Molecular Biology Final Exam - spring 1997

ALL EXAMS ARE DUE AT noon on MAY 10, 1997.

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 May 9. 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 noon on MAY 10, 1997. You may use a calculator, a ruler, and graph paper if you want. 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.
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How long did this exam take you to complete (excluding typing)?



8 pts.
1) Figure 1 has taken ER and Golgi membrane microsomes and analyzed the proteins as indicated. What can you determine about the proteins examined in this experiment?
IgG is a negative control to show that the other two immunoprecipitations (IMP) are due to the specificity ot the primary antibody. The anti-syntaxin 5 antibody IMP coprecipitates syntaxin 5 and rbet1, though not all of the rbet1 is pulled down. The rbet1 IMP pulls down all of the rbet1 but only about half of the syntaxin 5. syntaxin 6 and calnexin do not coprecipitate with either IMP. Therefore, about half of the syntaxin 5 and half of the rbet1 interact in a complex (dimer?) but about 50% of these two proteins are not in this complex.

8 pts.
2) Figure 2 shows the results of 3 experiments. KB cells were analyzed at various times (in hours) post-infection (hr p.i.) with either dl753 (a deletion mutant of a particular virus that is inactive inside cells) or rec700 (wt virus), or after treatment with Epidermal Growth Factor (EGF). The 170 kDa EGF-receptor (EGF-R) was analyzed after the membranes were treated with the detergent NP-40. DME is a particular kind of tissue culture medium, serum contains EGF, and cI21 is an anti-EGF-R antiserum.
What can you conclude from these data?
In the presence of the mutant virus, EGF-R has a very long half-life; it is not degraded within 10 hours after it is made. With the wt virus, the receptor is completely degraded between 6 to 8 hours after it is made. When EGF-R binds EGF, it is degraded very rapidly, in less than 1 hour. Perhaps the rec700 virus binds to the receptor in order to enter the cells and in binding, it causes the receptor to be degraded as if it has its normal ligand bound to it.

10 pts.
3) Figure 3 is merely to illustrate the kind of data one can get when using a particular P element to clone a novel gene. In this figure, we see that a reporter gene product is detected (light spots on the black background) in the photoreceptors of the fly (and nowhere else).
Use this strain of flies and the simplest method we have discussed in class to clone a gene that is expressed only in photoreceptors. You must briefly outline the steps you would need to perform in order to clone the wild-type allele of this gene. (There is more than one way to clone this gene but full credit will be given to the one method that is the most direct and requires the fewest steps. Partial credit will be given to all other methods.)
This looks like the results of an enhancer trap, therefore:
Digest the genomic DNA (gDNA) with the appropriate enzyme that cuts once in the P element. Ligate the gDNA.
Transform this ligation into bacteria and plate on antibiotic containing plates (plasmid rescue). Isolate the gDNA contained in the plasmid.
Use this gDNA as a probe on a wild-type genomic library and isolate the wild-type allele.
Verify this is the wild-type allele by RFLP analysis, in situ hybridization, or a functional assay.

12 pts.
4) a) Calculate the molecular weight of the bands in lanes 1 and 3 of figure 4 A (the line indicates the location of the resolving gel). You must use the graph paper provided in this test in order to get full credit.
Lane 1 = ~60 kDa
Lane 3 = ~24 kDa

b) A soluble form of CD4 (sT4) was made, as well as two shorter forms of CD4 (V1V2 and V1). The soluble form of CD4 is comprised of four domains called V1 through V4. Above each ABC lane marker is a label that indicates which CD4 construct was used in the 4 experiments (a construct expressing V1 and V2 fused together was used twice). In all "A " lanes, an anti-CD4 antiserum was used for immunoprecipitation; in all "B" lanes, HIV gp120 was incubated with the indicated forms of CD4 and then an anti-gp120 antiserum was used for immunoprecipitation; in all "C" lanes, the anti-gp120 antiserum was used for immunoprecipitation in the absence of gp120. What can you conclude about HIV gp120 and CD4?
First, we notice from the legend that this is an immunoblot and that lanes C are negative controls. Lanes A identify the molecular weight of the CD4 constructs used in this experiment (positive controls). Therfore, lanes B tells us that gp120 interacts with all of the soluble CD4 constructs and specifically, only the V1 region is enough of the CD4 to interact with gp120 (V1 is sufficent).

9 pts.
5) Figure 5 is an in situ hybridization of a fly egg before fertilization and it detects the mRNA from a single gene that is localized in the posterior end.
a) Hypothesize how a maternal mRNA can be localized to a particular region of an egg.
Some part of the mRNA sequence must be responsible for targeting the mRNA. Probably a portion of the mRNA is recognized by an RNA binding molecule and this molecule is also capable of binding to a certain region (posterior) of the egg. This RNA binding molecule may be a part of the cytoskeleton or it might bind to it.
b) Design an experiment to determine how the mRNA can be localized to any discrete area of the egg. You do not need to give any volumes of reagents, just OUTLINE how you would design an experiment to determine the mechanism for an mRNA to stay in one region and not diffuse across the egg.
I would use PCR to divide the cDNA that encodes the mRNA into three parts: 5' UT, coding, and 3' UT.
I would ligate these three regions to a reporter sequence - one that is not found in flies and can be detected easily by in situ hybridization.
I would then add these three constructs to three batches of eggs (microinjection?) and perform in situ hybridization. Presumably, only one of the three constructs would be localized to the posterior region.
I would continue to narrow down the region by making smaller PCR products from the correct region.

8 pts.
6) Beta tubulin is half of a dimer (a/b dimer) that helps form microtubules. From these immunofluorescence data, what can you conclude about the role of microtubules in organelle structure?
Nicodazol disrupts the normal structure of microtubles (fig 6B). TCP1 appears to lable the Golgi (figure 6C). Therefore, microtubles are at least in part responsible for maintaining the correct structure of the Golgi body since it is no longer in its normal shape as revealed by TCP1 labeling (figure 6D).

8 pts.
7) a) From figure 7, what can you conclude about this promoter's response to the transcription factor RBP3? (RBP3 expression is under control of the CM viral promoter in this experiment.)
The promoter which lacks the ATF region, appears to still be activated by the transcription factor RBP3. The promoter region that lacks the two E2F regions is not very responsive to the same transcription factor. We also see that at the higer doses of RBP3, the wt and -80/70 constructs become less activated while the ''E2F-less'' construct does not decrease its transciption as the amount of RBP3 increase.
b) What is the obvious next experiment to do if you want to define the promoter even more?
I would like to know if both E2F regions are required for transcription, or is one sufficent? So I would delete each E2F region separately and compare these two additional contructs to the other three constructs.

8 pts.
8) What can you conclude from the data concerning RBP3 in figure 8?
RBP3 is a expressed in humans.
RBP3 mRNA is about 3 kb long.
RBP3 is expressed in every tissue tested.
We cannot conclude how much is expressed in each tissue since there is no control for variations in loading of RNA in each lane.

10 pts.
9) a) What can you conclude about the location of the protein noggin ?
It appears to have a signal sequence so it is made in the rough ER. It does not have any obvious transmemebrane domains so it is probably a soluble protein. Since it does not end in KDEL, it must leave the ER. Since secretion is the default pathway, it is probably secreted outside the cell.
b) What can you conclude about the subcellular location of W1 during translation and later, when W1 finally reaches its proper subcellular location?
It appears to have many transmembrane domains so it must be made on the rER during translation. And, since we don't know if it is retain along the way, the default pathway suggests that it would be located on the plasma membrane.

9 pts.
10) Tell me the name of the cDNA that has the accession number M94130. What organism is it from and in what journal was the sequence first published?
1. transformer protein (tra-1)
2. Caenorhabditis elegans
3. Cell
In case you have not memorized the email method of doing this, you may use these directions:
1) Send a message to the following address:

2) Send a message that has the following format:
(Enter in the accession numbers for the sequence you found in the search above. Put only one accession number per line.) Hit return twice and then send the email.

10 pts.
11) Answer the following questions about the yeast 2-hybrid system:
a) What controls do you need to perform when cloning a cDNA by this method?
You need to make sure to do two negative controls - each plasmid by itself in the cells to make sure that the cells do not turn blue with just one of the GAL4 domains present.
b) List all of the parts of each plasmid used for this method. You do not need to give me the particular amino acids, just the functional domains that we discussed in class.
Both plasmids need: bacterial ori and antibiotic resistance marker; yeast ARS; different selectable markers (e.g. URA3 and LEU2). The DNA-binding domain plasmid needs only the bait cDNA. The activation domain plasmid needs a nuclear targeting sequence and it needs to be built with three variations so that the cDNA can be translated in all three reading frames.
c) Which of the two vectors/functional domains is used to make the genomic or cDNA library?
The activation domain plasmid.
d) Where within the yeast cell must the resulting fusion proteins go?
The nucleus.
e) How are these proteins targeted to the proper subcellular location?
They have nuclear targeting sequences.

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