How were the answers in red calculated? Set up a standard ligation reaction for

Question

How were the answers in red calculated?

Set up a standard ligation reaction for your PCR product (6.3 kb) with your new vector (3.8 kb) in a 10 1 total volume. For the sake of simplicity, pretend that you have access to a P2 pipette. Remember the following: a. You want a 6:1 (insert:vector) ratio b. Use the maximum amount of DNA possible (250 ng) c. The A26o reading for your purified digested vector is 0.07 (sample was diluted 1:30) d. The A26o reading for your purified digested insert is 0.3 (sample was diluted 1:15). 2X ligation buffer (with ATP) Vector (105 ng/uL) Insert (225 ng/uL) T4 DNA ligase (400U/ Water Total 0.22 L (22.7 ng) 1 L (227.3 ng) 2.78 10 You'll need 10X insert for every 1X vector (which equates to 11X parts of total DNA in the reaction). 250 ng/11 parts 22.7 ng. So you'll need 22.7 ng of vector and 227.3 ng of insert for this reaction.

Explanation / Answer

2X ligation buffer (with ATP) = 5 µl (This is used to maintain as standard protocol, the total reaction volume is 10 µl and ligation buffer is 2X, so to make it 1X solution, we have to use 5 µl 2X buffer in 10 µl).

Vector (105 ng/µl). We want to use 1:10 times dilution of vector:insert. So, if we consider 105 X 0.22 = 23.1 ng

So, we need .22 µl of vector insert where 23.1 ng of vector is present.

We need insert 231 ng (because insert should be 10 times as vector)

So, we need 1 µl of insert which is 225 ng that is nearly 10 times equals to the 231 ng of insert.

Water is needed for this reaction is (10 - (5 + 0.22 + 1 + 1) = 2.78 µl

So, we need 23.1 ng of vector and 231 ng of insert for this reaction. That will exactly make 1:10 (vector:insert).

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