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Recombinant DNA Practice Problem Solutions

Recombinant DNA Practice Problem Solutions


1) a) If the DNA was not digested, it would remain in chromosome-sized fragments. Chromosomes are usually at least 1000kbp - this size DNA fragment does not even enter the pores of a normal gel and would bunch up at the edge of the well. It would still be denatured, and transferred, and bind the probe, but in a single band at the edge of the well.

b) In this case, the DNA fragments would stay in the well and would probably be lost in the subsequent manipulations. If they were not lost, they would be denatured, transferred, and bind probe in a band at the well.

c) In this case, the double-stranded fragments in the gel will not be transferred to the nitrocellulose but the probe will not be able to base-pair with them because both strands will already be base-paired. This will result in a blank autoradiogram.

d) If the DNA fragments are not transferred to the nitrocellulose, there will be nothing for the probe to bind to, resulting in a blank autoradiogram.

e) The probe will bind all over the nitrocellulose, like the transferred DNA fragments do; and, like the fragments, it will not be washed off the filter during washing. This will result in a black autoradiogram.

f) With no probe, there can be no signal, giving a blank autoradiogram.

g) The nitrocellulose will be soaked with probe, some bound and some just soaked in. The whole filter will be radioactive, resulting in a black autoradiogram.

These are shown below (only one lane shown for simplicity):



2) The blots are as follows, explanations are below:



a) Southern: The probe can hybridize with all the EcoRI fragments in the above map. The fragments are (from left to right): 1.1kbp, 0.8kbp, 0.7kbp, and 3.9kbp.

Northern: The predicted mRNA is 2.2knt long = exon 1 (1.2knt) + exon 2 (1.0knt).

Western: The protein is 50kDa.

b) Southern: The 1.1kbp EcoRI fragment will become 0.2kbp shorter = 0.9kbp. All other bands will remain the same.

Northern: This mutation partially inactivates the promoter, so Q mRNA levels are 20% of wild-type. Because the promoter has not moved, transcription will start in the same place, so the Q mRNA will be the same size, but a lower quantity. This will result in a smaller band on the Northern (not a lighter band of the same size).

Western: Because the Q mRNA is normal length, the resulting Q protein will be 50kDa, but since there is less mRNA, there will be less protein, resulting in a smaller band.

c) Southern: Unless this mutation were to convert a sequence into a new EcoRI site, the EcoRI restriction map will not change. Since we did not say that a new EcoRI site say that a new EcoRI site was created, there will be no change.

Northern: RNA polymerase cannot read codons, so it cannot tell that there has been a mutation, so the Q mRNA size will be the same.

Western: The nonsense mutation will result in premature termination of translation, resulting in a substantially smaller protein product. Roughly 300nt of mRNA will be translated = 100 codons = roughly 10kDa (1 amino acid is roughly 0.1kDa).

d) Southern: Since EcoRI cannot cut at 2.6kbp, the 0.7kbp and 3.9kbp fragments will be joined together into a 4.6kbp fragment. All others will be the same.

Northern: the Q mRNA size and quantity will be unchanged (see c).

Western: A single amino acid change in a 50kDa protein is unlikely to make a noticeable change in its molecular weight.

e) Southern: The 0.7kbp fragment will be shortened by 0.5kbp to 0.2kbp. All other fragments will be the same.

Northern: Since the deletion is entirely within an exon, the Q mRNA will be shortened by the same amount as the DNA.

Western: This mutation will delete the ATG codon, preventing translation initiation at the normal site. Most likely, this will result in no protein being synthesized. Translation could start at another AUG, but any protein made will likely be read in another reading frame and not have the same antigenic sites as protein Q and therefore not bind the anti-Q antibody.

3) One way to think about this problem is to make a chart of what the two models predict for lac Z DNA, lac Z mRNA, and b-galactosidase protein levels with and without lactose. From this chart, we can then look for differences between the models that we can discern experimentally.

model (i) model (ii)




Note that in both models, the lac Z DNA (the lac operon) is always present and that b-galactosidase protein increases in both in response to growing with lactose.

The difference is in the level of lac Z mRNA in cultures grown without lactose:

model (i) says that is should be low

model (ii) says that it should be high

So, our objective is to determine the level of lac Z mRNA in cultures grown without lactose. This is best done using a Northern blot.

- The Theoretical Minimum Experiment would then be to take RNA isolated from a culture grown without lactose, and run it on a Northern blot probed with radioactive lac Z DNA. If model (i) is correct, the autoradiogram should be blank; if model (ii) is correct, the autoradiogram should show a band. In a perfect world, this would be all that you have to do.

However, this experiment is unsatisfactory for the following 'real-world' reason. How can we be sure that, if we see a band, that it is due to full mRNA levels (+) or just the background transcription (-)? Conversely, if we don't see a band, is that because there is no mRNA (-), or because full levels or mRNA (+) are too low to detect on the blot. Since we do not know the sensitivity of our measurement, we cannot be sure if what we see corresponds to a (+) or a (-) in the chart. In order to more carefully resolve this, we need to see if lac Z mRNA levels change or do not change in response to lactose. This leads to:

- The Practical Minimum Experiment would be to run mRNA from cultures +/- lactose on a Northern blot, probed with radioactive lac Z DNA. If model (i) was correct, the level of lac Z mRNA should increase in the culture grown in lactose. If model (ii) was correct, the level of lac Z mRNA should stay the same.


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