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 TH01 9.3 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 - 100-400 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 D3S1358FGA VWA AMELD8S1179 D21S11 D18S51 D5S818 D13S317 D7S820

7. ALLELIC LADDERS

8. Development of miniSTRs to Aid Testing of Degraded DNA

9. Profiler Plus  COfiler  SGM Plus  Green I Profiler  Blue TH01 Amel D16S539 D7S820 CSF1PO TPOX D3S1358 D16S539 D18S51 D21S11 Amel D3S1358 D18S51 D21S11 D8S1179 D7S820 D13S317 D5S818 D19S433 D2S1338 FGA vWA FGA TH01 D3S1358 vWA FGA D7S820 D5S818 D13S317 TH01 CSF1PO TPOX D8S1179 vWA TH01 CSF1PO TPOX Amel FGA D3S1358 Amel PCR Product Size (bp) Same DNA Sample Run with Each of the ABI STR Kits Power of Discrimination 1:5000 1:410 1:3.6 x 10 9 1:9.6 x 10 10 1:8.4 x 10 5 1:3.3 x 10 12

10. STR LOCI ALLELES TPOX –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 –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 D16S539D18S51 D8S1179 D3S1358 FGA VWA AMEL

13. CSF1PO D5S818 D21S11 TH01 TPOX D13S317 D7S820 D16S539D18S51 D8S1179 D3S1358 FGA VWA 13 CODIS Core STR Loci AMEL Sex-typing Position of Forensic STR Markers on Human Chromosomes Penta E Penta D D2S1338 D19S433

14. STR Allele Frequencies Exclusions don’t require numbers Matches do require statistics Caucasians (N=427) Blacks (N=414) Hispanics (N=414) TH01 Marker * Proc. Int. Sym. Hum. ID (Promega) 1997, p. 34 Number of repeats Frequency

15. Hardy - Weinberg Equilibrium frequency at one locus A1A1 A2A2 A1A1 A2A2 A1A2A1A2 A1A2A1A2 A2A2A2A2 A1A1A1A1 A1A1A1A1 A1A2A1A2 A2A2A2A2 freq(A 1 ) = p 1 freq(A 2 ) = p 2 p12p12 p22p22 2p 1 p 2 p12p12 p1p2p1p2 p1p2p1p2 p22p22 (p 1 + p 2 ) 2 = p 1 2 + 2p 1 p 2 + p 2 2

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 locus 1 x ƒ locus 2 x ƒ locus n = ƒ combined Criteria for Use of Product Rule Inheritance of alleles at one locus have no effect on alleles inherited at other loci

17. ItemD3S1358 D16S539 TH01 TPOX CSF1P0 D7S820 Q1 16,1610,12 8,9.3 9,10 12,12 8,11 Item D3S1358 vWA FGA D8S1179 D21S11 D18S51 D5S818 D13S317 D7S820 Q1 16,16 15,17 21,22 13,13 29,30 16,20 8,12 12,12 8,11 CoFIler ProfIler Plus

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

19. D3S1358 = 16, 16 (homozygote) Frequency of 16 allele = 0.3071 When same allele: Frequency = genotype frequency (p 2 ) (for now!) Genotype freq = 0.3071 x 0.3071 = 0.0943 This is the random match probability

20. ItemD3S1358 D16S539 TH01 TPOX CSF1P0 D7S820 Q1 16,1610,12 8,9.3 9,10 12,12 8,11 Item D3S1358 vWA FGA D8S1179 D21S11 D18S51 D5S818 D13S317 D7S820 Q1 16,16 15,17 21,22 13,13 29,30 16,20 8,12 12,12 8,11 CoFIler ProfIler Plus

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

22. VWA = 15, 17 (heterozygote) Frequency of 15 allele = 0.2361 Frequency of 17 allele = 0.1833 When heterozygous: Frequency = 2 X allele 1 freq X allele 2 freq (2pq) Genotype freq = 2 x 0.2361 x 0.18331 = 0.0866 Overall profile frequency = Frequency D3S1358 X Frequency vWA 0.0943 x 0.0866 = 0.00817 This is the combined random match probability

25. 1314151617181920 1300010010 141410101040 153714841 161127731 17112381 181660 1931 2000 Frequency of allele 13 = [(1 + 1)/(196*2)] x 100 = 0.510% i.e. total # of occurrences / total # of alleles Frequency of allele 15 = [(4+6+7+14+8+4+1)/(196*2)] x 100 = 11.224% i.e. total # of occurrences / total # of alleles NOTE: for the case of the homozygous occurrence (16,16) the frequency of allele 16 is twice the number of individual observations

26. 0.0050.1020.1120.2010.2630.2220.0840.010 0.0050.0050.2000.2200.3940.5150.4350.1650.020 0.1022.0394.4788.03710.5168.8763.3590.400 0.1122.4598.82511.5479.7473.6880.439 0.2017.91920.72217.4926.6190.788 0.26313.55722.8878.6601.031 0.2229.6607.3100.870 0.0841.3830.329 0.0100.020 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

27. 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 0.9798 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.

28. The  2 test first calculates a  2 statistic using the formula: where: A ij = actual frequency in the i-th row, j-th column E ij = expected frequency in the i-th row, j-th column r = number or rows c = number of columns A low value of  2 is an indicator of independence. As can be seen from the formula,  2 is always positive or 0, and is 0 only if A ij = E ij for every i,j. CHITEST returns the probability that a value of the  2 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  2 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 14.98 is returned, and this is equal to the  2 statistic