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 insidepositive 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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 Fall '07
 French
 MV, Vm

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