2) Explain why regulation of transcription frequently involves the promoter and
protein interactions with the promoter.
The RNA polymerase initially binds the promoter sequence. Once bound,
structural changes occur in the pro moter–RNA polymerase complex to initiate
transcription. To prevent transcription or enhance transcription of a specific gene,
it is most direct to inhibit or enhance initiation through the promoter.
3) Describe the three steps of transcription initiation that occur before the
elongation phase begins focusing on the key features of RNA polymerase at each
step.
1. Promoter Recognition (Binding)
Key Events:
o RNA polymerase binds to a specific DNA sequence called the promoter.
o In prokaryotes, the promoter contains conserved sequences, such as the -
10 (TATAAT) and -35 (TTGACA) regions, recognized by the sigma (σ)
factor of RNA polymerase.
o In eukaryotes, general transcription factors (e.g., TFIID) help RNA
polymerase II bind to the TATA box or other promoter elements.
Key Features of RNA Polymerase:
o In prokaryotes, the RNA polymerase holoenzyme (core enzyme + sigma
factor) is responsible for promoter recognition.
o In eukaryotes, RNA polymerase II relies on transcription factors and the
pre-initiation complex (PIC) to recognize the promoter.
2. Formation of the Transcription Bubble
Key Events:
o RNA polymerase unwinds a small portion of the DNA double helix near
the promoter, creating an open complex or transcription bubble.
o This bubble exposes the template strand of DNA, which will be used to
guide RNA synthesis.
o The DNA strands separate over a region of about 10-14 base pairs.
Key Features of RNA Polymerase:
o RNA polymerase has helicase activity (in prokaryotes) or works with
associated factors (in eukaryotes) to unwind the DNA.
o In prokaryotes, the sigma factor stabilizes the open complex.
3. Initiation of RNA Synthesis (Abortive Initiation)
Key Events:
o RNA polymerase begins synthesizing the RNA strand using the DNA
template strand.
o Several short RNA sequences (2–10 nucleotides) are synthesized and
released during this phase, a process known as abortive initiation.
o Once RNA polymerase successfully synthesizes a short RNA molecule
(long enough to stabilize), it transitions to the elongation phase.
Key Features of RNA Polymerase:
o RNA polymerase remains stationary at the promoter during this phase,
repeatedly attempting RNA synthesis.
o In prokaryotes, the sigma factor is still attached but will soon be released
after the transition to elongation.
o In eukaryotes, RNA polymerase II begins to shed some transcription
factors as it clears the promoter
4) Consider a bacterial promoter with – 35 and – 10 elements. What assay is best
to show that RNA polymerase binds at regions centered on the –35 and –10
positions upstream of the start site of transcription?
A DNA footprinting assay is the best choice. This assay will show a footprint
around those positions if RNA polymerase binds the promoter. An EMSA will
also test protein binding to DNA. You could see a result that RNA polymerase
binds the promoter with an EMSA, but you will not know that the binding is
centered on the –35 and –10 sites. ChIP is also possible, but DNA footprinting
is more relevant when considering only a specific DNA sequence.
DNase I Footprinting Assay:
This technique identifies the exact DNA region where a protein, such as RNA
polymerase, binds.
DNA is labeled at one end, and the DNA-protein complex is exposed to limited
digestion with DNase I, which cleaves DNA at regions not protected by the bound
protein.
If RNA polymerase binds specifically to the –35 and –10 elements, those regions will
be protected from DNase I digestion, creating a "footprint" when the fragments are
analyzed on a gel.
5) State whether the following statement is true or false, and explain your
conclusion: "The sequence of the –35 element is always 5'-TTGACA-3'."
The –35 element of bacterial promoters is a consensus sequence, which means it represents
the most common sequence found at that position but not the only sequence. The actual
sequence of the –35 element can vary between different bacterial promoters. The consensus
sequence 5'-TTGACA-3' is derived from aligning multiple bacterial promoters and
identifying the most frequent nucleotide at each position.
Consensus sequences serve as a guideline for RNA polymerase recognition but are
not fixed. Variations in the –35 element sequence can occur and may influence the
strength of promoter binding and transcription initiation efficiency.
Promoters with sequences closer to the consensus sequence tend to be stronger
because they interact more efficiently with RNA polymerase, while deviations can
lead to weaker promoters.
6) Given the three models for initial transcription in bacteria (transient excursion,
inchworming, and scrunching) which model represents the hypothesis most
supported by data? Describe the general conclusions of these experiments.
Scrunching: RNA pol makes a single-stranded bulge in DNA. RNA pol remains
bound to DNA. DNA is unwound and pulled into the RNA pol.
7) Describe the two proofreading functions of RNA polymerase in prokaryotes.
1. pyrophosphorylitic cleavage: uses PP to cleave the phosphodiester bond
at the 3'-end of the nascent RNA. This reaction removes the last nucleotide (or a
few nucleotides) from the growing RNA chain by reversing the polymerization
reaction. The removed nucleotide is released as a nucleotide triphosphate (NTP).
2. kinetic proofreading (backtracking and hydrolytic cleavage): RNA
polymerase detect mismatched nucleotides during elongation. If an incorrect
nucleotide is incorporated, it causes a delay in the forward translocation of the
RNA polymerase. This delay allows the enzyme to reverse its movement
(backtrack) and cleave the recently added mismatched nucleotide, removing it
from the RNA strand. After cleavage, RNA synthesis resumes with the correct
nucleotide.
10? What steps in the eukaryotic transcription cycle are stimulated by
phosphorylation of the carboxyl terminal (CTD) of the large subunit of RNA
polymerase II and beyond?
Phosphorylation of serine residues in the CTD tail of Pol II is required for promoter
escape and for efficient elonga tion. In addition, different patterns of
phosphorylation allow the tail to recruit factors required for RNA processing as
well. Thus, regulation of tail phosphorylation ensures these events are coor
dinated appropriately.
12? Howdoesthe function ofpoly-A polymerase differ from RNApolymerase
Poly-A polymerase does not require a DNA tem plate and adds up to 200 Asto the
30 end of mRNAs. RNA poly merase requires a DNA template and incorporates all
four NTPs for RNA
