8'5 potassium ion transport: K in 35K out

1
8.5 potassium ion transport Kin
35Kout a) what is DG at 37 oC for
transporting K into cells with no DV? ie. ignore
charge effects
first, write the equation for the transport
Kout Kin Keq will then product /
reactant Kin/Kout 35 DG RTln Keq DG
1.987310ln 35 DG 2190 cal/mol
b) for a cell with a resting potential of -60 mV,
what is DG?
DG RT ln Keq zFVp DG 2190
123062(-.060) DG 806 cal/mol ie. membrane
potential aids K entry into the cell
2
8.5 continued c) what is the maximum amount of K
ions that can be pumped in by the hydrolysis of 1
ATP
need to find the DG at the given concentrations
of ATP, P, and ADP DG' DGo' RTln
ADPP/ATP (product over reactant) DG
-7300 1.987310ln 0.0010.01/0.005 DG -7300
- 3828 -11,128 cal/mol for ATP hydrolysis at
these 11,128/806 13.7, so a maxium of 13 K
ions could be transported
3
8.7 sodium ion transport with ATP20 ADP2,
P1 mM, Vp-75mV at 25 oC and 150 mM external
sodium, pumping 3 molecules per ATP?
basically the amount of energy to move sodium
from inside to outside will be determined by the
DG of ATP under the given conditions DG' DGo'
RTln ADPP/ATP DG' -7300 1.987298 ln
.002.001/.020 (note decimals to balance) DG'
-7300- 5452 DG' -12,752 cal/mol for 3 Na
ions DG' -4250 cal/mol per ion to calculate
the minimum intracellular sodium ion
concentration, DG RTlnNain/Naout
zFVp -4250 1.987298ln (x/0.15)
123062(-.075) -4250 592ln(x/0.15) -
1730 -2520 592 ln (x/0.15) -4.26
ln(x/0.15) 0.014x/0.15 x.002 M 2mM minimum
Nain for 1 molecule of ATP pumping 3 Na
4
8.7 cont. b) If the molecule being pumped out
were uncharged, would the answer for part a be
higher or lower if all other conditions remain
constant? The inside of the cell is negatively
charged, so pumping positively charged molecules
out will take extra energy (ie. making the
inside more negative). Therefore the answer
would be lower. calculating the values from the
equation above DG RTlnNain/Naout
zFVp -4250 1.987298ln (x/0.15)
023062(-.075) -4250 592ln(x/0.15) - 0 -4250
592 ln (x/0.15) -7.18 ln(x/0.15) 0.00076x/0
.15 x.00011 M 0.11 mM uncharged molecule
5
8.8 Gastric juice of the stomach has a pH 2.0.
Inside the epithelial cells the pH7.0 and they
have a membrane voltage of -70 mV at 37oC. a)
what is the concentration gradient of protons
across the membrane? pH -log H therfore, H
in gastric juice is .01 M and H inside the
epithelial cells is 1x10-7 M b) calculate DG
secreting 1 mole of H at 37oC. DGin
RTln(in/out)zFVp DGout -RTln(in/out)
zFVp DGout - 1.987310ln(1x10-7/0.01)
123062(-0.07V) DGout -616ln(1x10-5)-1614 DG
out --7092-1614 DGout 8706 cal/mol
6
8.8 cont c) Do you think one molecule of ATP is
sufficient for moving 1 H across the membrane
under these conditions? It is a very close thing.
If you assume that ATP is hydrolyzed to ADP at
-7.3 kcal/mol, then it will NOT work. If ATP is
hydrolyzed to AMP OR the DG of ATP to ADP is more
negative, then it could occur. d) If protons
were free to move back into the cell, what would
be the membrane potential required to keep them
out of the cell? For the H ions NOT to move into
or out of the cell, the system must be in
equilibrium, or DG0. Then 0 RTln(in/out)z
FVp -zFVpRTln(in/out) -23062Vp1.987310ln(
1x10-7/.01) -23062Vp-7092 Vp0.308V 308mV