Question 1
a)
i) If genomic DNA from the wild type bacterium were digested with EcoRI, three different fragments would be detected. The approximate sizes of the fragments would be 1.625 Kb, 1.375 Kb and 0.25 Kb. This is because there are two different digestion sites for EcoRI within the region complementary to the probe, and therefore three different fragments would partially hybridize with it after the digestion.
ii) There are only two restriction sites for BamHI shown in the diagram, but neither of them is within the region complementary to the probe, and both of them are upstream of this region. This means that, even though a fragment of approximately 0.625 Kb would be generated after digestion with this enzyme, it would NOT be possible to detect this fragment using the probe shown in the diagram. It is not possible to predict the size of the fragment that would be detected using this probe,because that depends on the exact position of the first restriction site for BamHI downstream of the region complementary to the probe, which is not shown in the diagram.
iii) After digestion with both restriction enzymes, three different fragments would be detected using the probe. Two of them, with approximate sizes 1.375 Kb and 0.25 Kb, would be the same ones detected after digestion with EcoRI alone. The third one would be shorter, however, reflecting the fact that there is an internal BamHI restriction site. The size of this third fragment would be slightly above 1 Kb .
b)
In mutant A, the entire region complementary to the probe is absent, because it is included within the deletion that characterizes this mutant. The probe would be therefore useless to detect any fragments, since it would not be able to hybridize with any of them. In summary, regardless of the particular digestion regime applied (EcoRI, BamHI or both), no fragments would be detected using that probe.
c)
i) Mutant B has a 0.5 Kb insertion upstream of the region complementary to the probe. Therefore, after digestion with EcoRI, two of the fragments detected by the probe would have the same size as in the wild type case (1.375 Kb and 0.25 Kb) and the third one would be larger, with an approximate size of 2.125 Kb.
ii) The same rationale enunciated for the wild type case after digestion with BamHI applies here as well. Regardless of the presence of an insertion in this mutant, there are no BamHI restriction sites downstream of the region complementary to the probe shown in the diagram, and therefore it is not possible to predict the size of the fragment that would be detected by the probe based on this information alone.
iii) After digestion with both EcoRI and BamHI, two of the fragments detected would be identical as in the first case (corresponding to 1.375 Kb and 0.25 Kb) and the third one would be slightly above 1.5 Kb .
d)
i) Using genomic DNA from the wild type strain and primers W and Y, the approximate size of the band after the PCR experiment would be 3 Kb .
ii) W and X have the same orientation, meaning that they hybridize with the same strand of DNA and not with complementary ones. Therefore, the PCR experiment would not work and no bands would be visualized at the end.
iii) Using primers X and Y, the approximate size of the band corresponding to the amplification product would be 1.875 Kb .
e)
i) The deletion that mutant A exhibits appears to be of approximately 1 Kb according to the diagram. Since it is located within the region that would be amplified in a PCR experiment using primers W and Y, the band visualized at the end would be 1 Kb shorter, that is, 2 Kb .
ii) The same rationale used in the wild type case applies here. W and X have the same orientation, and therefore they are not suitable to be used together in a PCR experiment. No bands would be observed.
iii) Assuming that the region complementary to primer X is not included within the deletion that this mutant carries, the amplified band would be approximately 0.875 Kb .
f)
i) Since this mutant has an insertion, the amplified fragment using primers W and Y would be 0.5 Kb longer than in the wild type case, that is, 3.5 Kb .
ii) The same rationale explained before applies here. W and X have the same orientation, the PCR experiment would not work and no bands would be detected.
iii) Since the insertion carried by this mutant is outside the region that would be amplified using this set of primers, the size of the band would not be affected. As in the wild type case, the band would have an approximate size of 1.875 Kb .
Question 2
a)
b)
c) After the transformation, the yeast should be cultured in a medium which lacks leucine. Under this conditions, only the transformants that carry the LEU2 gene will be able to grow, whereas the parental strain will not. After certain amount of time time, only the transformants will be present in the culture medium and they can easily be selected then.
d) Gel A shows the migration pattern of total DNA without having been subjected to any digestion step with restriction enzymes. Therefore, all the lanes (regardless of whether the DNA was obtained from the parental strain or from a transformant) show the same pattern of migration, corresponding to long pieces of DNA. These become broken during preparation, and the randomly sheared pieces migrate at the same position of a fragment of approximately 25 kb. The gel shows this single band in all the lanes, running slightly behind the largest fragment of the lambda DNA ladder (23.1 Kb), which is exactly what would be expected.
Gel B shows the pattern of total DNA migration after it was digested with the restriction enzyme XhoI. Under this conditions, the total DNA of the transformants would include a 10.6 Kb fragment corresponding to the entire sequence of plasmid 2 (7.6 Kb) plus the flanking sequences in the parental chromosome (3 Kb, see previous diagram). Therefore, the lanes corresponding to the transformants show this band migrating behind the 9.4 Kb fragment of the lambda DNA ladder. However, the lane corresponding to the DNA obtained from the parental strain does not include this band, because the plasmid has not been inserted between the two XhoI restriction sites in this case. Therefore, only a 3 Kb fragment has been released after the digestion, and this fragment is too small to be observed in the gel. However, after the DNA has been transferred to a membrane for Southern blotting and probed using the 1.5 Kb EcoRI-BamHI fragment labelled with radioactive phosphorous, this 3 Kb fragment can be easily observed after the corresponding autoradiography, migrating at the expected position (at the lower end, faster than the 4.4 fragment of the lambda ladder). Notice that the larger 10.6 Kb fragments present on the lanes corresponding to the transformants were also labelled by this probe. This is because the plasmid included in the fragment includes this 1.5 Kb EcoRI-BamHI sequence (see diagrams above), and therefore the probe also hybridizes with them. The photograph of the original gel after the transfer does no longer show any bands, as expected, because all the DNA fragments have been transferred to the membrane.
Question 3
In mutant 1, it appears that the gene has been affected by a deletion, because the combined size of the two fragments observed in the Southern blot appears to be lower than the 3.5 Kb HindIII fragment present in the wild type strain and that includes the gene. Since this mutant does not produce the blue pigment, the deletion is likely to have been within the 1750 bp coding region. Moreover, the fact that there are two bands in this lane and not a single one as on the wild type lane indicates that the deletion must have caused the appearance of a novel HindIII restriction site somewhere between the two original ones present in the wild type genome
Mutant 2 shows a single band that migrated slower than the one in the wild type lane, indicating a greater size. Therefore, the mutation is likely to be an insertion within the 1750 bp coding region.
In the lane corresponding to mutant 3, no bands are observed. This suggests that a major deletion has eliminated the entire 3.5 Kb HindIII fragment containing the gene in this mutant, because otherwise the probe would have detected at least one fragment.
The lane corresponding to mutant 4 shows a single band of the same size as that observed in the wild type. This suggest that no insertions or deletions have occurred, but since the gene is non functional, a point mutation that has not altered the size of the fragment in any significant way must have occurred.
Finally, the lane corresponding to mutant 5 shows a single band migrating faster than the one seen on the wild type lane, indicating a lower size. This suggests that the mutation has been a deletion that encompasses the coding region, but that (unlike the case with mutant 2), this deletion has not caused the appearance of a novel HindIII restriction site.
Question 4
a) The 618 bp band on the second lane appears to show roughly the same brightness and width as the 4.4 Kb band in the DNA standards lane. Therefore, the amount of DNA on this band can be estimated to be around 34 ng. The other two bands (355 bp and 243 bp) are dimmer than the band corresponding to the 2.0 Kb standards, but brighter and wider than the one corresponding to the 0.56 Kb standard. Therefore, the amount of DNA in each of them can be roughly estimated to be between 4 ng and 16 ng. Notice that the 355 bp band is brighter than the 243 bp band. Therefore, regardless of the absolute amount of DNA present in each of them, it is certain that the value is higher in the longer band. If pressed to give a number, I’d say that the 355 bp is 10 ng and the 243 bp 6 ng.
b) One picomol of the largest band (0.618 Kb) has a mass of 407.9 ng (the actual product of 0.618 x 660, which is not 421 ng as stated in the text of the question). If 407.9 ng then corresponds to 6 x 1011 molecules, then in the 34 ng estimated to be in the band 0.5 x 1011 molecules can roughly be found. Using an analogous method, the following values can be estimated for the remaining two bands:
355 bp band:
660 x 0.355 = 234.3 (mass of 1 picomol) number of molecules in band: 0.26 x 1011
243 bp band:
660 x 0.= 160.4 (mass of 1 picomol) number of molecules in band: 0.22 x 1011
Source:
Lodish, H., Berk, A., Kaiser, C.A., Krieger, M., Scott, M.P., Bretscher, A. et al. (2007) Molecular Cell Biology (6th edition). New York: W.H.Freeman & Co.
