discussion Flashcards
How does the RNA polymerase holoenzyme initiate transcription in prokaryotes, and what are the functions of its subunits?
The RNA polymerase holoenzyme is required for transcription initiation in prokaryotes.
* It consists of core enzyme subunits (α, β, β’, ω) and a sigma (σ) factor.
Alpha (α) subunits:
* Help assemble the enzyme.
* Promote interactions with regulatory proteins.
Beta (β) subunit:
* Carries out the catalytic activity of RNA synthesis.
Beta prime (β’):
* Binds to the DNA template during transcription.
Omega (ω):
* Assists in enzyme assembly.
* Involved in regulation of gene expression.
Sigma (σ) factor:
* Recognizes and binds to the -35 and -10 promoter sequences.
* Positions the polymerase correctly for transcription initiation.
* Promotes unwinding of DNA strands at the start site.
* Different sigma factors allow the polymerase to bind and transcribe different sets of genes, enabling regulation under various conditions.
What are the two main mechanisms of transcription termination in prokaryotes, and how do they work?
1) Intrinsic termination (rho-independent):
* Occurs when RNA polymerase pauses over a weak DNA-RNA hybrid.
* Polymerase backtracks to stabilize the hybrid.
* It encounters a GC-rich stem-loop (hairpin) structure followed by a U-rich sequence.
* This causes the polymerase to fall off and release the RNA.
2) Rho-dependent termination:
* Requires the Rho protein, a hexameric ATP-dependent helicase.
* Rho binds to a C-rich sequence called the rut site on the RNA.
* It translocates along the RNA toward the paused RNA polymerase.
* Upon reaching the polymerase, Rho causes dissociation of the transcription complex and release of the RNA.
What are the key differences between transcription in prokaryotes and eukaryotes?
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RNA polymerases:
* Prokaryotes: One RNA polymerase.
* Eukaryotes: Three RNA polymerases (I, II, III) with distinct roles. -
Initiation factors:
* Prokaryotes: Use sigma factors for promoter recognition.
* Eukaryotes: Use multisubunit general transcription factors (e.g., TFIID, TFIIH). -
Transcription & translation:
Prokaryotes: Co-transcriptional translation (occur simultaneously in the cytoplasm).
Eukaryotes: Spatially separated—transcription in the nucleus, translation in the cytoplasm. -
RNA processing:
* Prokaryotes: Produce a simple transcript used directly.
* Eukaryotes: Transcripts undergo processing (5’ cap, splicing of introns/exons, 3’ poly-A tail).
The initial transcript in eukaryotes is called pre-mRNA. -
DNA packaging:
Prokaryotes: DNA is naked or loosely organized, no histones.
Eukaryotes: DNA is packaged into chromatin with histones.
How is transcription initiated in eukaryotes, and what roles do general transcription factors (GTFs) play?
General Transcription Factors (GTFs) are required to initiate transcription by RNA polymerase II.
* GTFs bind to the promoter before RNA polymerase can assemble.
TBP (TATA-binding protein) is a subunit of the TFIID complex, which is the first to bind to the promoter.
Other GTFs (e.g., TFIIA, TFIIB, TFIIE, TFIIF, TFIIH) sequentially join to form the pre-initiation complex.
* TFIIH has kinase activity and phosphorylates the C-terminal domain (CTD) of RNA polymerase II.
* CTD phosphorylation triggers the transition from initiation to elongation by weakening RNA polymerase’s interaction with the GTFs.
How does the spliceosome assemble and function during mRNA splicing?
The spliceosome assembles stepwise, involving several small nuclear RNAs (snRNAs) called U RNAs (e.g., U1, U2, U4, U5, U6).
U RNAs base pair with:
* The pre-mRNA at splice sites.
* Each other, to position catalytic residues precisely.
Some U RNAs are directly involved in catalysis (e.g., U6 acts like a ribozyme).
During splicing:
* U1 and U2 recognize the** 5’ splice site** and branch point.
* U4/U6 and U5 join, then U1 and U4 leave to activate the spliceosome.
A lariat structure forms when the branch point attacks the 5’ splice site.
* The lariat intron is released, and exons are joined together to form mature mRNA.
What are key features of microRNAs (miRNAs) and their role in gene regulation?
miRNAs are generated from long RNA transcripts transcribed by RNA polymerase II.
These precursors are processed into a 21–25 nucleotide hairpin structure.
miRNAs typically do not have perfect base pairing with their target mRNAs.
They bind mainly to the 3’ UTR of target mRNAs to repress translation or promote degradation.
This pathway is mainly used for the regulation of endogenous genes (within the organism).
What are the key characteristics of siRNA and how does it function?
Generated from long double-stranded RNA (dsRNA).
Processed into 21–25 nucleotide short double-stranded RNAs.
siRNAs usually have perfect homology with their target mRNA.
This allows for precise cleavage of the mRNA.
Functions primarily in gene silencing, especially in viral defense or silencing transposons and foreign genes.
How do the siRNA and miRNA pathways differ in terms of processing and gene silencing mechanisms?
🔹 Shared steps (both siRNA and miRNA):
* Dicer processes long RNA precursors into ~21–25 nt double-stranded RNAs.
* Both small RNAs are loaded into the RISC complex (RNA-induced silencing complex).
* One strand (the guide strand) is retained; the other is discarded.
🔹 siRNA Pathway:
* Originates from long dsRNA (e.g., viral, synthetic, or transposon-derived).
* Dicer cleaves it into perfectly complementary siRNAs.
* siRNA-loaded RISC binds perfectly matching mRNA.
* Leads to mRNA cleavage and degradation.
🔹 miRNA Pathway:
* Originates from endogenous Pol II transcripts that form a hairpin structure.
* Processed by Dicer in the cytoplasm.
* miRNA-loaded RISC binds imperfectly to the 3′ UTR of target mRNAs.
* Causes translational repression or mRNA destabilization, not direct cleavage.
How is the genetic code read, and what do insertion/deletion experiments reveal about its structure?
The genetic code is read sequentially in one direction (5’ → 3’), without skipping or backtracking.
The code is triplet-based, meaning each amino acid is encoded by a sequence of 3 nucleotides (codon).
- Single base substitutions change only one amino acid, supporting a non-overlapping code.
- Insertions or deletions (indels) of 1 or 2 nucleotides cause a frameshift, altering all downstream amino acids.
- Insertion or deletion of 3 bases restores the original reading frame (but with extra/missing amino acid[s]).
The genetic code is degenerate, meaning some amino acids are encoded by more than one codon.
Benefits of degeneracy:
* Material conservation: The same amino acid can be encoded in multiple genes without changing protein structure.
* Mutational robustness: Silent mutations (especially in the 3rd base of a codon) are less likely to alter the amino acid, reducing harmful effects.
* Helps buffer against point mutations, especially in coding regions.
