Lecture 6 (AT) Flashcards
(22 cards)
How is DNA damage prevented? Option 1.
Profreading by DNAP –> DNAP can go backwards and take out incorrect bases using its exonuclease activity.
How is DNA damage prevented? Option 2.
Polymerase can be switched to Pol V, which has a low fidelity and is less sensitive to errors.
How is DNA damage prevented? Option 3.
Hydrolysis of damaged nucleotides –> 8-oxoGTP can form 8-oxoGMP which cannot be used in replication and so will prevent DNA damage.
Direct reversal of DNA damage?
- Alkyltransferases remove methyl groups via enzymatic hydrolysis.
- Photoreactivation - reversal of pyrimidine dimers. Photolyases can remove crosslinked bonds of photodimers.
Name the three broad mechanisms of single strand break repair.
Base excision repair (BER), Nucleotide excision repair (NER) and Mismatch repair (MMR).
Explain the process of base excision repair.
DNA glycosylase takes the base out of the DNA. Creates an AP site. The AP is then cleaved by AP endonuclease, and the deoxyribosephosphate removed by DNA ligase. DNAP fills the space.
Explain the process of nucleotide excision repair.
This is slightly larger scale damage. An oligonucleotide is removed by a repair complex. The gap is filling using DNA ligase and DNAP.
Explain the process of mismatch repair.
The cell distinguishes between original and new bases through hemimethylation. Mismatch is recognised by MutS. Signals to the cell –> MutL and MutH bind to the site of mismatch, forming a complex with MutS. Conformational change causes the new strand to be cut.
Name the two types of double strand break (DSB) repair.
Non-homologous endjoining (NHEJ) and homologous recombination.
Non-homologous endjoining (NHEJ)
The DNA ends of the DSB are bound by protein complexes with exonuclease activity. Trims back overhangs. Ends are joined and so there is a LOSS OF BASES.
Homologous recombination
Relies on the presence of an intact sister chromatid to the DNA strand with a DSB. Can occur without any loss/alteration of nucleotides as the sister chromatid is used as a template. Results in the formation of Holliday junctions.
DNA damage in the leading strand in front of the replication fork can cause fork arrest. Mechanisms in which the cell deals with this?
Fork Regression, break induced replication and/or a combination of the two previous mechanisms.
Fork regression.
Replication fork can stop and regress if it comes across a lesion in the leading strand. The lesion is bypassed by extending the leading strand, using the lagging strand as a template, followed by reversal of fork regression, This results in the formation of a Holliday junction.
What is a Holliday junction?
Occur during recombination and result in the strand of one DNA template invading another. They are special intermediates that form: they contain 4 DNA strands between 2 DNA helices.
Issue with fork regression and the production of Holliday junctions?
Cleavage of the HJ by HJ resolvase produces an intact template and a broken ended molecule. Produces an abrupt end to the long chromosome.
How is the production of the broken ended molecule overcome?
By Break Induced Replication (BIR).
BIR
The leading strand of the loose end is trimmed and RecA binds to the ssDNA to begin the homology search in the template strand. Strand invasion then occurs and the D loop forms. Eventually the replication fork is reestablished.
The loose ends are not entirely gone. Branch migration occurs to allow gaps to be filled in. Another HJ forms. HJ resolution.
Search for homology can go wrong. Why is this an issue?
E.g. section of repetitive DNA, the resolution of HJ would result in large scale deletions, translocations etc.
The safer alternative is a combination of the two mechanisms. Explain this.
Fork regression, extension of the leading strand, invasion of the strand in front. Gives a much larger region for homology.
What does the CRISPR-Cas9 system use?
DSB repair to create mutations on guide RNA (gRNA).
gRNA is introduced into a cell along with? Function of this?
Cas9 protein. The gRNA binds to a complementary sequence of DNA; Cas9 will cleave the DNA and cause a DSB.
How does CRISPR-Cas9 exploit homologous recombination?
Introduce a template with a large region of homology and a sequence of interest. The homologous arms will bind to the DSB ends, inserting the sequence of interest.
