Q&A_Recombinant - Q1 How is the DNA amplified in...

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Q1 How is the DNA amplified in the shaking water-bath? There are two kinds of plasmids, Stringent plasmids only replicate when the bacterial chromosome replicates, the other kind which is what our plasmids were are relaxed plasmid and they replicate independently of the main chromosome and hence we will get anything from 10-200 copies per cell. Q2 How to calculate DNA mass? Since Concentration = Mass/Volume 0.01 μ g (conc. of DNA used)/ μ L* 25 μ L (vol of DNA used)* 564 bp(smallest band seen)/48502 bp (total size of λ phage DNA) This is equal to 2.9 x 10 -3 μ g or 2.9 ng (nano gram) Using the same formula and replacing 564 with other fragment, 125(smallest fragment that did not show on our gel), we get 6.4 x 10 -3 μ g or 0.64 ng (nano gram) Therefore, the minimum mass of DNA that is detectable with our system that we had is less than about 2.9ng but more than about 0.6 ng. Q3 What we are observing with electrophoresis? Electrophoresis is used to separate DNA molecule of differing sizes, for this we had a matrix (agarose gel) through which the molecules of DNA will move as influenced by the electrical field established and the size of the pores within the gel. Q4 Explain what the process of joining the pKAN and pAMP entails?/ The type of recombinant that we need to get is a little confusing. Ligation of the (fragment) plasmids that carry AMP r and KAN r gene will make those Ecoli cells immune to ampicillin and kanamycin (antibiotics) and hence those cells will be able to grow in the agar plates that did have one or both of those antibiotics. Our purpose was to create an E coli colony that has a ligated plasmid giving resistance to both hence carry both the fragments that posses AMP r and KAN r gene. The ligated plasmid mixture that you prepared and labeled as ‘L’ which we used
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Q&A_Recombinant - Q1 How is the DNA amplified in...

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