The ligation reaction is designed so that the vector and the insert DNA is at 1:

Question

The ligation reaction is designed so that the vector and the insert DNA is at 1:3 molar ratio. If we use 50 ng of vector DNA for the ligation reaction, how much insert DNA (in ng) do we need? (The vector is roughly 4200 bp and the insert is 500 bp.) If the concentrations of vector DNA and insert DNA are both 5 ng/ul, how much vector and how much insert do you need (in ul)?
The ligation reaction is designed so that the vector and the insert DNA is at 1:3 molar ratio. If we use 50 ng of vector DNA for the ligation reaction, how much insert DNA (in ng) do we need? (The vector is roughly 4200 bp and the insert is 500 bp.) If the concentrations of vector DNA and insert DNA are both 5 ng/ul, how much vector and how much insert do you need (in ul)?
The ligation reaction is designed so that the vector and the insert DNA is at 1:3 molar ratio. If we use 50 ng of vector DNA for the ligation reaction, how much insert DNA (in ng) do we need? (The vector is roughly 4200 bp and the insert is 500 bp.) If the concentrations of vector DNA and insert DNA are both 5 ng/ul, how much vector and how much insert do you need (in ul)?

Explanation / Answer

The ratio of vector and insert DNA is 1:3

insert DNA mass in ng = 3 X (insert DNA length in bp/ vector length in bp) X vector mass in ng

                                      = 3 X ( 500/4200) X 50

                                       =17.85

If both are 5 ng/ l, then we will need 1l of vector and 3 l of DNA insert.

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