09 Flashcards
You would like to study transcription in vitro (in the test tube). Think of everything you would need to add to the test tube to make transcription work. The test tube already has a buffer suitable to perform in vitro transcription.
DNA (something to copy from), RNA nucleotides, polymerase
Properties that are different between RNA and DNA
A ribose sugar instead of a deoxyribose
Uracil instead of Thymine and -OH at each bottom 2’ position
Single stranded instead of the double helix
RNA Polymerase
Composed of multiple protein subunits
Inner regions are responsible for RNA synthesis - highly conserved
Inside area is involved with the catalytic synthesis of RNA
Outer regions are involved in protein-protein or protein-DNA interactions - less conserved
Only 1 RNA polymerase in bacteria; bad if it mutates
3 Steps of Transcription Initiation
- Binding (closed complex): Initial binding of the polymerase (mediated by sigma factors) to the promoter region
RNA Polymerase does NOT bind directly to the DNA; it binds through the sigma factor - Promoter melting (open complex): DNA strands separate (called melting) to form a transcription bubble
- Initial transcribing complex: Inefficient incorporation of 10 of nucleotides. Often needs to repeat
Prokaryotic Promoters
+1 → The “start site”of transcription —The first nucleotides that are transcribed
The regions at –35 and -10 → -35 base pairs and -10 base pairs to the left of the transcriptional start site.
Where the RNA polymerase, through sigma factors, will bind.
-10 contains TATA box
Upstream
to the left of the promoter
Downstream
to the right of the promoter
The addition of an ____ _______ (sigma/σ-factors) allows the RNA pol to recognize consensus sequences (most likely of a nucleotide being at that region) in different promoters
initiation factor
If prokaryotes have only one RNA polymerase, how are promoters of different genes recognized?
There are different types of sigma factors. σ 70 is the most common in E. coli
Name the TATA box (-10) sequence
TATAAT
How do we know the consensus sequences of the -10 and -35 positions?
300 sequences known to function as σ70 promoters were aligned
Look for highest %’s
Looked at spaces between two (-35 and -10) boxes, all feel between 15 and 19 nucleotides
Closer together → makes protein flex more
Further apart → makes protein stretch more
RNA Polymerase Orientation
RNA pol synthesizes/moves in 5’ to 3’ direction
Reads template from 3’ to 5’
Top and bottom strand are complimentary
Template strand is determined by the promoter sequence
RNA pol moves left to right: bottom strand is template
RNA pol moves right to left: top strand is template
How do σ factors mediate polymerase binding to the promoter?
σ region 4:
Contains a helix-turn-helix motif
Recognizes the -35 sequence to secure RNA pol
One helix interacts with bases in the major grove (H-bonds), other helix interacts with DNA backbone
Very strong interaction
σ region 2:
Recognizes the -10 (TATA box) region
Initiates DNA melting (DNA strand separation) by base flipping on the non-template strand
Aromatic amino acids form a hydrophobic “pocket” and promote energetically favorable interactions with flipped out bases
Nothing enzymatic pulls the strands apart; happens on its own
A + T’s have 2 H bonds → weak area of interaction [as opposed to C-G’s 3 H- bonds], can be pulled apart easily
What type of experiments determined that aromatic amino acids are important for melting the DNA?
Made mutations in the sigma factor area that form pockets.
Changed the hydrophobic amino acids (Y,F,T,W) for alanine (non-polar) substitutions: tyrosine, phenylalanine, threonine, tryptophan
Measured β-gal activity (transcriptional read-out)
The idea is that successful transcription of a lacZ gene leads to the production of β-galactosidase, an enzyme that breaks down lactose into a detectable product.
If mutations in the σ factor reduce transcription, β-galactosidase activity decreases, indicating that DNA melting was impaired.
Initiation of transcription begins at promoter clearance
Once RNA polymerase binds to the promoter and forms an open complex, it must successfully transition into elongation by escaping the promoter.
For RNA polymerase to break free from the promoter and transition into elongation, it must synthesize at least ~10 nucleotides.
If it fails to do so, the transcript is released, and RNA polymerase starts over.
Early transcription attempts often fail, producing short, aborted RNA transcripts.
Once it reaches the threshold, it clears the promoter, releases the σ factor, and enters the elongation phase.
What happens after 10 nucleotides are copied?
Promoter Escape:
Polymerase-promoter interactions are broken
Polymerase-σ initiation factors interactions are weakened and eventually broken
Elongation phase can begin
RNA polymerase protein complex has different regions
- The DNA entry channel
- The DNA exit channel
- The rNTP channel
- The RNA exit channel
Rho Dependent Termination
ATP dependent rho helicase activity
The rho helicase recognizes a “rut” sequence in the RNA
The consensus sequence for rut has not been identified
The Rho protein will unwind the RNA from the DNA and stop transcription. Mechanism is fuzzy.
Rho Independent
Depends on two sequence motifs at the end of the RNA sequence:
1. Short inverted repeats that form a hairpin; hairpin boots off the RNA polymerase
2. Stretch of A:T base pairs
UUUUUU stretch is unstable enough so that the RNA can dissociate from the DNA
DNA Replication vs. Transcription
Similarities:
Polymerase
Direction; 5’ to 3’
Ability to correct [but not as much as DNA]
Differences:
Making/copying only 1 strand of DNA
Making RNA molecule, not DNA
