I am optimizing a ligation step of a plasmid using a vector: 14801bp, 149ng/µl and an insert: 403bp, 280ng/µl. I need to know the molar concentrations of these DNAs so as to understand the molar concentrations that I need for 1:3 ligation.
Thank you.
Leah
You would need to factor in the GC content of your sequence. If you know the sequence you can then calculate the percentage of your weight which is each of the different nucleotides and then calculate the molarity of each before adding the numbers together.
Here how I claculate the needed insert mass :
3/1=[Mass(insert)/Mass(Vector)]*[Length(vector)/Length(insert)]
Here, the optimal vector mass is 100ng
So, M(ins)=3[100*L(ins)]/L(vector)
In your case, M(ins)= 3[100*403]/14801 = 8.17ng
Meaning that you need for each 100ng of vector, 8.17ng of your insert.
You wrote that your insert: 403bp, 280ng/µl, meaning that for each 1ul of insert you need 23ul vector 14801bp, 149ng/µl.
However, I think you use a very little insert into avery big vector/ You may consider using smaller vector, unless you are required by protocol
Hi Haida,
Thank you for the insight.
Yes, my protocol needs that size of the insert into the plasmid.
Thank you for the calculations too.
I appreciate.
Keep in mind that since you have a large plasmid you'll need about 50ng plasmid for every 3 kb of plasmid size. Your plasmid is about 15KB (give or take) or 5 times that size. So, in your case you'll need about 250 ng plasmid for a reaction. Then, you calculate the ratio of insert needed. One way to think about the relative ratios is the size difference between the vector and the insert. Your vector is about 36 times larger than your insert. If you use roughly 2 ul of your vector (at 150 ng per ul), then that is a good amount of vector for 1 ligation. You'll need to dilute your insert a lot though. If you were to dilute your insert to also be 150 ng your will need 1/6 ul for a 1:3 ratio of vector:insert. Hope this helps and isn't just a math jumble.
- Badges
- Science topic