/639 Final Exam Solutions, 2007

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1 /639 Final Exam Solutions, 2007 Problem Part a) Larger conductance in the postsynaptic receptor array; action potentials (usually Ca ++ ) propagating in the orward direction in the dendrites; ampliication by subthreshold conductances (Na + or Ca ++ ). The increase in input impedance o the dendrites as the synapse moves away rom the soma would also be accepted. Part b) Theories o neural plasticity oten postulate a Hebbian learning rule in which synapses are strengthened or weakened depending on the association o presynaptic activity (activation o the synapse) and postsynaptic activation o the neuron. Backpropagation o the AP allows the postsynaptic state o the neuron to be communicated to synapses throughout the dendritic tree. Part c) Ca ++ admitted to the spine head is conined to the spine and thereore serves as a speciic signal associated with that spine, useul or plasticity. Depolarization o the spine head spreads with little attenuation to the dendrite and is thus a general signal or the dendritic tree. Part d) In order or the decay to be exponential, the current must be applied to the end o a single long cable. I the euivalent cylinder theorem applies to this neuron, then current I is applied to the end o that cylinder. The decay would be exponential to the extent that the euivalent cylinder is long. Current I 2 produces a complex decay that includes current into side branches and is unlikely to be exponential. Part e) The euilibrium point or this system is x k x 2 0 Using the hint, try the ollowing candidate or Lyapunov U(x, x 2 ) a(x k )2 + (x 2 0) 2 This unction is positive deinite (assuming a>0) everywhere except at the euilibrium point where it is zero. Its time derivative is du U dx x + U dx 2 x 2 2a(x k )x 2 + 2x 2 ( kx rx 2 ) (2a 2k)x x 2 (2 a k 2)x 2rx 2 2 2

2 2 I ak, then du/ -2rx 2 2 which is negative everywhere except at the euilibrium point. Thus U is a Lyapunov unction on the whole plane and the euilibrium point is a stable attractor. Part ) A linear perceptron can compute any classiication that is linearly separable, meaning that w T x + b is >0 or one class and <0 or another, or some weight vector w and a bias b. For the separable examples, many sets {b, w} are possible. Problem 2 () Separable {-,, } (2) Separable {,, } (3) Not separable (4) Separable {,, -} (5) Not separable Part a) Let G m and C m be the membrane conductance and capacitance per unit area and R i be the axoplasmic resistivity. Then the parameters (length constant λ and input conductance G ) o the cylinders are given by the ollowing: λ j a j G j 2G m 3/2 π a j 2G m R i R i j, 2, 3. The conditions or the euivalent cylinder are then L j l j λ j j, 2, 3 L + L 2 L + L 3 (or L 2 L 3 ) G G 2 + G 3. The third condition, matching loads, is met by the zero axial-current condition at the ends o the tree. Note that i the membrane parameters G m, C m, and R i are assumed to be uniorm throughout the tree, then they cancel rom the euations above, leaving only L 2 L 3 reuires that l 2 a 2 l 3 a 3 G G 2 + G 3 reuires that a 3/2 a 2 3/2 + a 3 3/2. The properties o the euivalent cylinder are L EC L + L 2 λ EC λ l EC λ (L + L 2 ) l + l 2 λ λ 2 a EC a G EC G

3 3 Part b) Adding the voltage-dependent channel just adds a battery/resistor pair G K E K in parallel with the circuit (C m, G m, E m ) or the resting membrane (going back to the irst step in the derivation o the cable euation). Thus the resting membrane is represented by the circuit A in the igure at right. As long as G K is a constant, which will be so i V is constant everywhere in the dendritic tree, then the batteries and resistors can be replaced by a new resting circuit consisting o Thevenin euivalents, as in circuit B, which holds i the zero o membrane potential is reset to E me, as is usually done in deriving the linear cable euation. E me G KE K + G m E m G k + E m and G me G K + G m Now the parameters o the dendritic tree are changed by replacing G m with G me in the euations o part a). Note that G me >G m so the length constants will be shorter, and the G s will be larger. V E me. Problem 3 Part a) The current I produces potentials as ollows V I K II I V j K Ij I V 0 I, and plugging these into the deinition o subunit in the problem statement gives K Ij I K II I > C K I I 0 K II I or K Ij > C Both o the ollowing parts reuire the calculation o K I0 which reuires propagation o the membrane potential through the branchpoint. At the branchpoint, the input admittances o the and 2 cylinders are Y G tanhl and Y 2 G 2 tanhl, where L is the common electrotonic length o all the cylinders and the euations are simpliied because o the closed-end boundary conditions. The transer impedance K I0 is given by K IB A B0, where B is the branchpoint and A B0 is the voltage gain rom B to 0. Using ormulas discussed in class

4 4 K IB ( G tanhl + G 2 tanhl)coshl + G 3 sinhl ( G + G 2 + G 3 )sinhl A B0 so cosh L K IB A B0 ( G + G 2 + G 3 ) sinhl coshl. () Part b) The euation to be solved is K IX impedance across the distance L X in cylinder 3. > C, where K IX is the transer K IX Y X cosh L X + G 3 sinh L X, (2) where Y X is the input admittance at the end o the subunit (L X rom the end o cylinder 3). It is llikely that at D.C. (), En. (2) is a monotonic decreasing unction o L X. Thus the euation K IX > C will have a uniue solution where K IX CK I0. Finding Y X is a mess, because it is the input admittance o the cylinder o length (L-L X ) between the end o the subunit and the branchpoint, and is given by Y X G 3 (G tanh L + G 2 tanh L) G 3 + tanh (L L X ) + (G tanh L + G 2 tanh L) tanh (L L X ) G 3. (3) So the condition to be solved or L X is assembled rom a combination o Ens. (), (2), and (3). Part c) In this case, there are two conditions to be solved K IY > C and K IZ > C K I0 is given by En. (). K IY and K IZ take the same orm or this problem, so only the irst will be written out below.

5 5 K IY K IB A BY K IB cosh L Y + G tanh(l L ) Y sinh L G Y K IB coshl Y + tanh(l L Y )sinh L Y (4) and K IB is given by the euation derived in part a). The euation or K IZ results ater replacing L Y in En. (4) by L Z, G by G 3, and (L-L Y ) by (L-L Z ). Problem 4 Part a) The barrier diagram is at right. The membrane potential terms in the energy diagram are negative because the Cl moves in the outward (-V) direction when it moves into prestin. From the diagram the parameters reuested in the problem statement can be written: G S G * +W λzfv k αe (G B λ /2 zfv )/ RT λ /2 zfv / RT k e k r αe (G B λ /2 zfv G* W +λzfv )/ RT αe (G B G* W +λ /2 zfv )/ RT k r0 e λ /2 zfv / RT, where k 0 and k r0 contain all the constant parameters. To simpliy urther, let β λ F / 2RT, then, using the act that z- k k e βv and k r k r0 e βv k αe G B / RT The usual dierential euation is dπ and k r0 αe (G B G* W )/ RT. k [Cl]Χ k r Π. To avoid conusion, Χ is the raction o prestins with no Cl and Π is the raction o prestins with Cl. As usual, Χ + Π, so that dπ k ( k + k r )Π. Where k k Cl ; because the chloride concentration is a constant, it can be absorbed into k.

6 6 Part b) In the steady state with VV 0 and ΠΠ 0 dπ k ( k + k r )Π 0 so that k ) ( k ) + k r ))Π 0 0, where the dependence o the rate constants on V is noted. Part c) Substituting the voltage dependence gives the steady-state relationship k e βv 0 e βv 0 + k r0e βv 0 Π 0 0. (*) Dierentiating this euation w.r.t V and Π gives the change in Π that occurs in response to a change in V to keep the system in steady state, k e βv 0 βv k e βv 0 βv + k r0 eβv 0 βv Π 0 e βv 0 + k r0 eβv 0 π 0. The dierentiation uses the notation dvv and dππ. The euation can be simpliied again using the steady-state relationship. Add to the irst term in brackets giving k r0 e βv 0 βv k r0 eβv 0 βv k e βv 0 βv + k e βv 0 βv + k r0 eβv 0 βv Π 2k 0 r0 eβv 0 βvπ 0 k e βv 0 + k r0 eβv 0 The irst two terms are again zero rom the steady-state relationship, so 2k r0 e βv 0 βvπ 0 e βv 0 + k r0 eβv 0 π 0. This is a relationship between the small-signal variables that can be recast as π 0. π 2k r0 e βv 0 β Π 0 e βv 0 + k v (**) r0 eβv 0 Part d) -dπ/ is the small-signal rate o transer o Cl into the membrane (moles/s/cm 2 ) and is thereore a current i multiplied by zf. Because chloride is negatively charged, the actual charge movement is Fdπ/. Dierentiating En. (*) in the exam problem, the same as En. (**) above, gives zf dπ I CapNL C NL ) dv, which is the current-voltage relationship o a capacitor

7 7 2F k C NL r0 e βv 0 β Π 0 e βv 0 + k r0e βv 0 (***) Part e) At large negative voltage, all the prestin is saturated with Cl so there is no charge movement or small v; similarly, at large positive voltage, there is no Cl in the membrane and again there is no charge movement or small v. To see the eect o membrane potential on the capacitance, note that the steadystate relationship (En. (*)) allows P 0 in En. (***) to be written as ollows C NL 2F k r0 e βv 0 β e βv 0 + k r0 eβv 0 2F k Π 0 r0 e βv 0 β e βv 0 + k r0 eβv 0 2F k k r0 β e βv 0 + k 2 r0 eβv 0 k e βv 0 e βv 0 + k r0 eβv 0. As V 0 or V 0 -, C NL 0 because o the exponentials in the denominator. This corresponds to the ualitative explanation given in the previous paragraph. The unction looks like the igure in the problem set, i graphed, with a peak capacitance at V 0 RT F ln( k o k ro )/λ.