|Index to this page|
The movement of any molecule or ion down — or up — a concentration gradient involves a change in free energy, ΔG ("Delta G")
- down releases energy so ΔG is negative
- up consumes energy so ΔG is positive
|Link to a discussion of free energy changes.|
The amount of free energy released or consumed can be calculated from the equation
- a concentration of glucose inside the cell of 0.5 millimolar (mM) and a
- concentration of glucose outside the cell of 5 millimolar (mM)
- a body temperature of 37°C, so an absolute temperature of 37 + 273 = 310°K, and
- the plasma membrane is permeable to glucose.
ΔG = (2)(273+37) x ln (0.5/5)
= (2)(310) x ln (0.1)
= (620)(−2.3) = −1426 cal/mole
= −1.4 kcal/mole
(If you prefer to work with log10, multiply the log10 by 2.303, thus log10 (0.1) = −1; −1 x 2.303 = −2.3)
Because the process proceeds with the release of free energy, it can proceed spontaneously. In this case, facilitated diffusion would be required because glucose needs transport channels to allow it to pass through the lipid bilayer of the plasma membrane.
Filtration of the blood in the glomeruli of the kidneys produces a nephric filtrate with a concentration of glucose the same as that of the blood (~ 5 mM).
|Link to discussion of kidney function.|
All of this glucose is normally reclaimed by transport
- from the fluid within the proximal tubule
- across the plasma membrane at the apical surface of the epithelial cells lining the tubule and
- across the plasma membrane at the basolateral surface of the cell into the interstitial fluid (Link to discussion of the apical and basolateral surfaces of epithelia) and
- on back into the blood.
As the process continues and more and more glucose is removed from the fluid, the concentration gradient up which the glucose must be pumped — by active transport — increases.
What is the free energy needed to move glucose back from the tubular fluid to the blood when the concentration in the tubular fluid has dropped to 0.005 mM?
The problem is to pump glucose into the cell (where it is about 0.5 mM) and then across the plasma membrane at the basolateral surface of the cell into the interstitial fluid, where the glucose concentration is 5 mM (the same as in the blood). So the total gradient through which the glucose must be pumped is 0.005 mM -> 5 mM.
ΔG = (2)(310) x ln (5/.005)
= 620 ln (1000) = (620)(6.91) = + 4284 cal/mole
= + 4.3 kcal/mole
Where is the needed energy to come from?
The active transport of glucose is mediated by the Na+/glucose transporter. (Link to discussion of the Na+/glucose transporter and other symport pumps.)
This is a symporter because both the sodium ion and the glucose molecule are passing through the membrane in the same direction:
- sodium DOWN its gradient of about
- 140 mM outside to
- 10 mM inside
- while glucose is going UP its gradient (0.005 mM -> 5 mM).
A mole of sodium ions (Na+) moving down this concentration gradient releases −1.6 kcal of free energy.
ΔG = (2)(310) ln (10/140)
= (620) ln (0.07) = (620)(− 2.64) = −1637 cal/mole
= −1.6 kcal/mole
Is this enough to move a mole of glucose?
No, but there is another force we must consider.
- Sodium ions carry a single positive charge and
- the interior of the cell is negatively charged [link to discussion of how this comes about].
- So the attraction between opposite charges provides a second force for bringing Na+ into the cell.
This, too, can be quantified.
- z = the charge on the ion (+1 in this case)
- F = 23,062 = the calories released as one mole of charge moves down a voltage gradient of 1 volt (1000 mV)
- Vm = the membrane potential, about − 70 mV in mammalian cells.
ΔG = (+1)(23,062)(- 0.07)
= −1,614 cal/mole (−1.6 kcal/mole) Adding this to the free energy available from the concentration difference ( −1637 cal/mole) gives us a total yield of 3.3 kcal/mole from this combined or electrochemical gradient.
Still not enough to move a mole of glucose, so at least two sodium ions are needed to bring one molecule of glucose into the cell.
So the driving force for the active transport of glucose (and other small organic molecules, e.g., amino acids) is the force provided by the movement of sodium ions following their electrochemical gradient.
But in the last analysis, the driving force is provided by the energy of ATP synthesized in cellular respiration. This is because the sodium gradient across the plasma membrane is created by the active transport of Na+ OUT of the cell by the Na+/K+ ATPase; the single most profligate user of energy in the our body.
|Link to discussion of the Na+/K+ ATPase|
The active transport of amino acids is also mediated by Na+ cotransport.
How many sodium ions are needed to provide the free energy to transport a molecule of glutamic acid from a concentration of 0.1 mM outside the cell to 20 mM inside the cell?
Again, assume a temperature of 37°C (310°K).
At pH of ~7, glutamic acid molecules carry a net charge of minus 1 [View].
So, once again, we have a problem of determining the movement of a molecule against an electrochemical gradient; that is, against both
- a concentration gradient (20/0.1 = 200) and a
- electrostatic gradient (moving a negative charge against a voltage of − 70 mV).
ΔG = (R)(T) x ln(20/0.1) + (z)(F)(Vm)
= [(2)(310) x ln(200)] + [(−1)(23,062)(− 0.070)
= (620) x (5.3) + 1614
= 3286 + 1614
= 4900 or 4.9 kcal/mole
Because sodium ions release only 3.3 kcal/mole (above), at least 2 Na+ are needed to cotransport one molecule of glutamic acid.
3 March 2015