Components and Processes within a Polymerase Chain Reaction - Assignment Example

Name : xxxxxxxxxxx Institution : xxxxxxxxxxx Course : xxxxxxxxxxx Title : Genetics Tutor : xxxxxxxxxxx @2010 GENETICS Q1. Describe in detail the components and processes within a polymerase chain reaction (PCR). Polymerase chain reaction is a form of scientific technique that is made use of in the amplification of pieces of DNA that result into a great number of copies of a type of DNA sequence. The technique relies on the complementarity of the bases of DNA. It is noted that in the event a DNA molecule is heated then the hydrogen bonds that hold the double helix structures are denatured and thus the molecule is unzipped into single strands (Chamberlain, 1990). However, in the event a DNA solution is cooled then the respective complementary base pairs can be joined and thus it results into the original double helix structures. It is noted that as PCR progresses the DNA generated is made use of as a replication template and thus a chain reaction is set into motion. PCR can therefore be modified and utilized to perform a wide range of genetic manipulations. The components of PCR are as follows: a DNA template that comprises of the DNA needs to be amplified, two primers that are situated complementary to 3’ ends of every sense and anti-sense strand that is the DNA target, Taq polymerase or DNA polymerase which has an optimum temperature of 70 °C, deoxynucleotide triphosphates (dNTPs) that forms the building blocks that DNA polymerase is synthesized to form a new DNA strand, Buffer solution that provides proper chemical environment for optimum stability and activity of the polymerase DNA, monovalent cation for instance potassium ions, divalent cations for instance manganese (Mn2+ ) or magnesium (Mg2+ ) (Chamberlain, 1990). Procedure Below is a diagrammatic representation of the PCR procedure: Denaturing takes place between 94-96 °C, annealing takes place at approximately 65°C and elongation at 72°C. The blue lines are a representation of DNA templates to which the primers (red arrows) anneal and are consequently extended by the use of DNA polymerase (light green circles), the DNA products are represented by green lines which are in turn used as templates with the progression of PCR (Allen, 1991). Q2. Describe the Hardy-Weinberg law, stating and discussing the assumptions on which it is based. Also describe one method one could use to test for deviation from Hardy-Weinberg equilibrium. Hardy-Weinberg law works on the principle that genotype and allele frequencies that are available in a population stagnate from generation to generation. This implies that the frequencies remain in equilibrium unless there is the introduction of specific disturbing influences. The influences that may disturb the balance are non-random mating, selection, meiotic drive, mutations, gene flow, limited population size, random genetic drift and overlapping generations (Sanchez, 2004). The law is based on the assumptions that static allele frequencies that exist in populations have; no mutation that is the alleles are constant, no emigration or migration which implies that there is no allele exchange between populations, a very large population size and lack of selective pressure that advocates for or against any particular genotypes that is lack of natural selection. It is noted that genotype frequencies are stable during random mating. The testing of deviation from Hardy-Weinberg law is mostly performed by Pearson’s chi-squared test. Below is an example of Pearson’s chi squared test: χ2 test for deviation The table shows data to be made use of in the calculation of Hardy-Weinberg principle. Table Genotype White-spotted (AA) Intermediate (Aa) Little spotting (aa) Total Number 1469 138 5 1612 From which allele frequencies can be calculated: and Hardy-Weinberg results are as stipulated below: Chi-square test is as stipulated: There exists 1 degree of freedom. The degrees of freedom that exist in the test for Hardy-Weinberg proportions refer to number of genotypes to the number of alleles. The 5% significance level that is attached to 1 degree of freedom is stipulated at 3.84 and since the chi square amount is less than this then it follows that the null hypothesis of the aforementioned population lies within Hardy-Weinberg frequencies and thus not rejected (Bonetta, 2004). Q3. Describe the structure and function of telomeres. Telomeres are recurring DNA sequences that are situated at the end of linear chromosomes of a large number of eukaryotic organisms and a small number of prokaryotes. Telomeres stand in for incomplete semi-conservative DNA copying at chromosomal ends (Bloom, 2004). The protection against non-homologous end joining (NHEJ) and (HR) homologous recombination make up the vital “capping” function of telomeres that separates them from DNA. More explicitly; double strand breaks (DSBs). Telomere It is noted that a great number of prokaryotes have circular chromosomes that do not suffer from untimely replication termination. Some bacterial chromosomes (for instance Borrelia and Streptomyces) have telomeres and are linear. These chromosomes are different in comparison to those of eukaryotic chromosomes in their functions and structure. Bacterial telomeres structures are in the form of proteins and they are attached to the ends of straight chromosomes and hairpin loops which are single-stranded DNAs located at the termini of linear chromosomes (Allen, 1991). In multi-cellular eukaryotic organisms there is active telomerase in stem cells, germ cells and various kinds of white blood cells. It has been noted that steady shortening of the telomeres which is accompanied by their replication in the body of cells have a vital role in senescence and thus in cancer prevention. This is due to the “fuse” role played by telomeres. However, their aforementioned role dies down after a large amount of cell divisions which results into loss of genetic information available in the chromosome that result with future divisions. Shorter telomeres can also be made use of as an energy sparing mechanism in energy limiting environment. Telomere length ranges from between 300 to 600 base pairs which exist in yeast to a number of kilobases which exist in humans and constitute guanine-rich long repeats of between six to eight base pairs of long repeats. Eukaryotic telomeres are usually known to terminate with 3′ of lone stranded DNA that is overhang and is necessary for the telomere capping and maintenance. Multiple proteins that bind double and single stranded telomere DNA are known to exist. The aforementioned have roles in telomere capping and maintenance. Q4. Describe real time PCR and some of its applications. Give one example of a method used for real time PCR. Real time PCR is a technique of simultaneous DNA amplification and quantification. DNA is mostly amplified by the use of polymerase chain reaction (PCR). Every round of amplification is followed by quantification. The common techniques of quantification are the utilization of fluorescent dyes which intercalate together with modified DNA oligonucleotides and double-strand DNA. These types of DNA fluoresce in the event of their hybridization with complementary DNA. Real-time PCR is integrated with reverse transcription-polymerase chain reaction (RT-PCR) to attach value to low abundance messenger RNA that enables a researcher to be able to quantify the relative gene expression at any given time or in any given tissue or cell type. The joined method is referred to as quantitative RT-PCR (Chamberlain, 1990). Real-time PCR is made use of in diagnosing microbial pathogens. Its high sensitivity, specificity, ease of performance and its short turnaround time for its results make it an efficient technique for antigen-based and conventional culture assays. It is made use of in diagnostic and basic research. In research it is made use of in the provision of quantitative measurements that are necessary for gene transcription. Diagnostic real-time PCR is made use of in the detection of nucleic acids that allow for the detection of cancer, infectious diseases and genetic abnormalities. It is also utilized in the detection of infectious diseases like flu. Real-time PCR methods are inclusive of the following types; RT-PCR (reverse-transcription PCR) and RTQ-PCR. Real-time PCR is joined with reverse transcription to assign value to messenger RNA and non-coding RNA that is available in tissues or cells. The development of PCR technologies that are based on fluorophores and reverse transcription allows for the measurement of the amplification of DNA during PCR that is available in real time which is measured at every PCR cycle. The generated data can be consequently analyzed by the use of computer software to aid in the calculation of relative gene expression. It can also be used in the detection and quantification of various DNA samples to determine the abundance of specific DNA sequences (Bloom, 2000). Q5. The following RFLP results were obtained at a locus in a population. 26/120 51/120 43/120 2.0 kb 0.8 kb Table Genotype 2.0kb 0.8kb 2.0kb/0.8kb Total Number 26 51 43 120 a. Determine the frequencies of the 2.0kb and 0.8 kb alleles. p = 2 × (2.0kb) + (2.0kb/0.8kb)/ 2× ((2.0kb) + (2.0kb/0.8kb) + (0.8kb) = (2 × 26) + 43/ 2 × (26 + 43 + 51) = 0.39 q= 1-p = 0.604 b. What are the expected frequencies of all possible genotypes in this population if you assume Hardy Weinberg equilibrium? Hardy–Weinberg expectation is as follows: Exp (2.0kb) = p2 n = 0.392 × 120 = 18.252 Exp (2.0kb/0.8kb) = 2pqn = 2 × 0.39 × 0.604 × 120 = 56.5344 Exp (0.8kb) = q2 n = 0.6042 × 120 = 43.77792 X2 = ∑ (0-E)2 / E = (26 – 18.252)2 / 18.252 +(43 – 56.5344)2 /56.5344 + (51-43.77792)2 / 43.77792 = 3.289 + 3.240 + 1.191 = 7.72 c. By carrying out a chi squared test, determine whether this population is in Hardy Weinberg equilibrium at this locus. It is therefore rejected since it is more than 1. Q6. A population has four times as many heterozygous individuals as homozygous recessive individuals. Work out the frequency of the recessive gene assuming HWE and a bi-allelic locus. Show your working in full. Let p2 (AA) be homozygotes q 2 (aa) homozygotes with recessive gene. Let it be x. 2pq Aa heterozygotes. Let it be 4x Frequency will therefore be: x/5x × 100 = 20% Frequency is therefore 20. References Chamberlain, J.S., et al. 1990. Multiplex PCR for the diagnosis of Duchenne muscular dystrophy. In PCR Protocols: A Guide to Methods and Applications (M.A. Innis, et al., eds.). Academic Press, New York, pp.272-281. Bloom, M (2000). Polymerase Chain Reaction. Cold Spring Harbor, New York. Allen, R., et al. 1991. Analysis of the VNTR locus AmpliFLP D1S80 by the PCR followed by high resolution PAGE. American Journal of Human Genetics 48:137. Sanchez, J.A et al. (2004). Real-Time Polymerase Chain Reaction. Proc. Natl. Acad. Sci. USA. Bonetta, L. (2004). National Methods. McGraw Hill; New York. Read More