Question

Calculate the volume of plasmid and insert needed for a ligation reaction using the specifications for NEBL T4 DNA ligase (0.020 pmol vector, 0.060 pmol insert). This reaction is ligating a 3kb vector (25ng/μL) and a 2kb (10ng/μL) insert together, and the fragments have complementary sticky ends. Round final answers to the nearest hundredths place.

Answer 1

Answer:

The correct answer -

1. volume of the plasmid: 2 ml

2. insert needed = 100 mg

Explanation:

Calculating the inserted by the formula:

$\dfrac{insert\ size}{vector\ size}\times\dfrac{moles\ of\ insert}{moles\ of\ vector}=\dfrac{insert}{vector}$

for a mormal vector size using 50 mg of vector DNA per ligation reaction.

take x as required insert:

$\dfrac{2}{3}\times\dfrac{0.060}{0.020}=\dfrac{x}{50}$

$\dfrac{2}{3}\times\dfrac{3}{1}=\dfrac{x}{50}$

$\dfrac{6}{3}=\dfrac{x}{50}$

3x = 300

$x=\dfrac{300}{3}$

x = 100 mg

Volume of plasmid vector is :

$\text{volume of plasmid}=\dfrac{\text{amount of vector needed}}{\text{conc. of vector}}=\dfrac{50}{25}$

volume of plamid vector = 2 ml.