8 Part Ⅰ Three Classic Cases 1.Case 1— Saddam’s identification2. Case 2—last Russian Czar ’s identification3. Case 3—Genetic legacy of Genghis Khan
9 Case 1 Saddam’s identification. Verifying the identity ofSaddam Hussein
10 Saddam Hussein （28 April 1937 – 30 December 2006） The former President of Iraqi
11 Case 1Saddam’s identification.Saddam Hussein was killed or captured bythe United States military .The United States military must verifiedthe identity of Saddam Hussein.
12 Saddam was known to have many ‘stunt doubles’ to protect his life Why?Case 1Saddam’s identificationSaddam was known to have many‘stunt doubles’ to protect his lifefrom assassins.
13 The ability to verify his identity through Case 1Saddam’s identification.The ability to verify his identity throughgenetic testing was essential to knowing thatthe United States in fact ‘had their man.’
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 profilesY 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 remainsshortly after they were killed for use as referencesamples 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.
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 analysisFive autosomal STR markers (VWA, F13A1,FES,ACTBP2,TH01,) were used to examine the nine skeletons.Approximately one gram of bone from each of the skeletonsyielded about 50 pg of DNA, just enough for PCRamplification 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 2last Russian Czar ’s identificationWhile 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 answerthis question.Mitochondrial DNA was extracted fromthe femur of each skeleton and sequenced. Bloodsamples were then obtained from maternally relateddescendants of the Romanov family and sequencedin 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 unbrokenmaternal descent from Tsarina Alexandra. Hisblood sample thus provided the comparison toconfirm the sibling status of the children and thelinkage of the mother to the Tsarina’s family.
34 WHO?Tsarina, childrenThe 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?TsarThe 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. PedigreeLineage of Romanov Family.
42 How to find the clue?In a study of more than 2100 males from Central Asia ,ChrisTyler-Smith from the University of Oxford used variation onthe Y chromosome to provide insights into aspects of humanhistory and evolution.He found that approximately 8% ofthose studied had a unique Y chromosome lineage.
43 What have they analysed? 1.Y-STR (short tandem repeats,STR)profile2.Y-SNPs(Single nucleotide polymorphisms ,SNPs)
48 What is the clue from the Y-STR&Y-SNPs analysis? The highest frequency was in Mongolialeading to the assumption that it was thesource of these particular male lineages.
49 Conclusionthe 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.
51 Part Ⅱ SummaryEvery 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 SummaryThe 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 hisown.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 missingpersons ?)11. mass disaster investigations.(9.11 events occurred in the United States ; Southeast Asia the tsunami )
60 male’s chromosome22 pairs of autosomala 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 Daughter22(pairs)+XXMother,22+XFather,22+YSon22(pairs)+XYMother,22+X
63 （Ⅲ）The basic knowledge of parentage testing The Mendelian inheritance law is the basis ofpaternity testing.The child inherits 23 chromosomes from themother and another set of 23 chromosomes fromthe biological father.
64 （Ⅲ） The basic knowledge of parentage testing Since, the mother contributes half of the child’snuclear DNA, the father must contribute the otherhalf of the child’s nuclear DNA. To human geneticmarkers, such as STRs, each person’s DNAcontains two copies of these markers.one copyinherited from the father and the other from themother.
65 （Ⅲ） The basic knowledge of parentage testing The laboratory performs a series of DNA tests eachfor a different genetic marker. First, the analystidentifies the alleles that are shared between themother and child, called maternal obligatory genesor maternal alleles. The child’s alleles that are notshared with the mother must come from thetrue biological father.
66 （Ⅲ） The basic knowledge of parentage testing These are called paternal obligatory genes orpaternal alleles. Theoretically, if the allegedfather does not share any one of the paternalalleles with the child, he will be excluded or foundnot to be the child’s biological father.
67 （Ⅲ） The basic knowledge of parentage testing If the alleged father shares all of the paternalalleles with the child, he cannot be excluded andcalculations can be made as to his likelihood ofbeing the father as well as the probability that heand the mother produced a child with the testedcharacteristics.
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 thefact that in the absence of mutation a childreceives one allele matching each parent atevery genetic locus examined.
71 Part Ⅲ Parentage Testing Paternity testing, uses results from a allegedfather ,a mother and a child to answer thequestion if the alleged father could haveFathered the child versus a random man.Who is my father?
72 Part Ⅲ Parentage TestPaternity 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 law2.The GM in Y- chromosome: Paternal inheritance3.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 heterozygousheterozygoushomozygoushomozygous－＋
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 paternalalleles with the child, he cannot be excluded andcalculations can be made as to his likelihood ofbeing 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
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 paternalobligatory genes with the child, he will be excluded or foundnot to be the child’s biological father.
84 （ Ⅰ ）The principle of parentage testing 2 （ Ⅰ ）The principle of parentage testing 2.The GM in Y- chromosome: Paternal inheritanceMale offspring Y - DNA typing must be the same as his father'sMale from one paternal have the same Y-DNA typing.
85 AmpFℓSTR® Y filer Kit Y-STR loci Father Son DYS456 14 DYS389Ⅰ 11 18DYS389Ⅱ26DYS45815DYS19DYS3858DYS3939DYS3917DYS439DYS63520DYS39210GATA-H4DYS437DYS43812DYS448
88 （Ⅰ）The principle of parentage testing 3 （Ⅰ）The principle of parentage testing 3.The GM in mtDNA: Maternal inheritanceThe mtDNA typing of the child is the same as his or her mother.All compatriots from one maternal have the same mtDNA typing.
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-loopSame DNA base sequenceAGTCmotherAAAGTCson
95 Mother,daughter,son have the Same DNA sequence.
97 （Ⅱ）Paternity inclusion M1AF1If 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 ，9D5S ，9D5S ，9Is the father
99 1. Paternity Index (PI) If the man tested cannot be excluded as the biological father of the child in question, thenstatistical calculations are performed to aid inunderstanding the strength of the match. The mostcommonly applied test in this regard is thepaternity index (PI).
100 1.Paternity Index (PI)The paternity index (PI) is the ratio of two conditionalprobabilities where the numerator assumes paternity and thedenominator assumes a random man of similar ethnicbackground was the father. The numerator is the probabilityof observed genotypes, given the tested man is the father,while the denominator is the probability of the observedgenotypes, given that a random man is the father.
101 1. Paternity Index (PI)The paternity index is a likelihood ratio of two probabilitiesconditional upon different competing hypotheses. Thislikelihood ratio reflects how many times more likely it is tosee the evidence under the first hypothesis compared tothe second hypothesis. When mating is random,theprobability that the untested alternative father will transmit aspecific allele to his child is equal to the allele frequency inhis race。
102 Combined Paternity Index (CPI) The PI is calculated for each locus and thenindividual PI values are multiplied together toobtain the combined paternity index (CPI) for theentire 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%
108 CPI:42059.5； RCP：99.99% STR lociI alleged father son paternity index D8S117914，1613，16D21S1132.230，32.2D7S8208，1111，12CSFIPOD3S135815，1615，18D5S8189，11D13S317D16S53912，1310，13D2S133818，2019，20D19S43314，15.214，14.2vWA1414，17D12S39118，2218，19D18S5115，17D6S104317，20FGA2524，25AmelogeninX，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％
111 （Ⅲ）Parentage Exclusion M2AF2If 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.D8S117914，14D8S117914，14D8S117914，16Exclusion
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.
119 STR locialleged fathersonpaternal indexD8S11791412，142.6302D21S1130，32.231，31.20.0008D7S8208，1283.2279CSFIPO10，1211，120.6275D3S135816，18150.0007D5S8181110D13S3178，111.8969D16S5391.8005D2S133824，2519，241.9113D19S43313，1613，140.8568vWA14，1916，170.0001D12S391185.2356D18S5115，210.0015D6S104314，1719FGA22，24242.8736AmelogeninX, YX, 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?