Question on digestion - (Sep/19/2005 )

Hi, guys:

I am a new student working on molecular biology. Today I digested a vector containing 4kb fragment with spe1 and xho1. I expected to get 2 bands, but I got 3. Does it mean I failed this time and why?
Another question: Will ligation inhibit transformation? What is the key point for ligation and transformation?

Kindest regards!

Yang

Are you sure the vector only had one cut site for each enzyme? If the vectors has two cuts sites for on of the enzymes that would lead to three bands. Did you use the enyzmes together or seperate? Either way, you need to make sure the buffer you used is compatible for both enzymes. If one of the enzymes isn't 100% efficient in the buffer you used that might lead to three bands (2 smaller bands representing the expected bands and 1 that is vector that hasn't been completely cut). Did you run a standard ladder on the gel? You should be able to use that to figure out what size the fragments are that you're seeing and determine from there what the problem is.

Hi, thanks for reply, I am so sorry I didn't describe my question clearly.

First of all, I am sure the vector has neither of these two restriction sites. And I use buffer 1 for digestion at the same time. The activities of sep1 and xho1 is 55 and 100 respectively. I think my vector has not been cut completely. Do you think it is possible to get correct result if I increase the amount of the enzyme and decrease the amount of the plasmid, and prolong the restriction time?

Kindest regards!

Yang

a good control is to add only linearized plasmid when you are looking at the gel to check for complete digestion

that will tell you if the upper band is the same as vector with one cut

good luck

i think you are right, 55 is pretty wimpy in comparison

I agree with the developing consensus here -- you likely have one band representing linear plasmid (cut only once, probably by XhoI), and two additional bands resulting from plasmid cut once by SpeI and once by XhoI.

You might be able to confirm this a bit more by looking at the size of the bands produced -- you did run a DNA ladder on the gel, right? If what we think occurred actually did, you'd have a 4 kb fragment (your insert released by SpeI/XhoI digestion), a band representing your vector only (what's left over after release of your insert), and a larger band the size of your vector plus 4 kb, representing the linearized plasmid.

Possibilities to overcome such difficulties would include incubating the double digestion longer, separating the digestions into two steps, or perhaps finding an isoschizimer of either of the two enzymes (PaeR7I, TliI, BssHI, Sfr274I, and SlaI are isoschizimers of XhoI, and AhlI and BcuI are isoschizimers of SpeI) that might have different optimum buffers.