Meriems lecture Flashcards
DNA replication is
semi conservative
DNA replication in E.coli has 3 steps
- Initiation, 2. elongation, 3. termination
Elongation is performed by
DNA polymerase III
The alpha subunit of RNApolII
is responsible for 5’ to 3’ polymerisation
The epsilon subunit of RNApolII
is responsible for proof reading exonuclease, it removes any mismatches via a repair system
To replicate the double helix,
the strands need to be separated, and this is an energy dependent process
The DNA polymerase acts like a
hand (attach slide pic)
DNA polymerase synthesises DNA strand in
5’ to 3’ direction
The lagging strand cannot be replicated
continuously, the DNApolII will have to go backwards as the DNA continues to get unwound.
The lagging strand forms a loop because
the DNApolII wants to move in the same direction
The lagging strand forms a loop because
the DNApolII wants to move in the same direction
RNA primers need to be removed
to complete replication
DNA polI is responsible for
Maturation of Okazaki fragments (and DNA repair)
Initiation of replication in E.coli occurs at
oriC
SSB proteins bind to the separated strands to ensure
that both strands don’t re-ligate automatically
SSB proteins bind to the separated strands to ensure
that both strands don’t re-ligate automatically
H.Sapiens have
60000 origins of replication
Eukaryotic DNA is assembled into
chromatin
If there is a free double stranded end for DNA, this would suggest that
there has been some DNA damage
End replication problem is basically where
the lagging strand will not completely replicate, resulting in a terminal gap
Telomerase extends the 3’ end of
the chromosome. This means that overall, no genomic information has been lost from each time DNA replication occurs.
Telomerase is activated in
85% of cancers
DNA damage does not necessarily
mean mutation
Rate of spontaneous mutations can be
5x10^-11 for mamallian systems
Mutations can result from errors in DNA replication
- failure of epsilon subunit of DNA polymerase to proof- read
- Tautomeric shift of bases causing mismatches in base pairing
Each of the 4 bases can exist in
alternative forms (tautomers)
A tautomeric shift can cause a point mutation
Where a CG pair turns to a CG+, and causes the G+ to bind with T, and replication to thereby occur, it would result in a CG base pair to convert to a TA pair
slippage of the template strand can cause
deletion
Slippage on the newly synthesised strand can cause
insertion
Mismatch repair system
repairs replication errors
Base excision error
repairs damaged bases. This occurs via DNA glycosylase which removes the base and leaves a gap. AP endonuclease recognises a sugar with missing base and cuts the phosphodiester backbone of DNA. DNA polymerase I fills in the gap and DNA ligase seals the DNA strand
Nucleotide excision repair
repairs unusual base structures which distort DNA such as pyrimidine dimers
Mechanism of mismatch repair system
see attach slide
What happens to your DNA upon DNA damage?
where there are two thymine bases beside each other, exposure to UV can cause Thymine dimers to form. This can be repaired using CPP photolyase and light. Placental mammals cant do this though so they just cut i.e. NER.
Synthesis of RNA from DNA template require
RNA polymerase as well as RNTP’s
RNA polymerase copies one strand
of duplex DNA into RNA
In prokaryotes, a single RNA polymerase
transcribes all genes
The three steps of RNA transcription
- initiation
- elongation
- termination
Transcription is initiated at specific regions in DNA
- gene promoters
There is a signal sequence on RNA which signals to end transcription
AAA
Formation of charged tRNA
occurs via a synthetase enzyme joining amino acyl and tRNA
tRNA’s decode the
mRNA message
Each amino acyl tRNA synthetase charges a tRNA
with the correct amino acid
Translation also occurs in 3 steps
- Initiation
- Elongation
- Termination
Prokaryotes have
polycistronic mRNA’s,
Eukaryotes have
monocistronic mRNA’s
The initiator tRNA is the only tRNA which can
bind to the P- site
Why regulate gene expression?
a) economical
b) some gene products are required at different times or in different environments
c) gene products are required in different amounts
d) In eukaryotes, differential gene expression occurs during development
How is gene expression regulated?
At the transcription level, initiation and elongation of the transcript
At the post transcription level, Transcription processing (folding of mRNA prevents translation) and protein modification
The E.coli sigma 70 sub unit recognises a specific set of gene promoters
and binds to the -35 and -10 regions
The initiation of gene transcription is controlled by regulatory proteins
These include positive regulators and negative regulators. In prokaryotes, Activator proteins bind to a DNA sequence close to the promoter and help recruit the RNA polymerase to the promoter.
Genes can be activated by either
binding the activator protein or removing the repressor protein
Prokaryotes generally
have polycistronic mRNA’s. This gene structure is called the operon.
Lactose is the inducer of
the lac operon
LacI creates the
lac repressor which binds to the operator to prevent transcription
With no inducer present,
the repressor makes contact with the operator
When inducer is present it binds
to repressor, altering its structure and it cannot bind to the operator
The lac operon is positively regulated by
CAP
Lactose + glucose =
low basal lac gene expression
Lactose no glucose
high lac gene expression
