Regarding the Articles: Should we concentrate on materials which were on the second exam, or read the articles over again?
--Study your notes from class. It shouldn’t be necessary to read over the articles again. Questions would be broader.
Current Two Articles?
--You shouldn’t worry about them too much.
Last day of class?
--Don’t worry about it.

Circle Question (4 on Exam 2). The replication “fork” appears at the point where the extra strand attaches to the circular strand. This is the “Leading Strand.” It is extended indefinitely. Then, the okazaki fragments appear as base pairs to the “tail,” which is the lagging strand template. The circle remains the leading strand template indefinitely.

G proteins (see Cell Biology). EFTU does the same thing, only in a different context. EFTU prevents peptide .... formation. The time before hydrolysis allows tRNA to dissociate.

Figure 28-29.
Gene: +1 on the right
TATA box is in the middle.

How does RNA Polymerase find it and bind stabily enough?
TBP binds the Tata box
TBP is part of TF2D
the UAS “Upstream Activatior Sites” bind proteins which interact with TF2D, stabilizing it.
Then, the RNA Polymerase finds TF2D. (TF II D)
All the interactions stabilize everything. There are also “Mediators” that are in-betweens, helping the activator proteins touch TF2D.
Then, the RNA Polymerase will get attached correctly.
HMG Proteins help bend DNA.
Long DNA is more flexible than short DNA.

LAC Repressor Protein: The Operator sequence? O1 is the main sequence in this case. The closer the inverted sequences are to each other, the tighter the LAC repressor protein bonds.

A List of Topics:
Part 1:
Monomer Structures
DNA and RNA Structure
denaturation/renaturation
supercoiling of DNA, topoisomerases
Part 2:
DNA replication -- The Basic Mechanism
(Bidirectional, semiconservative, etc.)
DNA Polymerases “Pol 1, and Pol 3” (Fidelity, Processivity)
Replication Initiation (in Bacteria)
Replication, Elongation, and Termination (a little)
DNA Ligase.
(Eukaryotes-->Telomerase.)
Part 3:
DNA Damage
DNA Repair: BER, NER, Mismatch “MMR.”
Recombination -- A couple of the enzymes.
SOS System, Lesion Bypass Polymerases.
Exam 2.
Part 4:
XP-V
Transcription - mechanism (bacteria).
Aspects of Regulation (LAC operon)
DNA-Protein Interaction
Eukaryotic -> Chromatization “Chromatin” and how it affects transcription.
(TF2D, TBP are Eukaryotic.)
The basic Eukaryotic method is similar to that of bacteria.
Translation--
The Genetic Code
Amino-acyl-tRNA synthetases (Proofreading and Fidelity)
The Ribosome.
Translation Initiation. (only in Bacteria), elongation [EF-G helps move], (termination).
EF-Tu, Kinetic Proofreading.
Part 6:
DNA Sequencing (making DNA libraries)
Genome Sequencing
--->End of Class<---

The SOS System:
recA is a gene, somewhere in the E. Coli chromasome.
Binding sites for LexA are nearby!
LexA binds to the site, and represses transcription of the recA gene. Under normal conditions, LexA is available, and it binds that gene. However, there is an equilibrium with Free LexA. There is a little RecA protein in the cell, but a low level under ordinary conditons. The same is true to UvrA, and UmuC,D (especially. There is hardly any UmuC,D in the cell unless the SOS Response is triggered). There remains a chance for any of the genes to be transcribed, however.

When there is no DNA damage, LexA is bound to most of the sites on the chromasome.

DNA damage: single stranded DNA accumulates in the cell. The single stranded DNA is the signal that activates the SOS Response.
Reason: Mix ssDNA with the RecA protein and ATP. The RecA proteins bind the ssDNA. If LexA is added at this point, the LexA protein becomes cleaved, and it is inactive. This is an in vitro experiment. So we can presume that the bound RecA proteins cut the LexA (or cause it to be cut).
Presume DNA replication with damage in the lagging strand.
The okazaki fragment will end early, ‘cause the replication fork will be stuck on the base error. This leaves a single-stranded gap, which RecA will eventually bind (there is a fair amount available at any one point). Once RecA is bound to the ssDNA in the chromasome, the free LexA will eventually end up cleaved by the attached RecA (an Active Protease). The concentration of free LexA will be greatly reduced. Then, there won’t be as much LexA left to bind the DNA. Once the LexA is lost, then the other genes (UvrA some, UmuC,D a little) will be produced in different quantities (predefined ratios).

If the damage is repaired, there will be no ssDNA left to activate the RecA. Then, the LexA gene (which has been expressed the whole time (repressed by LexA)). Then the gene will eventually be disabled again.

Note: the binding sites don’t have exactly the same sequences. This means that the genes can be transcribed at different levels with different concentrations of LexA available.

UmuC,D-: Sensitivity to UV light?
Umu is from genetics. Someone isolated a mutant E. Coli and named it UmuC-. Let’s say the protein is inactive in this cell.
The effect on the cell (phenotype) was that the mutant cells were a lot more sensitive to UV than the wild-type cells.
BUT: There are FEWER mutations in the surviving cells.
Wild Type Cells->UV-> the cells survive, but random mutations appear everywhere (in some of the cells).

Idea: Dna damage in template? Replication stuck? The only thing to do is have a DNA Pol that will replicate across the damage, putting a random nucleotide, ‘cause then the cell MIGHT get the right base (25% chance). UmuC,D are an error-prone DNA polymerase. Together, they are an enzyme that is a DNA Polymerase. It is used to replicate damaged DNA when there is nothing else the cell can do. (Philosophy: it’s better to live with a mutation than to just die, since a lot of mutations do not cause bad effects to the cell.)

Repression of umuC,D is very strong, because it would mutate the DNA a LOT if it made a lot of the enzyme often. It’s better to make UvrA and UvrB to repair the DNA before trying the UmuC,D enzyme to make a random copy. Note: RecA also cleaves UmuD protein, so it will take extra-extra long before UmuC,D has to repair the damage.
Transcription Mechanisms for Bacteria: (differences between Eukaryotic and ...) --> there’s a list on the website which shows the differences.
alpha2BetaBeta’Sigma binds the promoter and unwinds the DNA all by itself. If the promoter is strong, a high level of transcription will result.
If nucleotide triphosphates are present, it will start transcription, which is the elongation phase. The DNA stays unwound a little, RNA/DNA base pairs, the RNA is pushed from the enzyme making the copies. Note: The sigma subunit dissociates before replication starts, so alpha2 beta betaprime keeps its 18-basepair DNA unwound with several base pairs of RNA on the DNA. This will (slowly) transcribe the DNA until it gets to termination, which occurs at a bunch of AAAAAAAs in a row, producing a bunch of UUUUUs in the RNA. A hairpin needs to be in the RNA; the enzyme will stop for a little while, and the weak base pairs of UA will detach the RNA, and the enzyme will leave with it.

Note: A weak promoter needs some activator proteins to help out the RNA Polymerase(?) CAP activator protein was shown in class.

RNA Polymerase 2 consists of about 10-12 proteins. It is analagous to the CORE. By itself, it can synthesize RNA, but it can’t recognize the promoter. TF II’s are needed (A), B, D, F, and G, etc. bind the promoter and carry out the “Basal Transcription.” Basal transcription is always low. Activator proteins are always needed (they bind upstream sites and assist). This is the difference between Eukaryotes and Prokaryotes (bacteria).

Also note: Chromatin, depending on the structure, can cause general repression if it has compacted the DNA. Basically, every gene gets repressed with the Chromatin, then activator proteins turn on the relevant genes. In bacteria, there are a few proteins kept from synthesis by binding proteins, while the others are produced in different proportions depending on the strength of the bacteria.
Eukaryotes: Everything was repressed. Huge stretches are in super-compacted sections, no protein access.
Bacteria: Lac operand (one promoter) marker. There is no “Universal Repressor.”

Advice: Read the Class Notes.

Translation: Exit site (wasn’t discussed in class... the RNA as it’s leaving was bound there for a while...)

DNA Sequencing: Primers were added to ssDNA. The primers were radioactively labeled (with 32P) (or use flourescent TAG on the ddNTP to detect it at the other end) when the DNA Polymerase attached a deoxy-ribo-nuclease. Denature the strands and run it on a gel to sort the molecules by size.
A total of four lanes are required, using all four different ddNTPs (ddATP, ddCTP, ddGTP, and ddTTP).

10.4 base pairs per turn, no writhe=raw linking number. Use a slightly lower linking number to make the DNA easy to unwind during replication.

CAP activates transcription: How? The CAP protein binds at the CAP binding site “upstream” from the promoter. The RNA Polymerase binds both the promotor AND the CAP protein. If there were no CAP, then the RNA Polymerase wouldn’t bind as well (some of the “glue” is missing!) That’s how CAP helps the RNA Polymerase stay next to the promoter long enough to get started.
When are the exams going to be ready?
They will probably be graded by next Wednesday. You can come to the Office, or call to ask about them.