Exams 2007

Advanced Molecular Biology-Dr. Schleif

October 9, 2007

90 minute exam

Each question is worth 10 points.

1.  In principle, for the process of transcription elongation, why might you expect to find both topoisomerase I and topo II associated with a transcribing RNA polymerase molecule?

        As polymerase proceeds forward, it must go round and round the the axis of the DNA, but this soon becomes impossible due to the long  RNA product.  Instead, the segment of DNA under the polymerase must instead rotate.  In front of the polymerase positive supercoils must be removed, topo II activity, and behind, negative supercoils removed, topo I activity.  

2.  Suppose you had an antibiotic whose action on protein synthesis was analogous to the action of rifamycin on RNA synthesis.  How could you use the antibiotic to measure the elongation rate in protein synthesis?

        Induce a gene whose protein product you can easily assay, add the inhibitor, measure how long until there is no further increase in the levels of the gene product.

3.  The following schematic shows only the DNA, RNA primers and DNA pol III at one point in a replication cycle.  Make a similar drawing depicting the system at the time DNA synthesis of the next Okazaki fragment is about to begin.

Answer

4.  How would the structure of the alpha-helices of a coiled-coil have to be different in order that the helices of the coiled-coil become straight and parallel?

        Keep the stripes of hydrophobicity from wrapping around the alpha-helicesby winding the alpha helices a little tighter, 3.5 amino acids per turn instead of  3.6.

5.  How could you use RNA interference to identify introns in genes?

        RNA mediated interference takes place in the cytoplasm.  Because an intron is removed from the mRNA before it is exported to the cytoplasm, double-stranded RNA to regions that lie in introns will produce no interference.

6.  How would you separate linear doublse-stranded 1000 base pair DNA molecules from forked DNA molecules in which the three arms are each 333 base pairs long?

        2D gel electrophoresis as described in the book and reading.

7.  Why do all tRNA molecules possess roughly the same shape?

        In order that they all may function in the process of protein synthesis by utilizing the same interactions with the ribosome.

8.  Why might you expect to find some interesting interactions between eukaryotic RNA polymerases and histones?

        Possibly in “looking” for promoters, polymerases would have to displace histones.  More likely however, because particular histones are associated with the beginnings of genes, they need to be put there by RNA polymerase as it begins transcription, or put back there when a gene with such a histone is transcribed.

9.  Two rowboats that are caught away from shore in a hailstorm in which there is no wind, no propulsion of the boats, and the large hailstones falling straight down, will tend to stick together.  Why?

        Depletion attraction, reading from first lecture.

10.  Suppose a solution of DNA + ribonucleoside triphosphates you add transcription factor A, then B, then RNA polymerase, and you initially see no transcription.  Now suppose you add B, then A, then RNA polymerase to an identical mixture and transcription is initially strongly stimulated.  Explain what could produce such a difference.

        Factor A interferes with binding by factor B. Factor B stimulates RNA polymerase binding, and if it has bound, A cannot bind.

Advanced Molecular Biology-Dr. Schleif

November 13, 2007

90 minute exam

Each question is worth 10 points.

1.  Using nonmutant components from the lac and  ara regulation systems, design a genetic circuit that will turn on the activity of a promoter only in the simultaneous presence of IPTG and arabinose.

2.  Comment on the validity of arguing that because a temperature sensitive mutation can be isolated in a gene, the gene product of that gene must be a protein.

3.   Suppose you have a particular lacZ mutant and you wish to find a short, 10-20 ribonucleotide, that binds to b-galactosidase and causes the protein to regain enzymatic activity.  How would you proceed?

    Synthesize DNA coding for a promoter and in the transcribed region, randomize

    20 nucleotides followed by a termination sequence.

    Clone, transform into cells with the Z mutant.

    Select for lac⁺ cells

4.  How would you go about finding the tightest possible rho independent bacterial transcription termination sequence?

    Clone a stretch of 30 randomized nucleotides between the lac promoter and lacZ gene.  Induce the lac promoter and select for LacZ minus.

5.   If you hypothesize that the intervening sequence in a particular gene has no particular biological importance, how would you FIRST proceed to gain some confidence that perhaps your idea is correct?

    In sequence databases compare in related species the frequency of changed nucleotides in the coding and noncoding regions.

6.  By examination of the PAM 250 matrix one can see that evolutionarily well tolerated and poorly tolerated amino acid substitutions make biochemical sense.  Suppose instead that the tolerated and nontolerated substitutions appeared to make no sense biochemically.  What would you guess is the reason and where would you look for support for your idea?

    The amino changes that are found in proteins are found over evolutionary time reflect both the ease of generating the change by alteration of the DNA sequence and the ease of tolerating the amino acid change in the protein.   If the frequently found changes made no sense biochemically, perhaps it is because these are the changes that can be made with single nucleotide changes in the coding region and the rarely made amino acid changes require multiple nucleotide changes. A check of the codon table would verify or refute the idea. By the way, similar amino acids are encoded by similar codons.

7.  Typically DNA binding proteins possess selectivity for their correct binding site by factors of perhaps 10⁵.  A selectivity of this magnitude by a restriction enzyme would leave the host’s chromosome highly vulnerable to self-cleavage from occasionally cutting at a non site (and therefore not methylated) and cell death, particularly when the cells are subjected to long periods during which growth is not possible (like storage in the refrigerator).  How do restriction enzymes pump up their selectivity?

    Restriction enzyme bind, wait a while, cut one strand, wait a while longer and then cut the other strand, in effect, reading the DNA sequence two or three times.  Presumably then there are highly unlikely to cut both strands at a nonrecognition site.  

8.  If you are determining the sequence of a newly discovered bacterium by the pyrosequencing technique, what are the sequences for which you will have serious trouble, and how many such sequences do you expect to exist in the genome?

bases this comes out to about 200 times.

9.  If someone gives you 100 mg of a purified human protein (purified the hard way without clones or overexpression) outline the fastest way to obtain pure DNA of the gene coding for the protein.  Do not worry about introns.  They can be present in your product.

    Determine enough of the amino acid sequence (perhaps eight to ten residues) to identify the gene in the genomic DNA sequence.  From this design PCR primers to amplify the gene from crude genomic DNA.

10.  A lac operon in which the lacZ gene has been replaced by the gene for GFP and the lacY gne has been replaced by the gene for YFP was introduced into cells, strain A.  Another strain, B, was made by introducing a lac operon with the GFP gene replacing lacZ at a position 50,000 nucleotides clockwise from the replication origin and a copy of the lac operon with the YFP gene replacing lacZ at a position 50,000 nucleotides counterclockwise from the replication origin.  Individual cells of strains A and B were quantitated for both GFP and YFP.  In cells of strain A, whatever the number of GFP molecules, the number of YFP molecules were almost the same.  In cells of strain B, there was no correlation between the levels of GFP and YFP.  What do you conclude?

    Fluctuations in the the numbers of ribosomes translating a messenger are rather small, but fluctuations in the number of transcripts from a promoter are much larger and are the main contributor main source of the noncorrelation in strain B.

Each question is worth 10 points.

1.  Why is it difficult to isolate an a^- mutation lying in the MAT locus of  Sacchromyces yeast of mating type a?

Because the a mating type product has no function in type yeast of a mating type.

2.  Predict the phenotype of a protein consisting of the DNA binding domain of AraC connected by a 25 amino acid polypeptide chain to another copy of the DNA binding domain.

Constitutive as it could easily bind to I1-I2.

3.  How would you explain the (hypothetical) observation that at the location of the bite from a particular insect, one would eventually grow a new thumb?

The venom contains several transcription factors that can transform a somatic cell back to the embryonic stem cell state (a la the hottest science item in the news these past few months) and then a “thumb-specific” transcription factor starts the cell down the thumb developmental pathway with perhaps additional transcription factors also present to aid the process.

4.  Provide a concise explanation for the fact that about half of the time a particular site-specific recombination system inverts a particular segment of DNA and about half of the time the system deletes the same segment of DNA. 

Consider the example shown in class where B-B’ x P-P’ goes to B-P’ + P-B’ and how it can lead to deletion or inversion of the material between the  B-B’ and P-P’ sites depending on the orientation of one site with respect to the other. The situation described in the problem would arise if B could not be distinguished from B’ or P from P’, for example if B is the same as B’ or if the recombinase were blind to the difference.

5.  Please summarize your resolution of the Caesar’s Nose paradox that was defined in an early lecture and mentioned several times during the course—which is “How can a gene, or its exact replication products, remain intact over thousands of years?”

The fact that making a sperm or egg cell takes only one of the two chromosome copies, bad sperm or bad eggs that have taken or used defective chromosomes are discarded, and that intact and functional chromosomes can be reconstructed from two chromosomes with defects by multiple recombination events.  I think you might get by with the first two, and that having all three is as good as you could do.

6.  What is meant by a “strain” of a prion, and how might the existence of different prion strains of the same cellular protein be explained?

Different strains have different and characteristic rates of killing.  As suggested in class, the portion or amount of the prion protein molecule driven into the prion conformation as it is added to the prion aggregate could be templated by a protein molecule already in the prion aggregate.

7.  In normal cells p53 blocks DNA replication if there exists appreciable damaged DNA.  Why would p53 mutants tend to be found in cancers?

Blocking replication of damaged DNA allows time for DNA repair, and this reduces the mutation rate.  Since it is advantageous for a cancer to make mutations, they would tend to have inactive p53 protein.

8.  Suppose the administration of a particular antigen induces in some individuals the synthesis of antibodies against the antigen and, in addition, the synthesis of a second class of antibodies that show no activity whatsoever against the antigen.  How do you explain this second class and what experiment would you suggest to test your idea?

The antigen induced an antibody, which is itself a new shape not previously seen by the organism.  Thus, this induces the second class of antibody.

9.  A mutation in RNA polymerase made cells (bacterial) either tryptophan minus or constitutive for trp operon expression (your choice).  Provide a plausible explanation, i.e. what could be altered about the properties of RNA polymerase consistent with what you know about polymerases in general and about the trp operon?  To forestall your using a trivial explanation, I am not interested in proposals that include directly altering the ability of RNA polymerase to terminate transcription at the usual termination sequences.

RNA polymerase must pause to allow a ribosome to load onto the trp leader mRNA in order that the appropriate hairpin structure form depending on whether the ribosome can translate up to the correct position and then be stuck there or not. Thus, a polymerase that didn’t pause would almost always terminate at the attenuation site, as would a polymerase that transcribed considerably faster than usual (Mutants are known that transcribe three times faster than wild type, and are certainly plausible because DNA polymerase works more than ten times as fast as RNA polymerase.

10.  As usual with my last exam, I’ve only been able to come up with nine fun questions and therefore I’m coping out by awarding ten free points to anyone who writes “AraC Rules”.