DNA Metabolism

What is the biochemistry of DNA replication and what proteins are required?

• DNA replication is semiconservative, with each daughter molecule composed of one newly synthesized strand and one template strand. DNA replication begins at an origin and proceeds in both directions using two replication forks.

• Each replication fork produces one leading strand and one lagging strand. The leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously, producing Okazaki fragments. The presence of multiple β clamps allows for rapid lagging strand synthesis.

• In prokaryotes, DNA polymerase III is the main polymerizing enzyme; in eukaryotes, DNA polymerase δ and DNA polymerase ε fulfill that role. Other polymerases are used specifically during removal of primers, DNA repair, or polymerization of non-nuclear DNA, such as mitochondrial DNA.

• An active prokaryotic replication fork requires topoisomerase, DNA helicase, primase, single-stranded DNA binding protein, the core polymerase, and a spare β clamp for synthesis of the next Okazaki fragment. Binding of helicase is a critical part of the initiation process. Association of most of the elements at the replication fork allows for leading and lagging strands to be synthesized at a similar rate.

• High fidelity is needed to eliminate errors of replication. DNA polymerases discriminate by proper base pairing to ensure that only A-T and G-C base pairs fit the geometry of the active site. The 3′ to 5′ proofreading function of DNA polymerases removes incorrect nucleotides from the nascent DNA strand.

• The HIV reverse transcriptase (HIVRT) enzyme copies its RNA genome into DNA before inserting the viral genome into the host DNA. Viral reverse transcriptase enzymes are structurally similar to the E. coli Pol I enzyme. One key difference, however, is that reverse transcriptase enzymes use RNA as a template rather than DNA.

How is DNA replication initiated and terminated in prokaryotes and eukaryotes?

• Prokaryotic replication has one origin and one site of termination opposite the origin. Eukaryotic replication involves multiple origins on each chromosome, and origins are activated at different times within the S phase of the cell cycle.

• DNA replication in E. coli is initiated at the oriC region by the binding of ~20 subunits of DnaA protein to sequences very near the 9-bp repeats. The DnaA monomers form a right-handed helical filament that is crucial for proper unwinding at oriC to initiate DNA synthesis.

• The Tus–Ter complex in E. coli ensures that termination of bidirectional DNA replication occurs in the termination region of the circular bacterial genome. When the Tus–Ter complex is in the nonpermissive orientation, as would happen if the replication fork moves too quickly and passes through the termination region, then the helicase stops.

• Initiation of eukaryotic DNA replication requires the origin recognition complex (ORC) and the MCM2-7 protein. The ORC consists of six protein subunits and is similar to DnaA in that ORC loads the MCM2-7 helicase onto the eukaryotic chromosome.

• Replication fork activation in eukaryotes requires cyclin-dependent protein kinases (CDKs), which are activated by cyclin proteins. A variety of CDKs exist, each activated by a different cyclin, and specific cyclins are produced at certain points in the cell cycle.

• Eukaryotic chromosomes have telomeres at the chromosome ends to facilitate DNA synthesis of these linear DNA molecules. Telomeres will shorten with each replication process unless the enzyme telomerase is present.

• Telomeric DNA sequences are added to the ends of chromosomal DNA by the enzyme telomerase, which is a ribonucleoprotein that synthesizes DNA by using an RNA template. The combination of mutations that cause uncontrolled cell growth and high levels of telomerase contributes to cancer tumorigenesis.

What are the common causes of DNA damage and how does it lead to mutations?

• DNA damage can lead to mutations, depending on the severity of the damage. Common sources of damage include reactive oxygen species, UV light, alkylating agents, and errors of replication.

• Mutations to a single nucleotide can be silent (the codon changes but the amino acid does not change) or can cause changes to gene products through a single amino acid substitution called a missense mutation (conservative or nonconservative). Nonsense mutations result in the insertion of a termination codon resulting in a truncated protein.

• The Ames test determines whether a substance is mutagenic by using a bacterial strain that lacks the ability to produce histidine (his⁻). If, after exposure to a mutagenic compound, the his⁻ bacteria are able to form colonies in a medium that lacks sufficient supplemented histidine for the bacteria to grow normally, then the bacteria have acquired a compensatory DNA mutation.

• In the Ames test, exposure of the test compound to enzymes in rat liver extracts before adding the compound to the bacterial plate is used to determine whether chemical modification of the compound inactivates its mutagenic potential or, in some cases, activates its mutagenic potential.

• There are three primary causes of DNA damage: (1) biological damage caused by errors that occur during the processes of DNA replication, repair, and recombination; (2) chemical damage caused by environmental exposure to reactive compounds; and (3) physical damage to DNA caused by high-level exposure to ultraviolet, X-ray, and gamma radiation.

What are the mechanisms of DNA repair and how does faulty repair cause cancer?

• There are four mechanisms of DNA repair: (1) mismatch repair, (2) base excision repair, (3) nucleotide excision repair, and (4) direct repair.

• Mismatch repair in E. coli is initiated when a homodimer of the mismatch repair protein MutS recognizes and binds adjacent to a base pair mismatch. A second mismatch repair protein, MutL, forms a complex with MutS through an ATP-dependent association, followed by association with the third protein, MutH, to form the MutS-MutL-MutH mismatch repair protein complex.

• Base excision repair involves removal of the damaged base by DNA glycosylase enzymes that cleave the N-glycosyl bond, thereby creating an abasic site, followed by removal of the deoxyribose-5′-phosphate, replacement of the gap with the correct nucleotide, and ligation of the phosphodiester bond.

• Nucleotide excision repair in E. coli is similar to base excision repair and involves four steps: (1) recognition of the DNA damage, (2) excision of a segment of DNA surrounding the damage, (3) filling in the resulting gap, and (4) ligation of the single-strand nick. Detection of DNA lesions requires the E. coli UvrAB excinuclease complex that scans for damage.

• Direct DNA repair processes are mediated by the enzymes DNA photolyase and O⁶-methylguanine-DNA methyltransferase (MGMT). The DNA photolyase system uses visible light to reverse the formation of cyclobutane dimers by UV light, whereas direct repair by the MGMT enzyme involves dealkylation of O⁶-methylguanine (O⁶MeG).

• Single- and double-strand breaks in DNA are repaired by enzymes that recognize the damage and rejoin the DNA strands by using DNA ligase. Double-strand DNA breaks are repaired by homologous recombination.

• Mutations in human DNA repair genes, such as BRCA1/BRCA2 (which repair double-strand breaks) or the mismatch repair genes hMLH1/hMSH2, are associated with high cancer rates, such as occur in Lynch syndrome.

What is the biochemical mechanism of homologous recombination in meiosis?

• Recombination translocates genetic material from one place in the genome to another. Recombination is essential for natural selection because it gives rise to genetic diversity.

• Homologous recombination in eukaryotes occurs during meiotic cell division in which sister chromatids generated during prophase of the first meiotic division exchange DNA through chromosomal crossover. Homologous recombination during meiosis permits the exchange of genetic information between paternal and maternal chromosomes.

• DNA crossover events occur during meiosis I, when four sister chromatids are present. After the events in meiosis II, the gametes have either intact genetic material without recombination or a recombinant chromosome with a mixture of maternal and paternal DNA.

• Homologous recombination requires a double-strand break to allow the formation of a Holliday junction at the location of the DNA crossover. Resolving the Holliday junction requires the enzyme resolvase and produces either a crossover product or reversion to the original chromosomes.

How do viruses and transposons integrate DNA into host genomes?

• Bacteriophage λ integrates its genome into a specific site on the E. coli chromosome, allowing for passive replication of the viral genome each time the E. coli chromosome is replicated.

• HIV uses reverse transcriptase to convert its single-stranded RNA genome into double-stranded DNA prior to insertion into the host genome. The viral integrase enzyme is required for integration, along with several host proteins associated with double-strand break repair.

• A transposon is a segment of DNA that moves throughout the genome; transposition requires a transposase protein for both removal of the transposon from its current site and insertion into a new site in the genome. DNA sequences at the ends of the transposon segment are arranged as inverted repeats (IRs) and function in the recombination process.

Everyday example: CO-23: Defects in DNA Mismatch Repair