Molecular Biology Test #2 1998

Spring 1998 Molecular Biology Exam #2 - Real Data

Figures cannot be reproduced on the web version due to copy right restrictions - see originals on reserve in library.

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 noon Monday March 23, 1998. EXAMS ARE DUE AT 8:30 ON MONDAY, March 23. You must use the web as indicated on question #7 and only for #7. 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)?

Most of these questions have more than one correct answer, but here are the ones I thought of.

8 pts.
1) Figures 1 and 2 show the results when a receptor on the plasma membrane of mammalian cells (FcRII) interacts with its ligand (IgG). There are 2 naturally occuring alleles of FcRII (B1 and B2), and one engineered allele (Tail-) that has had its cytoplasmic tail deleted completely. Figure 1 is provided to give you some background information.

Interpret the data from figure 2.

The figure 1 immunofluorescence photomicrographs show the distribution of ligand either on the surface or internally. Figure one shows that the B2 allele can internalize the ligand, while the B1 allele cannot. From this, we look at figure 2 and see again the distribution of ligand IgG.
For B2, the labeling looks very different between permeabilized and non-permeablilized so we can conclude that the ligand can be internalized. B1 and Tail- receptors do not produce significantly different images, suggesting that they cannot internalize the ligand. Therefore, the carboxyl-terminus appears to play a significant role in the internalization of this ligand, and either the lack of a cytoplasmic tail (Tail-) or the wrong sequence (B1) prevents this internalization.

8 pts.
2) A recent discovery was made about individuals who are resistant to the onset of AIDS after they have been infected with HIV. As we all know, there is genetic variation in the population and this includes coding as well as non-coding portions of the genome. Some individuals have a chemokine (short protein used in cell-to-cell communications) that has an altered 3' UT region. In these individuals, the amount of chemokine is higher than in people who do not have the altered 3' UT region.

A. Formulate an hypothesis to explain what is being observed.
Since we know that the 3' UT region plays a role in mRNA stability, it seems likely that this "altered" allele produces a more stable mRNA which results in a higher amount of translated chemokine.

B. Devise an experiment to test your hypothesis.
To test this, I would build a chimeric cDNA in an expression system to be used in tissue cultered cells (or in yeast). The cDNA would encode for GFP and have the normal GFP 5' UT region. On the 3' end of the construct, I would delete the normal GFP 3' UT region, and ligate in the "normal" chemokine 3' UT region or the "altered" 3' UT region. Then I would look at the amount of GFP mRNA produced with the two constructs via an RNA blot.

8 pts.
3) As you can see in figure 3, you can now buy RNA blots all ready to probe from companies like Clonetech.

A. What assumption have the experimenters made with this figure? What control would you like to see?
They have assumed that equal amounts of good quality mRNA was loaded into each lane.
I would like to see a control probe for these blots, such as ERD2, to show that each lane contains the same amount of mRNA. I do not want to use actin as the probe since muscles will probably make more of this than the other tissues.

B. Let's assume that the proper controls have been performed, interpret figure 3.
Panel A shows that there are two sizes of mRNA that bind to this probe. If both are really adhalin, then there is probably an alternative splicing situation going on such that two different lenghts of mRNAs are produced. This does not look like smearing due to degradation of mRNA.
In adults, skeletal muscle makes much more mRNA than heart tissue, but adhalin is not detected in any other tissue (it seems to be muscle-specific).
The mRNAs are about 1.8 and 1.0 bases in length.

In panel B, we do not learn much else, except that in the fetus, there is much more expression in the heart (of two sizes). Maybe this is due to the stage of development and the heart is actively pumping.

7 pts.
4) Figure 4 was published in Cell . Interpret the results.
There are many problems with this figure as it stands. The top panel is trying to determine the molecular weight( MW) ,the tissue distribution, and the amount of mRNA in each tissue. As a part of this process, they have probed the same blot with a ß actin probe. As you can see in the bottom panel, WASP has two mRNAs one at about 3.4 kb and one at about 2.0 kb and this is detectable only in the thymus lane.
However, due to the poor quality of the ß actin bands, it is difficult to say if all of the WASP mRNA has been detected or not. For example, see the Liver lane, there is no ß actin band so we cannot be sure if there is any WASP mRNA in the tissue but the RNA on the blot was of poor quality. Likewize, we cannot determine relative expression levels since the intensities of the ß actin bands vary from lane to lane - especially note that the thymus lane has the most intense ß actin band, so probably more mRNA was loaded into that lane than any other. So MW is the only thing that can be resolved for sure.

7 pts.
5) Interpret figure 5B. Ignore the lines below the graph marked 1, 2, 3.
This Kyte-Doolittle plot is used to suggest if a protein is an integral membrane protein or not. Since most people use 2.0 as the cut off point, and there are no portions of the graph that reach this high, it looks like WASP is a cytosolic protein. Also noted is the absence of a strongly hydrophobic region at the amino terminus, suggesting that there is no signal sequence which is required for proteins to be located inside the ER, Golgi, or secreted.

7 pts.
6) Figure 6C shows a family pedigree and an agarose gel. Interpret these results if an open symbol means wild-type, a spot in the middle means carrier genotype, and a black shape means diseased status for this recessive disease.
The two women are carriers for the trait and the grandson is diseased. It appears from this agarose gel that there are two alleles in this family. When the alleles are amplified by PCR and cut with Mnl I, there is either a single band of about 240 bp, or two bands of 90 and 150 bp. If a person has two copies of the 240 bp allele, then he or she is homozygous wild-type (wt). If one carries a copy of each, as these women do, then the person is heterozygous. A homozygous recessive person will have the disease phenotype and carry only the 90/150 bp allele.

10 pts.
7) For this question, I want you to show off your computer skills that might be required of you for a job as a lab tech. You may use the web to figure out how to do this, including this site <http://bio.davidson.edu/Biology/Courses/Molbio/NIHsearch.htm>, but you may choose your favorite way.

A. You work in a lab that studies Okazaki fragments.

Find the RasMol image for Taq DNA polymerase and put it on your web page so that I can click on it from your main page.
Click here for an example.

B. You work for Monsanto and have obtained a peptide sequence (IEESQFAIVVFSENY) from a plant protein. Search Genbank and tell me two things.

1) What is the name of the full-length protein and what species did it come from?
TMV resistance protein N -from tobacco (Nicotiana glutinosa)

Score = 73 (33.5 bits), Expect = 0.013, P = 0.013

Identities = 15/15 (100%), Positives = 15/15 (100%)

Query: 1 IEESQFAIVVFSENY 15

IEESQFAIVVFSENY

Sbjct: 63 IEESQFAIVVFSENY 77

2) Are there any proteins from other species that have a high degree of sequence similarity. Explain your answer. (You might want to cut and paste some of your Genbank information to support your answer.)

Yes.

resistance protein RPP5 - like [Arabidopsis thaliana]

Score = 57 (26.2 bits), Expect = 2.1, P = 0.88

Identities = 11/15 (73%), Positives = 13/15 (86%)

Query: 1 IEESQFAIVVFSENY 15

I+ES A+VVFSENY

Sbjct: 1418 IKESSIAVVVFSENY 1432

TMV resistance protein N-like [Arabidopsis thaliana]

Score = 52 (23.9 bits), Expect = 10., P = 1.0

Identities = 10/15 (66%), Positives = 14/15 (93%)

Query: 1 IEESQFAIVVFSENY 15

IE+S+ +IVVFSE+Y

Sbjct: 73 IEQSKMSIVVFSEDY 87

These proteins come from another plant, but probably share some homology, thus similar sequence and similar function.

10 pts.
8) Figure 7 deals with a protein called Fgf5 that is involved in the growth of hair. A mutant strain of mice exists that has had this gene mutated so that normal Fgf5 protein is not made (recessive mutation).

A. Interpret the results from figure 7.
From the series of Southern blots, it appears that the normal protein cannot be made in the mutant because intron 1 has been deleted. Exons 2 and 3 seem to be fine, but the intron between exon 1 and 2 also appears altered, probably due to the same deletion. The HindIII blot is easily explained, but the XhoI blot is more problematic. The two bands indicate that the probe binds to two Xho fragments, suggesting that the probe spans the Xho site. Since both bands are altered in the go lane and yet the probe binds to two fragments, it is not clear how a single mutation could produce two bands each with different sizes. This suggests that another mutation has occurred as well.

B. Given what you have learned in figure 7, hypothesize how the mutant phenotype can be very long hair compared to wild-type.
If you do not have a functional protein and the hair is long, this suggests that the wt protein tells the hair when to stop growing (i.e. when the hair has reached full length). This would explain why different mammals have different length hair but individuals within a species/bread have hair of a similar length.

Questions 9 - 11 are related questions on the same topic. You might want to look at all 3 questions before answering any of them.

8 pts.
9) Figure shows two immunofluorescence micrographs from a familiar lab (see title and authors). Interpret the results from figure 8 if you know that p58 is a protein that is located in the salvage compartment.
The most striking result from this figure is that ERD2 does not reside in the "salvage compartment" as defined by p58, but appears to be located in a more Golgi-like structure (note the perinuclear staining and figure 10 of this test).

8 pts.
10) Figure 9 shows some experiments with lysozyme with modified carboxyl-termini. Three constructs were made that terminate in either KDEL, HDEL, or DDEL. Interpret these results.
We learn that these cells have an ERD2 that is capable of retaining lysozyme-KDEL, but not lysozyme ending in HDEL or DDEL. When the human ERD2 is expressed, we learn that DDEL works as a retention signal which means that the human ERD2 recognizes DDEL but not HDEL. What we cannot know is whether the hERD2 recognizes KDEL or not. However, the data in figure 10 on this test suggests it does recognize KDEL.
We are not told if the hERD2 is over-expressed or not, so we cannot guess whether the DDEL retention is due to over-expression retention of secondary recognition sequences.

8 pts.
11) What do we learn new about ERD2 from figure 10 if you know that GalT is a protein that is found in the Golgi compartment. (By the way, did you realize that the Golgi body was discovered 100 years ago in 1898?!)
What we learn is that the location of hERD2 (and others species' ERD2 too?) can be located in different organelles, depending on the amount of ligand present. When there is not an over-expression of a KDEL terminating protein (AARL), hERD2 is lcated in the Golgi, as we also saw in figure 8. But when KDEL-terminating proteins are over-expressed, hERD2 relocates into the ER. This is probably due to the large number of KDEL proteins that are being returned to the ER and so a higher proportion of ERD2 is in the ER compared to when there are fewer KDEL proteins. This relocation of ERD2 does not suggest that the structure of the Golgi has been lost or disrupted as can be seen by the normal labeling of GalT when KDEL is over-expressed.

9 pts.
12) Briefly answer these lab questions:

A) Tell me how you would make a 0.5% agarose gel that has a volume of 60 ml, is made of 0.5X TBE (you have 1 L of 5X TBE) and the MW of agarose is 89.
0.3 g agarose poured into
6 ml of 5X TBE + 54 ml water
microwave, cool, add ethidium bromide, and pour

B) What is calf intestine alkaline phosphatase used for (outside a calf, that is)?
We used it in lab to remove the 5' phosphates from the BamHI-cut plasmid so that it would not self-ligate. This helps ensure that the only ligation that can result in a colony after transformation is one that has an insert cloned into the plasmid.

C) What is the function of phenol/chloroform in a DNA prep?
This is used to remove protein contamination away from the DNA.

2 pts.
13) What unit of measure is used for the molecular weight of proteins? (This is a give-me question to bring the total up to 100 pts.)
daltons

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