What is a library?
First let us discuss the vectors that can be used - for example lambda
Before we get too involved in discussing the DNA insertions in a library, let us discuss the vectors that hold the foreign DNA. You could think of this as being the "bindings" and "covers" for the books in a real library - they carry the information by holding together the pages.
Almost any recombinant vector can be used to make a library, but some are easier to work with than others. Phage lambda derivatives are easy to use for genomic and cDNA libraries, because the plaques are a stable reference stock for your clones. When you lift a membrane from the collection of plaques, you always leave some plaques behind that can be picked out and propagated.
We have discussed this already for lambda gt11 - the idea that you can grow a "lawn" of bacteria with plaques
in it, induce expression of the fusion protein with IPTG, and collect an imprint
of the plaques on a circle of nitrocellulose. You could detect your specific phage
using an antibody probe (if an expression system) or a DNA probe (by hybridization).
For gt11 we'll assume that we've got an antibody probe.
| Here's how it works. First, you mix the E. coli host and the lambda gt11 library in a tube, wait about 30 minutes for the phage to adsorb to the bacteria, then mix in some molten agar-containing medium (cooled to 45 C), and "plate" the mixture onto a bacterial plate that already has a layer of solidified agar-containing medium. This first layer of agar may contain X-gal and IPTG (although it is usually the case that the IPTG is soaked into the nitrocellulose filter - we'll forget about that so it is easier to show you what happens) |
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After overnight growth the plate might look like this when you hold it up to the light. Plaques are visible, and some of the plaques are blue. As you recall, the blue plaques represent bacteriophage that do not have insertions, so the lacZ gene is uninterrupted. The colorless plaques have insertions into the lacZ gene. If this is a cDNA library, then we expect that each colorless plaque might represent a different inserted cDNA. |
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| If we apply a nitrocellulose membrane to this plate, allowing it to soak up the proteins that have been expressed and released from the dying cells, then probe the nitrocellulose with an antibody and detect the antibody binding with a secondary antibody and enzyme-linked detection system, we might see results as shown on the right. Note that it is important to align the nitrocellulose membrane with the plate so you know which end is up! |
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| If you have marked the nitrocellulose so that you know its orientation, you can go back to the plate and cut out (punch out) the plaque that generated the positive signal by antibody binding. There will still be some phage in the plaque that can be regrown, although some of the phage are stuck to the nitrocellulose (and lost) and others will have diffused out of the plaque. |
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Now you have a phage stock that is partially purified. Since the phage diffuse in
the plate, you can only expect that your specific phage will be enriched in the population
that you cut out. You need to repeat the plating experiment and detection system
several times to enrich for the phage that is the right one. Your first attempt at
replating might go something like this:
Note that a fraction of the plaques on the plate are positive with the antibody, but most are still other things. You then need to pick an isolated "positive" plaque from this second generation plate and try again with a third (and probably fourth) generation plate to increase the level of purity. |
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This is what you eventually hope to see. In this case, every plaque in the sample is detected as a "positive" with the antibody probe. That means that the phage you have picked is now "plaque purified", and all of the phage in the stock are of one kind. It is now a clone. |
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A lambda clone can be grown in large quantities and stored as a purified bacteriophage
stock, at a concentration of 109 to 1011 phage per ml. You
may add chloroform to the stock to preserve it. You may also preserve the lambda
clone as purified lambda DNA, but to infect cells you would need to "package"
the DNA into capsids again.
What next?
You now have a gt11 clone that expresses a fusion protein that reacts or cross-reacts with your antibody probe. It could be a good result, but then again it might not be the protein you are really looking for. You will need to sequence the inserted DNA to see what you've got, and it is often the case that you want to move the inserted DNA to a plasmid vector.
If the EcoRI sites are still intact in the gt11 vector, you could isolate the insertion
by digesting the purified lambda DNA with EcoRI and isolating the foreign DNA insert
on a gel. You could also use the polymerase chain reaction to isolate the insertion,
but this might be difficult if the insertion is large (remember that gt11 can clone
up to 9 kbp).
Plasmid vectors
Lucky thing that you remembered to replicate the plate before adsorbing the colonies to nitrocellulose - now you have a reference colony you can pick and grow!
A second problem with standard laboratory plasmids is that it is difficult to clone large insertions into them, and this is particularly a problem if the library consists of genomic DNA fragments rather than cDNAs. Most plasmids do not grow well if they have large (10 kbp) insertions. While that is similar to the limit for gt10 and gt11, we'll be discussing later, lambda phage that can handle up to 23 kbp of DNA. The problem of "insertion size" and representation in the library may become more clear from the next example.
How many different recombinants must a library have?
Since a library contains random fragments, you can never be 100%
certain that it contains a clone you desire.
Here's an example of the problem. Suppose you have a genome of 100 kbp, and you break
it into random fragments of about 10 kbp for cloning (of course a real genome would
be much larger). If you obtain a library with ten phage in it, would the entire genome
be represented?
The answer is no! Some of the clones would overlap.
Let's try the same problem, substituting umbrellas for DNA clones:
Suppose we want to protect a group of movie stars from the rain, while they're walking from their limos to the doors of the Dorothy Chandler Pavillion ( a journey of 100 killibradpitts (kbp) - a new measure of distance). We have 10 umbrellas on hand, each able to cover 1/10th of the distance from the street to the door. If we distribute them randomly, will they cover the entire path and keep the stars dry?
Of course, if we are thinking that umbrellas = clones and sidewalk = genome, then some sequences in the genome will not have been cloned, and will not be represented in your library. That's fine, so long as the sequences that you want to clone are not left out, but we are not always so fortunate. What would happen if we made the library twice as big? Let's go back to the Academy Awards, and see how we fare with twice the coverage:
As you can see, we got closer to complete coverage, but still have some unprotected areas. To be precise, only 86 killibradpitts of sidewalk are covered and 14 are uncovered.
With our analogy of clones, the uncovered regions represent uncloned sequences in the genome. The way we can calculate what's missing is pretty easy -- it's based on the Poisson distribution. If we have one "covering" of the genome (as in the ten clone example) the chance of a sequence being cloned is (1- 1/exp(1)) or about 0.63 (or 63%), where exp(x) = the base of the natural logarithm e to the x power, . If we have two "coverings" of the genome (as in the twenty clone example) the chance of a sequence being cloned is (1- 1/exp(2)) or about 0.86 (or 86%), where exp(2) means the square of e. The pattern continues as your library contains more independent clones. With ten "coverings", the chances reach 0.99995 (or 99.995%), but you can see that you never reach absolute certainty of cloning any particular sequence.