REVIEW AND PROBLEM SET

Solute Intracellular conc. (mM)

Extracellular conc. (mM)

A+ 7 104 B+ 110 8 C++ 1 0.01 D- 5 10 E- 10 100 F- 2 2

G (uncharged) 4 4 H (uncharged) 3 1

Review Question 1

A. If the membrane potential of a hypothetical cell is –60 mV (cell interior negative):a) Given the extracellular concentration listed on the table

above, what would the predicted intracellular concentration of each of the solutes A-H have to be for passive diffusion across the membrane.

b) Given the intracellular concentrations calculated in part a), what can we conclude about the transport mode of each of the solutes that are not passively distributed.

B. Calculate the Nernst equilibrium potential for each solute.

1. Here we are told that the cell membrane potential is -60mVWe are given a list of solutes (A-H) - some are charged with different

valences and some are uncharged.

The Nernst equation is used to calculate the answers to A and B.

1. To begin, we are asked to use the extracellular concentrations in column 3 to calculate the intracellular concentration of each solute at at equilibrium. Here we know Vm (-60mV) we know the extracellular concentration from column 3, our unknown is the intracellular concentration.

2. Now we use the information in the table for the ACTUAL intracellular concentration and compare the values that we calculate for the PREDICTED concentrations. This gives us information regarding whether the solute is indeed passively distributed across the membrane, actively pumped in or actively pumped out.

3. Finally, we are going to use the information given in the table for the ACTUAL intracellular and extracellular concentrations and calculate the Nernst equilibrium potential (Eion) for each of the solutes.

2

1

log3.2

S

S

CC

zFRTE

i

o

CC

zFRTE log3.2

Recall the above equation from Lecture 2. Now let’s refer the concentration on side 1 to the outside of a cell (extracellular conc.), Co and the concentration on side 2 to the inside of a cell (intracellular conc.), Ci. We can re-write the above equation as:

The Nernst Equation (Nernst Equilibrium Potential)

i

oion C

Cz

E log60

Now lets simplify the equation: 2.3RT/F at 37°C reduces to ‘60’ and lets keep the valence z in the equation. We can now use this equation to calculate the answers to parts A and C.

Also, note log (10/100) = -1 and log (100/10) = +1

i

o

CC

zE log60

iC104log

16060

iC

104log1 Ci = 1040 mM

The predicted concentration for A+ is 1040 mM but the actual (or measured) value is 7 mM. The value of 1040 mM is far greater than the actual value so this solute is not passively distributed, therefore it must actively pumped out of the cell.

A+

7104log

160

AE

Therefore, the Nernst equilibrium potential for A+ is +70.3 mV

i

o

CC

zE log60

iC8log

16060

iC8log1 Ci = 80 mM

The predicted concentration for B+ is 80 mM but the actual (or measured) value is 110 mM. The value of 80 mM is less than the actual value so this solute is not passively distributed, therefore it must actively pumped in to the cell.

B+

1108log

160

BE

Therefore, the Nernst equilibrium potential for B+ is -68.3 mV

i

o

CC

zE log60

iC01.0log

26060

iC01.0log2 Ci = 1 mM

The predicted concentration for C++ is 1 mM and the actual (or measured) value is 1 mM. Thus this solute is passively distributed.

C++

101.0log

260

2

CE

Therefore, the Nernst equilibrium potential for C++ is -60 mV

i

o

CC

zE log60

iC10log

16060

iC10log1 Ci = 1 mM

D-

510log

160

DE

Therefore, the Nernst equilibrium potential for D- is -18.1 mV

The predicted concentration for D- is 1 mM but the actual (or measured) value is 5 mM. The value of 1 m M is less than the actual value so this solute is not passively distributed, therefore it must actively pumped in to the cell.

i

o

CC

zE log60

iC100log

16060

iC

100log1 Ci = 10 mM

The predicted concentration for E- is 10 mM and the actual (or measured) value is 10 mM. Thus this solute is passively distributed.

E-

10100log

160

EE

Therefore, the Nernst equilibrium potential for E- is -60 mV

i

o

CC

zE log60

iC2log

16060

iC2log1 Ci = 0.2 mM

F-

22log

160

FE

Therefore, the Nernst equilibrium potential for F- is 0 mV

The predicted concentration for F- is 0.2 mM but the actual (or measured) value is 2 mM. The value of 0.2 mM is less than the actual value so this solute is not passively distributed, therefore it must actively pumped in to the cell.

G is an uncharged solute, therefore we only need to consider the chemical gradient NOT an electrical gradient. If the solute is passively distributed, at equilibrium, with an extracellular concentration of 4 mM, we would have an intracellular concentration of 4 mM. This is also the measured value for G, therefore this solute is passively distributed. Nernst does not apply.

H is an uncharged solute, therefore we only need to consider the chemical gradient NOT an electrical gradient. If the solute is passively distributed, at equilibrium, with an extracellular concentration of 1 mM, we would have an intracellular concentration of 1 mM. This is less than the measured value for H (3 mM), therefore this solute is actively transported in to the cell. Nernst does not apply

Solute Predicted

intracellular concentration

Transport mode

Calculated Nernst

equilibrium potential

A+ 1040 Active (out) +70.3 B+ 80 Active (in) -68.3 C++ 1 passive -60 D- 1 Active (in) -18.1 E- 10 passive -60 F- 0.2 Active (in) 0

G (uncharged) 4 Passive -

H (uncharged) 1 Active (in) -

Consider a closed system bound by rigid walls and a rigid membrane separated the two compartments. Assume the membrane is freely permeable to water and impermeable to sucrose.

A) If both compartments contain pure water and a pressure is applied to the piston establishing a hydrostatic pressure difference across the membrane, which direction will water flow in? What will the initial rate of water flow depend on?B) If no force is applied to the piston and 100 mM sucrose is placed in compartment A, which direction will the meniscus in compartment B move? What concentration of NaCl (also impermeant) would have to be added to compartment B to prevent volume displacement? What hydrostatic pressure must be applied to the solution in compartment A to prevent this volume flow?

Piston

BA

Review Question 2

Piston H2OBA

2A

The water will flow from side A to BThe initial rate of flow of water will depend on:Magnitude of the pressure differenceArea of the membrane (units of pressure – force/area)Hydraulic conductivity (or water permeability) of the membrane

If both compartments contain pure water and a pressure is applied to the piston establishing a hydrostatic pressure difference across the membrane, which direction will water flow in? What will the initial rate of water flow depend on?

100 mM sucrose

Piston H2O

BAThe osmotic pressure of side A > B. Water will flow from low to high osmotic pressure (but high to low water concentration).50 mM NaCl will be required to negate the osmotic effect of sucrose in side A (actually slightly more than 50 mM because of the incomplete dissociation of NaCl).Note: (each sucrose molecule = 1 osmolyte, NaCl has 2 osmolytes per mole….MgCl2 has 3 etc. so if the question asked for the concentration of MgCl2 instead of NaCl, your answer would be approx. 33 mM)

2BIf no force is applied to the piston and 100 mM sucrose is placed in compartment A, which direction will the meniscus in compartment B move? What concentration of NaCl (also impermeant) would have to be added to compartment B to prevent volume displacement?

What hydrostatic pressure must be applied to the solution in compartment A to prevent this volume flow?

Considering 100 mM sucrose, thus a concentration gradient of 100 mM, at equilibrium, = PH. The hydrostatic pressure that is required to negate the osmotic pressure ( ) required is determined from the van’t Hoff equation: = RTCS = (25.4)CS at 37°C. (Note: With appropriate conversions one can estimate osmotic pressure using the relation 1 osmole/liter = 19,300mm/Hg = 25.4 atm.) Therefore with an osmolarity difference of 100 mosmoles/liter = 0.1 osmoles/liter, the pressure required is 2.5 atm.

2B

Consider two compartments of equal volume separated by a membrane that is impermeant to anions and water

A) If in addition the membrane is not permeant to Na+, what is the orientation and the magnitude of the potential difference across the membrane at 37C? What is the composition of compartment B when the system reaches equilibrium?

B) If the properties of the membrane change and now the membrane is only permeant to Na+, what is the orientation and magnitude of the potential difference?

C) If both Na+ and K+ are permeable, but PNa>PK what will be the orientation of the potential difference initially? What will be the orientation of the potential difference and the composition of compartments A and B when electrochemical equilibrium is reached?

A B100 mM NaCl

10 mM KCl

100 mM KCl

10 mM NaCl

Review Question 3

3A

10 K 100 K-----

+++++

This is an example of an equilibrium potential for K+. At equilibrium there is no change in concentration of the BULK solution. (P.D. 60 mV negative on side B)

A B

If in addition the membrane is not permeant to Na+, what is the orientation and the magnitude of the potential difference across the membrane at 37C?

What is the composition of compartment B when the system reaches equilibrium?

3B

100 Na 10 Na-----

+++++

This is an example of an equilibrium potential for Na+. (P.D. 60 mV negative on side A)

A B

If the properties of the membrane change and now the membrane is only permeant to Na+, what is the orientation and magnitude of the potential difference?

10 K 100 K

100 Na 10 Na

*** Impermeable to anions and waterBUT remember …...because PNa>PK ………the potential that develops

will be governed by the Na+ permeabilityThis is an example of a diffusion potential (side B positive) that will collapse over time as the Na+ and K+ gradients collapse. We cannot calculate the precise value for the potential because the question gives us insufficient information….i.e., we do not know the value of the permeability. Electrochemical equilibrium will be reached when the concentration gradients, and consequently the P.D. (potential difference) have collapsed to zero. 55mM NaCl and 55mM KCl in compartments A and B (at zero electrochemical driving force).

3C

A B

If both Na+ and K+ are permeable, but PNa>PK what will be the orientation of the potential difference initially? What will be the orientation of the potential difference and the composition of compartments A and B when electrochemical equilibrium is reached?

AND LASTLY……………..

What happens to the movement of an ion if the Vm is different from Eion

100 X+ 10 X+-

-

+++

A B-

Nernst…..

mVAB

zEX 60

10010log

160

][][log60

100 X+ 10 X+-

-

+++

A/in B/out

-

Measured value+30 mV

EX

+-60

Out/B

In/A

+30

Even though the Nernst potential dictates the equilibrium to be at -60 mV, if the measured value is +30 mV, there is movement of X from A to B.

100 X- 10 X-

mVio

zEX 60

10010log

160

][][log60

Measured value+100 mV

EX-+60

Out

In +100

+ve outward current follows the movement of +ve charge – by convention. Here X- is moving IN to the cell.

100 X+ 10 X+

mVio

zEX 60

10010log

160

][][log60

Measured value-90 mV

EX--60

Out

In

-90

Here X+ is moving IN to the cell. Electrical gradient>chemical gradient.