2. 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. Police released surveillance image of the suspect.

3. „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)

4. 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

5. 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 C o 2 minutes, ProtK inactivated store on ice

6. ORGANIC Filter Paper CHELEX Blood stain PUNCH WASH Multiple Times with extraction buffer PERFORM PCR PCR Reagents SDS, DTT, EDTA and proteinase K INCUBATE (56 o C) Phenol, chloroform, isoamyl alcohol QUANTITATE DNA Apply blood to paper and allow stain to dry Blood stain VORTEX (NO DNA QUANTITATION TYPICALLY PERFORMED WITH UNIFORM SAMPLES) Water INCUBATE (ambient) 5% Chelex INCUBATE (100 o C) REMOVE supernatant INCUBATE (56 o C) QUANTITATE DNA PERFORM PCR Centrifuge REMOVE supernatant TRANSFER 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, 2 nd Edition © 2005 Elsevier Academic Press DNA-extraction protocols

7. 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 MgCl 2 finally: 1 ul Ultra Fast DNA Polymerase (store on ice) RUN THE PCR REACTION.

8. PCR Program: CSI 1. 95 C o 2 minutes 2. 95 C o 10 sec 3. 55 C o 15 sec 4. 72 C o 20 sec 30x: step2-step4 1 min 72 C o 4 C o (total time: ca. 30 min)

9. (+) 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 MUTATION WHAT TYPE OF MUTATION IS THIS?

10. (+) 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

11. (+) 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?

12. (+) 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)

13. (+) 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

14. (+) 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

15. (+) 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’

16. (+) 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’

17. +/+ -/- +/ - GEL ELECTROPHORESIS OF THE PCR PRODUCTS

18. TT CTCT CC CTCT CTCT 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, 2 nd Edition © 2005 Elsevier Academic Press Capillary electrophoresis of SNPs amplified with ASA

19. (A)Simultaneous amplification of three allels on a DNA template of homologous chromosomes 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 AC B smalllarge Modified Figure 4.3 J.M. Butler (2005) Forensic DNA Typing, 2 nd Edition © 2005 Elsevier Academic Press Locus A Locus C (t) (RFU) RFU: relative fluorescence unit t: time 50 bp70 bp 10min15 min Deletion of Locus B How does the chromatogram appear?

20. PCR product size (bp) Figure A7.1, J.M. Butler (2005) Forensic DNA Typing, 2 nd Edition © 2005 Elsevier Academic Press

21. DNA profile from mass disaster victim DNA profile from direct reference (toothbrush believed to have belonged to the victim) D5S818D13S317 D7S820 D16S539 CSF1PO Penta D (A) Direct comparison of STRs with related objects

22. (B) Indirect comparison: kinship analysis ? son wife victim D5S818D13S317D7S820 D16S539CSF1POPenta D 10,109,109,138,98,1411,13 son wife 10,128,108,98,12 11,13 10,109,1211,139,911,1412,13 ?,109,??,13 9,? ?,14 11,? or ?,13 victim (father) actual profile Predicted victim profile mass disaster victim profile Figure 24.1, J.M. Butler (2005) Forensic DNA Typing, 2 nd Edition © 2005 Elsevier Academic Press

23. 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

24. 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: gctagtaacggcattaccag Primer P2: catcgcataagaatttcacg 1 2 3 4 5

25. 75180587475 80282515 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. 340 100 120 152 272 336 360340 760 THE NUMBER OF TANDEM REPEATS: PRACTICAL TASK: VNTR ANALYSIS DETERMINE THE REPEAT NUMBERS IN THE AMPLIFIED VNTR ALLELES!

26. D1 = biological daughter of both parents D2 = child of mother & former husband Explain the basis of paternity testing! 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

27. VNTREvelyn JennaJamesLarsKirkJasonChris A7 & 82 & 73 & 72 & 72 & 83 & 73 & 2 B4 & 56 & 447646 C11 & 8 11 & 99 & 1013 & 8109 & 129 & 13 D 7 & 19 21 & 7 7 & 921 & 715 & 821 & 15 E10 & 612 & 66 & 912 & 917 & 10 6 & 1217 & 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.

28. VNTREvelynLinda JamesLarsKirkJasonChris A7 & 82 & 73 & 72 & 72 & 83 & 73 & 2 B4 & 56 & 447646 C11 & 811 & 99 & 1013 & 8109 & 129 & 13 D7 & 1921 & 7 7 & 921 & 715 & 821 & 15 E10 & 612 & 66 & 912 & 917 & 106 & 1217 & 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? CHRIS Why? Why VNTR locus B is peculiar? X-LINKED

29. SNP

30. Determine the putative eye and hair colour of the suspect, if the following SNPs are detected in the sample: Blue/greenBrownBlond/redBrown HERC2 Rs12913832 GG/CC AA/TTGG/CCAA/TT Oca2 rs1800407AA/TTGG/CCAA/TTGG/CC Tyr Rs1393350AA/TTGG/CCAA/TTGG/CC IRF4 rs12203592TT/AACC/GGTT/AACC/GG SLC24A5 rrs16891982GG/CCCC/GGGG/CCCC/GG ExoC2 rs49592270AA/TTCC/GGAA/TTCC/GG EYE HAIR Gene/SNP Results of sequencing: Rs12913832: AA Rs1800407: GG Rs1393350: GG Rs12203592: CT Rrs16891982: CC Rs49592270: AC Results of sequencing: Rs12913832: AA Rs1800407: GG Rs1393350: GG Rs12203592: CT Rrs16891982: CC Rs49592270: AC 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:

31. 420 620 90 140 180 270 340 1900 272 152 120 336 100 340360 victim 3 2 1 suspects marker Reference Gel 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

32. 500 250 272 152 120 336 100 340 360 victim 3 2 1 suspects marker Reference Gel 100