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.

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