Chapter 3 Essays

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Chapter 3 Essays
2015-02-05 11:33:55
Test One
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  1. 1)      How can modifications be done to an already folded and stable protein? 
    a.       Over evolutionary time, genetic mechanisms can duplicate genes, allowing one gene copy to evolve independently to perofrm a new function. This forms protein families. The structure of the different members of a protein family have been more highly conserved than has the amino acid sequence                
  2. 1)      Explain multidomain proteins? 
    a.       They are said to have originated from the accidental joining of the DNA sequences that encode each domain, creating a new gene. Novel binding surfaces have been created at the juxtaposition of domains, and many of the functional sites where proteins bind to small molecules are located there. In domain shuffling, many large proteins have evolved through the joining of preexisting domains in new combinations. 
  3. 1)      What is the structure of the domains? 
    • a.       A stable core formed from strands of beta sheets, from which less-ordered loops of polypeptide chain protrude. The loops form binding sites for other molecule. Their structure provides a convenient framework for the generation of new binding sites for ligands through small changes to their protruding loops.
    • b.      Protein domains can also be ingtegrated into other protein when the DNA encoding that domain undergoes tandem duplication. The duplicated modules with this “in-line” arrangement can be readily linked in series to form extended structures
    • c.       The same pairs of domains can occur repeatedly in the same relative arrangement in a protien
  4. 1)      Are all polypeptide chain structured? 
    • a.       No, elastin has relativey loose and unstructured polypeptide chains covalently cros-linked to produce a rubberlike elastic meshwork that can be reversibly pulled from one conformation to another.
    • b.      In general, proteins use short loops of polypeptide chains to bind other molecules. Unstructured regions may have interactions, undergoing structural transitions once bound. Unstructured regions may also connect the binding sites for proteins that function together to catalyze a biological reaction. These regions can act as “tethers” that concentrate sets of interacting proteins 
  5. 1)      What are the advantages of the use of smaller ubunits to build larger structures?
    • a.       A large structure built from one or a few repeating smaller subunits requires only a small amount of genetic information
    • b.      Both assembly and disassembly can be readily controlled, reversible processes, because the subunits associate through multiple bonds of relatively low energy
    • c.       Errors in the synthesis of the structure can be more easily avoided, since correction mechanisms can operate during the course of assembly to exclude malformed subunits.
  6. 1)      How do enzymes achieve extremely high rates of chemical reactions? 
    by increasing teh local concentration of substrate molecules at the catalytic site, holding the atoms in the correct orientation for the reaction, and using binding energy to overcome Ea and reach TS. They can use acids and bases simultaneously too.
  7. What are ways that enzyme can lower the activation energies and speed up the reactions they catalyze
    • a.       In reactions involving two or more reactants, the active site also acts like a template, or mold, that brings the substrates together in the proper orientation for a reaction to occur between them
    • b.      In addition, when a substrate bind to an enzyme, bonds in the substrate often bend, changing the substrate shape, which drive a substrate toward a particular transition state
    • c.       Finally, like lysozyme, many enzymes participate intimately in the reaction by briefly forming a covalent bond between the substrate and a side chain of the enzyme
    • d.      Binding of substrate to enzyme rearranges electrons in the substrate, creating partial negative and positive charges that favor a reaction
  8. 1)      Explain regulation. 
    • a.       At one level, the cell controls how many molecules of each enzyme it makes by regulating the epression of the gene that encode that enzyme.
    • b.      The cell alos controls enzymatic activities by confining sets of enzyes to particular subcellular compartments, enclosed by distinct membranes
    • c.       Enzymes can undergo covalent modification to control their activity
    • d.      The rate of protein destruction can occur via targeted proteolgyis
    • e.      The most common type is the binding of substrates to enzymes at special regulatory sites outside the active site, altering the rate at which the enzyme converts its substrates to product
    •                                                               i.      Ex: feedback inhibition
  9. 1)      How can a phosphorylation event affect the protein that is modified?
    • a.       First, because each phosphate group carries two negative charges, the enzyme-catalyzed addition of a phosphate group to a protein can cause a major conformational change in the protein by attracting a cluster of positively charged amino acid side chains
    •                                                               i.      This can, in turn, affect the binding of ligands elsewhere on the protein surface, dramatically changing the protein’s activity.
    • b.      Second, an attached phosphate group can form part of a structure that the binding sites of other proteins recognize.
    • c.       Phosphorylation is unidirectional because of the large amount of free energy released when the phosphate-phosphate bond is broken to produce ADP
  10. 1)      Describe the kinase family? 
    a.       It is a very large family, which hare a catalytic sequence of about 290 amino acids. The various family members contain different amino acid sequences on either end of the kinase sequence and often have short amino acid sequences inserted into loops within it. Some of these additional amino acid sequences enable each kinase to recognize the specific set of proteins it phosphorylates. Other parts of the protein regulate the activity of each kinase so that it can be turned on or off.
  11. Explain Cdks
    • a.       A cyclin-depenedent protein kinase phosphorylates serines and thereonines and they are central components of the cell-cycle control system in eukaryotic cells. Individual Ckd proteins turn on and off in succession, as a cell proceeds through the different phases of its division cycle.
    • b.      Cdk monitors a specific set of cell components—a cyclin, a protein kinase, and a protein phosphatase—and it acts as an input-poutput device that turns on if, and only if, each of these components has attained its appropriate activity state. 
  12. 1)      How does a Cdk protein become active? 
    • a.       A Cdk protein becomes active as a serine/threonine protein kinase only when bound to a second protein called a cyclin. But, that is only one of the three distinct “inputs” required to activate the Cdk.
    • b.      In addition to cyclin binding, a phosphate must be added to a specific Thr side chain, and a phosphate elsewhere in the protein (covalently bound to a specific tyrosine side chain) must be removed.
    • c.       Some cyclins rise and fall in concentration in step with the cell cycle, increasing gradually in amount until they are destroyed at a particular point in the cycle. The sudden destruction of a cyclin immediately huts off its partner Cdk enzyme, and this triggers a specific step in the cell cycle. 
  13. 1)      Explain the Src family of protein kinases?
    • a.       It was the first tyrosine kinase to be discovered.
    • b.      Structure: they contain a short N-terminal region that becomes covalently linked to a strong hydrophobic fatty acid, which holds the kinase at the cytoplasmic face of the plasma membrane. Next are two peptide-binding modules, a Src homology 3 (SH3) domain and a SH2 domain, followed by the kinase catalytic domain.
    • c.       These kinases are normally inactive, in which a phosphorylated tyrosine near the C-terminus is bound to the SH2 domain, and the SH3 domain is bound to an internal peptide in a way that distorts the active site of the enzyme and helps to render it inactive
  14. 1)      How is Src activated? 
    • a.       It involves at least two specific inputs: removal of the C-terminal phosphate and the binding of the SH3 domain by a specific activating protein, causing phosphorylation of the tyrosine that allows self activation.
    •                                                               i.      The activation of the Src kinase signals the completion of a particular set of separate upstream events.
  15. 1)      How are GTP proteins controlled? 
    • a.       They are controlled by regulatory protein that determine whether GTP or GDP is bound. Ras is inactived by GTPase-activating protein (GAP), which binds to the Ras protein and causes it to hydrolyze its bound GTP molecule to GDP—which remain tightly bound—and inorganic phosphate, which is released.
    • b.      The Ras protein stays inactive until it encounter a guanine nucleotide exchange factor (GEF), which binds to GDP-Ras and causes it to release its GDP.
    • c.       Because the empty nucleotide-binding site is immediately filled by a GTP molecule, the GEF activates Ras by indirectly adding back the phosphate removed by GRP hydrolysis. 
  16. 1)      How does the repositioning of the tRNA regarding EF-Tu occur? 
    • a.       The dissociation of the inorganic phosphate group, which follows the reaction GTPàGDP and Pi causes a shift of a few tenths of a nanometer at the GTP-binding site.
    • b.      This tiny movement causes a conformational change to propagate along a crucial piece of alpha helix, called the switch helix, in the Ras-like domain of the protein.
    • c.       The switch helix seems to serve as a latch that adheres to a specific site in another domain of the molecule, holding the protien in a “shut” conformation.
    • d.      The conformational change triggered by GTP hydrolysis causes the switch helix to detach, allowing separate domains of the protein to swing apart, through a distance of about 4 nm.
    • e.      This releases the bound tRNA molecule, allowing its attached amino acid to be used. 
  17. 1)      How do proteins ensure that they are not wiped out? (Way One)
    a.       Large proteins formed from many domains can perform more efficiently than small proteins. Several processes in the cell are catalyzed by a highly coordinated,linked set of 10 or more proteins. These protein machines use an energetically favorable reaction to drive an ordered series of conformational changes in one or more of the individual protein subunits, enabling the ensemble of proteins to move coordinately. 
  18. 1)      How do proteins ensure that they are not wiped out? (Way Two)
    a.       They have interchangeable parts that make efficient use of genetic information.
  19. 1)      Explain the Drosophila compound eye? 
    a.       It consists of 800 ommatidia, each composed of 8 photoreceptor cells and 12 accessory cells. Beginning with the development of the R8 photoreceptor, each differentiating cell induces its uncommitted immediate neighbors to adopt a specific fate and assemble into the developing ommatidium 
  20. 1)      Explain R7.
    • a.       The development of the R7 photoreceptor, required for detecting UV light, is deficieint in mutant flies. 
  21. Summarize the Sev process.
    Boss binds to Sevenless on the surface of the R7 precursor cell, causing phosphorylation of tyrosine in Sev, which then attracks an adaptor protien called Drk, which has an SH2 region that binds to the phosphorylated tyrosine and two Sh3 domains. This causes activation of the Ras-GEF Sos, which is bound via the SH3 domains. SOS stimulates inactive Ras to replace GDP with GTP, thus activating Ras and transmitting the signal for the R7 precursor cells to differentiate into photoreceptors.
  22. 1)      Compare phosphorylation with GTP phosphorylation. 
    • a.       Although one type is activated by phosphorylation and the other by GTP binding, in both cases the addition of a phosphate group switches the activation state of the protein and the removal of the phosphate switches it back again.
    •                                                               i.      A protein kinase covalently adds a phosphate from ATP to the signaling protein, and a protein phosphatase removes the phosphate. Some signaling proteins are activated by dephosphorylation rather than by phosphorylation
    •                                                             ii.      A GTP binding protein is induced to exchange its bound GDP for GTP, which activates the protein; the protein then inactivates itself by hydrolyzing its bound GTP to GDP