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

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

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

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

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

ALLELIC LADDERS

Development of miniSTRs to Aid Testing of Degraded DNA

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

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

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

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

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

*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

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

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

Item D3S1358 D16S539 TH01 TPOX CSF1P0 D7S820 Q1 16,16 10,12 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

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

Frequency = genotype frequency (p2) D3S1358 = 16, 16 (homozygote) Frequency of 16 allele = 0.3071 When same allele: Frequency = genotype frequency (p2) Genotype freq = 0.3071 x 0.3071 = 0.0943 This is the random match probability

Item D3S1358 D16S539 TH01 TPOX CSF1P0 D7S820 Q1 16,16 10,12 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

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

Frequency = 2 X allele 1 freq X allele 2 freq 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

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

13 14 15 16 17 18 19 20 13 0 0 0 1 0 0 1 0 14 1 4 10 10 10 4 0 15 3 7 14 8 4 1 16 11 27 7 3 1 17 11 23 8 1 18 16 6 0 19 3 1 20 0 0 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% NOTE: for the case of the homozygous occurrence (16,16) the frequency of allele 16 is twice the number of individual observations

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= (1-0.069) = .931

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.

h= 151/196 = .7704 H= (1-0.7704) = .2296 PE = (.77)2x(1-2x.7704x[.2296]2) PE = 0.545 Note difference

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

From the observed allele frequencies that we have just calculated 0.005 0.102 0.112 0.201 0.263 0.222 0.084 0.010 0.005 0.005 0.200 0.220 0.394 0.515 0.435 0.165 0.020 0.102 2.039 4.478 8.037 10.516 8.876 3.359 0.400 0.112 2.459 8.825 11.547 9.747 3.688 0.439 0.201 7.919 20.722 17.492 6.619 0.788 0.263 13.557 22.887 8.660 1.031 0.222 9.660 7.310 0.870 0.084 1.383 0.329 0.010 0.020 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

h= 158.96/196 = .811 H= (1-0.811) = .189 PE = (.811)2x(1-2x.811x[.189]2) PE = 0.611

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 14.98 is returned, and this is equal to the c2 statistic

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.