if I have a vector of 3421 bp and I want a 3molar excess of my sequence which is about 11kb, how do I go about

Answer 1

Good old ligation...

Let's assume you have Cv ng/microL vector and Ci ng/microL insert
How much volume of vector (Vv) and insert (Vi) do you use?
First of all choose how many ng of vector you want to have in the ligation reaction.
Let's say you want m.
Then you need Vv= m/Cv microL

You need to have 3:1 insert/vector molar ratio.
You would need to find the molecular weights to convert to moles. However there is the approximation that the average MW for a bp (2 bases, on on each strand) is 660.
So for your vector: MWv=660*bpv where bp is the number of base pairs of your vector (3421)
for your insert MWi=660*bpi where bpi the number of base pairs of the insert (11000)

moles vector = Cv*Vv/MWv
moles insert = Ci*Vi/MWi
You want 3/1=mole insert/mole vector =>
mole insert =3 mole vector =>
Ci*Vi/MWi = 3* Cv*Vv/MWv =>

Vi =3* (Cv/Ci)* (MWi/MWv)*Vv we substitute the MW

Vi =3* (Cv/Ci)* (660*bpi/660*bpv)*Vv =>

Vi =3* (Cv/Ci)* (bpi/bpv) * Vv microL

This is a general equation you can use, plugging in the appropriate values each time. NOTE that here the concentration is expressed in ng/microL (which is equivalent to microg/mL)

Trying to insert 11 Kb in a 3 Kb vector can be really tough, so actually you might want to try different ratios and not only 3:1.
Good luck with it.