Question

Question 3 1 pts Many nerves, for example, motor neurons used in muscle contraction, are wrapped in myelin, a sheath of repeated units of lipid bilayers and proteins. The effect of myelin is to reduce the amount of ions leaking out the membrane, which therefore reduces the degree of amplification of a signal required during de- and repolarization, and, hence, increases the voltage signal speed. With regard to our model circuit, a myelinated axon simply can be represented by an axon with a thicker membrane, which has the effect of increasing the membrane resistance but reducing the membrane capacitance. In the previous lab, we found we could estimate an axon's length constant by simply determining the length of an axon segment for which the axon resistance equals the membrane resistance. Thus, this occurs when Ra = = Rm = Pm 21SO Pmrt V 2p where -, wherer is the radius of an axon and t is the thickness of a membrane. Hence, , for a myelinated axon fiber will be larger than that for an axon fiber without myelin because of the effectively, thicker membrane t. But does the time constant change for this ideal case of a myelinated axon fiber as compared to an unmyelinated fiber? Hint: As you worked out in the pre-lab, we can treat the membrane as a parallel plate capacitor if we simply unroll it. A segment of length I will simply have a capacitance Cm = K€, 277, where kis the dielectric constant of myelin (which, again, is about the same as that for the membrane). In the ideal case, since the myelinated membrane is thicker, the membrane capacitance decreases, dominating the change in resistance, so the time constant would be smaller. In the ideal case, the geometric factors for the resistance and capacitance of the axon membrane cancel, and so there would be no change in the time constant. In the ideal case, the myelinated axon would have both a larger resistance and capacitance, so the time constant would be larger. In the ideal case, since the myelinated membrane is thicker, the membrane resistance is larger, dominating the change in capacitance, so the time constant would be larger.

Answer 1

(Note: Answered only first part, next question can be answered separately.)

Answer: 4th option

The length constant (or l) of aron fibre is a measure of how fejn the voltage foavels dowor the axon before it decays to zero. The length constant is directly proportional to e resistence of the reyson's membrane. Tu tilbly it is, Orore is the resistance. Thues for an axon with myclimated Sheet will have higher resistance than one without it. Now time constatova (z) is a treasure of how quickly a final value ou voltage Lovel decays to 63% of its The lower the time constant, the faster a memboar wel respond to a stimulus. The time constant (t) is given by the relation 1 T = Rm Cm Wlure, er is rembogoue resistence Cm is capacitance of neural membrane which is a measure of its ability to store charge. neuron's voltage level decay

Now, if a neurom is covered with myelin sheet, its capacitance will decrease due to increase in distance of separation between charges. But as explained above, ils resistant will increase for a myelinated axon. Thus, these two factors will cancel each other and resulting time constem will have no change. The increase in memboome resistance doesn't dominate the change in capacitance as explained above. This time constant wald an increase. Similarly. Change ion capacitance can't dominali change in resistance as well. Therefore, all first three options are incorrect and last option is the comect one the connect one (4th option is correct answer )