Experiment 7 Molecular Biology 18/20 PDF

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This is my assignment for Molecular Biology 1 in Spring Semester 2021...

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Experiment 6- Emma Brandham- Student ID: 13912946 Q1. Thermostable DNA polymerases exhibit activity profiles such as the one shown below for a Taq polymerase. The PCR tubes for Experiment 7 are, for convenience, set up at room temperature (RT) until being loaded into the thermal cycler for the amplification process, in which the protocol specifies that each extension step in the cycles be carried out at 72o C for 2.5 min. In one of the PCR cyclers, the technician accidently set the extension temperature at 65o C. Would the product yield be the same as that from the 72o C extended product? Explain your answer briefly. [2 marks] [Clue: Answer is in the graph below and extension time.] The product yield in these two different temperatures would not be the same as there will a lower speed in extension rates at 65 degrees in comparison to 75 degrees. As in the graph the extension rate at 65 degrees is around 100 s-1 in comparison to 150s-1 at the peak at 75 degrees. The reason why there will be different results is as a lower temperature than the optimal can effect the way the primers bind as primers need to find a compatible sequence. Q2. Two honours students have related projects that use the lmrA gene. Their supervisor asks them to begin their projects by producing a large amount of lmrA DNA, using PCR and the same bookended primers as used in this experiment. In a slightly confusing situation, one student uses the pAMG25 plasmid as the template, but the second student finds a small amount of the actual lmrA linear double-stranded DNA. Both proceed independently with their PCRs. Has one of the students acted incorrectly? What do you expect the two PCR outcomes to be when the students view their products in an agarose gel? [2 marks] [Clue: Check the source of the primers.] Both students are correct regardless of using pAMG25 as a template as the bookended primers will anneal to the ends of the lmrA insets. As the students are using bookended primers, these primers will bind to the corresponding target sequences regardless of the template used. However, the two PCR outcomes may different when viewing the agarose gel profile due to the relationship between the molecule size and the pore size of the agarose gel. Student 1 will see the pMAG25 plasmid further above in the gel as it is larger than the actual lmrA double-stranded DNA that student 2 uses, as a larger molecule will not move as far through the agarose gel when compared to a smaller molecule. Q3. You are planning a PCR experiment for which you have designed forward and reverse primers as shown here: Forward: 5'-ATGGCTTAAAGCTACATGAAAGAGGTCCACAAATGGCCAAT-3' Reverse: 5'TTGGAAGGCTAAGAATTGACCAACAGTCAATTGTTCTGAAAC-3' You now need to calculate the Tms of these primers to see if they are suitable for your PCR. A useful online tool for calculating primer Tms is the Promega Biomath Oligo site: https://www.promega.com.au/resources/tools/biomath/tm-calculator/ When you enter the site all you need to do is enter the forward oligo sequence in Step 1 and choose the first GoTaq Green buffer system in Step 3 and hit the Calculate button. Repeat for the reverse primer. From the calculated Tms, are your primers suitable or unsuitable for your PCR? Explain. [3 marks] [Clue: Primer Tms calculated from the Biomath site should enable you to answer this question.] -

Forward Oligo Sequence: Tm = 75 degrees Celsius, %GC: 41 Reverse Oligo Sequence: Tm = 73 degrees Celsius, %GC: 38

These primers would be suitable as they are both of a similar length (one base pair apart), have similar melting temperature and %GC content. As the forward oligo sequence has a higher amount of base pairs, the melting point is slightly higher than the reverse sequence, however the difference is small so is still suitable for the PCR.

Q4. This question concerns the gel photo on the Result Sheet. The PCR results in lanes 3-6, where the annealing temperatures were 50o C-65o C seem to have worked equally well. (a) What would you expect if we had used an annealing temperature of say 40o C? (b) Can you explain the results in lanes 7 and 8, given that Taq polymerase is still very active at these higher temperatures? [3 marks] [Clue: (a) Too low an annealing temperature will cause the primers to make mismatched bindings on the DNA template. (b) All about primer Tms.] A) If we had used a lower annealing temperature of 40 degrees Celsius, this would result in a smaller amount of the target gene being amplified, resulting in a lot of non-target DNA. This low amount of target gene will reduce overall reaction efficiency and the incorrect temperature will reduce the efficiency of primers. B) In lanes 7 and 8, we see these results as while Taq polymerase is still active at high temperatures, the primers will not work properly at these temperatures. The melting point (Tms) means that 50% of the hydrogen bonds within primers will broken and after this temperature, due to the continued breaking of hydrogen bonds, the primer bonds with the DNA will be melted away. This bod breakage results in the gel profiles lanes of 7 and 8 in the results. Q5. Calculate the theoretical product yield at the end of 15 cycles, that is halfway to the 30 cycles used in this PCR experiment. [1.5 marks] [Clue: A calculator will do this for you.] 40 nanograms x 215 = 1310720 nanograms, = 1.31072 milligrams Q6. In this experiment, the extension time was 2.5 mins, comfortably long enough to copy the entire lmrA sequence, at ~1.7 kb, bookended by the primers, give the rule-of-thumb of ~1 kb/min. A student who was sceptical of this PCR extension rate, wanted to do the same PCR but by doubling the extension time to 5 mins. The amplified product yields for the two different PCRs were examined in an agarose gel. What differences, if any, would you expect to see in the gel lanes? Explain your answer. [2 marks] [Clue: Shouldn’t need a clue, as the answer is more or less in the question.] In the experiment where the student doubled the extension time, there would have been more cycles resulting in more gene being amplified. In an agarose gel, the lengthened extension time would have been able to be seen as the band would look larger (due to more genetic material) and the band would be closer to the origin (due to the size of the gene and how molecules move through agarose gel). Q7. You are planning a PCR experiment and you are supplied with the set of primers shown below. What is wrong with this primer set? [1.5 marks] 5’-AACTGGATACACTAGCGACATGCCCAGAGACATACTCATGCGG-3’ 5’GAATCTACACACTGGGGGGGGGATCCCCCCCCCCTAAACTGCC-3’ [Clue: Check the primer sequences.] In the second primer of this primer set, internal secondary structure must be considered. If a primer has short, internal complementary sequences (in the second one: 9 guanine bases and 9 cytosine bases) will self anneal into a loop like structure as well as annealing to the target strand. This makes this primer set incorrect to use as the primer pair can be mainly unavailable to make bonds with the DNA as the primer will be competing with the internal secondary annealing of each primer and the complementary dimer formation.

Q8. The picture below is a composite of separate sets of PCRs. Lane 1 is the template control. Lanes 2 to 5 and lanes 6 to 9 are from PCRs that used 1, 2, 5, and 10 ng of template, respectively, in otherwise identical reaction mixtures. You can just make out small amounts of product in lanes 4 and 5, but lanes 6 to 10 have copious amounts of amplified DNA. (a) Can you say which PCR parameter is different between these two sets of PCRs, that is, in lanes 2-5 versus 6-9? (b) Why are there seemingly equivalent amounts of amplified DNA in lanes 6 to 9, given that the starting template masses were so different? [3 marks] [Clue: (a) The left set is low yield and the right set is high yield, yet both begin the PCRs with the same set of template masses. Only one parameter can cause the vast increase in yield seen in the second set of products (lanes 6-9)] A) The differences between lanes 1-5 and 6-9 is the amount of cycles the DNA has gone through and how many times the template has been amplified. As lanes 6-9 have gone through more cycles, there will be an increase in yield as seen in the gel profile. B) As there is a high number of cycles, this extends the DNA band in the gel profile as there is more genetic material and as a result tends to “mask” the starting masses, making it seem like there is equivalent amounts of DNA despite different starting templates. Q9. You are about to embark on a research project in which you begin by using PCR to amplify a gene of interest from a eukaryotic chromosomal DNA source. There are two types of thermostable DNA polymerases available in the laboratory, namely Phusion and Platinum Taq DNA polymerase and Pfu Ultra DNA polymerase. Check the graph below for the fidelity of thermostable DNA polymerases. Which one should you use and why? [2 marks] [Clue: Answer is in the graph. The type of thermostable DNA polymerase that should be used in this experiment is Phusion, as it has a higher fidelity rate in comparison to Platinum Taq polymerase and Pfu Ultra. Higher fidelity rate becomes increasingly important when high accuracy is needed (such when completing a research project)....