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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
• 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
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
Allelic Ladder Formation
Separate PCR products from varioussamples amplified with primers targetedto a particular STR locus
Combine
Re-amplify
Find representative allelesspanning population variation
Polyacrylamide Gel
Profiler Plus Allelic Ladders
D3S1358 FGAVWA
AMEL D8S1179 D21S11 D18S51
D5S818 D13S317D7S820
ALLELIC LADDERS
Development of miniSTRs to Aid Testing of Degraded DNA
Profiler Plus
COfiler
SGM Plus
Green I
Profiler
Blue
TH01
Amel D16S539
D7S820
CSF1POTPOX
D3S1358
D16S539 D18S51D21S11
Amel
Amel
D3S1358
D3S1358
D18S51D21S11
D8S1179
D7S820
D13S317
D5S818
D19S433 D2S1338
FGA
vWA
vWA
FGA
TH01
D3S1358 vWA FGA
D7S820D5S818D13S317
TH01CSF1POTPOX
D8S1179
vWATH01 CSF1PO
TPOXAmel FGAD3S1358
Amel
PCR Product Size (bp) Same DNA Sample Run with Each of the ABI STR Kits
Power of Discrimination1:5000
1:410
1:3.6 x 109
1:9.6 x 1010
1:8.4 x 105
1:3.3 x 1012
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
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
13 CODIS Core STR Loci with Chromosomal Positions
CSF1PO
D5S818
D21S11
TH01
TPOX
D13S317
D7S820
D16S539 D18S51
D8S1179
D3S1358
FGA
VWA
AMEL
AMEL
CSF1PO
D5S818
D21S11
TH01
TPOX
D13S317
D7S820
D16S539 D18S51
D8S1179
D3S1358
FGA
VWA
13 CODIS Core STR Loci AMEL
AMEL
Sex-typing
Position of Forensic STR Markers on
Human Chromosomes
Penta E
Penta D
D2S1338
D19S433
STR Allele Frequencies Exclusions don’t require numbers Matches do require statistics
0
5
10
15
20
25
30
35
40
45
6 7 8 9 9.3 10
Caucasians (N=427)
Blacks (N=414)
Hispanics (N=414)
TH01 Marker
*Proc. Int. Sym. Hum. ID (Promega) 1997, p. 34
Number of repeats
Fre
qu
ency
Hardy - Weinberg Equilibrium frequency at one locus
A1 A2
A1
A2 A1A2
A1A2
A2A2
A1A1
A1A1 A1A2 A2A2
freq(A1) = p1
freq(A2) = p2
p12 p2
22p1p2
p12 p1p2
p1p2 p22
(p1 + p2 )2 = p12 + 2p1p2 + p2
2
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 = ??
D3S1358 = 16, 16 (homozygote)
Frequency of 16 allele = 0.3071
When same allele:
Frequency = genotype frequency (p2)(for now!)
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 = ??
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 = 0.00817This is the combined random match probability
13 14 15 16 17 18 19 2013 0 0 0 1 0 0 1 014 1 4 10 10 10 4 015 3 7 14 8 4 116 11 27 7 3 117 11 23 8 118 16 6 019 3 120 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%i.e. total # of occurrences / total # of allelesNOTE: for the case of the homozygous occurrence (16,16) the frequency of allele 16 is twice the number of individual observations
0.005 0.102 0.112 0.201 0.263 0.222 0.084 0.0100.005 0.005 0.200 0.220 0.394 0.515 0.435 0.165 0.0200.102 2.039 4.478 8.037 10.516 8.876 3.359 0.4000.112 2.459 8.825 11.547 9.747 3.688 0.4390.201 7.919 20.722 17.492 6.619 0.7880.263 13.557 22.887 8.660 1.0310.222 9.660 7.310 0.8700.084 1.383 0.3290.010 0.020
From the observed allele frequencies that we have just calculateda 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
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 performedIn 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 asindicating that the the sample used is not statisticallydifferent from the general population.
The 2 test first calculates a 2 statistic using the formula: where:
Aij = actual frequency in the i-th row, j-th columnEij = expected frequency in the i-th row, j-th columnr = number or rowsc = number of columnsA 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 Aij = Eij 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
