Gene Expression III: Eukaryotic Post-Transcriptional Regulation

Question 1

Post-transcriptional control

Answer 1

All of the regulatory mechanisms that control gene expression following transcription. We can turn off a gene, but if we can’t get rid of the products of transcription, the mRNA and protein, then we haven’t really regulated anything.

Question 2

Identify the major steps leading from a gene to a functional protein.

Answer 2

The gene needs to be transcribed to mRNA, then the mRNA needs to be processed. Then the mRNA needs to be translated to protein. The protein is then processed to get correct folding.

Question 3

Identify post-transcriptional steps in gene expression that may be regulated.

Answer 3

The main ways are to regulate or intervene with mRNA processing control and alternative splicing. Capping and polyadenylation cleavage can prevent the mRNA from being translated. Can also look at the translatability of the mRNA as well as protein degradation. All of which will result in regulating the expression of that gene.

Question 4

Describe the major patterns of alternative RNA splicing

Answer 4

Alternative splicing is used to make isoforms or variant proteins from a single gene. There are many ways that it can occur and it is quite possible because out of the ~25,000 genes we can make ~100,000 proteins based on alternative splicing alone. Understand that the great variety of proteins can be produced from using alternative splicing. The 4 most common forms of splicing are 1) Exon skipping, 2) Alternative 3’ splice site selection, 3) Alternative 5’ splice site selection 4) Intron retention

Question 5

Exon Skipping

Answer 5

This is the most common form of alternative splicing (~40%) It is when an exon is skipped.

Question 6

Alternative 3’ and 5’ Splice Site Selection

Answer 6

Get extra material that will be part of the mRNA and product

Question 7

Intron Retention

Answer 7

This is where an intron is not spliced out and is kept in.

Question 8

Alternative Splicing of the fibronectin gene

Answer 8

Two variations of fibronectin can be produced via different splice patterns of the mRNA even though they are from the same gene! The cells produce specific guide factors to help it produce the correct form and alternative splicing for this particular mRNA.

Question 9

Alternative Splicing of the Axon Guidance Protein DSCAM in Drosophila

Answer 9

In the DSCAM gene, through alternative splicing, it can produce over 38,000 different splice variants. It is thought that DSCAM in humans is linked to Downs Syndrome.

Question 10

Explain how regulatory proteins aid in selection of splice sites

Answer 10

Alternative splicing can be regulated by proteins that either promote the use of a specific splice site or block the use of another. These proteins bind to regulatory sequences in the pre-mRNA called splice site enhancers and splice site silencers.

Question 11

Negative Control of Alternative Splicing

Answer 11

If there is a strong splice site, it will be used and we don’t need an SR protein to mark the exon-intron junction. In negative control, there is a silencer sequence in the intron or exon and a repressor protein will bind to it and discourage splicing. It is NOT a blockade by the repressor protein or that it is blocking the binding site. Rather, it is an encouragement.

Question 12

Positive Control of Alternative Splicing

Answer 12

Here there is a weak splice site that would not normally be used. But, by using an activator protein that binds to an enhancer sequence in the intron or exon, it tells it to splice.

Question 13

SR Proteins

Answer 13

They bind to the exon-intron junctions and influence splicing. They are serine and arginine rich proteins (hence SR). Through the expression of SR proteins certain cell know what to and what not to express.

Question 14

Explain how the regulation of polyadenylation site selection during the processing of the 3’ end of an immunoglobulin mRNA can result in the formation of membrane-bound and secreted forms of an antibody from the same gene

Answer 14

In a native B-Cell that has not seen antigen, it will produce antibody but the antibody will be bound to the membrane of the cell. When antigen becomes available it will bind to the antibody and activate the B-cell making it able to produce antibody that can now circulate and is not membrane bound. These membrane-bound and secreted antibodies are from the same gene. However, it is the choice of the polyadenylation site that chooses whether or not we will express a bound or secreted antibody. We have a DNA and we can see an intron with a weak polyadenylation signal and then one with a strong one downstream. When the cell uses the strong polyadenylation site, it will include some of the intron into the protein and that intron happens to be hydrophobic. This allows the cell to produce antibodies that are membrane bound. This is not regulation based on splicing! It is based on polyadenylation because we never reached the 3’ splice site, splicing didn’t even come into play. It can also choose the weaker polyadenylation site and produce a shorter mRNA product that does not have the hydrophobic region and thus is NOT membrane bound when high levels of CstF are produced. CstF allows it to choose the weak polyadenylation site.

Question 15

Describe the function of initiation factor eIF-2B in maintaining eIF-2 in its more active state, and explain the effect of eIF-2 phosphorylation on mRNA translation.

Answer 15

eIF-2B is a guanine nucleotide exchange factor that will catalyze the reactivation of the eIF-2-GDP to eIF-2-GTP which can now bind another tRNA. This is the “rate-limiting” step of translation initiation because it has to be reactivated before it can bind another tRNA. If the cell is stressed and wants to turn off protein synthesis, however, it will phosphorylate the eIF-2-GDP. Now, it will bind irreversibly to the eIF-2B thus creating a limiting amount of eIF-2B to reactivate the eIF-2 to continue initiation. When the conditions get better a phosphotase will remove the phosphate.

Question 16

Describe the effects of the phosphorylation of the eIF4E binding protein, eIF4E-BP, on mRNA translation and explain why eIF4E-BP can block the recruitment of the small ribosomal subunit to the capped end of the mRNA.

Answer 16

eIF4E is the cap binding protein and is associated with eIF4G. eIF4E is activated by growth factors and hormones resulting in phosphorylation of eIF4E-binding protein that is bound to the eIF4E and acts to inhibit eIF4E. eIF-4E-BP prevents 4E from interacting with 4G so now the cap binding protein cannot interact with eIF4G and the 4G is crucial for binding of the small ribosomal subunit. The eIF4E with the binding protein can still bind to the small ribosomal subunit, it is just that without eIF4G it cannot fully engage. Thus, a kinase would phosphorylate the binding protein so the inhibitor is now inactivated and it is released. No 4E and 4G can associate and we can get translation. Here, phosphorylation promotes translation.

Question 17

eIF4E

Answer 17

eIF4E, when expressed in high levels can also allow cells to overcome and express mRNAs with 5’ UTRs (untranslated regions). Usually there are oncogenes in this region and thus it is classified as a protooncogene because it can actually cause or be turned on in tumor cells.

Question 18

Describe the function of an Internal Ribosome Entry Site (IRES) in mRNA translation.

Answer 18

A small subset of eukaryotic mRNAs can be translated without a 5’ cap because they contain an IRES that allows the 40S ribosomal subunit to bypass the cap when forming a functional translation initiation complex with the corresponding mRNA. Inhibition of the cap binding protein eIF4E is thought to produce mRNAs with the IRES while inhibiting cap dependent translation. eIF4G (a modified version) can bind directly to the IRES and then the small ribosomal subunit can be recruited. IRES is also crucial when the cell needs to shut down protein synthesis but still needs a small subset of proteins “on”. Such an example is apoptosis.

Question 19

Predict the effect of phosphorylation and dephosphorylation of eIF4E-BP on the translation of mRNAs that contain an IRES.

Answer 19

If the eIF4E-BP is phosphorylated and thus allows for eIF4E to bind to the 5’ cap to initiate translation, then there would be less mRNAs produced containing and IRES to bypass this requirement. However, if the eIF4E-BP is dephosphorylated or not phosphorylated to begin with then we would have a lack of active eIF4E able to initiate translation. Thus, we would expect to see a higher amount of mRNAs produced containing an IRES.

Question 20

Explain the advantage to cells of unstable intermediates and products (e.g. mRNAs and proteins that are rapidly degraded)

Answer 20

If it is a protein that needs to be rapidly regulated, it will tend to have a high degradation rate (short half-life) because we want to be able to degrade it and fast! Also, we may need to make it fast and this would require the same!

Question 21

What are three pathways by which an mRNA can be degraded?

Answer 21

mRNAs can be degraded via 3 ways:

1) Deadenylation of the poly-A tail via exonucleases triggers its degradation
2) MicroRNAs (miRNAs) can bind to regions of the mRNA and can mark it for degradation
3) Nonsense-mediated mRNA decay

Question 22

Deadenylation of the Poly-A tail via exonuclease activity.

Answer 22

Deadenylation of the Poly-A tail via exonuclease activity triggers the mRNAs degradation. Initially, the poly-A tail is slowly degraded via exonuclease activity. Once it is degraded to its “critical length” (~25-30 nucleotides long) it recruits a decapping enzyme that decaps the mRNA and now the exonuclease can rapidly degrade the mRNA from both sides. The stability of an mRNA depends on how long its poly-A tail is and how quickly it is shortened. If a message is being actively translated there won’t be deadenylation occurring. But, when there is free mRNA waiting to be translated, the deadenylases will come in and start degrading the poly-A tail. Messages that are not being used will be targeted for degradation.

Question 23

MicroRNAs (miRNAs) and mRNA Degradation

Answer 23

miRNAs are involved in regulating gene expression. miRNAs are encoded within introns or the materials between genes. They will be capped and polyadenylated. What happens in this process is that the miRNA will first be cleaved in the nucleus and then it will enter the cytoplasm where DICER will come in and cleave it down to a double stranded RNA. Then, one of the two of the RNA strands will be chosen as the guide strand and the other will be degraded. Then RISC binds to the guide strand and this miRNA can go and bind to a mRNA. Then, there are two things that can occur.

1) If only a few bases will be complementary (need 7) this seed sequence will still recognize mRNA and binds. Then it will be shuttled to a P-body where it will be slowly degraded.
2) If a large amount of sequences are recognized, then this mRNA will be rapidly degraded through an Argonaute Nuclease. This is less common.

miRNAs can actually regulate hundreds of mRNAs. There have been over 2,500 miRNAs discovered and each can recognize multiple mRNAs (regulate 20-30% of human genes). Also, a variety of miRNAs have been shown to be involved in cancer when they are not regulated correctly.

Question 24

Nonsense-mediated mRNA decay

Answer 24

Nonsense-mediated mRNA decay is a special pathway for eliminating mutant or aberrant mRNAs that contain premature stop codon. It is very important for the cell to not translate these mRNAs so there are mechanisms to deal with this. With this, the pre-mRNA has an intron that contains a stop codon. In normal splicing, the intron would have been spliced out. Once splicing occurs accurately, EJCs (exon junction complexes) mark the exon junctions. When mRNA exits the nucleus it undergoes a test round. of translation to make sure it is correct. The ribosome will recognize EJCs and removes them. If there was incorrect splicing, and now there is an internal stop codon, when the test round occurs, the cell sees a STOP codon and an EJC which can’t normally occur! EJCs should be before the stop codon not after. Thus, Upf proteins are recruited and bind to EJCs to trigger degradation of the incorrect mRNA.

Question 25

Describe the relationship that exists between the translation and degradation of an individual mRNA

Answer 25

This ties in with the competition of the mRNA for translation vs. degradation. If an mRNA is being translated, it will not be degraded. However, if it is just floating there, it will be degraded slowly.

Question 26

Describe the source of miRNAs and their assembly in an RNA-induced silencing complex (RISC) complex

Answer 26

miRNAs are usually located within introns. They are then cleaved by DICER and form single strands of RNA that will associate with Argonaute and other proteins to form RISC. The RISC with the miRNA will then scan and complementary bind to the mRNA to regulate it.

Question 27

Describe how a RISC identifies the mRNA it regulates

Answer 27

the miRNA has a complementary sequence to the mRNA.

Question 28

Describe the ubiquitin conjugation system and the role of ubiquitin in the selective degradation of misfiled proteins and short-lived (regulatory) proteins

Answer 28

Ubiquitin tags proteins for degradation by an ATP-proteosome. Ubiquitin is a small 8.5 kilodalton protein (76 amino acids) present in all eukaryotic cells. The carboxyl-terminal glycine of ubiquitin becomes covalently attached to the epsilon-amino group of lysine residues (isopeptide bond) in proteins destined to be degraded. Additional ubiquitin molecules are then added to lysines to produce a linear chain of ubiquitin units attached to a target protein. The polyubiquitinated protein is then degraded by a proteosome. The attachment of ubiquitin requires ATP and is catalyzed by three enzymes 1) E1 (ubiquitin activating enzyme) 2) E2 (ubiquitin conjugating enzyme) and 3) E3 (ubiquitin ligase). The energy from ATP causes ubiquitin to be added to the E1 activating enzyme which then transfers ubiquitin to E2. E2 then binds to the E3 and the substrate and then place the ubiquitin on the substrate. This process occurs over again.

Question 29

Ubiquitin

Answer 29

Ubiquitin tags proteins for degradation by an ATP-proteosome. Ubiquitin is a small 8.5 kilodalton protein (76 amino acids) present in all eukaryotic cells. The carboxyl-terminal glycine of ubiquitin becomes covalently attached to the epsilon-amino group of lysine residues (isopeptide bond) in proteins destined to be degraded.

Question 30

E1

Answer 30

ubiquitin activating enzyme

Question 31

E2

Answer 31

ubiquitin conjugating enzyme

Question 32

E3

Answer 32

ubiquitin ligase

Question 33

Describe the function of E3 Ubiquitin Ligases in the recognition and ubiquitination of protein substrates.

Answer 33

The E3 Ubiquitin Ligases are what bind to the substrate and recognize the protein that needs to be degraded. They then recruit the E2 with the ubiquitin to tag the protein.

Question 34

Identify Properties of the Proteosome that distinguish it from other proteases and identify properties that allow it to distinguish between substrate and non-substrate proteins

Answer 34

The proteasome is composed of a central 20S cylinder supplemented with 2 19S caps. There are then proteases in the cylinder. The caps are important in regulating what will be allowed in for degradation. A unubiquitinated protein will not be able to bind to the cap and thus not be allowed in for degradation. however, a ubiquitinated protein can bind to the cap. Thus it will be unfolded and threaded into the cylinder where the proteases are to digest it. The ubiquitin is then recovered and recycled. It is specific for ubiquitinated proteins where other proteases can just digest proteins. Cell cycle proteins are quickly degraded via this system

Question 35

Explain the function of the ubiquitin-proteosome system in common neurodegenerative disorders like parkinson’s and Alzheimer’s disease.

Answer 35

There is a mutation of the E3 ligase and thus it cannot bind and recognize the protein to degrade it. Get accumulation of the protein.

Question 36

Describe the rationale for using proteasome inhibitors in the treatment of cancer

Answer 36

Some cells have I-Kappa B and NF-Kappa B. The I-Kappa B will bind to the NF-Kappa B to inhibit it. NF-Kappa B promotes cell growth and survival (most likely not regulated in cancers then). Inhibiting the I-Kappa B degradation maintains NF-Kappa B in an inhibited state. The result would be that some cancer cells would undergo apoptosis and other create immune responses. If I-Kappa B is phosphorylated, it will release the NF-Kappa B and allow NF-Kappa B to promote cell growth and survival.