Dynamical Systems in Neuroscience:

Solutions to ExercisesSolutions to Chapter 21. T = 20 ◦ C ≈ 293 ◦ F.E Ion = RTzF ln [Ion] out 8315 · 293 · ln 10=[Ion] in z · 96480when z = ±1. Therefore,log 10[Ion] out[Ion] inE K = 58 log(20/430) = −77 mVE Na = 58 log(440/50) = 55 mVE Cl = −58 log(560/65) = −54 mV= ± 58 log 10[Ion] out[Ion] in2.I = ḡ Na p (V − E Na ) + ḡ K p (V − E K ) = p{(ḡ Na + ḡ K ) V − ḡ Na E Na − ḡ K E K }()= (ḡ Na + ḡ K ) p V − ḡNa E Na + ḡ K E Kḡ Na + ḡ K} {{ } } {{ }ḡE3. The answer follows from the equation4. See Fig. 10.1.I − g L (V − E L ) = −g L (V − ÊL) , where Ê L = E L + I/g L .Function V 1/2 k Function V max σ C amp C basen ∞ (V ) 12 15 τ n (V ) −14 50 4.7 1.1m ∞ (V ) 25 9 τ m (V ) 27 30 0.46 0.04h ∞ (V ) 3 −7 τ n (V ) −2 20 7.4 1.2Remark: Hodgkin and Huxley shifted V 1/2 and V max by 65 mV so that the resting potential isat V = 0 mV.5. (Willms et al. 1999)Ṽ 1/2 = V 1/2 − k ln(2 1/p − 1) ,˜k =k2p(1 − 2 −1/p ) .The first equation is obtained from the condition m p ∞(Ṽ1/2) = 1/2. The second equation isobtained from the condition that the two functions have the same slope at V = Ṽ1/2.6. See author’s webpage.7. See author’s webpage.407