DNA replication, repair and methylation Flashcards
What is the difference in the origins of replication between eukaryotes and prokaryotes?
Prokaryotes only have one origin of replication (called OriC) because they have their whole genome in one long strand. Therefore, prokaryotic genome only consists of one replicon. Eukaryotes have many origins of replication, at least one on each chromosome. Origins of replications don’t always fire at the same time, so replication bubbles from two different origins can fuse together to make one bubble if they meet.
Describe the initiation of replication in E. coli cells, and the parts of DnaA, DnaB and DnaC.
The prokaryotic replicator sequence (OriC) consists of three 13mer AT rich sequences, followed give five 9mer sequences. The five 9mers act as binding sites for the DnaA. Five subunits of DnaA bind to each 9mer. Once a sixth DnaA subunit binds to ATP and joins the other DNaA subunits, the complete DnaA hexamer interacts with the 13mer sequences that results in separating the DNA into single strand.
DnaB is a helicase enzyme and DnaC is a helicase loading protein. DnaC loads DnaB onto the separated DNA strand (one on each strand). DnaB is then activated and start to unwind the strands to form a growing replication bubble. DnaB breaks the hydrogen bonds between the complementary bases which requires ATP hydrolysis. DnaB also recruits primase, which synthesises the RNA primer ready for the initiation of replication by DNA Pol.
Single strand binding proteins are also vital for keeping the two strands separate by binding to the ssDNA.
Describe the structure of the DNA Pol III holoenzyme.
The DNA Pol III of prokaryotes consists of the three DNA Pol III core, that is the main component necessary for synthesis of the growing DNA strand by adding nucleotides from the 5’ to 3’. The core consists of three subunits, alpha, epsilon and theta. The alpha subunit is the polymerase domain, epsilon is the 3’-5’ exonuclease activity domain and the theta has an unknown function.
These are attached to flexible linkers (tau subunits) that are bound to the sliding clamp (consists of beta subunits) and the sliding clamp loader, which are necessary for recognising the RNA primers (primer:template junction) and binding to them.
There are more subunits, such as gamma, delta, psi etc. and these are thought to play a part in enhance the processivity by assisting the replisome assembly.
Describe the process of DNA replication by the lagging strand by DNA Pol III.
The lagging strand must be synthesised in short Okazaki fragments, because the synthesis of a new strand must always be in the 5’ to 3’ direction. The sliding clamp binds to every primer:template junction. The two lagging strand DNA Pol III cores then take turns to bind the primers in the sliding clamp and begin the synthesis of the new strand. Synthesis continues until the Pol III core reaches the primer from the previous Okazaki fragment, whereby then it dissociates. This process then repeats. This is called the trombone model because it is almost a back-and-forth. movement, as if the DNA Pol is pulling the strands through.
What is the problem that is found when replicating a circular piece of DNA? And how does topoisomerase II solve this issue?
Two circular DNA will form a catenane, where the two are interlinked. Topoisomerase II nicks the double strand and unlinks the two copies. This is called decatenation.
What is the replication termination of linear DNA problem? What are the two possible solutions for the problem?
On the lagging strand, there needs to be nick translation because each Okazaki fragment has to have its RNA primer removes and replaced with the DNA equivalent. This is where the gap between the end of one Okazaki fragment is joined to the beginning of the next (where the primer was before). But this causes a problem at the 5’ end of the strands because the most 5’ Okazaki fragment has no fragment to join to. There is a short stretch of DNA that does not get covered in an Okazaki fragment, and it is too short to have a primer. So the RNA primer is removed and this leads a small gap where the lagging strand is a few nucleotides shorter than the template, leaving a 3’ overhang. This incomplete replication would create a problem as a few nucleotides would be lost every time the DNA is replicated.
Some prokaryotes use protein priming, where a protein (usually a tyrosine) is added to the 5’ ends of the new strands. This protein then provides the OH group needed for DNA Pol III to synthesise the last bit of the strands.
The other solution is telomeres - these are TG rich sequences at the 3’ end of DNA in eukaryotes that can be extended by telomerase in order to ensure that the lost nucleotides are within the telomere sequence not the important gene coding areas.
Describe how telomeres ensure that the ends of DNA are not lost.
Telomeres are TG rich repeat sequences at the ends of linear DNA that protect it from degradation. There is still an overhang but it is only the loss of a telomere sequence, and not protein-coding DNA. Telomeres are regulated by telomerase, a RNA-dependent DNA polymerase enzyme, that can synthesise a DNA strand using a RNA template. Telomerase is a riboprotein, which means that it includes RNA within its native structure - RNA that consists of the telomere repeat sequence. The telomerase reverse transcriptase domain (TERT) is used to extend the 3’ overhang sequence using the telomerase RNA as the template. It does this repeatedly by translocating, rebinding and polymerising.
This extended 3’ end is then used as the template for a new Okazaki fragment, whereby a primer is synthesised and the complementary strand is made. This still leaves a slight 3’ overhang but now it is only a loss of telomere sequence, instead of the actual genome.
How is telomerase regulated?
There are DNA binding proteins that help to regulate telomerase. They bind along the dsDNA telomeres and identify the end of the telomeres. Increasing DNA binding proteins is thought to inhibit telomerase.
Telomeres also form a T-loop structure at the end, where the end telomere sequence complementary binds to a sequence further down to form a loop structure. This protects the end of the telomeres.
What is the general process of DNA polymerisation?
DNA polymerisation requires a DNA template strand and the new strand is synthesised from the 5’ to 3’ end. Pol uses the 3’ OH group in order to polymerise the next nucleotides. If the bases don’t pair, the three-dimensional alignment does not fit to allow for the 3’ OH to initiate the nucleophilic attack on the triphosphate of the incoming nucleotide. A double phosphate, called a pyrophosphate, is produced as a result. Pyrophosphatase breaks this down into two phosphates, which generates a lot of energy.
Polymerisation on the leading strand is continuous, whereas on the lagging strand, it is done discontinuously in Okazaki fragment. These fragments are later joined together via nick translation.
How does nick translation occur and what proteins are involved?
Nick translation is catalysed by DNA Pol I. Pol I extends the 5’ end of the Okazaki fragment while its exonuclease activity degrades the RNA primer. This joins the two fragments together and gets rid of the nick (gap). DNA ligase is needed to catalyse the synthesise the phosphodiester bonds between the nucleotides.
Use the ‘right hand model’ to explain the structure of the alpha subunit of DNA Pol III.
The 3D structure of the alpha subunit resembles a right hand that is almost forming a circle around the DNA that passes through it. The palm of the hand is where the active site is - the active site is very small and is small enough that it rejects RNA nucleotides from entering so that only DNA nucleotides are used for polymerisation. This is due to the small different that there is a OH group on the 2’ of RNA, where there is only a H group on DNA.
The active site includes two magnesium ions. One facilitates the nucleophilic attach by pulling the H away and creating a polarity in the 3’ OH group. The other ion stabilises phosphate.
The specificity of what nucleotide binds next is determined by the ‘finger’ regions of the alpha subunit. This region creates a kink in the template strand so that only the next nucleotide in the template is exposed, to avoid confusion. When the correct nucleotide interacts with the amino acid residues in the ‘fingers’, they close in towards the active site to assist the catalysis.
The ‘thumb’ region is not involved in the catalysis but is known to stabilise the DNA strand.
How does the DNA Pol III correct its mistakes?
The epsilon subunit of the core has exonuclease activity. When it senses that the incorrect nucleotide has been added, it will move the strand to the exonuclease region, the last nucleotide will be removed, the strand is moved back to the active site and polymerisation can continue. The exonuclease activity is in the 3’-5’ direction.
What is the meaning of DNA Pol processivity? How can it be increased?
Processivity is determined by the number of nucleotides added per subunit every time it binds to a primer:template junction. Low processivity DNA Pol will only add a few nucleotides after the primer:template junction before releasing, whereas if it is high processivity, it will add the whole Okazaki fragment.
Processivity is facilitated by the sliding of the DNA polymerase along the DNA template. The polymerase interacts tightly with the DNA strand in non-sequence specific ways e.g. electrostatic attraction. Every time a nucleotide is added, it must partially release the DNA strand and rebind to it on the next nucleotide. If this is smooth and stable, there will be higher processivity.
Subunits such as the sliding clamp and accessory subunits can enhance processivity. E.g. protein-protein interactions with the clamp subunit prevents the DNA polymerase from dissociating from the template.
What are the three ways in which DNA polymerisation can be targeted and disrupted by chemotherapeutic agents? Give examples of each.
Diminishing the supply of correct nucleotides: this requires the use of nucleotide precursor analogues that inhibit DNA polymerisation. These inhibit the synthesis of nucleotides in the cells. For example, 5-fluorouracil (5-FU) is an analogue of uracil with a fluorine attached and 6-mercaptopurine (6-MP) is an analogue of adenine. 5-FU is used as a therapeutic against cancerous cells (esp. colorectal) and 6-MP is used to treat acute leukaemia.
Targeting nucleotide incorporation: These analogues are used to be able to be integrated into the growing DNA strand and then inhibit further polymerisation of the strand. For example, Ara-C is used to treat acute leukaemia and is an precursor analogue of deoxycytidine. It contains a arabinose sugar that disrupts the polymerisation.
Intra- and inter- strand links: Intra- and inter- strand links inhibit the translational machinery in the cell. Cisplatin is an example that forms intra-strand links between the N7 of adenine or guanine and then forms inter-strand link between opposite strands of the helix. It can be used to treat cancer. Another example is nitrogen mustard gas. It forms inter-strand links between the N7 of two guanines. It has been used as biological warfare but has potential uses to treat leukaemia and lymphoma.
What is the main role of methylation on DNA?
Methylation and methylation patterns allow the cell to know whether the genome has been replicated already and prevents re-replication. As the genome is usually methylated on both strands, the presence of hemi-methylated DNA (half methylated) shows that this is a newly synthesised chromosome.
