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Please help to chose Sample and Provide a reason why you chose it.

Question 9 After 2 hours, a DNA gel is run on 4 different ligation samples. Each sample is run twice, once after incubating i

After 2 hours, a DNA gel is run on 4 different lination camnlar Cacheamnla in Sample 1 Sample 2 Question 9 Sample 3 Sample 4

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Answer #1

Given the approximate size of pET 28a vector = 5400 bp or 5.4 Kb.

We are digesting four different samples, where a DNA fragment has been cloned in pET28a. The use of two enzymes Xho 1 and Nde 1 indicate directional cloning of the DNA fragment, as these two regions are found in the MCS of the cloning vector. This means that digestion with a single enzyme, Xho 1 , should linearlize the plasmid, and digestion with Xho 1 and Nde 1 should release the cloned fragment.

If we look at sample 1, digestion with single as well as double enzyme has resulted in the release of fragment of ~1.3 Kb. so this is not the right sample, unless a single enzyme is used for cloning. but this is not the case here.

Sample 2: This shows a linerarized vector at ~10 Kb, which is doube the size of pET 28a, but digestion with two enzymes resulted in a ~5 kb and a ~1.3 kb fragment, whcih does not add up to the ~10kb vector. This is not the right sample either.

Sample 3: This appears to the the right sample. Digestion with Xho 1 shows a linearized vector at ~6.3 kb and with both enzymes there are two bands , one at ~5 kb and another at ~1.3 kb. This does add up to the original size of the linearized vector when xho 1 alone was used.

Sample 4: This appears wrong, as digestion with xho 1 yields a different size fragment of ~2 kb along with the ~5kb fragment. The total size here is ~7 kb. whereas digestion with both enzymes yielded ~5 kb and ~1.3 kb fragment. The same vector cannot be of two sizes 7 kb and 6.3 kb!

The sample that correctly incorporated the DNA into the vector therefore, is sample 3.

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