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Unformatted text preview: Problem 1 Axon ion movements The squid axon has a specific membrane capacitance of 10-6 Farads/cm 2 . The membrane potential is -60 mV at rest and +40 mV at the peak of the action potential. When the membrane potential changes, the charge on the membrane capacitance changes. A. 1. How many Na + ions must enter the axon per square centimeter of axon membrane to change the charge from the resting potential to the peak action potential? The change in potential from rest to peak is +100 mV. Q=C V = 10-6 Farads x 0.1V = 10-7 coulombs. The Faraday constant is 9.65x10 4 C/mole, Avogadro's constant is 6.03x10 23 molecules/mole. ions = (10-7 coulomb)/(9.65x10 4 C/mole)x(6.03x10 23 molecules/mole) = 6.24 x 10 11 ions per cm 2 2. How many K + ions must leave to return the potential to rest? the same B. 1. For an axon which is 400 m in diameter, what is the surface area of a 1 cm length of axon? Surface per cm length: 0.126 cm 2 2. How many Na + ions must enter the axon per centimeter of axon length to change the charge from the resting potential to the peak action potential? Ions per cm length: (0.126 cm 2 ) x (6.24 x 10 11 ions per cm 2 ) = 7.84 x 10 10 ions per cm 3. How many K + ions must leave to return the potential to rest? the same C. 1. For an axon which is 400 m in diameter, what is the volume of a 1 cm length of axon? Volume per cm length: 1.26x10-3 cm 3 per cm (or 1.26 x 10-6 l per cm) 2. The intracellular ion concentrations for the squid axon are [Na + ] = 55 mM and [K + ] = 410 mM. What changes in the internal ion concentrations for Na + and K + occur during an action potential if the only ion movement is to charge and discharge the membrane capacitance? Na + entry: (7.84 x 10 10 ions per cm) / (6.03x10 23 molecules/mole) / (1.26 x10-6 l per cm) = 1.04 x 10-7 M per action potential. So fractional change is (1.04 x 10-7 M) / (5.5 x 10-2 M) = 1.88 x 10-6 K + exit is the same in terms of moles, and fractional change is (1.04 x 10-7 M) / (4.1 x 10-1 M) = 9.42 x 10-7 D....
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