Electrical Engineering 3BB3: Cellular Bioelectricity (2008) Solutions to Midterm Quiz #1

Transcription

1 Electrical Engineering 3BB3: Cellular Bioelectricity (2008) Solutions to Midter Quiz #1 1. In typical excitable cells there will be a net influx of K + through potassiu ion channels if: a. V Vrest >, b. the cell is repolarizing after an action potential, c. V Vrest =, or d. V < EK. (5 pts) The answer is d. is less than K V < EK. Fro Eqn. (3.26), the potassiu current will be negative only when V E, and a negative potassiu current corresponds to an influx of potassiu ions. 2. The intracellular and extracellular electrolytes of excitable cells have a priarily linear conductivity that arises because of phenoena described by: a. Fick s law of diffusion, b. Oh s law of drift, c. Einstein s equation, or d. the Goldan Hodgkin Katz Equation. (5 pts) The answer is b. Oh s law of drift. Eqn. (3.2), Oh s law of drift, states that the ionic flux in an electrolyte is proportional to the potential gradient ( Φ ). The resulting current will therefore be proportional to the potential gradient, such that the electrolyte s conductance is linear (see slides 13 & 14 of Lecture #3). 3. Active ebrane transport refers to: a. ion flow through the ebrane due to opening of activation particles, b. ion flow through the ebrane caused by a negative resting potential, c. ion flow through the ebrane producing the rising phase of the action potential, or d. puping ions through the ebrane against their concentration gradient. (5 pts) The answer is d. puping ions through the ebrane against their concentration gradient (see slide 4 of Lecture #4 and slides 5 8 of Lecture #12). 4. The capacitance per unit area of an excitable cell s plasa ebrane is: a. fairly constant across different types of excitable cells, b. ostly unaffected by opening and closing of ion channels, c. explained well by the lipid bilayer thickness and dielectric properties, or d. all of the above. (5 pts) The answer is d. all of the above (see slide 6 of Lecture #4, slide 12 of Lecture #9 and page 56 of Plonsey & Barr). Dr. I. Bruce 27/02/2008 1/6

2 5. When the ebrane is hyperpolarized, the Hodgkin Huxley odel s sodiu channel h particle is: a. activated, b. deactivated, c. inactivated, or d. deinactivated. (5 pts) The answer is d. deinactivated. This was discussed in Lecture #10. (See also slide 15 of Lecture #11.) 6. The best patch-clap configuration for studying the effects of changing the intracellular electrolyte coposition on single channel properties is the: a. inside-out patch configuration, b. outside-out patch configuration, c. cell-attached configuration, or d. whole-cell recording configuration. (5 pts) The answer is a. inside-out patch configuration. The various patch-clap configurations are shown on slide 9 of Lecture #7. As discussed in that lecture, creating a ebrane patch with the intracellular side of the ebrane outside of the patch pipette allows for the patch to be easily placed in different solutions siulating different electrolyte copositions. 7. Considering the effects of Nernst equilibriu potentials, increasing the teperature of an excitable cell with a negative resting ebrane potential will ake the ebrane: a. becoe even ore polarized (i.e., have an even lower resting potential), b. becoe less polarized (i.e., have a higher, but still negative, resting potential), c. becoe polarized with a positive resting potential, or d. have an unchanged resting potential. (5 pts) The answer is a. becoe even ore polarized (i.e., have an even lower resting potential). Fro the Nernst equation (3.21), the Nernst equilibriu potentials for any excitable cell are all proportional to the absolute teperature T. In the equation for the resting ebrane potential (3.31), all the Nernst potentials appear in the nuerator. Consequently, the teperature T can be taken out the front as a coon factor, and an increase in the teperature will increase the agnitude of the resting potential. Thus, if the resting potential is negative, then it will becoe even ore negative with an increase in teperature. Dr. I. Bruce 27/02/2008 2/6

3 8. The ain Frankenhaeuser Huxley odel potassiu channel: a. has a linear conductance, b. is outwards rectifying (i.e., allows uch larger outwards currents than inwards currents), c. is inwards rectifying (i.e., allows uch larger inwards currents than outwards currents), or d. has four activation particles. (5 pts) The answer is b. is outwards rectifying (i.e., allows uch larger outwards currents than inwards currents). The I-V relationship for this channel given by Eqn. (12.27) is a nonlinear, outwards rectifying conductance (see slide 12 of Lecture #12 for a plot of the I-V curve). Note that this channel has only two activation particles, as described by Eqn. (12.28). 9. Explain the cause of the shape of a stereotypical action potential (AP) wavefor. In your explanation, include a discussion of why an excitable cell always tends to fire APs with siilar wavefors but different types of excitable cells can have vastly different AP wavefor shapes and durations. (15 pts) A stereotypical AP wavefor is illustrated below. An AP is initiated by sufficient opening of voltage-gated sodiu channels due to activation of sodiu activation () particles. This causes an increase in the influx of sodiu, which depolarizes the ebrane, leading the further opening of particles. The onset of this positive feedback loop gives rise to the foot of the AP wavefor, followed by the rising phase during which there is a assive influx of sodiu through voltage-gating sodiu channels as practically all these channels open. The peak of the AP approaches (but does not exceed) the sodiu Nernst equilibriu potential. The falling phase is caused by closing of sodiu inactivation (h) particles (reducing the inward sodiu current) and opening of potassiu activation (n) particles (increasing the outward potassiu current), both of which lag behind sodiu activation because of their slower tie Dr. I. Bruce 27/02/2008 3/6

4 constants. Because of the slow dynaics of the n and h particles, in soe cells the potassiu current reains higher than the sodiu current as the ebrane repolarizes, such that it overshoots the resting potential and approaches (but does not exceed) the potassiu Nernst equilibriu potential. One or ore afterpotentials ay occur as the ebrane potential and the gating particles eventually return to their resting values. During the AP, the sodiu and potassiu currents grow so large that the changes in the ebrane potential are driven only by the gating particle dynaics, which are dependent in turn only on the ebrane potential, not the stiulus that initiated the AP. The AP wavefor is consequently deterined by the ion channel conductances, the gating particle dynaics, and the Nernst equilibriu potentials, which are properties of the cell not the stiulus. Thus the wavefor reains fairly consistently the sae for all APs fired by an excitable cell. Different excitable cell types can have different AP wavefors, however, because they ay have different Nernst equilibriu potentials (see slide 14 of Lecture #4), conductances (see slide 14 of Lecture #7), gating particle dynaics (see slides 3 & 4 of Lecture #12), or a range of different ionic channel types (slides of Lecture #12). 10. What is the relative refractory period, how is it easured, and what causes it? (15 pts) The relative refractory period is the period of tie following an action potential (AP) during which an action potential can only be generated by a stiulus with a greater aplitude than that required when the ebrane is at rest. It is easured by finding the resting threshold stiulus aplitude for a stiulus duration less than the absolute refractory period. Then a pair of stiuli with the resting threshold stiulus aplitude (and sae stiulus duration) is applied with various intervals between the stiuli to find the iniu inter-stiulus interval that produces an AP in response to each stiulus. The relative refractory period duration is typically then easured fro soe point on the first AP wavefor, such as the AP peak or the botto of the falling phase, to the onset of the second stiulus. Note that with intra-cellular or patch-clap recordings it is possible to see the whole AP wavefor, and therefore either the AP peak or the botto of the falling phase can be used to ark the start of the refractory period. However, for extracellular recordings it is typical for only the peak of the AP to be visible above the easureent noise, such that only the peak can be used in this case. The relative refractory period is caused by the ebrane potential and the voltage-gated ion channel gating particles taking tie to return to their resting values following an AP. Note that the relative refractory period can extend beyond the tie at which the ebrane potential returns to its resting value, since the gating particles will take soe tie to reach their steady-state values for the resting potential. If the ionic conductances are above their resting values, then the net ebrane resistance will be below its resting value (see slide 8 of Lecture #6), and consequently any injected current will produce a saller depolarization than if the conductances are at their resting values (see slide 4 of Lecture #6). Dr. I. Bruce 27/02/2008 4/6

5 11. An excitable cell has an intracellular chloride concentration of extracellular chloride concentration of 120 M e = i = 2.4 M, and a teperature of 40 C. a. What is the chloride Nernst equilibriu potential E for this cell?, an b. How uch chloride (in M) would need to ove fro the extracellular space to the intracellular space in order for E to increase by +30 V? (15 pts) a. The teperature 40 C corresponds to an absolute teperature of T = = 313 K, and chloride has a valence of Z = 1. Fro the given the ionic concentrations, the chloride Nernst equilibriu potential is: E RT i = ln = ln = V. ZF e b. If X M of chloride needs to ove fro the extracellular space to the intracellular space to increase E by +30 V, then: E RT + X i X = ln = ln = V ZF X X e ( X) ( 1 e ) 3 ( ) X ln = 120 X ( ) X = e 120 X X = 120 e X + = 120 e e 2.4 X = = M e Dr. I. Bruce 27/02/2008 5/6

6 12. An excitable cell has the parallel-conductance odel for a ebrane patch shown below. inside g Na = 0.1 g K = 0.8 S/c 2 S/c 2 C = 0.9 µf/c 2 E Na = +72 V E K = 90 V outside Consider the case where the ebrane is voltage-claped at a potential of V () t = 80 V for tie t < 0 and is released fro the voltage clap at tie t = 0. If the ebrane is then 2 subjected to an intracellular current injection of I 0 = 10.8 μa c for tie t 0, find the ebrane potential response V t for all t 0. (15 pts) The resting potential is: V g E + g E ( ) ( ) ( ) Na Na K K rest = = = gna + gk the ebrane resistance per unit area is: R = k c 1.111k c g + g = Ω = Ω, Na K and the ebrane tie constant is: τ = RC = kω = 1 For a constant current injection of 2 2 c 0.9 μf c s. I0R + Vrest = = = 60 V, or 20 V above the initial potential of ( ) 72 V, 2 I 0 = 10.8 μa c the steady-state ebrane potential is then: Consequently, the ebrane potential oves exponentially with a tie-constant of 1 s fro the initial potential of V t = 0 = 80V to the steady-state potential V ( t ) = 60 V according to: V () t ( ) ( ) t t = 20 1 e 80 V or 20e 60 V, where t is in units of s. V t = 0 = 80V caused by the voltage clap. Dr. I. Bruce 27/02/2008 6/6