Homework, Schleif's Section Graduate Biophysical Chemistry
Feb. 15
1. About how many protein molecules are in a typical bacterial cell, at pH 7.0, about how many H+ ions? Since proteins are exquisitely sensitive to the pH of the buffer they are in, what is going on here?
Around 60 H+, but a million or more protein molecules. Each one of them, however "knows" the pH. This means that the ionized form of H is moving around extradordinarily rapidly so that each protein is visited by an H+ many times in an interval much shorter than the time it would denature due to being in the wrong pH environment. Note that a string of water molecules is effectively a copper wire with respect to H+. An H+ can enter at one end and the hydrogens along the line each shift position and an H+ emerges at the other end.
2. When their numbers are small, some birds multiply by a factor of r from one year to the next. When their density becomes high however, competition interferes with reproduction. Let R(n) represent the level of bird population of the nth year, where R=1 is the absolute maximum. Then we will approximate bird growth from one generation to the next as R(n+1)=rR(n)x(1-R(n)). Starting from an R(0)=0.001 examine R(n) initially, and at around n=100 for r between 3 and 4. In particular, what happens when r changes from 3.58 to 3.59. One good way to do this problem is to use a spreadsheet.
For low values of r you should have found that the population rather rapidly approched some final and stable value. For r approaching 3, the population oscillates and slowly settles down. As r becomes larger, the population jumps first between two values, then with increasing r it jumps between four values, then eight, and finally becomes chaotic. This amazing behavior of a simple mathematical equation was only discovered relatively recently, perhaps around 1970.
Feb. 20
1. Suppose you have a sequence of five bases, say AAAGG, that are read from the major groove of the DNA. You wish to invert this sequence so that the protein reading this sequence is pointed in the opposite direction along the axis of the DNA cylinder. Conceptually, suppose you cut the DNA on opposite sides of this sequence at sites 10 base pairs apart, reverse the DNA by 180 degrees and reinsert. Alternatively, suppose you cut the DNA on opposite sides of the sequence at sites 5 base pairs apart, invert, make any necessary adjustments, and reinsert. How do the two resulting products differ from one another with respect to sites or sequences more than 10 base pairs away from the original AAAGG sequence. You may find the following drawing to be helpful.
If your cuts are exactly one helical turn apart, you can rotate the cut piece 180 degrees about an axis that is in the plane of the paper and perpendicular to the axis of the helix. Note that what was toward the viewer, that is, out of the plane of the paper toward you is now back after the rotation. This allows direct reconnection of the backbones and the 5' to 3' directions along the two phosphodiester backbones is retained. If your cuts are five base pairs apart. In order to reconnect the strands and retain the 5' to 3' polarity, after the rotation about the first axis as above, you must also rotate the DNA segment 180 degrees about an axis parallel to the DNA axis. Thus, the binding site ends up on the opposite face of the DNA.
2. Does negative supercoiling tend to assist or hinder the formation of short single-stranded bubbles in DNA?
It assists bubble formation, which is why DNA is negatively supercoiled in vivo. A way to understand this is to consider the example given in class of 1000 bp of DNA that is covalently closed, giving Lk=100, Tw=100, Wr (supercoiling) = 0. Instead if you close in the presence of ethidium bromide, at the moment of closure, Lk=90 (say), Tw=90, Wr=0. Then remove the ethidium bromide, Lk=90 (still, it can't change), Tw=100 (DNA really likes to be normally twised, and hence Wr=-10. In fact, DNA doesn't like to be wound up in supercoils, and it resists. Therefore, it wants to reduce supercoiling, and Wr goes to -9. As a result, to keep Lk=90, Tw must go to 99. That is, the existence of the negative supercoiling somewhat unwound the twist of the DNA.
Feb. 22
1.Suppose exposure of a eukaryotic cell to an external growth factor ultimately leads to dramatically increased expression of gene A and we find that the earliest detection of increased A synthesis occurs 60 minutes after administration of the growth factor. In an effort to track down what takes so long, we want to estimate the likely time required for the central regulatory factor responsible for A's synthesis to reach gene A once the factor has been activated (say by phosphorylation). Let us assume that there are 1,000 molecules of transcription factor randomly scattered throughout the cell, that they instantaneously become phosphorylated, and that once phosphorylated, they have completely free access to the nucleus. Under these assumptions, how long on average would it take for an activated transcripton factor to reach the enhancer for gene A? In light of your answer, what would your next step be in your work on understanding the delay?
This is analogous to the situation of diffusion of nucleotides to the active site of RNA polymerase. The most important difference from the nucleotide example given in the book is that the concentration of the transcription factor is 10-9M instead of 10-4. The estimated time to bind is around 10 seconds. Since this is much slower than the 60 minutes, something else must be eating up the time.
2.Consider two DNA replication forks that collide head on. What activities not normally associated with a replication fork would be needed to resolve the situation?
Reasonable guesses are proteins to disengage at least one of the PCN factors that encircle the DNA and hold the polymerases near the DNA and a special topoisomerase to separate the two DNA's as they likely will be entwined at the end of replication.
Feb. 27
1. In both prokaryotes and eukaryotes multiple polypeptides often are required to initiate transcription of a gene. After initiating transcription, do these polypeptides remain there on the DNA or must an initiation complex be entirely reconstructed for the next round of transcription? Design an experiment to determine the answer in the case of a prokaryotic system in which one protein in addition to RNA polymerase is required to activate transcription. Describe several of the controls that you would have to do in order to demonstrate that your main experiment worked the way you want.
In vitro to buffer lacking triphosphates, add the positive acting transcription factor and polymerase to template DNA containing the promoter. Allow time at 37 degrees for open complexes to form. Then add a large excess of DNA containing a binding site for the transcription factor, but no promoter. Then add triphosphates and quantitate the amount of RNA product made. Do the same experiment, but include rifamycin with the triphosphates. This limits the RNA synthesis to a single round. If more RNA is made in the transcription done without the rifamycin, then the transcription factor must have remained on the original template DNA and activated multiple rounds of transcription. You need to ensure that the competitor is capable of binding all the transcription factor added to the reaction. Do this by adding competitor DNA first. Also need to show the the rifamycin blocks all initiations. Do this by adding the rifamycin before the polymerase.
2. It is plausible that the rate of transcription across a gene depends on local sequence properties. Using measurements on growing cells of sequences from nucleic acid derived from growing cells, how could you identify speed-up and slow-down regions across a gene?
ChipSeq and examine the average polymerase density across the gene. Slow regions will have more polymerase.
March 1
1 & 2. It would be nice to be as smart as RNA polymerase and to be able to recognize promoters from sequence information only. The objective of this problem is to explore the feasibility of this objective. Most E. coli promoters deviate somewhat from the canonical -35 and -10 sequences. Look up the sequences of about ten promoters to gain an idea of promoter variability. Now, with a simple criterion approximating what you found, e.g. match 6 out of 10 bases, apply your criterion to some (at least 200 nucleotides) random sequence or sequence from E. coli. Although it would be great if you can program this on Python or do it on a spreadsheet, note that you can also do this quite quickly by hand by putting your canonical sequence on a piece of paper that you slide across your sequence and apply your simple criterion at each position of your sliding sequence. What do you conclude?
Six or seven bases matching the -35 and -10 regions is common. You should have found that you can match 6/10 bases around every one to two hundred bases. Real promoters are not found this frequently in DNA, therefore, something else must be going on.
March 29
1 & 2. In PyMol construct a ten alanine residue peptide. Use PyMol to set all the phi angles to -75 and all the psi angles to 150. Note that the phi and psi dihedral angles are defined by atoms along the polypeptide backbone and that the resulting dihedral does not depend on whether the atom order was along the backbone in the N to C or the C to N direction. Describe and comment on the resulting structure. For your own amusement, in a dipeptide set omega to cis and trans and look at the resulting structures.
It is, of course, a helix with very close to three residues per turn, but it is LEFT-HANDED. This is known as a polyproline II helix, and is the state of polypeptides as they emerge from the active site on ribosomes.
April 3
1 & 2. Historically how was the direction of protein elongation determined? What would be a good way to determine the direction now, using modern techniques and technology?
As most of you found, looking at hemoglobin synthesis. One modern way to do it would be to use any cells or tissues that are synthesizing lots of one particular protein, e.g. E. coli synthesizing beta-galactosidase, administer deuterium labelled amino acids for 30 seconds, then stop cells and purify completed beta-galactosidase. Digest with protease and using a mass spec, determine whether it is peptides from the N- or C-terminus that are deuterium labelled.
April 5
1. 2. Suppose the I2 site in ara is changed to I1. Why might one expect araBAD expression to become constitutive? Why might it not become constitutive?
Normally, AraC loops the DNA because its DNA binding domains are held in positions appropriate for looping and because O2 binds a DNA binding domain 10 x tighter than I2 does. Since I1 binds a DNA binding domain 10 x tighter than O2 does, changing I2 to I1 has a very good chance of forcing the protein to bind to the I1-I1, site even in the absence of arabinose, and because of its position, it would induce transcription. The reason that it might not comes from the fact that the I2 site partially overlaps the polymerase -35 site. Thus, when the I2 sequence is converted to an I1 sequence, the overlapped part of the -35 sequence is altered, and hence initiation by RNAP could be severely damaged. Note that the all purpose answer of "allostery" (thinking that the I2 sequence has special properties that put a DNA binding domain that has bound to it in an inducing conformation) doesn't fit the facts. It is worth noting (come by my office if you want to check it out on a 3D model) that two proteins can both simultaneously "read" at least three bases in common if they put alpha helices into the major groove of DNA from opposite sides of the helix.
April 10
1. DNA seems shockingly stable in that genes stay the same and functional for thousands of generations whereas almost all other material objects denature, oxidize, erode etc. Comment.
I assigned this problem for several partial reasons. One of which is that I was hoping that one of you might come up with a more satisfying answer than I have. Another is to illustrate that just because someone asks a scientific question doesn't mean that there is an answer. My own best, but not fully satifying answer, to the question is: a. Redundancy in the information permits error correction in the cases of damage that is detectable, b. The process of egg and sperm production as well as embryo development allows for some very stringent quality control measures that discard mistaken copies and retains only the good and functional copies of chromosomes.