Police seek suspect
CASE:
Police released surveillance
image of the suspect.
Police are looking for a people who robbed a convenience laboratory while
wearing mask early Sunday morning. He killed a medical student who was shot
multiple times in the face and back.
A police press release did not specify the weapons used. Police described the
suspect as a male in his early 20s standing perhaps 6 feet tall with a slim build
and light complexion who wore a mask, gray jeans and black hooded shirt.
"These animals need to be taken off the streets and hanged," sad the security
guard who found the body. The incident, which happened a week after a
schoolgirl was abducted, murdered and her body dumped on a roadside has
drawn widespread condemnation. Investigators took blood samples from 5
suspects.
MISSION:
Your task is to investigate the new crime scene and to compare the samples
with the other ones.
„People lie but evidence doesn’t lie”
Types of evidences:
- indirect (e.g. photo)
- direct (e.g. hair)
- „cold evidence” (hair, textile)
- „hot evidence” (DNA)
Practical tasks:
0. Sample collection on the crime scene
1. DNA extraction
2. DNA amplification (PCR)
3. DNA staining (gel electrophoresis)
4. Analysis of samples
1. task: DNA preparation from lymphocytes
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•
•
•
•
•
•
•
•
•
sample: blood on a piece of textile - cut a half blood patch out and take it in the
Eppendorf-tube
add 1000 ul sterile water to the prepared sample
vortex
incubation at room temperature for a few minutes
vortex
spin with 1500 rpm at 2 min
pipette out 950 ul
add 150 ul of 1% Chelex solution + 50 ul ProtK enzyme to the textile sample
incubate the tube in your hand for a few minutes
vortex
65 Co 2 minutes, ProtK inactivated
store on ice
ORGANIC
SDS, DTT, EDTA
and
Blood
stain
proteinase K
DNA-extraction protocols
Filter Paper
CHELEX
Blood
stain
Apply blood to
paper and allow
stain to dry
Water
INCUBATE (56 oC)
PUNCH
Centrifuge
INCUBATE (ambient)
Phenol,
chloroform,
isoamyl alcohol
Centrifuge
REMOVE supernatant
VORTEX
5%
Chelex
Centrifuge
TRANSFER aqueous (upper) phase
to new tube
REMOVE supernatant
INCUBATE (56 oC)
TE buffer
WASH Multiple Times with
extraction buffer
PCR
Reagents
INCUBATE (100 oC)
Centrifuge
CONCENTRATE sample
(Centricon/Microcon-100 or ethanol
precipitation)
Centrifuge
QUANTITATE
DNA
QUANTITATE
DNA
PERFORM PCR
PERFORM PCR
Figure 3.1, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
(NO DNA QUANTITATION
TYPICALLY PERFORMED WITH
UNIFORM SAMPLES)
PERFORM PCR
2. PCR
Add the following components to 5 ul DNA sample:
• 31,5 ul of water
• 5 ul PCR buffer mix (dNTP included)
• primer 1 (forward) 2.5 ul
• primer 2 (reverse) 2.5 ul
• 2,5 ul MgCl2
finally:
• 1 ul Ultra Fast DNA Polymerase (store on ice)
RUN THE PCR REACTION.
PCR Program: CSI
1. 95 Co 2 minutes
2. 95 Co 10 sec
3. 55 Co 15 sec
4. 72 Co 20 sec
• 30x: step2-step4
• 1 min 72 Co
• 4 Co
(total time: ca. 30 min)
PROBLEM 1.
THE UPPER SEQUENCE IS WILD TYPE (+), THE LOWER IS MUTANT (-).
FIND THE THE MUTATION
WHAT TYPE OF MUTATION IS THIS?
(+) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT.................. CTA GCG CCA TCC CTT CTA ACT
(-) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA..................... CTA GCG CCA TCC CTT CTA ACT
THE UPPER SEQUENCE IS WILD TYPE (+), THE LOWER IS MUTANT (-).
FIND THE THE MUTATION
WHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATION
(+) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT.................. CTA GCG CCA TCC CTT CTA ACT
(-) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA..................... CTA GCG CCA TCC CTT CTA ACT
THE UPPER SEQUENCE IS WILD TYPE (+), THE LOWER IS MUTANT (-).
FIND THE THE MUTATION
WHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATION
WHAT METHOD DO YOU KNOW FOR FINDING POINT MUTATIONS AND SINGLE
NULEOTID POLYMORPHISMS?
(+) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT.................. CTA GCG CCA TCC CTT CTA ACT
(-) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA..................... CTA GCG CCA TCC CTT CTA ACT
THE UPPER SEQUENCE IS WILD TYPE (+), THE LOWER IS MUTANT (-).
FIND THE THE MUTATION
WHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATION
WHAT METHOD DO YOU KNOW FOR FINDING POINT MUTATIONS AND SINGLE
NULEOTID POLYMORPHISMS?
AS-PCR (ALLELESPECIFIC)
(+) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT.................. CTA GCG CCA TCC CTT CTA ACT
(-) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA..................... CTA GCG CCA TCC CTT CTA ACT
THE UPPER SEQUENCE IS WILD TYPE (+), THE LOWER IS MUTANT (-).
FIND THE THE MUTATION
WHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATION
WHAT METHOD DO YOU KNOW FOR FINDING POINT MUTATIONS AND SINGLE
NULEOTID POLYMORPHISMS?
AS-PCR (ALLELESPECIFIC)
DESIGN PRIMERS FOR ALLELESPECIFIC AMPLIFICATION OF THE SEQUENCES
BELOW
(+) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT.................. CTA GCG CCA TCC CTT CTA ACT
(-) CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA..................... CTA GCG CCA TCC CTT CTA ACT
PRIMER ARRANGEMENT
(+) 5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT........... CTA GCG CCA TCC CTT CTA ACT 3’
3’
GGA CTG AAA TCG TAC GGC CCT AGC CAA GAA TTA
GAT CGC GGT AGG GAA GAT TGA 5’
(-) 5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA........... CTA GCG CCA TCC CTT CTA ACT 3’
3’ GGA CTG AAA TCG TAC GGC CCT AGC CAA GAA TTT
GAT CGC GGT AGG GAA GAT TGA 5’
PRIMER SEQUENCES
5’ ATG CCG GGA TCG GTT CTT AAT 3’
5’ AGT TAG AAG GGA TGG CGC TAG 3’
(+) 5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAT........... CTA GCG CCA TCC CTT CTA ACT 3’
3’ GGA CTG AAA TCG TAC GGC CCT AGC CAA GAA TTA
GAT CGC GGT AGG GAA GAT TGA 5’
(-) 5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA........... CTA GCG CCA TCC CTT CTA ACT 3’
3’ GGA CTG AAA TCG TAC GGC CCT AGC CAA GAA TTT
5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA3’
GAT CGC GGT AGG GAA GAT TGA 5’
5’ AGT TAG AAG GGA TGG CGC TAG 3’
PCR PRODUCTS
5’ ATG CCG GGA TCG GTT CTT AAT 3’
5’ AGT TAG AAG GGA TGG CGC TAG 3’
(+) 5’ ATG CCG GGA TCG GTT CTT AAT........... CTA GCG CCA TCC CTT CTA ACT 3’
3’ TAC GGC CCT AGC CAA GAA TTA
GAT CGC GGT AGG GAA GAT TGA 5’
(-) 5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA........... CTA GCG CCA TCC CTT CTA ACT 3’
3’ GGA CTG AAA TCG TAC GGC CCT AGC CAA GAA TTT
5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA3’
GAT CGC GGT AGG GAA GAT TGA 5’
5’ AGT TAG AAG GGA TGG CGC TAG 3’
GEL ELECTROPHORESIS OF THE PCR
PRODUCTS
+/+
-/-
+/-
Capillary electrophoresis of SNPs amplified with ASA
(A)
Sample 1
TT
CT
Sample 2
CC
(B)
CT
CT
TT
CC
TT
(TTTTT)–primer1(chromosome 20)-ddT/ddT
(TTTTT)–(TTTTT)-primer2(chromosome 6)-ddC/ddT
(TTTTT)–(TTTTT)–(TTTTT)-primer3(chromosome 14)-ddC/ddT
(TTTTT)–(TTTTT)–(TTTTT)–(TTTTT)-primer4(chromosome 1)-ddC/ddC
Figure 8.2, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
(A) Simultaneous amplification of three allels on a DNA template
of homologous chromosomes
Deletion of Locus B
Locus C
Locus A
Locus A
Locus C
Locus B
(B) Multiplex PCR products with size-based separation method in
capillary electrophoresis: instead of bands: peaks are the PCR products
(RFU)
A
B
C
How does the
chromatogram appear?
10min
15 min
small
large
50 bp
RFU: relative fluorescence unit
t: time
(t)
70 bp
Modified Figure 4.3 J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
PCR product size (bp)
Figure A7.1, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
(A) Direct comparison of STRs with related objects
DNA profile from
mass disaster victim
DNA profile from
direct reference
(toothbrush believed to have
belonged to the victim)
D5S818
D7S820
D13S317
CSF1PO
D16S539
Penta D
(B) Indirect comparison: kinship analysis
victim
?
D5S818
D13S317
11,13
8,12
11,13
Predicted
victim profile
mass disaster
victim profile
wife
son
D7S820
D16S539
CSF1PO
8,12
8,9
10,12
8,10
wife
8,14
8,9
9,13
10,10
9,10
son
11,? or
?,13
?,14
9,?
?,13
?,10
9,?
victim
(father)
12,13
11,14
9,9
11,13
10,10
9,12
actual
profile
Figure 24.1, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
Penta D
3. Gelelectrophoresis
• add 5 ul PCR sample to the prepared 5 ul
blue loading dye
• load the entire sample (10 ul) into the
GelRed stained agarose gel
• run it for 30 min, with 100V
PRACTICAL TASK: VNTR ANALYSIS
AMPLIFICATION AND ANALYSIS OF THE ”FGA” VNTR LOCUS USED BY THE FBI.
THE REPEATED SEQUENCE UNIT IS 4 bp LONG, ITS SEQUENCE IS: TTTC.
THE NUMBER OF THE REPEATS IS 5-250, THE EXPECTED DNA FRAGMENTS ARE
IN THE 60-1040 bp RANGE.
DNA WAS ISOLATED FROM 5 MEMBERS OF the suspects.
The PCR primers were designed to the flanking sequences at both sides of the repeats.
1
2
3
4
5
Primer P1: gctagtaacggcattaccag
Primer P2: catcgcataagaatttcacg
PRACTICAL TASK: VNTR ANALYSIS
DETERMINE THE REPEAT NUMBERS IN THE AMPLIFIED VNTR ALLELES!
1
2
3
4
5
1900
770
620
760
340
360
336
340
420
340
270
272
180
152
140
120
100
90
THE NUMBER
OF TANDEM
REPEATS:
75
180
58
74
75
75
80
28
25
15
CALCULATION
OF REPEAT
NUMBERS:
Subtract the
lenght of the
primers (40bp)
from the size of
the DNA
fragments,
divide the
remaining by 4.
Explain the basis of paternity testing!
D1 = biological daughter of both parents
D2 = child of mother & former husband
S1 = couple’s biological son
S2 = adopted son
All humans have some VNTRs and
VNTRs come from the genetic information donated by parents
– can have VNTRs from mother, father or a combination
– will not have a VNTR that is from neither parent
PRACTICAL TASK 2.
Evelyn and her daughter Jenna visits the forensic department. Evelyn would like to know
who is the real father of Jenna. The possible fathers are all members of the rock band
Happiness. You perform VNTR analysis and determine the identity of the father. Who is he?
Why?
Why VNTR locus B is peculiar?
VNTR
Evelyn
Jenna
James
Lars
Kirk
Jason
Chris
A
7 & 8
2 & 7
3 & 7
2 & 7
2 & 8
3 & 7
3 & 2
B
4 & 5
6 &
4
7
6
4
6
C
11 &
8
11 & 9
9 & 10
13 & 8
10
9 & 12
9 & 13
D
7 &
19
21 & 7
21 & 7
7 & 9
21 & 7
15 & 8
21 &
15
E
10 & 6
12 & 6
6 & 9
12 & 9
17 &
10
6 & 12
17 &
12
4
PRACTICAL TASK 2.
VNTR
Evelyn
Linda
James
Lars
Kirk
Jason
Chris
A
7 & 8
2 & 7
3 & 7
2 & 7
2 & 8
3 & 7
3 & 2
B
4 & 5
6 &
4
7
6
4
6
4
C
11 &
8
11 & 9
9 & 10
13 & 8
10
9 & 12
9 & 13
D
7 & 19
21 & 7
21 & 7
7 & 9
21 & 7
15 & 8
21 & 15
E
10 & 6
12 & 6
6 & 9
12 & 9
17 & 10
6 & 12
17 & 12
Evelyn and her daughter Jenna visits the forensic department.
Evelyn would like to know who is the real father of Jenna. The possible fathers are all
members of the rock band Happiness. You perform VNTR analysis and determine the
identity of the father. Who is he? CHRIS
Why?
Why VNTR locus B is peculiar? X-LINKED
SNP
SNP
Determine the putative eye and hair colour of the suspect,
if the following SNPs are detected in the sample:
Gene/SNP
HAIR
EYE
Blue/green
Brown
Blond/red
Brown
AA/TT
GG/CC
AA/TT
GG/CC
AA/TT
GG/CC
HERC2
Rs12913832
GG/CC
Oca2
rs1800407
AA/TT
Tyr
Rs1393350
AA/TT
GG/CC
AA/TT
GG/CC
TT/AA
CC/GG
TT/AA
CC/GG
GG/CC
CC/GG
GG/CC
CC/GG
AA/TT
CC/GG
AA/TT
CC/GG
IRF4
rs12203592
SLC24A5
rrs16891982
ExoC2
rs49592270
The HIrisPlex system for simultaneous prediction of hair and eye colour from DNA Forensic Science International: Genetics
Volume 7, Issue 1, January 2013, Pages 98–115
IrisPlex: A Sensitive DNA Tool for Accurate Prediction of Blue and Brown Eye Colour in the Absence of Ancestry Information Journal:
Results of
sequencing:
Rs12913832:
Rs1800407:
Rs1393350:
Rs12203592:
Rrs16891982:
Rs49592270:
AA
GG
GG
CT
CC
AC
Compare the reference samples with the amplified
samples and determine the number of repeats in the
amplified samples.
Reference
Gel
suspects
1
2
3
victim
marker
1900
620
272
336
340
152
360
420
340
270
180
120
140
100
90
1
2
3
Who is probably the perpetrator?
Reference
Gel
suspects
1
2
victim
3
marker
360
500
340
336
272
250
152
120
100
100