Ch 10 Final Review: essay

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DesLee26
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Ch 10 Final Review: essay
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2014-11-10 07:06:01
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Test Three
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  1. 1) What are the five forms of regulation?
    • a.       Allosteric control: allosteric proteins have distinct regulatory sites and multiple functional sites that small signal molecules can bind to. There is also an element of cooperativity involved ex: ATCase
    • b.      Isozymes: allow the varying of regulation of the same reaction at distinct locations or times to meet the specific needs to certain tissues at certain times; catalyze the same reaction but have different structure, Km and Vmax values
    • c.       Reversible covalent modification: addition of a group, usually phosphoryl groups, to activate or deactivate certain proteins ex: PKA
    • d.      Proteolytic activation: an irreversible conversion of an inactive enzyme into an active one through cleavage

    e. Controlling the number that is active
  2. 1)      What is the structure of ATCASE?
    • a.       It has three regulatory subunits, which are dimers, and two trimeric catalytic subunits
    •                                                               i.      The catalytic subunit is catalytic but is unresponsive to CTP and does not have sigmoidal kinetics
    •                                                             ii.      The regulatory subunit can bind CTP but is not catalytic
    •                                                           iii.      Two catalytic trimers are stacked one on top of the other, linked by three dimers of the regulatory chains; each r chain in a dimer interacts with a c chain in a trimer, making contact with a structural domain in the r chain
    •                                                            iv.      It has T and R states
  3. 1)      What are the two forms of ATCase? Explain your answer. 
    • a.       The T state is present in the absence of substrate; low affinity for substrate and low catalytic activity. Adding substrate has two effects: increases the probability that each enzyme will bind at least one substrate molecule and it increases the average number of substrate molecules bound to each enzyme. The presence of additional substrate will increase the fraction of enzyme molecules in the more active R state because the position of the equilibrium depends on the number of active sites that are occupied by substrate
    • b.      Conversion between the two states is concerted, all or none. An increase in substrate favors a transition from T to R
    • In the T state, the regulatory dimers hold the two catalytic trimers sufficiently close to each other that key loops on their surfaces collide and interfere with conformational adjustments necessary for high-affinity substrate binding and catalysis
  4. 1)      What are the effects of CTP and ATP binding to ATCase?              
    • a.       When CTP is bound, the enzyme is in the T state; a binding site for this nucleotide exists in each regulatory chain in a domain that doesn’t react with the catalytic subunit. The binding of CTP makes it more difficult for substrate binding to convert the enzyme into the R state. It will decrease the rate of the reaction.
    • b.      ATP will increase the rate of the reaction at a given aspartate concentration because it competes with CTP for binding to regulatory sites. So, high levels of ATP prevent CTP from inhibiting the enzyme. 
  5. 1)      What explains the increase in ATCase activity in response to increased ATP concentration?
    ATP concentrations has two explanations: (1) high ATP concentration signals a high concentration of purine nucleotides in the cell; the increase in ATCase activity will balance the purine and pyrimidine pools; and (2) a high concentration of ATP indicates that energy is available for mRNA synthesis and DNA replication and leads to synthesis of pyrimidines needed
  6. 1)      Explain isozymes in terms of lactate dehydrogenase.
    a.       Humans have two types: an H that is expressed in heart muscle and an M expressed in skeletal muscle. They have a primary sequence that is 75% the same. Each functional enzyme is tetrameric. The H4 isozyme in the heart has a higher affinity for substrate than does M4. Also, high levels of pyruvate inhibit the H4 but not the M4 isozyme. The M4 functions in the anaerobic environment of skeletal muscle; and, the H4 functions in the aerobic environment of the heart. When in the womb, the M isozyme takes form; and, slowly as birth approaches and even after birth, the H isozyme becomes more prominent until adulthood, when only the H is located in the heart. 
  7. 7) Why is phosphorylation a highly effective means of controlling the activity of proteins? (6 reasons)
    • a.       The free energy of phosphorylation is large; half of the energy produced is consumed in making it irreversible; the other half is conserved in the protein. Hence, phosphorylation can change the conformational equilibrium between different functional states by a large factor
    • b.      A phosphoryl group gads two negative charges to a modified protein, disrupting old electrostatic interactions and forming new ones.
    • c.       A phosphoryl group can form three or more hydrogen bonds.
    • d.      Phosphorylation and dephosphorylation can take place in less than a second
    • e.      Phosphorylation often evokes highly amplified effects. For example, a single activated kinase can phosphorylate hundreds of target proteins in a short interval
    • f.        ATP is the cellular energy currency
  8. Explain PKA
    • a.       It is activated by cAMP. PKA has regulatory subunits and catalytic subunits. In the absence of cAMP, the subunits form an inactive R2C2 complex. When cAMP binds, the subunits dissociate into R2 subunits and two C subunits, becoming active.
    • b.      PKA has two lobes. ATP and part of the inhibitor fill a deep cleft between the lobes. The smaller lobe contacts the ATP-Mg2+, whereas the larger lobe binds the peptide and contributes key catalytic residues. The two lobes move closer to one another on substrate binding. Residues 40 to 280 constitute a conserved catalytic core that is common to essentially all known protein kinases. 
  9. 1)      How does the binding of cAMP activate the kinase? 
    a.       Each R chain contains the sequence Arg-Arg-Gly-Ala-Ile, which matches the consensus sequence except for alanine in the place of serine. In the R2C2 complex, this pseudosubstrate sequence of R occupies the catalytic site of C, preventing the entry of protein substrates. The binding of cAMP to the R chains allosterically moves the pseudosubstrate sequences out of the catalytic sites, allowing the C chains to bind and phosphorylate substrate proteins
  10. 11) What is the structure of chymotrypsinogen? How is it activated?
    a.       Chymotrypsinogen exists as a single polypeptide chain of 245 residues. When, a bond joining arginine 15 and isoleucine 16 is cleaved by trypsin, pi-chymotrypsin is formed. Pi-chymotrypsin acts on other pi-chymotrypsin molecules by removing two dipeptides to yield alpha-chymotrypsin. The three resulting chains remained linked by disulfide bonds. 
  11. 1)      Specifically speaking, how does cleavage of a single peptide bond activate chymotrypsin?
    a.       The newly formed amino terminal group of isoleucine 16 turns inward and forms an ionic bond with aspartate 194. This then triggers conformational changes, one being methionine moving from a deeply buried position in the zymogen to the surface of the active enzyme, as well as residues 187 to 193 moving farther from each other to form the substrate specificity site. Hydrogen bonds between the negatively charged carbonyl oxygen atom of the substrate and two NH groups of the main chain stabilize the tetrahedral transition state. 
  12. 1)      What changes when trypsinogen is activated? 
    a.       The flexibility in the zymogen becomes well-defined in trypsin. The oxyanion hole in trypsinogen is too far from histidine to promote formation of the transition state
  13. 1)      Explain trypsinogen activation and subsequent activations after trypsinogen is produced.
    a.       Enteropeptidase will cleave a peptide bond in trypsinogen as it enters the duodenum, allowing the now activated trypsin to act on other trypsinogen and zymogens. 
  14. 1)      What are the two means of blood clotting? 
    • a.       Intrinsic pathway: activated by exposure to anionic surfaces on rupture of the endothelial lining of the blood vessels
    • b.      Extrinsic pathway: initiated when trauma exposes tissue factor, an integral membrane glycoprotein; after tissue factor is exposed, small amounts of thrombin, the key protease in clotting, are generated. Thrombin then amplifies the clotting process by activating enzymes and factors that lead to the generation of yet more thrombin
    • 2)      Explain the process fibrinogen activation.
  15. 1)      Explain the process of fibrinogen activation.
    • a.       It is activated into fibrin by thrombin, which cleaves four arginine-glycine- peptide bonds in the central globular region of fibrinogen, releasing an A peptide form each of the Aα chains and B from Bβ chains
    • b.      Structure: three globular units connected by two rods; consists of six chains Aα, Bβ, and γ. The rod regions are triple-stranded alpha-helical coiled coils. The homologous β and γ chains have globular domains at the carboxyl-terminal ends and binding holes that interact with the peptides. Knobs are also present; and, the knobs of the alpha subunits fit into the holes of on the gamma unit to form a protofibril, which is extended when the knobs of the beta subunits fit into the holes of beta subunits of other protofibrils
    • c.       Thus, peptide bond cleavage exposes new amino termini that can participate in specific interactions
    • d.      A soft clot is stabilized by formation of amide bonds between side chains of lysine and glutamine, which is cross-linked by transglutaminase. 
  16. 1)      Explain the process of thrombin formation.
    a.       Thrombin exists prematurely as prothrombin, which has four domains—serine protease domain at the C-terminus end, the gla domain, and two kringle domains. Each works to keep prothrombin inactive. Activation is begun by proteolytic cleavage of a peptide bond (between arginine and threonine) to release a fragment containing the first three domains. This cleavage yields active thrombin. 
  17. 1)      What is the importance of vitamin K for the synthesis of prothrombin?
    a.       Normal prothrombin contains gamma-carboxyglutamate, which is a strong chelator of Ca2+, allowing prothrombin to bind Ca2A+, anchoring the zymogen to phospholipid membranes derived from blood platelets after injury. The binding of prothrombin to phospholipid surfaces brings prothrombin into close proximity to two clotting proteins that catalyze its conversion to thrombin. The calcium-binding domain is then removed during activation, freeing thrombin from the membranes. 
  18. 1)      What are the mechanisms that normally limit clot formation at the site of injury?
    a.       The lability of clotting factors contributes to the control. Activated factors are short lived because they are diluted by blood flow, removed by the liver, and degraded by proteases. So, thrombin can both catalyze the formation of thrombin but also initiate a deactivation clotting cascade. Antithrombin III can also inactivate thrombin by forming an irreversible complex with it. 
  19. 1)      What happens to the clots themselves?
    They are dissolved. Fibrin is split by plasmin, a serine protease that hydrolyzes peptide bonds in the coiled-coil region. Plasmin molecules diffuse through aqueous channels to cut the accessible fibrin connector rods. Plasmin’s activation from plasminogen is activated by tissue-type plasminogen activator (TPA). When TPA is bound to fibrin clots, it activates adhering plasminogen, not free plasminogen

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