police seek suspect police are looking for a people who robbed a convenience laboratory while...
TRANSCRIPT
Police seek 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. Your task is to investigate the new crime scene and to compare the samples
with the other ones.
MISSION:
CASE:
Police released surveillance image of the suspect.
„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
• 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 Filter Paper
CHELEXBlood stain
PUNCH
WASH Multiple Times with extraction buffer
PERFORM PCR
PCR Reagents
SDS, DTT, EDTA and
proteinase K
INCUBATE (56 oC)
Phenol,chloroform,
isoamyl alcohol
QUANTITATE DNA
Apply blood to paper and allow
stain to dryBlood stain
VORTEX
(NO DNA QUANTITATION TYPICALLY PERFORMED WITH
UNIFORM SAMPLES)
Water
INCUBATE (ambient)
5% Chelex
INCUBATE (100 oC)
REMOVE supernatant
INCUBATE (56 oC)
QUANTITATE DNA
PERFORM PCRPERFORM PCR
Centrifuge
Centrifuge
Centrifuge
Centrifuge
REMOVE supernatantTRANSFER aqueous (upper) phase to new tube
CONCENTRATE sample (Centricon/Microcon-100 or ethanol
precipitation)
Centrifuge
TE buffer
Figure 3.1, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
DNA-extraction protocols
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)
(+) 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
PROBLEM 1.
THE UPPER SEQUENCE IS WILD TYPE (+), THE LOWER IS MUTANT (-). FIND THE THE MUTATIONWHAT 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 MUTATIONWHAT 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 MUTATIONWHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATIONWHAT 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 MUTATIONWHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATIONWHAT 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 MUTATIONWHAT TYPE OF MUTATION IS THIS? MISSENSE POINT MUTATIONWHAT METHOD DO YOU KNOW FOR FINDING POINT MUTATIONS AND SINGLE NULEOTID POLYMORPHISMS?
AS-PCR (ALLELESPECIFIC)
DESIGN PRIMERS FOR ALLELESPECIFIC AMPLIFICATION OF THE SEQUENCES BELOW
(+) 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 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’
5’ ATG CCG GGA TCG GTT CTT AAT 3’
5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA3’ 5’ AGT TAG AAG GGA TGG CGC TAG 3’
PRIMER SEQUENCES
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 GAT CGC GGT AGG GAA GAT TGA 5’
5’ ATG CCG GGA TCG GTT CTT AAT 3’
5’ CCT GAC TTT AGC ATG CCG GGA TCG GTT CTT AAA3’ 5’ AGT TAG AAG GGA TGG CGC TAG 3’
PCR PRODUCTS
5’ AGT TAG AAG GGA TGG CGC TAG 3’
+/+ -/- +/-
GEL ELECTROPHORESIS OF THE PCR PRODUCTS
TT CT
CC
CT
CT
CC
TT TT
(A)
(TTTTT)–(TTTTT)-primer2(chromosome 6)-ddC/ddT
(TTTTT)–primer1(chromosome 20)-ddT/ddT
(TTTTT)–(TTTTT)–(TTTTT)-primer3(chromosome 14)-ddC/ddT
(TTTTT)–(TTTTT)–(TTTTT)–(TTTTT)-primer4(chromosome 1)-ddC/ddC
(B)
Sample 1
Sample 2
Figure 8.2, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
Capillary electrophoresis of SNPs amplified with ASA
(A) Simultaneous amplification of three allels on a DNA templateof homologous chromosomes
Locus A Locus CLocus B
(B) Multiplex PCR products with size-based separation method in capillary electrophoresis: instead of bands: peaks are the PCR products
A CB
small large
Modified Figure 4.3 J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
Locus A Locus C
(t)
(RF
U)
RFU: relative fluorescence unitt: time
50 bp 70 bp
10min 15 min
Deletion of Locus B
How does the chromatogram appear?
PCR product size (bp)
Figure A7.1, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
DNA profile from mass disaster victim
DNA profile from direct reference
(toothbrush believed to have belonged to the victim)
D5S818 D13S317D7S820
D16S539CSF1PO
Penta D
(A) Direct comparison of STRs with related objects
(B) Indirect comparison: kinship analysis
?
son
wife
victim D5S818 D13S317 D7S820 D16S539 CSF1PO Penta D
10,10 9,109,138,98,1411,13 son
wife10,12 8,108,98,128,1211,13
10,10 9,1211,139,911,1412,13
?,10 9,??,139,??,1411,? or ?,13
victim (father)
actual profile
Predicted victim profile
mass disaster victim profile
Figure 24.1, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition © 2005 Elsevier Academic Press
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.
Primer P1: gctagtaacggcattaccagPrimer P2: catcgcataagaatttcacg 1 2 3 4 5
75 180 58 74 75
75 80 28 25 15
1 2 3 4 5
420
620
90
140
180
270
340
770
1900
CALCULATION
OF REPEAT
NUMBERS:
Subtract the
lenght of the
primers (40bp)
from the size of
the DNA
fragments,
divide the
remaining by 4.
340
100
120
152
272
336360340
760
THE NUMBER OF TANDEM REPEATS:
PRACTICAL TASK: VNTR ANALYSIS
DETERMINE THE REPEAT NUMBERS IN THE AMPLIFIED VNTR ALLELES!
D1 = biological daughter of both parentsD2 = child of mother & former husband
Explain the basis of paternity testing!
S1 = couple’s biological sonS2 = 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
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 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
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?
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 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
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? CHRISWhy? 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:
Blue/green Brown Blond/red Brown
HERC2
Rs12913832 GG/CC AA/TT GG/CC AA/TT
Oca2
rs1800407 AA/TT GG/CC AA/TT GG/CC
Tyr
Rs1393350 AA/TT GG/CC AA/TT GG/CC
IRF4
rs12203592 TT/AA CC/GG TT/AA CC/GG
SLC24A5
rrs16891982 GG/CC CC/GG GG/CC CC/GG
ExoC2
rs49592270 AA/TT CC/GG AA/TT CC/GG
EYE HAIRGene/SNP
Results of sequencing:
Rs12913832: AARs1800407: GGRs1393350: GGRs12203592: CTRrs16891982: CCRs49592270: AC
The HIrisPlex system for simultaneous prediction of hair and eye colour from DNA Forensic Science International: GeneticsVolume 7, Issue 1, January 2013, Pages 98–115IrisPlex: A Sensitive DNA Tool for Accurate Prediction of Blue and Brown Eye Colour in the Absence of Ancestry Information Journal:
420
620
90
140
180
270
340
1900
272
152
120
336
100
340 360
victim 3 2 1
suspects
marker
ReferenceGel
Compare the reference samples with the amplified samples and determine the number of repeats in the
amplified samples.
Who is probably the perpetrator?
1 2 3
500
250
272
152
120
336
100
340
360
victim 3 2 1
suspects
marker
ReferenceGel
100