Lectures 8-19 Flashcards
(189 cards)
1
Q
What is the delta G of a thermodynamically favored event?
A
Negative deltaG. The G (free energy) of the products is lower than that of the reactants.
2
Q
Why does a thermodynamically favored process not necessarily proceed spontaneously then?
A
Often there is a high-energy intermediate or transition state.
3
Q
What is the energy required to reach the transition state?
A
The activation energy.
4
Q
Enzymes do what?
A
They form transitory bonds with the reactants. These enzyme-reactant complexes have relatively low energy and make the reaction a more likely even by lowering the activation energy.
5
Q
Do enzymes affect the equilibrium of the reaction?
A
NO. Equilibrium depends solely on deltaG. Spontaneously, given enough time, product would accumulate, and the reverse reaction to form the initial reactants would be even less likely. At steady state, the relative concentrations of reactants and products would be the same, whether the reaction were catalyzed or not. The enzyme changes ONLY the reaction rate.
6
Q
What is Le Chatelier’s principle?
A
A change in the concentration of reactant or product will shift the equilibrium of the reactions involved. AKA if you have a reaction A->B->C , Continual depletion of B in the reaction catalyzed by Enzyme 2 increases the rate of A->B.
7
Q
Describe the Lock and Key method.
A
The lock-and-key model postulates that the binding site is a static, rigid structure that selects for substrates that fit optimally (not only with respect to shape, but also in the potential to form bonds with residues in the active site). This model has largely been superseded by the induced-fit model, which offers a better account of the thermodynamic basis for catalysis.
8
Q
Describe induced fit.
A
In the induced-fit model, the active site of the enzyme in its unbound state includes a low-affinity binding site for the substrate. When the substrate binds to this site, the enzyme undergoes a conformational change (see figure below) that increases the active site’s affinity for the substrate, but which requires that the substrate undergo steric changes that bring it closer to a transition state.
9
Q
Use metal rod binding to explain the advantage of induced fit over lock and key.
A
1st screenshot.
10
Q
When is a reaction first order?
A
The initial rate is directly proportional to [S]. Rate here is dependent on [S].
11
Q
When is a reaction zero order?
A
At very high substrate concentrations, the relationship flattens, and V0 becomes constant. In this region, where rate is independent of [S], the reaction is said to be zero-order.
12
Q
What is Vmax?
A
The highest possible reaction rate for a given catalyzed reaction.
13
Q
How do you determine which enzyme out of two has a higher affinity for the substrate?
A
Whichever enzyme reaches vmax at a lower substrate concentration.
14
Q
What is Km
A
The substrate concentration at which V0 is 50% of Vmax
15
Q
What is the Michaelis-Menten equation?
A
16
Q
When [S] is much lower than Km what can we do?
A
We can ignore [S] in the denominator, and since Vmax and Km are constants, V0 is directly proportional to [S] (first-order behavior).
17
Q
When [S] is much larger than Km what can we do?
A
When [S] is far larger than Km, we can ignore Km in the denominator, and V0 = Vmax (zero-order behavior).
18
Q
Rearrange the Michaelis Menten equation.
A
1/Vo=
19
Q
How does increasing the concentration of the substrate affect a competitive inhibitor?
A
increasing the concentration of the substrate can displace a competitive inhibitor. At sufficiently high substrate concentrations, which effectively deny the inhibitor access to the binding site (all of the binding sites are occupied by S), the enzyme catalyzes the reaction at Vmax.
20
Q
How is the Y-intercept on a Lineweaver-Burk plot (1/Vmax) affected by a competitive inhibitor?
A
It is not affected, it remains the same.
21
Q
What value changes with a competitive inhibitor?
A
The apparent Km because it takes more substrate (or higher [S]) to displace the inhibitor and occupy the number of binding sites that yield 50% of Vmax. As a result of this the Lineweaver-Burk plot (Km/Vmax) becomes STEEPER and the X intercept (-1/Km) shifts to the RIGHT. See second screenshot for plot.
22
Q
What is the equation used to describe the relationship of Vo and [S] in the presence of a competitive inhibitor?
A
23
Q
How does increasing the substrate concentration affect a non-competitive inhibitor?
A
It doesn’t. Even at saturating substrate concentrations, where all of the enzyme is substrate-bound, Vmax is reduced from its value in the absence of inhibitor.
24
Q
How is the Lineweaver-Burk plot altered for a non-competitive inhibitor?
A
Since Vmax is reduced, the y-intercept (1/Vmax) increases. Because substrate binding is unaffected, Km stays the same and therefore so does the X-intercept (-1/Km). This causes the slope of the line to become steeper or INCREASE. See third screenshot.
25
What is uncompetitive inhibition?
Uncompetitive inhibitors can only bind the enzyme after substrate has bound, since formation of the ES complex creates the site for the inhibitor.
26
How is the Lineweaver-Burk plot affected by uncompetitive inhibitors?
When the inhibitor binds, it stabilizes the ES complex, thus slowing the dissociation of the substrate from the enzyme
(ESE +S), and this reduces Km, shifting the x-intercept to the left. The inhibitor also interferes with the catalytic activity of the enzyme, so Vmax decreases as well. In the presence of an uncompetitive inhibitor, the Lineweaver-Burk plot is shifted UPWARD and PARALLEL to the no-inhibitor setting, since Km and Vmax decrease to similar degrees.
27
When a regulatory enzyme is regulated by the binding of a modulator this is called...
An allosteric enzyme.
28
What can modulate an allosteric enzyme?
Either the substrate itself (in which case the modulatory site is the same as the active site) or some other molecule that acts outside the active site.
29
What is homotropic modulation?
When the substate itself is the modulator of the enzyme. Such enzymes have multiple active sites, usually on different subunits, and the binding of the substrate to the first active site increases or decreases the affinity of other active sites for the substrate.
30
What is positive cooperativity?
When the binding of a substrate increases the affinity for more binding to the enzyme (O2 on hemoglobin).
31
What curve demonstrates positive cooperativity?
Sigmoidal instead of hyperbolic.
32
What is heterotropic cooperativity?
When the enzyme responds to a modulator other than its substrate. Often the product. This is negative feedback.
33
What is an example of negative feedback where the product is many steps beyond that?
Heme is the product and feeds all the way back to give feedback to gamma-aminolevulinic acid synthetase which catalyzes the rate-limiting step of the reaction 8 steps earlier.
34
Positive modulators affect Km and Vmax how?
Usually decrease Km with no effect on Vmax.
35
Negative modulators affect Km and Vmax how?
negative modulators can produce an apparently competitive inhibition (increased Km, no effect on Vmax; see middle panel, lower plot), or apparently non-competitive inhibition in which Vmax is reduced but Km unaffected (right panel).
36
Describe Thr286, how it gets activated and what happens when it does.
Ca2+ is a ubiquitous second messenger whose effects on proteins often depend on its complexing with calmodulin, a small, highly conserved protein that can bind as many as four Ca2+ ions. These Ca2+ ions induce a conformational change in calmodulin, exposing a hydrophobic surface that can bind to hydrophobic pockets in other proteins and acting as a modulator (usually positive).
One of the proteins to which Ca2+/calmodulin binds is CaMKII (see figure on next page), a serine/threonine protein kinase (that is, it phosphorylates those residues, provided that they are in a particular context with respect to neighboring amino acids and geometry). CaMKII, which is one of the most abundant proteins in the brain and implicated in memory formation, is comprised of 12 homologous subunits (it is a dodecamer). Each of the 12 subunits is a single polypeptide that contains a catalytic domain, a regulatory domain, a hinge that connects them, a phosphorylatable threonine (Thr286) within the hinge region, and a hydrophobic Ca2+/calmodulin binding pocket. In the inactive resting conformation, the hinge is closed and the regulatory domain makes the catalytic domain inaccessible to potential substrates.
A rise in intracellular Ca2+ produces an increase in the concentration of Ca2+/calmodulin, which binds to its hydrophobic site in CaMKII and causes CaMKII to adopt an open conformation that exposes the catalytic domain as well as Thr286. CaMKII now can phosphorylate its substrates, but the level of Ca2+/calmodulin usually drops quickly, so it quickly dissociates from CaMKII and the enzyme again is inactive. This is how modulators work, in a readily reversible manner. However, if the Ca2+ increase was particularly large, Ca2+/calmodulin levels stay high for longer than usual, and this gives CaMKII subunits the opportunity to phosphorylate the exposed Thr286 of a neighboring subunit. Once the Thr286 is phosphorylated, the hinge cannot close even after Ca2+/calmodulin has dissociated, and CaMKII remains active for much longer – until a protein phosphatase called PP1 can remove the phosphate group from Thr286.
37
What is a zymogen?
An inactive precursor that has to be cleaved to become an active enzyme.
38
What is a kinase?
enzymes that transfer a phosphate group from a high-energy donor (such as ATP) to a substrate are called kinases (for example, cAMP-dependent protein kinase)
39
What is a cofactor?
Many enzymes require an additional small molecule, called a cofactor, to perform their catalytic function. In some cases, the cofactor is covalently bound to the protein part of the enzyme, in which case the protein itself is called the apoenzyme and the full complex is the holoenzyme.
40
What is Kcat?
The number of operations that a single molecule of enzyme can perform per second. Kcat is referred to as the turnover number.
41
What is the equation for Kcat?
Kcat= Vmax/[Et]
Where [Et] is the molar quantity of enzyme present in the reaction.
42
What is the best estimation of the efficiency of an enzyme under physiological conditions?
The specificity constant, which is Kcat/Km. This value (with units M-1s-1) indicates how efficient the enzyme is when free binding sites are abundant.
43
If Kcat/Km is large what does that indicate?
If kcat/Km is large, then the reaction may proceed at a reasonable rate even if the enzyme is not highly expressed (that is, [ET] is small) or if substrate concentration is low.
44
What does the Na/K ATPase pump do?
Exports 3 Na+ molecules and imports 2 K+ molecules at the expense of one ATP.
45
What does the Na/K ATPase pump look like in the E1 conformation?
It has 3 high affinity Na+ binding sites facing the cytosol that become occupied. There are also two low affinity unoccupied K+ binding sites.
46
How does Na/K ATPase pump change conformation from E1 to E2?
ATP binds to the pump and is hydrolyzed by the pump’s ATPase activity, and a high-energy phosphate bond is formed with an aspartate residue on the cytoplasmic side of the pump.The energy of the phosphate is used to drive a conformational change in the protein to E2, a process called the “power stroke”. As a result, the Na+ ions move to low-affinity sites that are exposed to the extracellular space, where they are weakly bound. The E2 conformation also presents two high-affinity K+ binding sites to the extracellular side
47
How does Na/K ATPase pump change conformation from E2 to E1?
The three Na+ ions diffuse away from their low-affinity sites and into the extracellular space. At the same time, two K+ ions from the outside bind their high-affinity sites, causing hydrolysis of the aspartyl-phosphate bond. The loss of the phosphate group returns the pump to its E1 state, transferring the K ions to their low-affinity binding sites that face the inside of the cell, from which they dissociate into the cytoplasm.
48
How is the Ca2+ concentration kept low within the cell?
The SERCA P-class pump, which sequesters Calcium from the cytosol into the sarcoplasmic reticulum. It operates very similarly to the Na/K ATPase pump: when Ca2+ is bound to the high affinity cytosolic sites, ATP is hydrolyzed, the liberated phosphate group forms a high- energy bond with an aspartate on the cytoplasmic side, inducing a conformational change (to E2) that closes off the Ca2+ pocket from the cytoplasmic side, trapping Ca2+ in the protein.
49
What do P-class pumps move?
Only ions.
50
What do V/F pumps move?
ONLY protons.
51
What do V-class pumps do?
```
V-class pumps
contribute to the acidification of organelles such as lysosomes by pumping protons from the cytoplasm to the lumen of the organelle.
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52
What do F-class pumps do?
F-class pumps, one type of which (F0-F1 ATPase) is highly expressed in mitochondria, typically run “in reverse”, using the movement of protons down their gradients to synthesize ATP.
53
How do ABC pumps bind ATP?
Through conserved regions called ATP-binding cassettes (thus, “ABC”).
54
What do ABC pumps transport?
Unlike the P- and V/F classes, ABC pumps often transport uncharged and even hydrophobic molecules.
55
What are Multi-drug resistance (MDR) proteins?
A type of ABC pump. They are highly expressed in epithelial cells. They transport small, polar molecules, including some products of normal metabolism, but they can also pump a wide variety of drugs out of cells. Thus, tumors that overexpress MDR proteins are resistant to treatment by multiple and unrelated anticancer drugs.
56
What is the cystic fibrosis transmembrane regulator (CFTR)?
An ABC pump expressed in the lung and other organs. Although structurally an ABC-class pump, it has no known “pumping” function. However, it incorporates a channel that is permeable to Cl–, and that is regulated by protein kinase A. Cystic fibrosis has been linked to loss-of-function mutations in CTFR, which reduce Cl– transport across pulmonary epithelial cells. As a result, the mucus secreted by these cells becomes excessively viscous, compromising gas exchange and predisposing the lung to infection.
57
How do transporters differ from pumps?
Transporters, like pumps, bind solutes and undergo conformational changes to ferry them across the membrane. In contrast to the pumps, the transporters have no ATPase activity; rather, they rely on existing gradients to move solutes.
58
What does a uniporter do?
They conduct a single species of molecule down its gradient, facilitating a thermodynamically favored process by circumventing the hydrophobic barrier of the lipid bilayer (facilitated diffusion).
59
What do co-transporters do?
They couple the thermodynamically favorable movement of one type of molecule (down its gradient) to the unfavorable movement of another (secondary active transport).
60
Symporter
Transporter moving solutes in the same direction across the membrane.
61
Exchanger
Also called an anti porter is a transporter move two solutes in opposite directions.
62
Describe the action of a Glucose Unitransporter (GLUT)
GLUTs bind a single molecule of glucose at a time. A conformational change exposes the glucose-binding site alternately to the extracellular and intracellular sides, and the rate of cycling is accelerated by occupation of the binding site in either conformation. In most cells, the concentration of glucose is higher outside the cell than inside, so the extracellular binding site is more likely to become occupied. The intracellular concentration of glucose is kept low by the rapid phosphorylation of glucose to glucose-6-phosphate. Upon the conformational change, the glucose is exposed to the low concentration inside the cell, and it diffuses away from the binding site. GLUT spontaneously returns to its initial conformation, and another molecule of glucose can bind from the outside.
63
Describe the Na+/Ca2+ exchanger.
Na+/Ca2+ exchanger is an antiporter that couples Na+ entry to Ca2+ efflux. Thus, it serves a role in Ca2+ handling that is complementary to the sequestration performed by the SERCA pump, and in cardiac cells it contributes to Ca2+ clearance more than SERCA does. It has a stoichiometry of 3 Na+ : 1 Ca2+, so it is electrogenic, accumulating positive charges inside the cell.
64
Which SLGT Is expressed in the early part of the renal tubule?
SLGT2.
65
What does SLGT2 do?
In the early part of the tubule that is near the glomerulus, the concentration of glucose in the urine is relatively high, so the transporter that is expressed in this region (SGLT-2) doesn’t need to work against a large concentration gradient in taking up glucose into the cell. The stoichiometry of SGLT-2 is 1 Na+ : 1 glucose, which is sufficient to accumulate glucose against a 100-fold gradient.
66
Which SLGT is expressed later on in the kidney tubule?
SLGT1.
67
What does SLGT1 do?
There is so little glucose later in the tubule rabsorption becomes more of a challenge since the glucose gradient strongly resists movements from the lumen into the cell. SGLT-2 would be ineffective. Cells in this part of the tubule express the SGLT-1 symporter, whose stoichiometry is 2 Na+ : 1 glucose. By allowing the Na+ gradient to deplete by 2 molecules for each glucose molecule reabsorbed, SGLT-1 can work against a glucose gradient up to 30,000-fold. As a result, essentially no glucose is excreted in the urine.
68
Which channels provide predominant membrane conductance?
Channels that are selective for K+ usually provide the predominant membrane conductance.
69
What is equilibrium potential? (Veq)
Also known as reversal potential (Vrev), it is the value of Vm when the concentration gradient and the electrostatic force are equal and opposite.
70
Nernst equation:
Veq(S)=RT/ZsF*ln([So]/[Si])
Simplified...
Veq(S)=58/Zs*log([So]/[Si])
71
If there is 150 K+ inside the cell and 4K+ outside the cell calculate Vk.
Vk=58/1*log(4/150)= -91mV.
72
At resting -71 membrane potential if you allowed the flow of K+ ions what would happen in which direction?
If the resting membrane potential is –70 mV, the opening of K+ channels will cause the cell to hyperpolarize. The net movement of K+ in this case (the K+ current) is toward the outside of the membrane. The movement of cations out of the cell is referred to as an outward current, and outward currents hyperpolarize membranes.
73
Which kind of current depolarizes membranes?
Inward current, or flow of positive ions into the cell.
74
What is the driving force?
Driving force= Vm- Veq
So if membrane is -70 and Vk is -91, the driving force is +21 mV.
75
What is conductance?
The ability of a material to carry current.
76
What is the electrophysiological unit for conductance?
The picosiemen (pS)
77
What is the unit of resistance?
The ohm.
78
What is the electrophysiological unit for resistance?
Biological membranes usually have very high high resistance, and the total membrane resistance of a cell usually is expressed in megohms (MΩ; 106 ohms).
79
What is Ohm's law?
Ohm’s law describes the simple relationship between current, voltage, and conductance: current = conductance x voltage
or I=g*V
Where I represents the membrane current for a given species, g represents the conductance of the membrane for that ion, and V represents the driving force for that ion.
80
There is a linear relationship between the membrane current (I) carried by a specific ion species, and the driving force (Vm-Veq) for that ion. What is the slope of that line?
Conductance, or g.
81
Is conductance voltage dependent?
No. We know this because of the linear relationship of I to V.
82
How does Voltage relate to conductance?
Voltage is inversely proportional to conductance.
| V=I/g
83
How does Voltage relate to resistance?
Voltage and resistance are directly proportional. V=I*R.
So, the greater the resistance of the membrane to electrical current (e.g., the fewer open ion channels), the more the membrane potential changes in response to the application of a given current.
84
Why doesn't exhibit Ohm's law in a time course matching that of the applied current?
Because membranes have capacitance, or the ability to store charges.
85
What makes up an inward rectifying K+ Channel?
They're made up of tetramers of homologous subunits. The M2 helical regions are on the inside and taper down towards each other on the intracellular side. The M1 subunits are more lateral. In the pore between the subunits a loop projects down (p-loop) creating a narrowing that only allows for the passage of K+ ions.
86
What is the selectivity filter?
The selectivity filter comprises a signature sequence of 5 - 6 residues within the P-loop. In this region, carbonyl groups from the protein backbone of each subunit project into the pore. These form rings of dipoles, with the electronegative oxygen oriented towards the center.
87
How many K+ can fit into the selectivity filter at once?
1
88
What keeps Na+ out of the selectivity filter?
The carbonyl groups of the selectivity filter project into the pore with the electronegative oxygen facing outwards at just the right distance so that the K+ ions interact with them on either side giving them a good attachment. Na+ are smaller and therefore has a greater distance and weaker interaction with the carbonyls.
89
In order for K+ to pass via an inward rectifier channel it must be dehydrated. How is it stable enough to do that.
The carbonyl groups in the selectivity filter are key for this. They are optimally positioned so that a dehydrated K+ ion can interact with all four carbonyls within a ring, thereby compensating for the cost of dehydration.
90
What happens if a cation enters a Kir from the intracellular side?
They can't get past the selectivity filter but can enter the pore and interact with negative residues. When this happens the poor is blocked and conductance is low. Mg2+ and spermine are both capable of this.
91
Under normal physiological conditions which direction is K+ net flow?
Net outward. Because Vm is always positive to VK (remember that K+, like any ion, moves through the membrane in the direction needed to push Vm towards the ion’s Veq).
92
What is the average conductance of a Vir when a membrane is depolarized?
Relatively low because they remain blocked.
93
What is the average conductance of a Vir when a membrane is hyperpolarized?
Conductance is relatively high because there would be more K+ inward current unblocking the pore.
as an example...
What would happen if the membrane potential were set at -150 mV, which is very negative to VK? The K+ current would be strongly net inward, and this would keep the blocking cations out of the pore. Accordingly, these channels can carry large inward K+ currents – something that is seen only under experimental conditions.
94
What tends to make a voltage gated channel open? Membrane depolarization or hyperpolarization?
They tend to open on depolarization which is in contrast to the Vir.
95
Describe the structure of the K+ Voltage gated channels.
These channels are tetramers, with each subunits containing six transmembrane helices named S1 through S6 (Fig. 6). The most carboxy-terminal of these helices, S5 and S6, together with their linker region, are highly homologous to M1 and M2 of the inward rectifiers, with the same inverted tepee structure, P-loop, and characteristic K+ channel sequence within the selectivity filter. The S6 helices are the ones that line the pore. The four other transmembrane helices are mostly embedded in the lipid bilayer lateral to the channel pore. Among these, S4 is unique in containing multiple positively charged residues.
96
Which helices in voltage gated K+ channels are analogous to M1 and M2 in Kir channels?
S5 and S6. S6 resembles M2 and lines the pore.
97
Which helices in voltage gated K+ channels have multiple positively charged residues?
S4.
98
What is the function of the S4 helices?
It is the voltage sensor.
99
How does S4 work?
When Vm is well polarized, S4 is positioned near the intracellular side of the membrane, since it is attracted to the uncompensated negative charges in the cytoplasm. With the S4 helices in this position, the S6 helices are pinched together near the intracellular entrance to the pore, preventing ion flow. This point of convergence is the activation gate, which must be opened before the channel can pass a current. Upon membrane depolarization, there are fewer uncompensated negative charges near the inner face of the membrane to attract the S4 helices, which then move toward the outer leaflet of the bilayer. When all four of the S4 helices move to this outer position, the activation gate is pulled open and K+ ions can flow through the pore.
100
How do voltage gated K+ channels inactivate?
Voltage-gated K+ channels carry their own intracellular blocking cations, in the form of basic amino acids at the N-terminus of each subunit (the “ball-and-chain”). When the activation gate opens, binding sites within the pore for the positively charged ball are exposed, enabling the ball to lodge in the pore and block ion flow.
101
How does the Voltage gated K+ channel activation gate close?
even upon membrane repolarization and return of the S4 voltage sensors to the inner leaflet, the activation gate cannot close until the ball-and-chain vacates the pore. When the ball-and-chain does exit (probably attracted by the relatively negative interior of the membrane), a flicker of current can sometimes be detected in the brief interval before the activation gate closes.
102
What is the structure of voltage gated Na+ and Ca2+ channels?
the Na+ and Ca2+ channels are formed from a single polypeptide with 24 transmembrane helices that are arranged in four domains, each with six helices. These domains, numbered I – IV, are homologous to the separate subunits of voltage-gated K+ channels.
103
How do voltage gated Na+ and Ca2+ channels inactivate?
Similar to V gated K+ channels though instead of the ball and chain, positive charges on the long intracellular linkers can enter the pore and block it.
104
For example, if a depolarizing current causes inward rectifiers to close, would depolarization be smaller or larger than expected by Ohm’s law?
Larger because membrane conductance for K+ is reduced.
105
What does tetrodotoxin (TTX) do?
Blocks voltage gated Na+ channels.
106
What does tetraethyl ammonium (TEA) do?
Blocks most v-gated K+ channels.
107
When does the refractory period end?
The refractory period ends when the full complement of Na+ channels has been de-inactivated.
108
Describe the propagation of an action potential along a membrane.
The propagation of an action potential along the axon of a neuron. The potential was initiated to the left of the segment depicted, and is moving toward the right through adjacent regions of the membrane. Na+ entering the cytoplasm through voltage- gated Na+ channels during the rise of the action potential (blue arrows) results in a localized depolarization that spreads, passively for some distance from the site of channel opening. (Note that this spread does not depend on the movement of Na+ ions through the cytoplasm, but rather on a near-instantaneous redistribution of charges in response to the entry of uncompensated positive ions.) This + process brings the membrane to the immediate right of the Na entry to the threshold for initiating an action potential, which is identical in height and duration to its predecessor (propagation without loss). In the wake of the Na+ influx, K+ exits the cell through the delayed rectifiers, repolarizing the membrane and terminating the action potential in that region. Propagation cannot move toward the left because the recently-opened Na+ channels have yet to de- inactivate.
109
Why does heart muscle spend a substantial part of its time depolarized?
Heart muscle cells, unlike skeletal muscle and most neurons, spend a substantial portion of their lives in a depolarized state because the mechanical pumping action of the heart relies on contractions that last hundreds of milliseconds.
110
Describe and draw a ventricular action potential.
At rest (Phase 4; diastole), these cells are very well polarized at about -90 mV, and inward rectifying K+ (Kir) channels provide the dominant conductance. When the membrane is depolarized to the threshold for action potential initiation, voltage- gated Na+ channels are activated. This results in a rapid depolarization (Phase 0;upstroke), followed by a partial repolarization as most, but not all, of the Na+ channels inactivate (Phase 1; early/fast repolarization). Depolarization also causes the inward rectifiers to close, thereby eliminating what would otherwise be a very large outward current that would prematurely terminate the action potential. Phase 2 (plateau) reflects the delayed opening of voltage-gated Ca2+ channels that inactivate with very slow kinetics, as well as a smaller inward current through the remaining non-inactivated voltage-gated Na+ channels. The result is a sustained depolarization that doesn’t require much Ca2+ influx, because the conductance of the Kir channels is small at such a depolarized level. Importantly, several types of voltage-gated K+ channels slowly activate during the plateau phase. The termination of the plateau depolarization (Phase 3; repolarization) depends on two factors: (1) the building K+ current that is contributed by the various voltage-gated K+ channels, and (2) a diminishing Ca2+ current as the voltage-gated Ca2+ channels inactivate. Phase 4 (diastole) begins again after the membrane returns to its well-polarized resting potential.
111
What does the Q-T interval correspond to?
The Q-T interval corresponds to the duration of the ventricular action potential and contraction.
112
What is the major concern for a patient with LQTS?
The major concern in a patient with LQTS is the possibility that the delayed and relatively slow repolarization will give rise to premature action potentials during Phase 3, without waiting for the normal trigger from the atria (Fig. 9). This can lead to a dangerous arrhythmia called tornado de pointes, in which the ventricles contract in an uncoordinated manner, independent of the normal atrial rhythm, profoundly reducing the pumping efficiency of the heart.
113
What does the voltage clamp method do?
It keeps Vm constant and measures Im, even though the membrane conductance (gm) might change due to the behavior of voltage-dependent channels. With Vm held constant, any time-dependent variations in Im must reflect changes in underlying membrane conductance.
114
What value can we calculate from the voltage clamp method and why?
Since the values of Im and Vm are both known, the membrane conductance can be calculated from g = I / V. The voltage-clamp thus enables the active properties of the membrane, which change with Vm and time, to be analyzed.
115
What does it tell us if an I-V plot is linear?
That the conductance of the channel is not voltage dependent.
116
What does it mean when Im=0 on an I-V plot?
the voltage at which permeant ions are equally likely to move inward or outward through open channels. This, of course, is the equilibrium potential for the ions that carry the current. In our I-V plot, Im = 0 when Vm = -91 mV, as expected for a current carried by K+.
117
At more negative potentials than Keq (for a positive ion) the current is...
Inward.
118
At membrane potentials more positive than the Keq (for a positive ion) the current is...
Outward.
119
In Voltage clamp recordings, outward currents have BLANK values and inward currents have BLANK values.
Positive, negative. (Im) This reflects the direction of current that the amplifier injects into the cell to offset currents flowing through membrane channels.
120
According to the nernst equation a tenfold change in the outside or inside concentration of a monovalent ion will shift Veq by how much?
58 mV.
121
What would happen to the I-V relationship if we blocked half of the K+ leak channels with a drug?
The membrane conductance would decrease by half, exemplified in a shallower slope on the I-V plot. However Vk will remain unaffected and the plots will cross at the same point (for K -91)
122
Draw and compare the plots of inward rectifying and voltage gated K+ channels.
See 4th and 5th screenshots.
123
Draw the I-V Plot of a voltage gated Na+ or Ca2+ channel.
See 6th screenshot
124
Are Ligand-gated channels ion selective?
No. Unlike the voltage-gated Na+ and Ca2+ channels, excitatory ligand-gated channels are not selective for specific ions, and the reversal potential for these channels are determined by the relative conductances for the different ion species that they carry. Thus, an excitatory ligand-gated channel that is permeable to Na+, Ca2+, and K+ might have a reversal potential of 0 mV.
125
How can ligand-gated channels produce inhibitory stimuli?
They activate Cl- conductance.
126
What effect does Cl- have on Vm?
Often, the reversal potential for Cl- is close to the resting membrane potential so there is little change in Vm.
127
How is the increase in Cl- conductivity inhibitory?
The increase in Cl- conductance is inhibitory because of the decrease in total membrane resistance, which will attenuate the membrane depolarization that is produced by any excitatory current that occurs while the Cl- channels are open (remember, V = I x R).
128
Name two types of Cys Loop channels.
Nicotinic acetylcholine receptors and GABA receptors.
129
What are nicotinic acetylcholine receptors permeable to?
Nonselective cation permeability.
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What effect do nicotinic acetylcholine receptors have?
They depolarize the membrane (excitatory effect)
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What are GABA receptors permeable to?
Anions. Like Cl-
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What effect do GABA receptors have?
Inhibit neurons by hyperpolarizing the membrane and by increasing total
membrane conductance.
133
What are two types of Glutamate receptors?
AMPA and NMDA
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What do Glutamate receptors do?
Mediate fast synaptic transmission in the brain
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What are glutamate receptors permeable to?
Na+, K+ and sometimes Ca2+
136
What is the structure of a cys-loop receptor?
They are heteromeric or homomeric pentamers, and are characterized by a
signature loop in the large extracellular N-terminal domain that is formed by cross-linked cysteines. All subunits include four membrane-spanning (M) helices, and a large intracellular loop between M3 – M4 that associates with cytoskeletal partners. The walls of the ion channel are formed by the five M2 helices (one from each subunit). The channels of Cys-loop receptors are selective for either cations or anions, but otherwise are relatively non-discriminating. Each Cys-loop receptor has two binding sites for its ligand, both of which must be occupied for the channel to open.
137
The two extracellular acetylcholine binding sites on nAChRs are formed by the?
α−δ and α−γ interfaces.
138
How are nAChRs selective to cations?
Selectivity for cations is conferred by rings of negatively charged residues lining outer regions of the pore, on both the extracellular and cytoplasmic sides. Same thing in the reverse (positive residues) holds true for GABA receptors.
139
What ions are nAChRs permeable to?
Na+ (inward) and K+ (outward), but they also allow Ca2+ and Mg2+ ions to move through. The pore is large enough for most of these ions to move through with their hydration shells intact.
140
What's the reversal potential of nAChRs?
The reversal potential is about 0 mV.
141
How does the gating for nAChRs work?
In the absence of acetylcholine, each of the M2 helices has a hydrophobic kink directed into the pore, and together they block the channel. When both binding sites are occupied by acetylcholine, the M2 helices twist away from the center of the pore, permitting current flow. GABA binding is thought to work the same.
142
What's the reversal potential of GABA?
About -80mV.
143
What is the structure of Glutamate receptors?
Protein is composed of tetramers, with each subunit containing three membrane-spanning
regions called TM1, TM2, and TM3. The pore is formed from a loop that dips into the membrane from the cytoplasmic linker between the TM1 and TM2 helices, reminiscent of the P-loop that characterizes the voltage-gated channels and inward rectifiers.
144
What is the AMPA type GABA receptor permeable to?
AMPA-type is selective for Na+ and K+
145
What is the reversal potential of AMPA?
About 0mV
146
What is the NMDA type GABA receptor permeable to?
Ca2+
147
What is the reversal potential of NMDA?
Reversal potential is ~ +20 mV
148
What is the gating scheme for AMPA?
It requires glutamate to bind.
149
What is the gating scheme for NMDA?
It requires glutamate and extracellular glycine to bind. The NMDA-type channel is blocked by an extracellular Mg2+ ion at the resting Vm. The Mg2+ is expelled upon strong membrane depolarization.
150
How does glutamate reach AMPA and NMDA receptors?
It is released presynaptically and glutamate receptors are post-synaptic.
151
What happens initially when glutamate reaches the post synaptic neuron?
he released glutamate crosses the synapse and binds to and activates AMPA receptors,
depolarizing the postsynaptic neuron. Glutamate also binds to the NMDA receptors and opens their channel gates, but Mg2+ ions block the NMDA channels.
152
How are NMDA receptors activated and what is the result?
If the presynaptic neurons are very active, the resulting surge in synaptic glutamate activates a large number of AMPA-type channels, strongly depolarizing the postsynaptic neuron. As a consequence, the Mg2+ ions are expelled from the NMDA channels (their attraction for the inside of the membrane is reduced). Ca2+ now flows into the postsynaptic neuron through the NMDA channels, activating enzymes and producing long-lasting changes in the efficiency of communication between the presynaptic and postsynaptic neuron that are thought to underlie the formation of memories.
153
Where is the IP3 receptor predominantly found?
The ER.
154
What are the two receptors that lead to efflux of calcium into the intercellular space?
The IP3 receptor and the ryanodine receptor (which is found on both ER and SR)
155
Describe how the IP3 receptor works.
IP3 is generated in response to the activation of specific G-protein-coupled receptors or receptor tyrosine kinases that stimulate phospholipase C (PLC), which cleaves phosphatidylinositol-4,5-diphosphate (PIP2) into two different second messengers: diacylglycerol, which remains within the lipid bilayer, and IP3, which diffuses into the cytoplasm. When IP3 reaches the ER membrane and binds to its receptor, it induces a conformational change that opens the channel. Consequently, the cytoplasmic concentration of Ca2+ increases.
156
What is Ca2+- induced Ca2+ release (CICR)?
In many types of muscle cell, the brief Ca2+ influx that enters through nAChRs in response to neuronal stimulation is not sufficient for contraction, and the amplification provided by ryanodine receptors provides the additional required Ca2+.
157
How do additional ryanodine receptors get activated and how are they shut off?
The released Ca2+ in turn serves to activate nearby ryanodine receptors, accelerating the rate of release. Since the ryanodine receptor is inhibited by high Ca2+ concentrations, this positive feedback eventually is curtailed, preventing excessive Ca2+ release. Ca2+ stores are replenished during periods of relaxation.
158
What is an important consequence of Phospholipase C activation and why?
The depletion of PIP2 to form diacylglycerol and IP3. PIP2 is located on the inner leaflet and it's negatively charged head group is available to interact with positively charges species in the same area. Recall the positively charged "ball and chain" on the N terminus of voltage gated K + channels. Therefore when PIP2 is available it can inhibit the inactivation of voltage gated K+ channels.
159
Draw and explain a photoreceptor rod and cone.
Drawing is screenshot 7. Explanation is there as well.
160
How is the Na+/Ca2+ channel on the outer part of a photoreceptor gated?
It is gated by the second messenger cyclic guanosine monophosphate (cGMP), and light determines the level of cellular cGMP.
161
Describe phototransduction.
Phototransduction begins with a photon activating the visual pigment that is present in the outer segment (we’ll use the example of rods, which express the pigment rhodopsin). Rhodopsin is a G- protein-coupled receptor that stimulates GDP-GTP exchange on transducin, a G-protein that in turn activates cGMP phosphodiesterase to hydrolyze cGMP. Thus, light reduces cellular [cGMP], and decreases the current flowing through the cGMP-gated channels in the outer segment of the rod. The dark current is diminished, and the membrane hyperpolarizes as the contribution of the inner current to the membrane potential is reduced and Vm moves toward the K+ equilibrium potential. This response to a stimulus is graded (not all-or-none), in contrast to the depolarizing action potentials characteristic of nerve and muscle.
162
What is a GIRK channel?
Between action potentials, the membrane conductance of a pacemaker is dominated by an inward rectifier called G-protein-regulated inward rectifying K+ channel (GIRK). When the action potential ends and the cell begins to repolarize, GIRK channels returns to their high conductance state. The greater the conductance of the GIRK channels upon depolarization, the more negative Vm goes and the longer it takes before the cell spontaneously depolarizes to the threshold for triggering the next action potential.
163
What activates GIRK channels?
Acetylcholine, which is released by neurons of the parasympathetic system and binds to Gi-and GO- coupled receptors (muscarinic receptors), slows the heart rate by activating GIRK channels. This happens very quickly (one beat!) That this effect is mediated not by a cytoplasmic messenger, but rather by a direct effect of G-protein βγ subunits acting within the membrane on GIRK.
164
How can you tell if a drug signal uses a second messenger or membrane bound messengers?
If a drug affects a particular channel by generating a diffusible second messenger, such as cyclic AMP, then the drug should affect channels in the patch when it is applied outside. In contrast, a drug that acts through membrane- delimited messengers, such as G-protein βγ subunits, will only work when it is applied directly to the
patch.
165
What is intrinsic efficacy?
Full agonists differ with respect to intrinsic efficacy, which is the effect produced by the drug binding to a single receptor. To produce the maximum effect, a full agonist with relatively low intrinsic efficacy needs to bind more receptors than a full agonist that has higher intrinsic efficacy.
166
If there are spare receptors for a full agonist, which direction would you see the response curve shift when compared with the binding curve?
To the left.
167
What happens when a full agonist is challenged with increasing concentrations of a non-competitive antagonist?
When a full agonist is challenged with increasing concentrations of a noncompetitive antagonist, the dose-response curve shifts to the right with no loss of maximal effect until the antagonist blocks so many receptors that no “spare” pool remains. Beyond this concentration, the EC50 of the agonist remains constant but the maximum effect progressively declines.
168
Drugs with a small dissociation constant have a BLANK affinity for the receptor.
High.
169
What happens if a full agonist has a relatively low affinity for the receptor?
A higher concentration of this drug is required to occupy the same number of receptors as another agonist with higher affinity.
170
What is potency of a drug?
The potency of a full agonist is inversely related to the concentration needed to produce a given response: the lower the required concentration, the higher the potency. Potency is expressed as the EC50, which is the drug concentration that produces 50% of the maximum response or a therapeutically relevant criterion response .
171
What determines potency?
Affinity and intrinsic efficacy. A potent drug has both high affinity and high intrinsic efficacy.
172
What is a partial agonist?
A partial agonist is incapable of producing the full response of a system, even when its concentration is sufficient to bind all of the receptors, because its intrinsic efficacy is too low.
173
What do the binding and response curves look like for a partial agonist?
For partial antagonists, the binding curve and the concentration response curve are superimposed, because these drugs produce their maximum response only when they bind the full pool of receptors.
174
What is an inverse agonist?
Inverse agonists produce a response, but it is in the opposite direction to that of a full or partial agonist. For example, a full or partial β-adrenergic receptor agonist will increase the synthesis of cyclic AMP, while an inverse agonist will reduce cyclic AMP levels below baseline.
175
What is the equation for the number of receptors bound to drug? (B)
B= (Bmax*[D])/(Kd+[D])
Where Bmax is the total number of receptors, Kd is the dissociation constant for the drug, and [D] is the concentration of the drug. Essentially the same as the Michaelis-Menten equation.
176
If we assume that effect (E) is linearly related to binding, we get the equation...
E= (Emax*[D])/(E50+[D])
177
What is an antagonist?
Antagonists bind to receptors and do not change their activity, but they prevent an agonist from producing a response. “Pure” antagonists have no intrinsic efficacy, in contrast to partial agonists.
178
Antagonists that bind to the same site as an agonist?
Competitive. Those that bind somewhere else are noncompetitive.
179
Irreversible antagonists...
Irreversible antagonists covalently modify the receptor, rendering it permanently inactive; recovery depends on the synthesis of new enzyme.
180
What is the effect of a competitive reversible antagonist on the concentration-response curve?
If the antagonist is competitive and reversible, its antagonism can be overcome by increasing agonist concentration; the effect of the antagonist is surmountable. In effect, the antagonist shifts the concentration-response curve for the agonist to the right. A large enough dose of the agonist can restore maximum effect.
181
What does the extent of the antagonist curve shift depend on?
The extent of this shift depends solely on the concentration of the antagonist and its affinity for the receptor. Therefore, a given concentration of an antagonist will produce the same size shift for any agonist at that receptor.*
182
How do we measure the effect of reversible competitive antagonist on the binding of an agonist? (equation)
B=(Bmax*[D])/(Kd(1+([A]/Ka))+[D])
Again B is amount of agonist bound to receptor. Same equations with E's for the response to the agonist.
183
What is used to compare the abilities of different competitive antagonists to inhibit agonist-induced responses?
pA2. It is the negative log of the antagonist concentration that produces a two-fold shift in the EC50. This is, in fact, the negative log of Ka, since a two-fold increase in the EC50 for an agonist is seen when [A] = Ka:
184
If pA2 is relatively high, then what value of antagonist is needed to block the response to the agonist?
A relatively low concentration of antagonist would be needed.
For example: -9 -6
IftheKAforantagonist“Y”is1x10 M,andKAforantagonist“Z”is1x10 M(that is, “Z” has lower affinity for the receptor), then pA2 = 9 for drug “Y”, and pA2 = 6 for drug “Z”. Consequently, when present at the same concentration, “Y” will produce the greater shift in the response curve of an agonist.
185
Is pA2 the same for an antagonist across different receptor types?
No. pA2 is a property of the antagonist acting at a specific receptor. Its binding to a different receptor will be characterized by a different pA2, since its dissociation constant depends on the receptor type.
186
How do we discuss the potency of an antagonist?
We don't!! Antagonists have no intrinsic activity and therefore no potency.
187
Does adding more agonist affect Irreversible and noncompetitive (allosteric) antagonists?
No. The effects of these antagonists are non-surmountable; that is, increasing the maximal effect of an agonist does not restore its maximal response.
188
What determines the effect of irreversible antagonists?
Both the concentration of the antagonist and time of treatment. Over time, the number of covalently modified receptors accumulates.
189
What effect do irreversible and noncompetitive antagonists have on concentration-response curves?
If there are spare receptors (only full agonists) it is possible to lose receptors to the covalent modification without decreasing the maximum effect of the agonist. The more spare receptors (the higher the intrinsic efficacy of the agonist) the greater the number of receptors that must be modified before one sees a decrease in the maximal response
ii) If the agonist is a partial one, then by definition there are no spare receptors, and any loss of available receptors causes a decrease the maximum effect of the agonist.
