Post-transcriptional Gene Regulation Flashcards
Describe the different forms of alternative splicing that are regulated events following transcription.
Alternative splicing can lead to:
exon skipping
Alternative 3’ or 5’ splice site selection
Intron retention (i.e. not splicing)
Mutually exclusive exons (i.e. one or the other is included)
How can alternative splicing impact post-transcriptional processes including translation and mRNA packaging?
Alternative splicing can determine which promoter is included in the mRNA transcript
It can also dictate which cleavage and polyA site is included
Both of these splicing events are not the regulated event
How can 1 mRNA code for a family of proteins with different functions?
Alternative splicing of an mRNA can determine which exons are included in the transcript.
Example: Fibronectin can exist in the extracellular matrix of fibroblasts or in hepatocytes. These fibronectin forms differ by two integrin binding domains that are crucial for adhesion in the basement membrane, but are spliced out of the hepatocyte form.
Describe how DSCAM can have more than 38,000 different splicing patterns from 4 exons.
Splicing of this gene choses 1 base pair from each of 4 exons (A1-12, B1-48, C1-33 and D1-2)
12*48*33*2 = 38,016 possible splicing patterns
How can an mRNA with a strong splice site be prevented from splicing?
Negative control
A repressor protein can bind to a silencer sequence to inhibit splicing
Because of the strong interaction (near perfect base pairing), even in the absence of an SR protein, the spliceosome would be able to initiate splicing.
The repressor protein prevents this splicing from occuring
How can a weak splice site be recognized by a spliceosome?
Usually, the spliceosome will pass over weak splice sites
In the presence of an activator protein (i.e. SR proteins), the spliceosome will recognize the weak splice site and initiate splicing
What is the difference between constitutive splicing and alternative splicing?
In alternative splicing, the levels of the activator and repressor proteins vary between tissues leading to different mRNAs in different tissues
What regulatory proteins aid in the selection of splice sites?
SR proteins (serine, arginine rich proteins) bind to ESEs (exon splicing enhancers) along with the complex of U proteins that make up the spliceosome
Describe the differences in splicing between the membrane-bound antibody and the secreted antibody.
The antibody gene contains 2 potential polyA sites and 2 stop codons
The membrane-bound form of antibodies is produced when the second cleavage/polyA site is used (this is the optimal site). The intron contains the first stop codon, but is spliced out to form the antibody with a terminal hydrophobic peptide that can insert into the membrane.
The secreted antibody is produced when the first cleavage/polyA site is used. This causes part of the intron to be incorporated into the mRNA because the 3’ splice site is cleaved away. The final protein has a terminal hydrophobic peptide from the intron region.
What are the roles of eIF2, eIF4E and eIF4G?
These are involved in the initiation of translation
eIF2 bound with a GTP is attached to the initiator tRNA that is attached to the P-site of the small ribosomal subunit
eIF4E is a cap binding protein
eIF4G is a scaffold protein that binds to the ribosomal subunit to allow it to scan along the mRNA searching for the start codon
When the small ribosomal subunit reaches the AUG start codon, what happens?
GTP is hydrolyzed causing eIF2 and the other initiation factors to dissociate from the complex.
The large ribosomal subunit can then bind to the Methionine tRNA at the P-site.
How is eIF2 recycled and how does this regulate translation?
Following hydrolysis of GTP, eIF2 is bound to GDP
When there is enough energy in the cell, eIF2B (a guanine nucleotide exchange factor) removes the GDP and replaces it with a GTP to reform an active eIF2
If the cell is stressed, a protein kinase phosphorylates eIF2-GDP which keeps eIF2 in inactive form and dramatically slows protein synthesis
Describe the steps in the regulation of cap-binding proteins.
The cap binding protein eIF-4E is bound to inhibitor 4E-BP. It can still bind to the mRNA, but cannot interact with eIF-4G which prevents the ribosome from attaching
4E-BP is inactivated by phosphorylation which frees up the eIF-4E
How are inefficient transcripts translated properly?
eIF4E is able to process mRNA that have difficult-to-read conformations
These mRNAs are not properly translated when active eIF4E is present at low levels
What is an IRES?
Internal ribosome entry site
This allows for cap independent translation
Translation can be initiated without eIF4E by using an IRES. Still need the eIF4G to bind to the small ribosomal subunit
This is a way the cell can turn off general translation, but still keep a subset of proteins activated (for example, this occurs in apoptosis)
True or false: the time it takes to achieve the steady state level of mRNA depends on the rate of mRNA synthesis and the rate of mRNA degradation
FALSE
The time it takes to reach steady state is only affected by the rate of degradation
A cell with a degradation rate of 10 minutes will respond to changes in rate of synthesis ________ than a cell with a degradation rate of 1 minute
(Faster or Slower)
Slower
The rate of degradation of a cellular constituent determines how rapidly it responds to changes in synthesis rate
How are old mRNAs degraded?
Over time, the length of the poly-A tail decreases (deadenylation)
When the tail is <30 base pairs, the mRNA will be degraded from both ends (3’–>5’ by an exonuclease, and then 5’—>3’ after the cap is removed)
What determines how stable and mRNA molecule is?
How fast the poly-A tail is degraded
What process competes with de-adenylation?
Translaton
When an mRNA is being translated, the poly-A tail is protected from deadenylation by poly-A-binding proteins involved with the translation machinery
Describe how miRNAs regulate gene expression
miRNAs are short RNAs that form hairpins by folding on themselves
The hairpin is cleaved by Dicer
Argonaute discards one of the 2 strands leaving the RISC complex and one strand to basepair with mRNA
Depending on the extent of the basepair matching, the mRNA will either be rapidly degraded (extensive matching) or translation will be reduced (less extensive mathing)
What is a seed sequence?
The bound portion between miRNA and mRNA, which only needs to be 7 base pairs long
True or false: A single miRNA can regulate many genes
True.
Because only 7 base pairs need to match and they do not need to be perfect or sequential, one miRNA can regulate many genes
How are broken/unuseful RNA sequences degraded?
Nonsense-mediated mRNA decay
Splice errors often produce premature stop codons in the middle of coding regions due to frameshifts
Exon-junction complexes (EJC) bind following successful splicing
Upon entering the cytosol, a test translation is conducted
If no errors exist, then all the EJCs should be pushed off the mRNA
EJCs present on the mRNA following test translation is a red flag because a stop codon must have been reached, the mRNA will be rapidly degraded
What role does ubiquitin play in protein degradation?
Ubiquitin is the tag that marks proteins for degradation by the proteasome
Describe the ubiquitin pathway of protein turnover
E1 (ubiquitin activating enzyme) is ubiquitinated
The ubiquitin is transferred from E1 to E2 (ubiquitin conjugating enzyme)
E2 complexes with E3 (ubiquitin protein ligase) and binds to the substrate to ubiquitinate
The substrate is then polyubiquitinated
The proteosome cap binds to the ubiquitinated tail and the substrate is degraded within the cylinder
The ubiquitins are then recycled back to E1
Describe the structure of the proteasome
The proteasome is composed of a central 20S cylinder supplemented with two 19S caps
How are cell cycle regulators degraded?
The ubiquitin system is used to rapidly degrade cell cycle regulators.
Specific regulators are present at different stages of the cell cycle, and should not be present for the other cell cycle stages, so rapid degradation is necessary
E3 is expressed at the end of each stage, which causes ubiquitination of proteins marking them for degradation
How could proteasome inhibition be a target for cancer treatment?
NF-kappa B is normally contained within the cell as an inactive complex with its inhibitor I-kappa B
When I-kappa B is phosphorylated, it dissociates from NFkB allowing NFkB to enter the nucleus and cause cell growth and survival.
I-kappa B is normally degraded by the proteasome
By blocking proteasome activity, we may be able to keep the concentration of IkB higher leading to less gene activation by NFkB
