1. Hello, everybody!

2. The college of Forensic Medicine in Kunming Medical University
Paternity Test
The college of Forensic Medicine
in Kunming Medical University
Bingying Xu, Tel:

3. This is precisely what's going on ?

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

5. The course content Three Classic cases The overview of parentage test
The principle of parentage test

6. The course content
The principle of paternity
inclusion
exclusion

7. Part Ⅰ
Three Classic Cases

8. Part Ⅰ Three Classic Cases
1.Case 1— Saddam’s identification
2. Case 2—last Russian Czar ’s identification
3. Case 3—Genetic legacy of Genghis Khan

9. Case 1 Saddam’s identification.
Verifying the identity of
Saddam Hussein

10. Saddam Hussein （28 April 1937 – 30 December 2006） The former President of Iraqi

11. 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.

12. Saddam was known to have many ‘stunt doubles’ to protect his life
Why?
Case 1
Saddam’s identification
Saddam was known to have many
‘stunt doubles’ to protect his life
from assassins.

13. The ability to verify his identity through
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.’

14. ?

15. 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

16. Case Saddam’s identification In this case, DNA samples from Saddam’s two sons provided the family reference samples.
Uday Qusay

17. Case 1 Saddam’s identification
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.

18. Case Saddam’s identification 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.

19. Case Saddam’s identification 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.

21. Case 2 last Russian Czar ’s identification Identifying the remains of the last Russian Czar

22. Last Russian Czar：Nicholas II （18 May 1868 – 17 July 1918 ）

23. Hotchpotch in 1911

24. Case 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.

25. Case 2 last Russian Czar ’s identification Nicholas II and his family were removed from power and murdered during the Bolshevik Revolution of 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.

26. Case last Russian Czar ’s identification 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.

27. 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.

28. Case 2 last Russian Czar ’s identification
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.

29. BUT!! last Russian Czar ’s identification
Case 2
last Russian Czar ’s identification
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.

30. HOW? Case 2 last Russian Czar ’s identification
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.

31. Prince Philip, Duke of Edinburgh
Who can provide biological sample?
Prince Philip, Duke of Edinburgh

32. He is the husband of present British Queen Elizabeth

33. Tsarina,children WHO? maternal descent from Tsarina Alexandra. His
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.

34. WHO?
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.

35. WHO?
Tsar
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.

36. Lineage of Romanov Family.
Pedigree
Lineage of Romanov Family.

37. Case 3 Genetic legacy of Genghis Khan

38. 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 ）

39. 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.

40. By the end of his life, the
Mongol Empire occupied a
substantial portion of Central
Asia and China.

41. Map of Genghis Khan's expedition

42. How to find the clue?
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.

43. What have they analysed?
1.Y-STR (short tandem repeats,STR)profile
2.Y-SNPs(Single nucleotide polymorphisms ,SNPs)

44. 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)

46. 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.

48. 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.

49. 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.

50. Part Ⅱ
Summary of Parentage test

51. Part Ⅱ Summary
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.

52. Part Ⅱ Summary
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.

53. Part Ⅱ Summary （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.
因大多数情况下孩子的母亲已确定，而要求鉴定一般是可疑的“父亲”与子女间的关系，故有时又称为父权鉴定。此外，亲子鉴定技术还可以延伸到祖孙关系、同胞关系、叔侄关系、舅甥关系等血缘关系的分析鉴定。这些鉴定类型统称血缘关系或亲缘关系鉴定。

54. （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.

55. （ 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.

56. （ 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 )

57. 9.11 events
wenchuan earthquake

58. the tsunami of Southeast Asia

59. Fertilization （Ⅲ）The basic principle of parentage testing
Father’s Sperm
Mother’s Egg
Child’s Cell

60. male’s chromosome
22 pairs of autosomal
a pair of sex chromosomes(X,Y)

61. Female’s chromosome 22 pairs of autosomal
a pair of sex chromosomes(X,X)

62. Father,22+X Mother,22+X Father,22+Y Son 22(pairs)+XY Mother,22+X
Daughter
22(pairs)+XX
Mother,22+X
Father,22+Y
Son
22(pairs)+XY
Mother,22+X

63. （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.

64. （Ⅲ） The basic knowledge of parentage testing
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.

65. （Ⅲ） The basic knowledge of parentage testing
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.

66. （Ⅲ） The basic knowledge of parentage testing
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.

67. （Ⅲ） The basic knowledge of parentage testing
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.

68. (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)

69. (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.

70. Part Ⅲ
Parentage Test

71. Part Ⅲ Parentage Testing
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 father?

72. Part Ⅲ Parentage Test
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.

73. （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
常染色体遗传标记，按孟德尔遗传规律传递；
（2）Y-染色体遗传标记，按父系遗传方式传递，可检验样品是否来自同一父系，适用于父子单亲亲子鉴定或男性同胞之间、隔代或旁系的亲缘关系鉴定；
（3）线粒体DNA多态性，按母系遗传方式传递，可检验样品是否来自同一母系，适用于母子单亲亲子鉴定或同胞之间、隔代或旁系的亲缘关系鉴定；
（4）X-染色体上基因座或遗传标记，由于遗传定向性，适用于三联体亲子鉴定或除父子关系外的其他单亲的亲缘鉴定。

74. （ Ⅰ ）The principle of parentage testing
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）

75. child 1 Child 2 A locus STR B locus ＋ － STR father father mother
heterozygous
heterozygous
homozygous
homozygous － ＋

76. （ Ⅰ ）The principle of parentage testing
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. ，
即通过父权相对机会来作出定量估计。当父权相
对机会达到一定水平，我们可以认为他是孩子
的生父

77. Paternity Inclusion
father
son
mother

78. Paternity Inclusion

79. （ Ⅰ ）The principle of parentage testing
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.

80. Paternity Exclusion
father
son

81. Paternity Inclusion
Paternity Exclusion

82. F1 F2 C M
Parentage Testing

83. （ Ⅰ ）The principle of parentage testing 2. Y- chromosome

84. （ Ⅰ ）The principle of parentage testing 2
（ Ⅰ ）The principle of parentage testing 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.

85. AmpFℓSTR® Y filer Kit Y-STR loci Father Son DYS456 14 DYS389Ⅰ 1118
DYS389Ⅱ 26
DYS458 15
DYS19
DYS385 8
DYS393 9
DYS391 7
DYS439
DYS635 20
DYS392 10
GATA-H4
DYS437
DYS438 12
DYS448

86. Y-STR profile

88. （Ⅰ）The principle of parentage testing 3
（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.

89. Maternal Inheritance
father
mother
son
son
daughter

90. 克林顿 “拉链门事件”。莱温斯基蓝裙子上的污迹让克林顿险遭弹劾，因为克林顿知道DNA不会说谎，一点干涸的精斑就能指证矢口抵赖的罪犯。
一条沾了总统精斑的蓝色裙子那条裙子先是交给了独立检察官斯塔尔，证实上面的确是精液。随后，由白宫药物检验师，抽取了克林顿的血液样本，经过联邦调查局实验室的DNA比对实验，以及一项更为精确的RELP化验结果证实，裙子上的精液与克林顿的DNA相符，而巧合的可能性在白种人中只有千亿分之七点八七。也就是说，要将地球上几百年内存在过的白人加起来，才可能找到一个相同的。

91. fertilization

92. Sperm mitochondria stay outside fertilized egg.
在精卵结合的受精过程中，仅头部进入卵细胞，故受精卵中的线粒体几乎全部来自于卵细胞，所以线粒体遗传系统表现为母系遗传，受精过程中携带突变的mtDNA分子的母亲将把它传给她所有的子女，不管男女，但只有她的女儿再把这种突变传给下代。即同一母系祖先的个体，其线粒体序列是相同的。
Sperm’s nucleus go into the egg, the fertilized egg's mitochondria is from the mother.
Sperm mitochondria stay outside fertilized egg.

93. mtDNA——D-loop Same DNA base sequence A G T C A A A G T C mother
daughter

94. mtDNA——D-loop
Same DNA base sequence A G T C
mother A A A G T C
son

95. Mother,daughter,son have the Same DNA sequence.

96. (Ⅱ)The inclusion of Paternity

97. （Ⅱ）Paternity inclusionM1
AF1
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.
D5S ，9
D5S ，9
D5S ，9
Is the father

98. Case A：Paternity Exclusion
Loci
AF1 C1 M1
Result
ABO
blood typing A O
Can’t be excluded
D16S539
11, 11
12, 11
12, 12
D7S820
11, 10
D13S317
12, 9
13, 9
CSF1PO
13, 12
TPOX
11, 8
11, 9
TH01
9, 8
9, 9
F13A01
6, 3.2
3.2, 3.2
FES/FPS
vWFⅢ
17, 16
18, 17
HPRTB
13, 13
14, 13
FABP
10, 9
10, 10
LPL
12, 10
D8S1179
16, 10
16, 12

99. 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).

100. 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.

101. 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。

102. 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.

103. 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.

104. 2.Relative Chance of Paternity (RCP)
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%

105. STR loci
alleged father
son
mother
paternity index
D3S1358 16
16，18
15，18
TH01 9
D21S11
29，30
30，31.2
D18S51 15
13，15
13，14
PentaE 19
D5S818 11
11，13
D13S317
9， 11
8， 9
D7S820
8， 11
8， 13 13
D16S539
10，12
11，12
CSF1PO
10，13
PentaD
10，11
12，13
vWA
14，16
16，19
D8S1179 14
TPOX
8，11
FGA
21，22 22
22，23
D19S433
14.2
13，14.2
D2S1338 23
Amelogenin
X, Y
X, Y
X, X
CPI： RCP： %

106. STR loci
alleged father
daughter
mother
paternity index
D3S1358
15，17
14，17
14，15
TH01 7
7，10
9，10
D21S11
31.2，34.2
30，34.2
29，30
D18S51
13，23
13，14
PentaE
11，17 17 16
D5S818
7，11
7，12
11，12
D13S317 8
8，13
D7S820
9，12
8，9
8，11
D16S539 9
9，13
11，13
CSF1PO 12
PentaD
12，13
13，15
vWA
14，16
14，18
18，19
D8S1179
10，13
TPOX
9，11
FGA
22，24
23，24
22，23
D19S433
13，13.2
14，15.2
D2S1338
19，24
19，20
Amelogenin
X, Y
X, X
CPI： RCP： %

107. STR loci
alleged mother
daughter
paternity index
D8S1179 13
D21S11
32.2,30
29，30
D7S820 8, 8,
CSFIPO
10，12
11，12
D3S1358
15，16
15,16
D5S818
9， 12
9,11
D13S317 8,
8,10
D16S539 12
D2S1338
22，23 23
D19S433
13, 14
13， 15
vWA
17, 19
17, 19
D12S391
20, 24
19， 24
D18S51
13，22
13, 19
D6S1043 11
11，18
FGA
19, 23
23, 25
Amel X /
CPI： RCP：99.99%

108. CPI:42059.5； RCP：99.99% STR lociI alleged father son paternity index
D8S1179
14，16
13，16
D21S11
32.2
30，32.2
D7S820
8，11
11，12
CSFIPO
D3S1358
15，16
15，18
D5S818
9，11
D13S317
D16S539
12，13
10，13
D2S1338
18，20
19，20
D19S433
14，15.2
14，14.2
vWA 14
14，17
D12S391
18，22
18，19
D18S51
15，17
D6S1043
17，20
FGA 25
24，25
Amelogenin
X，Y /
CPI: ； RCP：99.99%

109. 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％

110. (Ⅲ)The exclusion of Paternity

111. （Ⅲ）Parentage ExclusionM2
AF2
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.
D8S1179
14，14
D8S1179
14，14
D8S1179
14，16
Exclusion

112. Case B：Parentage Exclusion
Loci
AF2 C2 M2
Conclusion
D7S1179
14，14
14，16
excluded
D16S539
10,11,
9，13
13, 13
D7S820
12, 11
9，10,
11, 10
D13S317
12，13
10，12
10，11
can’t be excluded
CSF1PO
10, 10
11, 11
TPOX
8, 8
8，11
TH01
9，10
9, 9
F13A01
3.2, 3.2
6, 4
4, 3.2
FES/FPS
vWFⅢ
19, 18
17, 16
16, 16
HPRTB
14, 14
13, 12
F13B
10, 9
CYAR04
11, 7
6.1, 6.1
11, 6.1
CD4
7, 7
12, 12
12, 7

113. （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.

114. 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.

115. 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.

116. As the existence of mutation and other factors:
3. The suggestions during the parentage testing:
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.

117. 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.

118. STR loci
alleged father
son
mother
D16S539
11, 10
13, 9
13, 13
excluded
D7S820
12, 11
10, 9
D13S317
13, 12
12, 10
can’t be excluded
CSF1PO
10, 10
11, 11
TPOX
8, 8
11, 8
TH01
9, 9
F13A01
3.2, 3.2
6, 4
4, 3.2
FES/FPS
vWFA31/A
19, 18
17, 16
16, 16
HPRTB
14, 14
F13B
CYAR04
11, 7
6.1, 6.1
11, 6.1
CD4
7, 7
12, 12
12, 7

119. 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
D13S317
8，11
1.8969
D16S539
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
5.2356
D18S51
15，21
0.0015
D6S1043
14，17 19
FGA
22，24 24
2.8736
Amelogenin
X, Y
X, Y /
CPI= ,RCP= %
PS: The loci that doesn’t follow the genetic law calculate PI according to the mutation rate at μ=0.002.

120. 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?

121. Thank you！

122. Thank you!