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PriyaSar's picture

An insert of 1 kb is ligated to plasmid of 4 kb in a molar ratio 4: 1 respectively in a final DNA concentration of 10 µg/ ml.The amount of insert and plasmid required in µg will be..?

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Here is the explanation

PROCEDURE - Determining DNA concentrations for ligations

As an example, for any given ligation, use the following formula:

amount of insert = insert size / vector size x 3 (molar ratio of insert / vector) x amount of vector to be used

In the case of a 4.0 kb sized vector (after proper digestion) and a 0.4 kb sized fragment:

In order to use 1 mg of vector (cohesive end ligation), the following calculations are needed:

insert size / vector size = 0.4 / 4 = 0.1 (1/10)

molar ratio of insert / vector (since this is a cohesive end ligation with two different ends, there is no problem with multiple inserts, so a high ratio of insert to vector is possible -for blunt end ligations this ratio should be much lower, even 1:1) = 3/1

Therefore, for a ligation between a 4.0 kb vector and a 0.4 kb fragment, starting with 1 mg of vector, we get:

amount of insert = 0.4/4.0 x 3 (3/1) x 1 mg

amount of insert for 1 mg of vector = 0.9 mg

If this had been a blunt end ligation, a 1:1 ratio would have been used and the amount of insert necessary would have been 0.3 mg

Note that these calculations were made for the equivalent of 10 units of ligase