advanced problem solving exam one cells + sys

Two dendrites, both 20 mV depolarization at tips. Dendrite 1: 1mm from initial segment, λ=1mm. Dendrite 2: 2mm from initial segment, λ=2mm. Will neuron spike?

Step 1: Calculate decay for each dendrite. Dendrite 1: V = 20 × e^(-1mm/1mm) = 20 × e^(-1) = 20 × 0.37 = 7.4 mV. Dendrite 2: V = 20 × e^(-2mm/2mm) = 20 × e^(-1) = 20 × 0.37 = 7.4 mV. Step 2: Sum at initial segment: 7.4 + 7.4 = 14.8 mV. Step 3: Compare to threshold: Need 15 mV to go from -70 to -55 mV. Result: NO SPIKE (14.8 < 15 mV)

Why do both dendrites in the previous problem contribute equally despite different distances and λ values?

KEY INSIGHT: Both dendrites are exactly 1 space constant away from initial segment. Dendrite 1: 1mm distance ÷ 1mm λ = 1λ. Dendrite 2: 2mm distance ÷ 2mm λ = 1λ. At 1λ distance, signal always decays to 37% regardless of absolute distance. This is why λ is the fundamental unit for measuring electrical distance in neurons.

CALCULATION DRILL: [K+]o changes from 3mM to 15mM. Calculate new EK, driving force change, and predict effect on excitability.

Step 1: New EK = -58 log(15/140) = -58 log(0.107) = -58 × (-0.97) = +56 mV. Wait - let me recalculate: EK = -58 log(15/140) = -58 × (-0.97) = +56 mV. Original EK ≈ -90 mV. Step 2: Change = +56 - (-90) = +34 mV more positive. Step 3: DF becomes 34 mV less negative → easier to reach threshold → MORE EXCITABLE

MECHANISM DETAIL: Describe the complete sequence of Na+ channel states during an action potential, including all gates.

Resting (-70mV): Activation gate CLOSED, Inactivation gate OPEN → Channel CLOSED. Depolarization to threshold: Activation gate OPENS rapidly, Inactivation gate still OPEN → Channel OPEN → Na+ influx. Peak AP (+30mV): Activation gate OPEN, Inactivation gate CLOSES slowly → Channel INACTIVATED → No Na+ flow. Repolarization: Activation gate CLOSES (deactivation), Inactivation gate still CLOSED → Channel INACTIVATED. Return to rest: Activation gate CLOSED, Inactivation gate RE-OPENS → Channel ready for next AP

COMPLEX CABLE PROBLEM: Axon segment has Rm=10kΩ·cm, Ri=100Ω·cm, diameter=2μm. Calculate λ and predict 50mV signal decay over 200μm.

Step 1: Convert units: d=2μm=0.0002cm, a=0.0001cm. Step 2: rm = Rm/(2πa) = 10,000/(2π×0.0001) = 159×10^6 Ω·cm. ri = Ri/(πa²) = 100/(π×(0.0001)²) = 3.18×10^9 Ω/cm. Step 3: λ = √(rm/ri) = √(159×10^6/3.18×10^9) = 0.0224cm = 224μm. Step 4: Distance = 200μm = 0.89λ. V = 50 × e^(-0.89) = 50 × 0.41 = 20.5mV

DRUG MECHANISM DEEP DIVE: Ouabain blocks Na/K ATPase. Explain why this doesn't affect AP shape but would eventually kill the neuron.

Immediate effect (seconds-minutes): NONE on AP shape. AP depends on electrochemical gradients maintained by high [Na+]o/[Na+]i and [K+]i/[K+]o ratios. Pump contributes only ~3-5mV to resting potential. Long-term effect (hours): Pump normally exports 3Na+, imports 2K+ per ATP. Without pump: Na+ accumulates inside, K+ depletes. ENa becomes less positive, EK becomes less negative. Eventually: Gradients collapse → no APs possible → osmotic swelling → CELL DEATH

SELECTIVITY FILTER MECHANICS: Why can't Na+ pass through K+ channels even though Na+ is smaller when dehydrated?

Key insight: It's about HYDRATED sizes and dehydration energy. Hydrated Na+ ≈ 7.2Å vs hydrated K+ ≈ 6.6Å. K+ selectivity filter process: 1) Hydrated K+ enters vestibule. 2) Carbonyl oxygens precisely positioned to strip water. 3) Dehydrated K+ (2.66Å) makes optimal contacts. 4) Low energy barrier for passage. For Na+: Dehydrated Na+ (1.9Å) too small for optimal carbonyl contacts → energetically unfavorable to dehydrate → REJECTED. Size complementarity is crucial.

TEMPORAL SUMMATION QUANTITATIVE: Dendrite τ1=10ms. Two 5mV EPSPs: first at t=0, second at t=5ms. Calculate peak voltage assuming exponential decay.

EPSP decay: V(t) = V₀ × e^(-t/τ). At t=5ms, first EPSP decayed to: V₁ = 5 × e^(-5/10) = 5 × e^(-0.5) = 5 × 0.61 = 3.05mV. Second EPSP adds 5mV at t=5ms. Total at t=5ms: 3.05 + 5 = 8.05mV. Peak occurs just after t=5ms since second EPSP is rising while first continues decaying. Maximum temporal summation when τ₁ >> inter-spike interval.

MYELIN QUANTITATIVE ANALYSIS: Unmyelinated axon (d=1μm, v=1m/s) gets myelinated. Myelin increases rm by 100×, decreases cm by 100×. Predict new velocity.

Conduction velocity depends on λ/τ₂. λ = √(rm/ri), so λ increases by √100 = 10×. τ₂ = ri×cm, so τ₂ decreases by 100× (cm decreases, ri unchanged). New v ∝ (10×λ)/(0.01×τ₂) = 1000× improvement? Not quite - real improvement ~50-100× due to other factors. Matches physiology: 1μm unmyelinated ≈ 1m/s, 1μm myelinated ≈ 50m/s.

INTEGRATION PROBLEM: Neuron receives 3 simultaneous inputs. Input 1: +8mV at soma. Input 2: +12mV at 0.5λ. Input 3: -6mV (inhibitory) at 0.3λ. Will it fire?

Calculate contribution of each input at initial segment: Input 1: 8mV (no decay). Input 2: 12 × e^(-0.5) = 12 × 0.61 = 7.32mV. Input 3: -6 × e^(-0.3) = -6 × 0.74 = -4.44mV. Sum: 8 + 7.32 - 4.44 = 10.88mV depolarization. New Vm: -70 + 10.88 = -59.12mV. Threshold: -55mV. Result: NO SPIKE (needs 4.88mV more). Shows importance of inhibitory input location.

CHANNEL KINETICS DETAIL: Voltage clamp to +20mV. INa peaks at 0.5ms then decays to zero. IK rises slowly, peaks at 2ms, stays high. Explain molecular mechanisms.

Na+ channel kinetics: Fast activation + slow inactivation = transient current. Mechanism: Depolarization → S4 voltage sensors move outward rapidly (<0.5ms) → activation gate opens → Na+ rushes in → IFM inactivation peptide swings into pore (1-2ms) → current stops despite continued depolarization. K+ channel kinetics: Slower activation, no inactivation. S4 movement slower than Na+ channels → gradual opening over 1-2ms → sustained current during depolarization. Different kinetics enable AP shape.

REVERSAL POTENTIAL ANALYSIS: I-V plot shows early current reverses at +52mV, late current at -85mV. Identify ions and explain deviations from predicted values.

Early current: Reverses at +52mV ≈ ENa(+56mV) → Na+ current. Late current: Reverses at -85mV ≈ EK(-90mV) → K+ current. Deviations from theory: 1) Na+ channels have slight Ca2+ permeability → ENa slightly less positive. 2) K+ channels may have slight Na+ permeability → EK slightly less negative. 3) Local ion depletion/accumulation during large currents. 4) Temperature differences from standard 25°C. This analysis was key evidence for ionic theory of AP.

REFRACTORY PERIOD MECHANISMS: Distinguish absolute vs relative refractory periods. A second stimulus 2ms after first AP requires 2× normal current. Which period and why?

Absolute refractory: Na+ channels inactivated → NO AP possible regardless of stimulus strength. Lasts ~1ms. Relative refractory: Na+ channels recovering from inactivation + K+ channels still open → LARGER stimulus needed. At 2ms post-AP: Some Na+ channels available but many still inactivated, plus delayed K+ channels open → relative refractory period. Need 2× current because: 1) Fewer available Na+ channels. 2) Outward K+ current opposes depolarization. 3) Membrane potential still hyperpolarized from K+ efflux.

CA2+ CHANNEL SELECTIVITY PROBLEM: Wild-type Ca2+ channel has EEEE selectivity filter. Mutant has EEKA (one glutamate→alanine). Predict permeability changes and explain.

Wild-type EEEE: 4 negative charges create strong Ca2+-selective filter. High PCa2+/PNa+ ratio (~1000:1). Mutant EEKA: Only 3 negative charges. Reduced Ca2+ selectivity because: 1) Fewer negative charges can't coordinate Ca2+ as effectively. 2) Larger effective pore allows Na+ passage. 3) Reduced Ca2+ binding affinity. Result: PCa2+/PNa+ ratio decreases to ~100:1. More Na+ current because [Na+]o >> [Ca2+]o. Current carried primarily by Na+ despite remaining Ca2+ selectivity.

SPACE CONSTANT MANIPULATION: Dendrite has λ=0.5mm. Drug X doubles membrane resistance. Drug Y halves internal resistance. Predict combined effect on λ.

Original: λ₀ = √(rm/ri). Drug X: rm → 2rm. Drug Y: ri → 0.5ri. Combined: λnew = √((2rm)/(0.5ri)) = √(4rm/ri) = 2√(rm/ri) = 2λ₀. New space constant = 2 × 0.5mm = 1.0mm. Functional consequence: Signals decay less over distance → better spatial summation → more excitable neuron. EPSP at 1mm now contributes 37% instead of 14% (e^(-2) vs e^(-1)).

EQUILIBRIUM POTENTIAL TEMPERATURE DEPENDENCE: Calculate EK at body temperature (37°C) vs room temperature (25°C) for standard ion concentrations.

At 25°C: EK = -58 log([K+]o/[K+]i) = -58 log(3/140) = -90mV. At 37°C: Use 61mV factor instead of 58mV. EK = -61 log(3/140) = -61 × (-1.67) = -102mV. Difference: 12mV more negative at body temperature. Physical basis: RT/F term increases with temperature. Biological significance: Neurons are more polarized at physiological temperature → may need stronger stimuli to reach threshold. Always specify temperature in experiments!

INACTIVATION REMOVAL EXPERIMENT: Shaker K+ channel loses inactivation when N-terminal "ball" is removed. Adding synthetic ball peptide restores inactivation. Explain the mechanism and significance.

Normal Shaker channel: N-terminal ball (20 amino acids, rich in R,K,H) connected by flexible chain. Mechanism: 1) Depolarization opens channel via S4 movement. 2) Ball swings into open pore within ~10ms. 3) Positive charges on ball interact with negative charges in pore. 4) Current blocked despite continued depolarization. Ball removal: Channel opens normally but can't inactivate → sustained current. Ball addition: Exogenous peptide can block open channels → inactivation restored. Significance: Proves "ball-and-chain" model, shows inactivation is separate from activation.

CABLE THEORY PREDICTION: Axon initial segment generates 100mV AP. Predict voltage 1λ, 2λ, and 3λ away if passive spread only (no voltage-gated channels).

Passive decay formula: V(x) = V₀ × e^(-x/λ). At 1λ: V = 100 × e^(-1) = 100 × 0.37 = 37mV. At 2λ: V = 100 × e^(-2) = 100 × 0.14 = 14mV. At 3λ: V = 100 × e^(-3) = 100 × 0.05 = 5mV. From rest (-70mV): Actual Vm would be -70+37=-33mV, -70+14=-56mV, -70+5=-65mV at respective distances. This demonstrates why active propagation (regenerative APs) is essential for long-distance signaling.