Transcript No Slide Title

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 various
samples amplified with primers targeted
to a particular STR locus
Polyacrylamide Gel
Combine
Re-amplify
Find representative alleles
spanning population variation
Profiler Plus Allelic Ladders
VWA
D3S1358
AMEL
D8S1179
D5S818
FGA
D21S11
D13S317
D18S51
D7S820
ALLELIC LADDERS
Development of miniSTRs to Aid Testing of
Degraded DNA
Same DNA Sample Run with
Each of the ABI STR Kits
PCR Product Size (bp)
D3S1358
vWA
FGA
Blue
TH01
Amel
Power of Discrimination
CSF1PO
TPOX
Green I
TH01
D13S317
D3S1358
Amel
D5S818 vWA TPOX FGA
D8S1179 vWA
D13S317
Amel
D3S1358 D5S818
D21S11 FGA
D3S1358
Amel
TH01
TPOX
D3S1358
vWA
Amel
D8S1179 TH01
D21S11
D19S433
1:5000
1:410
CSF1PO
D7S820
D7S820
D18S51
D7S820
CSF1PO
D16S539
D16S539 D18S51
D2S1338
FGA
Profiler
1:3.6 x 109
Profiler Plus
1:9.6 x 1010
COfiler
1:8.4 x 105
SGM Plus
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
TPOX
D3S1358
D8S1179
D5S818
FGA
CSF1PO
TH01
VWA
D7S820
AMEL
D13S317
D16S539
D18S51
D21S11
AMEL
Position of Forensic STR
Markers on
TPOX
Human
D3S1358
Chromosomes
D5S818
D2S1338
FGA
CSF1PO
D8S1179
D13S317
D16S539
VWA
D7S820
13 CODIS Core STR Loci
Penta E
TH01
AMEL
Sex-typing
D19S433
D18S51
D21S11
Penta D
AMEL
STR Allele Frequencies
Exclusions don’t require numbers
Matches do require statistics
45
40
TH01 Marker
Frequency
35
30
Caucasians (N=427)
Blacks (N=414)
Hispanics (N=414)
25
20
15
10
*Proc. Int. Sym. Hum. ID
5
(Promega) 1997, p. 34
0
6
7
8
9
9.3
Number of repeats
10
Hardy - Weinberg Equilibrium
frequency at one locus
A1A1 A1A2 A2A2
p12 2p1p2 p22
freq(A1) = p1
freq(A2) = p2
A1
p12
A1 A 1 A 1
p1p2
A2 A 1 A 2
A2
p1p2
A1A 2
p22
A2 A 2
(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
ProfIler Plus
Item
D3S1358 vWA
Q1
16,16
15,17
FGA
D8S1179 D21S11 D18S51 D5S818 D13S317 D7S820
21,22 13,13
29,30
16,20
8,12
12,12
8,11
CoFIler
Item
Q1
D3S1358
16,16
D16S539
10,12
TH01
8,9.3
TPOX
CSF1P0
9,10
12,12
D7S820
8,11
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)
Genotype freq = 0.3071 x 0.3071 = 0.0943
This is the random match probability
ProfIler Plus
Item
D3S1358 vWA
Q1
16,16
15,17
FGA
D8S1179 D21S11 D18S51 D5S818 D13S317 D7S820
21,22 13,13
29,30
16,20
8,12
12,12
8,11
CoFIler
Item
Q1
D3S1358
16,16
D16S539
10,12
TH01
8,9.3
TPOX
CSF1P0
9,10
12,12
D7S820
8,11
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 = 0.00817
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
14
0
1
15
0
4
3
16
1
10
7
11
17
0
10
14
27
11
18
0
10
8
7
23
16
19
1
4
4
3
8
6
3
0
20
0
0
1
1
1
0
1
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 alleles
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.
13
14
15
16
17
18
19
20
13
0
14
0
1
15
0
4
3
16
1
10
7
11
17
0
10
14
27
11
18
0
10
8
7
23
16
h= 151/196 = .7704
H= (1-0.7704) = .2296
PE = (.77)2x(1-2x.7704x[.2296]2)
PE = 0.545
Note difference
19
1
4
4
3
8
6
3
0
20
0
0
1
1
1
0
1
0
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.
Can exclude
everyone
except carriers
of allele A
A
B
A
AA
AB
C
AC
BC
A
A
B
AB
AB
C
AC
AC
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
0.005 0.102 0.112
0.005 0.005 0.200 0.220
0.102
2.039 4.478
0.112
2.459
0.201
0.263
0.222
0.084
0.010
0.201
0.394
8.037
8.825
7.919
0.263 0.222 0.084
0.515 0.435 0.165
10.516 8.876 3.359
11.547 9.747 3.688
20.722 17.492 6.619
13.557 22.887 8.660
9.660 7.310
1.383
h= 158.96/196 = .811
H= (1-0.811) = .189
PE = (.811)2x(1-2x.811x[.189]2)
PE = 0.611
0.010
0.020
0.400
0.439
0.788
1.031
0.870
0.329
0.020
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.