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..?

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

HI,

Here is the explanation

PROCEDURE - Determining DNA concentrations for ligationsAs 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