Controls in the construction of recombinant plasmids?

[Fanghur]

I really need help with this question; my teacher told us that he is not going to be collecting these questions but that they might show up on the final exam next week; and I was not in class when we took it up due to my previous class (which was an Immunology lab where we were doing an ELISA assay, so it wasn't my fault) running late. The question is this;

 

What is the purpose of the following transformation controls in the construction of Recombinant plasmids?

 

a) Uncut plasmid

b) Ligation of cut vector that hasn’t been phosphatased.

c) Ligation of cut, phosphatased vector.

 

 

I'm pretty sure that the uncut plasmid is used so that we can calculate the efficiency of transformation, but the other two have me stumped. Any help would be greatly appreciated.

 

And like I said; this is just exam review, not an assignment

[CharonY]

a) is not for finding the efficiency of the transformation, as it will have to evaluated for each construct differently. This is because the transformation efficiency is dependent on the cells as well as the construct. However it is a control for the competence of your cells.

 

For b and c, what do you think will be the difference between those two conditions. What does the phosphatase do and how does it affect the ligation?

[Fanghur]

Well I know that the phosphatase removes the 5' phosphates from the strands to prevent them from self-ligating to themselves; but wouldn't that also prevent the insert from being ligated into the plasmid at a later time? As for b, the only thing I can think of is that it would be used to check the frequency of self-ligation; but I don't see what good that would do.

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And I meant to say "transformation frequency" not efficiency; since we could use the number of transformants to determine what percent of the total number of cells plated took in the plasmid (only they could survive on the antibiotic media).

[CharonY]

For transformation frequency the same applies. The uncut plasmid will always have a higher frequency than any ligation. How would that be a control. What info would that give you regarding your ligation?)

Again, generally it is a control for the cells. In addition you can check ligation efficiency.

 

Regarding the dephosphorylation: you are right that it prevents self-ligation. But think about it a little more. What effect would that have on the cloning? I.e. why do you dephosphorylate in the first place? Following that line of though, what would you expect when comparing b and c and what would it tell you?

[Fanghur]

I guess the purpose of dephosphorylation is to ensure that the DNA ligase only ligates the foreign DNA inserts into the cut vectors rather than simply undoing the actions of the restriction enzymes by re-ligating the two ends of the cut vectors together; since the inserts would still have their phosphates attached, but the cut ends of the vectors wouldn't.

 

But I'm still drawing a blank as to what the point of the b control is.

[CharonY]

For the most part correct. The phosphorylation is indeed done to ensure that the vector really carries an insert. Of course, doing it alone does not yield much info, but it does if compared to other results. Most controls only do make sense in comparison with another experiment.

So if you compare b to c what outcome do you expect (in terms of colonies), assuming all steps of the cloning experiment worked perfectly. Under which condition would that not be the case?

Visualize all the steps of the experiments and then speculate.

[Fanghur]

But wouldn't the number of colonies be greater in the unphosphatased vector? Since it is the selectable marker, not the insert that makes the plasmid ampicillin resistant, I would think that if anything, unphosphatased vector would stand a better chance of making it inside the cells because the vectors which self-ligated would not contain the insert and would therefore be smaller than the ones that did.

 

If I'm overlooking something obvious it's only because when we actually did this experiment the only control we actually used was the uncut plasmid; I didn't even know what alkaline phosphatase was until some time after the experiment was finished.

 

I'm not being lazy, CharonY. We just haven't been taught this stuff before; or if we have it wasn't in any great detail.

[CharonY]

You are probably thinking too complicated. Let us take a step back. What are controls for? Answer: they allow us to check what happened at every step of the experiment. Now back to the controls at hand.

Non-dephosphorylated vectors can religate and therefore you are correct in assuming that normally you would have a higher number of clones transformed with non-phosphorylated vectors. However, you would also have to look longer to find a clone with the proper insert.

 

Now imagine you conduct experiments in the op and imagine the following scenarios:

1) a has the highest amount, followed by b, then c

2) a has the highest amount, but b and c are equal (though both lower than a)

3) a, b and c are equal

 

What happened at every step? It appears to me that you understood the individual reactions, but you just have to piece the things together. Again, remember that controls are supposed to work in a specific way, but if they don't they hint at something gone wrong.

[Fanghur]

Oh I think I get it now; the reason that you run both a transformation control with both the phosphatased and unphosphatased vector is so that you will be able to tell whether or not the transformation worked the way it was supposed to when you do the actual transformation of the vector plus your insert. Assuming that no self-ligation occurred, then the number of colonies for the (let's call it the unknown, for lack of a better term) should be roughly the same as the control where the vector was treated with alkaline phosphatase, in which case you'll know if the cloning procedure worked properly.

 

On the other hand, if there was self-ligating, then the unknown would have the same (roughly) number of colonies as the control that was not treated with alkaline phosphatase, which would indicate that the cloning procedure was not very efficient.

 

Am I at least on the right track?

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And just to take that logic a bit further, that's also why the uncut plasmid would have the most colonies of the three controls; because it would essentially be the same as the self-ligated plasmids in terms of size. So it would have more colonies than the unphosphatased (control b) because b would be a mixture of both self-ligated and inserted plasmids; so since the transformation rate is lower for the plasmids with the insert, there wouldn't be as many colonies in control b as in control a.

[CharonY]

Essentially you are right (I think), though it would be easier if you would be a bit more organized in formulating your thoughts.

 

Plasmid size does not figure in right now. You have the following steps starting from having the plasmid (and insert) purified:

restriction, dephposphorylation, ligation, transformation. Each of the steps can be incomplete and you make controls to ensure that that particular step worked.

 

Based on this you are correct to assume that if everything worked perfectly, the plasmids gained from b should be lower than a, but how does it compare to c?

Maybe you should give it a try and answer the hypothetical situations that I gave you. Also keep the subsequent steps of a cloning experiment in mind.