CHEM112B_Midterm (SP09)_Key

CHEM112B_Midterm (SP09)_Key - chem-112b-midterm-key-S2009...

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chem-112b-midterm-key-S2009.doc 5/2/2009 Page 1 of 4 Chem 112B: Molecular Biochemistry (DNA) Lab. Midterm Answer Key S2009 Instructor: T. Crowley Eight questions, 100 pts total. For questions requiring a numerical answer, show calculations for partial credit. 1) (12 pts) In the plasmid DNA purification procedure performed in exercise 6, a mixture of NaOH and the detergent sodium dodecyl sulfate (SDS) was applied to the resuspended bacteria. If the NaOH had been omitted but all subsequent steps performed as specified in the lab manual, which of the following would be the most likely result? Explain how you made your choice . a) The cells would not have been lysed, therefore no DNA would have been released. b) The yield of plasmid DNA would have been as expected but with significant chromosomal DNA contamination. c) Chromosomal DNA, but no plasmid, would have been purified. Correct answer is b . Purpose of NaOH is to raise the pH and disrupt hydrogen bonds between the bases in the complementary strands of all DNA molecules. In the next step the pH is brought back down by adding K-acetate to 0.7 M and the two strands can reanneal. Because the two strands in each plasmid molecule are interlocked rings, they reanneal very efficiently to their original configuration. However the chromosomal DNA has been broken into linear fragments during cell lysis, so the “partner” strands do not all reanneal properly and a tangled mass of chromosomal DNA, protein and SDS forms in the high salt. If the NaOH is omitted from the procedure, the chromosomal DNA will never be denatured and therefore remain soluble. This will result in contamination of the plasmid DNA. 2) (10 pts) Why is it necessary to resuspend bacteria in a solution containing KCl, MnCl 2 , CaCl 2 and dimethylsulfoxide to make them competent for transformation? The E. coli bacteria used for recombinant DNA work do not take up foreign DNA very efficiently in typical growth media. One reason is that the plasma membrane has a nonpolar interior that cannot be traversed by negatively-charged DNA molecules. The DMSO (much less polar than water) and metal ions make the membrane more permeable to the plasmid DNA. DMSO also increases the permeability of the membrane to water. This increases the survival rate of cells put through a freeze-thaw cycle. (Water expands upon freezing. Without a permeabilization agent, this expansion will crack the membrane and kill the cell.) 3) A Pst I fragment is ligated into the pGEM vector (3.2 kb) and the resulting construct introduced into E. coli by transformation. Plasmid DNA is purified and analyzed on the two gels shown below. (In each case, Hind III-lambda markers are in the leftmost lane and the construct is in the rightmost lane. Marker lengths given
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This note was uploaded on 05/26/2010 for the course CHEM chem 112b taught by Professor Crowley during the Spring '09 term at UCSD.

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CHEM112B_Midterm (SP09)_Key - chem-112b-midterm-key-S2009...

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