Molecular Biology 2nd Exam - spring 1997
ALL EXAMS ARE DUE AT 9:30 am ON MONDAY APRIL 7.

There is no time limit on this test. You may find it easier to take this test over several days, though if you are confident in your molecular skills, you could wait until Sunday night. However, I predict it will take many of you a bit longer to think of all the answers (just some friendly advice). You are not allowed to use your notes, any books or journals, nor are you allowed to discuss the test with anyone until all exams are turned in at 9:30 am on Monday, April 7. You may use a calculator and/or a ruler and graph paper. The answers to the questions must be typed, though you may want to supplement your text with hand drawn figures (write neatly for any labels in your figures).

Please do not write your name on any page other than this cover page. Staple all your pages (INCLUDING THE TEST PAGES) together when finished with the exam.
Name (please print):

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How long did this exam take you to complete (excluding typing)?

10 pts.
1) As you can see from this first abstract (figure 1), a lot of scientists find it difficult to write in plain English.
Please summarize this abstract in your own words using as little jargon as possible. Your summary should NOT be longer than the original summary.
CPE is a receptor for determining if a secretory protein is secreted all the time or under limited conditions. A mutant mouse lacks CPE, a trans-Golgi protein, so a pituitary hormone is secreted all the time and not secreted under limited conditions like it should be. A loss of CPE can lead to endocrine diseases.

10 pts.
2) Figure 2 shows some DNA footprint analysis. CC+20 is a particular sequence of DNA. CRP is a protein, as is RNA polymerase. Notice that the top line of + or - is indicating the presence or absence of CRP; the bottom line indicates the presence or absence of RNA polymerase. The numbers to the right of the data are intended to be used as a yard stick for your convenience.
Please interpret these data as completely as possible.
After the molecular weight markers (m), is a negative control lane. This shows what the banding pattern is like in the absence of any added proteins. The next lane shows that CRP1 binds to three regions of CC+20: bases 32 - 42, 23-30, and 16-21. When CRP1 and RNA polymerase are both mixed with the DNA, there are 6 areas that are bound with protein. In addition to the three regions mentioned above, 72-110, 55-65, 43-52. It seems likely that the previous three regions are bound with CRP1, while the latter three are bound with RNA pol. (but only in the presence of CRP1). Finally, when RNA pol is used alone, there are no areas that are bound with protein, which suggests that RNA polymerase cannot bind to its region in the absence of CRP1.

10 pts.
3) Figure 3 is examining the role of different alleles of p53 in tumor formation. Note that Giesma stain lables DNA and thus what you see are dark spots of foci; these spots are evidence of rapid cellular proliferation.
Please interpret these data. Note that p53dl has no functional p53.
When cells are given myc and ras oncogenes, the cells become transformed and form foci. If a non-functional allele of p53 is used (dl) then the cells continue to divide rapidly as revealed by foci, and there is no difference between the two temperatures. When the val-153 allele is used, a temperature sensitive phenotype is revealed, suggesting a ts mutant allele of p53. At 37.5š C, the foci seem more numerous and larger which suggests that at this temperature, val-153 enhances tumor formation. At the colder temperature, there are few if any foci which suggests that p53 is functioning as a dominant tumor suppressor. (Unfortunately, humans do not maintain this cooler body temp.) (Note that p53 val-153 is a point mutation from wt p53 with a amino acid substitution from wt to valine.)

10 pts.
4) Figure four shows a band shift assay in which the promoter DNA is radioactive (or ''hot'') and every lane has purified transcription factor E2F added to it. A ''-'' indicates that there is no addition of identical promoter DNA sequence that is non-radioactively labeled (no ''cold'' DNA added), while ''+'' indicates that there was addition of identical promoter DNA sequence that is non-radioactively labeled ('cold" DNA added). The molecular weight of E2F plus promoter is indicated by the arrow. The labels at the top indicate the source of additional cell extracts. For example, L means that the cell line ''L'' was homogenized and then these proteins were added to the tube for the pair of lanes (- and +) before the band shift gel was run.
a) Explain the data you see.
b) What is the purpose of the ''+'' lanes?
In all lanes, there are at least two bands: the one at the bottom is unbound probe and the one at the top appears to be non-specific since it appears uniformly. Near the arrow, there are varying numbers of bands in the '-" lanes and none in the '+" lanes. The arrow indicates the size of E2F plus DNA and we assume that all bands of this molecular weight are just that. Any bands above this size may be E2F plus other proteins bound to the probe, or other proteins without E2F (a control lane of proteins w/o E2F would have settled this issue). Bands just below the arrow may indicated other proteins binding to the probe without any E2F, or E2F that is slightly degraded still binding to the probe. In the NIH3T3 and dF9 lanes, there are some bands very low, near the probe. Since the same bands appear +/- cold competitor DNA, we can conclude that this is non-specific binding of ''junk'' to the probe. If it were specific, then the cold DNA should have out competed for this molecule and the probe would not have shifted. We can consider this to be artifact. The CV1 lane indicates that something in the CV1 cells can bind to an inhibit E2F from binding the promoter, or cannot bind alone to promoter in the absence of another factor which is present in all other cells.

10 pts.
5) Here is another band shift assay (figure 5) . This time we will compare three different binding proteins (HMG-1, MATH20, and MATH10) and two different promoters (SAR and SV40). Make the assumption that all these lanes were analyzed on a single get so there is no variation due to differences in different gels. Note that the first lanes in panels A and C are negative signs and not the number one; these lanes had no protein added to the assay.
Interpret these data.
This experiment is very well controlled since each lane has an experimental probe (SAR DNA) and a negative control DNA (SV) which never undergoes a band shift which demonstrates that any shifting of SAR is due to specific interactions with the added proteins.
In the HMG-I lanes, we see that SAR can bind anywhere from 1 (0.5 lane) to ~7 (25 lane) monomers of HGM-I. The number of monomers bound appears to increase with increasing concentration of HGM-I added.
In the MATH20 lanes, we see that SAR binds this protein at a low concentration (0.5) and only a monomer is bound through 10.0. At 25 ng, it looks like more than one MATH20 binds to SAR.
In the MATH10 lanes, it appears that a monomer binds at 0.25 ng, a dimer begins at 0.5 ng through 10 ng. In the 25 ng lane, perhaps a trimer of MATH10 is binding to SAR and there is no probe left with only monomers on it.
Math20 is the largest protein, then Math20, and HMG-1 is the smallest of the three.


10 pts.
6) Figure 6 is a set of results from CAT assays. NF-kB, IRF-1, and ATF2/cJun are three transcription factors. The two different reporter plasmids vary in their promoter sequences such that the DNA helix has been added to by half a rotation (panel B) compared to the promoter in panel A. As a result, all the sequences on one side of the double helix that were pointing up (panel A) are now pointing down (panel B).
What can you conclude from these data?
This is a CAT assay in which one of two promoters is incubated with different transcription factors and we can see the degree to which each transcription factor(s) activates the transcription of the reporter gene.
In panel A, we see that NF-kB can activate transcription slightly when 3 µg of plasmid was used. IRF-1 is a little better at activating, 1 µg gives a noticable signal and 3 µg increases transcription even more. ATF2/cJun has no activity. When all three are added, we see that there is a synergistic effect, even 30 ng is very effective at activating transcription.
When the helix has been rotated 180š, we see that NF-kB and IRF-1 are still affective at activating, but maybe there is a slight decrease compared to panel A. However, We notice that the synergistic effect is completely gone; lanes 16 and 17 look like 8 and 9 (IRF-1 alone). Therefore, it appears that the synergy requires the transcription factors to bind to a particular side of the promoter, relative to the coding region.

10 pts.
7) Figure 7C shows hydropathy plots for two related molecules (FRL1 and FRL2). Assuming that their structures are well conserved and that differences in amino acid sequence have no major impact on overall structure, what can you deduce about the structures/topology of FRL1 and FRL2?
FRL-1 and 2 appear to be made in the ER, as indicated by the first hydrophobic peak, which is indicative of a signal sequence. Furthermore, we see that both proteins have a terminal region which may span the membrane. FRL-1 may span the membrane two more times at around amino acid 100. If so, then both proteins would have their termini on opposite sides of the membrane, though FRL-1 would also have two small loops on either side of the membrane; it would look like a deformed letter N while FRL-2 would look like the letter I.

10 pts.
8) Given the data in figure 8A - 8C, what can you conclude about FRL1 and FRL2 expression?
A FRL-1 mRNA is barely detectable in the egg and is much more prevalent in stage 10 and 12 embryos, but is gone from stage 14 on. FRL-1 must be a maternal messenger in the egg but is also expressed by the new embryo.
FRL-2 is firsted detected in stage 10 (absent in the egg) and its presence increases from then on. It is transcribed only in the young embryo and not the mother's ovaries to contribute to the egg. The ODC band indicates that RNA was present in all lanes, and tells us that equivalent amounts were loaded into each lane so we can compare intensity of the bands to expression levels.

B FRL1 is present in all parts of stage 10 embryos, but gsc is not. gsc is present in the dorsal and vegetal halves only (and of course the entire embryo - lane Em). EF1a is a positive control which lets us know that the blank lanes for gsc is not due to a lack of RNA.

C This did not photocopy well, but we see that FRL2 is expressed in the upper and left most portions of the embryo. Given standard presentation of embryos, it looks like the head and back are positively labeled, suggesting a central nervous system expression pattern.

10 pts.
9) Determine the chromosomal location of XRCC. (It must be nice to have the data for an entire figure donated to you by another researcher!)
XRCC4 is located on chromosome 5, on the long arm (q), between regions 11.2 and 13.3.

10 pts.
10) And the final question...
Figure 10 shows a series of immunofluorescence micrographs using NIH3T3 cells (which is generic cell line that many researches grow in culture) that have been infected/transfected as described in the legend. HA stands for Hemagglutinin which is used as an epitope tag similar to the c-myc tag. The anti-p19 antibody was generated by injecting a mouse with a (cognate) peptide from the full-length p19. Panels D and E have the exact same NIH 3T3 cells grown in the presence of a modified form of dUTP (called BrdU in panel E) which can be incorporated into growing DNA strands in place of dTTP, and can be detected by a monoclonal antibody.
Please interpret panels A, B and C, and then tell me the conclusion from panels D and E.
The transfected NIH3T3 cells express p19, as indicated by the white spots in panel A. B shows that the peptide blocks the labeling, which indicates that the antibody is specific for p19 and not another protein. C shows that in these transfected cells, are expressing the HA-tagged p19.
In E, we see the location of some nuclei and D shows the same labeling.Therefore, p19 is targeted to the nuclei of NIH3T3 cells.

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