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Discuss how eukaryotic genes differ from bacterial genes. What aspect of translation (protein synthesis) in eukaryotic cells accounts for the observation that most eukaryotic mRNAs encode a single protein? What mechanism allows a single eukaryotic gene in metazoans to encode two or more related proteins?

The Cap is recognized in the cytoplasm by special initiation factors to allow assembly of a ribosome at the mRNA initiation site. Because the initiation of translation involves Cap recognition, each protein in eukaryotic cells is encoded on its own mRNA. The capped base of an mRNA corresponds to the start site of transcription of the precursor RNA from which it is processed.

Monocistronic Most eukaryotic genes are monocistronic, meaning their mRNAs encode a single protein. Often, a eukaryotic primary transcript forms a single mRNA that encodes a single protein. Most eukaryotic mRNAs have a 5′ cap structure that directs ribosome binding, with translation beginning only at the closest AUG codon.
Polycistronic Prokaryotic genes are mostly polycistronic, with one mRNA encoding multiple proteins involved in a biological process. Along the mRNA, there is a ribosome binding site near each coding region’s start site. Translation can initiate at any of these sites, allowing production of different proteins from one mRNA.

From the article on genes.

Besides transcription initiation, what other step in the transcription of a gene is regulated for some specific genes? Using the example of HIV, describe how the HIV Tat protein stimulates transcription of the integrated HIV genome.
Other than initiation, sometimes elongation is activated. Transcription of viral HIV TAR from the LTR promoter by the T cell RNA Polymerase II will stop, and only resume after: the HIV Tat protein has bound the TAR RNA; Cyclin T/CDK9 then binds the Tat protein; and the Pol II CTD is then phosphorylated.
Mobile elements gave rise to the most abundant repetitious DNA sequences in the genomes of multicellular organisms. Describe three different classes of this type of reiterated sequence and the mechanisms by which they have become so widespread in our genomes. Use diagrams.
A transposon (a kind of interspersed repeat) moves as shown below, and during S phase it can hop from a daughter strand into the parent strand and thereby increase its copy number.

A retrotransposon (another kind of interspersed repeat) is transcribed normally, then reverse-transcribes to form dsDNA that integrates the same way that a tansposon does: a staggered nick; insertion; filling in of the 5′ and 3′ ssDNA. The upstream LTR acts as a promoter and the downstream LTR contains a poly-A site to produce transcripts from the integrated element. Between them is a coding region that encodes proteins for transposition and also acts as a copy template.

Lastly, a non-LTR retrotansposon is the same as an LTR retransposon but has a different method for integration.

Mobile elements are thought to have had a profound influence on the evolution of genomes of multicellular organisms. How are mobile elements thought to have influenced the evolution of contemporary protein coding regions and transcription control regions? Give examples of mechanisms relating to mobile elements in general and to two different classes of mobile elements. Use diagrams. (You do not need to discuss or diagram mechanisms of transposition or retrotransposition.)
Mobile elements can shuffle exons around in a phenomenon called exon shuffling. Exon shuffling.

With non-viral retrotransposons, if a LINE has a weak poly(A) signal, then sometimes transcription will continue and include an adjacent 3? gene (eventually terminating at that gene’s strong poly(A) signal). ORF2 then reverse-transcribes the RNA transcript of the LINE and gene, eventually inserting the gene at a new location along with the SINE in a phenomenon known as exon shuffling.

Exon shuffling can occur double crossover between interspersed repeats.

Also, via DNA transposons.

How does the mechanism of integration of LTR retrotransposons differ from the mechanism of integration of non-LTR retroransposons? What mechanism generates the short direct repeats of the integration site sequence at each end of retrotransposons?
An LTR retrotransposon integrates in the same manner as a transposon:

A non-LTR retrotransposon, conversely, integrates as follows:

What is the basic building block of chromatin? What is it composed of? What parts of these basic building blocks are subject to post-translational covalent modifications to control the condensation of specific regions of chromatin? What influence does the degree of chromatin modification have on gene expression? What experimental method can be used to determine if a specific gene is present in condensed chromatin or decondensed chromatin in a specific type of cell? Briefly describe the method.
Structure Overview
Primary Two each of each histone ( H2A, H2B, H3 and H4, except H1) interlock to form a disc-shaped structure ~10 nm in diameter. In all eukaryotes, 147 bp of DNA wraps almost twice around the histone octamer to form the nucleosome. Nucleosome “beads” are each spaced by 15-55 bp (depending on species) “strings” of linker DNA. The basic and positively-charged histones bind tightly to DNA, protecting it from proteins. Careful nuclease treatment will digest linker DNA and release individual nucleosomes with their DNA still intact. However, linker DNA is somewhat protected by bound H1 and by inter-histone interactions. Newly replicated DNA in vivo assembles into nucleosomes shortly after the replication fork passes, nucleosomes do not spontaneously form in vitro at physiological salt concentration when histones are added to DNA. However, nuclear proteins have been characterized that bind histones and assemble them in vitro with DNA. These are thought to assemble new DNA into histones in vivo as well.
Secondary Nucleosomes (the 1° chromatin structural unit) will stack on top of each other at physiological conditions (~0.15 M KCl), and the stacks will then intertwine into an irregular spiral (solenoid arrangement) that is ~30 nm and contains ~6 nucleosomes per turn. H1, the 5th major histone, is bound to DNA on inside of solenoid with one H1 molecule associated with each nucleosome. The 30-nm solenoid is less uniform than a perfect solenoid; condensed chromatin may actually be quite dynamic, with regions occasionally partially unfolding and then refolding into a solenoid structure.
Tertiary At special scaffold attachment regions (SARs), the 30 nm fibers attach to a flexible protein scaffold; the unattached regions form chromatin loops. DNA can be released from the protein scaffold by treatment with detergent. In addition to this general structure, thousands of low-abundance regulatory proteins associate with specific DNA sequences. During meiosis, chromatin further folds and compacts into visible metaphase chromosomes.

DNAse sensitivity is an effective way to determine if a gene is condensed or decondensed form. A condensed gene will be more resistant to DNAse than a gene that is not condensed. Afterward, decondensation of the entire genome and PCR analysis will reveal what has and has not been digested.

Most human cells are diploid and contain the 22 types of human chromosomes and either two X-chromosomes (in females) or one X and one Y chromosome (in males). How many DNA molecules are in the nuclei of most human cells? How are these DNA molecules physically distributed in a cell nucleus during interphase? What accounts for the structure of the polytene chromosomes observed in some insect tissues?
There are often 46 DNA molecules, two for each chromosome. During interphase, these DNA molecules are diffused into loose non-overlapping regions within the nucleus. Polytene chromosomes arise the following way:

What experimental strategy using the yeast S. cerevisiae led to the functional characterization of the three critical elements of a eukaryotic chromosome required for normal chromosome replication and segregation? State what these three critical functional elements are and how the function of each one can be assayed or observed.
A replication origin (ARS), centromere and telomeres. These were identified using the yeast experiment:

Progeny of Transfected Cell
Plasmid Recipient Growth Mitotic Segregation Observation
LEU+ Circular LEU- Yeast None Transfection with a LEU+ plasmid does not alone restore LEU to a LEU- cell.
LEU+ ARS+ Circular LEU- Yeast Some Poor Replication occurs, but poor segregation means only ~10% of progeny carry the plasmid.
LEU+ ARS+ CEN+ Circular LEU- Yeast Yes Good A centromeric (CEN) genome fragment is needed for strong segregation.
LEU+ ARS+ CEN+ Linear LEU- Yeast None Linearization (via restriction enzymes) of a TEL- circular plasmid makes it unstable.
LEU+ ARS+ CEN+ TEL+ Linear LEU- Yeast Yes Good Linear plasmids must carry the telomeric (TEL) gene fragment at each each end to remain stable in progeny cells.
What problem does the enzyme telomerase overcome? How does it accomplish this?
DNA replication requires an RNA primer to initiate synthesis, which is degraded after priming. The loss of these primers on the lagging strand of the chromosome ends will result in a loss of information with each round of replication. Telomerase is a special enzyme that uses its own RNA template to add telomeric repetitive DNA to chromosome ends.
How do we know that transcription of a gene most often is what determines whether the encoded protein will be synthesized in a specific type of cell, say albumin in a liver cell or a neurotransmitter receptor in a neuron? Describe the strategy of an early experiment that proved this point.
If a protein is not expressed, its corresponding gene rarely is transcriptionally active. Thus, control of protein synthesis is at the level of transcriptional control.

The cDNA microarray experiment identified gene regulation at the transcriptional level. cDNA is single-stranded DNA that has been reverse-transcribed in vitro from cellular mRNA. By soaking a cDNA microarray with labeled RNAs, it is possible to determine which genes are being transcribed in a particular cell; only active genes were being transcribed. Thus, regulation is at the transcriptional level.

What are the two most important functional domains of a eukaryotic transcriptional activator protein? What aspect of an activator’s structure helps to explain why eukaryotic DNA sequence control elements function together even when the number of base pairs between the control elements is altered? What kinds of protein complexes do activation domains interact with? How do these protein complexes function to stimulate transcription?
The two most important elements are the: DNA-binding domain; and activator domain. The flexible protein domain allows the base pairs between DNA sequence control elements to be altered. Activation domains interact with coregulatory proteins, which can merely stabilize binding of the transcription factor (and even RNAP II) to the DNA or can even be bound to enhancer elements (DNA sequences) near or far, upstream or downstream that then interact with the mediator complex that recruits RNAP II in a productive manner.
How do enhancer elements function to stimulate transcription? What kinds of proteins are required for them to function? How are they thought to control transcription form a promoter even when they are 50 kb away from the promoter in the genome sequence?
Enhancers do not act on the promoter region itself, but are bound by activator proteins. These activator proteins interact with the mediator complex, which recruits polymerase II and the general transcription factors which then begin transcribing the genes. Enhancers can also be found within introns. An enhancer’s orientation may even be reversed without affecting its function. They can be so far or near because the genome can bend around, bringing the enhancer-bound activators close to the mediator complex.
Purified RNA polymerase II cannot initiate transcription from a promoter without the assistance of other proteins. What is the general term for the proteins that bind to a promoter along with RNA polymerase II and help the polymerase to initiate transcription? What are the specific names of these proteins? How does a TATA-box specify a transcription start site? How does an initiator element specify a specific start site?
The RNA polymerase II General Transcription Factors (Pol II GTFs): (TFIIA); TFIIB; TFIID; TFIIE; TFIIF; TFIIH. These are required along with Pol II to transcribe virtually all eukaryotic protein coding genes. Within TFIID is a TATA-box binding domain that binds the TATA-box; next, TFIIA binds upstream to TFIID and then TFIIB downstream to TFIID. This multi-protein complex binds RNA Pol II (forming the basal transcription complex) and transcription initiates. In genes without a TATA box, an initiator element binds TBP instead but without sequence specificity.
Describe a biochemical method for analyzing the extent of acetylation of histones in nucleosomes associated with a specific gene of interest. How was this method used to help understand the mechanism by which eukaryotic repressors inhibit transcription?
Step Overview
Isolation Chromatin is isolated from cells that were treated with formaldehyde to cross-link proteins to DNA.
Shearing Chromatin is mechanically sheared .
Labeling This sheared chromatin is mixed with an antibody specific for acetylated (unrepressed) N-terminal histone tails.
Precipitation Nucleosomes with their antibody-bound histone tails are immunoprecipitated.
Release Immunoprecipitated DNA is released and examined by PCR, then run on a gel; bands indicate gene of interest was acetylated..
Conclusion A good positive control — just to verify your gene of interest is present — is to run PCR on sheared chromatin without immunoprecipitation; your gene should undergo PCR regardless of its acetylation state. A good negative control is to immunoprecipitate with antibody from non-immunized animals, like a rabbit or mouse. Lastly, you can assay for active transcription of a gene by immunoprecipitating with an antibody for RNA Polymerase II.
Describe three mechanisms by which the activity of a transcription factor can be regulated?
Synthesis Regulation of TF synthesis (transcription of the TF gene).
Activity Regulation of transcription factor (TF) activity (activators and repressors). Regulation of TF activity by interaction with small molecules (ligands), and post-translational modifications, especially phosphorylation and dephosphorylation. Regulates Nuclear transport, import and export; DNA binding to cognate DNA site; Interactions with co-activators.
Degradation Regulation of TF degradation.
What mechanism can account for the fact that RNAs transcribed by RNA polymerase II are capped and usually polyadenylated, while these types of RNA processing do not occur on RNAs transcribed by RNA polymerase I or III? (Be sure to discuss a distinguishing feature of RNA polymerase II and the function of general transcription factors in this process).
To ensure that only RNAP II transcripts are capped, the RNAP II CTD is phosphorylated by TFIIH, and the phosphorylated CTD recruits capping enzymes. After ~25-30 nucleotides have been transcribed and the nascent pre-mRNA is emerging from RNAP, the first transcribed nucleotide (the 5′ end) is bound via a 5′-5′ triphosphate linkage to a 7-methylguanosine. The 2’-hydroxyls of the first two transcribed nucleotides are also often methylated.

The Cleavage/Polyadenylation reaction is carried out the polyadenylation complex, a large multi-protein complex assembled from CPSF, CStF, CF1, and CF2. Some pre-mRNA’s have multiple poly(a) signals, and cleavage and polyadenylation at these different signals can include or exclude various 3′ terminal exons. Ser2 on the CTD is phosphorylated

Step Overview
Upstream Binding Upstream AAUAAA bound by Cleavage and Polyadenylation Specificity Factor (CPSF; 4 subunits).
Downstream Binding Downstream G/U rich element is bound by the Cleavage Stimulatory Factor (CStF; 3 subunits).
Additional Factors RNA cleavage requires Cleavage Factors (CF1 and 2); the endonuclease is the 73kd subunit of CPSF.
Binding of PAP The PolyA polymerase (PAP) binds, activating cleavage of the 3′ end.
Fragment Digested The free cleaved 3’ piece is rapidly degraded by 5’-3’ exonucleases.
Start of Poly(A) Tail Poly(A) Polymerase slowly adds ~12 adenosine residues to the 3’ end of the 5’ product RNA.
Poly(A) Binding Prtn The 12 adenosine residues are a binding site for PolyA Binding Protein (PABPII).
PAP Stimulation Bound PABPII stimulates PAP to quickly add ~250 adenosines, forming a poly(A) tail bound by many PABPII’s.

TFIIH is a large General Transcription Factor (GTF) of 9 subunits, nearly as large as Pol II. Two of the subunits are homologous to DNA helicases, enzymes that use energy from ATP hydrolysis to separate the strands of a DNA double helix. One of the subunits is a kinase that phosphorylates Ser5 of the Pol II large subunit C-terminal domain (CTD) heptapeptide repeat (Tyr-Ser-Pro-Thr-Ser-Pro-Ser).

What signals (i.e. RNA sequence) in a pre-mRNA specify the site of polyadenylation? How is the site of polyadenylation related to transcription termination by RNA polymerase II? What is the mechanism that controls the change in the polyadenylation site of immunoglobulin heavy chain µ pre-mRNA as immature B lymphocytes develop into IgM secreting plasma cells?
Step Overview
Upstream Binding Upstream AAUAAA bound by Cleavage and Polyadenylation Specificity Factor (CPSF; 4 subunits).
Downstream Binding Downstream G/U rich element is bound by the Cleavage Stimulatory Factor (CStF; 3 subunits).

Coupling 3′ cleavage & transcription termination:

Theory Overview
Antitermination Model The polyA sequence is recognized as it leaves the polymerase and causes loss of elongation factors and/or gain of termination factors.
Torpedo Model The free 5’ end after cleavage of the polyA site recruits exonucleases that rapidly degrade the RNA and induce release from the template when they reach the polymerase.

B lymphocytes carry IgM, and upon exposure to their complementary antigen become antibody-secreting plasma cells. This switch is accomplished by a change in the IgM mRNA poly(A) site. There are two poly(A) sites, with the upstream site requiring more CStF. In B cells, CstF64 levels are low; in plasma cells, CStF64 expression is unregulated.

What unusual chemical linkage in RNA is generated during the process of pre-mRNA splicing? What are the nucleotides in the RNA that participate in this unusual chemical bond? What kind of chemical reaction results in the splicing of exons in pre-mRNA? How many of these reactions occur when one intron is spliced out of a pre-mRNA?

1st Transesterification

Within the intron, the 2′ hydroxyl of the branchpoint A attacks the 5′-end phosphate, releasing the 5′ exon and forming an intron lariat. The new phosphodiester bond of the A nucleotide makes it a branched nucleotide bound to its attack-ee and two adjacent bases.

2nd Transesterification

The 3′ hydroxyl of the detached 5′ exon attacks the phosphate at the 3′ end of the intron. This causes release of the intron lariat and ligation of the exons.

rna splicing lariat formation transesterification branchpoint

The observation that the sequence near the 5’-end of U1 snRNA is complementary to the consensus 5’ splice site sequence of pre-mRNAs led to the hypothesis that the U1 snRNP is the cellular molecule that recognizes 5’ splice sites during splicing. Describe an experiment that proved that this base pairing is indeed required for RNA splicing. What term is used to refer to this type of experimental approach to test the functional significance of base pair interactions between RNAs?
This is a complementary mutation experiment. It involves making a mutation to eliminate a function, followed by reversing the mutation restores function. This identifies conserved sequences needed for functionality. For example, the 5′ splice site was formed with a point mutation and splicing activity was lost; the mutation was reversed and splicing activity was restored. Obviously, splicing is regulated at this sequence.
Consensus splice site sequences occur several times in the long introns of many mammalian genes, yet the spliceosomal snRNPs and splicing factors splice the correct splice sites together with high fidelity. What is the current model for how the RNA splicing machinery identifies the correct splice sites? (Discuss the sequences in the pre-mRNA that are important in this process, the proteins that interact with those sequences, and the important snRNPs and splicing factors that function in the process.) What is a succinct phrase used to refer to this mechanism?

Exon Definition (aka Exon Recognitino) is performed by the Cross Exon Recognition Complex (CERC).

U1 snRNP Recognizes and interacts with pre-mRNA consensus 5′ splice site.
U2 snRNP Recognizes and interacts with Branch Point A.
SR Protein Serine-Arginine Rich Protein binds to exonic splicing enhancers.
U2AF Recognizes 3′ splice site AG (65 kDa subunit) and pyrimidine-rich region (35 kDa subunit)
Mutations in the splice sites of human genes generally induce exon skipping, but sometimes result in the use of alternative “cryptic” splice sites rather than exon skipping. What mechanisms can account for this? Alternatively, a mutation within an exon of the SMN2 gene involved in the disease spinal muscle atrophy results in the skipping of this exon in most Smn2 mRNAs. What mechanism can account for this?
Alternative RNA splicing is an important aspect of gene control in metazoans (multicellular animals). What two general molecular mechanisms can regulate the use of alternative splice sites in complex transcription units? Give an example of each type of mechanism.

Two general molecular mechanisms are to: repress exon inclusion (as Sxl does); and promote exon inclusion (as Tra does). Sxl sits on L3 and blocks assembly of spliceosome components, thus causing the entire exon to get skipped. Tra stabilizes the exon recognition complex by binding SR proteins, thus causing an exon to get included that otherwise would not have been.

Sxl sits on L3, forming a functional L1-L2-L4 Sxl protein. This Sxl sits on Tra’s 2nd exon and a function Tra protein is expressed composed of Exons 1 and 3. Rbp1 and Tra2 sit at the 5′ end of the 4th Dsx exon, promoting splicing at that region and thus inclusion of Exon 4 (to encode a functional Dsx protein).

Most proteins have a diameter that is much smaller than the diameter of the central transporter of nuclear pore complexes (NPCs). Yet proteins >30 kDa cannot diffuse through a nuclear pore? What is thought to prevent large proteins from diffusing through the central transporter of NPCs.
In the NPC, there are FG-nucleoporins that line the channel. FG-nucleoporins have long stretches of hydrophilic amino acids as well as hydrophobic sections called FG repeats composed of phenylalanine (F) and glycine (G). The FG repeats associate with each other to form a molecular web through which ions, small proteins, small metabolites and proteins less than 30 kDa can diffuse through.
Explain the current model for how large nuclear proteins are translocated through nuclear pore complexes into the nucleus. What proteins are critical for determining the direction of transport from the cytoplasm to the nucleus and how is this accomplished? What is the function of NTF2?

Transport is unidirectional because of rapid dissociation of importin. Ran-GTP is only hydrolysed to Ran-GDP in the cytoplasm, favoring release of Ran and binding of cargo.

Importin Importins α and β form a heterodimer together.
Importin α Binds to the basic NLS (nuclear localization signal) of the cargo.
Importin β Has low affinity interactions with FG domains in the nuclear pore.
Ran-GTP Upon reaching the nucleoplasm, importin binds tightly with Ran-GTP. The cargo is released.

Cytoplasm The importin/Ran-GTP complex diffuses back into the cytoplasm.
Ran-GAP Ran-GTP interacts with Ran-GAP while in the cytoplasm.
Ran-GDP Ran-GTP/GAP interactions convert Ran-GTP to Ran-GDP and cause release of importin.
Nucleus Ran-GDP diffuses back into the nucleus
Ran-GFP In the nucleus, Ran-GDP encounters Ran guanine nucleotide exchange factor (Ran-GFP) which converts Ran-GDP→Ran-GTP.

Nuclear transport factor 2 (NTF2) binds Ran-GDP and FG-nucleoporins to return Ran-GDP back to the nucleus.

Some hnRNP proteins that appear to be exclusively nuclear by fluorescently staining fixed cells with antibodies (immunofluorescence) actually shuttle rapidly between the nucleus and cytoplasm, spending most of their time in the nucleus. What experimental approach led to the discovery of this type of nuclear-cytoplasmic shuttling of hnRNP proteins? What was done to prevent the observation of newly synthesized protein during the course of the experiment?
How can you tell if it shuttles if it is so rapid? Well, you fuse the labeled cell to an unlabeled cell. But you need to block new protein synthesis, because new proteins may shuttle into the cytoplasm. A good chemical inhibitor of transcription is cyclohexamide. Then if the unlabeled nucleus exhibits fluorescence then you know that the hnRNPs have shuttled out and made their way over.
Some mRNAs in the cytoplasm of egg cells are not translated until the egg is fertilized. The same process regulates the translation of some mRNAs in the dendrites of neurons so that they are not translated until that dendrite receives synaptic input from an associated axon of another neuron. What mechanism is responsible for this form of translational control? How does it operate?

Translation can be controlled by cytoplasmic polyadenylation. Certain mRNAs are made to have short poly(A) tails and thus no translation initiation. These are localized at synapses, far away from that cell’s nucleus. Synaptic activity will stimulate polyadenylation in the immediate region of the synapse.

This mechanism has been determined for the egg, and is likely similar for neurons:

Translational Dormancy The dormant state is maintained by binding of cytoplasmic element binding protein (CPEB) to the cytoplasmic polyadenylation element (UUUUAU) which causes polyadenylation to halt. CPEB binds Maskin, which then binds EIF4E.
Translational Activation Phosphorylation of CPEB causes Maskin to fall off. All the poly(A) factors are now free to bind, extend the poly(A) tail and thus make the mRNA translationally active.
What is RNA interference? What are the two possible outcomes for an mRNA that is targeted by the pathway? What seems to determine one form of control or the other? What proteins are involved in this process? Which are involved in generating the small RNAs, and which mediate the targeting of those RNAs to a particular mRNA?

RNA interference pathways (RNAi) are mediated by small sequence-specific RNA molecules that shut down gene expression by one of three mechanisms: translational repression; mRNA degradation; and heterochromatin formation. Cleavage is a sort of rapid degradation by both 5′-3′ and 3′-5′ exonucleases. A targeted mRNA has two possible outcomes:

RISC Complex w/ siRNA Base-pairs extensively with target mRNA and induces cleavage. Argonaut is responsible for cleavage of target RNA when perfect base-pairing occurs. Argonaut has two domains: PAZ (binds end of miRNA) and PIWI (nuclease activity).
RISC Complex w/ miRNA Imperfect base-pairing with target mRNA results in inhibition of translation. If base-pairing is imperfect, several miRNA-RISC complexes must associate with the target mRNA. These inhibit translation, and the mRNA gradually becomes associated with P bodies that lead to its degradation by a unique and slow mechanism.

Dicer, which cleaves long dsRNA into short fragments of ~20 nucleotides. One of the two strands of each fragment, known as the guide strand, is then incorporated into the RNA-induced silencing complex (RISC) (promoted by Dicer).

A pre-miRNA consists of the RNA, Drosha and Pasha. Primary miRNA transcript cropped in nucleus by enzyme Drosha. Exported by exportin 5. dsRNA portion of this pre-miRNA is bound and cleaved by Dicer to produce the mature miRNA molecule that can be integrated into the RISC complex

Thus, miRNA and siRNA share the same cellular machinery downstream of their initial processing.

What is the role of the exon junction complex in determining whether an mRNA will be subject to nonsense mediated RNA decay (NMD)? What determines whether a termination codon will be interpreted as premature? Which of the general mRNA decay pathways are involved in NMD?
Human disease mutations that create nonsense mutations do not always produce a truncated protein. Often they lead to rapid mRNA degradation via Nonsense-Mediated Decay. During splicing in the nucleus, the Exon Junction Complex, containing 4 proteins, is deposited ~20 nt upstream from each exon/exon junction in the final mRNA. This marks the junctions and stays on the RNA as it is transported to the cytoplasm and goes through the first round of translation. The EJC is thought to be displaced from the mRNA during translation.

If translation terminates >50 nt upstream of an EJC, it is not displaced and signals for that mRNA to be degraded. Nonsense mutations generally lead to an early stop codon, so this means that most exonic nonsense mutation will be degraded. The EJC recruits a protein called UPF1 to the mRNA. This in turn stimulates binding of decapping enzymes and deadenylation, leading to rapid degradation. Nonsense mediated decay is advantageous because it prevents the production of truncated and potentially dominant negative mutant proteins in the heterozygous state.

Note that the normal termination codon is almost always in the last exon of a gene. Nonsense mutations in this exon do not lead to nonsense mediated mRNA decay and produce slightly truncated proteins.

Cyclin-dependent kinases regulate two important transitions in the eukaryotic cell cycle: entry into S-phase and entry into mitosis. Describe three general molecular mechanisms used to regulate the activity and substrate specificity of cyclin-dependent kinases to regulate these cell cycle transitions.

CDKs are controlled at every level: presence (controlled degradation); activity (inhibitors); and even specificity (cyclins). Examples are shown below:

Cyclins The specificity of cyclin-dependent kinases is regulated by their cyclins. For example, S. cerevisiae uses the same CDK throughout the entire cell cycle. However, different cyclins bind to the CDK and direct it to different substrates for phosphorylation.
Inhibitors S-cyclin+CDK expression is activated by G1-cyclin+CDK, but S-cyclin+CDK is immediately bound by the inhibitor Sic1. In late G1, G1-cyclin+CDK reaches a high enough concentration to marks Sic1 for degradation. In this manner, S-cyclin+CDK is inactive until the cell is ready for S-phase.
Degradation Mitotic-cylin+CDK (MPF) concentrations peak in metaphase, and its elimination must be very carefully regulated. Mitotic cyclins carry a conserved destruction box, which is bound by APC/C-Cdh1 as soon as all sister chromatids have separated.
Ubiquitin ligases also play a fundamental role in coordinating the eukaryotic cell cycle. What ubiquitin ligases are involved in controlling the cell cycle? How are their activities regulated?

SCF and APC/C are ubiquitin protein ligase complexes that that control three major transitions in the cell cycle: onset of S-phase through degradation of Sic1 by SCF; initiation of anaphase via degradation of securin by APC-Cdc20; and exit from mitosis via degradation of mitotic cyclins by APC-Cdh1. APC has several substrates that must be degraded at different times in the cycle; thus, its activity is directed by specificity factors that bind it. SCF only degrades Sic1 and thus its activity is regulated only by phosphorylation of its substrate.

Complex Overview
SCF Degradation of phosphoryated Sic1 or p27 to activate S-phase cyclin. SCF is a ubiquitin protein ligase needed for polyubiquitination and proteasomal degradation of phosphorylated Sic1. In contrast to the APC/C, the SCF ubiquitin-protein ligase is not regulated by phosphorylation or other modifications of specificity factors, but rather by phosphorylation of its substrate, Sic1.
APC/C The anaphase promoting complex/cyclosome (aka APC/C) is a E3 ubiquitin protein ligase that is bound by various specificity factors that direct it to degrade different substrates at different times in the cycle.

S.F. Overview
Cdc20 Cdc20 directs APC/C to degrade Securin, initiating anaphase. Induces partial degradation of mitotic cyclins.
Cdh1 Initiates telophase via: complete degradation of mitotic cyclins; or, in metozoans, by degradation of geminin. This allows formation of prereplication complexes on DNA replication origins. Inactivated by G1 -cyclin+CDK.
What kind of mutation in a mitotic cyclin can lead to entry into mitosis but failure to exit mitosis, i.e. failure to decondense the chromosomes and reassemble the nuclear envelope?
MPF (mitotic-cyclin+CDK) drives cells into metaphase but is not needed to continue the cell cycle. In fact, mitotic cyclins must be degraded in order for chromosomes to decondense. Mitotic cyclins all have a conserved sequence at their N-terminus (Arg-X-X-Leu-Gly-X-Ile-Gly-X) that is recognized by APC/C-Cdh1. If this sequence is mutated then mitotic cyclins will not be degraded. This will result in normal mitotic events until the sister chromatids separate in telophase, at which point the sister chromosomes will fail to decondense and the nuclear envelope will not reassemble.
How does the check point control of chromosome segregation prevent exit from mitosis until chromosomes have been brought to the proper locations during late anaphase?
How does the metaphase checkpoint prevent sister chromatid separation at the onset of anaphase until every kinetochore has become associated with spindle microtubules? (To answer this question thoroughly, you will have to describe what holds sister chromatids together at their centromeres and how this association is broken at the onset of anaphase.)

In the spindle assembly checkpoint (aka metaphase checkpoint), mitotic arrest deficient 2 (aka Mad2) blocks metaphase until every single kinetochore has properly attached to spindle microtubules. Mad2 exists in an open conformation (Mad2O) and a closed conformation (Mad2C).

  1. Mad1 and Mad2C form a tetramer that binds unattached kinetochores via the Mad1 subunit.
  2. Mad2C of the kinetochore-bound tetramer can transiently bind Free Mad2O.
  3. This transient interaction causes Mad2O to bind and inactivate Cdc20, and to convert from Mad2O→Mad2C.
  4. Mad2C-Cdc20 transiently interacts with additional free Mad2O.
  5. This causes Mad2O to bind and inactivate additional Cdc20, forming a second Mad2C-Cdc20.
  6. The cycle repeats and free Mad2O is quickly converted to Mad2C-Cdc20.
  7. Binding of microtubules to the kinetochore displaces Mad1-Mad2C tetramers.
  8. Free tetrameric Mad2C cannot bind Mad2O. Instead, free tetrameric Mad2C binds p31comet.
  9. p31 then binds the Mad2C of the Mad2C-Cdc20 complexes, resulting in release of active Cdc20.

Just a few Mad1-Mad2C tetramers bound to kinetochores can generate enough Mad2C-Cdc20 to overcome p31 activity. Once all kinetochores have attached to microtubules (thus releasing all Mad1-Mad2C tetramers), p31 activity predominates and active Cdc20 is released from Mad2C. The active Cdc20 binds APC/C, and the APC/C-Cdc20 degrades securin, the inhibitor of separase. Free separase digests the Kleisin subunit of cohesin, breaking open the Smc1-Smc3-Kleisin ring and allowing sister chromatids to separate. Cohesin is a Smc1/Smc3/Kleisin heterotrimer that holds together sister chromatids, with Kleisin acting like a clasp.

How is the dismantling of the nuclear lamina during prophase and its reassembly during telophase accomplished? Where does the nuclear envelope go during mitosis?
Long and fibrous lamin proteins form a layer of structural support for the nuclear envelope. Lamin is phosphorylated in prometaphase, causing a conformational change and the loss of laminal structural properties. Without laminal support, the nuclear membrane breaks apart and absorbs into the smooth endoplasmic reticulum. The endoplasmic reticulum breaks apart, but is bound to the lamin via the inner nuclear membrane Lamin B Receptor; and the lamin binds to chromatin.

As anaphase ends, dephosphorylation of existing lamin begins. Once the genetic material has fully segregated at the completion of anaphase, production of new lamin is well underway. The new lamin drags tubes of smooth endoplasmic reticulum across the surface of the chromatin; these tubes flatten and merge, forming a solid nuclear membrane. The endoplasmic reticulum and lamin detach themselves from the chromatin. In the mature daughter cell, the lamina is a continous layer that is bound to the inner membrane of the nuclear envelope by emerin proteins.

What mechanism accounts for the sudden onset of DNA synthesis in S. cerevisiae?
Active S-cyclin+CDK phosphorylates and activates proteins that initiate DNA synthesis at origins of replication. However, S-cylin+CDK is inhibited by Sic1 while S-phase cyclin and S-phase CDK is being produced. The inhibitor is then precipitously degraded by Late-G1-Cyclin+CDK. This de-repression unleashes a massive wave of active S-cyclin+CDK, as opposed to a slow rise in activity that would have occurred without the repressor. This permits the sudden activation of large numbers of DNA replication complexes and thus the sudden onset of DNA synthesis.
In eukaryotes, DNA replication at an origin initiates only one time during S-phase. What mechanism accounts for this? In proliferating cells, how is this block to re-initiation at a replication origin overcome in preparation for replication during the next S-phase?
The Origin Recognition Complex binds the origin of replication and then recruits replication proteins. Once S-cyclin+CDK activates the complexes, replication can commence. The DNA replication proteins are deactivated and/or fall off the genome, or actively move along the DNA (for example, the DNA Polymerase). These DNA replication complexes are not expressed so the DNA replication complex cannot reform. The problem of reassembling the pre-replication complex is resolved later via mitotic cyclins, which activate expression of DNA replication proteins but leave the DNA replication complexes inactive. Mitotic cyclins are quickly degraded during telophase and thus do DNA replication complexes are not expressed during S-phase.
Why do inactivating mutations in the RB (retinoblastoma) gene contribute to the generation of cancer cells? Why do mutations in INK4 genes contribute to the generation of cancer cells?
Rb binds to E2F and represses its activation function. Rb is deactivated upon phosphorylation by mid G1 cyclin-CDKs (and, eventually, late G1 cyclin-CDKs). E2F is a transcriptional activator for a litany of genes required for DNA synthesis. If Rb is mutated, then E2F is not properly inhibited and the genome replication can get out of control. Cancer cells proceed through the cell cycle very rapidly and thus need this sort of high levels of DNA synthesis.

G1 cyclin-CDK inactivates human Cdh1 and human inhibitors of cyclin-dependent kinase 4 and 6 (INK4s), important tumor suppressors that inhibit passage through G1 by inhibiting the mid G1 cyclin-CDKs. Both genes encoding INK4a are mutated in many human tumors, as they are less able to inhibit passage into G1.

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