The Polymerase Chain Reaction: - Essay Example

Deadline: To handed in 2nd of March The Polymerase Chain Reaction: Its use in genetic testing. Results. Attach a photograph or copy of the class data below as Fig. 3. The gel must be clearly labelled.Fig. 3. Genotypes of individuals in the class for the TPA-25 gene, Alu insertion polymorphism.Discussion/Questions: 1). If the reaction has worked, what is your genotype for the polymorphism? My gel did not show any reaction.If your DNA has migrated to the band designated at 100 bp position, then your genotype is homozygous for the absence of TPA-25 gene Alu insertion polymorphism (-/- or short/short).

However if it migrates to the 400 bp band position you are considered homozygous for the presence of TPA-25 gene Alu insertion polymorphism(+/+ or long/long). Presence of bands in both 100 bp and 400 bp position shows you are heterozygous showing polymorphism in one allele and normal gene in the other (+/- or long/short).2). Be clear that you understand the PCR. In the temperature cycling what is the purpose of a). the 94C step The 94C step in the procedure results in denaturation of the DNA.

At this temperature the hydrogen bonds holding the two strands of the double helix are broken and the DNA separates into single strands. b). the 50C step The 50C step lets the primers anneal to the complementary sequences specific to them in the template DNA. The temperature for this step is carefully chosen so as to allow the primers to bind to the correct complements. The exact annealing temperature depends on the length and sequence of the primers used for the procedure.

A lower annealing temperature might result in amplification of non-specific sequences due to incorrect binding of primers. c). the 72C step? 72C is the temperature at which the Taq DNA polymerase is most active. At this temperature the polymerase binds and adds nucleotides to the 3′ ends of annealed primers resulting in the extension of complementary DNA.3). Ideally, what degree of amplification would be achieved in 30 cycles? Show your logic. Under ideal conditions, the amount of PCR products doubles at each cycle and the primer directed DNA extension amplify the region between the primers in a geometric manner.

Hence after 30 cycles it shows a 2^30 amplification which theoretically is an amplification factor of a billion approximately.4). From the class data, determine the genotype distribution - count how many students are long/long, long/short or short/short for the polymorphism. long/long = 6 long/short = 5 short/short = 75). Calculate the allelic frequencies for the long and short alleles. (This is simply means the proportion of the TPA-25 gene that is in the long form in the population studied but NB each homozygote has two copies of that particular allele.

For example if the data gave 25 long/long; 32 long/short and 43 short/short, then the frequency of the long allele is (25 x 2) + 32 (Total number of long alleles) divided by the Total number of alleles (long and short) i.e. 200.Frequency Long = (25 x 2) + 32 = 0.41 200Frequency for long allele= (6 x 2) + 5 = 0.47 36Frequency for short allele = (7 x 2) + 5 = 0.52 366).

In this case the polymorphism is harmless but for other genes, which can be screened just as easily, this may not be the case. Certain polymorphisms are indicative of certain genetic diseases and PCR provides a very effective way of screening in many cases. Suppose that in this case the insertion of the Alu element disrupts the function of the gene and prevents a functional protein being produced. For the samples shown in the class data who would be ‘normal’ and who would suffer from the genetic disease?

Would carriers exist and if so who would be so-classed? Justify your answers with brief explanation.The homozygotes for the short gene in both allele or normal TPA-25 gene will be normal as the gene is not disturbed in both of the alleles. The homozygotes for the long gene in both alleles having inserted Alu element will show the genetic disease due to the disrupted gene function that prevents the functional protein from being produced. If the trait in question is recessive the heterozygotes showing a short allele and a long one namely a normal gene and a mutated gene with the Alu element inserted respectively would be carriers who do not show the genetic disease themselves but carry it forward to the future generation. 7). There are a number of ways of analysing a person’s DNA for a harmful mutation, for example RFLP analysis using Southern Hybridisation.

What advantages does PCR have over some of the older methods? PCR has a number of advantages over some of the other methods used in analysing DNA for harmful mutations which includes its speed and reliability. The sensitivity of PCR makes it possible to obtain amplification products even with degraded DNA samples or even from individual cells. PCR is not only a high fidelity process with lower error rates, the fidelity can be further improved by using novel thermostable enzymes. The expiry of key patents of PCR has also allowed it to be used in a number of applications without licensing fee.

References: List any references that you may have used below. Ensure that they are fully and correctly cited.1. Alberts, Bruce, et al. Molecular Biology of the Cell, 4th ed. New York: Garland Publishing, 2002.2. Brown, T.A. Gene Cloning and DNA Analysis An Introduction, 4th ed. Blackwell Publishing3. Lodish, Harvey, et al. Molecular Cell Biology, 4th ed. New York: W. H. Freeman, 2000.4. Micklos, David A., and Greg A. Freyer. DNA Science: A First Course in Recombinant DNA Technology.Cold Spring Harbor, NY: Cold Spring Harbor Laboratory Press, 1990.5. Watson, James D., et al. Recombinant DNA, 2nd ed.

New York: Scientific AmericanBooks, 1992.