Spring 1999 Molecular Biology Exam #1 - Learning the Tools





Name (please print here):


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Average for this test in 1999 was 85% (range 100% to 60%)


6 pts.


guanine on the left
cytosine on the right

cytosine on the 5 ' end
guanine on the 3 ' end

6 pts.

DNA ligase was used to clone the PCR product into the pNoTA plasmid.

16 pts.

3. The following notes were found in the kitchen of a work study student and a certain teacher has been accused of leaking classified information to students who wanted to take his class but were obstructed from doing so. Please help clarify the situation by showing that this course is not so hard and telling the congress how to do these calculations:

0.146 g NaCl + 150mL 50% SDS + 0.61 g Tris
Put about 200 mL of water in a flask, add the above and dissolve. pH to 7.5 with HCl, then add water to 500 mL.

178.6 µL of this stock (which has a concentration of 4.2 ng/µL).

c. What does OD 260 mean anyway??
This means that we measured the absorbance of light (260 nm wavelength) of a solution. If this were a DNA solution, we could determine the concentration of the DNA, as in problem "b" above.

0.01 g NaCl + 1.5 µL KOH + 4.32 g SDS
Put about 100 mL of water in flask. Dissolve the above and then bring the volume up to 150 mL with water.

18 pts.



methionine serine, asparagine, arginine, serine, valine, arginine, histidine, cysteine, tyrosine, alanine, isoleucine, leucine, alanine, ...proline....








6 pts.

5 ' untranslated region (UT)
3 ' UT
Both of these provide stability for the mRNA in the cytoplasm and can influence translation too.

6 pts.

To convert a Golgi-localized protein to a cytoplasmic one, the gene sequence would have to be altered. One would need to modify the encoding nucleotides of the first exon so that the portion which encoded the signal sequence was deleted. This would cause the protein to be translated on free ribosomes and stay in the cytoplasm.

6 pts.

I would use PCR with primers that amplified the ampR gene but also contained Kpn linkers. The PCR product and pB would be digested with Kpn, and then ligated together. The ligation reaction would be transformed into E. coli cells and plated on amp containing LB. If the cells grew, they must have contained the ampR gene. I would then either test them on Kan plates and look for cells that failed to grow (negative result) or digest minipreped DNA with Kpn and look for the appropriate banding pattern.

8 pts.


RFLP analysis would be the most obvious choice. DNA would be extracted from the cloth and then subjected to analyses at several loci (different restriction enzyme and probe combinations).

One should test the two consenting adults' DNA, as well as the technicians' DNA. A positive control (known standards) for each locus would be nice too.

14 pts.


Either sequence some of the protein and design a probe to screen a genomic library, or use a mAb and an expression library. If you do the latter, you would have to use the resulting cDNA to screen a genomic library, so plan A seems to be the better choice (one screening).

If you had a mAb, you could see if it bound the mouse homolog. If you get no signal, then you need to determine if mice make said dumb protein by screening a mouse library using the human sequence as a probe. Eventually, you could perform an RNA blot using RNA from a range of tissues to see where the gene is transcribed. If the original mAb works, then you could do a protein blot on a variety of mouse tissues. This mAb approach would detect protein expression directly while the Northern blot would be an indirect measurement.

I would put the human DUMB promoter upstream of GFP. Then I would secretly get this construct into the muscles of the big toe by injecting the DNA (not noticed by someone so dumb) into the toe. THen I would examine the toes daily with a UV lamp to find out the level of GFP expression.

4 pts.

A seems to be the most problematic since a hydrophobic R group has been changed to a hydrophillic one, and one that could be phosphorylated even. This occurs within a ß-pleated sheet. The substantial change in amino acid at position C is not in an area of certain structure, so it does not seem as critical.

10 pts.

This would follow the traditional route with the main exception that you cannot make a mAb against a mouse protein in a mouse. You must use another organism. Hybridomas would have to be screened to see which ones actually make antibodies to the DUMB protein - you cannot guess which ones will.

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