/ 4[(1-)a+] [(1-)b+]

/ 2[(1-)a+2]

(1-)(1-a-b)[1-(1-)(a+b)]

/ (1+)(1+2)

Paternity Index

Formulas: with Population Substructure

/ 2[(1-)a+2]

/ [(1-)a+3]

(1-)(1-a)[(1-)(1-a)+]

/ (1+)(1+2)

D8S1179

PENTA E

D18S51

D21S11

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

PI = X / Y

X = 0.5b + 0.5a

Y = 2ab

PI = (0.5b + 0.5a) / 2ab

PI = (a + b) / 4ab

9.3 7 6 6

PI = X / Y

X = 0.5b

Y = 2ab

PI = 0.5b / 2ab

PI = 0.25 / a

9.3 10

PI = X / Y

Y = 2ab

PI = b / 2ab

PI = 0.5 / a

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

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

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

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

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

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

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

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

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

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

21 25 19 19

PI = X / Y

X = 0.5b

Y = 2ab

PI = 0.5b / 2ab

PI = 0.25 / a

46.2 44.2

D18S51

PI = X / Y

X = 0.5a

Y = a2

PI = 0.5a / a2

PI = 0.5 / a

D21S11

PI = X / Y

X = 0.5a

Y = a2

PI = 0.5a / a2

PI = 0.5 / a

D3S1358

18 17 15 15

PI = X / Y

X = 0.5b

Y = 2ab

PI = 0.5b / 2ab

PI = 0.25 / a

D8S1179

PI = X / Y

X = 0.5b + 0.5a

Y = 2ab

PI = (0.5b + 0.5a) / 2ab

PI = (a + b) / 4ab

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

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

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

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

(prior=0.5)

Motherless Paternity 13 Core CODIS Loci (=0)

Combined Paternity Index 867

Motherless Paternity 13 Core CODIS Loci (=0.01)

Combined Paternity Index 309

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

BODY IDENTIFICATION

ALLEGED EVIDENCE ALLEGED

MOTHER FATHER

Three genotypes:

• Alleged Mother

• Child (missing)

• Alleged Father

Missing child

Reverse Paternity Index

X = is the probability that (1) a woman

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

Reverse Paternity

Analysis

Reverse Paternity Index

Y = is the probability that (1) a woman

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,

is BC.

RPI = X / Y

Denominator

Missing child scenario

Reverse Paternity

Analysis

c) Their child is type BC

Hypothetical DNA Example First Hypothesis

Numerator

Person

Alleged Mother

Alleged Father

Numerator

BC 0.5 0.5

2pApB 2pCpD

Probability = 2pApB x 2pCpD x 0.5 x 0.5

Reverse Paternity

Analysis

c) A child, randomly selected and unrelated to either

woman or man, is type BC

Hypothetical DNA Example Second Hypothesis

Denominator

Person

Alleged Mother

Alleged Father

2pApB 2pCpD

Probability = 2pApB x 2pCpD x 2pBpC

Paternity Analysis

Reverse Paternity

Analysis

2pApB x 2pCpD x 0.5 x 0.5

2pApB x 2pCpD x 2pBpC

LR = 0.25

Reverse Paternity

Analysis

Reverse Paternity

Analysis

2pApB pC

Probability = 2pApB x pC2 x 0.5 x 1

Numerator

Reverse Paternity

Analysis

Probability = 2pApB x pC2 x 2pBpC

Paternity Analysis

pBpB x 2pCpD x 0.5 x 1

pBpB x 2pCpD x 2pBpC

LR = 0.5

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

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

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

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!

Ranajit Chakraborty, PhD

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