Wilson Chan U1021014H GP05
POLYMERASE CHAIN REACTION PCR Amplification 1. Why is it necessary to have a primer on each side of the DNA segment to be amplified? The oligonucleotide primers on each side of the DNA segment binds specifically to the 3’ end and the 5’ end of the target sequence on the appropriate strands of DNA. It directs
and serves as a starting point for the DNA polymerase to attach to for DNA Taq replication.
2.
How did Taq DNA polymerase acquire its name? Taq DNA polymerase is an enzyme originally isolated from the bacterium Thermus aquaticus. It is a heat-stable DNA polymerase and its enzyme heat resistance is favoured since it can withstand the high temperature in the PCR process.
3.
Why are there nucleotides (A, T, G and C) in the master mix? What are the other components of the master mix, and what are their functions? The nucleotides, also called DNTPs (Deoxynucleoside ( Deoxynucleoside triphosphates) are free nucleotides and the raw materials utilized in the replication of new stands of the target DNA The other components of the master mix include: Taq DNA polymerase: The heat-resistant enzyme enzymatically assembles a new DNA strand from DNA building blocks, the nucleotides, by using single-stranded DNA as a template and oligonucleotides primers, which are required for initiation of DNA synthesis. Oligonucleotide primers: Pieces of DNA complementary to the template that serves as a starting point for the DNA polymerase to start making copies. Magnesium ions: A cofactor (catalyst) required by DNA polymerase to create the DNA chain. Salt buffer: Provides the optimum ionic environment and pH for the PCR reaction.
4. Describe the three main steps of each cycle of PCR amplification and what reactions occur at each temperature. Four cycles are shown here. The blue lines represent the DNA template to which primers (red arrows) anneal that are extended by the DNA polymerase (light green circles), to give shorter DNA products (green lines), which themselves are used as templates as PCR progresses.
Step 1: Denaturation o
o
By heating the sample to 94 C - 96 C for 1 minute, the DNA strands are denatured and are separated. During denaturation, the hydrogen bonds between the complementary base pairs are broken, and thus the DNA strands are separated. Step 2: Annealing of primers o
This step occurs at 60 C for 1 minute. In annealing, the specially-designed oligonucleotide primers recognize their specific complementary sequences on the template strands and anneal to the DNA template to allow the taq polymerase to attach to them. Step 3: Extension step o
This step occurs at 72 C for 2 min. Taq polymerase incorporates dNTPs and extends the
primers from the 3' end, generating newly synthesized complementary DNA strands. As PCR progresses, the DNA generated is itself used as a template for replication, setting in motion a chain reaction in which the DNA template is exponentially amplified.
5.
Explain why the precise length target DNA sequence doesn’t get amplified until the
third cycle. You may need to use additional paper and a drawing to explain your answer.
During the first cycle of polymerization, a new strand of DNA is fo rm from each primer. Taq polymerase lengthens this new DNA strand further than the target sequence. As a result, a new longer than usual DNA strand is produced, it contains both downstream/upstream primer. During the second cycle of polymerization, the new DNA strands produced in the first cycle and the original target DNA are used as templates for replication. At the end of this step, a mixture of small target sequences and larger hybrid DNA sequences will be produced. As shown in the figure above, there are already 2 target sequences ready to be duplicated. From the third cycle of polymerization onwards, the target DNA is replicated exponentially. Each subsequent round of PCR doubles the number of defined target sequences, giving rise to exponential amplification. Thus, a 30 cycle PCR would amplify a target DNA segment to more than a billion copies, whereas there are only 60 copies of the longer DNA strands, making the final product virtually pure target sequence.
Gel Electrophoresis of Amplified PCR Samples 1. Explain the difference between an intron and an exon. Exons encode the important information of the gene and are does not change very readily over time or between species. Any disruption to this code would lead to alter function of the polypeptide they encode or possibly even preve nt its expression. They are usually short, containing 150 to 200 base pairs in length on average. Introns are only well conserved at their starting and ending sequence at their ‘exon intron’ and ‘intron-exon’ boundaries only. The start of an intro specially contains the
nucleotide sequence GT and ends with AG. They do not code for protein sequences and are spliced out of mRNA molecules before the mRNA leaves the nucleus. They often vary in size and sequence among individuals.
2. Why do the two possible PCR products differ in size by 300 base pairs? The PCR primers amplify 641 base pairs within the PV92 re gion (TPA gene). However, some individuals contain a 300 base pairs Alu amino acid repeat within this region of TPA gene and amplification from there individuals a 941 base pairs fragment. This results in the formation two PCR products to differ in size by 300 base pairs due to the insertion of a 300 base pairs Alu repeat.
3. Explain how agarose electrophoresis seperates DNA fragments. Why does a smaller DNA fragment move faster than a larger one? The speed of the separation of the PCR fragment depends on its ne gativity and size. The DNA molecules are pulled to the positive end by the current. The agarose gel acts like a sieve to separate the charged DNA molecules according to size. The smaller molecules are able to navigate the agarose gel faster than the larger one, and therefore move further down the gel than the larger molecules. As DNA molecules move through the pores in the matrix, the larger molecules encounter more resistance and move slower.
4.
What kind of controls is run in this experiment? Why are they important? Could others be used? The controls used in this experiment are homozygous +/+, homozygous -/- and heterozygous +/- samples. The controls aid us by comparison of the unknown student samples with their known base pair lengths.
A negative control in the Master Mix can be used to run the experiment to ensure that no foreign/ unwanted DNA contaminates the Master Mix.
5.
What is your genotype for the Alu insert in your PV92 region? My genotype for the Alu insert in the PV92 region is homozygous +/+ as there is only one band that matches the homozygous +/+ control.
6. What are the genotypic frequencies of +/+, +/- and -/- in your previous group population? Fill in the table below with your previous group data.
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Lane 1: Homozygous (+/+) control Lane 2: Homozygous (-/-) control Lane 3: Heterozygous (+/-) control Lane 4 -7: Other groups sample Lane 8: My sample
Table 1. Observed Genotypic Frequencies for the Class Category Number Homozygous (+/+) Heterozygous (+/-) Homozygous (-/-)
2 1 2 Total
Frequency (# of Genotypes/Total) 0.40 0.20 0.40 1.00
