Student Reader

Genetics and Genomics Questions

References: Molecular Cell Biology 5th Edition, H. Lodish et al., W. H. Freeman & Co., New York
Arizona State; PSU; Rutgers

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?

Alternative RNA splicing is an important aspect of gene control in metazoans (multicellular animals). Two molecular mechanisms can regulate the use of alternative splice sites in complex transcription units: repression of splicing (or negative regulation of splicing) and activation of splicing (or positive regulation of splicing). In Drosophila sex determination, the Sxl protein represses splicing of a splice site in the Sxl pre-mRNA. Or, Sxl protein represses splicing of a splice site in Tra pre-mRNA. Tra protein activates a splice site in the Dsx pre-mRNA.

What kind of sequences are analyzed in “DNA fingerprinting.” What mechanism is thought to account for the variation in these sequences between every individual. Use a diagram to describe the mechanism.

Simple sequence repeat sequences are highly conserved. However, their lengths are highly variable from person-to-person. Simple sequence repeat length variations are used in DNA fingerprinting. This variation arises from unequal crossing over during meiosis.

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.

The three classes of mobile repetitive sequences are DNA transposons, LTR transposons, and Non-LTR transposons. DNA transposons replicate by a cut-and-paste mechanism. When excised during S-Phase, it is replicated by DNA synthesis. LTR (long terminal repeat) retrotransposons transpose through an RNA intermediate utilizing reverse transcriptase derived from the retrotransposon itself. Long terminal repeats contain promoter, polyA site and important elements for replication and insertion. Genes encode reverse transcriptase and DNA integrase, but not envelope proteins necessary for virulence. Therefore, it remains intracellular. dsDNA of LTR retrotransposon integrtes like a DNA transposon: staggered cleavage of target DNA followed by ligation of 5' ends to transposon to fill in gaps.

This figure shows how the 5' end is copied from a tRNA bound to primer binding sites (PBS), then priming and extensinos at the 3' end to copy the body of the virus, then making a 2nd strand to copy 3' end of virus, this second strand reprimes at the 4' end to make a complete virus with both LTR's.
  • Non-LTR retrotransposons also transpose through an RNA intermediate utiliziing reverse transcriptase. Non-LTR retrotransposons also encode a reverse transcriptase but use a different mechanism for insertion. ORF1 = RNA binding Protein; ORF2 = Reverse Transcriptase and Endonuclease. Most LINE elements are truncated due to incomplete copying of element during insertion. This makes them inactive for transposition, but they can still be mutagenic upon insertion and can still induce aberrant recombination events. Short Interspersed Elements (SINE's) do not encode proteins but transpose by the same mechanism as LINE's, presumably using the LINE proteins. SINEs carry within them a promoter for RNA Pol III allowing new RNA copies to be made.
    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?

    Chromatin immunoprecipitation allows a researcher to determine what proteins are associated with a particular sequence of DNA in vivo. First cells are treated with a cross-linking agent like formaldehyde to cross-link proteins to DNA. Then, chormatin is isolated from the treated cells and sheared into fragments of two to three nucleosomes by sonication. The fragments are then immunoprecipitated with a specific antibody. The protein-DNA cross-linking is reversed, and the immunoprecipitated DNA is isolated. Finally, the DNA sequence of interest is assayed by PCR. This method was applied using antibodies specific for acetylated histone tails. It was found that repressed genes are associated with hypoacetylated histone tails.

    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.)

    There are two general kinds of recombination events mediated by transposable elements that have allowed the generation of new coding sequences and transcriptional regulatory regions: homologous recombination between repetitive elements during meiotic recombination, and transduction (carrying) of adjacent sequence during transposition. Homologous crossing over between different elements during recombination can lead to duplication of genes or duplication of exons within a gene. Diagram 1 below shows how this kind of event between LINE elements led to the duplication of Globin genes in the beta Globin cluster:

    Paternal chromosme:

    If repeated elements are present in the introns flanking exons in each of two genes. Double crossover recombination will lead to the swapping of exons between the two genes, as is shown for Alu elements in diagram 2 below:

    Other double crossover recombinants can simply move an exon into an intron of another gene.

    In the transduction mechanism, DNA transposons flanking an exon can act as one long transposable element and carry the exon to a new site in the genome. If this adds an exon into the middle of an intron from another gene, that gene can gain a new exon in its mRNA, as shown in diagram 3 below:

    Similarly, transcripts from a non-LTR retrotransposon can extend into the downstream DNA to be polyadenylated at a cellular 3' exon. This long transcript can be transposed to a new site, inserting a new 3' exon and polyA site into a gene, as shown in diagram 4 below:

    Also, simple Alu insertions can evolve into new exons with relatively few point mutations. This also leads to exon creation during evolution.

    All of these same mechanisms can act on transposable elements inserted near promoters and enhancers. In this way, regulatory elements can be added to other promoters and enhancers causing them to be up or down-regulated in new tissues or under new conditions.

    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?

    LTR retrotransposons are reverse-transcribed into double stranded DNA in the cytoplasm. A tRNA functions as the primer for reverse transcription. The double-stranded DNA is then imported into the nucleus with the integrase protein that functions like the transposase of DNA transposons to integrate the ends of the linear retroviral DNA into chromosomal DNA. The reverse transcription process generates LTRs at each end of the integrated retroviral DNA. The upstream LTR functions as a transcription control element ("promoter" is acceptable here). The downstream LTR functions as the polyA signal for the full-length retroviral RNA.

    Non-LTR retrotransposons are bound by the ORF2 protein at their polyA-tail in the cytoplasm and imported into the nucleus as RNA. Once in the nucleus, the ORF2 protein cleaves chromosomal DNA in a T-rich region. The T-rich DNA strand hybridizes to the polyA tail of the retrotransposon RNA and primes synthesis of the first strand of retroviral DNA. Chromosomal DNA serves as the primer for reverse transcription. The promoter for non-LTR retrotranposon transcription is internal to the non-LTR retrotransposon DNA. The weak polyA site is also internal.

    For both types of retrotransposons, the duplication of sequence at the insertion site results from a staggard cut in the double-stranded DNA at the integration site. Synthesis of the complementary strands of the DNA at the insertion site results in the short direct repeat of the target site sequence at each end of the inserted retrotransposon.

    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.

    The basic building blocks of chromatin are nucleosomes attached to each other by 15-55 kb linear DNA. Nucleosomes are made of 147bp of DNA wrapped 2.5 times around histone octamer proteins (2 copies each of H2A, H2B, H3, and H4). Histone proteins have the structure shown in the image to the left. The histone tails are made of flexible 11-37 amino acid sequences. The H2A N-terminal tails are on the bottom, and the C-terminal tails are on top. Each H2A has a flexible C-terminal tail. H3 and H4 histone tails can be reversibly acetylated and deacetylated. When acetylated, their positive charge is neutralized and they no longer interact with DNA phosphate groups. Acetylation of histone N-termini therefore prevents formation of condensed 30nm fibers. Methylation maintains the postive charge, encouraging condensed 30nm fiber formation. Arginine can also be methylated. Serine and threonine can be phosphorylated, conferring a negative charge. 76-amino acid ubiquitonin protein can also add intself to some lysines, causing the histone to reconfigure.

    Southern Blotting following digestion can assay whether a gene is in condensed or decondensed chromatin as follows:

    • Isolated nuclei are digested with DNase I, which cleaves DNA at random location.
    • Condensed DNA is resistant to DNase, while decondensed DNA is not.
    • The DNA is removed from chromatin protein and digested completely with a restriction enzyme.
    • The DNA is analyzed by Southern blotting.
    • A gene protected from DNase yields characteristic fragments, while a gene digested by DNase will yield fragments too tiny to detect with Southern blotting.

    The basic building block of chromatin is the nucleosome. Chromatin is composed of a histone octomer of two subunits each of histones H2A, H2B, H3, and H4 plus 147 bp of DNA. The N-terminal histone tails are subject to post-translational modifications that control the condensation of specific regions of chromatin. Transcription is usually inhibited in chromatin with hypoacetylated histone tails. Chromatin with hyperacetylated histone tails is permissive for transcription.

    Assaying the DNase I sensitivity of a gene of interest can indicate whether a gene is assembled into condensed or decondensed chromatin. Nuclei are isolated from the cells being studies and incubated for varying periods of time with a constant amount of DNase I. The DNA is then isolated, cleaved with a restriction enzyme, and subjected to Sourthern blotting using the gene of interest as probe. Restriction fragments from genes in decondensed chromatin are lost at earlier time points in the DNase I digestion than restriction fagments from genes in condensed chromatin.

    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 is one DNA molecule per chromosome. Since there are 44 autosomal and 2 sex chromosomes, that makes 46 chomosomes = 46 DNA molecules. The DNA molecules are one component of a chromosome which fills a territory, or small domain in the nucleus that does not contain DNA from another chromosome. The DNA replicates many times without separating because there is no replication at the centromere or telomeres.

    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.

    The three functional elements required for correct replication and segregation of chromosomes are (a) replication origis, where DNAP's and other proteins initiate synthesis of DNA (b) centromeres, which is where the chromosomes separate during division and (c) two ends, or telomeres, to ensure the DNA remains intact during successive cell divisions. (page 434) Yeast transfection experiments were used to identify these three functional elements. The order of the steps is imperative.

    • Leu- cells are transfected with Leu+ plasmids. If the plasmid is maintained, the yeast are tranformed to Leu+ cells and form colonies on Leu- media.
    • Autonomous Replication Sequences (ARS) were identified because when inserted into a Leu+ plasmid, a high percentage of transfected cells were transformed to Leu+. However, due to poor segregation in meisosis, the plasmid did not appear in all daughter cells.
    • A yeast genome cDNA library was made with Leu+ ARS plasmids. Some of these plasmids transformed the cells and large colonies formed. This indiciated good segregation. The cloned sequences were found to be from the yeast centromere and were labeled CEN sequences.
    • Linearized LAU ARS CEN plasmids did not transform Leu- yeast. Addition of telomerase sequences (TEL) to the 3' ed resulted in big colonies. The linearized plasmid behaved like a chromosome in meiosis and mitosis.
    What problem does the enzyme telomerase overcome? How does it accomplish this?

    Enzyme telomerase overcomes the problem of shortening chromosomes by adding telomeric sequences. Telomeric sequences are repetitive oligomers with a high G content in strand with its 3' end at the end of the chromosome. (page 436) All human DNAP's elongate DNA chans at 3' end and all require an RNA or DNA priemer. As the replication fork approaches the end of the linear chromosome, synthesis of leading strand continues to the end of DNA template strand to complete one daughter DNA double helic. However, the lagging strand template is copied discontinuously and is not entirely replicated. When the final RNA primer is removed, there is no upstream strand onto which DNAP can bind to fill resulting gap. The daughter DNA from the lagging strand would shorten with each cell division (resulting in a detrimental loss of genes) without telomerase sequences. Telomerase elongates telomerase sequences by GOOGLE.

    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.

    Regulation is crucial because (a) gene control allows a single cel to adjust to the environment so growth and division is optimized and (b) resources are conserved. Regulation at the transcriptional level is logical because resources are not wasted producing mRNA that will not be translated. The experiment to verify transcriptional regulation predominates uses nascent-chain analysis (aka run-on transcription analysis) (page 449):

    • Isolated nuclei incubated with 32P ribonuclease triphosphates for ≤5 mins.
    • RNAP incorporates labeled nucleotides into nascent (growing) RNA chain.
    • Total radioactive label indicates overall transcription rate. Hybridize the labeled RNA to a membrane containing cloned DNA of genes of interest.
    • It is important to have a positive control (tRNA gene) and negative control (pV gene).
    • It was found that proteins specific to a certain cell type are not transcribed in other cell types.
    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 domains are the DNA-binding domain and the activation domain. This was discovered when these two domains were fused directly together and were still able to work in vivo. Also, flexible domains between activation and DNA-binding domains may explain why altered spacing between DNA elements is well tolerated. The DNA binding domain binds to specific DNA sequences, while the activation domain interacts with other proteins to stimulate transcription from a nearby promoter. These proteins can bind to each other or to RNAPII. THere are 4 common kinds of DNA-binding domains (page 464-5):

    • Homeodomain, a 60-residue DNA binding domain.
    • Zinc-finger where a polypeptide folds around a zinc ion to form a compact binding domain. There are C2H2 and C4 zinc-fingers.
    • Leucine-zipper, which binds to DNA as dimers held together by hydrophobic interactions. Every other amino acid is a leucine. There are basic residues in extended α-helical regions of monomoers, and interact with DNA at adjacent major grooves.
    • Basic Helix-Loop-Helix is the same as a leucine-zipper, except a nonhelical polypeptide separates two α-helical regions in each monomer.
    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 from a promoter even when they are 50 kb away from the promoter in the genome sequence?

    Contains multiple individual elements. Each element is a protein binding site. A protein kinase phosphorylates a transcription factor, thereby activating the DNA binding domain so it binds to the enhancer. It then interacts with RNAP by looping out intervening DNA and stimulating RNAP to transcribe. The proteins must have (a) flexible regions of polypeptide between DNA-binding and activation domains, allowing flexibility in distance between binding sites and (b) when several transcription factors bind to a small length of DNA, the interactions of multiple DNA-binding domains causes looping of intervening DNA.

    Enhancers function by being the binding sites for transcription factors (or activators). Transcription factors bound to an enhancer can interact with Mediator and RNA polymerase II and general transcription factors bound to a promoter by looping out the chromatin between the enhancer and the promoter.

    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?

    General transcription factors is the name for the proteins that bind to promoters with RNA polymerase II, helping it to initiate transcription. There names are TFIIB, TFIID, TFIIE, TFIIF, and TFIIH. (No points off for mentioning TFIIA. But TFIIC and TFIIG are not general transcription factors.) A TATA-box is a high affinity binding site for the TBP subunit of TFIID which initiates assembly of a pre-initiation complex composed of the general transcription factors and RNA polymerase II. An initiator element binds to another subunit of TFIID and initiates the assembly of a pre-initiation complex.

    Describe three mechanisms by which the activity of a transcription factor can be regulated?

    Any three of the following:

    • Regulation of transcription of the transcription factor (TF) gene.
    • Regulation of the translation of the TF mRNA.
    • Regulation of the degradation of the TF.
    • Regulation of the nuclear import and export of the TF.
    • Regulation of the TF by small, hydrophobic molecules, such as for the nuclear receptors.
    • Regulation of TF DNA binding activity by cooperative DNA-binding with a neighboring TF.
    • Regulation of TF DNA-binding by interaction with another polypeptide, as for MyoD and Id.
    • Regulation of the activation domain activity by post-translational modifications such as phosphorylation.
    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.

    Transcription termination is also regulated. When HIV is being transcribed by RNAPII, the RNAPII terminates after transcribing approximately 50 kb of the HIV genome. The HIV Tat protein is an anti-termination sequence-specific RNA-binding protein. Tat binds cooperatively with the cellular cyclin T protein to the RNA copy of the TAR sequence. The TAR sequence is folded into a hairpin loop, and cyclin T activates protein kinase CDK9. The substrate of CDK9 is CTD of RNAPII, preventing RNAPII from terminating every 50 kilobases. (page 485)

    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 is one DNA molecule per chromosome. Since there are 44 autosomal and 2 sex chromosomes, that makes 46 chomosomes = 46 DNA molecules. The DNA molecules are one component of a chromosome which fills a territory, or small domain in the nucleus that does not contain DNA from another chromosome. The DNA replicates many times without separating because there is no replication at the centromere or telomeres.

    What mechanism is thought to account for the observation that RNAs transcribed by RNA polymerase II are capped and polyadenylated, while these types of RNA processing do not occur to 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.

    The required enzymes for capping are found bound to the RNA polymerase II before transcription starts. As soon as the 5' end of the new transcript emerges the enzymes transfer to it and begin the capping process (this is a similar kind of mechanism to ensure capping as for polyadenylation). The enzymes for capping can only bind to RNA polymerase II ensuring specificity to only these transcripts, which are almost entirely mRNA.

    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? (A diagram of the alternative RNA processing in the production of secreted and membrane bound μ heavy chain mRNA would be helpful.)

    The site of polyadenylation is specified by a poly A site composed of an AAUAAA upstream and a G/U-rich sequence downstream of the cleavage-polyadenylation site. Polyadenylation does not occur at the site of transcription termination, but at a cleavage site upstream. Cleavage and polyadenylation of the pre-mRNA signals RNA polymerase II to terminate transcription at one of multiple possible sites. The secreted μ heavy chain mRNA is generated by polyadenylation at a site upstream of the poly A site for the mRNA encoding the membrane bound form of μ heavy chain mRNA. See diagram below. The affinity of the G/U-rich sequence at the upstream secreted μ heavy chain (μs) mRNA polyA-site for the CStF polyadenylation factor is very low, while the downstream membrane bound μ heavy chain (μm) mRNA is a strong polyA site with high affinity for polyadenylation factors. In immature B lymphocytes, the concentration of CStF is sufficiently low that the μs polyA site is not used, but the strong μm polyA site is used. When immature B lymphocytes mature into plasmacells, the concentration of CStF increases to a high enough level that the weak μs polyA site is used.

    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?

    A 2'-5' linkage is formed. Two transesterifcation reactions occur, with one phosphoester bond being exchanged for another. Spliceosomal and self-splicing transesterification reactions occur in a specific order. First, a specific branch-point nucleotide within the intron reacts with the first nucleotide of the intron, forming an intron lariat. Second, the last nucleotide of the first exon reacts with the first nucleotide of the second exon, joining the exons and releasing the intron lariat.

    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?

    A synthetic oligonucleotide that is complementary to and thus hybridizes with the 5'-end region of U1 snRNA was found to block splicing, indicating that this is the region that binds to pre-mRNA during the splicing reaction. Further evidence for the importance of bse pairing between the 5' end of U1 snRNA and the conserved 5' splice-site sequence of pre-mRNA came from experiments with genes that were mutated in the 5' splice-site consensus sequence of an intron. When genes containing these mutations were transfected into cells, splicing of the corresponding pre-mRNAs was blocked. However, when a mutant gene was cotransfected with a mutant U1 snRNA gene containing a compensating sequence change that restored base pairing with the mutant 5' splice site, splicing was restored (Figure 12-22). This result argued strongly that base pairing between the 5' splice site of a pre-mRNA and the 5' region of U1 snRNA is required for RNA splicing.

    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 recognition (also called exon definition) involves RNA-binding SR proteins which interact with exonic sequences called exon splicing enhancers. SR proteins have one ore more RRM RNA-binding domains and several protein-protein serine- and arginine-rich interaction domains. Once bound to exon splicing enhancers, SR proteins mediate cooperative binding of U1 snRNP to a true 5' splice site and U2 snRNP to a branch point through a network of protein-protein interactions spanning an entire exon. The complex of SR proteins, snRNPs, and other splicing factors (like U2AF) is called a cross-exon recognition complex and permits precise exon definition in long pre-mRNAs.

    The 65-kD subunit of U2AF binds to the pyrimidine-rich region enar the 3' end of introns and to the U2 snRNP. The 35-kD subunit of U2AF binds ot the AG dinucleotide at the 3' end of the intron & also itneracts with the larger U2AF subunit bound nearby. These two U2AF subunits specify the 3' splice site by promoting interaction of U2 snRNP witht he branch point.
  • Spliceosome-mediated splicing of pre-mRNA
    Mutations in the splice sites 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? Mutations in splice sites in human α- and β-globin genes generally result in the use of alternative “cryptic” splice sites rather than exon skipping. What mechanism can account for this?

    This mechanism is called aternative splicing. The mutation in the SMN2 exon eliminates binding by an SR-protein. This interferes with the formation of a cross-exon recognition complex. As a result, U2 snRNP fails to bind to the 3' splice site at the 5'-end of the exon and U1 snRNP fails to bind to the 5' splice site at the 3'-end of the exon. Since splicesomes do not assemble at these splice sites, the exon is skipped. In contrast, in the globin genes, mutations in splice sites do not interfere with the binding of SR-proteins to exonic splicing enhancers in the associated exon. The SR-proteins bound to exonic splicing enhancers stimulate the binding of U1 or U2 snRNPs to alternative cryptic splice sites that are utilized in place of the mutant splice site.

    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.
    • Repression of splicing (or negative regulation of splicing). In Drosophila sex determination, the Sxl protein represses splicing of a splice site in the Sxl pre-mRNA. OR Sxl protein represses splicing of a splice site in Tra pre-mRNA
    • Activation of splicing (or positive regulation of splicing). Tra protein activates a splice site in the Dsx pre-mRNA.
    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.

    The central transporter of the NPC is proposed to contain a molecular meshwork of FG-nucleoporins that act like a sieve holding back proteins >60 kDa. These FG-nucleoporins are proposed to contain extended chains of hydrophilic amino acids punctuated by hydrophobic FG-domains that associate with each other through hydrophobic interactions, generating the molecular meshwork.

    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?

    Nuclear pores are large protein complexes that cross the nuclear envelope, which is the double membrane surrounding the eukaryotic cell nucleus. There are about 3,000-4,000 nuclear pore complexes in the nuclear envelope of an animal cell, depending on the number of transcriptions.

    Nuclear pores allow the transport of water-soluble molecules across the nuclear envelope. This transport includes RNA and ribosomes moving from nucleus to the cytoplasm and proteins (such as DNA polymerase and lamins), carbohydrates, signal molecules and lipids moving into the nucleus. Although smaller molecules simply diffuse through the pores, larger molecules may be recognized by specific signal sequences and then be diffused with the help of nucleoporins into or out of the nucleus. This is known as the RAN cycle. Each of the eight protein subunits surrounding the actual pore (the outer ring) projects a spoke-shaped protein into the pore channel. The center of the pore often appears to contains a plug-like structure. It is yet unknown whether this corresponds to an actual plug or is merely cargo caught in transit. The whole pore complex has a diameter of about 150 nm, and the diameter of the opening is about 50 nm wide. Small particles (smaller than 17,000 daltons) are able to pass through the nuclear pore complex by passive diffusion. Larger particles are also able to pass through the large diameter of the pore but at almost negligible rates. Efficient passage through the complex requires several protein factors. Karyopherins, which may act as importins or exportins are part of the Importin-β super-family which all share a similar three-dimensional structure.

    Export of proteins: Some nuclear proteins have to be exported to the cytoplasma sometimes. Pre-ribosomal particules and mRNAs as well. There is therefore an export mechanism similar to the import. In the classical export scheme, proteins with an NES (Nuclear Export Sequence) can bind in the nucleus to form a heterotrimeric complex with an exportin and RanGTP (for example the exportin CRM1). The complex can then diffuse to the cytoplasm where GTP is hydrolysed and the NES-protein is released. CRM1-RanGDP diffuses back to the nucleus where GDP is exchanged to GTP by RanGEFs. This process is also energy dependent as it consumes one GTP. Export with the exportin CRM1 can be inhibited by Leptomycin B.

    Import of proteins: Any cargo with a Nuclear Localization Signal (NLS) exposed will be destined for quick and efficient transport through the pore. There are several kinds of NLS sequences but they are usually conserved polypeptide sequence with basic residues such as PKKKRKV. Any material with an NLS will be taken up by importins to the nucleus. The classical scheme of NLS-protein import is as this : Importin-α binds first to the NLS sequence, and acts as a bridge for Importin-β to attach. The complex importinβ-importinα-cargo is then directed towards the nuclear pore and diffuses through it. Once the complex is in the nucleus, RanGTP binds to Importin-β and displaces it from the complex. Then CAS (Cellular Apoptosis Susceptibility) protein, an exportin which in the nucleus is bound to RanGTP, displaces Importin-α from the cargo. The NLS-protein is thus free in the nucleoplasm. The Importinβ-RanGTP and Importinα-CAS-RanGTP complex diffuses back to the cytoplasm where GTPs are hydrolysed to GDP leading to the release of Importinβ and Importinα which become available for a new NLS-protein import round.

    Although cargo passes through the pore with the assistance of chaperone proteins, the translocation through the pore itself is not energy dependent. However, the whole import cycle needs the hydrolysis of 2 GTPs and is thus energy dependent and has to be considered as active transport. The import cycle is powered by the nucleo-cytoplasmic RanGTP gradient. This gradient arises from the exclusive nuclear localization of RanGEFs, proteins that exchage GDP to GTP on Ran molecules. Thus there is an elevated RanGTP concentration in the nucleus compared to the cytoplasm.

    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?

    HeLa and Xenopus cells were fused using polyethylene glycol to form heterokaryons containing nuclei from each cell type. These hybrids were treated immediately with cycloheximide to prevent protein synthesis. After two hours, the cells were fixed and stained with fluorescent-labeled antibodies specific for human hnRNP C and hnRNP A1 proteins. These antibodies do not bind to homologous Xenopus proteins.

    Under fluorescence microscopy, hnRNP C was found in HeLa cells and the heterokaryon HeLa nuclei, but not in the Xenopus cells nor nuclei. However, hnRNP A1 was found in HeLa cells and both heterokaryon nuclei, but not in the Xenopus cells.

    Since protein synthesis was blocked after fusion, it can be concluded that in the heterokaryon some human hnRNP A1 protein left the HeLa nuclei, moved through the cytoplasm, and entered the Xenopus nuclei.

    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?

    Cytoplasmic polyadenylation promotes translation of some mRNAs. Protein-mediated translational control helps regulate expression of some genes. Regulatory sequences (elements) in mRNAs interact with specific proteins to control translation are usually in the 3' or 5' UTR. Sequence-specific translation-control proteins may bind cooperatively to neighboring sites in 3' UTRs and function in a combinatorial manner similar to the cooperative binding of transcription factors to regulatory sites in an enhancer region of a gene. In most cases, translation is repressed by a protein biding to 3' regulatory elements; oftentimes, the mRNAs must then undergo cytoplasmic polyadenylation before translation. This is a critical aspect of gene expression in the early embryo. The egg cells (oocytes) of multicellular organisms contain many mRNAs, encoding numerous different proteins, not translated until after fertilization by a sperm cell. Some of these stored mRNAs have only 20-40 nucleotide poly(A) tails and only a few PABPI proteins can bind. Since many PABI are necessary to interact with eIF4G initiation factor, which stabilizes the interaction the mRNA 5' cap with EIF4E, translation is unable to occur with these stored mRNAs. In response to an external signal however, they will be polyadenylated and translation will commence.

    Two sequences are needed for polyadenylation in the cytoplasm:

    • AAUAAA poly(A) signal required for nuclear polyadenlyation
    • One or more upstream U-rich cytoplasmic polyadenylation elements (CPE), which is bound by a highly conserved CPE-binding protein (CPEB) containing an RRM and zinc-finger domain.

    In the absence of stimulation, CPEB bound to the U-rich CPE interacts with Maskin, which bidns to eIF4E on the mRNA 5' cap and prevents it form interacting with other intitiation factors (thereby blocking translation).

    Phosphorylation of Maskin allow cleavage and polyadenylation specificity factor (CPSF) and poly(A) polymerase to catalyze the addition of A residues. Sufficient amounts of PABPI are then able to bind to the poly(A) tail and stabilized interaction of initiation factors allows translation to proceed.

    In the caase of Xenopus oocyte maturation, the protein kinase which phosphorylates CPEB is activated in response to progesterone. A similar mechanism is involved in learning and memory. As postsynaptic neurons are stimulated, the postsynaptic neurons are able to "remember" which axon carried the action potential; the next time that axon stimulates the postsynaptic neuron, the response will be stronger. This is due to local sythesis of new proteins which increase the size and alter the neurophysiological characteristics of the synapse. Synaptic activity, rather than a hormone, is the signal inducing phosphorylation of CPEB in this case.

    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 is antisense inhibition where a sequence of RNA degrades another polynucleotide with exactly the same sequence.

    A long dsRNA mediating interfrence is recognized by Dicer ribonuclease and processed into a double-stranded intermediate referred to as short interfering RNA (siRNA) with 2123 nucleotide strands hybridized such as two bases at the 3' end of eachs trand are single-stranded.

    dssiRNA is processed into a multiprotein complex containing only one RNA strand. This RNA-induced silencing complex (RISC) cleaves target RNAs precisely complementary to their corresponding sssiRNAs. RISC complexes also appear to function in inhibition of translation by miRNAs. They can have an:

    • siRNA function (RNA interference) leading to cleavage of precisely complementary mRNAs
    • miRNA function, leading to translational repression of mRNAs with a few base-pair mismatches.

    RNA interference is an ancient cellular defense against certain viruses and mobiel genetice elements in both plants and animals. In plants, dsRNA also induces DNA methylation of genes with the same sequence; this inhibits transcription of the gene via binding of histone deacetylases. This does not occur in animals, however, and details about it are unknown. The RNA interference machinery, once it finds a double-stranded RNA molecule, cuts it up with an enzyme known as Dicer, separates the two strands, and then proceeds to destroy other single-stranded RNA molecules that are complementary to one of those segments. dsRNAs direct the creation of small interfering RNAs (siRNAs) which target RNA-degrading enzymes (RNases) to destroy transcripts complementary to the siRNAs.

    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?

    Nonsense-mediated decay, an mRNA surveillance mechanism, causes degradation of mRNAs in which one or more exons have been skipped during splicing. Except in pre-mRNAs alternatively spliced, exon skipping will alter the open reading frame of the mRNA 3' to the improper exon junction. This results in introduction of an incorrect stop codon. In nearly all properly spliced mRNAs, the stop codon is in the last exon. Nonsense-mediated decay results in the rapid degradation of mRNAs with stop codons before the last splice junction in the mRNA. Proteins in the exon junction complex function in nonsense-mediated decay. They interact with a deadenylase that rapidly removes the poly(A) tail from an associated mRNA to rapidly decap and degrad by a 5-3 exonuclease. Decaping pathway (deadenylation-independent). In properly spliced mRNA, exon-junction complexes are dislodged from the mRNA by passsage of the first pioneer ribosome to translate the mRNA to protect it form degradation; mRNAs with a stop codon before the final exon junction will have one or more exon-junction complexes associated with the mRNA, resulting in nonsense-mediated decay.

    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.

    There are three mechanisms of regulation (activation or repression) of cyclin-dependent kinase complexes:

    • Cyclin binding activates CDKs by modifying kinases for ATP positioning, substrate binding, and active site accessibility.
    • Loop phosphorylation activates CDKs by mutually strengthening cyclin-CDK binding.
    • CDK phosphorylation represses CDKs. Inhibitory phosphorylation represses CDKs. CDK feeds back on Wee and Cdc to inhibit and enhance their respective activities.
      • Wee adds an inhibitory phosphate
      • CDK-activating kinase (CAK) adds an activating phosphate
      • Cdc is a phosphatase activating CDK by removing the inhibitory phosphate added by Wee
    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?

    A Ubiquitin ligase (also called an E3 ubiquitin ligase) is a protein which covalently attaches ubiquitin to a lysine residue on a target protein. The ubiquitin ligase is typically involved in polyubiquitination: a second ubiquitin is attached to the first, a third is attached to the second, and so forth. Polyubiquitination marks proteins for degradation by the proteasome.

    However, there are some ubiquitination events which are limited to mono-ubiquitination, in which only a single ubiquitin is added by the ubiquitiin ligase to a substrate molecule. Mono-ubiquitinated proteins are not targetted to the proteasome for degradation, but may instead be altered in their cellular location or function, for example via binding other proteins that have domains capable of binding ubiquitin.

    The ubiquitin ligase is referred to as an E3 and operates in conjunction with an E1 ubiquitin-activating enzyme and an E2 ubiquitin-conjugating enzyme. There is one major E1 enzyme, shared by all ubiquitin ligases, which uses ATP to activate ubiquitin for conjugation and transfers it to an E2 enzyme. The E2 enzyme interacts with a specific E3 partner and transfers the ubiquitin to the target protein. The E3, which may be a multi-protein complex, is generally responsible for targeting ubiquitination to specific substrate proteins. In some cases it receives the ubiquitin from the E2 enzyme and transfers it to the target protein; in other cases it acts by interacting with both the E2 enzyme and the substrate but never itself receives the ubiquitin.

    The Anaphase-promoting complex (APC) and the SCF complex (for Skp1-Cullin-F-box protein complex) are two examples of ubiquitin ligase protein scaffold involved in recognition and ubiquitination of specific target proteins for degradation by the proteasome.

    There are several "families" of E3 ubiquitin ligases; these families are comprised by members that have particular protein domains that are capable of binding the E2 conjugase, and thus forming a ubiquitination complex. For example, the Skp1-Cullin-Fbox protein complex referenced above is formed by a family of E3 ubiquitin ligases that have an "F-box" domain. Thus the full complex is calles an "SCF" complex, for Skp1, Cullin, and F-box containing protein. Other E3 ubiquitin ligase proteins include those that have a RING domain (for Really Interesting New Gene). RING domains bind the E2 conjugase, and may be required for the enzymatic activity of the E2-E3 complex. Other E3 ubiquitin ligase proteins include those that have a HECT domain.

    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?

    A mutation in the mitotic cyclin destruction box that prevents the protein from being recognized by and polyubiquitinated by the APC ubiquitin ligase (with the help of the unphosphorylated Cdh1 specificity factor).

    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?

    Once chromosomes have segregated properly, telophase commences. This requires inactivation of MPF. Dephosphorylation of APC specificity factor Cdh1 by Cdc14 phosphatases leads to degradation f mitotic cyclins and loss of MF activity late in anaphase. During interphase and early mitosis, Cdc14 is sequestered in the nucleus and inactivated. The chromsome-segregation checkpoint monitors the location of segregating daughter chromsomes at the end of anaphase and determines whether active Cdc14 is availabel to promote exit from mitosis.

    This depends on the mitotic exit network. A key component is a small monomeric GTPase (Tem1) which is part of the GTPase superfamily of switch proteins. It controls the activity of a protein kinase cascade similarly to the way Ras controls MAP kinase pathways. During anaphase, Tem1 becomes associated witht eh spindle pole body (SPB) closest to the daughter cell bud. The SPB, from which spindle microtubules originate, is analogous to the centrosome in higher eukaryotes. At the SPB, Tem1 is maintained in the inactive GDP-bound state by a specific GAP (GTPase accelerating protein). The GEF (guanosine nucleotide-exchange factor) activating Tem1 is localized to the cortex of the bud and is absent from the mother cell. When spindle microtubule elongation has correctly positioned segregating daughter chromosomes into the bud, Tem1 comes into contact with its GEF and is converted intot the active GTp-bound state.

    The terminal kinase in the cascade triggered by Tem1•GTP then phosphorylates the nucleolar anchor that binds and inhibits Cdc14, releasing it into the cytoplasm and nucleoplasm of both the bud and the mother cell. Once active Cdc1 is avialable, a cell can proceed through telophase and cytokinesis. If daughte chromosomes fail to segregate into the bud, Tem1 remians inactive and Cdc14 is not released from the nucleolus, thereby blocking mitotic exit.

    PAGE 872 & 888. When daughter chromsomes have segregated properly in late anaphase, the Cdc14 phosphatase is activated and dephosphorylates Cdh1, allowing it to bind to the APC. This interaction leads quickly to APC-mediated polyubiquitination and proteasomal degradation of B-type cyclins and hence MPF inactivation. Since MPF is active when Cdc14 is first activated in late anaphase, it potentiall could compete with Cdc1 by rephosphorylated Cdh1. However, Cdc14 also induces expression of Sic1 by an inhibitory phosphate on a transcription factor that activates transcription of the SIC1 gene. Hence, Sic1 binds to and inhibits all B-type cyclin-CDK complexes. Thus, in late mitosis, the inhibition of MPF by Siz1 allows Cdc14 phosphatase to prevail; B-type cyclic APC specificity factor Cdh1 is dephosphorylated and directs precipitous degradation of all the B-type cylcins.

    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.)

    Structure. Each sister chromatid of a metaphase chromosome is attached t microtubules via its kinetochore, a complex of proteins assembled at the centromere. The opposite ends of these kinetochore microtubules associate with one of the spindle poles. At metaphase, the spindle is in a state of tension with forces pulling te two kinetochores twoad the opposite spindle poles balanced by forces pushing the spindle poles apart. Sister chromatides remain together at centromeres and multiple positions along the chromosome arms by multiprotein complexes called cohesins. Cohesin SMC proteins bind to each sister chromatid; other subunits of cohesin, including Scc1, then link the SMC proteins to firmly associate the two chromatids. This cross-linking activity of cohesin depends on a universal eukaryotic protein called securin.

    Prior to anaphase, securin binds and inhibits separase, a ubiquitous protease realted to caspace proteases which regulate apoptosis. Once all chromosome kinetochores have attached to spindle microtubules, the APC (Cdc20 and Cdh1) is directed by a specifciity factor called Cdc20 to polyubiquitinate securin, leading to the onset of anaphase. Polyubiquinated securin is rapidly degraded by proteasomes to release separase. Free from inhibition, separase cleaves Scc1 to break the protein cross-link between sister chromatids. At this point, poleward forces on kinetochores can move sister chromatids toward opposite spindle poles. PAGE 872

    How is the dismantling of the nuclear lamina during prophase and its reassembly during telophase accomplished? Where does the nuclear envelope go during mitosis?

    The nuclear envelope is a double-membraned extension of the rough endoplasmic reticulum. It contains numerous protein complexes. The lipid bilayer of the inner nuclear envelope is supported by the nuclear lamina, a meshwork of lamin filaments adjacent to the inside face of the nuclear envelope. The three nuclear lamins (A, B, and C) in vertebrate cells are cytoskeletal proteins, intermediate filaments, that are critical in supporting cellular membranes. Lamins A and C, encoded by the same transcription unit, are produced by alternative splicing of a single pre-mRNA. They are identical except for a 133-residue region at the C-terminus of lamin A absent in lamin C. Lamin B, encoded by another transcription unit, is modified post-transcriptionally by the addition of a hydrophobic isoprenyl group near the carboxyl-terminus. This fatty acid becomes embedded in the inner nuclear membrane to anchor the nuclear lamina in the membrane. All three nuclear lamins form dimers with a rodlike α-helical coiled-coil central section and globular head and tail domains. Polymerizing these dimers through head-to-head and tail-to-tail associations generates the intermediate filaments composing the nuclear lamina. Early in mitosis, MPF phosphorylates specific serine residues in all three nuclear lamins, thereby depolymerizing lamin intermediate filaments. The phosphorylated lamin A and C dimers are released into solution, whereas the phosphorylated lamin B dimers remain associated with the nuclear membrane via their isoprenyl anchor. Depolymerizing nuclear lamins leads to disintegration of the nuclear lamina meshwork and contributes to disassembly of the nuclear envelope. This all depends on phosphorylation of lamin A.

    The nuclear membrane and nuclear lamina then retract into the reticular endoplasmic reticulum, which is continuous with the outer nuclear membrane.

    Once MPF is inactivated in late anaphase by degradation of mitotic cyclins, the unopposed action of phosphatases reverses the action of MPF. The dephosphorylated inner nuclear membrane proteins are thought to bind to chromatin once again. As a result, multiple projections of regions of the ER membrane containing these proteins are thought to associate with the surface fo the decondensing chromsomes and then fuse with each other to form a continuous double membrane around each chromosome. Dephosphorylating nuclear pore subcomplexes allows them to reassemble nuclear pore complexes traversing the inner and outer membranes soon after fusion of ER projections. Membrane-associated SNARE proteins direct fusion of nuclear envelope extensions from ER not identified, but syntaxin 5 is both a V-snare and T-snare during reassembly of the Golgi. Ran-specific guanosine nucleotide-exchange factor (Ran-GEF) is associated with chromatin, so a high concentraion of Ran-GTP is prduced around the chromosomes. This directs membrane fusions at the chromosome surface. The reassembly of nuclear envelopes contianing nuclear pore complexes around each chromosome forms individual mininuclei called karyomeres. Subsequent fusion of karyomeres with each spindle pole generates two daughter-cell nuclei each containing a full set of chromsomes. Dephosphorylate lamins A and C appear to be imported through the reassembled nuclear pore complexes during this period and reassemble into new nuclear lamina. Reassembly of the nuclear lmaina is initiated on lamin B molecules, which remain associated with ER membrane via their isoprenyl anchors throughout mitosis and become localized to the inner membrane of reassembled nuclear envelopes of karyomeres.

    What mechanism accounts for the sudden onset of DNA synthesis in S. cerevisiae?

    When the S-phase cyclin-CDK complexes are synthesized in late G1, they are immediately bound by the S-phase inhibitor (Sic1). The activity of the S-phase Cyclin-CDK complexes increase precipitously when the S-phase inhibitor is degraded suddenly. This allows the S-phase Cyclin-CDKs to phosphorylate DNA replication initiation factors in pre-replication complexes bound to DNA replication origins resulting in the initiation of DNA synthesis. The S-phase inhibitor is polyubiquitinated by the SCF ubiqutin ligase after it is phosphorylated by late G1 Cyclin-CDKs.

    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?

    In eukaryotes, DNA replication at an origin occurs once per cell cycle (once during S-phase). Yeast replication origins contain an 11-bp conserved core sequence to which is bound a hexameric protein, the origin recognition complex (ORC) required for initiation of DNA synthesis. ORC remains associated with origins during all phases of the cycle. An MCM hexamer and several other replication initiation factors act to unwind the DNA. These complexes assemble in early G1 while B-type cyclin activity is low, then an active S-phase cyclin CDK and heterodimeric protein kinase (expressed in G1) are required for initiation of DNA replicaiton. This S-phase cyclin-CDK activation also causes CDc45 to bind to the prereplicaiton complex. It binds RPA, which stabilizes ssDNA and promotes binding of primase and DNA polymerase α to begin replicaiton. DNAP δ and accessory proteins are required for continued DNA synthesis and synthesis of most of the laging strand. In addition, Pol ε is required for chromosomal DNA synthesis but it has unknown function.

    Numerous proteins are phosphorylated, thereby disallowing them from forming another pre-initiation complex. Thus, phosphorylation both initiation and prevents reinitiaiton of DNA synthesis. ORC complexes immediately bind to the origin sequence in the replicated daughter duplex DNAs and remain bound.

    The initiaiton factors (Cdc6, Cdt1, and MCm10) get dephosphorylated in late anaphase and telophase when APC triggers degradation of all B-type cyclins. This allows the reassembly of pre-replication complexes during G1. Inhibition of APC throughout G1 sets the stage for accumulation of S-phase cyclins needed for onset of S-phase. This has two consequences, both of which mean chromosomal DNA is replication only once per cell cycle:

    • Pre-replication complexes assembled only during G1, when activity of B-type cyclin-CDK complexes is low
    • Each origin initiaties replication one time only during S-phase, when S-phase cyclin-CDK complex activity is high.
    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 for retinoblastoma. Rb gene is mutant in a rare human genetic disease in which afflicted invidiuals with a mutation that inactivates Rb gene invariably develop tumors of the retina (retinablastomas) early in childhood. In tumor cells, both copies of Rb gene, one on each chromosome, are mutant. Since wild-type Rb gene protects humans from developing retinoblastomas, it was labeled a tumor-suppressor gene.

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