Some help with the Nernst equation?

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The discussion centers around the application of the Nernst equation to determine the maximum concentration gradient for glucose transport across a membrane, specifically when using ATP hydrolysis with a free energy change of -12 kcal/mole. Initially, a calculation suggested a concentration ratio of 1.3, which was deemed incorrect as it implied a higher concentration outside the cell than inside, contradicting the active transport concept. After re-evaluation, the correct ratio was found to be approximately 0.0026, indicating that for every glucose molecule outside, there could be up to 385 molecules inside the cell. The conversation highlights the importance of careful calculation and understanding of thermodynamic principles, particularly in distinguishing between equilibrium states and active transport dynamics. Ultimately, the participants emphasize the need to critically assess computational results to avoid systematic errors.
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Nernst equation:

\DeltaG = -2.3 RT log10 \frac{C_o}{C_i} + zFV

R = 1.98x10-3 kcal/°K mole
T = absolute temperature in °K (37°C= 310 °K)
Co= concentration dissolved compound outside of cell
Ci= concentration dissolved compound inside of cell
z = the charge of the dissolved compound to be transported
F = Faraday constant 23 kcal/V mol
V = membrane potential in volt (V)

The question is:

What is the maximum concentration gradient that can be achieved when an uncharged molecule is pumped across the membrane with a \DeltaG of -12 kcal/mole?

I get the answer of \frac{C_o}{C_i} = 1.3 which just doesn't sound right to me! :(
 
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btw, this is an example of active transport. Think for instance that the transport utilizes 12 kcal/mole of energy from ATP to ADP hydrolysis to transport a single molecule of glucose across the membrane.

The membrane potential is -60 mV, btw. But that shouldn't matter, since the molecule is uncharged.

The Co/Ci ratio should be smaller than 1, since it is transport INTO the cell..
 
Your answer sounds feasible. Since the free energy change is negative this means that the reaction should favor the product. Your could mean that one side of the membrane will have 1.3 times greater concentration.

When the product has lower free energy than the reactant it means that the system itself will favor the products. Since -12 is a rather small value this means that at equilibrium (when free energy equals zero) will only slightly favor the products. Nevertheless, I suggest that you recheck your work; since this problem does not involve a lot of steps it should not be too difficult.

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But a Co/Ci of 1.3 means that for every 7 molecules of glucose outside of the cell, 3 will be transported inside the cell (leaving 4 outside at the end).

That means that sugar can never be taken up completely and that just sounds wrong to me..

What if there are a lot of pumps actively pumping the glucose into the cell, the gradient will always be 4 outside : 3 inside?


I can do the calculation again, but if I am making a systematic mistake in using the formula, it won't be of much use..

btw, you say yourself that the system favors the product, so isn't the ratio supposed to be <1 by default?
 
AH! I WAS making a systematic mistake in my calculation.. a very stupid mistake actually

Now I get an answer that makes much more sense:

-12 = -2.3 * 1.98x10^-3 * 310 * ln (Co/Ci)

where Co/Ci = 0.002595801

which means that for every molecule of glucose outside the cell,
there can be a maximum of 385 molecules inside the cell.

yeahh! lesson learned: always mistrust the answer that comes out of your calculator (good thing I did)
 
An equilibrium of 1.3 would mean that the concentration of the product is higher than the reactant at equilibrium. For example, for every ten molecules of glucose on one side of the cell there will be 13 on the other. Nevertheless it seems that you have found the source of your problem. The Nernst equation can easily result in calculation errors and thus it is important to consider the answer and recheck your work.

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Originally posted by GeneralChemTutor
An equilibrium of 1.3 would mean that the concentration of the product is higher than the reactant at equilibrium. For example, for every ten molecules of glucose on one side of the cell there will be 13 on the other. Nevertheless it seems that you have found the source of your problem. The Nernst equation can easily result in calculation errors and thus it is important to consider the answer and recheck your work.
Co/Ci = 1.3 = 13 outside / 10 inside.

That is not active transport into a cell, it should at least go past equilibrium.
 
Originally posted by Monique
AH! I WAS making a systematic mistake in my calculation.. a very stupid mistake actually

Now I get an answer that makes much more sense:

-12 = -2.3 * 1.98x10^-3 * 310 * ln (Co/Ci)

&Delta;Go = -2.3RTlog10K , or
&Delta;Go = -RTlnK , but NOT
&Delta;Go = -2.3RTlnK .

Recall that lnx = 2.3log10x.
where Co/Ci = 0.002595801

Second nitpicking point: when writing an "equilibrium constant," the term(s) in the denominator represent reactant(s) (or initial conditions), and the numerator the product(s) (or final conditions). That is, for sugar initially "outside" the cell wall, and "inside" at the end of the process, K = ci/co.
which means that for every molecule of glucose outside the cell,
there can be a maximum of 385 molecules inside the cell.

yeahh! lesson learned: always mistrust the answer that comes out of your calculator (good thing I did)

12/0.6 = lnK = 20; the ratio of the final (inside) concentration to the outside is 108 to 109.

Couple other points: 12kcal is a huge standard free energy change for processes which involve no more than a change in state of substances (no change in actual chemical composition --- phase changes, dissolution, and the like); "equilibrium" is used to describe the state of a system for which any change results in an increase in the free energy --- please don't confuse it with relative magnitudes of concentrations inside and outside cell walls in this system.
 
Originally posted by Bystander
Recall that lnx = 2.3log10x.
Right, thanks for correcting me :)

Second nitpicking point: when writing an "equilibrium constant," the term(s) in the denominator represent reactant(s) (or initial conditions), and the numerator the product(s) (or final conditions). That is, for sugar initially "outside" the cell wall, and "inside" at the end of the process, K = ci/co.
Well, it is not an equilibrium constant, all it says in my book is: "The free-energy change per mole of solute moved across the plasma membrane is equal to -RT ln Co/Ci."

12/0.6 = lnK = 20; the ratio of the final (inside) concentration to the outside is 108 to 109.
Ok, how did you go from ln Co/Ci = 20 to 108 : 109??

"equilibrium" is used to describe the state of a system for which any change results in an increase in the free energy --- please don't confuse it with relative magnitudes of concentrations inside and outside cell walls in this system.
You are saying concentration gradients have nothing to do with equilibrium states? But it does, normally it would be 1:1 across the membrane, if there is a potential it will reach equilibrium at a different point, if there is an active transport, there will be another equilibrium point.

And why is there a maximum concentration gradient? Hydrolysis of the ATP is causing glucose to be pumped over the membrane, hydrolysis could just go on forever until all the glucose is gone right?

I can understand the equation if the supply of ATP is the rate limiting factor, where the transport then depends on the equilibrium constant of the hydrolysis.

It doesn't make sense.
 
  • #10
Again, the exact question is:

What is the maximal concentration gradient that can be achieved by ATP mediated active transport into the cell, of an uncharged molecule such as glucose, assumed that the hydrolysis of 1 ATP results in the transport of 1 glucose.

Assume that the rate of hydrolysis of ATP to ADP occurs with a dG of -12 kcal/mole.

Code: 
      glucose
         ^
         |
        *|*
========***=========
        *|*
      ___|___
   ATP       ADP      dG of ATP to ADP = -12kcal/mol


I think the question is wrong, since the process doesn't depend on the glucose concentration, and thus a maximal concentration gradient can't be calculated for that molecule.
 
  • #11
Assumptions:

Without the energy released from the conversion of ATP to ADP the relative concentration of glucose across the membrane should be isotonic.

12kcal/mol input of energy is needed to transport one molecule of glucose across the cell. Divide this by avogado's number and you will get kcal/molecule (for the purpose of conceptualization).

Assuming that ATP is in excess relative to glucose we can say that K (the eq. constant) = [ADP][P][ATP]/[ATP] which equals [ADP][P]. Assuming no La Chatelier disturbance we can say that [ADP]= the square root of K. The concentration of ADP reflects the excess concentration of glucose inside the cell at equilibrium. There's your answer.

You should simplify the Nernst equation before you perform any calculations. Be sure to solve for K correctly: The free energy variable should be 0, the standard free energy should be -12 kcal/mol, and I believe you will need to convert kcal/mol to kilojoules/mol.

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  • #12
Originally posted by Monique
Well, it is not an equilibrium constant, all it says in my book is: "The free-energy change per mole of solute moved across the plasma membrane is equal to -RT ln Co/Ci."

Anytime &Delta;Go = - RTln (argument) shows up, "argument" is an equilibrium constant --- it might be called a partition coefficient, ionization constant, solubility product, concentration ratio, or some other term generated for a specific class of system, but the bottom line is that they are all equilibrium constants. If your book is actually saying "CO/Ci, there are some proofreading/typesetting problems with the author/publisher --- it's ALWAYS final state over initial state.
Ok, how did you go from ln Co/Ci = 20 to 108 : 109??

lnx = m; exp(lnx) = exp(m), or elnx = em; more specifically, exp(ln(c/c)) = Ci/Co = e20 = 1020/2.3. Fair 'nuff?
You are saying concentration gradients have nothing to do with equilibrium states? But it does, normally it would be 1:1 across the membrane, if there is a potential it will reach equilibrium at a different point, if there is an active transport, there will be another equilibrium point.

Without actually looking at the context in which the problem has been posed (your next post), it's a little tough to say what the author(s) is(are) trying to demonstrate with this exercise. I'm a physical chemist, and biochem baffles me --- equilibrium of this system from my standpoint is that the cell is dead, and everything is an uninteresting, homogeneous, foul-smelling mush. Okay, life processes occur at a slow enough rate that initial and final states can be defined for some steps, and the system regarded as being in some approximation of equilibrium in each state --- this might be the object of the problem, to demonstrate the magnitude and effect of chemical potential on this transport process --- I'm guessing on this. There is ONE equilibrium condition if we restrict things to just the thermodynamic approximation.
And why is there a maximum concentration gradient? Hydrolysis of the ATP is causing glucose to be pumped over the membrane, hydrolysis could just go on forever until all the glucose is gone right?

If it's just a thermodynamic model of the transport process, the ATP-ADP hydrolysis being nothing more than a "black box" to establish or maintain a -12kcal difference in chemical potential for sugar on each side of the membrane, you are correct, and all that can be calculated is a ratio of the concentrations on either side of the membrane (108). Hydrolysis can go on forever and if the chemical potential difference remains -12kcal/mol, the concentration ratio remains 108 --- ln(0) = - infinity, an unreachable condition for real systems.
I can understand the equation if the supply of ATP is the rate limiting factor, where the transport then depends on the equilibrium constant of the hydrolysis.

It doesn't make sense.

Gotta agree with you here --- it looks an awful lot as if the book you're working from is unnecessarily sloppy with its distinctions between thermodynamics and kinetics.

Last thing that occurs to me in this stage of the discussion is that the author may be misusing the word "gradient" --- and, therefore, is really interested in the "equilibrium" ratio of concentrations on the two sides of the cell membrane. You're the one with the book in her hand, so you're the one who's going to have to make the assessment.
 
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  • #13
Nothing is wrong with the problem. Most advanced biology classes are not so forthright in presenting the problem; they expect you to make some assumptions of your own. Only by making some assumptions does the problem make sense; for example, having an excess of ATP. In solving this problem in any angle, one needs to consider the concentration of ADP at equilibrium which will be the excess concentration of glucose inside the cell. The rest is explained in my last post.

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  • #14
Originally posted by Bystander
12/0.6 = lnK = 20; the ratio of the final (inside) concentration to the outside is 108 to 109.
But HOW can that be right.

If you assume that glucose is isotonic, and it is pumped INTO the cell, the maximal concentration INSIDE will in the end be LOWER than outside?

and lnK = 20, K = 0.000000003 (not 0.1)

K = 0.000000003, would mean 1 inside the cell, 333333333 outside..
now how can that be right? seriously, I wish I knew the correct answer..
 
  • #15
Originally posted by Monique
Nernst equation:

\DeltaG = -2.3 RT log10 \frac{C_o}{C_i} + zFV
I give up, I can't even find the same formula anywhere on the internet..
 
  • #16
Originally posted by Monique
But HOW can that be right.

If you assume that glucose is isotonic, and it is pumped INTO the cell, the maximal concentration INSIDE will in the end be LOWER than outside?

and lnK = 20, K = 0.000000003 (not 0.1)

K = 0.000000003, would mean 1 inside the cell, 333333333 outside..
now how can that be right? seriously, I wish I knew the correct answer..

Ratio of "this" to "that" equals "this" OVER "that" --- not "that" over "this." Inside conc. is on the order of one billion times greater than that outside. Ditch the "isotonic" assumption --- "isotonic" means same inside as out, which ain't the case.

K = 3x108 --- which is one half order of magnitude greater than 108 and one half less than 109 --- specifying much more than an order of magnitude in values for equilibrium constants is misleading in the sense that it implies far less uncertainty in the free energy measurements than is ordinarily possible (there are exceptions, but we'll leave the e-chem out of the discussion for the moment).

Can't find the Nernst eqtn as written in the problem? That's 'cause there ain't no such critter as it is written.

What you're after is &Delta;Go = -RTlnK = nF(script f)E(script e). The net crawls with various presentations, and they are all equivalent to what I just wrote --- that's given a few odd sub- and superscripts, couple definitions of standard states, and one or two sign conventions to distinguish European e-chem from "damned yankee" e-chem. If this is really bugging you, you might go to the attic/garage/closet, and dig out and dust off that p-chem book you had so many nightmares with as a junior --- no hour exams or frustrating labs with junk someone salvaged from a dumpster to distract you --- just give the relevant chapters a read once or twice a week, and focus on following the developments of the concepts ONE step at a time rather than ALL at once.

You are working from a text that is preparing people for careers as clinical technicians? And this problem is meant to illustrate the physical-chemical explanation for relating blood/serum sugar assays to other cellular or metabolic functions or disorders? If my guess is halfway correct, it explains the sloppy notation (med types are horrible that way), and the point of the exercise.
 
  • #17
OK, we are going somewhere

1. yes, the formula of outside to inside was not explained properly and is misprinted in the exercise.

2. I misunderstood your answer of 10^8 to 10^9 as a ratio, stupid..

I also dusted off an old biochem book which actually explains the terms and how the formula is derived.

Right now I am trying to figure what the difference is between dGo = -[/color]RTlnK and dG=RTlnK..
 
  • #18
Originally posted by Bystander
You are working from a text that is preparing people for careers as clinical technicians? And this problem is meant to illustrate the physical-chemical explanation for relating blood/serum sugar assays to other cellular or metabolic functions or disorders? If my guess is halfway correct, it explains the sloppy notation (med types are horrible that way), and the point of the exercise.
Well, no, I am not sure why I need to know this. The only thing I know for sure is that this is going to turn up as an exam question I'm fine with calculating electrical potentials and ion flow (which are actually way more complicated) but somehow this one didn't make sense.

But it all makes sense now thanks to you guys!


No, not a clinical technician *shudders* RESEARCH in the biomedical field is my thing. I guess this equation might be usefull to investigate the type of transport across a membrane is taking place.. simple diffusion/ ion channels/ facilitated/ active.. how much ATP would be required for transport to take place..

For instance (if someone likes to test themselves) another example:

ATP hydrolysis is directly coupled to a K+/Na+ pump, about 2 K+ ions are pumped INTO the cell and 3 Na+ ions are pumped OUT of the cell for every ATP hydrolyzed. You can ask yourself the question: is that estimate reasonable from a thermodynamic point of view.

So you make some measurements..
Na+ in=10mM out=140mM
K+ out=5mM in=100mM
V=0.07V

dG for Na+ = 13.55 kJ/mol times 3 = 40.65 kJ
dG for K+ = 0.97 kJ/mol times 2 = 1.95 kJ

dGt=42.59kJ

At first glance this seems to be too much energy for ATP hydrolysis to supply, but ATP concentrations in the cell are sufficiently high that the free energy change is much higher than the standard one.. thus it IS reasonable.



I also figured out how to calculate fractions from an equilibrium constant: fraction product = K/(1+K). So I am all set to go! Thanks :)
 
  • #19
&Delta;G(or F)o, the standard (Gibbs) free energy (change) for a process, sometimes the standard free energy of formation; the free energy change associated with whatever a balanced reaction requires in the way of reactants to form whatever products, the reactants and products all being in their standard states, often the standard states are "fictitious," physically unrealizable ---STANDARD is the key word --- you got to check glossaries for stars, filled dots, open circles, zeros, and whatnot among all the non-standard sub- and superscripts people use.

&Delta;G (no superscript), the free energy change, calculated as the sum of free energies of products in the actual final state minus the sum of free energies of reactants in the actual initial state.

G (or delta G) = Go + RTln(activity or fugacity)/(activity or fugacity of the standard state, which is one (1) by the definition of the standard state).

A set of r reactants and their stoichiometric coefficients, r1R1 ... rrRr, and products, P, (fill in all the italics and subscripts) at equilibrium constitute a system for which no further change is possible, and therefore, for which &Delta;G (no superscript) = 0. Then,

&Delta;G = 0 = GP-GR = (&Delta;)GoP1 + RTln aP1 (or f for fugacity)+ ... - reactants, where "a" is a product of concentration and an activity coefficient (usually "c," the molarity of aqueous solutions, and sometimes pressures, or 1 for the reactants and products that actually appear in their standard states).

Rearranging, 0 = sum of Gos + RT times the sum of lna , the ln terms rearranging to K =products/reactants (plus appropriate exponents for stoichiometry).

Yes, I'm aware you're a biochemist --- I'm not above working my way through some of the "applied" chemistry training material at times just to keep myself aware of what's involved in training people who've never been inside a lab --- thought you might be doing the same for whatever purpose you might have.
 
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  • #20
Yes, Go is a standard state at which the concentrations of the products and reactants are both 1 M.

dGo for ATP hydrolysis might be a negative number, but cellular concentrations of ATP are very high, thus dG will be even smaller.

Also, I derived my own formula, which I find easier to work with (especially when calculating gradients):

e^{(\frac{\Delta G - (zF\Psi)}{RT})} = K
 
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  • #21
OK, one more thing:

\Delta G = \Delta G^o + RT ln K is the formula..

The question is asked: what is the maximal gradient that can be reached. The answer is when the reaction has reached equilibrium and dG = 0, thus:

\Delta G^o = -RT ln K


I can understand that.


But there is also the question: what is the energy requirement of an reaction. In which case the formula is:

\Delta G = RT ln K


Let me think about that..
 
  • #22
Originally posted by Monique
Yes, Go is a standard state at which the concentrations of the products and reactants are both 1 M.

dGo for ATP hydrolysis might be a negative number, but cellular concentrations of ATP are very high, thus dG will be even smaller.

Also, I derived my own formula, which I find easier to work with (especially when calculating gradients):

e^{(\frac{\Delta G - (zF\Psi)}{RT})} = K


http://www.corrosion-doctors.org/References/images/D35.gif

Q = K.
 
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  • #23
K is at equilibrium, Q is at any point
 
  • #24
So propably

e ^-(\frac{\Delta G - (zF\Psi)}{RT}) = K

e ^(\frac{\Delta G - (zF\Psi)}{RT}) = Q
 
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