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notes are attached below for further information
Define the term competent cells as it relates to this laboratory.
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- In DNA isolation techniques, a washing step is always done prior to the final resuspension. What is the purpose of this step? In DNA isolation from blood samples, why does the vial for blood storage contain EDTA? In the preparation for DNA isolation in plants, the plant source is refrigerated and ground prior to extraction. Why is this so? Why are DNA pellets air-dried before resuspension in buffer?Electrophoresis is an extremely useful procedure when applied to analysis of nucleic acids as it can resolve molecules of different sizes with relative ease and accuracy. Large molecules migrate more slowly than small molecules in agarose gels. However, the fact that nucleic acids of the same length may exist in a variety of conformations can often complicate the interpretation of electrophoretic separations. For instance, when a single species of a bacterial plasmid is isolated from cells, the individual plasmids may exist in three forms (depending on the genotype of their host and conditions of isolation): superhelical/supercoiled (form I), nicked/ open circle (form II), and linear (form III). Form I is compact and very tightly coiled, with both DNA strands continuous. Form II exists as a loose circle because one of the two DNA strands has been broken, thus releasing the supercoil. All three have the same mass, but each will migrate at a different rate through a gel. Based on your…DNA extraction, PCR, gel electrophoresis, and DNA sequencing of PCR products. Give a description of these methods and the materials required to perform each of these molecular techniques
- Coomassie dye is the equivalent of what component used in analysis of DNA by agarose gel electrophoresis? How do you expect a Coomassie-stained gel to look compared to the result you might expect once you have completed your Western blot?Recombinant protein production by a genetically modified Escherichia coli strain is proportional to cell growth. Ammonia is used as a nitrogen source for aerobic glucose respiration. The recombinant protein has the general formula CH1,55O0,31N0,25, while that of the cellular biomass is CH1,77O0,49N0,24. The biomass yield from glucose equals 0.50 g/g, while the recombinant protein yield from glucose corresponds to 20% of the cell yield from substrate.a) How much ammonia is required? What is the oxygen demand? (b) If the biomass yield remains the same, what are the ammonia and oxygen requirements for a wild-type strain of E. coli, with cell biomass of the same elemental composition, but unable to synthesize the recombinant protein? (c) On an industrial scale, cultivation takes place in a continuous fermenter at 28°C and the desired recombinant protein production rate is 7 g/h. Since the viscosity of the culture broth is considerable, the energy input due to agitation cannot be neglected.…Below is the result of PCR products running on agarose gel electrophoresis. You expect to see one band, but the results show two bands, as shown below. Why is there a lower band? What are they? How can we avoid such a band?
- In a cotransformation experiment DNA was isolated from a donor strain that was proA+ and strC+ and sensitive to tetracycline. (The proA and strC genes confer the ability to synthesize proline and confer streptomycin resistance, respectively.) A recipient strain is proA− and strC− and isresistant to tetracycline. After transformation, the bacteria werefirst streaked on a medium containing proline, streptomycin, andtetracycline. Colonies were then restreaked on a medium containingstreptomycin and tetracycline. (Note: Each type of medium hadcarbon and nitrogen sources for growth.) The following resultswere obtained:70 colonies grew on the medium containing proline, streptomycin,and tetracycline, but only 2 of these 70 colonies grew whenrestreaked on the medium containing streptomycin and tetracyclinebut lacking proline.What would you expect the cotransformation frequency to be ifthe average size of the DNA fragments was 4 minutes and thetwo genes were 1.4 minutes apart?In the very first round of PCR using genomic DNA, the DNA primers prime synthesis that terminates only when the cycle ends (or when a random end of DNA is encountered). Yet, by the end of 20 to 30 cycles - a typical amplification - the only visible product is defined precisely by the ends of the DNA primers. Explain how.In a PCR-based crime scene investigation, similar to the one presented in the lab module with Brother Y and Brother X, there is a sample of DNA from a crime scene that is likely to belong to the guilty party. Based on the gel photo below, which shows the results of an electrophoresis gel following PCR amplification at one locus of 5 DNA samples - one crime scene sample and 4 suspects - which suspects can be excluded from this investigation? [Keep in mind that it is not possible for a heterozygous person to leave only one allele at a crime scene. If any one allele does not match, then that suspect is eliminated.] choose all that apply
- A 22-kb piece of DNA has the following restriction sites:A batch of this DNA is first fully digested by HpaI alone, then another batch is fully digested by HindIII alone, and finally, a third batch is fully digested by both HpaI and HindIII together. The fragments resulting from each of the three digestions are placed in separate wells of an agarose gel, separated by gel electrophoresis, and stained by ethidium bromide. Draw the bands as they would appear on the gel.Our PCR samples already contain loading dye, but sometimes this isn’t the case. If your samples didn’t already contain dye and you wanted to load your PCR sample onto an agarose gel, you’d need to add loading dye to the proper concentration. There is a 6X loading dye available for use; how many µl of this loading dye will you add to 10 µl of your sample so that it is at a 1X working concentration? Show your work.A student is trying to add 15.0 ng of DNA template to a 20.0 µL PCR. The DNA template is at a concentration of 65.0 ng µL-1, and the student determines that a serial dilution is required because directly adding the DNA template would require a volume less than 1.00 µL. The student wants to prepare an intermediate solution at a concentration of 15.0 ng µL-1. If the DNA template stock will be mixed with 13.0 µL of ultrapure H2O, calculate the volume (in µL) of the DNA template required to prepare the intermediate solution.