Ch 7A Essays

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Ch 7A Essays
2015-03-11 04:21:22
Test Three
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  1. 1)      How do the cell types of multicellular organisms become different from one another?
    • a.       Because they synthesize and accumulate different sets of RNA and protein molecules without altering their DNA.
    • b.      Experiment on frogs: nucleus of fully differentiated frog injected into frog egg, the normal tadpole develops since no important DNA sequences were lost.
    • c.       This can also be done by studying the banding patterns in condensed chromosomes, showing that they’re both identical and that DNA is not lost or rearranged during vertebrate development.
  2. 1)      What has DNA technology confirmed? 
    a.       They confirmed that changes in gene expression that underlie the development of multicellular organisms do not rely on changes in the DNA sequences of the corresponding genes. There are, however, a few cases where DNA rearrangements of the genome take place during the development of an organism. 
  3. 1)      What are the general statements that can be made regarding cell types?
    • a.       Many processes are common to all cells, leading to similar proteins, such as structural chromosomal proteins, RNA polymerases, DNA repair enzymes, etc.
    • b.      Some proteins are abundant in the specialized cells in which they function and cannot be found elsewhere
    • c.       A typical human cell expresses 30-60% of its 25,000 genes. The level of expression of almost every active gene varies among cell types
    • d.      Although differences in mRNAs among specialized cell types are striking, they underestimate the full range of differences in the pattern of protein production. Gene expression can be regulated in several places after transcription. Also, proteins can be covalently modified after they are synthesized
  4. 1)      What are most speciailized cells capable of doing? 
    They are capable of altering their patterns of gene expression in response to extracellular cues
  5. 1)      At what points can a cell control gene expression? 
    • a.       Transcriptional control: when and how often a gene is transcribed
    • b.      RNA processing control: control of splicing and processing of RNA trancripts
    • c.       RNA transport and localization control: selecting which completed mRNAs are exported from the nucleus and where in the cytosol they are localized
    • d.      Translational control: selecting which mRNAs in the cytoplasm are translated by ribosomes
    • e.      mRNA degradation control: selectively destabilizing certain mRNA molecules in the cytoplasm
    • f.        protein activity control: selectively activating, inactivating, degrading, or locating specific protein molecules after they have been made
  6. 1)      What was once proposed about gene regulatory proteins; and, what is now the belief?                
    • a.       It was thought that these proteins might require direct asccess to the hydrogen bonds between base pairs in the interior of the double helix to distinguish between strands. It is now clear that the outside of the double helix is studded with DNA sequence information that gene regulatory proteins can recognize without having to open the double helix.
    •                                                                i.      The edge of each base pair is exposed at the surface of the helix, presenting a distinctive pattern of hydrogen bond donors, hydrogen bond acceptors, and hydrophobic patches for proteins to recognize in both the major and minor groove. Only in the major groove are the patterns markedly different for each of the four base-pair arrangements. For this reason, gene regulatory prtoeins make specific contacts with the major groove. 
  7. 1)      What do short DNA sequences function as? 
    a.       Because specific nucleotide sequences can be read as a pattern of molecular features on the surface of the DNA double helix, they act as genetic switches by serving as recognition sites for the binding of specific gene regulatory proteins. 
  8. 1)      What does a gene regulatory protein recognize? 
    a.       It recognizes a specific DNA sequence because the surface of the protein is extensively complementary to the special surface features of the double helix in that region. The protein contacts the DNA with hydrogen, ionic, and hydrobhopic bonds, several of which form a tight and specific protein-DNA interface. Gene regulatory proteins have DNA-binding structural motifs that use alpha helices or beta sheets to bind to the major groove of DNA, which contains enough information to distinguish among DNA sequences.
  9. 1)      Explain the outside structure of proteins that contain helix-turn-helix motifs.
    a.       It is unique and thus enhances the versatility of the helix-turn-helix motif by increasing the number of DNA sequences that the motif can be used to recognize. Parts of the polypeptide chain outside the helix-turn-helix domain make important contacts with the DNA, helping to fine-tune the interaction. 
  10. 1)      What is a common feature of many sequence-specific DNA-binding proteins? 
    a.       They bind as symmetric dimers to DNA sequences thata re composed of two very similar “half-sites,” which are also arranged symmetrically. This arrangement allows each protein monomer to make a nearly identical set of contacts and enormously increases the binding affinity: doubling the number of contacts doubles the free energy of the interaction and thereby squares the affinity constant. 
  11. 1)      Though alpha helices are the primary mechanism used to recognize specific DNA sequences, what is another recognition motif? 
    a.       A two-stranded beta sheet, with amino acid side chains exptending from the sheet toward the DNA, reads information on the surface of the major groove and can recognize many different DNA sequences.
  12. 1)      What else can DNA binding proteins use to recognize DNA? 
    a.       They can use protruding peptide loops to read nucleotide sequences rather than alpha helices and beta sheets. An example is p53.
  13. 1)      Why are homodimers often used to recognize DNA?
    a.       This is a simple way of achieving strong specific binding; and, the portion of the protien responsible for dimerization is distinct from the portion that is responsible for DNA binding. One exception is the leucine zipper motif, so named because of the way the two alpha helices, one from each monomer, are joined together to form a short coiled-coil
  14. 1)      Aside from homodimers, what can also form? 
    a.       Heterodimers composed of two different subunits with distinct DNA-binding specificities can form, expanding the amount of specificities that the proteins can display. Thee are limits, however. If all types of leucine zipper proteins formed heterodimers, for example, there would be chaos. As a result, formation of heterodimers depends on how well the hydrophobic surfaces of the two leucine zipper alpha helices mesh with each other, which in turn depends on the exact amino acid sequences of the two zipper regions. 
  15. 1)      Do we know the DNA sequences recognized by gene regulatory proteins? 
    a.       Because protein surfaces of virtually any shape and chemistry can be made from 20 different amon acids, the same base pair can be recognized in many ways depending on its context. 
  16. 1)      What do many approaches for detecting sequence-specific DNA binding proteins rely on? 
    • a.       They rely on the detection in a cell extract of a DNA-binding potein that specifically recognizes a DNA sequence known to control the expresseion of a particular gene. One of the most common ways is based on the effect of a bound protein on the migration of DNA moelcules in an electric field.
    •                                                                i.      A DNA molecule is highly negatively charged and will therefore move rapidly toward a positive electrode when subjected to an electric field. When analyzed by PAGE, DNA moelcuels are separated according to their size because smaller molecules are able to penetrate the fine gel meshwork more easily than large ones. Protein molecules bound to a DNA molecule cause it to move more slowly through the gel; in general, the larger the protein, the greater the retardation of the DNA molecule. This phenomenon provides the basis for the gel-mobility shift assay. 
  17. 1)      What can affinity chromatography do? 
    a.       It can isolate enough pure protein to obtain a partial amino acid sequence and thus identify the gene, which both provides the amino acid sequence of the protein and provides the means to produce the protein in unlimited amounts through genetic engineering techniques. 
  18. 1)      What can be determined experimentally? 
    a.       The DNA sequence recognized by a gene regulatory protein can. Because they were discovered via isolation, the genes encoding the protiens were identified, overexpressed, and purified. DNA footprinting can then be used to determine the DNA sequence that gene regulatory protein recognizes. Another method can be using purified prtein to select from a large pool of different short DNA fragments only those that bind tightly to it. A consensus DNA recognition sequence for the gene regulatory protein can then be formulated. Computerized genome searches then identify candidate genes whose transcription the gene regulatory protein of interest might control 
  19. 1)      What does the availability of complete genome sequences allow? 
    a.       It allows identification of important regulatory sites on DNA. In this approach, genomes from several closely related species are compared. If chosen properly, the protein-coding portions are similar, but regions between sequences that encode protein or RNA molecules will have diverged considerably, except the regulatory sequences that control gene transcription. This is phylogenetic printing and is good for identify DNA seqeucnes that control gene expression. 
  20. 1)      What are the ways in which a GRP will be prevented from occupying all of its potential DNA-binding sites in the genome? 
    • a.       The protein may not be synthesized and thus be absent
    • b.      It may be present but lacking a heterodimer partner
    • c.       It may be excluded from the nucleus until a signal is received.
    • d.      Components of chromatin or other gene regulatory proteins that can bind to the same or overlapping DNA seqeucens may occlude its potential bidning sites on DNA
  21. 1)      What is within the promoter that directs transcription of tryptophan biosynthetic genes?  
    a.       Within the promoter is a regulator element called an operator.
  22. 1)      How is the block of gene expression regulated? 
    a.       To bind to its operator DNA, the repressor protein has to have two molecules of the amino acid tryptophan bound to it. Tryptophan binding tilts the helix-turn-helix motif of the repressor so that it is presented properly to the DNA major groove; without tryptophan, the motif swings inward and the protein is unable to bind to the operator. Thus, the tryptophan repressor and operator form a simple device that switches production of the tryptophan biosynthetic enzymes on and off according to the availability of free tryptophan. This is negative control. 
  23. 1)      Why are bacterial promoters only marginally functional on their own? 
    a.       This is because they are recognized poorly by RNA polymerase or because the polymerase has difficulty opening the DNA helix and beginning transcription. IN either case, these poorly funciotning promoters can be rescued by gene regulatory proteins that bidn to a nearby site on the DNA and contact the RNA polymerase in a way that increases the probability that transcription will be initiated. They can do so by providing an additional contact surface for the polymerase or contacting RNA polymerase and facilitating its transition from the initial DNA-bound conformation of polymerase to the actively transcribing form by stabilizing a transition state of the enzyme
  24. 1)      Compare activators and repressors. 
    a.       Both use a helix-turn-helix motif and both require a small cofactor in order to bind DNA. Some bacterial proteins (lie CAP and lambda repressor) can act as either activators or repressors, depending on the exact placement of the DNA seqeucne they recognize in relation to the promoter: if the binding site for the protein overlaps the promoter, the polymerase cannot bind and protein acts as a repressor.
  25. 1)      Explain the Lac operon.
    a.       It is under both negative and positive transcriptional controls by the Lac repressor protein and CAP. The Lac operon codes for proteins required to transport lactose into the cell and break it down. CAP enables bacteria to use alternative carbons sources such as lactose in the absence of glucose. The Lac repressor shuts down the Lac operon in the absence of lactose, allowing the control region of it to respond to and integrate two different signals, so that the operon is highly expressed only when two conditions are met: lactose is present and glucose is absent. In any of the other three possible signal combinations, the cluster of genes is held in the off state. See figure 7-39.
  26. 1)      Aside from using regulatory proteins to control RNA polymerase and its beginning transcription, what is another way that transcription initiation can be controlld?
    a.       Another method is based on interchangeable subunits of RNA polymerase. A sigma subunit is required for the bacterial RNA polymerase to recognize a promoter. Most bacteria produce several sigma subunits, each of which interacts with RNA polymerase and directs it to different sets of promoters. This scheme allows one large set of genes to be turned off and a new set to be turned on by replacing one sigma subunit with another. This strategy is efficient because it bypasses the need to deal with genes one by one. 
  27. 1)      What is the advantage of eucaryotes in having inceased complexity? 
    a.       It is the fact that dozens of signals can converge on a single promoter, with the transcription machinery integrating all these different signals to produce the appropriate level of mRNA.
  28. 1)      Distinguish eukaryotic gene regulation from prokaryotes.
    • a.       Eukaryotic RNA polymerase II requires five general TFs, whereas bacterial RNA polymerase needs only a single general transcription factor, the sigma factor. The stepwise assembly of the general TFs provides multiple steps at which a cell can speed up or slow down the rate of transcription initiation in response to gene regulatory proteins.
    • b.      Eukaryotes lack operons.
    • c.       Each bacterial gene is controlled by one or a few gene regulatory proteins, but eukaryotic genes can be contolled by many of different regulatory proteins.
    • d.      Eukaryotes have a mediator (24-subunit complex), which serves as an intermediary between gene regulatory proteins and RNA polymerase. Meiatory provides an extended contact area for gene regulatory proteins compared to that provided by RNA polymerase alone.
    • e.      The packaging of eukaryotic DNA into chromatin provides many opportunities for transcriptional regulation not available to bacteria
  29. 1)      How are the gene regulatory regions arranged in eucaryotes?
    a.       DNA sequences that control the expression of a gene are spread over long stretches of DNA. The gene control region is the whole expanse of DNA involved in regulating and initiating transcription of a gene, including the promoter,w here the general transcription factors and the polymerase assemble, and all of the regulatory sequences to which gene regulatory proteins bind to control the rate of the assembly processes at the promoter. Regulatory sequences can have spacer sequences between them that GRPs on’t recognize, but provides flexibility needed for efficient DNA looping. 
  30. 1)      What do GRPs allow genes to do? 
    a.       It allows them to be turned on or off individually. There are not simple rules for gene regulation. But, there are generalizations about how GRPs, once bound, set in motion the train of events that lead to gene activation or repression. 
  31. 1)      Explain activator proteins. 
    • a.       They bind to enhancers that can be located thousands of nucleotides down from the promoter. The simplest have a modular design with two domains. One usually contains a common structural motif; and, the second, called an activation domain, accelerates the rate of transcription initiation. The ultimate function of activators is to attract, position, and modify the general transcription factors, Mediator, and RNA polymerase II at the promoter so that transcription can begin. They do this by acting directly on these components and, indirectly, by changing the chromatin structure around the promoter.
    • b.      Some activator proteins bind directly to one or more the general TFs, accelerating their assembly on a promoter that is linked through DNA to that activator. Others interact with Mediator and attract it to DNA where it can then facilitate assembly of RNA polymerase and the general TFs at the promoter.
  32. 1)      What can’t general TFs, Mediator, and RNA polymerase do? 
    a.       They can’t aseembly on a promoter that is packaged in nucleosomes on their own. This is done to prevent leaky transcription. Gene activator proteins promote transcription initiation by changing the chromatin structure of the regulatory sequences and promoters of genes. 
  33. 1)      What are the ways of locally altering chromatin? 
    a.       Covalent histone modifications, nucleosome remodeling nucleosome removal, and nucleosome replacement. Gene activatory proteins use all four by attracting histone modification enzymes, ATP-dependent chromatin remodeling complexes, and histone chaperones to alter the chromatin structure of promoters they control. These local alterations make NA more accessible and assembly possible, as well as allow gene regulatory proteins to bind to the control region of the gene. These modifications provide favorable interactions for the binding of a large set of proteins that read a “histone code,” which include histone-modifying enzymes (reader-writer compelxes), chromatin remodeling complexes, and one of the general TFs. 
  34. 1)      How long do alterations of chromatin structure last? 
    a.       They can persist or, in some cases, rapidly reverse as soon as the gene regulatory protein dissociates from DNA. The rapid reversal allows the cell to switch on and off rapidly in response to external signals. The persistence of some modifications can be passed to the next cell generation.
  35. 1)      What is a special type of chromatin modification? 
    a.       It occurs as RNA polymerase II transcribes through a gene. In most cases, the nucleosomes just ahead of the polymerase are acetylated by writer complexes carried by the polymerase, removed by histone chaperones, and deposited behind the moving polymerase. They then are rapidly deacetylated and methylated, also by reader-writer complexes that are carried by the polymerase, leaving behind nucleosomes that are resistant to transcription. This is evolved to prevent spurious transcription re-initiation behind a moving polymerase, which is clearing a path through chromatin.
  36. 1)      How are genes poised to become activated? 
    a.       The regulatory regions for many genes are marked by a short, nucleosome-free region flanked by nucleosomes that contain the histone variant H2Az. This arrangement allows free access of gene regulatory proteins to the nucleosome-free region; in addition, the H2AZ-containing nucleosomes are thought to be easily disassembled, further facilitating transcription initiation. 
  37. 1)      How do gene activator proteins work synergistically? 
    a.       Where several factors work together to enhance the reaction rate, the joint effect is not merely the sum of the enhancements that each factor alone contributes, but the product. They can increase the reaction multiplicatively. Thus, gene activator proteins exhibit transcriptional synergy, where several activator proteins working together produce a transcription rate that is much higher than that of the sum of the activators working alone. There is no single order of steps that this occurs. 
  38. 1)      Explain efficient DNA binding.
    a.       It typically requires several sequence-specific DNA proteins acting together. For example, two genere regulatory proteins with a weak affinity for each other might cooperate to bind to a NA sequence, neither protein having a sufficient affinity for DNA to bind to the DNA site on its own. The DNA-bound protein dimer creates a distinct surface that is recognized by a third protein that carries an activator domain that stimulates transcription, in on example. This illustrates the point that protein-protein interactions that are too weak to form complexes in solution can do so on DNA, with the DNA sequence acting as the seed for the assembly of the protein complex. 
  39. 1)      Explain the ability of gene regulatory proteins to bind to more than one type of regulatory complex.
    a.       Individual eukaryotic gene regulatory proteins are not dedicated to activators or repressors; instead, they function as regulatory parts that are used to build complexes whose function depends on the final assembly of all of the individual compoents. This final assembly, in turn, depends both on the arrangement of control region DNA sequence and on the particular gene regulatory proteins present in active form in the cell. Each eukaryotic gene is therefore regulated by a “committee” of proteins, all of which must be present to express the gene at its proper level. 
  40. 1)      What can the precise DNA sequence to which a regulatory protien binds do? 
    • a.       It can affect the conformation of this protein and thereby influence its subsequent transcriptional activity.
    • b.      Typically, a few relatively short stretches of nucleotide sequence guide the assembly of a group of regulatory proteins on DNA. However, in some extreme cases of regulation by committee a more elaborate protein-DNA structure is formed. Since the final assembly requires the presence of many gene regulatory proteins that bind DNA, it provides a simple way to ensure that a gene is expressed only when the cell contains the correct combination of these proteins. The assembly of complexes of gene regulatory proteins on DNA provides a second important mechanism for combinatorial control. 
  41. 1)      What is an example of a complex, multicomponent genetic switch?
    a.       It is in Drosophila Even-skiped (Eve) gene, whose expression plays an important role in the development of the Drosophila embryo. If the gene is inactivated by mutation, many parts of the embryo fail to form, and the embryo dies early in development. When Eve begins to be expressed at the stage of development, the embryo is a single giant cell with multiple nuclei in a common cytoplasm that contains a mix of GRPs distributed unevenly, providing positional information. The nuclei begin to express different genes because of exposure to different GRPs. The regulatory DNA sequences controlling the Eve gene read the concentrations of GRPs at each positionalong the length of the embryo and interpret it in such a way that the Eve gene is expressed in seven stripes. 
  42. 1)      Explain mammalian gene control regions. 
    a.       8% of the coding capacity of a mammalian genome is devoted to the synthesis of proteins that serve as regulators of gene transcription, refleciting the complex network of controls governing expression of mammalian genes. Each gene is regulated by a set of GRPs; each of those proteins is a product of a gene that is regulated by a whole set of other proteins. The regulatory protein moelcuels are themselves influenced by signals from outside the cell, which can make them active or inactive. Gene expression is the result of a complicated molecular computation that the intracellular gene control network performs in response to info from the cell’s surroundings. 
  43. 1)      Explain gene regulation in the human beta-globin gene.
    a.       A complex array of gene regulatory proteins controls the expression of the gene, some acting as activators and others as repressors. The concentrations of many of the GRPs change during development, and only a particular combo of all the proteins triggers transcription of the gene. The human beta-globin gene is part of a cluster of globin genes that are transcribed exclusively in RBCs, but at different stages of mammalian development. The epsilon-globin gene is expressed in early embryo, gamma in later embryo and fetus, and delta and beta in the adult. The gene products differ in their oxygen-binding properties, suiting them for the different oxygenation conditions in the embryo, fetus, and adult. 
  44. 1)      How are the globin genes unusual? 
    a.       They are unusual in that, at the appropriate time and place, they are transcribed at extremely high rates; indeed, red blood cells are little more than bags of hemoglobin that was synthesized by precursor cells. To achieve the high level of transcription, the globin genes share a control region called the locus control region, which lies far upstream from the gene cluster and is needed for the proper expression of each gene in the cluster. The importance of the LCR includes a barrier sequence that prevents the spread of neighboring hetermochromatin into the beta-globin locus. This dual feature distinguishes the globin LCR from many other types of regulatory sequences in the human genome; hwever, the globin genes are not alone in having an LCR, as LCRs are also present in upstream of other highly transcribed, cell-type-specific genes. 
  45. 1)      Explain the insulator.
    a.       It prevents enhancers from running amok and activating inappropriate genes. It can block the communincation between an enhancer and a promoter, but, to do so, it must be located between the two. The distribution of insulators and barrier sequences in a genome is thought to divide it into independent domains of gene regulation and chromatin structure. 
  46. 1)      What are the circumstances in which an enhancer can activate a gene located on a second chromosome?
    a.       An example is in the regulation of the mammalian olfactory receptors. The olfactory receptor genes are dispersed among many different chromosomes, but there is only a single enhancer for all of them. Once this enhancer activates a receptor gene by associating with its regulatory region, it remains stably associated thereby precluding activation of any of the other receptor genes. 
  47. 1)      Explain the evolution of gene switches. 
    a.       Close-packed arrangement of bacterial genetic switches developed from more extended forms in response to the evolutionary pressure on bacteria to maintain a small genome size. This compression restricts the complexity and adaptability of the control device. IN contrast, the extended form of eukaryotic control regions—with discrete regulatory modules separated b long stretches of spacer DNA—faciliates the reshuffling of regulatory modules during evolution, both to create new regulatory circuits and to modify old ones. 

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