Electrical Engineering 3BB3: Cellular Bioelectricity (2013) Solutions to Midter Quiz #1 1. In a typical excitable cell, the ion species with the ost positive Nernst equilibriu potential is: a. potassiu, b. sodiu, c. calciu, or d. chloride. (5 pts) The answer is c. calciu. This is due to the substantial buffering of intracellular calciu, greatly reducing the free calciu in the intracellular space (see slide 14 of Lecture #4). 2. Currents through single ionic channels are stochastic predoinantly because of: a. the rando walk of ions in the diffusion process, b. rando fluctuations in the transebrane potential, c. rando changes in the ebrane capacitance due to oveent of lipids in the bilayer, or d. theral noise affecting the opening and closing of gating particles. (5 pts) The answer is d. theral noise affecting the opening and closing of gating particles. See slides 10 12 of Lecture #7 and slides 2 4 of Lecture #8. 3. 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). 4. When the ebrane is hyperpolarized, the Hodgkin Huxley sodiu channel h particle: a. activates, b. deactivates, c. inactivates, or d. deinactivates. (5 pts) The answer is d. deinactivates. This was discussed in Lecture #10. (See also slide 13 of Lecture #11.) Dr. I. Bruce 14/04/2014 1/6
5. The selectivity of an ion channel for anions versus cations is due priarily to: a. whether or not it has an inactivation particle, b. whether it is voltage-gated or ligand-gated, c. the sign of the charge of the protein residues in the narrowest part of the channel pore, or d. the diaeter of the narrowest part of the channel pore. (5 pts) The answer is c. the sign of the charge of the protein residues in the narrowest part of the channel pore. This was discussed in Lecture #7. 6. The experiental technique physiologists use to easure ionic currents while reoving the effects of the ebrane s capacitive current is: a. capacitance-clap recording, b. current-clap recording, c. inductance-clap recording, or d. voltage-clap recording. (5 pts) The answer is d. voltage-clap recording. This was discussed in Lecture #10 (see slide 3 of Lecture #10). 7. The sodiu-potassiu pup: a. pups 3 oles of Na out of a cell for every 2 oles of K that it pups in, b. pups 2 oles of K out of a cell for every 3 oles of Na that it pups in, c. generates action potentials, or d. is a for of passive ebrane transport. (5 pts) The answer is a. pups 3 oles of Na out of a cell for every 2 oles of K that it pups in. See slide 8 of Lecture #12. 8. The phenoenon in which an excitable cell does not fire an action potential if its ebrane is depolarized too slowly fro rest is referred to as : a. facilitation, b. accoodation, c. sleepiness, or d. refractoriness. (5 pts) The answer is b. accoodation. See slides 10 12 of Lecture #11. Dr. I. Bruce 14/04/2014 2/6
9. Describe the typical strength-duration tradeoff for a pulse of stiulating current to generate an action potential in an excitable cell and briefly explain the cause of this tradeoff. (15 pts) The strength-duration tradeoff refers to the fact that shorter current pulses need a larger current aplitude to reach the threshold for action potential generation. As the pulse duration increases, the threshold current drops until it approaches the asyptotic threshold current, referred to the rheobase current. The ajor cause of the tradeoff is the fact that it takes soe tie for the ebrane to charge up in response to a current pulse (due to the parallel RC-circuit nature of the ebrane and typically quantified by the ebrane tie-constant τ = RC ). Therefore, the threshold current will depend on how uch the ebrane is depolarized by the tie the current pulse is turned off. Consequently, shorter pulses need a larger current aplitude to produce sufficient depolarization, and thus sodiu channel activation, while the pulse is still on. The rates of ion channel activation and inactivation are secondary factors. 10. 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 Dr. I. Bruce 14/04/2014 3/6
of their slower tie 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 10 15 of Lecture #12). 11. Consider a patch of ebrane described by the parallel conductance odel shown below. a. If the resting transebrane potential V rest = 65 V, what is the sodiu Nernst equilibriu potential E Na? b. If the intracellular potassiu concentration [ K] what is the extracellular potassiu concentration [ ] = 400 M and the teperature is 37 C, i K e? (15 pts) a. For the given circuit, the resting ebrane potential will be deterined by: V g E + g E ( E ) ( ) 0.075 + 0.9 75 Na Na K K Na rest = = = gna + gk 0.075 + 0.9 Thus, the sodiu Nernst equilibriu potential is given by: ( E ) ( ) 0.075 Na + 0.9 75 = 65 0.975 65 0.975 + 75 0.9 ENa = = 55 V. 0.075 65 V. Dr. I. Bruce 14/04/2014 4/6
b. The potassiu Nernst equilibriu potential is deterined by: E K [ ] i [ K] 1 96487 [ K] RT K 8.314 310 400 = = = ZKF e e ln ln 75 10 V. Therefore, the extracellular potassiu concentration [ K] e is given by: 400 96487 75 10 ln = [ K] 8.314 310 e 400 = e [ K] [ ] e 96487 75 10 8.314 310 96487 75 10 8.314 310 K = 400 e M = 24.1 M e. 12. Consider a spherical cell with diaeter d = 7 μ and the parallel-conductance odel for a ebrane patch shown below. Note that the potential between the inside and the outside of the ebrane in the given circuit is the absolute transebrane potential V. If the ebrane is at rest for tie t < 0 and for tie t 0 a current of density 2 I = 10 μa c is injected into the intracellular space, deterine an expression for the inj relative transebrane potential v ( t ) for all ties t 0. (15 pts) For the given circuit, the net resistance for a square centietre of ebrane is: R 1 1 1 2 2 = = = kω c = 1 kω c, G g + g 0.1+ 0.9 Na and the ebrane tie constant is thus: K 2 2 ( ) ( ) τ = RC = 1 kω c 0.9 μf c = 0.9 s. Dr. I. Bruce 14/04/2014 5/6
As given in the equation sheet, the response of the absolute ebrane potential will be of the for: t τ ( ) = 0 + rest V t IR 1 e V. Consequently, the relative ebrane potential, which is defined as v = V V, will be equal to: rest t τ t 0.9 t 0. ( ) = 1 = v t IR 9 = 0 1 e 10 1 e V 10 1 e V, for 0 where t is in units of illisecond. t, Note that because the injected current is given as a density (i.e., per square centietre) and the ebrane circuit is given for a unit area of ebrane (i.e., for a square centietre of ebrane), it is not necessary to work out the ebrane surface area. However, knowing that the cell is sall indicates that the entire cell ebrane should be isopotential. Dr. I. Bruce 14/04/2014 6/6