·        An error is when one there is damage that is repaired; like a G gets coded where a T should be but is corrected to a T. No replication of the damage occurs

·        A mutation occurs when the damage is replicated during DNA replication and the G does not get corrected to a T but remains and the corresponding changes to from a C to an A

What is photoreactivation?

·        Fixes pyrimidine dimers

o   DNA photolyase detects and binds to the dimer (damaged DNA site)

o   Photolysase absorbs near-UV to blue region of light

o   Blue light cuts between the dimers; breaking the bonds and thus returning them to their normal state

o   Photolyase is released

o   DNA returns to normal

·        Not found in placental mammals

·        Occurs in both eukaryotes and prokaryotes

·        Removes one base

·        In E. coli

o   DNA glycosylase recognizes damaged base and breaks the glycosidic bond between the damaged base and its sugar leaving an AP site

o   AP site (apurinic or apyrimidinic; sugar without its purine or pyrimidine base)

o   AP site is recognized by AP-endonuclease that cuts the sugar phosphate backbone on the 5’ side of the AP site

o   DNA phosphodiesterase removes the Phosphate and the sugar

o   DNA polymerase I degrades DNA in the 5’→3’ direction, while filling in with new DNA

o   DNA ligase fixes the nick

·        Occurs in both eukaryotes and prokaryotes

·        Removes many nucleotides

·        Removes bulky lesions created by dimers that distort the Phosphate backbone

·        In E. coli

o   a pyrimidine dimer damages the DNA

o   endonuclease and helicase open and unwind and remove a section of DNA

o   DNA polymerase I adds new DNA

o   Ligase fixes the nicks

·        Human global nucleotide excision repair

o   Occurs anywhere on the DNA

o   XPC and XPA recognize the damage caused by a pyrimidine dimer (bulky lesion)

o   XPB and XPD are part of the TFIIH and act as helicases unwinding the section of DNA

o   XPG (endonuclease) cuts at the 3’ side

o   XPF (endonuclease) cuts at the 5’ side

o   DNA polymerase lays down new DNA

o   DNA ligase repairs the nicks

·        In eukaryotes

·        ds breaks are broken chromosomes that , if not repaired can lead to cell death or cancer

·        repaired in two ways

o   homologous recombination (yeast cells)

§  uses unbroken sister chromatid as the recombining partner

§  dominant mechanism in replicating cells during S and G2 phases as only one DNA copy is broken and the other is available to align the breaks properly

o   non-homologous end-joining (mammalian cells)

§  in G1 phase as the DNA has not replicated and no second, homologous chromosome is yet available to serve as a template for repair

·        Happens only directly after DNA replication

·        In E. coli

o   Due to incorporation of the wrong base and a failure of the proofreading system

o   Parental strand is recognized by its methylated adenines in GATC sequences

o   Corrects the mismatch in the complementary (progeny) strand

·        How?

o   MutH, MutL, MutS, and ATP recognize a base mismatch and identify the newly synthesized strand by the absence of methyl groups on GATC sequences, and introduces a nick into that new strand, across from the methylated GATC and upstream of the incorrect nucleotide

o   exonuclease 1, MutL, MutS, DNA helicase, and ATP removes DNA downstream of the nick including the incorrect nucleotide

o   DNA polymerase III holoenzyme with SSB fills in the gap

o   DNA ligase seals the nick

o   Methyltransferase methlylates GATC sequences in the progeny strand across from the methylated GATC sequence in the parental strand

·        In eukaryotes (mammalian cells)

·        in G1 phase as the DNA has not replicated and no second, homologous chromosome is yet available to serve as a template for repair

o   DNA ends attract Ku (binds)

§  Ku is a dimer of two polypeptides that protect the DNA ends from degradation until end-joining is complete

o   DNA-PK (DNA protein kinase) binds to Ku

o   The two ends of DNA-PK complexes find regions of micro-homology (synapsis) and then phosphorylate each other

§  Phosphorylation promotes the dissociation of the DNA-PK complex and activates the  helicase activity of Ku to unwind the DNA ends so the micro-homology regions can base-pair

o   Nucleases remove the flaps, gaps are filled and the DNA strands are ligated together

·        Dealing with damage without really repairing it

·        Path is induced by DNA damage including UV damage

·        Called error prone by-pass because the DNA the SOS response causes DNA to replicate even though the damaged region cannot be read correctly resulting in errors in the newly made DNA

·        HOW?

o   UV light (or other mutagenic treatment) activates RecA coprotease whose target is the product of the lexA gene, a repressor for a number of repair genes

o   RecA coprotease causes lexA to cleave itself

o   loss of lexA repression causes umuC and umuD to form a polymerase called polV

·        polV can do error prone bypass for pyrimidine dimers, AP sites and other UV induced lesions

·        Must have an error prone polymerease

·        Synthesizing across a lesion

·        Pol eta

·        Pol eta 60% of the time is right.

·        Means that DNA is synthesized on a “something is better than nothing”. Results in errors in transcription

·        polymerases that insert random nucleotides across from a pyrimidine dimer (can be: Pol IV, Pol eta, Umu

RNA polymerase

·        Core with subunits

o   β¹, β, α, α, ω

o   responsible for elongation

o   beta has the enzymatic activity

o   forms the phosphodiester bonds

·        Sigma (σ)

o   Required for it to be an intact holoenzyme

o   Mobile portion of the RNA polymerase (can come on and off)

o   Responsible for finding  and binding to the promoter and initiating transcription

o   σ factor allows initiation of transcription by causing the RNA polymerase holoenzyme to bind tightly to a promoter. This tight binding depends on local melting of the DNA to form an open promoter complex and is stimulated by σ. The σ factor can therefore select which genes will be transcribed.

·        Initiation

o   Holoenzyme comes in and finds the promoter sequence

o   Forms the open complex

§  Melting is required between the -10 box and the transcription start site (+1)

§  Holoenzyme must be bound for these steps (sigma)

o   RNA syntheses is initiated

§  Takes some time with several aborted starts but once it gets past 10 nucleotides then it has cleared the promoter

o   Once promoter is cleared sigma is released

·        Elongation

·1969 Travers and Burgess

oAllowed RNA polymerase holoenzyme to initiate and elongate RNA chains on a T4 DNA template at low ionic strength so that the polymerases could not dissociate from the template to start new RNA chains

oAfter 10 minutes when most initiation had stopped new, rifampicin-resistant core polymerase was added in the presence and absence of rifampicin

oAn immediate rise in both curves was observed indicating that sigma had associated with the new cores

oThat sigma associated with both the rifampicin resistant and rifampicin sensitive cores indicated that the core determines rifampicin resistance

·        Melting at the template strand has to happen in order for transcription to occur and will not happen at 15°C but better at 25°C and 35°C sigma is important and temperature affects it

·        Experiment

o   Labeled DNA is incubated at different temperatures with the holoenzyme to see at which temperature binding will occur

o   Generic bacterial promoter

§  -10 box which is AT rich

§  -35

§  Transcription start site (+1) is 10bp downstream, from -10 box

§  When the RNA polymerase binds at the promoter, it will unwind a portion near the -10 box to the transcription start site

·        Unwinding is what is necessary to have strong binding of the RNA polymerase (holoenzyme) to the DNA (affected by temperature)

§  TATAT sequence is easier to melt apart than GC

·        Combine RNA polymerase with a radio active promoter region where the RNAP will bind and induce melting

o   No nucleotide are added so no RNA synthesis will occur

·        While the DNA is melting, DMS (dimeltylsulfate) is added which adds methyl groups to adenines (if they are in a single trionic region, if adenine are paired then no methyl groups are added)

·        Then the protein is removed so that the DNA can come back together; Where melting occurred if the Adenines got methylated will not come back together cannot base pair to Ts so will stay single stranded

·        S1 (DNAse) comes in and chops any single stranded region (methylate)

·        Run these strands on a polyacrylamide gel

·        Examine the bands on the gel and the large bands is where melting occurred

·        Lⁱ gene makes a repressor protein (active form)

o   Lⁱ binds as a tetramer to the operator

§  Prevents the RNAP from binding and reading through

§  Occurs when there is no lactose and repression of the protein molecules that break down lactose is necessary (no wasted energy)

§  When lactose is present Lⁱ is still made but allolactose binds to the repressor  which causes an allosteric change causing it to fall off so that RNAP can bind and transcription can read through

o   Mutants

§  O^c  an operator constituent mutant that changes DNA sequence that the operon is on (cis-mutation) cannot bind repressor

§  I^- protein mutates the repressor so that it cannot bind the operator; as it affects multiple copies of the operon is called a (trans-mutation)

§  I^s super repressor cannot bind allolactose; can never come off the operator (trans-mutation)

·        Under high glucose and high lactose you do not want lac operon to function

·        Under low glucose and high lactose you do  want lac operon to function

·        No lactose no operon function

·        cAMP becomes high when glucose is low

·        cAMP binds to a protein called CAP activator protein

·        CAP and cAMP are both necessary for β-gal production

Explain the trp operon and transcription attenuation.

·        trp operon (weak so there is still some level of transcription)

o   synthesizes tryptophan

o   turn off is there enough on if not enough

o   has a repressor (the binding of the inducer causes the repressor to become a DNA binding protein)

§  made in a non-active state

§  forms a dimer and can only bind when there is plenty of tryptophan around

§  makes an allosteric changes turning it into a repressor

§  binds to the DNA preventing RNA polymerase to bind and stopping transcription

o   not binding enough; still leaky and a cell only wants to make tryptophan when it really needs it so has additional level of contro

·        attenuation

o   a process involved in transcription control

o   a sequence that causes pre-mature termination of transcription

o   trpL (leader) makes a non-functional polypeptide

§  trpL has two trp codons

o   Low tryptophan conditions

§  Transcription will occur but there will be a pause at translation while waiting for tRNA to bring a trp codon into the ribosome (transcription and translation are simultaneous in prokaryotes)

§  Ribosome stalls

§  This will cause the attenuator to form a  hair-pin loop where regions 2&3 will (interact-base pairs) hybridize

§  4 does not form base pairs and terminations does not occur

o   High tryptophan conditions

§  Hair-pin loops form as there is no stalling and the ribosome moves quickly; 1&2 base pair and 3&4 base pair making a terminator hair-pin and polymerase falls off and transcription stops

·        Pol I – rRNA

·        Pol II – mRNA

·        Pol III – tRNA and 5S rRNA

·        Column chromatography

o   Uses permeable solid matrixes is a column immersed in a solvent to separate proteins

o   Proteins can be separated according to their charge, hydrophobicity, size, or ability to bind to particular chemical groups

·        Chromatography separation

o   Uses A280 to show where protein is

o   Uses increasing gradient of salt to elude protein from the column

o   Uses radioactive UMP to show RNP activity

o   Three peaks separate at different salt concentration

o   Showed that there are 3 different RNAP (I,II,III)

·        Localization in the cell

o   Used the fraction number to know where in the cell the RNAP was located and the salt concentration to know which RNAP it was

o   Fraction 65 sald <2 = RNAP II is nucleoplasmic

o   Fraction 50 salt >2 = RNAP 1 is nucleolus (rRNA)

·        Response to alpha amanitin

o   Produced by death cap mushrooms

o   RNAPI is completely resistant to α-amanitan

o   Can be used to isolate RNAPI from RNAPII and III as II and III transcription will stop

o   Low dose knocks out RNAP11 (synthesizes proteins)

o   Higher concentration knocks out RNAIII

Epitope tagging

·        A method whereby the subunits of a particular protein can be determined

·        How?

o   Take one sub-unit of a RNAP; Clone and alter the gene and put it in a plasmid and make it form a couple of more amino acids on the end of the protein that the WT wouldn’t

o   These amino acids are immunoreactive (strong antibodies develop easily against them) but are not found in the wild type organism

o   Express the gene in the organism and then use an antibody to the epitope to pull down the whole protein complex

o   Denature with SDS

o   Then do an electrophoresis

·        A promoter is where RNAP binds

·        Every promoter needs TFIID

·        How RNAP binds to the promoter determines the direction in which transcription will occur

·        Needs to be near the start of transcription

·        Are not interchangeable

o   TATA box (binds with TBP((tata binding protein) which is part of TFIID that functions with RNAPII)

o   Class I promoters

§  rRNA precursor

§  has core and upstream element

o   RNAPIII Promoters

§  Three types

·        tRNA

·        5S rRNA

·        Human snRNA (small RNA involved in slicing

·        Enhancers

o   Increase transcription

o   Does not bind RNAP

o   Sequence is orientation independent

o   Can be up to 50kb away from where transcription is occurring can be upstream, in the middle of a gene or far downstream

o   Transcription factors will bind to these sequences

·        RNAPIIa

o   215 kD

o   Bound to DNA

o   Joins pre-initiation complex

·        RNAPIIo

o   240 kD

o   Elongates RNA

·        Has 9 subunits

·        Has kinase and helicase activities

·        Phosphorylates RNAPII when it is bound to DNA (RNAPIIa→RNAPIIo)