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CN2011 Problems 1 to 4 2011 Key

# CN2011 Problems 1 to 4 2011 Key - Problem set 1 2011 page 1...

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Problem set 1 2011 page 1 Problem 1 Axon ion movements 10 points total; 1 for each question and partial credit for correct equation, incorrect values. The squid axon has a specific membrane capacitance of 10 -6 Farads/cm 2 . The membrane potential is -65 mV at rest and +45 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 +110 mV. Q=C V = 10 -6 Farads x 0.11V = 1.1x10 -7 coulombs. The Faraday constant is 9.65x10 4 C/mole, Avogadro's constant is 6.03x10 23 molecules/mole. ions = (1.1x10 -7 coulomb)/(9.65x10 4 C/mole)x(6.03x10 23 molecules/mole) = 6.87 x 10 11 ions per cm 2 2. How many K + ions must leave to return the potential to rest? B. NOTE: either diameter was OK for B, C, D as long as it was stated! 1. For an axon which is 400 μm 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 (6.87 x 10 11 ions per cm 2 ) = 8.63 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 350 μm in diameter, what is the volume of a 1 cm length of axon? Volume per cm length: 9.62x10 -4 cm 3 per cm (or 9.62 x 10 -7 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: (7.55 x 10 10 ions per cm) / (6.03x10 23 molecules/mole) / (9.62 x 10 -7 l per cm) = 1.30 x 10 -7 M per action potential. So fractional change is (1.30 x 10 -7 M) / (5.5 x 10 -2 M) = 2.37 x 10 -6 K + exit is the same in terms of moles, and fractional change is (1.30 x 10 -7 M) / (4.1 x 10 -1 M) = 3.18 x 10 -7 D. 1. If we inhibited the Na/K ATPase with ouabain, how many action potentials could the axon carry before the internal [Na + ] doubled? Assuming that the change in ionic gradients (and hence equilibrium potential) did not make significant changes in the ion fluxes, a change of 55 mM would require about 422,100 action potentials. Actually, since E Na will get less positive, and so entry will be reduced, it will take more. E. Actual measurements from isotope fluxes per impulse in squid axon give the following values: net influx of Na + = 3.6x10 -12 moles of Na + per cm 2 of axon membrane per impulse net efflux of K + = 4.0x10 -12 moles of K + per cm 2 of axon membrane per impulse 1. Why are these values larger than the net flux values you calculated from the amount of charge required to charge and discharge the capacitance? 2. With these actual flux measurements, how many impulses could the axon carry for question 1D?

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CN2011 Problems 1 to 4 2011 Key - Problem set 1 2011 page 1...

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