1999 Molecular Biology Exam #2 - Using Molecular Tools

There is no time limit on this test, though I have tried to design one that you should be able to complete within 2.5 hours, except for typing. You are not allowed to use your notes, or any books, nor are you allowed to discuss the test with anyone until Monday March 22, 1999. EXAMS ARE DUE AT 10:30 ON MONDAY, March 22. 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.


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:




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


6 pts.
1. Figure one shows a Southern blot using genomic DNA from four mice. Tell me the genotype for these four mice if you know that the goal was to convert wild-type mice to mice that lack the gene for craving cheese. The mutation was to be created by deleting a large portion of the appropriate chromosome.
From left to right: heterozygous, homozygous wt, homozygous deletion, heterozygous

9 pts.
2. In figure 2, the stringency of the conditions for a Southern blot were altered.

a) What is the difference between high and low stringency as far as the results are concerned?
At high stringency, the probe only bound to one restriction fragment per lane. At low strengency, the probes bound to many bands, creating the smear. This must be due to the fact that there are many genes with some low level sequence similarity that can be detected with low stringency conditions.

b) What experimental conditions were manipulated to alter the stringency?
No formamide, higher salt concentrations, and lower temperature.

c) Tell me "in plain English" what formamide does given that it was left out in the low stringency condition.
It must play a role in breaking hydrogen bonds at a given temperature. Since stringency was lower without it, that means that fewer hydrogen bonds are needed to hold the probe onto the different bands at this temperature. Therefore, formamide is a chemical way to raise the melting temperature of a probe to its target. If the melting temperature is raised, then it will hold on longer, and thus provide higher stringency.

9 pts.
3. Figure three shows a series of 2 control dots and 7 experimental dots on a dot blot. In this experiment, The dots on the membrane are copies of DNA that will bind to one of two alleles (either the M. musc. allele or the M. spr. allele). mRNA was isolated from the 7 experimental individuals and converted to radioactive cDNA by RT-PCR. The PCR products were made single stranded and half of each cDNA was incubated with the two types of dots. The seven individuals were all heterzygous for the two alleles.

a) Interpret these results.
Since all individuals are heterozygous for these two alleles, you would expect each to bind to both probes (M and S) but they do not. Therefore, each individual expresses either one allele or the other. . The M probe seems to bind better in this experiement.

b) What is the function of the M and S lanes in this figure ?
These are the positive and negative controls and contain either the M or S allele DNA. Each spot only bound its own probe indicating that the probes are working (+) and the spots do not bind the other probe (-) indicating the probes are specific.

c) Do you think that function (from question b) was successfully carried out? Explain your answer.
Yes, the controls seem to work since spots are only seen when M was probed with M and S probed with S.

6 pts.
4. Figure 4 shows the results of a FACS experiment comparing two cell populations: wild-type and a mutant called 13D. Tell me what is going on in these two cell types if ethidium bromide was used to label the cells. Do not break your answer down to an account for each hour but just summarize what is happening over the 13 hours shown here. You might want to use a ruler.
Overall, we know that ethidium bromide binds to DNA and will fluoresce when excited by a UV light. Therefore, this FACS shows that wt cells dubled their DNA content during this time period while many of the mutant cells more than doubled their DNA (4-8 times as much DNA in some cells). This mutation must have destroyed the cells' ability to stop after one round of replication.

8 pts.
5. Figure five is a well controlled experiment, but the MW standards did not photocopy well. Please interpret figure 5 D for me.
Even numbered lanes: The actin is used as a positive control for RT-PCR. As long as the actin bands are there, we know that each RNA preparation was good enough to perform the Ich experiemnts. All the controls worked perfectly.
Odd numbered lanes: Ich-1
S is expressed in thymus, adult heart, and embryonic day 15 brain of mice. Ich-1L is expressed in adult kindey, embryonic day 15 brain, and adult brain of mice. It is interesting to note the brain starts off expressin both but converts to just Ich-1L.

12 pts.
6. Figure 6, believe it or not, shows one line of sequence that was left out of a paper. You must use the WWW to answer this question and you may use any pages from the Davidson Biology web site that might help. Please tell me:

a) the name of the protein
polycystic kidney disease 1 protein (PKD1)

b) what journal published the sequence first

c) what species this sequence is from
Homo sapiens

d) the first three and last three amino acid names from this line in figure 6
lysine, glycine, leucine ...... glutamine, threanine, alanine

e) how big the mRNA is that encodes this protein
14136 bp

f) what chromosome this gene is on
on chromosome 16

10 pts.
7. Figure 7 did not photocopy perfectly. In the original, TRAF1 does not touch hTNF-R2.

a) Design an experiment to determine whether TRAF1 does or does not bind to hTNF-R2.
I would make a monoclonal antibody to hTNF-R2. We would need a cell that does not make TRAF2 since it appearst to bind both hTNF-R2. and TRAF1. Finally, I would metabolically label the appropriate cells and immunoprecipitate hTNF-R2.. I would run the resulting pellet on an SDS-PAGE and obtain a fluorograph. If TRAF1 bound directly to hTNF-R2., then I would see two bands on the fluorograph. If not, there would be only one band. As controls, I would want to perform the same experiment with cells that did make TRAF2 (+ control), as well as cells that did not make hTNF-R2 (- control).

b) Design a second experiment to clone the cDNA for the protein that TRAF1 interacts with in the cytoplasm (indicated by a dark question mark below the white arrow).
I would probably make another mAb but this one would bind TRAF1. I would do an immunoprecipitation of TRAF1, and copurify the next molecule in the pathway (we'll call it X). I would then get some amino acid sequence for X, make an oligo, probe a cDNA library, and isolate the X clone.

10 pts.
8. On a related topic, Figure 8 shows an immunprecipitation experiment. GST is a generic protein that was used to fuse onto hTNF-R2 (ignore the icd part). Various fusion proteins were made that deleted a number of different amino acids of the cytoplasmic tail of hTNF-R2, as indicated in the figure. For example, "-16" means that the first 16 amino acids were deleted. D304-345 means that only those amino acids were deleted, while 384-424 means that only those amino acids were used in the experiment. All variations of hTNF-R2 were fused to GST and the antibody recognized GST only. The arrows indicated bands that are not the GST fusion proteins.

Interpret these results.
In this experiement, there are three bands that coprecipitate with the full length (wt) contruct (about 45 kDa, 55 kDa, and 68 kDa in size).
All three proteins are absent in lanes -37 and -59 which suggests that amino acids 17-37 are neccessary for binding to all three proteins.
Amino acids 304-345 do not appear to play any role in binding since all 3 bands are present when these amino acids are deleted.
Amino acids 384-424 appear to be sufficient to bind to the 45 kDa protein since it is the only band in the last lane.

8 pts.
10. Here's an unusual question. Figure 10 shows the sequencing gels for a wild-type mom and a mutant dad. On your copy of figure 10, below the lanes, label each lane properly for each nucleotide.
For both gels, from left to right: G A T C

8 pts.
12. The last question is related to an abstract I saw. A gene that causes a disease has been localized to a small region of the X chromosome that you have cloned in cosmid (let's say 40 kb in length). This disease is caused by a lack of an unknown protein and results in immunodeficiency (loss of B and T cell function even though the cells are present). Tell me how you might clone this disease causing cDNA. Be sure to devise a way to demonstrate that you have the correct cDNA.
Step 1:
First, you would want to use this cosmid DNA as a probe on a Northern blot to see how many mRNAs it recognized. If it recognized only one band, you would be ready for the next step. If it recongized more than one band (i.e. several different genes), then you would have to narrow down your probe. This would require you to use smaller fragments of the cosmid as probes on blots to see which fragments bound to mRNA made in B and T cells. This would allow you to find the right piece to use in the next step.

Step 2:
Use the appropriate piece of DNA as a probe to screen a cDNA library made from B and T cells. This will allow you to isolate the appropriate clones.

Step 3:
This clone would have to be used on a Northern blot using B and T cells from wt and diseased individuals. The clone should show the appropriate bands in the wt lanes but be absent from diseased cell lanes. Or, you could use complimentation to fix B and T cells from a diseased individual.

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