Foundations and Applications of Molecular Biology
October 4, 2017
80 Minute Exam
Each question is worth 10 points
1. Why would you expect that negative supercoiling would assist formation of an open complex by RNA polymerase? Describe the principle of an optical method that was used to directly show this.
If there were an excessive number of negative superhelical turns in the DNA, say Wr = - 20, the DNA would try to reduce them, that is, try to go to, Wr = -15. In order the Lk remain constant, Tw would be reduced, and reducing Tw is untwisting or melting the DNA. The optical method is that described in the assigned paper by Ebright and Strick, introduction of a superhelical turn in a short piece of DNA dramatically shortens it, to an extent that the shortening can be observed microscopically.
2. About what fraction of the volume of a typical cell is constituted by water?
80% to 90%
3. A population of 300 base pair double stranded DNA molecules was synthesized of random sequence (except for one six base stretch that is recognized by a restriction enzyme). The linear DNA was incubated with DNA ligase until 5% of the molecules had circularized. The circles were isolated, opened at the restriction enzyme site and sequenced. Analysis of the sequences revealed that almost every circle contained at least one instance of a specific 6-base sequence which is not that of the restriction enzyme recognition site. Comment. (The more insightful your comments, the higher your score.)
The sequences are likely to be intrinsically bent sequences, dA-6, and they would tend to be in phase, that is, separated by multiples of 10-11 base pairs.
4. RNA A has no homology to gene B or regions before or after it. Only when RNA A is present is messenger of gene B translated. What is likely to be going on?
Most people answered (correctly) that B could be something like rRNA, for which they got full credit. I had in mind quite a different answer. Translation of B messenger is inhibited by a folded RNA C that hybridizes to it. RNA A can hybridize to C, using completely different segments of C. Thus, in the presence of A, molecule C is inactivated.
5. Topoisomerase activity is clearly needed ahead of a DNA replication fork. What about behind a fork? Why?
Yes. Even though the single-stranded regions behind the replication fork are single-stranded and can swivel, the long twisted duplex(es) behind the fork are really not free to rotate to introduce the necessary twists in the new duplex. If the question had been phrased better, it would have asked about the twists in the newly synthesized DNA, thus forestalling the answer (yes, to reintroduce negative supercoiling) that got full credit, but did not address the question which was in my mind.
6. Describe the basic principle of the most definitive experiment you know of that the U1 snRNP is involved in messenger splicing, i.e. do you know a good experiment, and do you know how the experiment worked?
Almost no one appears to have read the book! A base of the U1 that was postulated to base pair with a nucleotide of the 5' splice recognition site was altered and splicing at the site was reduced. Upon changing the corresponding base in the splice site recognition sequence to restore Watson-Crick base pairing restored splicing.
7. What was the key realization concerning their experimental procedure that ultimately led Fire and Mello to the discovery of silencing?
After my admonition in class that you look specifical at the paper to see what key piece of knowledge Fire and Mello knew that allowed them to make their discovery, very few bothered to do it.
That the in vitro transcription that they were using to generate their so-called antisense RNA was not fully specific, and their RNA also contained transcripts from the other strand.
8. What in vitro experiment would you do to demonstrate the existence of error correction by an RNA polymerase?
Look for the production of low levels of nucleotide monophosphates being produced during transcription. Their existence would indicate they had been excised from growing RNA chains in the process of error correction.
9. The transcription of many genes, mostly eukaryotic, requires the require the binding of multiple copies of the same transcription factor to the enhancer. What is gained by this multiplicity requirement?
To make the response of the system more sharply dependent (more like a binary on and off switch) on whatever signal the transcription factor is responding to.
10. What is the apparent "energy source" that drives RNA self splicing reactions near to completion? Why even ask the question?
The question arises because the reactions do not utilize the usual biochemical sources of energy for reactions such as ATP. The "energy" is the fact that at the start there is one molecule and at the end, there are multiple molecules. Their combined entropy is greater than that of the original molecule.
Foundations and Applications of Molecular Biology
November 1, 2017
80 Minute Exam
Each question is worth 10 points
1. Suppose you had a yeast or bacterium that possessed a property that was essential for your research, that the yeast or bacterium does not transform, and that your research absolutely requires that you be able to transform the cells using standard transformation techniques. How would you go about isolating a mutant of your cells that could be transformed with the standard transformation techniques?
Mutagenize, attempt a standard transformation using a plasmid carrying a drug resistance, and select for transformants carrying the drug resistance.
2. What does the best evidence for DNA looping in eukaryotic cells look like?
Contact maps from Hi-C experiments show enhanced contact between well separated chromosomal locations and look like the following:
3. Why do most of the mutations found in human genetic diseases turn out to be nonsense?
Generally, only when there is a significant deficiency is a human genetic disease recognized. Many missense mutations hardly change the activity of the mutated protein whereas most nonsense mutations almost completely abolish activity, and therefore, these are the ones most often identified.
4. In a procedure to clone a fragment of DNA in a plasmid, the following four DNA samples containing identical amounts of plasmid DNA were prepared and then transformed into competent cells with the results indicated. What is the explanation for the numbers of colonies observed from each sample? Also, was the cloning likely to have been successful?
A. Plasmid cut with restriction enzyme EcoRI that has a single recognition site within the plasmid ‐ 6 colonies.
B. Plasmid cut with the restriction enzyme then treated with DNA ligase ‐ 200 colonies.
C. Plasmid cut with restriction enzyme, treated with alkaline phosphatase, purified, and then treated with DNA ligase ‐ 0 colonies.
D. As in C, but with the addition of DNA fragment that had been amplified by PCR, the ends cut off with the restriction enzyme, and purified away from the cut ends and enzyme, and then the mixture of cut plasmid plus purified fragment ligated ‐ 30 colonies.
A. Although the plasmid was completely cut, as shown in sample C, it religated closed within the transformed cells because of the stickky ends left from RI cutting.
B. Ligase resealed the cut plasmid.
C. Removing the 5’ phosphates by phosphatase left ends of the DNA strands that cannot be ligated, and zero colonies means that all the plasmid had been cut by the RI
D. The the incoming fragment of DNA provided a phosphate at each end that allowed ligation of one strand, leaving a nick at each end of the fragment that was then repaired in the recipient cells.
5. When constructing complicated self-assembling structures by annealing DNA fragments, why are prolonged annealing times required?
Annealing for a long time allows time for dissociation of improperly hybridized structures, even those that may have only one mispaired base out of 50 or more bases, and their replacement with perfectly hybridized structures that are just a little bit more stable.
6. What is the best evidence that a segment of U1 RNA base pairs with sequence at the donor site in mRNA splicing?
As described in the book, using a base change in U1, and a compensating base change in the mRNA site.
7. Why is or is not a topoisomerase activity required for homology-based genetic recombination?
It is required. Branch migration for any distance requires rotation of the DNA duplex to which the single-stranded DNA is hybridizing to. This rotation will require topoisomerases to relieve the excess or deficiency in twist.
8. Write a command that will insert a 1 with a probabililty p (<= 1) into a spreadsheet cell or into location i in a one dimensional array in a computer program.
if (rand() < p, 1, 0)
or, depending on the language, something like
if rand() < p {
x(i) = 1 }
else {
x(i) = 0 }
9. Name as many amino acid pairs a and b as you can for which their R groups make it sensible that the tRNA synthetase for residue b have an editing capability that hydrolyzes tRNAb misacylated with amino acid a. (Points will be subtracted for wrong answers, thus it would be unwise to list all 190 amino acid pairs.)
Glycine-alanine, valine-isoleucine, cysteine-serine, serine-threonine
10. Why are there no restriction-modification enzyme recognition sites containing more than eight specific base pairs?
Many of the DNA molecules that threaten a bacterium would be unlikely to possess an eight base recognition site, and would therefore be immune to the restriction-modification system.
Foundations and Applications of Molecular Biology
December 6, 2017
80 Minute Exam
Each question is worth 10 points
1. What is the key component that allows the lac system to display the maintenance phenomenon in which exposure to a certain concentration of inducer induces half the cells. Then during prolonged growth in the presence of lower concentrations of inducer, those cells which had been induced remain induced, and those which were not induced remain uninduced. (An example of stochastic regulation and "differentiation" in prokaryotic cells!)
The lac active transport system. Once induced by threshold concentrations of IPTG, it takes the concentration of inducer from subinducing concentrations to concentratiions in the cell that fully induce and keep the system fully induced.
2. Name several mechanisms which act to "sharpen" the response of the arabinose operon to increasing concentrations of arabinose, that is, to make it a nonlinear?
The active transport system, that the protein is dimeric and each subunit responds to arabinse, and when the DNA is looped, the promoter is inaccessible to RNA polymerase.
3. For most DNA binding proteins, it is much easier to measure their dissociation rate from their DNA binding sites than to measure their equilibrium dissociation constants. Why, in fact, in most cases is the dissociation rate an excellent proxy for the equilibrium dissociation constant?
The equilibrium dissociation constant is the ratio of the dissociation rate to the association rate, k-1/k1. Experimentally it has been found that the association rate is the same for almost all DNA binding proteins. Thus Kd is proportional to k-1
4. We all know that lactose is not the true inducer of the lac operon. How do we know or could we know that arabinose is the true inducer of the arabinose operon.
Mutations in the enzyme catalyzing the first step of the arabinose utilization pathway are still inducible for the other enzymes. Alternatively, one could look for arabinose induced transcription in vitro with purified components. This is very much harder to do.
5. When cloned lac operator-promoter is used to drive expression of a cloned gene, the induction of the cloned gene by the addition of IPTG is only about 50-fold. The induction ratio of beta-galactosidase in the normal lac operon is about 1000-fold. Why the difference?
Sequences outside the promoter-operator region act to reduce the uninduced basal level. In fact, they act by allowing lac repressor to loop from one of two outside sites to the operator. One could also have hypothesized that a DNA sequence outside the operator-promoter region is required for full induction--an enhancer.
6. In bacteria, stretches of barren mRNA, as occur beyond translation termination codons, are substrate for the rho transcription termination factor. On the basis of what you now know (or should know) predict why transcription of ribosomal RNA is not similarly terminated by rho.
Shortly after initiating transcription at a ribosomal RNA promoter, a protein binds to the RNA polymerase-and emerging rRNA making use of a specific sequence near the 5' end of the rRNA, and this protein interferes with rho function. One could also postulate that there never are extended stretches of barren rRNA because ribosomal proteins bind to the RNA while it is being synthesized.
7. How do we know without DNA sequence information, that lambda phage integration and excision is catalyzed by a site-specific mechanism that operates precisely, to the base pair, in contrast to the recombination occurring anywhere within a region of ten to hundreds of nucleotides of sequence homology?
Lambda can be forced to integrate in sites within genes. When it is then induced and excises, the gene is fully functional showing that excision was precise to the nucleotide. This would not be the case if the integration were not the precise opposite of integration.
8. In VDJ recombination, the following occurs.
What is the product when the starting material is the following?
Answer
9. Suppose the following: 1. You are absolutely convinced that in the process of development of nearly every nerve cell of the brain, DNA rearrangements occur. 2. Repeatedly, high coverage sequencing DNA from brain tissue yields only unrearranged DNA sequence. How would you proceed?
A sample of tissue taken for sequencing would contain many many different DNA rearrangements. You need the DNA from a single cell. Take a brain tumor, grow a sample of cells sufficient for DNA sequencing from a single cell isolated from the tumor. Alternatively, long read sequencing might give suggestive evidence, but you need to believe the one read out of 20 that shows a sequence discontinuity.
10. How does Tn10, which transposes strictly by cutting itself out of one location and pasting itself into another location, manage to replicate faster than the DNA of its "host"?
By cutting itself out of the chrommosome just after passage of a replication fork. No matter where it then pastes itself, it will be replicated sooner than it would have been replicated if it had stayed put.
Haiku summarizing the course
Sorry, I can't list the author's names.
always remember
not all who wander are lost
it's a random walk
Memorize few facts
Logically you must think
Success will be yours
Questions are challenging
Approximation could help you out
Be smart, good luck
Brownian motion
Intracellular chaos
The life of a cell
Add five to the turn,
Lose the chance to bind.
Add eleven, Ok.
the life of lambda
lysogenic or lytic
CII(two) levels choose
DNA looping.
Twist, Writhe, add more Supercoil.
Think about the scale!
AraC and lac -
operons with different
gene regulation.
You use footprinting-
Definitely not CHIP-seq,
That's way too new age.
Negative when loops,
Positive when unloops, we
see for AraC
Brownian motion?
Perhaps the chelate effect...
Did you read the book?
Logical thinking
The only way to survive
As an undergrad