1. (a) Imagine you are using a phage lambda vector to make a cDNA library. When you plate out the
library you will likely do so in a manner that allows several plaques to overlap partially (because you
need to pack a lot of clones into a reasonable amount of space). You want to make a DNA prep from
one particular plaque that does appear to overlap with others. You start by picking a plug of agar from
the plate that likely includes 90% correct phage but also 10% contamination with other phage.
(i) How would you make a pure culture of just the single recombinant phage without contamination
from neighbors (don’t worry about how you check which pure phage is correct and which is the
contaminant- just think about how to get a pure culture from the original mixed phage sample)?
Explain the steps.
(ii) Once you have the purified phage and you amplified it further by infecting, say, a 10ml culture of
growing bacteria, you want to purify phage DNA largely uncontaminated by E. coli genomic DNA.
What key step(s) in the purification procedure allows you to do this?
(b) Imagine you have a recombinant plasmid, “A”, of 5kb (a 2kb insert in a 3kb vector). Let’s say
there is one EcoRI site and one BamHI site, 1.5kb apart (3.5kb apart the long way around). You have a
second plasmid, “B” which is identical except that it has a G-C base-pair at one position instead of an
A-T base-pair. You cut plasmid A with EcoRI and plasmid B with BamHI. Imagine for the sake of
this question that cutting is 100% efficient. You denature EcoRI-cut A and BamHI-cut B and mix
equal amounts before hybridizing at high stringency. You then transform competent E. coli.
(i) what type of DNA molecule in the DNA mixture after hybridization would you expect to give rise
to the majority of transformants?
(ii) you pick one colony (assume it is a typical one, not an exception) and do a plasmid DNA prep
followed by sequencing. You sequence the DNA and deduce that there appears to be a roughly equal
representation of plasmids A and B in the DNA sample. Explain what you think happened soon after
DNA entered the bacterium (during transformation) that gave rise to the colony you picked.
(iii) Given the above finding (regardless of how it came about) how would you recover a pure
population of only plasmid A (or B) from the colony that contained both A and B?
2. I explained some important issues about constructing a genomic library using a phage lambda
vector as the cloning vehicle. However, for many years BACs have been the most common choice for
constructing good genomic libraries from large genomes such as the human genome because a BAC
can tolerate inserts up to about 300kb generally with very good reliability (i.e. stable propagation).
Imagine you want to make a good library of human genomic DNA in a BAC vector and you already
decided that you are going to generate the genomic DNA fragments via a partial Sau3A digest (leaving
sticky 5’ GATC 3’ overhangs).
(a) Before ligation you size-select cut genomic DNA fragments in the 150kb-300kb range. What are
the benefits of doing this? You should aim to mention the TWO most important benefits.
(b) You consider how to obtain a library (set of colonies) that has as few “double-inserts” as possible
and if possible, but less important, as few BACs with no inserts as possible.
(i) What are the pros and cons of using alkaline phosphatase prior to ligation to achieve those
objectives? Explain what could be treated with alkaline phosphatase.
(ii) Is there a potentially better strategy (not involving alkaline phosphatase) to achieve the objectives
set out in (b)?
(ii) If you had a linearized BAC vector with 5’ GATC 3’ overhangs and TOPO (vaccinia
topoisomerase, as in TOPO cloning) linked to those ends (in the usual way: i.e. attached to the 3’
phosphate of the recessed 3’ends) could you use that vector to clone the Sau3A-cut genomic DNA
fragments? Mention any extra steps that would be necessary.
(c) The key property of a BAC that allows large inserts to be stably maintained is that BACs are
maintained at 1 to 2 copies per cell. However, DNA preps for BACs present at just 1-2 copies per cell
are not very good (low yield: literally a hundred times worse than for ColE1 plasmids). Consequently,
modern BACs have an additional conditional multi-copy origin of DNA replication. How do you
think the activity of such an origin of replication can be made conditional?
(d) Why does a YAC vector typically contain a ColE1 type origin of DNA replication and an antibiotic
3. Making exactly specified junctions
You have a cloned cDNA for a mouse gene in an ampicillin resistant plasmid vector, as diagrammed
below. A full-length cDNA is a complete copy of an mRNA and therefore has sequences
corresponding to 5’ untranslated sequence (5’ UTR) preceding the translation initiation codon, coding
region (starting with the ATG triplet [AUG in mRNA] translation initiation codon and ending with a
stop codon, say TAG [UAG in mRNA]) followed by 3’ untranslated sequence (3’ UTR), which will
include a signal for polyadenylation of the mRNA within mouse cells.
You have a second DNA clone (also an ampicillin resistant plasmid) that could be called a “mouse
expression vector”. It has a DNA segment that acts as a strong promoter in commonly used mouse
tissue culture cells and directs transcription to start immediately after the region designated
“promoter”. After a very short 5’ UTR there is a good translation initiation codon, followed by a
sequence that encodes a short sequence of amino acids known as an epitope tag (because good
antibodies are available to recognize this short peptide sequence). Beyond the epitope tag segment are
some restriction sites and then a segment that will form the 3’ UTR in the final construct and direct
polyadenylation of the transcribed RNA (polyA signal sequences).
This type of expression vector is generally used to make a fusion protein between the epitope tag and a
protein of choice. This is accomplished by inserting the coding region for the gene of choice between
the epitope tag and the polyA signal segments (discarding most or all of the 3’ UTR of the original
gene’s cDNA- the presence of some sequences from the 3’-UTR region are generally neither a
problem nor an advantage). If the final DNA is introduced into mouse cells the promoter will direct
transcription to start before the epitope tag sequence. Transcription will continue through the inserted
cDNA coding sequence segment and terminate (with polyadenylation) after the 3’ UTR segment.
Protein will be translated from that mRNA starting from the AUG in the epitope tag sequence (written
as ATG DNA sequence in the diagram) to produce an amino acid sequence of MDYKD….. The
triplets of codons read from the tag segment will continue into the cDNA segment. You will thereafter
only get the proper amino acids normally encoded by the cDNA if the connection between the epitope
tag and the cDNA “aligns” the normal reading frame of the cDNA with that set by the epitope tag. At
the junction between the epitope tag and the cDNA it is generally fine to encode a few amino acids
that belong neither to the tag nor to the normal protein encoded by the cDNA.
You can assume that the restriction enzymes named in the diagrams cut only at the sites shown,
and that the DNAs do not have sites for restriction enzymes that are not shown. Sites of cutting for key
enzymes are indicated by a slash (all written 5’ to 3’ for one strand; the complementary strand will be
cut in the analogous position):
The questions below are about the process of successfully creating a cloned correct expression
construct from the two starting DNAs.
(a) The most common strategy for inserting the required cDNA segment into the expression vector to
make a cloned product would be to generate a convenient cDNA fragment from the supplied plasmid
by PCR and then insert that fragment (after suitable manipulation) between the epitope tag and 3’
UTR segments of the expression vector.
To do this you have to decide first exactly what sequences you will use for PCR primers.
Considering the downstream primer (closer to the 3’ end of the gene), you could design a primer that
hybridizes anywhere downstream of the stop codon in the cDNA and later cut the PCR product with
XbaI, which cuts cDNA sequence just a few bps downstream of the stop codon.
The precise design of the upstream primer is more complicated.
Draw out and explain EXACTLY the primer you would design.
You should make clear how your design will accommodate a simple ligation strategy to make the
desired product and show clearly that the product you make will have appropriate reading fame
alignment (by drawing both strands of the final sequence in this critical region).
(b) If the expression vector did not have any convenient restriction enzyme sites (such as HindIII,
EcoRI) how, in outline could you modify your strategy (precise sequence details not required)?
(c) Instead of using PCR you could instead cut the cDNA with KpnI and ligate to an adapter that
converts the site into a HindIII site, allowing subsequent ligation to the HindIII site of the expression
vector. (Assume it is OK if the first 4 or 5 amino acids normally encoded by the cDNA are missing
from your final product).
(i) What sequence of adapter would you use? You should draw some neighboring expression vector
and cDNA sequence also and show both strands- that is the only way you can clearly show where
ligations will take place and that the resulting product should have a correct reading frame alignment.
If you used this strategy you have some additional choices to make:
(ii) would you use phosphorylated or unphosphorylated adapter? Explain.
(ii) you need to cut the cDNA with KpnI before ligating to adapter. You will also need to cut with
XbaI at some point and gel purify a cDNA fragment at some point. What is the ideal order of events,
i.e. when do you cut with XbaI and when do you gel purify a cDNA fragment? Explain. (The final
step would be ligation to HindIII-XbaI cut expression vector).
5. The Gateway cloning system is often used to create molecules similar to the products of question 4.
In fact, it is particularly good for doing that on a massive scale, for example constructing 10,000
different constructs each with a different cDNA connected to an expression vector. In that situation an
expression vector would be a Destination Vector and each of the initial cDNA clones would be in a
separate ENTRY vector. For this scenario…
(a) Why is the Gateway system particularly good for making these 10,000 clones efficiently,
compared, for example, to the methods discussed in Q4?
Your answer should both
(i) consider the relative volume of work involved (think of all of the steps)
(ii) consider whether the precise strategy can be applied universally without worrying about the
precise sequence of each cDNA one by one (this would affect efficiency and automation).
(b) If you were making a fusion protein as described in Q4
(i) where would epitope tag sequences lie in the Destination vector relative to attR1 sequences?
(ii) where would epitope tag sequences lie in the Entry vector relative to attL1 sequences?
For the above you need to describe the positions with enough landmarks to make your answer clear or
(iii) In the final product would epitope tag and cDNA sequences be as close together as in the product
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To obtain a pure plaque we can perform a serial dilution. The phage stock obtained from the agar is diluted and then applied to a bacterial lawn so that only a single phage particle infects a bacterium when its progeny come out of bacterial after lysis; they infect the neighbouring host cells to make a clear individual plaque.
(ii) We take the phage lysate and filter it with bacteriological filters to get a solution of phage particles. To remove the bacterial genomic DNA and other RNA contaminants, we incubate it with DNAse and RNAse. The phage DNA would be intact because it is inside the capsid coat.
Now the phage particles are precipitated using NACL/PEG solution. Now we dissolve the phage particle with a resin containing guanidium thiocynate, denature the capsid proteins and isolate the phage DNA by 80% isopropanol precipitation.
The question does not mention where the EcoR1 and BamH1 sites are present on the plasmid and where the GC base pair is present. When we digest one plasmid with EcoR1 it is linearised to give a 5Kb linear DNA molecule. So is the case with the other plasmid digested with BamH1. However, since the stringency is high during hybridization procedure, most of the molecules that we would obtain would be of the parental types. Hence in the transformants we would get the parental plasmids of 5Kb mostly. Hence we would get plasmid A and plasmid B in most of the transformed E coli.
(ii) The linear molecules of both the plasmids A and B must have entered the bacterial and then they would have recircularized to produce both the plasmids that replicated and remained as a separate identity in that bacterial cell. Hence, we have got and equal representation of both the plasmids....