Question

Glucose is pumped into some animal cells by a symporter powered by the simultaneous entry of Na+. The entry of Na+ provides a free-energy input of 10.8kJ/mol under typical cellular conditions (extracellular [Na+] =143mM, internal [Na+] =14mM, and membrane potential = 50mV). How large a concentration of glucose can be generated by this free-energy input ( what is the ratio of glucose in / ratio of glucose out) ?

I know that the answer is that the concentration of glucose inside the cell is 66 times as great as the outside but I don't know how to get this number. Please help me

Answer 1

I don't understand what I'm solving for because everything is given in the problem. The temperature is supposed to be 298.15 K because that is "under typical cellular conditions" so this is basically given in the problem. I don't understand how to relate what is given to glucose and I don't know how to get the answer which is 66. I have tried everything and I still don't get 66.

Answer 2

Oh, I see. That really does baffle me then, the equation I found doesn't seem to lead to any new information.
I deleted the incorrect reply to save any better helpers the time of reading through here. I failed to find anything meaningful on researching similar problems too, so I may see if my teachers are at all familiar with a problem like this tomorrow if it is worth anything by then. For now I will sleep. Good luck on the problems!

Answer 3

Thanks a lot for your help! I still have a week to complete this so hopefully I can get some help by then. Thank you :)

Answer 4

teh movement of something going against some concentration gradient should tell you that you're dealing with \(\Delta\)G here.
It's a formular very similar to the gibbs RT\(\times\)ln(ratio concentration of Na\(^+\) going in over Na\(^\+\)being pumped out).

Answer 5

I know it's \(\Delta\)G = charge \(\times\)energy \(\times\) (membrane potential) I believe?

Answer 6

\[\Delta G= RTLn \frac{ [Na+] In}{ [Na+]out }+ ZF \Psi\]
This is the equation where
G=10.8kL/mol
R= 8.314 J*K-1*mol-1
T=298.15 K
[Na+]in =14mM
[Na+]out= 143mM
Z = ionic charge
F= Faradays constant ( 96485JV-1mol-1
And the last symbol is the charge = -0.05V

I have all the info but how do I find the ratio of glucose in/ratio of glucose out?
Help! :(

Answer 7

Do you know what a ratio \(is\)???

Answer 8

Well the question is asking how large a concentration of glucose can be generated by this free-energy input. I have the answer which is:
The concentration of glucose inside the cell is 66 times as great as the outside.

I don't get how to figure this out with all the info that was given in the problem.

Answer 9

http://books.google.com/books?id=K3JbjG1JiUMC&pg=PA466&lpg=PA466&dq=free-energy+symporter&source=bl&ots=apKbWPyj4C&sig=r5HmEy52WOc6zbp0yq1xA2p5nrc&hl=en&sa=X&ei=LE9OU6dAwbPIBMKlgYgI&ved=0CEAQ6AEwAg#v=onepage&q=free-energy%20symporter&f=false

I found this resource earlier which provides an equation to use. In essence, it uses the energy of both the sodium transport and glucose transport. (tried to link earlier but tex did not cooperate and I had to leave before I could fix it).

Answer 10

Thank you! That helps a bit even though I still can't figure it out. I have tried several ways and I still don't get 66 :(

Answer 11

It sounds to me like you're missing some data here.

Answer 12

I tried to do this problem using two equations and got close answers. Calling the ratio of glucose Q.
\( \rm \Delta G = RT \ln Q - 10.8 = 0\) Obtained \(Q \approx 78.0159 \)
\( \rm \Delta G = RT \ln Q + RT \ln \dfrac{[Na^+]_{in}}{[Na^+]_{out}} + zFV = 0\) \(Q \approx 71.5199 \)
In either case it is safe to say I have no idea what I'm doing anymore but if I'm close to anything, 66 is not that far off.

Answer 13

This looks right, at least is close to 66. Thank you sooooo much for your help! I really appreciate you taking the time to help me out. You were a big help! :)

Answer 14

Glad to help! Later I'll try and see if there's any way to resolve the error in a logical way, but otherwise that's the best I got. lol

Answer 15

So my teacher gave us this hints today:
\[\Delta G Na^+ negative\]
\[\Delta G glucose positive\]
\[T=37 degrees C\]

Answer 16

The first two make sense just because the movement of Na is greatly favorable, while the movement of Glucose is not.
But whoa. T = 37 degrees C. I wonder where that comes from. Maybe we did have to solve for T initially from -10.8 = RT ln [Na]in/[Na]out + zFV... I'll be right back and see if that makes a good difference.

Answer 17

This is pretty close:
-10.8 = 8.314*T/1000*ln(14/143)-1*96.485*0.05
-10.8+1*96.485*0.05 = 8.314*T * ln(14/143)
T = (-10.8 + 1*96.485*0.05)/8.314 /ln(14/143)*1000 = about 309 K

Answer 18

Using the new value of T in my 2nd equation written, I just about hit it on the head:
T = 309 exp((0.05*96.485-8.314*0.309*ln(14/143))/(8.314*0.309)) = 66.7
T = 310: exp((0.05*96.485-8.314*0.310*ln(14/143))/(8.314*0.310)) = 66.3

Answer 19

The accepted average body temperature is at 37 degrees C compared to 36.3 C given by solving. Although it comes a little strange, then, that you actually didn't even need a lot of the information given to solve this. (Replace all sodium input with -10.8, set \(\Delta G\) to 0 so that all the energy generated from sodium transport balances the glucose using energy through work)

Answer 20

Wow you're a genius!!! Can't believe you figured it out, thank you so much! :)
I used the 37C or 310.15 (0.31015) and yeah I finally got the 66 I was looking for lol
You've been such a great help. Thank you sooo sooo much for taking the time out of your life to help people like me. I wish I was as smart as you! You're gonna get far in life.
Thanks a lot :)

Answer 21

Lol, I'm glad to help! I really like Chemistry/Biochemistry, but I still got a ways to go with the subject to truly understanding systems like this. I am only saved by knowing a good amount of Math that is useful for Chemistry and patience to research things... :P

Answer 22

I hate anything chemistry related lol but you're doing great! Trust me, I'm in a college biochemistry class, I'm "supposed" to know this stuff, but you, wow I'm amazed by your knowledge and persistence and you're only in HS!
I can send you a pdf of my biochemistry book as a thank-you gift if you want it since you're interested in the subject :)

Answer 23

That'd be great, actually! I have some on biology but not biochemistry. :D

Answer 24

Great! Let me see how I can do this lol

Answer 25

Biochemistry was one of the courses I forgot about because there was so much information to retain. 9 weeks was too fast.