1. PV92 PCR Informatics Chromosome 16 Day #1: What is PCR? Day #2: Alu Insertion & PCR

2. PCR Polymerase chain reaction (PCR) enables researchers to produce millions of copies of a specific DNA sequence in a relatively short period of time The specific sequence is called the “target sequence”

3. PCR 1 st described in 1985 by Kary Mullis He won the Nobel in 1993 for this Made it possible for researchers in a variety of biological fields to incorporate molecular biology (genetics) into their research. Pathology Botany Zoology Pharmacology

4. What Is It & Why Did it Revolutionize Research PCR produces exponentially large amounts of DNA from trace amounts. Drop of blood Single hair follicle One cheek cell

5. What’s Needed For PCR? Free nucleotides – building blocks DNA primers A strand of nucleic acid that serves as a starting point for DNA replication Complementary to target sequence Taq polymerase Comes from hot springs bacteria Can tolerate high heat of PCR Discovery of this bacterium made PCR possible

6. PCR Steps 3 Main Steps to PCR Denaturing at ~94 °C Annealing at ~54 °C Binding of a primer to DNA strand. Extension at ~72 °C The bases (complementary to the template) are coupled to the primer on the 3' side.

8. Lots of Copies!

9. PCR Animation http://www.sumanasinc.com/webcontent /animations/content/pcr.html

10. Any Questions?

11. PV92 PCR Informatics Chromosome 16 Day #1: What is PCR? Day #2: Alu Insertion & PCR

12. What is PV92? PV92 – A human-specific Alu insertion on C16 Alu’s are a type of transposon Recall that transposons are pieces of DNA that can move around within the genome of a cell Barbara McClintock discovered transposons and won the Nobel in 1983.

13. More About Alu Transposons Also known as ‘jumping genes’ Copies itself and inserts into new locations on the chromosome No evidence that ‘parent’ Alu segments are ever excised…what does this mean in terms of evolution & human population genetics?

14. Alu is a SINE as well as a Retrotransposon Alu is ~300 bp in size Therefore known as a SINE (short interspersed repetitive element) Highly Conserved Alu endonuclease is called so because it was isolated from Anthrobacter luteus Retrotransposon because it uses reverse transcriptase to copy itself

15. But Alu a Defective Transposon Can’t make it’s own reverse transcriptase (RT)…so…it hijacks another gene Hijacked gene is known as L1 (a LINE) and is basically a non-functional retrovirus that can make RT Say WHAT??????

16. It’s Like This… 1. Alu is transcribed into mRNA via RNA polymerase 2. mRNA hijacks L1 which converts Alu to ds DNA by way of RT 3. DNA copy of Alu is integrated into new chromosome site

17. PV92 Alu Insertion 5’ 3’ Alu Amplified Region A member of Alu repeat family Human-specific Alu insertion Found in a non-coding region of your DNA Not diagnostic for any disease or disorder (usually)

18. Introns, Exons and Alu Inserts Only ~5% of our genome consists of coding DNA (exons). This is where our functional proteins come from. Out of the remaining 95%, ~40% is intron based, or non-coding. We pass introns and exons on from one generation to the next.

19. Introns, Exons and Alu Inserts Introns are full of SINES, Alu being one of them. Origins of Alu unknown. We are targeting a specific locus on C16 that is known to carry the Alu sequence, or not. Some of you will have the Alu insert and others will not.

20. The Alu Insert is Dimorphic The PV92 Alu is dimorphic so there are two possible PCR products: 641 bp and 941 bp If you have the Alu insert on both of your homologous chromosomes = +,+ If you have Alu on one chromosome = +, - If you don’t have the insert = -,-

21. To Alu Or Not… Approximately 500,000 Alu copies per haploid genome, representing about 5% of the human genome. No insertion: 641 bp 300 bp Alu insert 641 bp With Alu: 941 bp

22. Possible PCR Products 941 bp 641 bp - +/- + We will use primers that are specific to the Alu region No Alu = 641 bp fragment Alu = 941 bp fragment S1S2S3S4

23. In the Lab… 1. We will harvest some of your cells… 2. Incubate them with Chelex resin (extract DNA)… 3. Use PCR to amplify the Alu gene… 4. Separate Alu fragments on 2% agarose gel… 5. Use Chi-Square or Hardy Weinberg to calculate population frequency of (+,+), (+,-) and (-,-).

24. Any Questions?

25. Calculating Observed Genotypic Frequencies

26. Hardy-Weinberg Equilibrium The HW equilibrium describes what happens to alleles in an ‘ideal’ population. 1. No selection (= rate of survival for all) 2. No mutation 3. No immigration/emigration 4. Large population 5. Random mating

27. Hardy-Weinberg Equilibrium We know that when we cross two individuals who are heterzygous for Alu we will see: + + - - + + - - So there is a 25% chance of being ++, 25% being - - and 50% + -. Hardy Weinberg ?

28. +/+ = p 2 +/- = 2pq -/- = q 2 -/- = q 2 pp pq qqpqppqq p 2 + 2pq + q 2 = 1 Hardy-Weinberg Equation p 2 + 2pq + q 2 = 1 p = frequency of + alleles q = frequency of - alleles

29. p = frequency of + allele q = frequency of - allele p 2 = frequency of ++ q 2 = frequency of – 2pq = frequency of +- p2 + 2pq + q2 = 1 Hardy-Weinberg Equation p2 + 2pq + q2 = 1

30. Hardy-Weinberg Equilibrium Example Genotype+/+ +/– -/- Total (N) # of People 255838 Observed 0.660.130.211.00 Frequency +/+ Genotypic frequency Number with genotype Population total (N) = 25 38.66 = =

31. Calculating Allelic Frequencies Number of + alleles 25 individuals with two + alleles = 50 + alleles 5 individuals with one + allele = 5 + alleles Total = 55 + alleles Total number of alleles 2N = 2(38) = 76 p Frequency of + alleles Number of + alleles Total number alleles = 55 76 = 0.72 = = p = 0.72; therefore q = 0.28 since p + q = 1.00

32. p2 + 2pq + q2 = 1 +/+ = p 2 +/- = 2pq -/- = q 2 -/- = q 2 pp pq qqpqppqq This should make sense now!

33. (p 2, 2pq, q 2 values) p2p2 + 1.00 = 2pq + q2q2 (0.72) 2 + 1.00 = 2(0.72)(0.28) + (0.28) 2 0.52 + 1.00 = 0.40 + 0.08 p 2 = 0.52 2pq = 0.40 q 2 = 0.08

34. Chi-Square Test Ok…so… +/+ (p 2 ) +/– (2pq) –/– (q 2 ) 0.52 Genotype frequency x = Population total (N) Expected number x = x = x = 3820 0.403815 0.08383 Genotype

35. X2X2 = ∑ ( Observed – Expected) 2 Expected +/+ +/– –/– 25201.25 5156.67 838.33 Genotype ObservedExpected (O–E) 2 E X 2 = 16.25

36. Allele Server (1 of 17) Cold Springs Harbor Laboratory DNA Learning Center Web site: http://www.dnalc.org/

37. Allele Server (2 of 17) Scroll through DNALC internet sites until BioServers Link appears

38. Allele Server (3 of 17) Click on Bioservers

39. Allele Server (4 of 17) Enter the Allele Server

40. Allele Server (5 of 17) Click on Manage Groups

41. Allele Server (6 of 17) Select Group

42. Allele Server (7 of 17) Scroll Down to Select “Your Group”

43. Allele Server (8 of 17) Fill Out Form

44. Allele Server (9 of 17) Click on Edit Group

45. Allele Server (10 of 17) Edit Your Group Information

46. Allele Server (11 of 17) Click on Individuals Tab

47. Allele Server (12 of 17) Add Each Student’s Information Add as much information as possible: Genotype (+/+, +/–. –/–) Gender Personal Information

48. Allele Server (13 of 17) Click on Done

49. Allele Server (14 of 17) Select and then Click OK

50. Allele Server (15 of 17) Analyze Data 2: Then Click Here 1: Click Here First

51. Allele Server (16 of 17) Click on the Terse and Verbose Tabs to Review Data Results

52. Any Questions?