1. Statistical Analysis of DNA
Simple Repeats
Identical length and sequence
agat agat agat agat agat
Compound Repeats
Two or more adjacent simple repeats
agat agat agat ttaa ttaa ttaa
Complex Repeats
Variable unit length & possible intervening seq
agat agat aggat agat agat ttaacggccat agat agat

2. STR NOMENCLATURE Microvariants Alleles that contain incomplete units
aatg aatg aatg aatg aatg aatg aatg aatg aatg aatg - 10
aatg aatg aatg aatg aatg aatg atg aatg aatg aatg - 9.3

3. STRs Used In Forensic Science
Need lots of variation - polymorphic
Overall short segments bp
Can use degraded DNA samples
Segment size usually limits preferential amplification of smaller alleles
Single base resolution
TH01 9.3
TETRANUCLEOTIDE REPEATS
Narrow allele size range - multiplexing
Reduces allelic dropout (stochastic effects)
Use with degraded DNA possible
Reduced stutter rates - easier to interpret mixtures

4. ALLELIC LADDERS Artificial mixture of common alleles
Reference standards
Enable forensic scientists to compare results
Different instruments
Different detection methods
Allele quantities balanced
Produced with same primers as test samples
Commercially available in kits

6. Profiler Plus Allelic Ladders
D3S1358
VWA
FGA
AMEL
D8S1179
D21S11
D18S51
D5S818
D13S317
D7S820

7. ALLELIC LADDERS

8. Development of miniSTRs to Aid Testing of Degraded DNA

9. Power of Discrimination
Same DNA Sample Run with Each of the ABI STR Kits
TH01
Amel
D16S539
D7S820
CSF1PO
TPOX
D3S1358
D18S51
D21S11
D8S1179
D13S317
D5S818
D19S433
D2S1338
FGA
vWA
PCR Product Size (bp)
Power of Discrimination
1:5000
1:410
1:3.6 x 109
1:9.6 x 1010
1:8.4 x 105
1:3.3 x 1012
Profiler Plus
COfiler
SGM Plus
Green I
Profiler
Blue

10. STR LOCI ALLELES TPOX TH01 THYROID PEROXIDASE Chromosome 2 AATG repeat
6 to 13 repeats
TH01
TYROSINE HYDROXYLASE
Chromosome 11
TCTA repeat (Bottom strand)
4 to 11 repeats
Common microvariant 9.3

11. STR LOCI ALLELES vWA D3S1358 von Willebrand Factor Chromosome 12
TCTA with TCTG repeat
10 to 22 repeats
D3S1358
Chromosome 3
AGAT with AGAC repeat
12 to 20 repeats

12. 13 CODIS Core STR Loci with Chromosomal Positions
CSF1PO
D5S818
D21S11
TH01
TPOX
D13S317
D7S820
D16S539
D18S51
D8S1179
D3S1358
FGA
VWA
AMEL

13. Position of Forensic STR Markers on Human Chromosomes
CSF1PO
D5S818
D21S11
TH01
TPOX
D13S317
D7S820
D16S539
D18S51
D8S1179
D3S1358
FGA
VWA
D2S1338
D19S433
13 CODIS Core STR Loci
AMEL
Sex-typing
Penta E
Penta D

14. *Proc. Int. Sym. Hum. ID (Promega) 1997, p. 34
STR Allele Frequencies Exclusions don’t require numbers Matches do require statistics 5 10 15 20 25 30 35 40 45 6 7 8 9
9.3
TH01 Marker
Number of repeats
Frequency
Caucasians (N=427)
Blacks (N=414)
Hispanics (N=414)
*Proc. Int. Sym. Hum. ID (Promega) 1997, p. 34

15. Hardy - Weinberg Equilibrium frequency at one locus
A1A1
A1A2
A2A2 A1 A2
A1A2
A2A2
A1A1
p12
2p1p2
p22
p1p2
p12
freq(A1) = p1
p22
freq(A2) = p2
(p1 + p2 )2 = p12 + 2p1p2 + p22

16. Product Rule frequency at one locus
The frequency of a multi-locus STR profile is the product of the genotype frequencies at the individual loci
ƒ locus1 x ƒ locus2 x ƒ locusn = ƒcombined
Criteria for Use of Product Rule
Inheritance of alleles at one locus have no effect on alleles inherited at other loci

17. Item D3S1358 D16S539 TH01 TPOX CSF1P0 D7S820
Q ,16 10, , , , ,11
Item D3S vWA FGA D8S D21S11 D18S51 D5S818 D13S317 D7S820
Q , , , , , , , , ,11
CoFIler
ProfIler Plus

18. D3S1358 = 16, 16 (homozygote)
Frequency of 16 allele = ??

19. Frequency = genotype frequency (p2)
D3S1358 = 16, 16 (homozygote)
Frequency of 16 allele =
0.3071
When same allele:
Frequency = genotype frequency (p2)
Genotype freq = x =
This is the random match probability

20. Item D3S1358 D16S539 TH01 TPOX CSF1P0 D7S820
Q ,16 10, , , , ,11
Item D3S vWA FGA D8S D21S11 D18S51 D5S818 D13S317 D7S820
Q , , , , , , , , ,11
CoFIler
ProfIler Plus

21. VWA = 15, 17 (heterozygote)
Frequency of 15 allele = ??
Frequency of 17 allele = ??

22. Frequency = 2 X allele 1 freq X allele 2 freq
VWA = 15, 17 (heterozygote)
Frequency of 15 allele =
Frequency of 17 allele =
When heterozygous:
Frequency = 2 X allele 1 freq X allele 2 freq
(2pq)
Genotype freq = 2 x x =
Overall profile frequency =
Frequency D3S1358 X Frequency vWA
x =
This is the combined random match probability

24. Population database
Look up how often each allele occurs at the locus in a population (the “allele” frequency)

26.
Frequency of allele 13 = [(1 + 1)/(196*2)] x 100 = 0.510%
i.e. total # of occurrences / total # of alleles
Frequency of allele 15 = [( )/(196*2)] x 100 = %
NOTE: for the case of the homozygous occurrence (16,16) the frequency
of allele 16 is twice the number of individual observations

27. Match probability (MP) is calculated as the square frequency of the
most common allele to provide the most conservative estimate
of a random match for a given individual.
The power of discrimination (PD) is one minus MP.
MP= (.26276)2= .069
PD= ( ) = .931

28. Heterozygosity is also called the frequency of heterozygotes and is represented by h in the following equation. Where nh is the number of individual observations with two alleles and n is the total number of individuals. Since one is either a homozygote or a heterozygote, the frequency of heterozygotes (h) plus the frequency of homozygotes (H) is equal to one.
The power of exclusion, PE, is defined as the probability of excluding a
random individual from the population as a potential parent based on the genotype of one parent and offspring,. The average for a given locus
is represented by the following equation:
The greater the heterozygosity (h), the greater the value of PE, and the greater the effectivenness of this locus as a means of excluding a
random individual from the population as a potential parent of a given individual.

29. h= 151/196 = .7704
H= ( ) = .2296
PE = (.77)2x(1-2x.7704x[.2296]2)
PE = 0.545
Note difference

30. A B C A B C AC AB BC AA AC AB Can exclude everyone except carriers
The greater the heterozygosity (h), the greater the value of PE, and the greater the effectivenness of this locus as a means of excluding a
random individual from the population as a potential parent of a given individual.
The more heterozygous allele distribution gives less variable allele distribution for offspring, allowing us to exclude more individuals as potential parents. A B C AC AB BC AA A B C AC AB
Can exclude
everyone
except carriers
of allele A

31. From the observed allele frequencies that we have just calculated
From the observed allele frequencies that we have just calculated
a table of expected observations is calculated.
Each entry is calculated as the allele frequency for that pair but the
result must then multiplied by the total number of individuals
When heterozygous:
2 x (allele 1 freq) x ( allele 2 freq) x N = (2pq) x 196
When homozygous:
(allele freq)2 x N = (p)2 x 196

32. h= /196 = .811
H= ( ) = .189
PE = (.811)2x(1-2x.811x[.189]2)
PE = 0.611

33. The c2 test first calculates a c2 statistic using the formula:
where:
Aij = actual frequency in the i-th row, j-th column
Eij = expected frequency in the i-th row, j-th column
r = number or rows
c = number of columns
A low value of c2 is an indicator of independence. As can be seen from the formula, c2 is always positive or 0, and is 0 only if Aij = Eij for every i,j.
CHITEST returns the probability that a value of the c2 statistic at least as high as the value calculated by the above formula could have happened by chance under the assumption of independence.
To find the c2 statistic value for the reported value of p:
Step 1.Select a cell in the work sheet, the location which you like the CHI-SQUARE statistic to appear.
Step 2. From the menus, select insert then click on the Function option, Paste Function dialog box appears.
Step 3.Refer to function category box and choose statistical, from function name box select CHIINV and click on OK.
Step 4.When the CHIINV dialog appears:
Enter the cell containing the p-value (0.9798) and then enter 28 for the degrees of freedom , and finally click on OK.
A value of is returned, and this is equal to the c2 statistic

34. We now have a table of observed and a table of expected values.
To compare the observed values with the expected values a a CHI-SQUARE test is performed
In EXCEL .
Step 1.Select a cell in the work sheet, the location which you like the p value of the CHI-SQUARE to appear.
Step 2. From the menus, select insert then click on the Function option, Paste Function dialog box appears.
Step 3.Refer to function category box and choose statistical, from function name box select CHITEST and click on OK.
Step 4.When the CHITEST dialog appears:
Enter the actual-range and then enter the expected-range , and finally click on OK.
The p-value will appear in the selected cell.
Since the p-value of is greater than the level of
significance (0.05), it fails to reject the null hypothesis.
This verifies the independence of the alleles, as well as
indicating that the the sample used is not statistically
different from the general population.