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Exercise 12 Analysis - Exercise 12 Analysis Giselle...

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Log of DNA marker size (bp) Rf Exercise 12 Analysis Giselle Zornberg 12-6-07 1. The colonies with vector DNA only should only have one band, whereas the colonies transformed with a recombinant vector should have two bands (or 3 because of contamination). The (-) colonies represented the colonies with vector DNA only because no restriction enzymes were present. The (+) colonies represent the colonies with a recombinant vector. Wells 2 and 4, which have T-white (+) and C2 blue (+), respectively, reveal two bands because restriction enzymes are present to separate the insert from the rest of the plasmid. Wells 1 and 3, which represent T-white (-) and C2 blue (-), contain one band because no restriction enzymes were present to cut the plasmid. Several of wells, namely well 1, have extra bands that were caused by contamination. Rf Log of DNA Marker size (bp) 0.982 2 0.912 2.301 0.8596 2.477 0.807 2.602 0.6842 2.699 0.614 2.913 0.526 2.929 0.456 3 0.421 3.217 0.386 3.301 0.368 3.699 0.351 4 0.333 4.079 2. a. plasmid pBluescript (in bp) = -2.519 (.526) + 4.508 = 3.183 (MW band 7) 10 3.183 = 1524.1bp = 1.524kb data taken from well with C2 Blue (-) b. DNA fragment that was inserted into pBluescript = -2.519(.614) + 4.508 (MW band 6)
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= 2.961 10 2.961 = 914.82bp = .914kb data taken from well with T-white (+) c. PCR product in control DNA sample: -2.519(.333) + 4.508 = 3.67 (MW band 13) 10 3.67 = 4668.45bp = 4.668kb d. PCR product in unknown DNA sample: -2.519(.386) + 4.508 = 3.536 (MW band 10) 10 3.536 = 3432bp = 3.432kb 3. a. C 1 = 20ng/μl V 1 = 5μl C 2 = ? V 2 = 1000μl C 1 V 1 = C 2 V 2 (20ng/μl)( 5μl) = C 2 (1000μl) C 2 = 0.1 ng/μl The final DNA concentration was 0.1 ng/μl b. 1:10 dilution: 0.01 ng/μl 1:100 dilution: 0.001 ng/μl 1:1000 dilution: 0.0001 ng/μl c. 1:10 dilution: (.01 ng/μl)(100 μg) = 1ng = .001 μg 1:100 dilution: (.001 ng/μg)(100 μg) = .1ng = .0001 μg 1:1000 dilution: (.0001 ng/μg)(100 μg) = .01ng = .00001 μg d. Transformation efficiency: 1:10 Test plate: (221+3)/.001 μg = 224000 (2.24 X 10 5 ) colonies per μgDNA 1:100 Test plate: (13)/.0001 μg = 130000 (1.3 X 10 5 ) colonies per μgDNA 1:1000 was told to disregard because no data was collected regarding the number of colonies in 1:1000 test plate e. The efficiency of the different plates was not the same. They should be the same because the conditions in preparing the plasmid and bacteria were the same. A possible reason for this difference is that samples that were not mixed well during the dilution process and thus could have had an uneven distribution of bacteria, resulting in a disproportional concentration of bacteria on each Petri dish.
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4. SKIP 5.
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