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The college of Forensic Medicine in Kunming Medical University Paternity Test The college of Forensic Medicine in Kunming Medical University Bingying Xu, Tel:13769175605 Email:bingying_xu@126.com

This is precisely what's going on ?

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

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

The course content The principle of paternity inclusion exclusion

Part Ⅰ Three Classic Cases

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

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.

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.

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.’

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

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.

Case 1 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.

Case 1 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.

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 last Russian Czar ’s identification 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 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.

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 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.

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.

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.

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

He is the husband of present British Queen Elizabeth

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.

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.

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.

Lineage of Romanov Family. Pedigree Lineage of Romanov Family.

Case 3 Genetic 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

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.

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 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.

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.

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。

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

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

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

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

（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.

（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.

（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.

（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.

（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.

(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 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?

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.

（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。

（ 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）

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

（ 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

Paternity Inclusion father son mother

Paternity Inclusion

（ 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.

Paternity Exclusion father son

Paternity Inclusion Paternity Exclusion

F1 F2 C M Parentage Testing

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

（ 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.

AmpFℓSTR® Y filer Kit Y-STR loci Father Son DYS456 14 DYS389Ⅰ 11 18 DYS389Ⅱ 26 DYS458 15 DYS19 DYS385 8 DYS393 9 DYS391 7 DYS439 DYS635 20 DYS392 10 GATA-H4 DYS437 DYS438 12 DYS448

Y-STR profile

（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.

Maternal Inheritance father mother son son daughter

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。

fertilization

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.

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

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

Mother,daughter,son have the Same DNA sequence.

(Ⅱ)The inclusion of Paternity

（Ⅱ）Paternity inclusion M1 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. D5S818 9，9 D5S818 9，9 D5S818 9，9 Is the father

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

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.

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%

STR loci alleged father son mother paternity index D3S1358 16 16，18 15，18 13.06287948 TH01 9 4.200861504 D21S11 29，30 30，31.2 5.679876697 D18S51 15 13，15 13，14 8.552095948 PentaE 19 60.37918126 D5S818 11 11，13 5.42122013 D13S317 9， 11 8， 9 2.742086175 D7S820 8， 11 8， 13 13 39.17340972 D16S539 10，12 11，12 4.066179181 CSF1PO 10，13 9.810754471 PentaD 10，11 12，13 6.605368844 vWA 14，16 16，19 7.707483134 D8S1179 14 6.377775544 TPOX 8，11 1.575832727 FGA 21，22 22 22，23 8.06160095 D19S433 14.2 13，14.2 9.05656659 D2S1338 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 3.222078818 PentaE 11，17 17 16 88.88888889 D5S818 7，11 7，12 11，12 30.36973328 D13S317 8 8，13 24.6008754 D7S820 9，12 8，9 8，11 16.04318569 D16S539 9 9，13 11，13 9.98109182 CSF1PO 12 6.300301608 PentaD 12，13 13，15 6.122049172 vWA 14，16 14，18 18，19 2.482606856 D8S1179 10，13 4.818820075 TPOX 9，11 4.215043423 FGA 22，24 23，24 22，23 3.483951526 D19S433 13，13.2 14，15.2 1.69653703 D2S1338 19，24 19，20 4.773513948 Amelogenin X, Y X, X CPI：10569379562111000 RCP：99.9999%

STR loci alleged mother daughter paternity index D8S1179 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 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 / CPI：1125624.4970 RCP：99.99%

CPI:42059.5； 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 1.712579 D3S1358 15，16 15，18 0.722335 D5S818 9，11 4.090996 D13S317 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 3.105590 D6S1043 17，20 4.545455 FGA 25 24，25 5.341880 Amelogenin 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

（Ⅲ）Parentage Exclusion M2 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

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

（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: 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.

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 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

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=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?

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Thank you!