Chapter 13 Final Review: Essays

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  1. 1)      What are the two types of ATP-driven pumps? 
    a.       They are Ptype ATPases and ATP binding cassette transporters, which undergo conformational changes on ATP binding and hydrolysis that cause a bound ion to be transported acros the membrane. This energy is used to drive the movement of ions against their concentration gradients, called primary active transport. 
  2. 1)      Why is the collection of fixed transporters important? 
    a.       It is important because it determines the ionic composition inside cells and the compounds that can be taken up from the cell’s environment. It defines the cell’s characteristics because a cell can execute only those biochemical reactions for which it has taken up necessary substrates. 
  3. 1)      What two factors determine whether a molecule will cross a membrane? 
    • a.       The permeability of the moelcuel
    • b.      The availability of an energy source
  4. 1)      How does sodium enter the cell? 
    a.       It passes through specific channels in the hydrophobic barrier formed by membrane proteins. It crosses via facilitated diffusion and can also be passive because energy is not really needed. 
  5. 1)      How is the sodium gradient established in the first palce? 
    Sodium must be pumped against a concentration gradient. Because moving the ion froma low to high concentration decreases entropy, energy is required. And so, it is active transport
  6. 1)      What are the two conformations a pump can exist in?
    a.       One with the bidnign site open to one side of the membrane and the other with the ion-binding sites open to the other side. To pump ions in a single direction across a membrane, the free energy of ATP hydrolysis must be coupled to the interconversion between these conformational states. 
  7. 1)      Explain SERCA.
    a.       This protein plays a role inrelaxation of contracted muscle, which is originally triggered by an abrupt rise in the cytoplasmic calcium ion level. Subsequent muscle relaxation depends on the rapid removal of Ca2+ from the cytoplasm into the sarcoplasmic reticulum. This pump maintains a Ca2+ concentration in the cytoplasm that is different than the SR. 
  8. 1)      What is the structure of SERCA?
    • a.       It is a single 110-kd polypeptide with a transmembrane domain consisting of 10 alpha helices. The transmembrane domain includes sites for binding two calcium ions, each of which is coordinated to seven oxygen atoms coming from a combination of side chain glutamate, aspartate, threonine, and asparagine residues, backbone carbonyl groups, and water molecules. A large cytoplasmic headpieces consists of three distinct domains, each with a different function.
    •                                                               i.      N domain: binds the ATP nucleotide
    •                                                             ii.      P domain: accepts the phosphoryl group on a conserved aspartate residue
    •                                                           iii.      A domain: serves as an actuator, linking changes in the N and P domains to the transmembrane part of the enzyme.
  9. 1)      How does SERCA’s conformation change upon binding ATP (or an analog)?
    a.       The N and P domains close around the phosphorylaspartate [analog], and the A domain rotates substantially relative to its position in SERCA with Ca2+ bound and without the phosphoryl [analog]. Furthermore, the transmembrane part of the enzyme rearranges substantially and the well-organized Ca2+-binding sites are disrupted, becoming accessible. 
  10. 1)      What is the overall mechanism of SERCA (and Na+-K+ ATPase)?
    • a.       The cycle begins with the enzyme in its unphosphorylated form with two calcium ions bound (E1); with Ca2+ bound, its E1-(Ca2+)2. In its E1-(Ca2+)2 state, SERCA can exchange calcium ions only with calcium ions from the cytoplasmic side of the membrane
    • b.      In the E1 conformation, the enzyme can bind ATP. When it does so, the N, P, and A domains undergo substantial rearrangements as they close around the bound ATP, but there is no substantial conformational change in the transmembrane domain. The calcium ions become trapped inside the enzyme.
    • c.       The phosphoryl group is then transferred from ATP to Asp 351.
    • d.      Upon ADP release, the enzyme again changes its overall conformation, including the membrane domain. This new conformation (E2 or E2-P if phosphorylatd) interconverts the E1 and E2 conformations, called eversion.
    •                                                               i.      In the E2-P conformation, the Ca2+-binding sites become disrupted and the calcium ions are released to the side of the membrane opposite that at which they entered; ion rnsport has been achieved
    • e.      The phosphorylaspartate residue is hydrolyzed to release inorganic phosphate
    • f.        With the release of phosphate, the interactions stabilizing the E2 conformation are lost, and the enzyme everts to the E1 conformation.
    • g.       Two more Ca2+ bind, completing the cycle
    • h.      This cycle can coincide with Na-K ATPase, which has an a2B2 tetremer; B does not directly take part in binding. 
  11. 1)      Explain inhibition by certain steroids.
    • a.       Digitoxin and ouabain are cardiotonic steroids that inhibit the dephosphorylation of the E2-P form of the ATPase when applied to the extracellular face of the membrane.
    • b.      Digitlis increase the force of contraction of heart muscle and is a good drug for congestive heart failure.
    • c.       Inhibition of the Na+-K+ pump by digitalis leads to higher levels of Na+ in the cell. The diminished Na+ gradient results in slower extrusion of Ca2+ by the Na-Ca2+ exchanger. The increase in intracellular levels of Ca2+ enhances the ability of cardiac muscle to contract
  12. 1)      What is the mechanism of active transport for ABC protiens? 
    • a.       The catalytic cycle begins with the transporter free of both ATP and substrate. The transporter can interconvert between closed and open forms.
    • b.      Substrate entes the central cavity of the open form from inside the cell, inducing a conformational change in the ABC proteins that increases their affinity for ATP
    • c.       ATP binds to the ATP binding casettes, changing their conformations so that the two domains interact strongly with each other
    • d.      The strong interaction between the ATP-binding casettes induces a change in the relation between the two membrane-spanning domains, releasing the substrate to the outside of the cell
    • e.      The hydrolysis of ATP and the release of ADP and inorganic phosphate reset the transporter for another cycle
  13. 1)      What is the mechanism for symporter action? 
    • a.       In the protonated form, the permease bidns lactose from outside the cell.
    • b.      The structure everts
    • c.       Lactose and a proton are both released to the inside of the cell
    • d.      The permease everts to complete the cycle
  14. 1)      Explain action potential.
    a.       In a resting state, the membrane potential is -60 mV. A nerve impulse, or action potential, is generated when the memrbaen potential is depolarized beyond the critical threshold value (-60 to -40 for example). The membrane potential becomes positive and attains a value of about +30 beffore repolarizing. 
  15. 1)      What does depolarization fo the membrane cause?
    a.       Depolarization beyond the threshold level leads to an increase in permeability to sodium ions, which begin to flow into the cell due to a large electrochemical grident across the membrane. The entry of sodium ions further depolarizes the membrane, leading to further increase in sodium ion permeability. This leads to a very large change in embrane potential. The membrane eventually becomes less permeable to Na and more to K+, which flows outward so the membrane potential returns toa negative value. 
  16. 1)      What is the structure of the sodium ion channel?            
    a.       It is a single 260 kd chain made of four internal repeats, each with a similar amino aci sequence. Each repeat contains five hydrophobic segments (S1,S2,S3,S5, and S6) and one positively charged S4 segment. Segments 1 throguh 6 are membrane spaning alpha helices and the positively charged residues in S4 are votage sensors. 
  17. 1)      What is the structure of the potassium ion channel? 
    a.       The channel contains only the pore-forming segments S5 and S6. It is a tetramer of identical subunits, each of which includes two membrane-spanning alpha helices. The four subunits come together to form a pore in the shape of a cone that runs through the center fo the structure. 
  18. 1)      How are the channels (sodium and potassium) exclusive?
    a.       K+ channel: beginning from inside the cell, the pore starts with a diameter of 10 angstroms and then constricts to a smaller cavity. Both the opening and the central cavity of the pore are filled with water, and a K+ ion can fit into the pore without losing its shell of bound water molecules. About two thirds of the way thorugh the membrane, the pore becomes restricted, causing potassium ions to give up their water moelcuels and interact directly with groups from the protein. The structure allows the solvated ions to penentrate into the membrane before the ions interact with the channel. For K+ to relinquish their water molecules, other polar interactions must replace those with water. The restricted part is built from residues contributed tby the two transmembrane alpha helices. In particular, a selectivity filter has the sequence TVGYG, which lies in an extended conformation and is oriented such that the peptide carbonyl groups are directed into the channel. Though K+ can do this, sodium ions are hindered because they lose more energy releasing their water than potassium does, which is compensate with the carbonyl oxygen atoms lining the selectivity filter. 
  19. 1)      Explain potassium ion transport. 
    a.       Four K+-binding sites are crucial for rpid ion flow int eh constricted region of the K+ channel. A hydrated potassium ion proceeds into the channel through the unrestricted part of the channel. The ion then gives up its coordinated water molecules and binds to a site within the selectivity filter region. The ion can move between the four sites within the selectivity filter because they have similar ion affinities. As each potassium ion moves into the selectivity filter, its positive charge will repel the potassium ion at the nearest site, causing it to shift to a site farther up the channel and in turn push upward any potassium ion already bound to a site farther up. Thus, each ion that binds anew favors the release of an ion from the other side of the channel. 
  20. 1)      What is the structure of a voltage-gate channel? 
    a.       The segments S1 through S4 form domains termed paddles hta textend from the core of the channel. These paddles include the segment S4, the voltage sensor itself, which forms an alpha helix lined with positively charged residues. Segments S1 through S4 are not enclosed withint he protein but are positioned to lie in the membrane. 
  21. 1)      What is a mechanism for voltage gating? 
    a.       In the closed state, the paddles lie in a down position. On membrane depolarization, the cytoplasmic side of the membrane becomes more positively charged, and the paddles are pulled through the membrane in an up position. In this position, they pull the four sides of the base on the pore apart, increasing access to the selectivity filter and opening the channel 
  22. 1)      Briefly explain what happens when a nerve impulse reaches the end of an axon.
    a.       It leads to a synchronous export of the contents of some 300 vesicles of acetylcholine into the cleft. The binding of acetylcholine to the postsynaptic membrane changes its ionic permeability, triggering an action potential. Acetylcholine opens a single kind of cation channel, the acetylcholine receptor, which is almost equally permeable to Na+ and K+
  23. 1)      What is the structure of theacetylcholine receptor? 
    a.       It is a pentamer fo four kinds of membrane-spanning subunits—alpha2, beta, gamma, and sigma—arranged in the form of a ring that creates a pore through the membrane. Acetylcholine binds at the alpha-gamma and alpha-sigma interface. 
  24. 1)      What is the basis of the channel opening upon acetylcholine binding? 
    • a.       Binding to the extracellular domains causes a structural alteration that initiates rotations of the alpha helical rods lining the membrane-spanning pore. The amino acid sequences of these helices point to the presence of alternating ridges of small polar or neutral residues and large nonpolar ones.
    • b.      In the closed state, the large residues occlude the channel by forming a tight hydrophobic ring. Bindng of acetylcholine allosterically rotates the membrane-spanning helices so that the pore would be lined by small polar residues rather than by large hydrophobic ones.
  25. 1)      What happens in the generation of an action potential? 
    • a.       A neurotransmitter is released, binds to receptor, causing it to open. Because the receptor is a nonspecific cation channel, sodium flows in and potassium flows out.
    • b.      The membrane potential moves to a value corresponding to the average of the equilibrium potentials of sodium and potassium.
    • c.       As membrane potential approaches -40 mV, the voltage sensing paddles of Na+ channels are pulled into the membrane, opening the sodium channels, allowing sodium to flow in and the membrane potential to rise toward the Na equilibrium potential.
    • d.      The slower pulled voltge sensing paddles of K strt to open at the same time that Na is inactivated by the ball.
    • e.      With the Na+ channels inactivated and only K+ open, the membrane potential rops rpidly owrd the K_ equilibrium potential.
    • f.        Eventually, K channels are inactivated by their ball domains and they become blocked.
    • Membrane potential returns to cintial value and inactivation domains are released and the channels return to their original closed states
  26. 1)      What are gap jucntions?? 
    a.       They are cell-to-cell channels that serve as passageways between the interiors of contiguous cells. They have a central hole, the lumen of the channel; and they span the gap between apposed cells. Small hydrophilic molecules pass through; and, so do ions. Proteins, amino acids, etc. are too large. They are important for intercellular communication since they can be coupled by the rapid flow of ions through the junctions, which ensures a rapid and synchronous response to stimuli. 
  27. 1)      What is the structure of gap junctions?
    a.       They are made of 12 molecules of connexin, which has four membrane spanning helices. Six connexin molecules form a connexon (hemichannel); and two connexons join end to end to form the channel between the cells. 
  28. 1)      How do gap junctions differ from other membrane poroteins? 
    • a.       They traverse two membranes rather than one
    • b.      They connect cytoplasm to cytoplasm
    • c.       The connexons forming a channel are made by different cells
  29. 1)      How are gap junctions closed?
    a.       They are closed by high concentrations of calcium ion and by low pH. It serves to seal normal cells from injured or dying neighbors
  30. 1)      What is the structure of aquaporins? 
    a.       It consists of six membrane-spanning alpha helices. Two loops containing hydrophilic residues line the actual channel and wter molecuels pass through single file. Specific positively charged residues toward the center prevent the transport of protons. 
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Chapter 13 Final Review: Essays
2014-11-13 04:46:27
Test Three
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