DNA Replication and Recombination III Flashcards
DNA synthesis in Eukaryotes
- Similar to synthesis in prokaryotes
> Unwinding at origin, creating replication bubbles with 2
replication forks that move bidirectionally ( semi-conservative
replication) - under control of DNA polymerases
> Same requirements: dNTPs, template, and primer.
Due to the larger size and complexity (linear chromosomes) of the eukaryotic genome, the process is much more complex.
- Considerably more DNA
- Complexed with nucleosomes
- DNA consists of many linear chromosomes and not as one
circular molecule
- DNA must dissociate from its histones prior to replication, but re-
associate with them after replication is done. - The doubling of the histone count during the replication process (as is a new strand being made)
indicates that DNA synthesis and histone production are two tightly
coupled events.
Eukaryotic DNA synthesis vs Prokaryotic:
- Large differences in amount of DNA
(Drosophila has 40X more than E. coli) - replication speed (DNA polymerase) is faster
(100 kb/min vs. 0.5-5 kb/min, but E. coli takes 20-40 min while Drosophila needs 3 min).
How is this possible if the genome is bigger?
- There’s considerably more origins of replication in euk. than in prok. ( E.coli only has one) - Synthesis is also bidirectional, but many origins of replication exist (~25 000 replicons found in mammals, each about 100-200 kb in size).
- Many more copies of polymerase than found in prokaryotes
(50K vs. 15).
By combining the large numbers of origins of replications and large numbers of DNA polymerase enzymes available in each cell, eukaryotes are able to replicate huge amounts of DNA in each cell in a very short time
Genomes and sequences
In yeast:
Yeast genomes have +/- 250-400 origins
- called autonomously replicating sequences (ARSs)
- approx. 120bp in length, and contains an 11bp consensus
sequence.
consensus sequence:
this sequence is highly similar or conserved in all these genomes
In mammals:
- origins are not determined according to sequence, but rather by chromatin structure
- Euk origins of replication also control the timing of replication through a complex of >20 proteins, called the pre-replication complex (pre-RC).
Eukaryotic DNA synthesis
Replication process:
- During early G1-phase of cell cycle, the origin is recognized by “origin recognition complex” (ORC; 6-protein complex)
- More proteins associate with ORC to form pre-RC
pre-RC very important for activating many of the proteins that are NB for the initiation of DNA synthesis - DNA polymerase engages to initiate synthesis.
- pre-RC disassembles until next G1-phase
- Functions of helicases and SSBP are the same
- RNA primers also present
- Okazaki fragments (10x smaller: 100-200 bp) present as well
Main difference is in polymerases
Eukaryotic DNA polymerases
14 different polymerases
Pol. α, δ, and ε are the major forms involved in initiation and elongation
Pol α – [4 sub-units]
subunit 1 synthesizes RNA primer ( on leading and lagging strand); subunit 2 adds the first 10-20 nucleotides of RNA primer (low processivity)
Then other alpha subunits add about another 20 dNTPs
THUS, the primers in euk. consist of ribo- and-deoxyribo-nucleotides.
Once the primer is in place: polymerase switching occurs:
- Pol. α dissociates from the strand and is replaced with Pol. δ, and ε.
- They both have higher processivity and 3’-5’ exonuclease activity
Pol δ synthesizes lagging strand
and Pol ε synthesizes leading strand.
- Both of these involved in many functions in cell- such as DNA repair
Pol γ
– replication of mitochondrial DNA
Others mostly involved in DNA repair
All other polymerases are involved in repair of DNA
DNA synthesis at the ends of linear chromosomes (1)
- Unlike the closed circular chromosomes found in prokaryotes, the chromosomes of eukaryotes are linear – two problems:
> Resemble double-strand breaks (DSBs) - DNA fragmented
internally. Such ends would be vulnerable to nucleases and
fusion with other double-stranded ends = bad for genome
> DNA polymerases cannot synthesize new DNA at the tips of
single strand 5’-ends
- The problems arise when the last RNA primer of the lagging strand
is removed.
- When RNA primer is removed- leaves a gap
- Because no free 3’-OH group is available for the polymerase
enzyme to attach a new nucleotide to, the gap cannot be filled by
DNA polymerases.
- The section at the end of both lagging strands would be lost with every replication cycle, and the chromosome would become shorter and shorter = eventually losing genetic info.
Problems solved with the unique telomere regions at the ends of chromosomes and the unique enzyme telomerase
Euk chromosomes contain short tandem repeats at the end of one strand (G-rich strand) and the complementary strand = C-rich strand
TTGGGG for Tetrahymena
TTAGGG for vertebrates
The strand ending with the 3’ hydroxyl (lagging strand template during DNA replication) is always the G-rich strand
DNA synthesis at the ends of linear chromosomes (2)
Telomerase
Telomerase- unique eukaryotic enzyme
> It is a ribonucleoprotein - within its structure is a section of RNA
which is essential for its catalytic activity
Has a dual functions at the telomere:
- In order to solve the problem of the chromosome ends being vulnerable to nucleases –> Telomerase’s form large loop structures
= T - Loops –> Telomere loops back on the chromosome
-And the single stranded DNA, at its end, is inserted into the double helix to form a triple stranded structure.
(1.) - these chromosome ends are stabilised by hairpin by forming a hairpin-loop of G-G bonds (G-quartets)
- Formation is facilitated by telomerase
(2.) -DNA replication is also facilitated by telomerase at the telomeres
(3.) - The telomerase RNA component (TERC) acts as a guide to attach the enzyme to the telomere and as a template for the synthesis of a DNA compliment
TERT -
Synthesis of DNA from a RNA template = reverse transcription
- The catalytic subunit of telomerase is TERT
= Telomerase reverse transcriptase
At the end of the telomere - a part of the RNA component will recognize the 3’ end of the single-stranded DNA
- It then attaches the telomerase at the end of the overhang
- The remainder of the RNA component is now free to act as a template for the reverse transcriptase to add DNA nucleotides to extend the 3’ tail
- Once this process is complete- telomerase can translocate again to the end of the chromosome and the process repeats.
THUS, the chromosome end is considerably extended.
- Once it is sufficiently extended, telomerase is released and standard DNA replication can take place.
[Summary]
TERC contains sequence complementary to TTGGGG, which binds 3’ G-rich tail and TERT synthesises more TTGGGG repeats
Telomerase translocates and repeats the process
Telomerase dissociates.
AND THEN:
- DNA pol α adds an RNA primer fills the gap
- DNA ligase repairs the phosphodiester bonds
Although the final gap- resulting from removal of last primer will still not be filled
- This gap is located far beyond the original end of the DNA in the chromosome
= chromosomal shortening is prevented
Only certain adult cell types express telomerase- such as adult stem cells.
- most somatic cells don’t express this enzyme.
- Somatic cells can only divide a certain amount of times until their telomeres are lost
- Chromosome damage either kills the cell or it becomes senescent
- Accumulation of senescent cells = Cell ageing
DNA Recombination
- Can occur between any 2 homologous double-stranded molecule
- Common instances are between viral and bacterial chromosomes:
In eukaryotic chromosomes, during crossing over in meiosis and also during the repair of damaged DNA
All models that explain this process rely on the complementarity between DNA strands to explain the precision of the exchange and the series of enzymatic processes that actually accomplish recombination.
Holliday Model
1.) 2 homologous duplexes of DNA align with each other
2.) An enzyme which is able to cut the DNA internally
(endonuclease ) –> nicks both strands at an identical position
3.) The loose ends are displaced
Subsequently pair to their complementary base pair sequences on the opposite duplex.
4.) A ligase joins the loose ends to their neighboring nucleotide polymers - forms a phosphodiester bond
= Results in Hybrid heteroduplex DNA molecules
5.) Formation of hybrid heteroduplex DNA molecules
= Strand displacement= results in a cross-bridged configuration
6.) These cross-bridges move down the chromosome
PROCESS = Branch migration
–>Increased length of heteroduplex DNA on both homologs.
When migration ceases, the duplexes bend…
7.) Separation of duplexes initiated by a 180º rotation of one-half of the duplex
this creates an intermediate planar structure
= chi-form (χ), or Holliday structure
8.) Two homologous strands that were previously uninvolved in the process are nicked by an endonuclease.
9.) Ligation completes the process of recombinant duplex formation.
The most important proteins in homologous recombination are
RecA (in E. coli) and Rad51 (in eukaryotes)
- These proteins are loaded onto ssDNA ends after endonuclease nicking and then search for a homologous DNA sequence in another molecule
- Many other enzymes regulate the actions of RecA and Rad51
Gene conversion
– non-reciprocal DNA exchange
= Pieces are not exchanged between the two homologs, one of
them grabs a piece of material from the other one.
- Appears as if one allele has been converted to another
