vw-,\\ : faAX raAX _. 1 _. 10:, t) \"(XI t) ax 10: + AX, t) Figure 6: Circuit diagram for two segments of length As): in the cable modelled dendrite. The parameters in our circuit system, being the values for the dierent resistances and the capacitance are dened per unit length, therefore when discussing resistance in a segment of length Ax, we need to include An: in our parameters to correctly describe the segmented circuit. These parameters are shown in Figure 5. The total membrane resistance R: is the resistance of the entire membrane of the dendrite. The membrane resistance is the strength of the membrane to resist the out ow of charge from within the dendrite. As you increase the surface area of the dendrite you allow more charge to escape and thus lower the resistance. This make the membrane resistance inversely proportional to the membrane surface area. The resistivity rm is dened below where L is the length of the dendrite. rm = a1. (7) As the length is increased by stretching the dendrite, the ion channel density decreases and the amount of charge able to leave the dendrites decreases, thus the resistivity increases. The axial resistance increases as the length of the dendrite is increased and thus \"Fa can be dened as axial resistance per unit length, where Ra is the total axial resistance. H To. = Ta (8) This difference in proportionality between the axial resistance and membrane resistance is why A1: is being divided to nd the membrane resistance of the segment and multiplied to nd the axial resistance of the segment. The membrane current 22m and injected current il- are currents per unit length aswell. Question 12: Use Ohm's law to write the change in potential AV from x to x + Ax. (Only the axial resistance will be needed) Question 13: With slight manipulation, your equation for the change in potential should be giving flashbacks to your math class. By taking the limit as Ax - 0, rewrite your equation for the potential by employing the definition of a partial derivative. You will need to divide both sides of your equation by something first for this to become clear. A similar process can be done with the Kirchhoff's current equation for our circuit diagram. After our partial derivative trick the current equation is as follows. im(x, t) - ii(x, t) = - ax (9) Using the equations for the potential and the current, along with the current equation from part 1, the following equation can be reached. av ra Ox2 2 = I'mCm at -+V -ii(x, t) rm (10) In its complete form this equation is very complicated and its solution is by no means trivial. As we said, approximation can be applied to simplify it. Before doing such, we will define our characteristic constants. Question 14: Calculate the units of Tm/ra and define a characteristic constant based on the units. Rewrite our cable equation defined in equation 4 using the characteristic constants you defined. Just as we begin to first understand our voltage equation in part 1 by taking the steady state solution. We will approximate our cable equation by assuming the steady state solu- tion to be true. This means that our potential is constant in time and the partial derivative with respect to time can be taken as zero. We will also assume that our cylinder is of infinite length and the current is injected, and only injected, perfect in the center of the length. Therefore our current is a constant of Io (These conditions will simplifying the steps in solv- ing our differential equation). Taking the steady state solution our cable equation becomes the following 120 - V 2x 2 = V (11) Where A corresponds to the constant you defined in question 14, though you may relabel it as you wish. Equation 5 has the following general solution V(2) = Ae-lz1/> + Belz//x (12) The constants A and B must be determined from the limiting conditions of our model.Question 15: Given that the current is injected at :r: = 0, determine what the values of A and B are by considering the limiting values of 11. As a gets really large what must the potential do? look to the behaviour of the eaponential functions to determine which constant is what. Remember, this is the steady state solution. How does that help in determining the constants? Also, in addition to the general equation above, we will need to add a shift to our potential so that the potential at zero is the resting potential value (TUmV). Question 16: Plot your solution to the potential as a function of w for the following values of A = 0.3, (15,07. Discuss the dependence of the potential and neuron signaling on A? Think about how /\\ e'ects the change in potential along with dendrite. It can be shown that the constant rm, and ra can be written as follows, Rm r, = (14) Where Rm is membrane resistance per unit area, a is the radius of the dendrite, and p is the resistivity of the cytoplasm which is the inverse of conductivity 0. Question 17: Redene /\\ in terms of a, Rm, and p using the equations above. Discuss how the dendrite radius and cytoplasm resistivity eect the potential in dendrites and ulti- mately neuron signalling. Question 18: Alzheimer's disease is characterized by the accumulation of the B-amyloid peptide {Ali} in the brain. It if formed from the breakdown of amyloid precursor proteins (APP). The build up of Ali in the brain effects the ion channels in neurons in multiple ways. In one study, it was shown that the A build up causes a decrease in sodium channels present in dendrites. As shown in gure 1, sodium ion channels control the inua: of current into the dendrite. How would the leak ion channels (say potassium) resistance and thus the number of potassium channels need to change such that the dendrites ability to activate an action potential is not changed due to the build up of A