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prob_set1_solns

prob_set1_solns - prob_set1_solns Page 1 of 3 Solutions to...

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Solutions to Problem Set #1 for week of April 6 1. a. If the value of component A becomes smaller, so will the value of component B. The positive sign connecting the two boxes indicates that the value of component B will change in the same way as does the value of component A. b. If the value of component D becomes smaller, the value of component E will get larger. The negative sign connecting the two boxes indicates that the value of E will change in the opposite way to the value of D. c. Loop A->B->C->D->E->F->A is a positive feedback loop because it contains 2 (an even number) of negative signs (signal inversions). d. Loop B->C->B is a negative feedback loop because it contains 1 (an uneven number) negative sign. 2. a) E Na = 61 [log (150/10)] mV = 61 X (1.176) mV = +71.7 mV E K = 61 [log (3/112)] mV = 61 X (-1.572) mV = -95.9 mV E Cl = -61 X log([Cl-] out /[Cl-] in )= -61 X (1.342) = -81.9 mV. Note: E Cl could also be calculated as 61 [log (4/88)] mV = 61 X (-1.342) = -81.9mV b) Depolarizations in action potentials approach (but do not necessarily reach) E Na . Therefore, in this organism, during an action potential V m may become as inside-positive as +71.7 mV, but it could not become more positive than this value. In fact, most action potentials are over too fast to allow the
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