Biology Assignment Help

Harvard-MIT Division of Health Sciences and Technology HST.131: Introduction to Neuroscience Course Director: Dr. David Corey

HST 131/ Neuro 200

Problem Set 1 – Ion Channels and Gating (due Monday, Sept 12, 2005) You will need access to two simulation programs for this problem set: IV.exe and IVVG.exe.

IV.exe will be useful to study IV plots, but is not necessary for this problem set. IVVG.exe will be used for questions 5 and 6. Lab View Run Time (lvrt.dll) must also be downloaded and stored in the same folder as IV.exe and IVVG.exe. Unfortunately, they don’t run on Mac computers.

1.

a.

The following are typical mammalian ion concentrations in mM. Calculate the Nernst potential for each ion species.

K+ Na+ ClCa2+

Internal 137 9 8 0.0001

External 5 140 115 1.8

b.

The following are ion concentrations in mM for the squid giant axon. Again, calculate the Nernst potential for each ion species. Although the ion concentrations for squid giant axons are quite different , you’ll note that the Nernst potentials are remarkably similar. How can this be the case?

K+ Na+ ClCa2+ 2.

Internal 380 32 38 0.001

External 15 455 524 9.3

Use the GHK equations and the mammalian concentrations above, and assume a wholecell permeability of 10-7 amps/M for each ion species. a. What is the inward and outward sodium current at +60 mV? (Show your work) b. What is the total chloride current at -60 mV? Recalculate the total chloride current at -60mV with an internal Cl- concentration of 25 mM. Does the current increase, decrease or stay the same? Explain.

3.

Explain the difference between the Nernst potential for an ion, the reversal potential for a single channel type, and the reversal potential (“resting potential”) for a whole cell.

When might all three be the same?

4.

At the reversal potential for a channel, does ion flow cease? Explain.

5.

In a typical action potential, the membrane potential of a neuron may go from –70 mV to +45 mV. a. Based on the expression V=Q/C, for a spherical neuron with a diameter of 23 µm and membrance capacitance of 1 µF/cm2, how much charge (in coulombs) must move across the membrane to effect this potential change? b. How many moles of Na+ would this represent? c. What mole fraction of the total intracellular Na+ is this? d. Assume that the extracellular space around this neuron has a width of 0.1 micrometer, and that the amount of potassium that leaves the cell during an action potential is equal to the sodium that entered. How does the extracellular potassium concentration change?

6.

Suppose a neuron has only chloride channels. The intracellular and extracellular concentrations are such that the Nernst potential for chloride is -35 mV. If you experimentally change the membrane potential to 0 mV, you measure 4.2 nA flowing through the channels. a. Is this current flowing inward or outward? Is the chloride flux inward or outward? b. Approximating the I(V) relation with a straight line, what is the equivalent conductance of the cell membrane? What is the equivalent resistance? c. Using the capacitance you calculated in question 5a, calculate the time constant for the neuron.

The program IVVG is a good model for voltage-gated channels that don't inactivate; for them it represents the steady-state current as a function of voltage. It's not so good for sodium channels, which have no steady-state current after they've inactivated. However, you could think of the sodium channel plot as a peak I(V) relation--what you would observe after the channels have activated but before inactivation kicks in.

Two main plots are shown, the I/V relationship and the P(V) function. Each ion species can be turned on or off individually, and is represented on the plots by a unique color. Ion concentrations, permeability, valence, gating charge, and gating midpoint can be adjusted for each ion species.

7.

Set all values to physiological values (Physiological Values button). Now turn on the Na+ channels. a. Observe the peak I(V) curve. If the Na+ gating charge (‘s’ value) is halved, how does this affect the peak I(V) curve? b. What physical change in the Na+ channel structure would this represent, and why would a cell want a larger or smaller gating charge for a channel?

Reset to the physiological values. Turn the Na+ channels off and K+ channels on. c. Change the midpoint of the probability function (‘midpt’) from –50 mV to –25 mV. How does this affect the peak IV curve for K+? What advantage could this give a neuron? d. Reset to physiological values once again. Under what general conditions is there inward current through potassium channels?

Harvard-MIT Division of Health Sciences and Technology HST.131: Introduction to Neuroscience Course Director: Dr. David Corey

September 09, 2005

HST 131/ Neuro 200

Solution

Problem Set 1 – Ion Channels and Gating (Answers)

Ex =

RT [X o ] ln zF [X i ]

1a.

Let RT/zF = 25.4 mV, assuming z=1 and T=22C. (Values for R and F can be found in the course guide, as well as this common assumption.)

Mammals if T=37C

Internal

External

Eq. Pot.

K+

137

5

-84.1 mV

-88.4 mV

Na+

9

140

+69.7 mV

+73.3 mV

Cl-

8

115

-67.7 mV (since z=-1)

-71.1 mV

Ca2+

.0001

1.8

+124.4 mV (since z=+2)

+130.8 mV

1b. Let RT/zF = 25.4 mV as above. Students may assume a lower temperature for seawater; this is acceptable.

Squid Giant Axon if T=10C

Internal

External

Eq. Pot.

K+

380

15

-82.1 mV

-78.8 mV

Na+

32

455

+67.4 mV

+64.7 mV

Cl-

38

524

-66.6 mV (since z=-1)

-63.9 mV

Ca2+

.001

9.3

+116.0 mV (since z=+2)

+111.3 mV

The Nernst potentials are similar because they depend on the ratio of the concentrations, which are themselves similar. Furthermore, they depend on the natural logarithm of the ratio, thus small changes in the ratio have little impact on the Nernst potential.

2a.

I i = [ X o ]zPx

Vm

zF RT

zF 1 − exp(Vm ) RT

Amps = -3.44 nA.

Let external sodium = 0.140 moles, valence =+1, V=0.060 volts, RT/zF= 0.0254 volts, and permeability = 10^-7 amps/mole. Remember the sign!

IO = +2.35 nA

2b.

At V= -60mV,

For Cl

⎡ [Xo] − [Xi ]exp(Vm zF )⎤⎥ ⎢ zF zF RT (Vm ) = (-0.06V) x (-39.37V −1 ) = 2.36 ⎥ It = z PCl (Vm )⎢ RT RT ⎢ zF ⎥ 1 - exp(Vm ) ⎥ [Cl-]I = 8 mM, It = +0.74 nA(outward current; inward ⎢⎣ RT ⎦ flux)

For [Cl-]I = 25 mM, It = -3.69 nA (inward current; outward Cl flux) Thus, the current reverses from outward (hyperpolarizing) to inward (depolarizing). Higher [Cl-]I ⇒ more outward flux of Cl-. Since z = -1, this means more inward current ⇒ (outward current remains the same). Thus, there is more total current.

3.

Nernst Potential for an Ion: The Nernst Potential for an ion is derived from a hypothetical situation where there is a concentration difference of that ion across the membrane, and the membrane is permeable to that ion species alone. The voltage at which the the chemical potential causing diffusion across the membrane is offset by the electrical potential across the membrane is known as the reversal potential (see question 4 for explanation that current continues to flow). Thus, the Nernst Potential is also the potential at which net ionic current for that ion switches from net inward flux to net outward flux.

Reversal Potential for a Single Channel: This is the potential at which net ion flux through the channel is zero. If a single channel were permeable to only one species of ion, then the reversal potential for the channel would be equal to the Nernst Potential for that species of ion. In practice, however, channels tend to be more permeable to some types of ions than others. Thus, current through the channel will reverse at a potential which reflects the relative permeability to all ions capable of passing through the channel. For example, assume that the Nernst Potential is -80mV for potassium and +40mV for sodium. If a potassium channel is 10x more selective for potassium than sodium, then the reversal potential for the channel will be close to -80mV. If the channel is relatively nonselective between sodium and potassium, the reversal potential will be closer to 0 mV. Resting Potential for the Cell: Again, another level of complexity. The cell’s resting potential is the potential at which the net ionic current into/out of the cell is zero. This is a balance between many different channel types. So, the net current through all channels open at rest is zero. Normally when we speak of resting potential, we mean the state of the cell between action potentials, so most voltage gated sodium channels are shut and the potassium channels mediating the after hyperpolarization are closed as well. The resting membrane potential consists of some potassium leak channels, and any other small conductances. Thus, resting potential is often near reversal for potassium (since this is the predominant conductance open at rest).

All three could be the same if: -channel is perfectly selective to one ion -cell has only that one sort of channel 4. At the reversal potential for a channel: First, let’s assume the channel only conducts one species of ion. The reversal potential is the potential at which inward current equals outward current. Thus, there are still ions flowing in and out of the cell, so ion flow does not cease. The two currents merely offset one another and the potential does not change.

Next, making the situation more complicated, the channel may be permeable to more than one type of ion. The reversal potential for the channel will be the point at which the net ion flux is zero. Assuming the Nernst Potentials for the ions are different (as is the case for common cations such as potassium and sodium), this means the reversal potential could be different from the Nernst potential for either channel.

So, for a cation channel that does not distinguish between potassium and sodium, the reversal potential may be about zero, even though potassium (-80mV) and sodium (+40mV) may have different Nernst potentials. So there could be a net flux of sodium in and potassium out at zero millivolts. Again, we conclude that ion flow does not cease at the reversal potential.

5.

Concentration change with an action potential.

Use Mammalian Concentrations 2 Area of a sphere A = 4πr ; if r = 11.5µm then A= 1661.9µm2 Since capacitance is ~ 1µF/cm2, Cm= 16.62 pF

5a. Charge Q = CV; if ∆V = 115mV then Q = 1.91 pC. 5b. Moles Q/zF; z=1 and F=96485 coul/mol so ∆mol = 1.98x 10-17 mol Na+ 5c. Mole Fraction Volume of a sphere

v=

4 π r3 3 3 ; volume =6370.6µm = 6.37 pL

Total sodium = concentration x volume = 9 mM x 6.37 pL= 5.73 x 10-14 mol Fractional change = (Sodium change)/(Total Sodium) = 1.98x10^-17/5.73x10^-14.

This is about =0.00034553.

Intracellular Na+ will rise by about 1.98x10^-17 mol/6.37x10^-12 liter, or about 3.11 µM, which is negligible.

5d. Volume of extracellular space: For the difference of volumes between spheres of r = 11.5 µm and r= 11.6 µm: 6538.3 µm3 – 6370.6µm3 = 167.6 µm3 or 168 fl.

Moles of Potassium: ∆ KO =1.98x 10-17 mol K+ Extracellular K+ will rise by about 116 µM, which may be significant, particularly if the cell were to fire multiple action potentials in rapid succession.

6.

6a. Since the reversal potential is -35mV, we know that the internal concentration is less than the external concentration. The osmotic pressure for chloride to come into the cell (since it is higher on the outside) is only offset by the negative potential at reversal (-35mV). Thus, when we change the potential to 0 mV, more chloride is diffusing into the cell (higher outside concentration) than into the cell (lower internal concentration. This is a net inward flux of negative ions, which is the same current as a net outward flux of positive ions. Thus, current is “outward” which chloride ion flux is “inward”. Conveniently, “outward” current is by convention positive, and the current stated in the equation (4.2 nA) is positive.

6b. The I(V) relation is approximated by a straight line. This is the same as an equivalent circuit with a battery to represent the reversal potential (Erev) and a resistor to represent the chloride channels. Ohm’s Law tells us that the slope between current and voltage is resistance. We know two points on the line: current is zero at -35mV, and current is 4.2 nA at 0mV. Thus, change in voltage is: +35mV and change in current is +4.2 nA. Equivalent resistance is: (35x10^-3 V)/(4.2x10^-9A) = 8.3 MOhms. Equivalent conductance is the inverse of equivalent resistance, so G= 1/R = 1/(8.3x10^6) = 0.00000012 Siemens = 0.12 uSiemens = 120 nS. 6c. The time constant, tau, is simply the product of R (Ohms) and C (F). R = 8.3x10^6 Ohms. C = 16.62x10^-12 F. Thus, tau = 137.9x10^-6 seconds or 137.9µs.

Simulation Questions (Require IV.exe, IVVG.exe, and lvrt.dll):

7a.

The steady state current passes inward current at a more hyperpolarized potentials, and its slope is less negative.

7b.

This would represent halving the charge in the S4 voltage sensor. A larger gating charge would

provide the cell with a channel that is more sensitive to potential changes, over a shorter range of voltages.

7c.

The channel opens at a more positive voltage, resulting in less outward current until higher

potentials are reached. This could result in a cell that is more excitable given that much less K current can flow to counteract the depolarizing Na current, and thus neurons would reach action potential threshold faster.

7d.

Inward current through potassium channels occurs when the cell is hyperpolarized (sufficiently negative potentials) that membrane voltage is below potassium reversal potential.

This rarely occurs in vivo, since it is potassium currents that tend to return the cell to rest following an action potential.