There is no time limit on this test, though I have tried to design one that you should be able to complete within 3 hours, except for typing. You are not allowed to use your notes, or any books, any electronic sources, nor are you allowed to discuss the test with anyone until all exams are turned in at 11:30 am on Monday March 27, 200. EXAMS ARE DUE AT 11:30 AM ON MONDAY, March 27. You may use a calculator and/or ruler. The answers to the questions must be typed on a separate sheet of paper unless the question specifically says to write the answer in the space provided. If you do not write your answers on the appropriate pages, I may not find them unless you have indicated where the answers are.
-3pts if you do not follow this direction.
Please do not write or type 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 here):
Write out the full pledge and sign:
Here is the honor code
"On my honor I have neither given nor received unauthorized information regarding this work, I have followed and will continue to observe all regulations regarding it, and I am unaware of any violation of the Honor Code by others."
How long did this exam take you to complete (excluding typing)?
You have called the 1-800 number for a new game show called "Who wants to be a Molecular Biologist?!" The host, Reagent Philament has created some screening questions that will separate the "wanna' be's" from the hard core molecular biologists.
Press 1 to continue.
Many of the figures are on the last pages of this test, while others are embedded in the questions. Good luck and remember in this game you are not allowed to use any life lines.
2. For $400. This question has some figures that are not necessary to answer the question, but help you understand the importance of this work. Go to the same URL to see the four additional figures.
a) Read the abstract.
b) Angiogenesis is the biological process of forming new blood vessels. Remember that blood vessels are lined with endothelial cells.
c) Design a series of procedures that would allow you to purify angiostatin for the first time. Up to this point (for the sake of the test) no one has any idea what molecule(s) is/are inhibiting tumor growth. You job is to find that/those molecules. Your only hint is that the molecule(s) does not work if it is boiled first. I will tell you that this group started with 20 liters of mouse urine, so you know it is not found in high concentrations.
First, I know it is probably a protein since boiling it destroys its activity.
So I am looking for a protein. I would use column chromatography (or non-SDS gels and purifying the separated proteins) to find out the molecular weight of the active molecule. I would continue to purify the active protein away from the non-active proteins. By this time, I would know its molecular weight.
Once I had it down to a single protein, I would make a mAb and see if adding the mAb to the urine blocked its activity.
3. For $800. A new Drosophila protein has been identified via its cDNA. The cDNA was generated from flight muscles. This fly cDNA was used to probe a blot using muscle tissues from three species. On the gel was loaded: 1 µg of Rat mRNA, 2 µg of Chicken mRNA, and 5 µg of Drosophila mRNA. See figure 2.
An electronic version for this is also available if you want to see it (same URL as in question 1)
a) Interpret figure 2 as fully as you can.
b) What control (if any) has been run on this blot?
c) What do you think the purpose was for figure 2?
a) Figure two shows that the fly cDNA binds to one mRNA from rat and chicken. The chicken mRNA homolog is the largest in MW, followed by the rat mRNA homolog. The fly mRNA has the smallest MW.
b) The fly mRNA is a positive control.
c) The purpose was to determine if there are homologs in other animals and to determine the MW of these mRNAs. The purpose was NOT to determine amounts of mRNA in each tissue.
4. For $1600.
The figure here (no.3) shows the results of an experiment using mRNA and RT-PCR. RT-PCR uses reverse transcriptase and then PCR to produce bands on a gel. The frog embryos were dissected at stage 8 and broken into three parts, or used whole (embryo lane). EF1 a is an elongation factor used during translation.
a) What is the purpose of using EF1a?
b) What is the purpose of leaving out RT in the lane marked "-RT" ?
c) What is the bottom line for this figure?
a) EF1a was used as a control to determine the relative
amount of RNA used in each RT-PCR experiment. It should produce
bands of approximate intensities in all lanes. It also served
as a positive control for the RT-PCR conditions.
b) This negative control insures that the template was RNA and not genomic DNA.
c) The bottom line is that BMP Rec is transcribed highest in the animal portion of the embryo, second highest in the marginal portion, and very weakly transcribed int eh vegetal portion.
5. For $8000. Figure 4 shows a summary for a lot of work. The question being asked in this experiment is "Which amino acids of mouse BRG1 are necessary to bind to mouse retinoblastoma (RB) protein [which is a powerful tumor suppressor]?"
The figure shows the amino acid numbers along the top and boxes indicating which portions of the full length protein (which is 283 amino acids long) are being expressed in each experiment. The black boxes (first 3 constructs from the top) showed full binding; the next 6 constructs are shaded gray and indicate partial binding to RB; the 5 white boxes indicate which constructs failed to bind to RB at all.
a) Does BRG1 have one linear sequence that binds to RB? Explain your answer.
b) Which amino acids play a direct role in binding to RB?
a) No. It appears that there are biding sites in the first 17 amino acids, as well as amino acids 17-118 and a third binding domain in the 118 - 215 stretch of amino acids. None of these parts seems to be sufficient for complete binding since the white boxes (non-binding parts) cover two sections but not the middle one.
b) It appears that amino acids 1-17, 51-118, and 118 - 215 bind to RB.
6. For $16,000. The abstract by Yin et al. (1994) show why Drosophila is a great model system for neurobiology. Cycloheximide is a drug that blocks translation of mRNA into protein. CREB is a transcription factor that must bind to cAMP before it can be functional. A dominant negative protein means that it is a dominant protein that blocks the wild-type proteins from functioning properly. ARM can be thought of as short-term memory.
a) How can a dominant negative allele work?
It must produce a protein that can bind where wt CREB binds but it is unable to activate transcription. By binding, it blocks wt CREB from being able to bind and thus activate transcription. The dominant negative protein also probably out competes the wt protein by being more abundant or having a higher affinity for the binding site.
b) What can you conclude about LTM since it is cycloheximide sensitive?
New protein production is required for long term memory.
c) What can you conclude about LTM since it was completely blocked when the dominant negative transgene of CREB was expressed?
LTM also requires new transcription.
d) From these experiments, what can you conclude about the differences between ARM and LTM?
ARM does not require the production of new proteins but LTM does. The LTM genes must be transcribed when needed and the mRNAs are not stored for use when needed with LTM.
7. For $32,000. One paper was entitled "The cytoplasmic domain of the X receptor is Sufficient to couple the receptor to its signal transduction pathway".
a) What type of experiment did the authors do to say it was sufficient?
They must have cut this portion off of receptor X and spliced it on to another receptor (Y) and shown that ligand binding to Y was able to start the X signal transductin pathway.
b) What type of experiment would the authors need to do to say it was necessary?
They would have to delete this same portion from receptor X and show that it no longer is able to signal.
8. For $64,000. The figure on the below is asociated with the figure legend # 5 at the end of this test. Are there any statements made in the figure legend that are not supported by the data?
You MUST explain your answer(s) to receive full credit.
A) It is very difficult to see the 7
kb mRNA in panel A. It is hard to understand why they discounted
the largest band in lane EP.
B) It is also hard to see the 5 kb band in the adult lane. It looks like there is a 7 kb band in the EP lane, especially since there was so little RNA loaded in that lane.
C) Very little EP RNA was loaded on this gel. It is not clear why the LP lane rp49 standard is shifted up.
Is that your final answer?
Sadly, the sponsors of the contest must confess that they were
unable to raise enough money to continue this fun game. However,
stay tuned for the newest idea, "Who wants to marry a Molecular
Biologist?!" (the correct answer being "who wouldn't