Biotechnology Explorer Program Serious About Science Education 5/17/09 1
Chromosome 8: PCR TM PCR Workshop Kirk Brown,, Tracy High School; Tracy, Ca Stan Hitomi,, Monte Vista High School; Danville, CA 5/17/09 2
PCR Procedures Day 1 Day 2 Day 3 5/17/09 3
What is PCR? DNA replication gone crazy in a tube! Makes many copies target sequence from template Uses Taq Polymerase heat-resistant DNA polymerase from Thermus aquaticus 5/17/09 4
Protocol Highlights Genomic DNA extraction Instagene = Chelex (cation exchange resin) binds cellular MgCl 2 56 o C loosens connective tissue and inactivates DNAses 100 o C ruptures cell membranes and denatures proteins 5/17/09 5
Instagene extraction Cell membrane Nuclear membrane Genomic DNA Mg ++ Mg ++ Mg ++ Heat disrupts membranes Mg ++ Mg Mg ++ ++ Instagene matrix binds released cellular Mg ++ 5/17/09 6
Micropipet Use 1 2 3 4 5 6 7 5/17/09 Twist dial to desired volume Pick up pipet tip Press plunger to first, soft stop Insert pipet tip into solution to be transferred Slowly release plunger to retrieve liquid Move pipet tip into desired tube Press plunger past first stop to second, hard stop to transfer liquid 7
Why is the Master Mix ORANGE? Master mix contains reagents needed for PCR amplification red and yellow loading dyes glycerol to make samples sink Amplified samples can be loaded directly onto agarose gels 5/17/09 8
What is needed for PCR? Template (the DNA you are exploring) Sequence-specific primers flanking the target sequence Forward Reverse Nucleotides (datp, dctp, dgtp, dttp) Magnesium chloride (enzyme cofactor) Buffer,, containing salt Taq Polymerase 5/17/09 9
How does PCR work? Heat (94( o C) ) to denature DNA strands Cool (64( o C) ) to anneal primers to template Warm (72( o C) ) to activate Taq Polymerase, which extends primers and replicates DNA Repeat multiple cycles 5/17/09 10
Denaturing Template Heat causes DNA strands to separate Denature DNA strands 94 o C 5/17/09 11
Annealing Primers Primers bind to the template sequence Taq Polymerase recognizes double-stranded substrate Primers anneal 64 o C 5/17/09 12
Taq Polymerase Extends Taq Polymerase extends primer DNA is replicated Extend 72 o C Repeat denaturing, annealing, and extending 40 cycles 5/17/09 13
The target product is made in the third cycle Cycle 1 Cycle 2 Cycle 3 5/17/09 14
The target sequence Chromosome 8 Intron of tissue plasminogen activator (TPA) gene Alu-TPA25 Amplified Region Exon 8 Alu Exon 9 Exon 10 Intron 8 5/17/09 15
Alu-TPA25 In a non-coding region of your DNA NOT diagnostic for any disease or disorder A member of Alu-repeat family One of 28 Alu repeats within the TPA gene Amplified Region Exon 8 Alu Exon 9 Exon 10 Intron 8 5/17/09 16
PCR Results. Alu-TPA25 is dimorphic so there are two possible PCR products: 660 bp 960 bp No insertion: 660 bp Alu insertion: 960 bp 330 bp each 300 bp Alu insert Amplified Region Exon 8 Alu Exon 9 Exon 10 Intron 8 5/17/09 17
Actual Alu-PCR Results + - +/- 960 bp 660 bp + - +/- 5/17/09 18
Alu repeats Short (282 bp) ) repetitive elements flanked by direct repeats Mobilized by an RNA polymerase-derived intermediate (retroposition( retroposition) Occur >500,000 times in the human haploid genome Named for the Alu I restriction site within the element 5/17/09 19
Evolutionary Significance of Alu-TPA 25 Highly conserved Inserted in the last 1,000,000 years Dimorphic (+/+, +/-, -/- ) Used in population genetics, paternity analysis, and forensics 5/17/09 20
To estimate frequency of Alu within a population: Amplify Alu-region from representative sample population Calculate the expected allelic and genotypic frequencies Perform Chi-squared Test 5/17/09 21
Alu and Population Genetics Hardy-Weinberg Equilibrium p q p 2 + 2pq + q 2 = 1 p q pp pq pq qq +/+ = p 2 +/- = 2pq -/- = q 2 5/17/09 22
Calculating Observed Genotypic Frequencies Genotype +/+ (p 2 ) +/- (2pq) -/- (q 2 ) Total (N) # of people 9 16 13 38 Observed frequency 0.24 0.42 0.34 1.00 Calculation: +/+ genotypic frequency = # with genotype total number of people (N) = 9/38 = 0.24 5/17/09 23
Calculating Allelic Frequencies Frequency of p = number of p alleles = 34 = 0.45 total alleles 76 Number of p alleles = +/+ = 9 with two + alleles = 18 + alleles +/- = 16 with one + alleles = 16 + alleles Total = 34 + alleles Total number of alleles = 2N = 2(38) = 76 5/17/09 24
Using Hardy-Weinberg Determine p 2, 2pq, and q 2 values= Expected genotypic frequencies since p + q = 1 p = 0.45, so q = 0.55 since p 2 + 2pq + q 2 = 1 (0.45) 2 + 2 (0.45)(0.55) + (0.55) 2 = 1 0.20 + 0.50 + 0.30 = 1 p2 = 0.20 2pq = 0.50 q2 = 0.30 5/17/09 25
Calculate Expected Number of Genotypes Expected number of genotype = Genotypic frequency x population number (N) Genotype Expected number +/+ 0.20 x 38 = 8 +/- 0.50 x 38 = 19 -/- 0.30 x 38 = 11 5/17/09 26
Chi Squared Test Observed Expected (O-E) 2 E 0.13 0.47 0.36 Total 0.96 +/+ 9 8 0.13 +/- 16 19 0.47 -/- 13 11 0.36 X 2 Critical Value (from statistics table) = 5.9 0.96 falls below 5.9 so the ratio is accepted. 5/17/09 27