The relationship between the distance and time of diffusion is d ; that is, the distance diffused is proportional to the square root of time (e.g., for three-dimensional diffusion, ). One way to solve the problem is to plug numbers in: first, second, . The first equation allows us to solve for D : squaring the first equation gives 25 μm ² = 6 D sec, which means that D = 25/6 μm ² /sec. Substituting D into the second equation gives . Squaring both sides gives 100 μm ² = (25 μm ² sec ⁻¹ ) · t , which means that t = 4 sec. We see that, because of the square root dependence on time, to go two times the distance by diffusion takes four (2 ² ) times as long.

The expressions should be J _inward = P _K × [K ⁺ ] _o and J _outward = P _K × [K ⁺ ] _i . A net flux is positive when it brings material into the cell. Therefore the correct expression for the net flux of K ⁺ is J _K = J _inward – J _outward = P _K ([K ⁺ ] _o – [K ⁺ ] _i ).

In solving all these problems, “very large volume of plasma” implies the “infinite bath” condition. In other words, the extracellular fluid volume is so large that a bit of solute or water entering or leaving the cell has essentially no effect on either the extracellular volume or the concentrations of solutes in the extracellular fluid.

• a.
  

  Answer: Water will move out of the cell. The final cell volume will be one half the initial volume. Explanation: The cell was initially equilibrated, which must mean that the intracellular concentration of permeant solute was C _P = 300 mM and the intracellular concentration of impermeant solute was C _NP = 10 mM. When considering the situation at equilibrium, we need not worry about the permeant solute because it can cross the cell membrane (and the intracellular and extracellular permeant solute concentrations will always become equal automatically). When the extracellular impermeant solute concentration is increased to 20 mM, an osmotic imbalance occurs, with the inside of the cell having a deficit of impermeant solute. Because the impermeant solute concentration in the cell cannot be changed by solute movement (impermeant solute cannot cross the cell membrane), the only way the cell can cope with the osmotic imbalance is to lose water and shrink its volume . The reduction in cell volume would increase the impermeant solute concentration inside until the intracellular and extracellular impermeant solute concentrations become equal. To calculate the volume change, we note that the initial equilibrium condition is C _{NP, Cell, Initial} = C _{NP, Bath, Initial} (impermeant solutes in the cell and bath are balanced) and the final equilibrium condition is C _{NP, Cell, Final} = C _{NP, Bath, Final} (impermeant solutes in the cell and bath are once again balanced). Concentration is just the number of moles per volume of solution; therefore the two equalities may be written as

  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='nNP, CellVCell, Initial=CNP, Bath, Initial=10 mM and’>?NP, Cell?Cell, Initial=?NP, Bath, Initial=10 mM andnNP, CellVCell, Initial=CNP, Bath, Initial=10 mM and
  n NP, Cell V Cell, Initial = C NP, Bath, Initial = 10 mM and
  
  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='nNP, Cell, FinalVCell, Final=CNP, Bath, Final=20 mM’>?NP, Cell, Final?Cell, Final=?NP, Bath, Final=20 mMnNP, Cell, FinalVCell, Final=CNP, Bath, Final=20 mM
  n NP, Cell, Final V Cell, Final = C NP, Bath, Final = 20 mM
  
  
  
  
  
  • 
    

    We know that the number of moles of impermeant solute inside the cell must remain constant because any impermeant solute originally inside the cell must remain there at all times. In other words, n _{NP, Cell, Initial} = n _{NP, Cell, Final} = n _NP,cell . Substituting this back into the two foregoing equations and rearranging gives

    
    
    <SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='VCell, Initial=nNP, Cell10mMandVCell, Final=nNP, Cell20mM’>?Cell, Initial=?NP, Cell10mMand?Cell, Final=?NP, Cell20mMVCell, Initial=nNP, Cell10mMandVCell, Final=nNP, Cell20mM
    V Cell, Initial = n NP, Cell 10 mM and V Cell, Final = n NP, Cell 20 mM
    
    
    
    
    
  • 
    

    Solving these two equations (e.g., by dividing the second equation by the first) gives

    
    
    <SPAN role=presentation tabIndex=0 id=MathJax-Element-4-Frame class=MathJax style="POSITION: relative" data-mathml='VCell, FinalVCell, Initial=12′>?Cell, Final?Cell, Initial=12VCell, FinalVCell, Initial=12
    V Cell, Final V Cell, Initial = 1 2
    
    
    
    
    
  • 
    

    In other words, the final cell volume will be half the initial volume.

  
  
  
  
• b.
  

  Answer: Water initially moves out of the cell (so the cell volume decreases), but then the volume gradually returns to the initial volume as equilibrium is reestablished (i.e., the final cell volume is the same as the initial cell volume). Explanation: When the extracellular permeant solute concentration is increased to 400 mM, initially an osmotic imbalance will occur. Because the membrane is more permeable to water than solutes, water will respond to the osmotic imbalance first by leaving the cell, so the cell should start to shrink. Even as water leaves the cell, the permeant solute, driven by its concentration gradient across the cell membrane, gradually permeates into the cell. Water follows osmotically, causing the cell volume to grow. Osmotic equilibrium can be reestablished only when the impermeant solutes are balanced across the cell membrane. Because the extracellular impermeant solute concentration is constant at 10 mM, equilibrium requires that the intracellular impermeant solute concentration return to 10 mM, which was its initial value. This can happen only if the cell swells back to its initial volume.

  
  
• c.
  

  Answer: Water initially enters the cell and causes an increase in cell volume. Then, in response to permeant solute movement, water leaves the cell and cell volume decreases. The final cell volume is one half the initial cell volume. Explanation: As soon as the extracellular solute concentrations change, an osmotic imbalance occurs; initially the total solute concentration inside is greater than the total solute concentration outside. Because water is the most permeant species, it crosses the cell membrane the fastest and is the first to respond to the osmotic imbalance by moving into the cell. This causes the cell to swell initially. Even as water enters the cell, the permeant solute, driven by its concentration gradient across the cell membrane, gradually permeates out of the cell. Water follows osmotically, causing the cell to shrink. As before, when a new osmotic equilibrium is established, the final volume of the cell is determined only by the presence of impermeant solutes. Because the outside impermeant solute concentration was changed from 10 to 20 mM, the situation concerning impermeant solutes is identical to that in part a of this problem—the final cell volume is one half the initial cell volume.

E _Na is defined by the Nernst equation:

• 
  

  After substituting in RT/F = 26.7 mV, and [Na ⁺ ] _i = 5 mM, we can solve for [Na ⁺ ] _o = 10.6 mM.

V _m can be calculated through the GHK equation:

• b.
  

  Use the Nernst equation:

  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-7-Frame class=MathJax style="POSITION: relative" data-mathml='ECl=RT(-1)Fln[Cl-]o[Cl-]i=26.7-1ln1055=-81.3mV’>?Cl=??(–1)?ln[Cl–]o[Cl–]i=26.7–1ln1055=-81.3mVECl=RT(-1)Fln[Cl-]o[Cl-]i=26.7-1ln1055=-81.3mV
  E Cl = R T ( – 1 ) F ln [ Cl – ] o [ Cl – ] i = 26 .7 – 1 ln 105 5 =- 81.3 mV
  
  
  
  
  
• c.
  

  If Cl ⁻ is allowed to move, it will move so as to drive V _m toward E _Cl . Because V _m = –4.8 mV, Cl ⁻ must move into the cell to drive V _m toward E _Cl = –81.3 mV. By definition, an inward flux is a positive flux ( Box 4.5 ).

  
  
• d.
  

  The Cl ⁻ flux carries negative charges into the cell, which has the same electrical effect as bringing positive charges out of the cell. A positive current is defined as the flow of positive charges out of the cell. Therefore the Cl ⁻ current is a positive (or outward ) current .

The Na ⁺ pump normally counteracts the osmotic consequences of the Donnan effect and thus helps maintain cell volume. When the Na ⁺ pump is inhibited, the osmotic imbalance arising from the Donnan effect would tend to cause the cell to be hyperosmotic relative to the extracellular fluid. This means that water would enter and cause the cell to swell.

The primary function of gated ion channels is to increase the permeability of the membrane to a specific ion. To increase the membrane K ⁺ permeability, the β-cell could alter the K ⁺ channel properties in the following ways:

• •
  

  The number of open channels could be increased or decreased. The overall permeability of the membrane will be directly related to the number of open channels.

  
  
• •
  

  The amount of time the channel is opened could be varied: the longer the channels are open, the higher the membrane permeability.

  
  
• •
  

  The permeability (or conductance) of a single open channel could be increased. Typically, cellular mechanisms exist to regulate the number of open channels and the amount of time the channel is open. The permeability of a single open channel is usually constant.

The most likely location of this amino acid would be in the segment of the P region that forms the selectivity filter of the Na ⁺ channel. The selectivity filter is a narrow region in the permeation pathway that determines which ionic species are allowed to pass. The selectivity filter works, in part, by making it energetically more favorable for a specific ion to enter the pore.

• 
  

  A single amino acid substitution in the selectivity filter could significantly alter its structure or chemical properties. For example, the amino acid substitution described in this question could make it energetically more favorable for the divalent Ca ²⁺ ion to enter the pore.

The concentration gradient for Cl ⁻ causes Cl ⁻ ions to enter the cell through the open channel. Thus a slight excess of negative charges builds up inside the cell and moves V _m in the negative direction. The developing V _m progressively impedes Cl ⁻ entry until V _m = E _Cl = –70.5 mV, at which point net Cl ⁻ flux becomes zero and V _m stops changing.

• b.
  

  The V _m follows a single-exponential time course ( Fig. E.1 ) with a time constant, τ _m ,

  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-8-Frame class=MathJax style="POSITION: relative" data-mathml='τm=Rm×Cm’>τm=?m×?mτm=Rm×Cm
  τ m = R m × C m
  
  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-9-Frame class=MathJax style="POSITION: relative" data-mathml='Rm=1γCl=110-11S=1011ohms’>?m=1γCl=110–11S=1011ohmsRm=1γCl=110-11S=1011ohms
  R m = 1 γ Cl = 1 10 – 11 S = 10 11 ohms
  
  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-10-Frame class=MathJax style="POSITION: relative" data-mathml='Cm=10-6F/cm2×10-7cm2=10-13F’>?m=10–6F/cm2×10–7cm2=10–13FCm=10-6F/cm2×10-7cm2=10-13F
  C m = 10 – 6 F / cm 2 × 10 – 7 cm 2 = 10 – 13 F
  
  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-11-Frame class=MathJax style="POSITION: relative" data-mathml='τm=1011ohms×10-13F=10-2sec’>τm=1011ohms×10–13F=10–2secτm=1011ohms×10-13F=10-2sec
  τ m = 10 11 ohms × 10 – 13 F = 10 – 2 sec
  
  
  

  
  

  
  

  

  
  

  
  

  When the Cl ⁻ channel opens (at time = 0), V _m goes from an initial value of 0 mV to a final value of –70.5 mV along an exponential time course. The time constant of the exponential (the membrane time constant, τ _m ) is 10 ⁻² seconds.

  
  

The change in V _m will decrease as a function of distance away from the site of current injection according to the following equation:

• 
  

  In this problem Δ V ₀ = –30 mV, and the length constant, λ, can be calculated:

  
  
  <SPAN role=presentation tabIndex=0 id=MathJax-Element-13-Frame class=MathJax style="POSITION: relative" data-mathml='λ=rmri=2.5×104ohm·cm1×105ohm·cm-1=2.5cm2=0.5cm’>λ=?m??‾‾‾√=2.5×104ohm·cm1×105ohm·cm–1‾‾‾‾‾‾‾‾‾‾‾‾√=2.5cm2‾‾‾‾‾‾‾√=0.5cmλ=rmri=2.5×104ohm·cm1×105ohm·cm-1=2.5cm2=0.5cm
  λ= r m r i = 2.5 × 10 4 ohm · cm 1 × 10 5 ohm · cm – 1 = 2.5 cm 2 = 0.5 cm
  
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