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 membrane to become this positive, so at the peak of the action potential the transmembrane potential is
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prob_set1_solns - prob_set1_solns Page 1 of 3 Solutions to...

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