Consider that I cut my vector backbone with a blunt end cutter (EcoRV) and Iam not performing the alkaline phosphate treatment which will preserve the 5' phosphate on the vector. Is it necessary for the PCR amplified insert (blunt end) to have 5' phosphate as well. Practically I know that it is not necessary as the above cloning has worked for me in the absence of 5' phosphate in the insert.
I also read somewhere that one of the phosphates ensure that the plasmid gets circularised and the nick on the other strand can be repaired by E.coli cells following transformation. Can anyone provide information on the exact repair process that goes inside the E.coli in this context (which would involve sealing the 3' and 5' OH groups).
Ligase enzyme will repair the nick in the strands. Any basic molecular biology textbook will show the details. A fresh PCR product does have a 5' phosphate. Phosphotase treatment will increase the efficiency of a ligation reaction, but it's not strictly necessary.
Ligation of two DNA fragments (eg vector + insert) needs at least one of the DNA ends to have 5' phosphate for the nucleofilic attack of 3'OH end to the phosphate group. If you have just one fragment with 5'phosphate, this will be sufficient for the ligation of that end with the 3'end of the other fragment. As you wrote, the other nick that is not ligated due to the absence of 5'phosphate, is repaired in cells.
Usually, primers are provided without 5'phospho group, unless you specifically request it or you add it on your own with a kinase reaction; thus PCR products usually don't have the 5' phosphate group. As soon as you cut the PCR product with restriction enzymes (RE) (blunt ended or sticky one) you will produce the 5' Phosphate you may need for ligation.
A general procedure for cloning is to remove 5'phosphate group on the linearized pladmid backbone with phosphatases to reduce self-circularization (especially when you single cut the plasmid or cutting with 2 enzymes with compatible ends) while using the 5'phosphate of the insert for the ligation.
Hi there,
5'OH can be removed by 5'exonuclease activity which generates 5'P then 5' to 3' polymerase activity at 3'OH will fill the gap. Finally ligation between 3'OH and 5'P will close the circle.
- Badges
- Science topic
- Similar topics
- Biological Science
- Molecular Biology
- Molecular Methods
- Cloning