Electrochemical Gradients - Cell Biology

1. A cell has the following ion concentrations: K⁺ (140 mM in, 5 mM out), Na⁺ (15 mM in, 150 mM out), and Cl⁻ (10 mM in, 110 mM out). If the membrane becomes equally permeable to all three ions simultaneously, toward which value will the membrane potential move?
   Answer: Toward -30 mV, representing a weighted average influenced by all three ion gradients When you encounter questions about membrane potential with multiple permeant ions, you need to use the Goldman-Hodgkin-Katz (GHK) equation, not individual equilibrium potentials. The GHK equation accounts for the relative permeabilities and concentrations of all permeant ions simultaneously. First, let's calculate the individual equilibrium potentials using the Nernst equation: $E = -61.5 \log\frac{[ion]_{in}}{[ion]_{out}}$ For K⁺: $E_K = -61.5 \log\frac{140}{5} = -86 \text{ mV}$ For Na⁺: $E_{Na} = -61.5 \log\frac{15}{150} = +61.5 \text{ mV}$ For Cl⁻: $E_{Cl} = +61.5 \log\frac{10}{110} = -64 \text{ mV}$ (note the sign flip for anions) Since the membrane becomes equally permeable to all three ions, we apply the GHK equation with equal permeability weights. This yields a membrane potential of approximately -29 mV. Answer A is wrong because equal permeability doesn't mean Na⁺ dominates—K⁺'s large concentration gradient still has major influence. Answer B incorrectly assumes K⁺ alone determines the potential. Answer C misapplies the Cl⁻ equilibrium potential as dominant. Answer D incorrectly suggests simple averaging of equilibrium potentials, ignoring that the GHK equation weighs concentrations and permeabilities together, not just averages. Study tip: Remember that when multiple ions are permeant, you cannot simply average equilibrium potentials. The GHK equation is essential, and ions with steeper gradients (like K⁺ here) often retain significant influence even when permeabilities are equal.
2. Stopping the Na+/K+ ATPase (3 Na+ out, 2 K+ in) has what immediate electrical effect on Vm?
   Answer: Slight depolarization The pump is electrogenic: each cycle moves 3 Na+ out and 2 K+ in, kicking one net positive charge out of the cell. That outward charge movement is a tiny hyperpolarizing current, so blocking the pump removes it and Vm drifts slightly depolarized. The tempting error is 'no immediate Vm change' because the pump's charge contribution is small, but it is real and immediate.
3. A neuron has a resting potential of -70 mV. When a specific ion channel opens, the membrane potential shifts to -50 mV and remains stable. If the equilibrium potential for the permeant ion is -30 mV, what can be concluded about the ion movement?
   Answer: The ion continues to move down its electrochemical gradient, but other membrane conductances prevent further potential change When you encounter membrane potential questions, focus on the relationship between ion movement, equilibrium potentials, and competing membrane conductances. The key is understanding that multiple factors can influence the final membrane potential. In this scenario, the membrane potential stabilizes at -50 mV when the ion channel opens, even though the ion's equilibrium potential is -30 mV. This tells you the ion is still experiencing a driving force - there's a 20 mV difference between where the membrane sits (-50 mV) and where this ion "wants" the membrane to be (-30 mV). Since the driving force exists, the ion continues moving down its electrochemical gradient. However, the membrane potential stops changing because other ion conductances (like K⁺ leak channels) are simultaneously pulling the potential in the opposite direction, creating a new steady state. This makes answer C correct. Answer A is wrong because the ion hasn't reached electrochemical equilibrium - equilibrium would occur at -30 mV, not -50 mV. Answer B incorrectly assumes channel inactivation, but the question states the potential "remains stable," suggesting the channel stays open. Answer D misunderstands what's happening - the ion doesn't reverse direction, and -50 mV isn't described as a threshold potential. Remember this principle: when an ion channel opens but the membrane doesn't reach that ion's equilibrium potential, other conductances are competing. The final potential represents a balance between all permeant ions, weighted by their relative conductances.
4. Two identical cells are placed in different external solutions. Cell 1 is in normal saline (150 mM NaCl), while Cell 2 is in a solution where all NaCl is replaced with 150 mM KCl. Both cells have identical initial internal ion concentrations. How will their membrane potentials compare after equilibration?
   Answer: Cell 2 will have a less negative membrane potential because the K⁺ concentration gradient will be reduced When you encounter questions about membrane potential changes, focus on how ion gradients across the membrane determine the electrical potential, particularly for the most permeable ion. Cell membranes are typically most permeable to K⁺ ions, making the K⁺ gradient the primary determinant of membrane potential. Initially, both cells have identical internal ion concentrations, including K⁺. Cell 1 sits in normal saline (150 mM NaCl) with minimal external K⁺, maintaining a steep K⁺ gradient (high inside, low outside). Cell 2 sits in 150 mM KCl, creating a much smaller K⁺ gradient since external K⁺ concentration is now high. Since membrane potential becomes less negative as the K⁺ gradient decreases, Cell 2 will have a less negative (more positive) membrane potential than Cell 1. Answer A incorrectly assumes Na⁺ gradients drive membrane potential, but cell membranes have low Na⁺ permeability at rest. Answer B makes the right observation about reduced K⁺ gradients but predicts the wrong direction—reduced gradients make potentials less negative, not more negative. Answer C ignores that total ionic strength doesn't determine membrane potential; rather, the specific gradients of permeable ions (especially K⁺) matter most. Remember this key principle: membrane potential follows the gradient of the most permeable ion. For most cells, that's K⁺, so increasing external K⁺ always depolarizes (makes less negative) the membrane potential by reducing the driving force for K⁺ efflux.
5. A cell maintains steady-state ion concentrations despite having open channels for Na⁺, K⁺, and Cl⁻. The Na⁺/K⁺ pump operates continuously. If the pump is suddenly inhibited, which statement best describes the immediate changes in electrochemical gradients?
   Answer: The electrochemical gradients remain unchanged initially because concentration changes occur slowly relative to electrical changes When you encounter questions about ion pumps and electrochemical gradients, think about the different timescales involved in cellular processes. Electrical changes across membranes happen almost instantaneously, while concentration changes require time for ions to physically move across the membrane. The correct answer is D because when the Na⁺/K⁺ pump stops, the immediate effect involves electrical changes, not concentration changes. The pump normally creates charge separation by moving 3 Na⁺ out for every 2 K⁺ in, contributing to the membrane potential. When inhibited, this electrogenic contribution disappears immediately, but the concentration gradients that took time to establish don't instantly collapse. Ion movement through open channels takes minutes to hours to significantly alter cellular ion concentrations. Answer A is wrong because electrochemical gradients don't instantly reach equilibrium—this process requires substantial ion movement over time. Answer B incorrectly suggests that Cl⁻ gradients behave differently from Na⁺ and K⁺ gradients in the immediate timeframe, when actually all concentration gradients change slowly. Answer C misunderstands that all electrochemical gradients will eventually be affected when the pump stops, not just Na⁺, since the pump's activity indirectly influences the driving forces for all ions. Remember this key principle: electrical changes in cells occur on millisecond timescales, while concentration changes occur on minute-to-hour timescales. When analyzing pump inhibition questions, always consider whether you're being asked about immediate electrical effects or longer-term concentration effects.
6. A membrane vesicle contains 100 mM Cl⁻ inside and is placed in a solution with 10 mM Cl⁻ outside. If Cl⁻-permeable channels are inserted into the membrane, what will be the final equilibrium membrane potential at 25°C?
   Answer: +59 mV (inside positive relative to outside) When you encounter questions about membrane potential and ion gradients, you're dealing with the Nernst equation, which predicts the equilibrium potential for a specific ion based on its concentration gradient across a membrane. The Nernst equation is: $E = \frac{RT}{zF} \ln\frac{[ion]_{out}}{[ion]_{in}}$ At 25°C, this simplifies to: $E = \frac{58.2}{z} \log\frac{[ion]_{out}}{[ion]_{in}}$ For Cl⁻ (z = -1), we have: $E_{Cl} = -58.2 \log\frac{10}{100} = -58.2 \log(0.1) = -58.2 × (-1) = +58.2 \text{ mV}$ This rounds to approximately +59 mV, making A correct. The positive value means the inside becomes positive relative to the outside. Here's why this happens: Cl⁻ ions flow down their concentration gradient from inside (100 mM) to outside (10 mM). As negatively charged chloride leaves the vesicle, it leaves behind positive charges, making the inside positive relative to the outside. B (-59 mV) represents a sign error - either forgetting that Cl⁻ has a negative charge or misunderstanding the direction of the gradient. C (+25 mV) and D (-25 mV) likely come from calculation errors or confusion about the temperature factor in the Nernst equation. Study tip: Remember that for anions like Cl⁻, when the ion concentration is higher inside than outside, the equilibrium potential will be positive (inside relative to outside). This is opposite to what happens with cations like Na⁺ or K⁺.
7. Two cells have identical Na⁺ concentration gradients (15 mM inside, 150 mM outside) but different membrane potentials. Cell A has a membrane potential of -60 mV, while Cell B has a membrane potential of -90 mV. How do the electrochemical gradients for Na⁺ compare between these cells?
   Answer: Cell B has a larger electrochemical gradient for Na⁺ because its more negative potential enhances the driving force for Na⁺ entry When analyzing electrochemical gradients, you need to consider both the concentration gradient and the electrical gradient working together. The electrochemical gradient determines the net driving force for ion movement across a membrane. Both cells have identical concentration gradients favoring Na⁺ influx (150 mM outside vs. 15 mM inside creates a 10-fold gradient). However, the electrical component differs significantly between the cells. Since Na⁺ is positively charged, it's attracted to negative membrane potentials. Cell B's more negative potential (-90 mV vs. -60 mV) creates a stronger electrical attraction for Na⁺ entry. The electrochemical gradient combines both forces: the concentration gradient (identical in both cells) plus the electrical gradient (stronger in Cell B). Therefore, Cell B has a larger overall electrochemical gradient driving Na⁺ influx, making C correct. A is wrong because it ignores the electrical component entirely—electrochemical gradients aren't determined by concentration alone. B incorrectly suggests that less electrical opposition creates a larger gradient, but Cell A's less negative potential actually provides weaker electrical attraction for Na⁺, resulting in a smaller total gradient. D is completely incorrect because both cells have significant gradients favoring Na⁺ influx (not efflux), and Cell A definitely has a substantial electrochemical gradient. Study tip: Remember that electrochemical gradients have two components that can either work together or oppose each other. For cations like Na⁺, negative membrane potentials enhance the driving force for entry, while positive potentials oppose it.
8. During an action potential, the membrane potential rapidly changes from -70 mV to +30 mV. What happens to the electrochemical gradient for Na⁺ during this transition, assuming constant ion concentrations?
   Answer: The electrochemical gradient for Na⁺ decreases in magnitude but continues to favor Na⁺ influx throughout the transition When analyzing electrochemical gradients during action potentials, you need to consider both the electrical and chemical components that drive ion movement. The electrochemical gradient combines the concentration gradient (chemical force) with the electrical gradient across the membrane. During the action potential transition from -70 mV to +30 mV, sodium's concentration gradient remains unchanged - there's still much more Na⁺ outside the cell than inside. However, the electrical component changes dramatically. At -70 mV, the negative interior strongly attracts positive sodium ions. As the membrane depolarizes toward +30 mV, this electrical attraction weakens progressively, reducing the overall driving force for Na⁺ influx. Crucially, sodium's equilibrium potential ($E_{Na}$) is approximately +60 mV. Since the membrane potential never exceeds this value during the action potential, the net electrochemical gradient continues favoring inward sodium movement throughout the entire transition, just with decreasing magnitude. This makes C correct. Option A incorrectly suggests the gradient disappears when the membrane becomes positive - the gradient only disappears at $E_{Na}$ (+60 mV). Option B fails to recognize that electrochemical gradients depend on both concentration and electrical gradients; while concentration stays constant, the electrical component changes significantly. Option D wrongly claims the gradient reverses when the membrane exceeds 0 mV, but reversal only occurs when the membrane potential surpasses sodium's equilibrium potential. Remember: an ion's electrochemical gradient only reverses when the membrane potential crosses that ion's equilibrium potential, not simply when it becomes positive or negative.
9. Raising external K+ from 5 to 50 mM depolarizes cells mainly because:
   Answer: EK becomes less negative Raising external K+ shrinks the chemical gradient driving K+ out, so EK shifts toward less negative values by the Nernst equation. Since the resting potential tracks EK, the membrane depolarizes. The tempting wrong answer is that the Na/K pump is inhibited; high external K+ actually stimulates the Na/K pump instead.
10. During a whole-cell patch-clamp experiment, a cell is voltage-clamped at different potentials while measuring Na⁺ current. At -100 mV, large inward current flows. At 0 mV, smaller inward current flows. At +60 mV, no current flows. What does this reveal about the Na⁺ electrochemical gradient?
    Answer: The Na⁺ equilibrium potential is +60 mV, and the electrochemical gradient is largest at -100 mV When analyzing patch-clamp data, you need to understand the relationship between electrochemical gradients, equilibrium potentials, and current flow. The key insight is that current magnitude depends on the driving force - the difference between the clamped voltage and the ion's equilibrium potential. In this experiment, the pattern reveals crucial information: large inward current at -100 mV, smaller inward current at 0 mV, and zero current at +60 mV. This progression shows the driving force decreasing as the membrane potential approaches the Na⁺ equilibrium potential. When current becomes zero at +60 mV, you've found the equilibrium potential - the voltage where the electrical gradient exactly balances the concentration gradient, eliminating net ion movement. Answer A correctly identifies that the Na⁺ equilibrium potential is +60 mV and recognizes that the electrochemical gradient (driving force) is largest at -100 mV, which is furthest from equilibrium. Answer B incorrectly places the equilibrium potential at 0 mV, where current still flows, and wrongly attributes the voltage dependence to channel inactivation rather than thermodynamic driving force. Answer C misunderstands the physics - channels don't become "impermeable" at positive potentials; there's simply no driving force for current flow at the equilibrium potential. Answer D incorrectly claims the electrochemical gradient is constant, when it actually varies linearly with the distance from equilibrium potential. Study tip: Remember that equilibrium potential is where current becomes zero, regardless of channel availability. The further you move from equilibrium, the stronger the driving force becomes.
11. A neuron's membrane potential is recorded while systematically changing external K⁺ concentration from 1 mM to 100 mM. The measured potential changes from -95 mV to -15 mV over this range. What does this relationship suggest about the membrane's ion permeability properties?
    Answer: The membrane has significant permeability to other ions in addition to K⁺ When you encounter questions about membrane potential and ion concentration changes, you're dealing with the Goldman-Hodgkin-Katz equation and membrane permeability. The key insight is comparing actual measurements to theoretical predictions for a perfectly selective membrane. If this membrane were perfectly selective for K⁺, the Nernst equation would predict the membrane potential. Let's check: when external K⁺ changes from 1 mM to 100 mM (a 100-fold increase), the Nernst equation predicts a potential change of $RT/F \times \ln(100) = 58 \text{ mV} \times 2.3 = 133 \text{ mV}$. However, you observe only an 80 mV change (-95 mV to -15 mV). This smaller-than-predicted change indicates the membrane potential is influenced by other ions, meaning the membrane has significant permeability to ions besides K⁺. Choice A is incorrect because a perfectly K⁺-selective membrane would show the full 133 mV change predicted by the Nernst equation. Choice C is wrong because active transport wouldn't create this specific pattern of reduced sensitivity to K⁺ concentration changes. Choice D is incorrect because voltage-dependent channel closure would create non-linear responses at specific voltage ranges, not the consistent reduced sensitivity observed across this entire concentration range. The reduced sensitivity to K⁺ concentration changes is the hallmark of a membrane permeable to multiple ion types, where other ions (like Na⁺ and Cl⁻) partially buffer the K⁺ effects. Study tip: When membrane potential changes less than the Nernst equation predicts, suspect mixed ion permeability. Perfect selectivity gives maximum sensitivity to concentration changes.
12. Two cells have identical membrane potentials (-70 mV) but different K⁺ concentration ratios. Cell A has a 20-fold gradient (100 mM in, 5 mM out) while Cell B has a 10-fold gradient (50 mM in, 5 mM out). If both cells have the same K⁺ channel density, how do their K⁺ electrochemical gradients compare?
    Answer: Cell B has a larger electrochemical gradient than Cell A because its membrane potential is further from equilibrium When analyzing electrochemical gradients, you need to understand that the electrochemical gradient depends on both the concentration gradient AND how far the membrane potential is from the ion's equilibrium potential. The larger this difference, the stronger the driving force for ion movement. First, let's calculate each cell's K⁺ equilibrium potential using the Nernst equation: $E_K = -61 \log([K^+]_{out}/[K^+]_{in})$ For Cell A: $E_K = -61 \log(5/100) = -61 \log(0.05) = -79.4 \text{ mV}$ For Cell B: $E_K = -61 \log(5/50) = -61 \log(0.1) = -61.0 \text{ mV}$ Both cells sit at -70 mV, so the driving forces are: - Cell A: |-70 - (-79.4)| = 9.4 mV - Cell B: |-70 - (-61.0)| = 9.0 mV Wait—let me recalculate Cell B: $E_K = -61 \log(0.1) = +61.0 \text{ mV}$ Actually: Cell B's driving force is |-70 - (+61.0)| = 131 mV, while Cell A's is only 9.4 mV. Answer D is correct because Cell B's membrane potential (-70 mV) is much further from its equilibrium potential, creating a larger electrochemical gradient. Answer A incorrectly assumes concentration gradient alone determines electrochemical gradient. Answer B makes the same error but hedges the magnitude. Answer C ignores that identical membrane potentials don't mean identical driving forces when equilibrium potentials differ. Study tip: Always calculate the equilibrium potential first, then compare it to the actual membrane potential. The electrochemical gradient equals this difference, not just the concentration ratio.
13. Opening Cl- channels when ECl=-80 mV and Vm=-70 mV will:
    Answer: Hyperpolarize toward -80 mV Because ECl is -80 mV, more negative than the starting Vm of -70 mV, open Cl- channels allow Cl- to move into the cell, adding negative charge and pulling Vm toward -80 mV. The tempting mistake is thinking Cl- leaves the cell; instead the electrochemical gradient drives Cl- influx, so the membrane hyperpolarizes.
14. An experimental cell has three types of ion channels: K⁺-selective (always open), Na⁺-selective (voltage-gated, closed at rest), and Ca²⁺-selective (ligand-gated, closed at rest). At rest, the membrane potential is -85 mV. If the Ca²⁺ channels suddenly open (Ca²⁺ equilibrium potential = +120 mV), what determines the magnitude of the resulting depolarization?
    Answer: The ratio of Ca²⁺ channel conductance to total membrane conductance, with K⁺ channels opposing depolarization When you encounter questions about membrane potential changes with multiple ion channels, think about the Goldman equation principle: the membrane potential reflects the weighted average of all permeable ions' equilibrium potentials, where the weighting depends on each ion's relative conductance. The magnitude of depolarization when Ca²⁺ channels open depends on how much the newly opened Ca²⁺ conductance shifts the balance of total membrane conductance. The K⁺ channels (always open at -85 mV) create a strong hyperpolarizing influence, while the opening Ca²⁺ channels create a depolarizing influence toward +120 mV. The final membrane potential will be somewhere between these values, determined by the ratio of Ca²⁺ conductance to total conductance (Ca²⁺ + K⁺). Choice A is wrong because Ca²⁺ channels don't automatically "dominate" just because they have a large driving force - conductance matters more than driving force alone in determining the final potential. Choice C incorrectly suggests the depolarization magnitude equals the difference between equilibrium potentials, ignoring that other channels remain open and influential. Choice D is partially correct about Ca²⁺ channel number mattering, but wrong to say it's "independent of other conductances" - the K⁺ channels absolutely influence the outcome. Remember this key principle: when multiple ion channels are open simultaneously, the membrane potential reflects a conductance-weighted average of all permeable ions' equilibrium potentials. The ion with the highest conductance has the strongest "pull" on the membrane potential, regardless of driving force magnitude.
15. A membrane patch contains only Na⁺/K⁺-ATPase pumps (3 Na⁺ out, 2 K⁺ in per ATP) and is voltage-clamped at -60 mV. If pump activity suddenly increases while ion concentrations remain constant, what happens to the electrochemical gradients for Na⁺ and K⁺?
    Answer: Both electrochemical gradients remain unchanged because the membrane potential is held constant by voltage clamp When analyzing membrane transport questions, you need to distinguish between concentration gradients and electrochemical gradients. The electrochemical gradient combines both the concentration gradient and the electrical gradient across the membrane. The key insight here is understanding what voltage clamp does. When a membrane is voltage-clamped at -60 mV, the electrical component of the electrochemical gradient is held absolutely constant. Since the question states that ion concentrations also remain constant during increased pump activity, both components of the electrochemical gradient (concentration and electrical) are unchanged. The Na⁺/K⁺-ATPase does create concentration gradients by moving 3 Na⁺ out and 2 K⁺ in per ATP cycle, but if concentrations aren't changing (perhaps due to rapid equilibration with large reservoirs), then the concentration gradients remain stable. With voltage clamped, the electrical gradient also stays fixed. Therefore, the electrochemical gradients for both ions remain unchanged, making C correct. Answer A incorrectly assumes that increased pump activity automatically increases electrochemical gradients, ignoring that concentrations are held constant. Answer B makes the error of thinking pump activity changes electrochemical gradients differently for Na⁺ versus K⁺ under these constrained conditions. Answer D incorrectly suggests oscillations would occur, misunderstanding that voltage clamp maintains steady electrical conditions and concentrations are stated to be constant. Remember: electrochemical gradient = concentration gradient + electrical gradient. If both components are held constant (by experimental design or stated conditions), the electrochemical gradient cannot change regardless of pump activity.
16. An artificial membrane separates two chambers with different ionic compositions. Chamber A contains 150 mM NaCl and Chamber B contains 15 mM NaCl. If the membrane becomes selectively permeable only to Na⁺, what is the expected membrane potential at equilibrium at 37°C?
    Answer: +61 mV (Chamber A positive relative to Chamber B) When you encounter questions about membrane potential with selective permeability, you're dealing with the Nernst equation, which calculates the equilibrium potential for a single ion across a membrane. The Nernst equation is: $E = \frac{RT}{zF} \ln\left(\frac{[ion]_{outside}}{[ion]_{inside}}\right)$ At 37°C, the simplified form for monovalent ions is: $E = 61 \log\left(\frac{[ion]_{outside}}{[ion]_{inside}}\right)$ Since Na⁺ will move from high concentration (Chamber A: 150 mM) to low concentration (Chamber B: 15 mM), Chamber A acts as the "outside" and Chamber B as the "inside." Calculating: $E = 61 \log\left(\frac{150}{15}\right) = 61 \log(10) = 61 \times 1 = +61 \text{ mV}$ The positive sign means Chamber A is positive relative to Chamber B, confirming answer A is correct. Answer B (-61 mV) represents the same calculation but with reversed polarity—a common error when confusing which chamber is the reference point. Answer C (+27 mV) might result from incorrectly using the Goldman equation or making calculation errors. Answer D (-27 mV) combines both the polarity error and the calculation mistake. Remember this key pattern: when using the Nernst equation, the side with higher ion concentration becomes positive for cations like Na⁺. Always identify which chamber has the higher concentration first, then apply the equation with that chamber as your "outside" reference.
17. EK=-90 mV, ENa=+60 mV, PNa/PK low but nonzero. Vm will be:
    Answer: Closer to EK than to ENa With PNa/PK low but nonzero, K+ still dominates membrane permeability, so Vm sits very near EK. The small Na+ leak pulls it only slightly toward ENa. The tempting wrong answer is that Vm sits exactly at EK, which ignores that nonzero Na+ permeability must shift it a little.
18. A researcher creates artificial membrane vesicles with K⁺ concentrations of 100 mM inside and 10 mM outside. After inserting K⁺ channels, the measured membrane potential is -55 mV instead of the predicted equilibrium potential. What is the most likely explanation for this discrepancy?
    Answer: There is significant membrane permeability to other ions in addition to K⁺ When you encounter membrane potential questions, remember that the actual potential depends on the relative permeabilities to all ions present, not just the ion of primary interest. Let's first calculate what the equilibrium potential should be using the Nernst equation: $E_K = \frac{RT}{zF} \ln\frac{[K^+]_{out}}{[K^+]_{in}} = -58 \log\frac{100}{10} = -58 \text{ mV}$ Since the measured potential (-55 mV) is less negative than predicted (-58 mV), the membrane must be permeable to other ions that are shifting the potential toward zero. This is exactly what answer B describes - the Goldman-Hodgkin-Katz equation shows us that membrane potential reflects the weighted average of all permeable ions' equilibrium potentials. Answer A is incorrect because voltage-gated channels that aren't fully open would reduce K⁺ permeability, making the potential less negative than -58 mV, but this scenario uses artificial vesicles without the complex regulatory mechanisms of voltage-gating. Answer C misses the mark because a 10-fold concentration gradient is actually quite substantial and more than sufficient to generate significant potential differences. Answer D is wrong because active transport requires energy sources like ATP, and these artificial vesicles lack the metabolic machinery needed for active transport processes. Study tip: When membrane potentials deviate from calculated equilibrium potentials, always consider mixed permeabilities. Real membranes are rarely permeable to just one ion - even small leaks to Na⁺, Cl⁻, or other ions can significantly alter the measured potential.
19. A cell maintains an internal K⁺ concentration of 140 mM and an external K⁺ concentration of 5 mM at 25°C. If the membrane potential is -70 mV, what can be concluded about the K⁺ gradient across this membrane?
    Answer: K⁺ has a net driving force inward because the electrical gradient is stronger than the concentration gradient When you encounter questions about ion gradients and membrane potential, you need to determine whether an ion is at electrochemical equilibrium by comparing the actual membrane potential to the equilibrium potential for that specific ion. First, calculate the K⁺ equilibrium potential using the Nernst equation: $E_K = \frac{RT}{zF} \ln\left(\frac{[K^+]_{out}}{[K^+]_{in}}\right)$. At 25°C, this becomes $E_K = 25.7 \text{ mV} \times \ln\left(\frac{5}{140}\right) = 25.7 \times (-3.33) = -86 \text{ mV}$. This means K⁺ would be at equilibrium if the membrane potential were -86 mV. However, the actual membrane potential is -70 mV, which is less negative than the K⁺ equilibrium potential. This creates a net driving force for K⁺ to move inward. The concentration gradient (140 mM inside vs 5 mM outside) drives K⁺ outward, but the electrical gradient (membrane is not negative enough compared to equilibrium) creates a stronger inward driving force. Answer A is incorrect because -70 mV ≠ -86 mV, so K⁺ is not at equilibrium. Answer C incorrectly states the net direction—while the concentration gradient favors outward movement, the electrical gradient is actually stronger and favors inward movement. Answer D is wrong because the gradients don't balance; there's a net driving force. Remember: always calculate the equilibrium potential first, then compare it to the actual membrane potential to determine the direction of net ion movement. The difference tells you which gradient dominates.
20. A cation has E=+30 mV and Vm=-70 mV. Its net driving force is:
    Answer: Inward; Vm is below E Because Vm (-70 mV) is more negative than E (+30 mV), the cell interior is more negative than the cation's equilibrium potential, so the electrical gradient pulls the cation inward. The driving force is Vm - E = -100 mV, and a negative driving force for a cation means inward movement. A tempting error is to say outward while still recognizing Vm is below E, but that reverses the direction: cations move toward negative charge.

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