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Module 12 Human DNA Fingerprinting and Population Genetics p 2 + 2pq + q 2 = 1.

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Presentation on theme: "Module 12 Human DNA Fingerprinting and Population Genetics p 2 + 2pq + q 2 = 1."— Presentation transcript:

1 Module 12 Human DNA Fingerprinting and Population Genetics p 2 + 2pq + q 2 = 1

2 GOALS Understand what an Alu sequence is and how it can be used to understand population equilibrium. Understand the importance of the PCR and its use in the “DNA fingerprint”. Understand the use of the Hardy-Weinberg theorem. Use the Chi-square analysis and see whether our data fits in Hardy-Weinberg theorem

3 Genes and DNA Gene is a piece of DNA that codes for a particular protein. Only 5% of the total DNA is present as genes. Other 95% DNA is called as non-coding DNA. Intron - In eukaryotes, the non-coding sequence is within the genes are transcribed into RNA but are not coded into the protein. Exon – the region of the transcribed RNA coded for the protein.

4 exonExon 2 Exon 1Exon 2 Exon 3 1 transcription Exon 1Exon 2 Exon 3 splicing translation protein Genomic DNA Pre-mRNA mRNA Intron 1 Intron 2 RNA SPLICING

5 Alu sequence The sequence of interest for this experiment is a short repetitive sequence seen within an intron sequence. This sequence is referred to as an Alu sequence after a restriction enzyme site that is located within this 300 base pair length of DNA. Alu sequences occur in the human genome about 500,000 times. ALU 3’ 5’ INTRON

6 Experiment outline In our experiment, PCR will be used to amplify a short nucleotide sequence from human PV92 locus on chromosome 16. The object is to create a personal DNA fingerprint that shows the presence(+) or absence (-) of the “Alu” DNA sequence on the paternal and maternal chromosomes. We will use primers for the Alu sequence that will generate a 550 bp fragment if the Alu sequence is not present and an 850 bp fragment if the Alu sequence is present through the application of the polymerase chain reaction.

7 PV92genotype DNA size of PCR products Homozygous (+/+)850 bp Homozygous (-/-) 550 bp Heterozygous (+/-) 850 + 550 bp

8 Polymerase Chain Reaction (PCR) Technique developed by Kary Mullis in 1983 (not very old!). Used in gene mapping, cloning, DNA sequencing and gene detections. Powerful tool for the human genome mapping. Widely used in forensic science, therapeutic, pharmaceutical and medical diagnostic purposes.

9 Contd… PCR produce a large amount of DNA in a test tube from a very small amount of starting material. e.g. A small amount of DNA can be taken from drop of blood, a hair follicle or cheek cells to generate millions of copies. A specific stretch of DNA can be amplified, such as an Alu sequence.

10 PCR Steps (See your handout) 1.Denaturation step- sample is heated to 94°C and at this temperature template DNA strands separate. 2.Annealing step- temperature is dropped to 60°C. primers will anneal to the template strands. 3.Extension step- sample is heated to 72 °C and polymerase extends the primers to make a complete copies of DNA template strands. All these steps are automated in the PCR machine

11 DNA polymerase used in PCR Thermally stable polymerase is needed. Why? Taq polymerase is used. Stable even at 94°c. Obtained from thermophilic bacteria, Thermus aquaticus.

12 PCR Requires 1.Reaction buffers 2.Four DNA bases (deoxynucleotides of A,G,T,C) 3.DNA polymerase 4.Two DNA primers (a small stretch of DNA that will recognize a particular sequence in the target DNA sequence) 5.Minute amounts of the desired template strands.

13 PCR MOVIE http://www.bio- rad.com/LifeScie nce/pdf/ExpPCR kit.pdf PCR movie

14 Contd… we will separate these fragments by gel electrophoresis to observe the DNA Fingerprint of each individual with respect to this Alu sequence at the PV92 locus. Homozygous(+/+) for the presence of the Alu repeat gives a 850 product, Homozygous(-/-) for the absence gives a 550 product and a heterozygous(+/-) gives two products of 850 and 550.

15 PV92genotype DNA size of PCR products Homozygous (+/+)850 bp Homozygous (-/-) 550 bp Heterozygous (+/-) 850 + 550 bp

16 Gel Electrophoresis Separates Shorter DNA Fragments from Longer DNA Fragments.

17 Gel for Alu marker

18 Hardy-Weinberg Theorem Once we determine the genotype distribution for the class (that is count how many students are ++, +-, and --), we apply the Hardy-Weinberg Theorem to this data set. The Hardy-Weinberg Theorem says that the sexual shuffling of alleles due to meiosis and random fertilization has no effect on the overall genetic structure of a population over generations unless acted upon by agents other than sexual recombination. What are these agents? (see handout)

19 Hardy-Weinberg (cont.) The Hardy-Weinberg equation allows us to calculate the frequencies of alleles in a gene pool if we know the frequencies of genotypes (which we know from our experiment) Population geneticists use: –The letter p to represent the frequency of one allele –The letter q to represent the frequency of the other allele p + q = 1

20 Hardy-Weinberg Equation p 2 + 2pq + q 2 = 1 Frequency of AA genotype Frequency of Aa + aA genotype Frequency of aa genotype One application of this is to calculate the percentage of the human population that carries an allele for a particular inherited disease. See example in Campbell textbook.

21 Hardy-Weinberg Equation p 2 + 2pq + q 2 = 1 Frequency of homozygous (+/+) genotype Frequency of heterozygous genotype Frequency of homozygous (-/-) genotype

22 Chi-Square Analysis Lastly, we can make sure that our class is in Hardy-Weinberg equilibrium by comparing our observed genotypic data in class to expected genotypic data –Observed data: class genotypes –Expected data: See website for www.dnalc.org. May be provided on lab report. www.dnalc.org

23 On 04/17,18…… Isolation of cheek cell DNA. PCR reaction (to be set up by the TA’s) Practice gel electrophoresis.

24 Next Class (on 04/22,23..) Gel electrophoresis of the amplified cheek cell DNA (done by TA’s) Calculate Hardy-Weinberg Equilibrium equation. Calculate the Chi-square value.


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