Molecular Biology Final Exam - spring 1998




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Questions based on figures from page A
8 pts.

Top band = ____
640 bp_____

Bottom band = ____
450 bp______

8 pts.

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.
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.

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.
.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.
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.


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.

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.
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.

+ = wt allele; - = knockout allele
1 = + / -
2 = + / -
3 = + / +
4 = - / -
5 = - / -

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.

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.

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.


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



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.

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.


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