Molecular Biology Final Exam - spring 1998

ALL EXAMS ARE DUE AT 5 pm on MAY 8, 1998.

I will be out of town all day May 8 so do not wait until Friday to ask questions.

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 7. 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, the WWW, any books or journals, nor are you allowed to discuss the test with anyone until all exams are turned in at 5 pm on MAY 8, 1998. You may use a calculator, a ruler. 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):

Write out the full pledge and sign:

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


Questions based on figures from page A
8 pts.
1) Use the photo of the agarose gel and the graph paper provided here to calculate the molecular weight of the two bands in lane A as marked. Notice the molecular weights on the left side of 1.6, 1.0, 0.51, 0.40, 0.34, and 0.30 kb. You must draw a correct graph to get credit. Write your answers here:

Top band = ____
640 bp_____

Bottom band = ____
450 bp______

8 pts.

2) Look at figure 2C and 2D. tk is a weak promoter and SV40 is the 3' UT portion that allows the CAT mRNA to be spliced and processed properly.
a) What is Oct-4? What is 6W?
Oct 4 is a transcription factor. 6W is an enhancer.
b) What do we learn about the location of 6W?
This enhancer can be placed upstream of the promoter tk or downstream and it is almost equally effective in either location. Transcription works better when Oct4 is present.

6 pts.
3) Look at Figure 3 A - F. In E and F we see the plasmids used in this experiment. p6WtkCAT is the reporter plasmid from figure 2. pE1A is a piece of the Adenovirus that encodes the E1A protein. pCMV-Oct4 plasmid uses a strong viral promoter (CMV) to drive the expression of Oct4.
a) What do we learn about Oct-4 and E1A as demonstrated by these CAT assays (A-D)?
There is an optimum ratio of E1A and Oct4 that must be present for the 6W enhancer to work properly. We see that as the amount of E1A is decreased, the optimum amount of Oct4 also must be decreased in order to maximize CAT expression. However, Oct4 cannot be a repressor directly or else the lanes with zero Oct4 would always have the greatest expression.
b) Why do you think panel A looks so funny?
It is smiling at us. As we saw in lab, gels (or in this case TLCs) do not always run in a straight line, no matter how much we think they should.

Questions based on figures from page B
16 pts.

4) Figures 2 - 4 deal with the Drosophila gene oskar (osk ). Look at all 3 figures before you answer any questions. These should be short answers.
a) When is osk transcribed in Drosophila (assume the positive control was positive for figure 2)?
This appears to be a maternal transcript, therefore it is transcribed when the egg is being made in the mother (see female and ovary lanes).
b) What role does osk play in Drosophila development?
.Given its localization to the posterior part of the early embryo, it seems likely that osk is involved in the development of posterior body parts of the embryo.
c) Briefly describe the localization of osk in wt embryos as shown in figure 4A - C. Stage 5 is at about 2.5 hrs while stages 2 and 3 are in the 0 - 2 hrs time period.
.There must be osk mRNA in all three embryos since in figure 2, we see that the mRNA is present through the first 5-8 hours. osk mRNA is located in the posterior most portion of the embryo in the region that will be come pole cells (germline cells). By 2.5 hours, the mRNA must be dispursed enough to be undetectable in one location suggesting that it might move into more anterior cells later in development.
d) What would you predict the phenotype to be for figure 3 if this shows an in situ hybridization for osk of a stage 3 embryo?
I would predict a bicaudal (two tails) phenotype which would be lethal.

8 pts.
5) What can you tell me about this epitope tag?
In this figure we learn that the epitope must be about 10 or 11 amino acids long in order for the mAb P5D4 to bind. We also learn that both the amino terminal protion and the carboxyl portion are important in binding, though the carboxyl protion is not absolutely required. We can see this since the rt construct does allow the mAb to bind, but not as well as the 11 mer construct (assuming equal amounts of protein loaded as appears to be the case with the a-P4 antiserum.

Questions based on figures from page C
6 pts.

6) Interpret these results from figure 4. PAB II is not a polymerase.

PAB II appears to enhance the rate of reaction of the polyadenylation by the poly(A) polymerase. As we can see in panel B, the amount of polyadenylation from 4 - 20 minutes is substantially greater than in panel A. We also note that in panel A, the rate of polyadenylation is linear with time while in B, there seems to be an exponential increase in activity after two minutes.

Questions based on figures from page D
8 pts.

7) Figures 2, 3 and the abstract are all related. What aspects of these two blots indicates the mutant genotypes? Notice that figures 2 and 3 are different kinds of blots so there might be more than one answer to this question.
We notice that in figure 2, those who have Charcot-Marie-Tooth disease (CMTD) have 3 bands and one of them is allele A (panel A) or allele E (panel B). wt individuals do have additional "shadow bands", but they are not correlated with the phenotype.
In figure 3, we are looking for the intensity of bands rather than allelic migration differences. The duplication appers in individual 302 (the mother) as an AB locus duplication and also carries a wt A allele. The children of this couple who have CMTD all have an A and a B band but have also inherited an extra B allele from their father.

10 pts.
8) Figure 4 shows the results of an experiment where we are monitoring a truncated protein called SC which is a mutant form of a plasma membrane receptor but it has had its transmembrane portion deleted. This mutant protein has about 800 amino acids while the full-length would have had about 1000 amino acids. WT means that of the remaining 800 amino acids, the sequence is wild type, while 669t has an amino acid substitution at position 669, and
D 655-668 is a short deletion mutation. The assay takes advantage of the fact that the cells used in this study form a tight monolayer of cells which secrete some proteins up, away from the attachment side (apical), and other proteins are secreted down (basolateral) towards the attachment side. It is possible to collect these two separated media (apical and basolateral) because the cells are grown on a fine mesh filter instead of solid plastic and the cells and mesh can be lifted out of the culture disk at any time.
Interpret these results.
WT: 1) It is secreted in the apical direction after 3 and 5 hours of chase time.
2) It takes over 5 hours for all of this molecule to leave the cells.

669t: 1) It is secreted in the basolateral direction within 1 hour and all of it is secreted by 3 hours.
2) It leaves the cell completely within within 3 hours.

D 655-668: 1) It is secreted in the apical direction within 1 hour and is completely secreted by 3 hours.
2) All of this construct is secreted by 3 hours.

Conclusions: It appears that amino acids 655 - 669 are crucial for secretion of this protein and in determining which direction it will go. Amino acid 669 determines the direction while 655-668 determines the rate of secretion. The wild type sequence goes to the apical surface but goes slowly compared to the other construct. It is worth stating that this protein would be inside vesicles and organelles and not in the cytoplasm so the lumen of these membrane compartments holds the targeting mechanism.

Questions based on figures from page E
12 pts.

9) Figure 1 shows a series of related panels related to a strain of knockout mice with the p27 gene being tested.
a) What are the genotypes for mice 1 - 5 in panel B?
+ = wt allele; - = knockout allele
1 = + / -
2 = + / -
3 = + / +
4 = - / -
5 = - / -

b) Describe the phenotypes from panels C and D and hypothesize an explanation.
In the absence of p27, the mouse appears bigger. Even a heterozygote appears to be bigger than the +/+ individual, though smaller than the -/- individual. The reason for this overall increase in size appears to be due to an increase in size of hte various tissues, especially noteable with the splean and thymus.
One wonders if this is due to uncontrolled cell division (perhaps of lymphocytes) or a lack of normal cell death with normal cell division still occurring.

c) Does this knockout mouse use the cell-specific mechanism for deleting a gene?
It appears that this is a complete knockout since no loxP / Cre loci are mentioned and we see normal positive (neo) / negative (tk) selection used to prove homologous recombination. There is also no use of Tet ON/OFF control of expression.

8 pts.

10) p53 has long been known to be involved in cancer as a tumor suppressor. In figure 2, we learn a new aspect about p53 - it is also an exonuclease. In this experiment, a 30mer oligonucleotide was end labeled either on the 3' end (left lanes) or the 5' end (right lanes) and incubated with 100% pure p53. Which end does p53 attack? Explain your answer. This one will take some time to think about but the answer should be short. Also, don't worry if you cannot read all of the legend, there is nothing important missing that I have not told you above, but if you want to try to read it you can.
It appears that p53 attacks the 3' end as can be determined by the fact that the 3' label is cut off first and the band goes from a 30mer to a 1mer within the first time point. On the right lanes, we see that the 5' label stays on the oligo as it is gradually chewed up (seen as smaller banding sizes) from the 3' end. The exonuclease activity seems to stop once the oligo is 2 nucleotides long.

Questions based on figures from page F
8 pts.

11) And the final question on this final is some what related to the paper we just finished on BMP and tld. Its great how our discussion of cloning the tld receptor was so close to what actually was done!

Ignore the top part (labeled EARLY) of panel A.

First, the frog (Xenopus ) homolog for tld was cloned. Then in flies, the tld receptor was cloned and called mad. This figure was performed with the frog homologs to mad (or Xmad1 and Xmad2). Two Xmad genes were cloned. E marks the lane where an entire embryo was used. The wedges indicate how much of the appropriate gene was injected into the frog eggs.

There are various mRNAs being used here to figure out which part of the embryo is being induced with the two Xmads. M. actin = back muscles Globin = ventral tissue N-CAM = notochord/CNS

a) Interpret these results. You do not need to describe in detail what is in each panel, simply summarize your conclusions for all panels.
This is very interesting. The same ligand (tolloid) binds to two different receptors in frogs and the different receptors induce different genes. Xmad1 induces Globin, a sign of ventral tissue induction. Xmad2 induces M. actin and N-CAM, signs of dorsal tissue formation. So these two receptors must have the same/similar extracellular ligand binding site amino acid sequence but have different intracellular domains to initiate two different pathways of gene expression. The interesting aspect is that the same ligand can give two diametrically opposed signals to the developing embryo depending on which receptor is present.
There is some conflicting information between panels A and C. In A, 2 ng is enough Xmad2 to initiate NCAM, but in panel C it is not. This is not explained in this figure and is troubling.

b) What do we learn about Xmads 1 and 2 that might lead to future research in flies and possibly humans?
We learn that there are two receptors in frogs, and we might want to look for two receptors in flies and humans. Also, we learn that each receptor must interact with different intracellular proteins and we might want to clone those different proteins. It might also be interesting to examine the tissue distribution of the two different mad receptors.

2 pts.
c) Final easy two points to bring the total up to 100 pts. How many bands did you see on your zooblot from lab?

Now go out into the world and clone, sequence, etc. and publish your results so I will have test questions for future classes!

Go to Original Test

Return To Molecular Biology Main Page

© Copyright 2000 Department of Biology, Davidson College, Davidson, NC 28036
Send comments, questions, and suggestions to: