Links for Maureen Hoatlin's CSF 2011 Class
- Enjoy some excellent animations. The virology animation includes viral replication. Viral styles of replication are complex and fascinating, also providing a target for therapeutic intervention.
Questions and Answers
- Can you explain the difference between XP-V and a mutation in XPA-XPG?
- XP patients have a defect in one of the many genes involved in NER (nucleotide excision repair) e.g., XPA, XPC, XPC, XPG etc. Thus, XP patients cannot repair DNA damage (e.g., photoproducts like cyclobutane pyrimidine dimers) caused by exposure to sunlight. The XP-V patients have a different defect but the same phenotype. They have a defective bypass polymerase, called Pol eta. Normally, Pol eta can synthesize (error-free) past a thymine dimer by inserting two A residues. In XP-V patients, other bypass Pols are substituted that are not error-free for this lesion. Also, in XP-V patients, the upstream XPA-XPG proteins are functional and NER repair is intact. Only the bypass polymerase is defective.
- I was reading over your notes and can't quite seem to understand why telomeres have high C-T content. Would it not necessarily be true that they would have high C-A content as well?
- Telomeres usually contain some version of tandem copies of sequences like 5'-CCCCAA-3' on one strand and 5'-TTGGGG on the complementary strand. The GT-rich strand comprises the 3'-end and **sticks out** longer than the CA strand (and forms a loop, sealing the end). Specifically for human telomeres, there are 300-8,000 sets of repeats of the sequence CCCTAA /TTAGGG, then a 100-200 nucleotide extension of single-stranded TTAGGG repeats, hence the comment that telomeres have high GT content. But why? The best way to understand this beautiful system is to watch the very simple but very revealing short animation in the link above called telomere animation of how telomerase works at the telomere (see especially step 5 and later). Note that it is all about the requirements of polymerase for a free 3' -OH (and a ss DNA template strand)! It might also help to scan quickly through the very good article from Nature Network listed above as well.
- I am a bit confused about the DNA polymerases that you mention on slide 56 of your lecture. Are the leading and lagging strand synthesied by different polymerases? Secondly, you mention polymerase eta and its role in synthesis of the leading strand. I thought eta was only involved in the bypass mechanism? Thanks for your clarification and help.
- I think I see the confusion. I believe it is partly confusion over greek letters used to name the pols. Pol eta (looks like an italics small n) is a translesion pol---so you are correct about that. However, slide 56 mentions pol epsilon (Greek letter looks like like an italics e) which is the replicative pol that has recently been shown to primarily replicate the leading strand in eukaryotes. The lagging strand is synthesized primarily by pol delta. This division of labor at the replication fork in eukaryotes is a recent discovery, so many texts will have it backwards. Note that in eukaryotes there are two replicative pols: polymerase delta and epsilon, whereas in prokayotes there is only one main replicative polymerase, Pol III. Primase is required in both systems. Hope this helps.
- I am still having trouble understanding PARPi's mechanism. Can you explain to me the normal interactions and mutated interactions of PARP and BRCA along with PARPi's function?
- The essence of the idea in the PARP/BRCA example, very briefly, is that the BRCA-deficient tumor cell has a defect in repair of DNA double strand breaks by a repair process called homologous recombination (HR). PARP1 is in charge of repairing single strand DNA breaks, which become DNA double strand breaks during replication. Thus, if PARP1 is inhibited, the DNA ss breaks convert to DNA ds breaks and must be repaired by HR. If HR is defective, the cell accumulates so much damage that it cannot survive. HR is defective in the tumor cell...but not in the wild-type cell!
That's how PARP1 inhibitors are relatively selective in killing the HR deficient tumor cell but not the wild-type cell (which is competent for HR)
- 1) When you talk about the number of base pairs in a genome, the number 3 x 10^9 was mentioned and I wanted to double check, is this for the haploid genome? And just out of curiosity, why is it reported for the haploid genome?
- 2) When you were talking about DNA replication a "licensing factor" was mentioned, and I was wondering, is this a protein? And what is it's function?
- 1. Yes, for haploid. Diploid cells have two homologous copies of each chromosome (in humans, one from the mother and one from the father) so you would be "counting twice." You could do that, but it is important to indicate haploid/diploid. I think the haploid number is reported b/c it is the total number of unique sequences.
- 2. Yes licensing factor is the old name for a set of proteins that bind to the origins. Licensing factor is now thought to include the proteins Cdc6 and Cdt1. These proteins bind to the origin recognition complex proteins, and are synthesized only in a specific phase of the cell cycle (G1). Once replication origins "fire" (or start) these proteins are degraded or exported and the origin can't be "licensed" for firing again until the proteins are synthesized and enter the nucleus in the next cell cycle. That's how the cell controls replication so that the DNA is replicated once and only once.
- there is a typo in lecture notes. the haploid human genome is 3x10^9 (billion)