Paternity Test The college of Forensic Medicine in Kunming Medical University Bingying Xu, Tel:13769175605 Email:[email protected]

Paternity Test

The college of Forensic Medicine in Kunming Medical University

Bingying Xu, Tel:13769175605

Email:[email protected]

This is precisely what's going on ?

.

This is your son!!!This boy is not my son.

The course content

• Three ClassicThree Classic casescases

• The overview of parentage The overview of parentage

testtest

• The principle of parentage The principle of parentage

testtest

• The principle of paternity The principle of paternity

inclusion inclusion

• The principle of paternity The principle of paternity

exclusion exclusion

The course content

Part Ⅰ

Three ClassicThree Classic CasesCases

Part Ⅰ Three ClassicThree Classic CasesCases

1.Case 1— Saddam’s identification

2. Case 2—last Russian Czar ’s identification

3. Case 3—Genetic legacy of Genghis Khan

Case 1 Saddam’s identification.

Verifying the identity of

Saddam Hussein

Saddam Hussein

（ 28 April 1937 – 30 December 2006 ）

The former

President of Iraqi

Case 1

Saddam’s identification.

Saddam Hussein was killed or captured by

the United States military .

The United States military must verified

the identity of Saddam Hussein.

Case 1

Saddam’s identification

Saddam was known to have many

‘stunt doubles’ to protect his life

from assassins.

Case 1

Saddam’s identification.

The ability to verify his identity through

genetic testing was essential to knowing that

the United States in fact ‘had their man.’

?

Case 1 Saddam’s identification

What methods they use ？

Forensic DNA testing using short tandem repeat

(STR) markers played an important role in the

identification effort.

Forensic DNA testing : autosomal STR profiles

Y chromosome STR profiles

Case 1


In this case, DNA samples from Saddam’s two

sons provided the family reference samples.

Uday Qusay

Case 1


Uday and Qusay were killed in a gunfight before.

DNA samples were collected from their remains

shortly after they were killed for use as reference

samples in verifying the identity of their Father.

Case 1


Scientists extracted DNA from Saddam’s 、 Uday’s

and Qusay’s biological samples and then amplified

the DNA samples using the autosomal STR kit to

obtain a full 13 loci STR profile. Saddam’s STR

profiles possessed alleles in common with STR

profiles of Saddam’s two sons.

Case 1


Additionally, the Y chromosome STR kit also

showed full allele sharing between Saddam and

his two sons indicating that the sample in question

was from their same paternal lineage. Saddam and

his two sons have common Y chromosome STR

loci.

Case 2 last Russian Czar ’s identification

Identifying the remains of the last

Russian Czar

Last Russian Czar ：

Nicholas II

（ 18 May 1868 – 17

July 1918 ）

Hotchpotch in 1911

Case 2

last Russian Czar ’s identification

Nicholas II, his wife, his son, his four daughters

, the family's medical doctor, the Emperor's

footman, the Empress' maidservant, and the

family's cook were executed in the same room

by the Bolsheviks on the night of 17 July 1918.

Case 2


Nicholas II and his family were removed from

power and murdered during the Bolshevik

Revolution of 1918. They were shot by a firing

squad , doused with sulfuric acid to render their

bodies unrecognizable, and disposed of in a

shallow pit under a road.

Case 2


Their remains were lost to history until July 1991 when nine

skeletons were uncovered from a shallow grave near

Ekaterinburg, Russia. A number of forensic tests were

attempted involving computer aided reconstructions and

odontological analysis, but as the facial areas of the skulls

were destroyed, classical facial identification techniques

were difficult at best and not conclusive.

Case 2

last Russian Czar ’s identification • DNA analysis

Five autosomal STR markers (VWA, F13A1,FES,ACTBP2,

TH01,) were used to examine the nine skeletons.

Approximately one gram of bone from each of the skeletons

yielded about 50 pg of DNA, just enough for PCR

amplification of several STR markers.

Case 2


What have scientists discovered ？

The remains of the Romanov family members

consisting of the Tsar, the Tsarina, and three

children were distinguishable from those of three

servants and the family doctor by their STR

genotypes.

Case 2


While the STR analysis served to establish

family relationships between the remains

through comparing matching alleles, a link still

had to be made with a known descendant of

the Romanov family to verify that the remains

were indeed those of the Russian royal family.

Mitochondrial DNA(mtDNA ) analysis was used to answer

this question.Mitochondrial DNA was extracted from

the femur of each skeleton and sequenced. Blood

samples were then obtained from maternally related

descendants of the Romanov family and sequenced

in the same manner.

Case 2 last Russian Czar ’s identification

Prince Philip, Duke of Edinburgh

Who can provide biological sample?

He is the

husband of

present British

Queen Elizabeth

Prince Philip is a grand nephew of unbroken

maternal descent from Tsarina Alexandra. His

blood sample thus provided the comparison to

confirm the sibling status of the children and the

linkage of the mother to the Tsarina’s family.

Tsarina,children

The sequences of all 740 tested nucleotides

from the mtDNA control region matched

between Prince Philip and the putative Tsarina

and the three children.

Tsarina, children

The mtDNA sequence from the putative Tsar

was compared with two relatives of unbroken

maternal descent from Tsar Nicholas II’s

grandmother, Louise of Hesse-Cassel. The two

relatives had the same mtDNA sequence.

Tsar

Lineage of Romanov Family.

Pedigree

Case 3Genetic legacy of Genghis Khan

Genghis Khan was the founder and Great Khan

(emperor) of the Mongol Empire, which became the

largest contiguous empire in history after his demise.

Genghis Khan

（ 1162—1227 ）

He came to power by uniting

many of the nomadic tribes of

northeast Asia. After founding

the Mongol Empire and he

started the Mongol invasions

that resulted in the conquest of

most of Eurasia.

By the end of his life, the

Mongol Empire occupied a

substantial portion of Central

Asia and China.

Map of Genghis Khan's expedition

In a study of more than 2100 males from Central Asia ,Chris

Tyler-Smith from the University of Oxford used variation on

the Y chromosome to provide insights into aspects of human

history and evolution.He found that approximately 8% of

those studied had a unique Y chromosome lineage.

How to find the clue?

What have they analysed?

1.Y-STR (short tandem repeats,STR)profile

2.Y-SNPs(Single nucleotide

polymorphisms ,SNPs)

What have they analysed?

1.Y-STR profile : 15 Y-STR loci

DYS389I (10), DYS389II (26), DYS390 (25),

DYS391 (10), DYS392 (11), DYS393 (13),

DYS388 (14), DYS425 (12), DYS426 (11),

DYS434 (11), DYS435 (11), DYS436 (12),

DYS437 (8), DYS438 (10), DYS439 (10)

What have they analysed? 2.Y-SNPs(Single nucleotide polymorphisms ,SNPs)

They analysed at least 16 Y-SNPs loci.

They are placed all of these samples in

haplogroup C*(xC3c), which is common in

Asia.

What is the clue from the Y-STR&Y-SNPs analysis?

The highest frequency was in Mongolia

leading to the assumption that it was the

source of these particular male lineages.

Conclusion the geographical distribution of these populations

closely matches the area of Genghis Khan’s

former Mongol Empire. The evidence that this Y-

lineage was from Genghis Khan and his close

male-line relatives was strengthened by a match

to a group in Pakistan who by oral tradition

consider Themselves direct male-line

descendants of Genghis Khan.

Part Ⅱ

Summary of Parentage test

Part SummaryⅡ

Every year in China ， more than 300,

000 paternity cases are performed where

the identity of the father of a child is in

dispute. These cases typically involve the

mother, the child, and one or more alleged

fathers.

Part SummaryⅡ

The determination of parentage is made

based on whether or not alleles are shared

between the child and the alleged father

when a number of genetic laws of

Inheritance.

Part SummaryⅡ

（Ⅰ） Parentage test

a kind of identification that

determines the blood

relationship between alleged

parents and the child according

to genetic law via genetic

markers analysis.

（Ⅱ） The reasons for parentage test

in China usually include:

1. Illegitimate child, the mother accused that a man

was the biological father of her child.

2. The husband suspected the child was not his

own.

3. It is suspected that the newborn baby was

confused in hospital.

（ Ⅱ ） The reasons for parentage test

in China usually include:

4. Confirmation of missing children or relatives.

5. Identification of the children beyond family

planning.

6. Inheritance disputes.( Children can inherit the

parents' heritage .)

7. Immigration cases.

（ Ⅱ ） The reasons for parentage test in China usually include:

8. Children abduction cases.

9. Pregnancy caused by rape.(Who is the fetus ’father?)

10. Identification of cadaver source(Who is missing

persons ?)

11. mass disaster investigations.(9.11 events occurred in the

United States ; Southeast Asia the tsunami )

9.11 events wenchuan earthquake

the tsunami of Southeast Asia

Fertilization

Father’s Sperm Mother’s Egg

Child’s Cell

（Ⅲ） The basic principle of parentage testing

male’s chromosome

22 pairs of autosomal

a pair of sex chromosomes(X,Y)

Female’s chromosome

22 pairs of autosomala pair of sex chromosomes(X,X)

Father,22+X

Mother,22+X

Daughter

22(pairs)+XX

Son

22(pairs)+XY

Father,22+Y

Mother,22+X

（Ⅲ） The basic knowledge of parentage testing

The Mendelian inheritance law is the basis of

paternity testing.

The child inherits 23 chromosomes from the

mother and another set of 23 chromosomes from

the biological father.


Since, the mother contributes half of the child’s

nuclear DNA, the father must contribute the other

half of the child’s nuclear DNA. To human genetic

markers, such as STRs, each person’s DNA

contains two copies of these markers.one copy

inherited from the father and the other from the

mother.


The laboratory performs a series of DNA tests each

for a different genetic marker. First, the analyst

identifies the alleles that are shared between the

mother and child, called maternal obligatory genes

or maternal alleles. The child’s alleles that are not

shared with the mother must come from the

true biological father.


These are called paternal obligatory genes or

paternal alleles. Theoretically, if the alleged

father does not share any one of the paternal

alleles with the child, he will be excluded or found

not to be the child’s biological father.


If the alleged father shares all of the paternal

alleles with the child, he cannot be excluded and

calculations can be made as to his likelihood of

being the father as well as the probability that he

and the mother produced a child with the tested

characteristics.

(Ⅳ)The basic principle of parentage testing

1.Child has two alleles for each autosomal marker.

(one from mother and one from biological father)

2.Child will have mother's mitochondrial DNA

haplotype. (barring mutation)

3. Child, if a son, will have father's Y chromosome

haplotype. (barring mutation)

(Ⅳ)The basic principle of parentage testing

The basis of paternity comes down to the

fact that in the absence of mutation a child

receives one allele matching each parent at

every genetic locus examined.

Part Ⅲ

Parentage Test

Part Parentage TestingⅢ

Paternity testing, uses results from a alleged

father ,a mother and a child to answer the

question if the alleged father could have

Fathered the child versus a random man.

Who is my Who is my father?father?

Part Parentage TestⅢ

Paternity testing laboratories often utilize the

same short tandem repeat (STR) multiplexes

and commercial kits to examine markers of child

,one or both parent .The outcome of parentage

test is simply inclusion or exclusion.

（Ⅰ） The principle of parentage test

1.The GM in autosome : The Mendelian inheritance law

2.The GM in Y- chromosome: Paternal inheritance

3.The GM in mtDNA: Maternal inheritance

1.The GM in autosome : Following the Mendelian

inheritance law

(1) The child inherits 23 chromosomes from the

mother and another set of 23 chromosomes from

the biological father. ——Paternity Inclusion （ Is

the father ）

（ Ⅰ ） The principle of parentage testing

STR

A locus

STR

B locus

child 1 Child 2

－

＋

father

mother

father

mother

father

father

mother

mother

homozygous heterozygous homozygous heterozygous


1.The GM in autosome ： Following the

Mendelian inheritance law

(1)Paternity Inclusion （ Is the father ）

If the alleged father shares some of the paternal

alleles with the child, he cannot be excluded and

calculations can be made as to his likelihood of

being the father.

Paternity Inclusion

father

son

mother

Paternity Inclusion

1.The GM in chromosome: Following the Mendelian inheritance

law

(2)Paternity Exclusion(Not the father)

The child can not take the allele not existing in parents.

If the alleged father does not share any one of the paternal

obligatory genes with the child, he will be excluded or found

not to be the child’s biological father.


Paternity Exclusion

father

son

Paternity Inclusion

Paternity Exclusion

FF11 F2F2 CC MM

Parentage Testing


2. Y- chromosome


2.The GM in Y- chromosome: Paternal inheritance Male offspring Y - DNA typing must be the same as his

father's

Male from one paternal have the same Y-DNA typing.

Y-STR loci Father Son

DYS456 14 14

DYS389Ⅰ 11 11

DYS390 18 18

DYS389Ⅱ 26 26

DYS458 15 15

DYS19 11 11

DYS385 8 8

DYS393 9 9

DYS391 7 7

DYS439 9 9

DYS635 20 20

DYS392 10 10

GATA-H4 9 9

DYS437 14 14

DYS438 12 12

DYS448 18 18

AmpFℓSTR® Y filer Kit

Y-STR profile

（Ⅰ） The principle of parentage

testing

3.The GM in mtDNA: Maternal inheritance The mtDNA typing of the child is the same as his or

her mother.

All compatriots from one maternal have the same

mtDNA typing.

Maternal Inheritance

father mother

daughter son son

fertilization

Sperm’s nucleus go into the egg, the fertilized egg's mitochondria is from the mother. Sperm mitochondria stay outside fertilized egg.

mtDNA——D-loop

Same DNA base sequence

A G T T C A A T

A AA G T C TTdaughter

mother

mtDNA——D-loop

Same DNA base sequence

A G T T C A A T

A AA G T C TTson

mother

Mother

Daughter

Son

Mother,daughter,son have the Same DNA sequence.

(Ⅱ)The inclusion of Paternity

D5S818 9 ， 9 D5S818

9 ， 9

D5S818 9 ， 9

M1 AF1

Is the father

（Ⅱ） Paternity inclusion

If the alleged father shares

some of the paternal

obligatory genes with the

child, he cannot be excluded

and calculations can be

made as to his likelihood of

being the father.

Loci AF1 C1 M1 Result

ABO

blood typing

A O A Can’t be excluded

D16S539 11, 11 12, 11 12, 12 Can’t be excluded



CSF1PO 12, 9 12, 12 13, 12 Can’t be excluded

TPOX 11, 8 11, 11 11, 9 Can’t be excluded

TH01 9, 8 9, 9 9, 9 Can’t be excluded

F13A01 6, 3.2 3.2, 3.2 6, 3.2 Can’t be excluded

FES/FPS 11, 11 11, 11 11, 11 Can’t be excluded

vWFⅢ 17, 16 17, 16 18, 17 Can’t be excluded

HPRTB 13, 13 14, 13 14, 13 Can’t be excluded

FABP 10, 9 10, 10 11, 10 Can’t be excluded

LPL 12, 10 10, 10 12, 10 Can’t be excluded


Case ACase A ：： Paternity ExclusionPaternity Exclusion

1. Paternity Index (PI)

If the man tested cannot be excluded as the

biological father of the child in question, then

statistical calculations are performed to aid in

understanding the strength of the match. The most

commonly applied test in this regard is the

paternity index (PI).

1.Paternity Index (PI)

The paternity index (PI) is the ratio of two conditional

probabilities where the numerator assumes paternity and the

denominator assumes a random man of similar ethnic

background was the father. The numerator is the probability

of observed genotypes, given the tested man is the father,

while the denominator is the probability of the observed

genotypes, given that a random man is the father.

1. Paternity Index (PI)

The paternity index is a likelihood ratio of two probabilities

conditional upon different competing hypotheses. This

likelihood ratio reflects how many times more likely it is to

see the evidence under the first hypothesis compared to

the second hypothesis. 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 。

Combined Paternity Index (CPI)

The PI is calculated for each locus and then

individual PI values are multiplied together to

obtain the combined paternity index (CPI) for the

entire set of genetic loci examined.

2.Relative Chance of Paternity (RCP)

combined paternity index (CPI) is a real number,

and it is difficult to judge the chance of paternity

through this value, Therefore, the CPI is usually

converted into a probability of paternity value, which

specifies the probability.

Probability of paternity is also called relative

chance of paternity (RCP), and in fact it is the

posterior probability of paternity that the tested

man is the father.

RCP=[PI/ （ PI+1 ） ]×100%

2.Relative Chance of Paternity (RCP)

STR loci alleged father son mother paternity index

D3S1358 16 16 ， 18 15 ， 18 13.06287948

TH01 9 9 9 4.200861504

D21S11 29 ， 30 30 ， 31.2 30 ， 31.2 5.679876697

D18S51 15 13 ， 15 13 ， 14 8.552095948

PentaE 15 ， 18 15 19 60.37918126

D5S818 11 11 ， 13 11 ， 13 5.42122013

D13S317 9 ， 11 8 ， 9 8 ， 9 2.742086175

D7S820 8 ， 11 8 ， 13 13 39.17340972

D16S539 10 ， 12 10 ， 12 11 ， 12 4.066179181

CSF1PO 10 ， 13 11 ， 13 11 ， 12 9.810754471

PentaD 10 ， 11 10 ， 12 12 ， 13 6.605368844

vWA 14 ， 16 16 16 ， 19 7.707483134

D8S1179 14 13 ， 14 10 ， 13 6.377775544

TPOX 8 ， 11 8 ， 11 8 ， 11 1.575832727

FGA 21 ， 22 22 22 ， 23 8.06160095

D19S433 14.2 13 ， 14.2 13 ， 15 9.05656659

D2S1338 23 23 23 27.38284864

Amelogenin X, Y X, Y X, X

CPI ： 3824607936701020 RCP ： 99.9999%

STR loci alleged father daughter mother paternity index

D3S1358 15 ， 17 14 ， 17 14 ， 15 2.696633674

TH01 7 7 ， 10 9 ， 10 29.59651661

D21S11 31.2 ， 34.2 30 ， 34.2 29 ， 30 70.43047104

D18S51 13 ， 23 13 ， 14 13 ， 14 3.222078818

PentaE 11 ， 17 17 16 88.88888889

D5S818 7 ， 11 7 ， 12 11 ， 12 30.36973328

D13S317 8 8 ， 13 8 ， 13 24.6008754

D7S820 9 ， 12 8 ， 9 8 ， 11 16.04318569

D16S539 9 9 ， 13 11 ， 13 9.98109182

CSF1PO 12 12 12 6.300301608

PentaD 11 ， 12 12 ， 13 13 ， 15 6.122049172

vWA 14 ， 16 14 ， 18 18 ， 19 2.482606856

D8S1179 11 ， 13 10 ， 13 10 ， 13 4.818820075

TPOX 9 ， 11 8 ， 9 8 4.215043423

FGA 22 ， 24 23 ， 24 22 ， 23 3.483951526

D19S433 13 ， 13.2 13 ， 14 14 ， 15.2 1.69653703

D2S1338 19 ， 24 19 ， 24 19 ， 20 4.773513948

Amelogenin X, Y X, X X, X

CPI ： 10569379562111000 RCP ： 99.9999%

STR loci alleged mother daughter paternity index

D8S1179 13 13 4.849661

D21S11 32.2,30 29 ， 30 24.752475

D7S820 8, 11 8, 11 2.291084

CSFIPO 10 ， 12 11 ， 12 0.627510

D3S1358 15 ， 16 15,16 1.470838

D5S818 9 ， 12 9,11 3.360215

D13S317 8, 13 8,10 0.865951

D16S539 11 ， 12 12 2.151463

D2S1338 22 ， 23 23 2.616431

D19S433 13, 14 13 ， 15 0.856751

vWA 17, 19 17, 19 4.038960

D12S391 20, 24 19 ， 24 10.000000

D18S51 13 ， 22 13, 19 1.419648

D6S1043 11 11 ， 18 4.545455

FGA 19, 23 23, 25 1.212415

Amel X X /

CPI ： 1125624.4970 RCP ： 99.99%

STR lociI alleged father son paternity index

D8S1179 14 ， 16 13 ， 16 2.958580

D21S11 32.2 30 ， 32.2 3.681885

D7S820 8 ， 11 11 ， 12 0.677140

CSFIPO 11 ， 12 11 ， 12 1.712579

D3S1358 15 ， 16 15 ， 18 0.722335

D5S818 9 ， 11 9 ， 11 4.090996

D13S317 11 ， 12 8 ， 11 1.030928

D16S539 12 ， 13 10 ， 13 2.670940

D2S1338 18 ， 20 19 ， 20 2.106150

D19S433 14 ， 15.2 14 ， 14.2 0.990099

vWA 14 14 ， 17 1.988072

D12S391 18 ， 22 18 ， 19 1.308901

D18S51 15 ， 17 14 ， 17 3.105590

D6S1043 17 ， 20 19 ， 20 4.545455

FGA 25 24 ， 25 5.341880

Amelogenin X ， Y X ， Y /

CPI:42059.5 ； RCP ： 99.99%

3.The standardization of parentage inclusion

After the parentage test, if the AF still can’t be

excluded, then the rate should be calculated. If the

conclusion meet two identified indicators at the

same time, there should have the predication that

AF is the biological father of the child.

For example of the conclusion:

CPI > 2000, while RCP > 99.99 ％

(Ⅲ)The exclusion of Paternity

M2 AF2

D8S1179

14 ， 14

D8S1179

14 ， 14

D8S1179

14 ， 16

Exclusion

（Ⅲ） Parentage Exclusion

If the alleged father does not

share any one of the paternal

obligatory genes with the

child, with the precondition of

none mutation exist, he will

be excluded or found not to

be the child’s biological

father.

Loci AF2 C2 M2 Conclusion

D7S1179 14 ， 14 14 ， 16 14 ， 14 excluded

D16S539 10,11, 9 ， 13 13, 13 excluded

D7S820 12, 11 9 ， 10, 11, 10 excluded

D13S317 12 ， 13 10 ， 12 10 ， 11 can’t be excluded

CSF1PO 10, 10 10 ， 11 11, 11 can’t be excluded

TPOX 9 ， 10, 8, 8 8 ， 11 excluded

TH01 9 ， 10 9, 9 9, 9 can’t be excluded

F13A01 3.2, 3.2 6, 4 4, 3.2 excluded

FES/FPS 12, 11 11, 11 11, 10 can’t be excluded

vWFⅢ 19, 18 17, 16 16, 16 excluded

HPRTB 14, 14 13, 12 13, 13 excluded

F13B 9, 9 10, 10 10, 9 excluded

CYAR04 11, 7 6.1, 6.1 11, 6.1 can’t be excluded

CD4 7, 7 12, 12 12, 7 excluded

Case BCase B ：： Parentage ExclusionParentage Exclusion

（Ⅲ） Parentage Exclusion

1. Two Situation ：

The child take the allele neither existing in

mother or AF.

The child doesn’t take the allele that the AF must

delivered to his biological child.

2.The standardization of parentage exclusion

If the alleged father does not share any one of the

paternal obligatory genes with the child, with the

precondition of none mutation exist, he will be

excluded or found not to be the child’s biological

father.

2.The standardization of parentage exclusion

Attention ：

As to avoid the influence of mutation, the cases

only have one or two loci that don’t follow the

genetic law still can’t make the exclusion, more

than three non-shared genetic markers must

occur before an alleged father is reported as

excluded.

As the existence of mutation and other factors:

1. The cases only have one or two loci that don’t

follow the genetic law still can’t make the exclusion,

only if there add more other genetic markers .

2. More than three non-shared genetic markers must

occur before an alleged father is reported as

excluded.

3. The suggestions during the parentage testing:

3. The suggestions during the parentage testing:

If three genetic loci do not match between an alleged

father and a child, the alleged father cannot be

excluded as being the true biological father. It is

important to keep in mind that the more genetic

systems examined the greater the chance of a random

mutation to be observed.

STR loci alleged father son mother

D16S539 11, 10 13, 9 13, 13 excluded

D7S820 12, 11 10, 9 11, 10 excluded

D13S317 13, 12 12, 10 11, 10 can’t be excluded

CSF1PO 10, 10 11, 10 11, 11 can’t be excluded

TPOX 10, 9 8, 8 11, 8 excluded

TH01 10, 9 9, 9 9, 9 can’t be excluded

F13A01 3.2, 3.2 6, 4 4, 3.2 excluded

FES/FPS 12, 11 11, 11 11, 10 can’t be excluded

vWFA31/A 19, 18 17, 16 16, 16 excluded

HPRTB 14, 14 13, 12 13, 13 excluded

F13B 9, 9 10, 10 10, 9 excluded

CYAR04 11, 7 6.1, 6.1 11, 6.1 excluded

CD4 7, 7 12, 12 12, 7 excluded

STR loci alleged father son paternal index

D8S1179 14 12 ， 14 2.6302

D21S11 30 ， 32.2 31 ， 31.2 0.0008

D7S820 8 ， 12 8 3.2279

CSFIPO 10 ， 12 11 ， 12 0.6275

D3S1358 16 ， 18 15 0.0007

D5S818 11 10 0.0007

D13S317 8 ， 11 8 ， 11 1.8969

D16S539 11 ， 12 11 ， 12 1.8005

D2S1338 24 ， 25 19 ， 24 1.9113

D19S433 13 ， 16 13 ， 14 0.8568

vWA 14 ， 19 16 ， 17 0.0001

D12S391 18 18 5.2356

D18S51 15 ， 21 14 0.0015

D6S1043 14 ， 17 19 0.0007

FGA 22 ， 24 24 2.8736

Amelogenin X, Y X, Y /

CPI=0.00000000000000002,RCP=0.000000000000002%

PS: The loci that doesn’t follow the genetic law calculate PI according to the mutation rate at μ=0.002.

Questions：

1.The principle of parentage testing?

2.The Genetic Marker usually used in

parentage testing?

3.The standardization of parentage inclusion?

4.The standardization of parentage exclusion?

Thank you!

Documents

Paternity Test The college of Forensic Medicine in Kunming Medical University Bingying Xu, Tel:13769175605 Email:[email protected]