Biotechnology
Homework 2 Answers Fall 2009
1. (i) Two strands from the same parent molecule can re-anneal to form linear DNAs (ending at
BamHI or HindIII sites) but also complementary strands from different parents can anneal to firm
circular molecules with a nick in each strand. These three types of molecule should be in roughly
equal proportion since there is full base-pairing in each.
Some answers postulated that there would be multimers formed from staggered hybridization of
Bam and Hind-cut fragments. You might expect that sort of product to form in competition with
circularization of a B-cut plus H-cut heteroduplex by analogy to the kinds of product that form
ion ligation reactions. However, hybridization is not exactly the same as ligation (here,
principally because many sites along a single-stranded region can nucleate association of two
strands. Hence, once, B-cut and H-cut strands associate through the 1.5kb or 2.5kb segment there
is very fast hybridization of the remaining complementary sections of the same molecule- much
faster than colliding with a new molecule to form a bigger structure. So, in practice, you find
almost all circles and very few longer molecules.
(ii) Linear DNA will not be competent to replicate inside the cell and will yield no
transformants. The circular DNAs will replicate well and produce many transformed colonies
(the nicks are readily repaired inside the cell).
Some answers said that linear DNA can be circularized inside a cell by ligation. That does not
happen at any appreciable frequency- circles must be formed before transfection. The reason for
this is probably that the activities of exonucleases are far greater than ligases; note also that E.coli
DNA ligase normally just seals nicks (& for ligations in vitro we use DNA ligase from T4
phage). I don’t believe it normally ever joins separate DNA molecules together in a cell
(generally a dangerous activity for cells).
Remarkably, some answers to (i) and (ii) claimed or implied that 4nt overhangs would hybridize
to each other (which does not happen at any appreciable steady-state level under any of the
conditions encountered in this experiment) to form circular DNA, while ignoring the far more
realistic possibility of annealing of 1.5kb or 2.5kb complementary overhangs.
(iii) You can mix, denature and anneal as described above, where one parent molecule has the
EcoRI site and the other does not. Then, after transforming competent cells you can pick several
colonies, grow each up, isolate plasmid DNA and test for the presence of an EcoRI site by cutting
with EcoRI (together with BamHI or HindIII because linear vs circular DNA is not as clear a
distinction as linear DNAs of different sizes). In this way, you can see what is the distribution of
outcomes- plasmids of only one type or the other, or a mixture of plasmids of the two types in a
single cell.
Many answers did not attempt to explain how to make the appropriate heteroduplexes to
transform into cells. That was understandable from the original phrasing of the question but an email was broadcast specifically explaining that you need to design the experiment from the basic
given starting plasmids to the end.
Perhaps a bigger problem was simply understanding what a mis-match is. As the question
explained by example of G-T it is a double-stranded DNA in which one or more positions have
non-complementary bases. Hence, the whole focus of the question is to make such a molecule
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and see what happens to it when introduced into E.coli. If your answer doesn’t test such a
molecule you cannot speculate sensibly on the outcome in part (iv).
Some answers devised a satisfactory way to create mis-match heteroduplexes without using the
often-missed insights of circular molecules formed in (i). The suggestion was to isolate 1.5kb
Bam-HindIII fragments from the two plasmids, mix, denature, anneal & ligate back ino BH-cut
vector (could equally well be done more simply with full-length Bam-cut molecules & just
circularizing at high dilution at the end). This does create good heteroduplexes but also creates
slightly more of plasmids with homoduplexes, which we do not want to test. In reality that poses
a big problem. When you examine 20 colonies after transformation you do not know which
originate from the heteroduplex. You therefore have to look at more colonies in total to make a
statistical inference about the real exp’t & you are less certain that some of the colonies indeed
derive from the heteroduplex.
In some cases a crucial missing component was the realization that this is an experiment- a way
of generating results that will allow you to make a deduction. This requires looking at many
colonies, knowing that they came from the appropriate molecules and being open-minded to
seeing a full range of possible outcomes (sometimes you cannot see something if you don’t look
for it specifically).
(iv) Without experience, you might just guess that each strand is copied during DNA replication
and hence you will generate copies of each type of plasmid. Since ColE1 plasmids are generally
present at at least 100 copies per cell you would therefore expect to see good representation of
each type of plasmid in each colony. You may, however, know that cells have mechanisms that
detect and repair anomalous DNA sequences including mis-matches. Normally, in a bacterial
cell DNA errors are introduced by DNA replication. Also, a dam methylase methylates GATC
sequences on the A. Shortly after DNA replication only the parent strand is methylated and the
DNA repair machinery uses a protein that detects methylated GATC to collaborate with mismatch detection proteins and initiate repair of the newly synthesized DNA according to the parent
template. In the example here both plasmid strands arrive in the cell at the same time and there is
no reason for the cell to consider one strand more likely correct than the other. Thus, repair could
alter either strand (in fact the position of nicks may influence this). If DNA repair happens faster
than DNA replication the result will be cells harboring only one type of plasmid; otherwise we
will recover two types of plasmid, as outlined earlier. If we were talking about genomic DNA
(for E. coli or humans) you would expect that DNA repair systems are fast enough to precede
DNA replication (indeed in mammalian cells DNA damage detection leads to arrest of DNA
replication to give DNA repair enzymes time to complete their jobs). For a plasmid DNA the
evolutionary pressures are different. I happen to have done these types of experiments and for the
cells and plasmids that I used it was very rare to find two different plasmids in the same cell.
Thus, most colonies either have only plasmid with EcoRI or only plasmid without EcoRI
(roughly 50:50). To do the experiment well it is good to gel purify the starting linear DNAs to
eliminate any final contribution from a small proportion of uncut circular DNAs (whose absence
could be shown by transforming the original linear DNAs).
Many answers were not focused on the real biological experiment of finding out what happens
inside the cell by controlling the DNA species entering the cell and then measuring what is there
afterwards. Instead, there were discussions just about interpretation of restriction digests (i.e.
about the analytical method, not the biology itself).
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2. (i) We want to have a plasmid with (in sequence) promoter, epitope tag, cDNA, polyA signal,
in which the junction between the tag and the cDNA maintains the appropriate reading frame.
That junction is also the only one where restriction enzyme sites are not compatible, so it
deserves the most attention. One could use an adapter or a linker to make that critical junction
and ensure the correct number of base-pairs are inserted. Either way, the simplest is to alter the
cDNA fragment so that it can be cloned into plasmid 1.
For an adapter cut with NcoI and XbaI, ligate to non-phosphorylated adapter, gel purify 2.2kb
fragment and ligate to plasmid 1 cut with BamHI and XbaI (also gel-purified- to remove any
uncut DNA, though not essential). Transform, pick colonies and test according to Pst-Xba or
Bam-Xba digests. Sequence candidates that look correct over the BamHI region by using a
primer from the cDNA or promoter sequence. The most likely erroneous product revealed by
sequencing would be in the adapter region. Here the oligos are short, so errors should be very
rare and the oligos are unphosphorylated in the first place so ligation of multiple adapters seems
impossible. An appropriate adapter might be as shown below.
5’
3’
(lower case denotes arbitrary sequence)
GA TCg cgg ag
3’
c gcc tcG TAC 5’
to link
5’ ….AAA G 3’
3’ …..TTT CCTAG 5’
and
5’
3’
C ATG G…… 3’
C….. 5’
The adapter could have 3bp (or some multiple of 3bp) more if required and the specific sequence
linking the epitope tag and cDNA need only avoid stop codons. The sequence is written in
triplets to indicate the reading frame set by the Met of the Flag linker, showing that the ATG of
the NcoI site falls in the same reading frame (…AAA GGA TCg cgg agC ATG….)
If using a linker, first cut plasmid 2 with NcoI, then fill in the end to be blunt with DNA
polymerase. Then ligate to linkers (phosphorylated for efficiency; non-phosphorylated will avoid
any chance of more than one linker in the final product). Inactivate the ligase (perhaps purify the
DNA by phenol extraction & ethanol ppt’n) and cut with BamHI (included in the linker). Use a
lot of enzyme because there will be a huge concentration of BamHI sites, especially if ph’d
linkers were used. Now cut with XbaI (cutting earlier would expose this end to DNA pol and
linkers), purify 2.2kb fragment (removes cut and unligated linkers at the same time) and ligate to
plasmid 1 cut with BamHI and XbaI as above. Cloning and checks as above.
A suitable linker would be:
5’
3’
ct GGA TCC ag 3’
ga CCT AGG tc 3’
so, looking at the top strand only, the final product would read AAA GGA TCC agC ATG …
keeping the correct reading frame. Again, the linker could be longer (3bp multiples downstream
of the BamHI site), and it could be non-symmetrical. If the linker is longer it may have the
advantage of BamHI cutting better, since all restriction enzymes actually need DNA either side of
their recognition sequence (and some need more than others to be efficient).
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Using either method, the final step will be to cut out and purify (from a gel) the polyA signal
fragment, ligate that to the new product above that has been cut with XbaI and EcoRI (gelpurified), transform cells and screen products by, for example, BamHI-EcoRI digestion.
Alternatively, this Xba-RI fragment could have been cloned into plasmid 1 first. It would also, in
this case be reasonable to ligate all three fragments (BamHi-EcoRI from plasmid 1, BamHI
(modified)-XbaI cDNA and XbaI-EcoRI polyA) at the same time and screen transformants.
Usually, it is not practical to do two steps at once but if all ends are sticky-ended and only one set
of ligations can give a viable product this can work. Often, however, the moral of the hare and
the tortoise (turtle) applies.
Some answers ignored reading frames, some acknowledged them but did not spell out sequence
or made an incorrect generalization that the linkers/adapters used must be of length that is a
multiple of 3bp. The only way you can make the right oligo and know you have it right is to
write out the entire junction sequence, showing both strands. Use the first initiator ATG to
determine which triplets are read as codons and proceed marking triplets until you are into the
cDNA and can compare with the desired codons (in this case you know simply that the ATG in
the NcoI site should be read as a triplet codon).
Many answers included unrealistic conceptions of what can be achieved by purifying fragments
after ligation. With very few exceptions, nothing is accomplished by doing that because you do
not produce enough of a desired product. In practice you almost always follow a ligation step
with a cloning step (transforming competent cells and then finding a colony with your desired
product). The cloning step is your main means of purification in a complicated procedure. That
means that whenever you design a ligation you should plan on forming a circular product with
suitable vector sequences to produce a viable plasmid. The final product also must be cloned.
The requirement for cloning is probably the most basic of the several ideas tested in this question.
A small point- the polyA signal is a sequence (often AAUAAA in RNA followed by a further
short consensus) that causes proteins to bind, cleave the RNA and add polyA. It is not itself a run
of A residues.
Many answers used the word “anneal” instead of “ligate” (only the latter involves phosphodiester
bond formation & only that can join molecules with short regions of complementary sequence,
insufficient for a stable hybrid).
(ii) PCR gives you two very important flexibilities- (a) you can pick the exact start and endpoints of an amplified fragment no matter what its sequence or distribution of restriction enzyme
sites, (b) you can add restriction enzyme site sequences and anything else of moderate length at
either end of a PCR product. Thus, it is simple to clone the polyA signal into plasmid 1 (or a
derivative of plasmid 1) by amplifying the polyA signl segment using one primer with XbaI
sequences at its 5’ end, and the other with 5’ EcoRI sequences. Here there is no reading frame to
preserve, so the exact position of these enzyme sites is not critical. You do normally put at least
4-5nt upstream of the site to make sure the restriction enzyme can cut efficiently at its site.
Since you have no template with the epitope tag coding sequence you have to make it
synthetically. The coding sequence is short enough that you can make one oligo with a restriction
enzyme site (BamHI) followed by the epitope coding sequence, followed by enough sequence to
allow hybridization to the cDNA template. Since you have pure template with relatively few
sequences for chance hybridization you actually could use only about 12 nt for hybridization but
most researchers would prefer to be much safer and include about 20nt in the type of primer
being proposed, giving a total length of roughly 60nt. That is a perfectly reasonable length for
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efficient synthesis of a good oligonucleotide, relatively cheaply. You would use this primer and
another downstream of the cDNA stop codon and with an XbaI site at its 5’ end.
As above, you could make your desired clone in two steps (insert the BamHI-XbaI cDNA
fragment into plasmid 1, then clone, then insert the XbaI-EcoRI polyA fragment and clone).
Alternatively, you could ligate all three fragments at once and then clone. The PCR products
would not necessarily have to be gel-purified but that is likely to improve yields of the correct
product and you would in any case need to run a gel to check that the PCR was effective. In
general, more purification prior to ligations means fewer tests of transformed colonies to find the
right one; that is generally worthwhile in saving labor and sometimes is essential. Checks of
clones are much the same as in (i) except that here you would want to be sure that the entire
coding region has been sequenced. You might expect errors due to PCR (hence use a proofreading enzyme) and some of your long primers might have errors. Nevertheless, you would be
unlucky if more than 2 of 3 sequenced products had either type of error.
A possible sequence for the long PCR primer would be
ctagtcGGATCC ATG GAC TAC AAG GAT GAT GAC GAC AAA ATG (XXX)6
where XXX matches cDNA sequence following the ATG in the cDNA
One mis-understanding that caused some to postulate unrealistic primers was a failure to assume
that we know the sequences of the plasmids we are dealing with. When making detailed genespecific constructs, such as this, one would always know the sequences of components (and could
find out or confirm as necessary by further DNA sequencing). This is the only way you could
hope to design primers that will hybridize at specific sites. A 6nt match is not good enough;
invariably, at least 17 nt of match is used in these situations and often more than 20nt to be safe.
To add restriction sites to PCR products one could ignore the template sequence (as above) or
you could use the near similarity to restriction sites (such as TCGAGA) by making a primer that
overlaps such sections but has a mis-match (here TCTAGA). If the primer is long enough (say
24nt or longer) the mis-match will not prevent good hybridization but the resulting PCR product
will include TCTAGA sequence (only). Hence, this method is a slight economy (five positions
of the enzyme site you are creating actually contribute to hybridization- compared to zero when
you simply add a site to the 5’ end of a primer).
While linkers/adapters can be used to introduce the FLAG sequence it is important to design the
simplest and most efficient way of constructing new molecules. The same argument applies to
using linkers or blunt-ended ligations to join PCR fragments rather than adding enzyme sites
directly to the ends of PCR products. Similarly, always make the junctions of different fragments
to join together unique if possible, so that only one set of ligations will produce a product that
gives transformed colonies.
3.
(i) The 0.9kb fragment appears to have good enough matches to your primers to be amplified.
Hence, if you just removed oligos and then sequenced the mixture of templates with one of your
original primers you would likely get two superimposed sequences. So, you want to avoid
priming the 0.9kb fragment during sequencing. You might do this by
(a) purifying the 1.6kb band from a gel before sequencing,
(b) using a third primer for sequencing (it will quite likely not hybridize to the 0.9kb band, unless
that segment represented a closely related gene or sequence over its entire length- perhaps
unlikely since it has a different size),
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(c) use internal primers to PCR a small portion of your mixed product- likely only the 1.6kb will
amplify,
(d) choose a different pair of primers for PCR and start over.
(ii) (a) A product that does not require reverse transcriptase does not derive from RNA. Hence,
the 1.4kb band derives from copying DNA, most likely genomic DNA present in the RNA
preparation. Note that this could happen even with a small DNA contamination because (i) PCR
is so sensitive and (ii) the mRNA signal might be fairly low iof that mRNA happens to be
expressed very poorly (or not at all in the cell type examined).
(b) This result implies that both products derive from RNA. Some pattern of alternative splicing
(for example, optional inclusion of a 0.3kb exon) could explain the two different RNA forms and
different RT-PCR products. Sequencing would show whether this idea is correct. It is also, of
course, possible that (as in (i)) there was unintended priming of a second, completely unrelated
RNA. Since priming really only requires a transient association and excellent match at the 3’ end
of the primer aberrant priming occurs more frequently than stable hybridization properties would
suggest. Also, once a couple of templates (in each direction) have been copied you generate
more templates, now with perfect primer matches.
4. (i) Antibiotic resistance is used to identify cells that have taken up a plasmid. Cells with viable
lambda phage DNA inside them reveal themselves by lysing, followed by diffusion of phage and
further rounds of infection to produce a macroscopic plaque in a lawn of cells.
(ii) When lambda DNA integrates it does not kill the cell or produce any major phenotype.
Hence, a clear marker of the presence of lambda DNA will be absent. Integrated lambda DNA
will be amplified linearly as the cell population grows but it will not be in high copy number and
will not be abundant. Also, there is no easy way to separate the lambda DNA from the rest of the
E.coli genomic DNA when isolating recombinant DNA.
(iii) When cells lyse their DNA is released into the medium and will not spin down with unlysed
cells. Hence, the supernatant will contain cellular DNA and lambda phage particles. One could
remove the cellular DNA by adding DNase (which cannot enter the phage particle) or by
precipitating the phage particles and discarding the supernatant (or both).
(iv) In lambda cloning you use a packaging reaction to produce phage particles. You then infect
cells. You choose the number of cells and assembled phage particles so that the multiplicity of
infection is much lower than one. In other words, you use a lot of cells so that most cells are not
infected and most of the rest are infected by only one phage particle (because two co-incident rare
events are very unlikely).
(v) You take some material (top agar) from the overlapping plaque region, allow phage to diffuse
out into an aqueous solution (Magnesium ions are essential to stabilize the phage), dilute
appropriately, infect more cells and spread on a plate to produce individual non-overlapping
plaques.
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