BIOMG 3320 Group 10 (Lectures 24-25-26)

  1. Common types of DNA damage:
    • Depurination and deamination: hydrolysis of DNA. Deamination converts a cytosine to a uracil. This is the reason why uracil is not in our DNA. If uracil was a normal base in our genome, we would not be able to detect it as damage!
    • Oxidation: Reactive oxygen species from metabolic processes
    • Alkylation: through chemicals. N-nitroso compounds in processed meats damage DNA.
    • Pyrimidine dimers; UV light induces thymine dimers.
    • Mismatches: Replication errors.
    • Double strand breaks: Ionizing radiation, replication fork encountering lesions.
    • Cells have specific ways to deal with different damages.
  2. Explain how DNA is constantly damaged
    It has been estimated that every cell may experience up to 10^5 spontaneous DNA lesions per day. The most common endogenous source is depurination. The most common exogenous source of DNA is sun damage.
  3. How to measure the mutagenic capacity of a compound? How do we know that a carcinogen is mutagenic?
    • The potential of a chemical to induce mutation can be measured using a specialized bacterial strain (Ames test)
    • Salmonella bacteria culture that requires histidine to grow: it has missense or frameshift mutation in one or more genes requiring histidine biosynthesis. Thus, an increase in mutational load may restore histidine synthesis.
    • Cells are added to agar with nutrients but no histidine. Some (but few) reverse revertant colonies arise spontaneously due to mistakes in DNA replication.
    • The chemical is added to another agar plate. The more bacteria colonies of drug-induced revertants that arise, the more mutagenic the chemical is.
  4. Explain mismatch repair
    • Type of lesion: Replication errors that escape proofreading.
    • Mismatch repair increases fidelity by 10^2.
    • One every 10^7 nucleotides added.
    • If mismatch errors are not fixed quickly and cells undergo a second round of replication, there will no longer be a mismatch error. (Cells can detect A-G mismatch, but not AT)
  5. Explain steps of mismatch correction
    • Scanning: Dimer of MutS protein frequently scans the genome to look for distortions that are produced by mismatch.
    • Nicking: Once MutS finds a distortion, it locks in. It hydrolyzes ATP and recruits MutL, which recruits MutH to the site. MutS/MutL coordinates the action of MutH, a nuclease, which creates a nick far away (up to 1k) from the mismatch at the closest GATC (to find out which strand is the parent strand).
    • Repairing: An exonuclease comes and removes all the DNA between the site of nick and the mismatch. The single-strand gap is filled in by DNA polymerase.
  6. Why is nicking so far away from the place of mismatch?
    • This is related to how the machinery tells which strand contains the mistake due to misincorporation.
    • In E.coli, the Dam methylase methylates A residues on both strands of the sequence 5’-GATC-3. You will not find more than one GATC in base 1000 pairs.
    • Once the region involving the methylated GATC is replicated, the newly synthesized strand will not be methylated (hemimethylated). Through this, MutH knows which strand is the parent strand.
    • MutH searches for the closest GATC to get information on which strand is the parent strand.
    • MutH binds at such hemimethylated sites, and is activated when it is bound by MutL/MutS located at a nearby mismatch.
    • MutH only nicks the unmethylated strand.
  7. Explain mismatch repair and cancer
    • All eukaryotic cells have proteins analogous to MutS and MutL (Example: MSH2 in humans, which is named MutS homolog 2)
    • Defects in human MSH2 and MLH1 predispose patients to familial colon, endometrial and ovarian cancers. Defects in such genes could account for up to 15% of all human colorectal cancers.
  8. Base excision repair
    • Type of lesion: modified bases (example: deaminated cytosine turning into uracil)
    • DNA glycolase cleaves N-glycosyl bond, removing the base and creating an AP site (no base).
    • AP endonuclease and exonuclease cuts the phosphodiester bond in the DNA strand containing the AP site and removes the abasic site.
    • DNA polymerase 1 and DNA ligase repair gap.
    • In the case of guanine depurination or any other damage that would remove the base, the steps may involve the last two steps, or only the last step.
  9. Nucleotide excision repair:
    • Type of lesion: “bulky lesion” that cause large distortions in the helical structure
    • One example is pyrimidine dimers caused by UV light.
    • In E.coli, UvrA and UvrB scan the genome for distortions.
    • UvrB melts DNA to form a single-stranded bubble around the lesion.
    • UvrB recruits the UvrC nuclease, which makes 2 nicks around lesion (creates 12 nt ssDNA segment)
    • UvrD helicase removes ssDNA and DNAPol1 and ligase fill hap.
    • Similar in eukaryotes, but much more complex.
  10. Explain translesion synthesis:
    • Type of lesion: Replication fork encountering a damaged DNA that cannot base-pair (so it cannot serve as a template for replicative DNA polymerase)
    • In translation synthesis, the replication machinery utilizes a special class of DNA polymerases that can bypass DNA lesions.
    • TLS is catalyzed by a special class of DNA polymerases - the Y family (In E.coli, Pol4 and Pol5)
    • TLS polymerases incorporate nucleotides in a manner that is independent of base-pairing but still needs a template.
    • TLS is error-prone, while TLS is likely to introduce mutations, it spares the cell the worse fate of an incompletely replicated chromosome.
  11. DNA recombination:
    • A process in which DNA molecules are broken and the fragments are rejoined in new combinations
    • Occurs during DNA repair and meiosis.
    • Regulation of gene expression
    • Assembly of immunoglobins
    • Cloning
    • CRISPR
  12. Types of recombination
    • Site-specific recombination: requires a defined sequence motif
    • Homologous recombination: between identical or almost identical sequences
    • Transposons: DNA sequences that jump at specific sequences or randomly.
  13. What are homologous sequences?
    DNA sequences (at least 50 bp) similar or identical in nucleotide sequence.
  14. What are the roles of RecA
    • It is a strand exchange enzyme that
    • mediates the search for matching sequences and
    • catalyzes the pairing of homologous DNA molecules. RecA circles around ssDNA and creates nucleoprotein filaments,
  15. Explain homologous recombination
    • Genetic exchange between a pair of homologous DNA sequences
    • Homologous sequences: DNA sequences (at least 50 bp) similar or identical in nucleotide sequence.
    • DNA breaks are required to initiate recombination.
    • Resection: RecBCD helicase/nucleases process 5; ends to generate 3’ ssDNA. In eukaryotes this is done by the MRX complex.
    • Strand invasion: A key step in HR. Strand invasion establishes the stable pairing between DNA molecules and initiates the exchange of DNA strands between two parental DNAs.
    • RecA protein promotes strand invasion. It is a strand exchange enzyme that mediates the search for matching sequences and catalyzes the pairing of homologous DNA molecules. RecA circles around ssDNA and creates nucleoprotein filaments, increasing the distance between adjacent bases (length of DNA is extended by 1.5 fold). RecA homologs are present in all mutations (Rad52 in eukaryotes)
    • A heteroduplex is formed between one strand from the invading sister chromatid and the sister chromatid that is providing information. Heteroduplex formation checks sequence homology. The crossing of the strand forms a Holliday junction.
    • The invading strands end with 3’ termini, so they can serve as primers for DNA synthesis. Both 3’ ends form heteroduplexes and synthesize DNA that capture the information in the break.
    • Newly synthesized DNA pairs with top strand.
    • DNA ligase ligates. DSB is accurately repaired.
    • Holliday junction resolution occurs through the action of structure-specific endonucleases called resolvases. (RuvC)
    • Resolvases have a few different ways to clip the DNA, which could result in crossovers (whole chromosomal segments are exchanged) or no crossovers (little piece exchanged).
  16. What are sister chromatids?
    Pair of sister chromatids are created as the replication fork passes through the DNA.
  17. When there is a DSB, why doesn’t the cell directly ligate the ends?
    • If you use a template, you can do it in an error-free manner.
    • Eukaryotes have ligases for joining blunt DNA ends (Non-homologous end joining)
    • NHEJ is very error-prone.
    • In many cases, this leads to diffusion between one end of a chromosome that fuses to another end of another chromosome.
  18. Protecting the telomere
    • The cell interprets the presence of DNA end as a sign of DNA break.
    • Attempts to repair telomeres would lead to chromosome fusion events
    • Proteins that bind to the telomere can protect it from nuclease action and repair mechanisms.
    • Telomeres form a t-loop to mask the end of the telomere.
  19. Meiosis
    • Meiosis leads to gene conversion (az) and crossovers (cok).
    • For genetic diversity, recombination must NOT occur between sister chromatids.
    • Exchange of information needs to concur between non-sister homologs.
    • The resolution step will determine if an HR event will lead to gene conversion or crossover.
    • Two homologous chromosomes have the same linear array of genes. Some sequences may differ slightly (one difference in a million)
  20. How do cells start HR in meiosis? How do they make the breaks?
    • Spo11: enzyme that introduces double strand breaks during meiosis.
    • Resembles topoisomerase 2, it is a dimer that introduces 2 breaks.
    • A tyrosine residue in Spo11 attacks the phosphodiester backbone to cut the DNA.
  21. Explain yeast transformation by HR
    • The aim: delete one gene out of 6k genes in the yeast genome.
    • Design DNA fragment that can be synthesized in-vitro. The DNA fragment should be homologous to the regions close to the gene. We will exchange information in the region between the homologous regions.
    • This region should have a marker - such as resistance to a specific chemical.
    • Use In vivo HR apparatus: We add the DNA fragment into cells. If there is a break occurring in the region of homology, there is a chance that the yeast will use the DNA fragment as a template in homologous recombination during cell division.
    • Select cells with selectable marker.
    • Efficiency in mice and humans was very low, which is how CRISPR came into play.
  22. How to introduce a site-specific DSB?
    • You need a sequence-specific DSB.
    • Old nucleases: ZFN, TALEN. DNA binding sequences that can be engineered to target specific genes. However, they’re very expensive, as sequence specificity is very difficult to engineer.
    • CRISPR: A mechanism for bacterial adaptive immunity through RNA-guided targeting of viral elements.
    • sgRNA (20 bp guide RNA), PAM (NGG), Cas9. Only requirement is PAM
    • Cas9 is a programmable nuclease.
Author
pelinpoyraz
ID
358513
Card Set
BIOMG 3320 Group 10 (Lectures 24-25-26)
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Updated