1.    There are 3 regulatory regions in eukaryotic DNA that control the level of transcription of a gene. Name them and where they are located relative to the start point of transcription.

2.    What role in transcription is played by the following molecules?  TBP, the tafs, TFIIH.

3.    Describe the chemical steps that remove an intron from a primary RNA transcript. Do not talk about the spliceosomes here, just the chemistry of the reaction.

4.    What type of methylation pattern on CG islets preceeding the beta-globin gene would be expected in the erythroblasts that synthesize hemoglobin?

5.    Choose either 3' cleavage and polyadenylation site selection or anti-termination of transcription and describe one system we discussed that is controlled in this way.

6.    If histones are only synthesized in the S phase of the cell cycle, how can their mRNA stability be altered to allow more translation then but almost none at other times? Suggest a possible mechanism that could explain this, based on our discussions of mRNA stability and its regulation.

7.    Illustrate three ways that a protein can interact with the cell membrane.

8.    Describe the role of cholesterol in the maintenance of membrane fluidity. In your answer illustrate how cholesterol can do this due to its orientation in the membrane.

I.    Transcription
    A.    Chemistry of the reaction to make an RNA polymer
        1.    Use NTPs as energy source. Remove Pyrophosphate.
        2.    Attach remaining monophosphate in a 5'-3' linkage, similar to DNA linkages.
    B.    RNA Polymerases
        1.    Do not need a preexisting 3'OH. Can attach two nucleotides together to start an RNA
                polymer.
        2.    Has three activities
            a.    unwinding activity: melts ds DNA to expose the template
            b.    polymerizing activity: catalyzes formation of the RNA transcript
            c.    winding activity: displaces the small (10-15 nucleotide) transcript from the template and
                   allows the DNA double helix to reform as the transcription moves forward.
        2.    Types
            a.    bacterial
                -consists of a core enzyme complex and a sigma factor
                -sigma factor needed for initiation
                -sigma factor recognizes the promoter
                -sigma factor takes the place of general transcription factors in eukaryotes
            b.    RNA pol I: transcribes the large rRNAs in eukaryotes
            c.    RNA pol II: transcribes the heterogenous RNA (primary RNA transcript) that
                   will become mRNA in eukaryotes
            d.    RNA pol III: transcribes small RNAs in eukaryotes (tRNAs, snRNAs, etc.)
    C.    Regulatory Sequences
        1.    Promoter
            a.    Loads RNA polymerase II onto the DNA where it can then find the startpoint of
                   transcription.
            b.    Located approximately 30-50 nucleotides upstream of the +1 startpoint.
            c.    Most common type is the TATA box.
                -Consists of a consensus sequence TATAA/TA that is the site of assembly of the
                  general transcription factors.
            d.    Alternative promoter types exist but are not included on the exam.
        2.    Upstream Promoter Elements
            a.    Located approximately 200-500 basepairs upstream from +1 startpoint.
            b.    Many different types have been identified.
            c.    Bind the specific transcription factors.
            d.    Believed to modify the ability of the general transcription factors to load on RNA
                       pol II.
        3.    Enhancers (Silencers)
            a.    Unusual sequences that can modify transcription rates at large distances from the +1
                   startpoint (1-5kb) and on either side of the gene and in either orientation.
            b.    Bind additional specific transcription factors that can either upregulate or downregulate.
            c.    Believed to interact with the general transcription factors by looping the DNA
                   to bring the enhancer(silencer) sequence to the vicinity of the promoter.
        4.    How RNA polymerase gets loaded on:
            a.    TFIID consists of TBP (TATA binding protein) and a complex of tafs.
            b.    TFIID complex binds to the TATA box through the TBP.
            c.    Inhibitors of TBP can bind to it and stop the assembly process.
            d.    If no inhibitor binds TPB, the remainder of the general transcriptions assemble.
            e.    This includes TFIIF bound to RNA polymerase II.
            f.    Once all are assembled, TFIIH acts.
                -Its protein kinase activity attaches phosphates to RNA polymerase II .
                -Its helicase activity melts the hydrogen bonds of the DNA to expose the template.
            g.    RNA polymerase II will move to the +1 startpoint and begin transcription.
    D.    Processing the primary RNA transcript
        1.    Capping the 5' end
            a.    Phosphohydrolase removes the gamma phosphate from the first RNA nucleotide.
            b.    Pyrophosphate is removed from GTP and the GMP is attached to the diphosphate
                   end of the transcript through an unusual 5'-5' linkage.
            c.    This step is catalyzed by guanyl transferase.
            d.    Then guanine methyl transferase catalyzes the attachment of a methyl group to the
                   7 position of the base of the attached G niucleotide, using S-adenosyl-methionine
                   (SAM) as a co-enzyme and the source of the methyl group.
            e.    Then 2'-O-methyl transferase attaches methyl group to the 2'OH of the first and second
                  nucleotides of the transcript if it is appropriate to do so.
                -Some eukaryotes do not methylate these nucleotides, some only methylate the first
                 and vertebrates methylate both. Again, SAM is the coenzyme.
            f.    This CAP is used during the initiation of translation as we discussed previously.
        2.       3'-end processing (cleavage and polyadenylation)
            a.   In the primary transcript, a poly-adenylation signal is made that reads AAUAA.
            b.   The actual cleavage and poly-A addition site is about 20-30 nucleotides downstream.
            c.    RNA pol II continues transcription past these sites and makes a G/U rich sequence in
                    the transcript. It continues transcribing.
            d.   CPSF binds to the AAUAA signal.
            e.   This facilitates the binding of the cleavage factors (CFI and CFII)and CstF, which binds
               to the G/U-rich sequence.
            f.   This allows PAP (poly-A polymerase) to bind.
            g.   The cleavage factors cleave the transcript at the poly-A site and PAP attaches the
                  poly-A tail.
            h.  At first the poly-A addition is slow. Then the CFI, CFII, and CStF release.
            i.  PABII binds to the poly-A tail and rapid polyadenylation proceeds. About 200 A
               residues get attached. The rest of the transcript beyond the cleavage site is degraded.
        3.   Removal of introns
            a.   Eukaryotic genes contain intervening sequences called introns.
            b.   RNA polymerase II transcribes these introns.
            c.   They must be removed from the transcript and the exons  (what is left) connected into a
               continuous mRNA sequence.
            d.   This is done by a mechanism known as splicing.
            e.  Evidence for splicing comes from a technique known as R-looping
                -mRNA is hybridized to single-stranded DNA strands (hydrogen bonds have
                  been melted).
                -Complementary sequences will hybridize.
                -Intron sequences will have no complementary sequence in the mRNA.
                -They "loop out" and can be seen in electron micrographs.
            f.    The splicing signal in the primary transcript has three major features.
                -There is a 100% conserved GU at the 5'end of the intron.
                -There is a 100% conserved AG at the 3' end of the intron.
                -There is an A nucleotide about 20-30- nucleotides from the 3'end of the intron that
                  is called the branch point and is 100% conserved.
            g.    The intron is removed by two transesterification reactions.
                -First, the 2'OH on the ribose of the branch point A attacks the 3' end of exon 1.
                Tthis breaks the phosphodiester bond between the last nucleotide of exon 1 and the
                 first nucleotide of the intron.
                -A bond forms between the 5'P of the first intron nucleotide and the 2'OH of the A.
                -Then, the 3' end of exon 1 (the 3'OH) attacks the 5' end of exon 2.
                -This breaks the bond between the 3' end of the intron and the 5'end of exon 2.
                -The 3' end of exon 1 links to the 5'end of exon 2 by a phosphodiester linkage.
                -The intron (in lariat form) is removed and degrades.
                -See page 502 in the text for diagrams of this reaction.
            h.    These reactions are catalyzed and controlled by nuclear particles called snurps.
                -consist of small nuclear RNAs complexed to proteins.
                -several kinds are involved, called U1, U2, etc.
                -U1 has snRNA that is complementary to the GU and flanking sequences at the exon/
                 intron junction. It binds first during the splicing reaction
                -U2 has snRNA complementary to the consensus sequence surrounding the A
                 branchpoint but not the A itself which bulges out. This facilitates the first reaction.
                -After U2 binds, the remaining snurps bind forming the spliceosome.
                -The RNA-RNA interactions between the transcript and the snurps and between the
                 snurps rearrange as the reactions continue, facilitating the process.
                -Finally, the spliceosomes, still attached to the intron, leave at the end of the splicing
                 reactions and the snurps disassemble.

II.    Control of Gene Expression
    A.    Control at the level of transcription
        1.    chromatin structure
            a.    The density of chromatin is greater in a gene that is not expressed than one that is.
            b.    Probed by DNAse sensitivity assays.
                -DNAse can degrade DNA if it is accessible to it.
                -The more densely packed the chromatin structure, the less accessible the DNA.
            c.    Studied using the beta-globin gene in erythroblasts (express the gene) and MSBcells
                    (do not express the gene).
            d.    Used the Southern Blot procedure.
            e.    Protocol of the assay
                - Extract DNA from the two cell types.
                -Subject it to increasing concentrations of DNAse.
                -Cut it into fragments with a restriction enzyme.
                -Run the fragments out on an agarose gel.
                -Blot the DNA onto a nitrocellulose filter.
                -Use a radioactively labeled probe of complementary sequence to beta-globin.
                -Probe for the presence of the beta-globin sequences if there.
                -Probe will hybridize to the sequences if there.
                -Expose filter to x-ray film and develop and analyze.
                -Results showed that the sequences disappeared quickly with increasing DNAse.
                  in the erythroblasts but did not disappear in the MSB cells.
                -Conclude that the chromatin is more dense for the beta-globin gene in the cell type
                 that does not express that gene.
        2. Methylation of CG islets
            a.    Actively transcribed genes tend to be undermethylated in the sequence CCGG.
                -The C next to the G would be methylated.
                -These sequences are called CG islets and are interspersed within regulatory regions of
                 genes.
            b.    Methylated CG islets can be detected by Hpa II restriction enzyme analysis.
                - Hpa II cuts in the sequence only if it is not metylated.
                -Msp I is another restriction enzyme that cuts this same sequence always, methylated
                 or not.
                -Therefore, if the beta-globin gene in erythrocytes was tested in this way, Hpa II would
                 cut in the CG islet surrounding this gene, as would Msp I. In a cell type that did not
                 express beta-blobin, Hpa II would not cut the sequence but Msp I would.
        3.    Inducible genes
            a.    Some genes are only transcribed when signalled to do so by some inducing molecule
                    often a hormone
                -Example is the glucocorticoid-responsive genes.
                -The hormone crosses into the cytoplasm of the responding cell because it is lipid-
                  soluble.
                -It interacts with a cytoplasmic receptor that is bound to an inhibitor, displacing the
                  inhibitor and activating the receptor.
                -This receptor-hormone comples enteres the nucleus and activates specific genes by
                   binding to sequences in the promoter-proximal elements.
        4.    Antitermination
            a.    RNA polymerase II often pauses during elongation of transcription and will only
                    complete transcription when signalled to do so.
            b.    One example is the tat protein of HIV.
                    - Tat binds to a sequence near the 5' end of the transcript called TAR, part of a stem-
                    loop structure.
                    -Cellular factors bind to another stem-loop structure.
                    -Together they coordinate interaction with the paused RNA polymerase II, somehow
                     causing it to renew transcription.
            c.    Another example is the control of the hsp70 transcript which responds to stress in cells.
                    -The non-stressed cell begin transcription but the RNA polymerase II pauses.
                    -In response to stress, a protein called HSTF changes shape from an inactive to an
                      active form.
                    -It now can bind near promoter-proximal elements called GAGA.
                    -In some way this causes the stalled RNA polymerase II to resume transcription as
                     well as stimulating more rounds of transcription by other RNA polymerase IIs.
    B.    Control at the level of mRNA processing
        1.    3' end processing control
            a.    Choice of cleavage and polyadenylation site can change subsequent patterns of
                of splicing and/or translation, making different proteins or forms of a protein.
            b.    Our example was the antibody molecule IgM that can exist as either a membrane-
                bound or secreted molecule, depending on the differentiation state of the B cell.
                -Earlier in the differentiation pathway, the B cell makes the membrane-bound molecule.
                -This requires that the COOH terminal contain two exons that code for a hydrophobic
                  region that can associate with the cell membrane.
                -To do this, an early poly-A signal is ignored and a later one recognized on the primary
                   RNA transcript.
                -Cleavage and polyadenylation using this second signal gives rise to a mRNA that
                 contains the necessary exons and is translated into the membrane bound protein.
                -Later, when the cell is ready to secrete the antibody, it is signalled to recognize the
                   the first poly-A signal and uses it.
                -Following cleavage and poly-adenylation, the subsequent splicing events make a
                mRNA that does not contain the exons for the hydrophobic region and the protein
                made will not associate with the cell membrane but will be secreted.
        2.    Control at the level of splicing is covered in Problem 3.
    C.    Control at the level of mRNA stability
        1.    The 3' untranslated region of eukaryotic mRNAs can contain repeated sequences of
              AUUUA or similar sequences that signal the mRNA to be degraded by nucleases.
            a.    This causes these mRNAs to have very short half-lives
            b.    The proteins coded for by these mRNAs are usually central to growth regulation for
                    the cell.
            c.    If a long-lived mRNA like b-globin has its 3'UTR replaced by the AUUUA sequences
                    from a short-lived mRNA, it becomes short-lived also.
            d.    Histones are regulated in a cell-cycle manner.
                    -Their mRNAs are unstable except during S phase when they become stable.
                    -due to stem-loop in the 3'UTR.
        2.    Example we looked at was the mRNA for the transferin receptor.
            a.    Has several AUUUA-like repeats in its 3'UTR.
            b.    Allows the formation of stem-loops that contain an iron-response element (IRE).
            c.    An IRE binding protein exists in two possible conformations.
                   -when iron amounts are low in the cell, it is active and binds to the IREs.
                   -This protects the mRNA from degredation.
                   -Therefore, translation will occur and more transferin receptor will be made.
                   -This will allow more iron to be brought into the cell.
    D.    Control at the level of translation
        1.    The 5' UTR also contains one or more stem-loop structures which must be melted
                for translation to be possible.
            a.    These can be the site for regulation.
            b.    Our example was the mRNA for ferritin, an iron-storage protein.
            c.    Ferritin mRNA has 5'UTR stem-loops with IRE elements.
            d.    The same IRE binding protein operating above can bind to these stem-loops.
            e.    Therefore, if iron amounts are low in the cell, the protein binds to the IREs.
            f.    This prevents the melting of the stem-loops and prevents translation initiation.
            g.    Therefore, no ferritin is made when it is not needed to store iron.
            h.    If it is needed, when iron amounts are high, the stem-loops will not be bound by the
                    inactive form of the binding protein and translation will initiate.
        2.    Another translational control is through eIF2.
            a.    This translation initiation factor is only active if it is not phosphorylated.
            b.    Phosphorylation of eIF2 causes it to bind irreversibly to eIF2B in a GDP-bound
                    form.
            c.    This depletes eIF2B, preventing it from activating other non-phosphorylated eIF2
                    proteins to their GTP bound and active forms.
            d.    This lowers the initiation of translation rate.
            e.     One protein kinase that can phosphorylate eIF2 is called the hemin-controlled
                    inhibitor (HCI).
                -HCI can be active or inactive depending on the amount of hemin in a developing red
                  blood cell.
                -If hemin is available for hemoglobin formation, the hemin will bind to the HCI
                 and inactivate it, preventing it from phosphorylating eIF2. This allows translation
                 initiation of the globin genes to continue at a fast rate.
                -If hemin is in low amounts, the HCI will be active and will phosphorylate eIF2,
                 lowering the rate of translation initiation as explained above.
Note: Many other control points exist in eukaryotic cells but we do not have enough time to
           review them all.

III.    Membranes
    A.    Structure of the cell membrane
        1.    Fluid-Mosaic Model
            a.    Dynamic, fluid phospholipid bylayer containing other lipids and proteins.
            b.    Made up of two leaflets, the E(external ) leaflet and the P or C (internal) leaflet.
            c.    C leaflet is more negatively charged and more fluid than E.
            d.    Phospho and glycolipids have at least one double bond in one of the two fatty acid tails.
            e.    This helps keep the fatty acid tails from getting to close. Helps keep them from tangling.
            f.    This helps maintain fluidity.
        2.    Lipid composition of the cell membrane
            a.    phospholipids
            b.    glycolipids
            c.    cholesterol
        3.    Types of phospholipids
            a.    phosphatidyl ethanolamine, mainly in C leaflet.
            b.    phosphatidyl serine, mainly in C leaflet.
            c.    phosphatidyl choline, mainly in E leaflet.
            d.    sphingomyelin, mainly in E leaflet.
        4.    Glycolipids
            a.    Contain sugar groups.
            b.    Always on the E leaflet.
            c.    Participate in cell-cell recognition, etc.
            d.    Includes the neutal glycolipids and the gangliosides (contain negatively charged
                    sialic acid moieties.
        5.    Synthesis of the phospho and glycolipids
            a.    Occurs on the cytoplasmic leaflet of the endoplasmic reticulum.
            b.    Then, flippases selectively "flip" particular phospho or glycolipids to the other
                    leaflet.
            c.    Once in a particular leaflet, the phospho or glycolipid cannot flip back.Only lateral
                    movement is allowed.
            d.    Then they pass through the endomembrane system to the cell membrane.
        6.    Cholesterol
            a.    Has mainly hydrophobic structure but one small hydrophilic section.
            b.    Can associate with the phospholipids by having its hydrophobic section interact with
                   the fatty acid tails and its hydrophilic section be near the phosphates.
            c.    Helps prevent the phospholipids from approaching too closely together.
            d.    Helps maintain the fluidity of the cell membrane.
        7.    Membrane proteins
            a.    Can associate with the membrane in a number of ways.
                - Single or multiple-pass transmembrane proteins that have a hydrophobic membrane-
                   spanning sequence of amino acids in a right-handed alpha helix.Called integral proteins.
                - Anchoring by attachment to a lipid anchor in the C leaflet, either a farnesyl anchor or
                   a myristate anchor.
                - Anchoring to the E leaflet by attachment to glycosyl phosphatidyl inositol.
                - Indirect association by interacting with another membrane-anchored protein.Called
                   peripheral proteins.
            b.    Are free to diffuse in the plane of the membrane.
                - Shown experimentally by studying heterocaryons(fusions between mouse and
                    human membranes).
                - Antibodies were directed against either mouse or human membrane proteins.
                - Human tagged with one colored-tag; mouse with another.
                -At time of fusion, the colors segregated to only one side of the heterokaryon,
                 representing the protein's original location in the original membrane.
                -After about 30-40 minutes, staining showed a homogenous mixture of the colors,
                  showing that the proteins freely moved in the plane of the membrane.
            c.    Membrane proteins can be solubilized using detergents.
                   - Studies of solubilized sodium-potassium ATPases indicated that not only the
                     proteins themselves are needed for activity but the lipid component is also required.

Link to exam 2 key and graders