1. Applications of DNA Profiling

2. Applications of DNA Profiling
Criminal identification
Paternity testing
Mass disaster victim identification
Immigration & inheritance dispute
Ancestry & racial identification
Can you think of anything else???

3. Comparison: Match/ No Match
3 possible outcomes:
Inclusion (Match): Peaks between the compared STR profiles have the same genotypes and no unexplained differences exist between the samples
Exclusion (Non-match): The genotype comparison shows profile differences that can only be explained by the two samples originating from different sources
Inconclusive: The data does not support a conclusion as to whether the profiles match
When utilizing data comparisons with DNA databases that may have data coming from many sources, it is important to recognize that different PCR primer sets can give rise to allele dropout
If any STR locus fails to match when comparing the genotypes between two or more samples, then the comparison of profiles between the questioned and reference sample is usually declared a non-match
Paternity testing is an exception to this ‘Single mismatch leads to exclusion’ rule because of the possibility of mutational events

4. Exclusion Crime case Parentage test
No positive association exists between unknown profile and reference DNA profile
Does not always involve statistics
Crime case
The DNA profile from the suspect did not match with the DNA profile obtained form the crime scene/victim
The suspect is therefore excluded as the possible donor of - Sample deposited at the crime scene/obtained from the victim
No statistical issue
Parentage test
The DNA profile of the disputed child did not match with the DNA profile obtained from Mr. X (Putative/accused father).
The putative/accused father is therefore excluded as being the biological father of child Y, as he lacks the genetic marker that must be contributed to the child by a biological father
Other situation (Sibship/Kinship test)
Not 100% conclusive as parentage test
It gives a probability
Reported as Likelihood Ratio ( Ratio of two probabilities)
Statistics involved

5. Inclusion Crime case Parentage test
Positive association exists between unknown profile and reference DNA profile
Always involves statistics
Crime case
The DNA profile from the suspect matches with the DNA profile obtained form the crime scene/victim
The suspect therefore can not be excluded as the possible donor of- aSample deposited at the crime scene/obtained from the victim
Parentage test
The DNA profile of the disputed child matches with the DNA profile obtained from Mr. X (Putative father).
The putative/accused father is therefore can not be excluded as being the biological father of child Y, as he possess all the genetic marker that must be contributed to the child by a biological father
Other situation (Sibship/Kinship test)
Not 100% conclusive as parentage test
It gives a probability
Reported as Likelihood Ratio ( Ratio of two probabilities)

6. Inconclusive No DNA profile obtained Partial DNA profile
Mixed DNA profile (more than one contributor)

7. Who committed the crime?
Crime Solving
Who committed the crime?
Murder
Theft
Burglary
Robbery
Matching suspect with the evidence

8. Example of Inclusion : Crime Case
Locus name
Crime scene
Suspect 1
Suspect 2
D3S1358
 12, 13
 13, 13
 12, 13
vWA
 14, 15
 15, 17
 14, 15
D16S539
 9, 11
 13, 15
 9, 11
D2S1338
 19, 22
 22, 23
 19, 22
D8S1179
 13, 15
 9, 10
D21S11
 30, 31.2
 30, 32.2
 30, 31.2
D18S51
 7, 13
 11, 13
 7, 13
D19S433
 15, 15
 13, 13.2
 15, 15
TH01
 4, 7
 9, 9.3
 4, 7
FGA
 29, 30
 21, 23
 29, 30
AMEL
X, Y

9. Example of Exclusion : Crime Case
Locus name
Crime scene
Suspect 1
Suspect 2
D3S1358
 12, 13
 13, 13
 11, 13
vWA
 14, 15
 15, 17
 12, 15
D16S539
 9, 11
 13, 15
D2S1338
 19, 22
 22, 23
 20, 22
D8S1179
13, 15 
 9, 10
D21S11
 30, 31.2
 30, 32.2
D18S51
 7, 13
D19S433
 15, 15
 13, 13.2
12, 15 
TH01
 4, 7
 9, 9.3
 7, 7
FGA
 29, 30
 21, 23
AMEL
X, Y

10. Statistics Involved! Since his DNA had a match, he did it !!!
He did it not do it !! His DNA matched by chance
Judge
Prosecution Lawyer
Defense Lawyer
Statistics Involved!

11. Random Probability of Match: Playing Cards
Probability of picking ace of Hearts = 1 52
The possibility of picking up four cards with same combination= × 1 52 × 1 52 × 1 52 =

12. Random Probability of Match
Probability of finding a random individual with the matching DNA Profile OR
Estimated frequency at which a particular DNA profile would be expected to occur in a population (Profile frequency)
What is Profile frequency? GF of locus 1 x GF of locus 2 x GF of locus 3 ……. X GF of locus n (Here, GF = Genotype frequency)
What is Random probability of match?
1/Profile frequency
What is GF? GF = 𝑃 2 if the individual is homozygous at that particular locus GF = 2pq if the individual is heterozygous at that particular locus Here, p = frequency of allele 1 q = frequency of allele 2
What is Allele frequency?
Observed no of allele at a particular locus / 2N Here, N = total number of individuals typed from a particular population

13. Locus
DNA Profile
Genotype frequency
D3S1358
16, 16
(0.30) = 0.090
vWA
15, 18
2 x 0.27 x = 0.075
D16S539
10, 11
2 x 0.19 x = 0.041
D2S1338
16, 19
2 x 0.04 x = 0.02
D8S1179
8, 9
2 x 0.21 x = 0.088
D21S11
24, 25
2 x 0.06 x = 0.011
D18S51
7, 10
2 x 015 x = 0.007
D19S433
13, 14
2 x 0.27 x = 0.14
TH01
4, 7
2 x 0.21 x = 0.05
FGA
25, 27
2 x 0.19 x = 0.091
Profile frequency= 2.39 x Random probability of match = 4 in 10 13

14. Likelihood Ratio
When matching STR profiles are obtained between a suspect (K) and the crime scene evidence (Q), it is necessary to quantify the evidentiary value of this match
Weight of evidence is expressed as Likelihood ratio
Likelihood Ratio involve a comparison of the probabilities of the evidence under two alternative propositions
The suspect matches because he left his biological sample at crime scene (Prosecution’s position)
Someone else committed the crime, the DNA profile matches by chance (Defense’s position)
Likelihood Ratio (LR)= 𝐻𝑝 (𝑃𝑟𝑜𝑠𝑒𝑐𝑢𝑡𝑖𝑜𝑛 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠) 𝐻𝑑 (𝐷𝑒𝑓𝑒𝑛𝑠𝑒 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠)
Hp = 1 (assuming 100% probability)
Hd = 7.66 x 10 −21 (Profile frequency)
(Hd is generally calculated for each locus using Genotype frequency)
Likelihood Ratio (LR)= x 10 −21 = 1.3𝑥10 20
It is 𝟏.𝟑 𝒙 𝟏𝟎 𝟐𝟎 times more likely that the suspect committed the crime

15. Verbal interpretationLR
Verbal predicate
1–10
Limited support for prosecution hypothesis
10–100
Moderate support for prosecution hypothesis
100–1000
Moderately strong support for prosecution hypothesis
1000–10 000
Strong support for the prosecution hypothesis
>10 000
Very strong support for prosecution hypothesis

16. What if …… Allele is not present in the database
Alleles that are extremely rare
What is allele frequency? Observed no of allele at a particular locus / 2N [Here N = total number of individuals typed from a particular population]
Under the current situation, Allele frequency = 5 / 2N (Here N= total number of individuals typed from a particular population)

17. Post conviction DNA testing
Krick Bloodsworth was convicted and sentenced to death for the murder and sexual assault of a 9 year old girl in 1985
The prosecution’s evidence included:
5 separate witnesses testified that they had seen him with the victim just prior to her death
A shoe impression found near the victim matched the size of Bloodsworth
Other circumstantial evidence
In 1993, the Court reviewed his case using DNA analysis
His DNA profile did not match with the semen sample recovered from the victim
He was acquitted from the charges in 1995 nn

18. Parentage Testing Who is the father ? Who is the mother ?
Pregnancy as a result of love affairs
Pregnancy as a result of rape
Pre-marital affairs
Extra-marital affairs
More than 1 mother claims a child
Baby swap in hospital wards
Child dislocated from the parents
Child trafficking

19. Paternity Test Results
2 possible outcomes of the test:
Exclusion:
Obligate parental alleles in the child DO NOT have corresponding alleles in the alleged father
Minimum 3 inconsistencies have to be observed
The tested man is EXCLUDED as the biological father of the child
No statistics involved
Inclusion:
Obligate parental alleles in the child all have corresponding alleles in the alleged father
The tested man CAN NOT BE EXCLUDED as the biological father of the child
Several statistical values are calculated to assess the strength of genetic evidence 19

20. Paternity Test (Inclusion)
Locus
Mother
Child
Alleged Father
D3S1358
15, 17
15, 15
vWA
16, 18
D16S539
9, 11
11, 13
12, 13
D2S1338
24, 26
17, 24
17, 23
D8S1179
14, 15
10, 14
10, 11
D21S11
30, 31.2
31.2, 31.2
31, 31.2
D18S51
14, 17
13, 17
D19S433
14, 14
13, 14
TH01
8, 8
6, 8
6, 9.3
FGA
22, 22.2
22, 24
23.2, 24
Amelogenin
X, X
X, Y

21. Paternity: Inclusion
Both paternal and maternal alleles should be accounted in the child’s DNA profile at all the loci tested
Mother
Tested man
Child 21

22. Paternity Test (Exclusion)
Locus
Mother
Child
Alleged Father
D3S1358
15, 17
15, 15
vWA
16, 18
16 18
D16S539
9, 11
11, 14
12, 13
D2S1338
24, 26
27, 24
17, 23
D8S1179
14, 15
10, 14
10, 11
D21S11
30, 31.2
31.2, 31.2
31, 31.2
D18S51
14, 17
15, 14
13, 14
D19S433
14, 14
TH01
8, 8
10, 8
6, 9.3
FGA
22, 22.2
22, 23
23.2, 24
Amelogenin
X, X
X, Y

23. Paternity: Exclusion Minimum three inconsistencies ! Mother Tested man
Obligate paternal alleles in the child DO NOT have corresponding alleles in the alleged father
Child 23

24. Language of the Paternity Testing
PI: Paternity Index
CPI: Combined Paternity Index
W: Probability of Paternity 24

25. What is Paternity Index (PI)?
PI is generally represented by the formula X/Y, where,
X = Chance of passing the obligate allele form the tested man
Y = Chance of passing the obligate allele from a random man
PI= 𝐶ℎ𝑎𝑛𝑐𝑒 𝑜𝑓 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑜𝑏𝑙𝑖𝑔𝑎𝑡𝑒 𝑎𝑙𝑙𝑒𝑙𝑒 𝑓𝑟𝑜𝑚 𝑡𝑒𝑠𝑡𝑒𝑑 𝑚𝑎𝑛 𝐶ℎ𝑎𝑛𝑐𝑒 𝑜𝑓 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑜𝑏𝑙𝑖𝑔𝑎𝑡𝑒 𝑎𝑙𝑙𝑒𝑙𝑒 𝑓𝑟𝑜𝑚 𝑅𝑎𝑛𝑑𝑜𝑚 𝑚𝑎𝑛

26. There are 16 possible combination of genotypes for a paternity trio26

27. Situation 1 PI= 𝟎.𝟐𝟓 𝟎.𝟓×𝑷𝒄 = 𝟏 𝟐𝑷𝒄 Tested man is the father
If the tested man is the biological father then the child has four possible genotypes:
Therefore, the probability of having a child with BC genotype from tested man and the mother is 0.25
PI= 𝟎.𝟐𝟓 𝟎.𝟓×𝑷𝒄 = 𝟏 𝟐𝑷𝒄
A random man is the father
If a random man is the father, the mother must pass allele B to the child.
The probability of passing the allele B from the mother is 0.5
The probability of passing allele C from a random man is the frequency of that allele in the population ? CD
Tested man AB
Mother BC
Child CZ
Random man C D A AC AD B BC BD

28. Situation 2 PI= 𝟎.𝟓 𝟎.𝟓×𝑷𝒄 = 𝟏 𝑷𝒄 A random man is the father? CC
Tested man AB
Mother AC
Child CZ
Random man
A random man is the father
If a random man is the father, the mother must pass allele A to the child.
The probability of passing the allele A from the mother is 0.5
The probability of passing allele C from a random man is the frequency of that allele in the population
Tested man is the father
If the tested man is the biological father then the child has four possible genotypes:
Therefore, the probability of having a child with AC genotype from tested man and the mother is 0.5
PI= 𝟎.𝟓 𝟎.𝟓×𝑷𝒄 = 𝟏 𝑷𝒄 C A AC B BC

29. Situation 3 PI= 𝟎.𝟐𝟓 𝟎.𝟓𝑷𝑨×𝟎.𝟓𝑷𝑩 = 𝟏 𝑷𝒄 ? AC AB
A random man is the father
If the mother passed on allele A (probability 0.5) the random man would have to pass allele B (probability 0.5 x PB)
Alternatively, if the mother passed on allele B (probability 0.5) the random man would have to pass allele A (probability 0.5 x PA) ? AC
Tested man AB
Mother
Child
Random man
Tested man is the father
If the tested man is the biological father then the child has four possible genotypes:
Therefore, the probability of having a child with AB genotype from tested man and the mother is 0.25
PI= 𝟎.𝟐𝟓 𝟎.𝟓𝑷𝑨×𝟎.𝟓𝑷𝑩 = 𝟏 𝑷𝒄 A C AA AC B AB BC

30. Paternity Index (Trio)GC GM
GTF
Numerator
Denominator PI AA 1 PA
1/PA AB
1/2PA
PA/2 AC BB PB
1/PB
1/2PB BC
PA+PB/2
1/PA+PB
1/2(PA+PB)

31. Paternity Index (Trio)GC GM
GTF
Numerator
Denominator PI AB AC BB
PB/2
1/PB
1/2PB BC BD

32. Paternity Index (Motherless)GC
GTF PI AA
1/PA AB
1/2PA BB
1/2PB
PA+PB/4PAPB AC
1/4PA

33. Paternity Index Formula (Trio)
Locus M C AF
Formula PI
D8S1179
13, 15
13, 13
10, 13
1/2PA
3.0303
D21S11
29, 32.2
30, 32.2
30, 31.2
2.2779
D7S820
7, 8
7, 11
11, 11
1/PA
4.1701
CFS1PO
11, 12
1.5029
D3S1358
16, 18
15, 16
15, 18
1.6281
TH01
9, 9.3
6, 9
6, 7
2.3202
D13S317
9, 11
9, 12
4.6253
D16S539
11, 13
13, 14
8, 14
49.505
D2S1338
17, 23
19, 23
½(PA+PB)
1.6491
D19S433
14.2, 15
15, 15.2
2.4826
vWA
16, 19
16, 17
1.3024
TPOX
8, 11
1.3768
D18S51
15, 15
15, 17
6.8027
D5S818
12, 12
1.5106
FGA
20, 24
23, 24
22, 23
2.5113
Single locus, no null alleles, low mutation rate, codominance 33 33

34. Combined Paternity Index
When multiple genetic systems (loci) are tested, PI is calculated for each system.
If the genetic system is inherited independently, the Combined Paternity Index (CPI) is the product of the system PI’s.
CPI = PI1 x PI2 x PI3 x ………… PIn
Considering the PI values calculated in the earlier table…..
CPI = x x x x x x x x x x x x x x = 70,89,344
CPI > 1000 favors paternity
CPI < 1000 do not favor paternity 34

35. Probability of Paternity
Probability of paternity denoted as “W”
W must be based on ALL the evidence
Genetic: Genetic evidence comes from the DNA paternity testing
Non-genetic: Measure of strength of one’s belief in the hypothesis that tested man is the father. Non-genetic evidence comes from the testimony of mother, tested man and other witnesses. Prior probability of paternity (p) is the strength of one’s belief that the tested man is the father based on the non-genetic evidence
Typical value used allover the world = 0.5 35

36. Calculation of W
W= 𝐶𝑃𝐼×𝑝 {𝐶𝑃𝐼×𝑝+ 1−𝑝 } , CPI = Genetic, P = Non-genetic When p = 0.5, W= 𝐶𝑃𝐼×0.5 {𝐶𝑃𝐼×0.5+ 1−0.5 } = 𝐶𝑃𝐼×0.5 (𝐶𝑃𝐼× ) = 𝐶𝑃𝐼 𝐶𝑃𝐼+1 Considering the PI values calculated in the earlier table, W= 7, 089,344 7, 089, = % 36

37. Verbal interpretationW
Verbal predicate
>99.9
Paternity practically proven
Paternity highly likely
95-99
Paternity very likely
90-95
Paternity likely
80-90
Certain indication of biological paternity
70-80
Formal indication of biological paternity 50
No verbal predicate
5-10
Paternity unlikely
1-5
Paternity very unlikely
<1
Paternity excluded

38. Strength of Evidence can be Expressed in Both Ways
Combined Paternity Index (CPI)
Probability of Paternity (W)
A CPI of > 1000 and
W of > 99.9%
Provides strong evidence in favor of paternity

39. Paternity Index Formula (Trio): Exclusion
Locus M C AF
Formula PI
D8S1179
13, 15
13, 13
10, 13
1/2a
3.0303
D21S11
29, 32.2
30, 32.2
29, 31.2
No match
D7S820
7, 8
7, 11
11, 11
1/a
4.1701
CFS1PO
11, 12
1.5029
D3S1358
16, 18
15, 16
19, 18
TH01
9, 9.3
6, 9
6, 7
2.3202
D13S317
9, 11
10, 12
D16S539
11, 13
13, 14
8, 14
49.505
D2S1338
17, 23
19, 23
½(a+b)
1.6491
D19S433
14.2, 15
15, 15.2
2.4826
vWA
16, 19
17, 17
TPOX
8, 11
1.3768
D18S51
15, 15
15, 17
D5S818
12, 12
1.5106
FGA
20, 24
23, 24
22, 23
2.5113
Minimum three inconsistencies are required to exclude paternity
Single locus, no null alleles, low mutation rate, codominance
Combined Paternity Index (CPI) = 0.000 39 39

40. What if? The number of inconsistencies are less than three?
Consider the possibility of mutation!
Mutation is a rare event
A single or double allele mismatch is due to a gain or loss of repeats caused by replication slippage in paternal or maternal meiosis
Mutation rate for paternal meiosis is higher than maternal meiosis
It has been estimated that approximately 1000 paternal offspring allele transfer would have to be observed before a mutation would be seen

41. Paternity Index Formula (Trio): Exclusion
Locus M C AF
Formula PI
D8S1179
13, 15
13, 13
10, 13
1/2a
3.0303
D21S11
29, 32.2
30, 32.2
30, 31.2
2.2779
D7S820
7, 8
7, 11
11, 11
1/a
4.1701
CFS1PO
11, 12
1.5029
D3S1358
16, 18
15, 16
19, 18
AMPI
TH01
9, 9.3
6, 9
6, 7
2.3202
D13S317
9, 11
9, 12
4.6253
D16S539
11, 13
13, 14
8, 14
49.505
D2S1338
17, 23
19, 23
½(a+b)
1.6491
D19S433
14.2, 15
15, 15.2
2.4826
vWA
16, 19
16, 17
1.3024
TPOX
8, 11
1.3768
D18S51
15, 15
15, 17
D5S818
12, 12
1.5106
FGA
20, 24
23, 24
22, 23
2.5113
Single locus, no null alleles, low mutation rate, codominance 41 41

42. AMPI Average Mutation Paternity Index
AMR = No of mutations per 1000 allele transfers or generations
APE = ℎ 2 (1- 2×h× 𝐻 2 )
Where,
h = heterozygotes
H = homozygotes nn

43. APE Locus Paternal mutation Maternal mutation APE D3S1358 0.001691
0.549
vWA
0.610
D16S539
0.701
D2S1338
0.664
D8S1179
0.729
D21S11
0.692
D18S51
D19S433
TH01
0.583
FGA
0.758

45. Paternity Index Formula (Trio): Exclusion
Locus M C AF
Formula PI
D8S1179
13, 15
13, 13
10, 13
1/2a
3.0303
D21S11
29, 32.2
30, 32.2
30, 31.2
2.2779
D7S820
7, 8
7, 11
11, 11
1/a
4.1701
CFS1PO
11, 12
1.5029
D3S1358
16, 18
15, 16
19, 18
0.0030
TH01
9, 9.3
6, 9
6, 7
2.3202
D13S317
9, 11
9, 12
4.6253
D16S539
11, 13
13, 14
8, 14
49.505
D2S1338
17, 23
19, 23
½(a+b)
1.6491
D19S433
14.2, 15
15, 15.2
2.4826
vWA
16, 19
16, 17
1.3024
TPOX
8, 11
1.3768
D18S51
15, 15
15, 17
0.0035
D5S818
12, 12
1.5106
FGA
20, 24
23, 24
22, 23
2.5113
Single locus, no null alleles, low mutation rate, codominance
Combined Paternity Index (CPI) = 67,20,748 45 45

46. Before Doing the Mutation Calculation
One has to explore the following:
Alternative set of STR markers
Y-Chromosome STRs
X-Chromosome STRs
Mitochondrial DNA

47. Why Mother Included in a Paternity Test?
Locus
Mother
Child
Alleged Father
D3S1358
15, 17
15, 15
vWA
18, 18
16, 18
17, 18
D16S539
9, 14
9, 12
D2S1338
24, 26
17, 24
17, 23
D8S1179
14, 15
10, 14
10, 13
D21S11
30, 31.2
31.2, 31.2
31, 31.2
D18S51
14, 17
13, 17
D19S433
14, 14
TH01
8, 8
6, 8
6, 9
FGA
22, 22.2
22, 24
23.2, 24
Amelogenin
X, X
X, Y
** Exactly the same situation may arise in a maternity test conducted in absence of a father

48. What to Do if The Father/Mother is Not Available?

49. Paternity Test in Absence of Mother
Father-child DNA profiles
50% match found
Y Chromosome profile (if the child is male)
X Chromosome profile (if the child is female)
50% match not found!
Excluded

50. Paternity Test in Absence of Father
Y Chromosome profile from FoAF (Male baby)
Y Chromosome profile from BoAF (Male baby)
X Chromosome profiles from MoAF/ (Female baby)
X Chromosome profiles from DoAF/ (Female baby)

51. Maternity Test in Absence of Father
Excluded
Mother-child DNA profiles
50% match found
50% match not found!
mtDNA Analysis

52. Is the declared relationship authentic?
Immigration Dispute
Is the declared relationship authentic?
Establishing family relationship
Father vs C1-C2-M
Father vs C1-C2 (Mother is not available)
Full sibling (Share both the parents)
Half sibling (Share a common parent)
Alleged grandson to a person
Alleged nephew to a person nn

53. Inheritance Dispute Who gets the money?
Dispute may arise after the death of someone as to who is the legal heir
Non availability of close or first degree relatives
Presence of fake or false claimant(s) nn

54. Identification of Mass Disaster Victims
Who was the victim?

55. Dead Body Identification
Direct comparison of DNA profiles
Comparing DNA profile from the dead body with the DNA profile from the personal effects (e.g. cloths, shaving razor, tooth brush etc.)
Carrying out a paternity test
Kinship analysis
Comparing DNA profile from the dead body with the DNA profiles from close relatives
In motherless situations: C-F-FoM, C-F-MoM, C-F-C2oM etc.
In fatherless situations: C-F-FoF, C-F-MoF, C-F-FoF-MoM, C-F-C2oF etc.
In absence of 1º relatives: Available close relatives ?
Classical paternity test
Reverse paternity test

56. Sibling DNA Testing

57. DNA Sharing between Nearest Kins
Relationship
DNA sharing (%)
Identical twins
100
Parent-child 50
Full siblings
Fraternal twins
Half siblings
Grandparent-grandchild
Uncle-nephew
Aunt-niece 25
First cousins
Great grandparent-great grandchild
Great uncle-great nephew
Great aunt-great niece
12.5

58. Sibling DNA Testing
DNA test to know whether two individuals are biologically related to each other without testing the parents.
Unlike DNA parentage test it is not 100% conclusive and gives a prediction only
Expressed as likelihood of the tested relationship.

59. Sibling DNA Testing Method
Sibling DNA test is carried out in 3 steps
Calculating the Sibling Index at each loci assuming 3 different relationships: A, B and C
A = Tested individuals are full siblings of each other
B = Tested individuals are half-siblings of each other
C = Tested individuals are unrelated
CSI or Combined Sibship Index is calculated by multiplying the SIs across
Calculating the likelihood of the relationship

60. Sibling DNA Testing Allele sharing by tested individuals Genotype A
Full sibling B
Half Sibling C
Unrelated
Share two alleles
Both are pq
1+p+q+2pq
P+q+4pq
8pq
Share two allele doubly
Both are pp
(1+p)2
2p(1+p)
(2p)2
Share one allele doubly
pp and pq
(1+p)
(1+2p) 2p
Share one allele
pq and pr
(1+4p) 8p
Share no allele
pq or pp vs.
rs or rr 1 2 4 nn

61. Sibling Index Locus Child-1 Child-2 SI for A SI for B SI for C SI 1A
15, 15
15, 16
SI 1A
SI 1B
SI 1C
vWA
18, 19
16, 18
SI 2A
SI 2B
SI 2C
D16S569
12, 12
9, 12
SI 3A
SI 3B
SI 3C
D2S1338
21, 25
SI 4A
SI 4B
SI 4C
D8S1179
13, 15
SI 5A
SI 5B
SI 5C
D21S11
29, 32.2
29, 30
SI 6A
SI 6B
SI 6C
D18S51
13, 16
13, 17
SI 7A
SI 7B
SI 7C
D19S433
12.2, 12.2
13.2, 15.2
SI 8A
SI 8B
SI 8C
TH01
9, 10
6, 9.3
SI 9A
SI 9B
SI 9C
FGA
19, 24
19, 22.2
SI 10A
SI 10B
SI 10C
Combined Sibship Index (CSI)
85.5
12.7
0.02
Combine Likelihood
4,242
634
88.5

62. Supplementary DNA Tests
Other STR markers
Lineage Markers
Y-Chromosome STRs
X-Chromosome STRs
mtDNA SNP Analysis
Autosomal DNA markers, such as the 13 core short tandem repeat (STR) loci, are shuffled with each generation because half of an individual’s genetic information comes from his or her father and half from his or her mother
Y chromosome (ChrY) and mitochondrial DNA (mtDNA) markers that represent lineage markers, Tare passed down from generation to generation without changing (Except for mutational events)
Maternal lineages can be traced with mitochondrial DNA sequence information, whereas paternal lineages can be followed with Y-chromosome markers

63. Y-Chromosome STR Analysis

64. Facts About Y-chromosome STRs
Analytical output gives a haplotype rather than genotype
Transmitted to male descendants unchanged (except the event of mutation)
Useful only when tested individuals are male
Do not recombine with dissimilar X chromosome almost its entire length
All paternal relatives therefore share the same Y-haplotype

65. Y-Chromosome STR Analysis
Locus
Child-1
Child-2
Child-3
Child-4
DYS456 16 17
DYS389I 14 13
DYS390 25 24
DYS389II 32 30 31
DYS458 15 18
DYS19
DYS385
11, 14
12, 14
DYS393
DYS391 10 11
DYS439
DYS635 23
DYS392
Y GATA H4 12
DYS437
DYS438
DYS448 19 20 21

66. Conclusion: Y-chromosome STR
If two profiles do not match then individual under suspicion can be excluded
If two profile matches there are several possibilities:
Tested individuals are Father-Son
Tested individuals are Brother-Brother
Any paternally related individual

67. X-Chromosome STR Analysis

68. Facts About X-Chromosome STRs
Analytical output gives a haplotype for males and genotype for females
In females, the two X chromosome behaves like autosome and recombine with each other
In females, X-Chromosome is present as homologous pair
Resemble autosome in this respect
Recombine with each other during meiosis
Like autosomal STRs, X-Chromosome STRs have been used in forensics under certain situations

69. X-Chromosome STRs: Use
In solving deficiency cases
Father-Daughter relationship (Motherless situation)
Mother-Son relationship (Fatherless situation)
Disputed child-MoAF (Proxy DNA test)

70. X-Chromosome STR Analysis
Locus
Father
Mother
Child
DXS8378 10
10-12
12-13
DXS9898 12
13-14
12-14
DSX8377 50
49-53
52-53
HPRTB 14
13-13
GATA172D05 9
8-8
8-9
DXS7423
14-15
14-14
DXS6809 33
33-33
32-33
DXS7132 15
13-15
DXS101 26
26-30
26-26
DXS6789
17-20
17-21
DXS9902
11-11
DXS6807 13
11-14
DXS7424
15-16
14-16

71. Applications Deficiency paternity/maternity testing of a female child
Father – daughter relationship
Mother - son relationship XY ? XX XY

72. Applications
If two alleged fathers are father-son ? XY XX

73. Proxy DNA Test Mother-Child vs. Mother of alleged father XX XY
Alleged father not participating XY XX

74. Proxy DNA Test Mother-Child vs. Daughter of alleged father
Alleged father not participating XX XY

75. Supplementary Sibling/Half-sibling TestXY XX
Sisters sharing same father
Half-sisters sharing same father

76. Amelogenin
Amelogenin gene encodes for protein in tooth enamel: Gene on X chromosome
But also on the part of the Y chromosome that is homologous to X chromosome: Pseudo Autosomal Region (PAR)
Therefore this gene is actually on both X and Y chromosomes
X chromosome has 6 bp deletion and Y chromosome doesn’t

77. Additional Markers
There are additional markers that are commonly used for Forensic other than these 14

78. SNP
Formed when errors (Mutations) occur in DNA replication during meiosis
Some regions of the genome are richer in SNPs than others, chromosome 1 contains 1 SNP/1.45 kb compared with chromosome 19, where 1 SNP/2.18 kb
Just two alleles, not highly polymorphic
Computational analyses have shown that on average SNP loci are needed to yield equivalent random match probabilities as the 13 core STR loci
SNP typing:
Restriction digestion: Laborious, inconclusive, too much DNA needed
Sanger sequencing: For mtDNA
Primer extension
Allele specific hybridization

80. mtDNA DNA results from bones, hair, and teeth
Greater number of copies per cell
Analysis involves DNA sequencing
The process involves the polymerase incorporation of dideoxyribonucleotide triphosphates (ddNTPs) as chain terminators followed by a separation step capable of single-nucleotide resolution

81. mtDNA Analysis
All maternally related individuals share same mtDNA

82. mtDNA Analysis
Mother
Female
Male

85. FBI A1 (L15997)
GAAAAAGTCT TTAACTCCAC CATTAGCACC CAAAGCTAAG ATTCTAATTT AAACTATTCT
CTTTTTCAGA AATTGAGGTG GTAATCGTGG GTTTCGATTC TAAGATTAAA TTTGATAAGA
CTGTTCTTTC ATGGGGAAGC AGATTTGGGT ACCACCCAAG TATTGACTCA CCCATCAACA
GACAAGAAAG TACCCCTTCG TCTAAACCCA TGGTGGGTTC ATAACTGAGT GGGTAGTTGT
ACCGCTATGT ATTTCGTACA TTACTGCCAG CCACCATGAA TATTGTACGG TACCATAAAT
TGGCGATACA TAAAGCATGT AATGACGGTC GGTGGTACTT ATAACATGCC ATGGTATTTA
ACTTGACCAC CTGTAGTACA TAAAAACCCA ATCCACATCA AAACCCCCTC CCCATGCTTA
TGAACTGGTG GACATCATGT ATTTTTGGGT TAGGTGTAGT TTTGGGGGAG GGGTACGAAT
CAAGCAAGTA CAGCAATCAA CCCTCAACTA TCACACATCA ACTGCAACTC CAAAGCCACC
GTTCGTTCAT GTCGTTAGTT GGGAGTTGAT AGTGTGTAGT TGACGTTGAG GTTTCGGTGG
CCTCACCCAC TAGGATACCA ACAAACCTAC CCACCCTTAA CAGTACATAG TACATAAAGC
GGAGTGGGTG ATCCTATGGT TGTTTGGATG GGTGGGAATT GTCATGTATC ATGTATTTCG
CATTTACCGT ACATAGCACA TTACAGTCAA ATCCCTTCTC GTCCCCATGG ATGACCCCCC
GTAAATGGCA TGTATCGTGT AATGTCAGTT TAGGGAAGAG CAGGGGTACC TACTGGGGGG
TCAGATAGGG GTCCCTTGAC CACCATCCTC CGTGAAATCA ATATCCCGCA CAAGAGTGCT
AGTCTATCCC CAGGGAACTG GTGGTAGGAG GCACTTTAGT TATAGGGCGT GTTCTCACGA
Revised Cambridge Reference Sequence (rCRS) – formerly known as the “Anderson” sequence
HV1
Hypervariable Region I
HV1
C-stretch
342 bp examined
Figure 10.6
Adapted from Figure 10.6, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
HV1
FBI B1 (H16391)

86. FBI C1 (L048)
Revised Cambridge Reference Sequence (rCRS) – formerly known as the “Anderson” sequence
GATCACAGGT CTATCACCCT ATTAACCACT CACGGGAGCT CTCCATGCAT TTGGTATTTT
CTAGTGTCCA GATAGTGGGA TAATTGGTGA GTGCCCTCGA GAGGTACGTA AACCATAAAA
CGTCTGGGGG GTATGCACGC GATAGCATTG CGAGACGCTG GAGCCGGAGC ACCCTATGTC
GCAGACCCCC CATACGTGCG CTATCGTAAC GCTCTGCGAC CTCGGCCTCG TGGGATACAG
GCAGTATCTG TCTTTGATTC CTGCCTCATC CTATTATTTA TCGCACCTAC GTTCAATATT
CGTCATAGAC AGAAACTAAG GACGGAGTAG GATAATAAAT AGCGTGGATG CAAGTTATAA
ACAGGCGAAC ATACTTACTA AAGTGTGTTA ATTAATTAAT GCTTGTAGGA CATAATAATA
TGTCCGCTTG TATGAATGAT TTCACACAAT TAATTAATTA CGAACATCCT GTATTATTAT
ACAATTGAAT GTCTGCACAG CCACTTTCCA CACAGACATC ATAACAAAAA ATTTCCACCA
TGTTAACTTA CAGACGTGTC GGTGAAAGGT GTGTCTGTAG TATTGTTTTT TAAAGGTGGT
AACCCCCCCT CCCCCGCTTC TGGCCACAGC ACTTAAACAC ATCTCTGCCA AACCCCAAAA
TTGGGGGGGA GGGGGCGAAG ACCGGTGTCG TGAATTTGTG TAGAGACGGT TTGGGGTTTT
ACAAAGAACC CTAACACCAG CCTAACCAGA TTTCAAATTT TATCTTTTGG CGGTATGCAC
TGTTTCTTGG GATTGTGGTC GGATTGGTCT AAAGTTTAAA ATAGAAAACC GCCATACGTG
TTTTAACAGT CACCCCCCAA CTAACACATT ATTTTCCCCT CCCACTCCCA TACTACTAAT
AAAATTGTCA GTGGGGGGTT GATTGTGTAA TAAAAGGGGA GGGTGAGGGT ATGATGATTA
HV2
Hypervariable Region II
HV2
268 bp examined
73-340
Figure 10.6
HV2
C-stretch
FBI D1 (H408)
Adapted from Figure 10.6, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press

87. mtDNA Analysis Result
The DNA sequence data is compared between the known and questioned samples
The DNA sequence must match in its entirety to be called a match
The result may be reported in terms of their variation from Anderson sequence (rCRS)
Useful when nuclear DNA analysis fails
When DNA molecules are highly degraded
Establishing maternal lineage

88. Differences from Reference Sequence
mtDNA sequences from tested samples are aligned with the reference rCRS sequence (e.g., positions )
ACCGCTATGT ATTTCGTACA TTACTGCCAG CCACCATGAA TATTGTACGG TACCATAAAT
rCRS
ACCGCTATGT ATCTCGTACA TTACTGCCAG CCACCATGAA TATTGTACAG TACCATAAAT Q K
16093
16129
Sample Q
16093C
16129A
Sample K
Differences are reported by the position and the nucleotide change (compared to the rCRS)

89. Now What? Questioned GCATATTGCGCCTA GCATATTGCGCCTA
Compare sequences of questioned items to known sequences
Are they different?
Case #1 Case #2
Questioned GCATATTGCGCCTA GCATATTGCGCCTA
Known GCACATTACGTCTA GCATATTGCGCCTA
*EXCLUSION *CANNOT EXCLUDE
*This is a simplification, the regions scrutinized are larger, and
analysis is far more complicated

90. Non Human DNA Testing Cats: ‘ MeowPlex’ with 11 STRs Dogs: STR+ mtDNA
Species identification: mtDNA regions- cytochrome b gene, 16S rRNA
Plants: Linking plant material to suspects or victims & Linking marijuana to aid in forensic drug investigations
Bacterial DNA in soil: Forensic soil evaluation
Wildlife DNA testing nn

91. References Fundamentals of Forensic DNA Typing- John M. Butler
Forensic DNA Typing, Biology, Technology, and Genetics of STR Markers- John M. Butler- 2nd edition
An Introduction to Forensic Genetics- William Goodwin