Question

in the membrane of axon, sodium potassium pump pumps sodium ions outside of the cell and potassium ions towards inside of the cell.
since sodium and potassium are ions they must carry some charge. either positive or -ve.
what charge is carried by potassium ion and what charge is carried by sodium ion.

Answer 1

@Preetha

Answer 2

@MagaliH

Answer 3

@nisa

Answer 4

@sam

Answer 5

@nicolebears

Answer 6

You can find the answer in the periodic table of the elements. What main group does sodium and potassium belong to?

Answer 7

You can find the answer in the periodic table of the elements. What main group does sodium and potassium belong to? From there use the octet rule/"principal" in order to decide what ion it want to be.

Answer 8

@lilyswan since both sodium and potassium ions are positively charged then in the case of Resting potential why and how positive charge is created outside the membrane and negative charge inside the membrane.

Answer 9

Outside the membrane of an axon, Neurons pump out positively charged sodium ions. In addition, they pump in positively charged potassium ions. Thus there is a high concentration of sodium ions present outside the neuron, and a high concentration of potassium ions inside. The neuronal membrane also contains specialised proteins called channels, which form pores in the membrane that are selectively permeable to particular ions. Thus sodium channels allow sodium ions through the membrane while potassium channels allow potassium ions through.

Answer 10

Sense we just give answers, I can say it have something to do with the conductivity of ion-channels along with Na/K-ATPase.
Some of the ion-channels are special designed in a such way, that the conductivity of ions becomes higher or lower depending on the membrane potential.
We can also describe this somehow mathematical:
We use Ohm's law for the electric current:
\[(E_{ion}-E_{m})=I_{ion}*R_{ion} \leftarrow \rightarrow I _{ion}=g_{ion}*(E _{ion}-E_{m})\]
This is a variation of Ohm's law where we look at the conductivity g. In the resting potential the electric current most equal 0 (equilibrium condition), we write:
\[\sum_{i}^{}(I_{ion})_{i}=I_{Na}+I _{K}+I _{Cl}=0\]
We choose only to work with the 3 ions as the conductivity of Ca is every low compared to other ions, along with that you should do this for all charged molecules in the cell.
Substitute and we get:
\[g _{Na}*(E _{Na}-E _{m})+g _{K}*(E _{K}-E _{m})+g _{Cl}*(E _{Cl}-E _{m})=0\]
\[g _{Na}*E _{Na}+g _{K}*E _{K}+g _{Cl}*E _{Cl}=E_{m}(g _{Na}+g _{K}+g _{Cl})\]
The conductivity of the ions on the left side of the equation most equal the full conductivity of the membrane:
\[g _{m}=g _{Na}+g _{K}+g _{Cl}\]
Substitute:
\[E_{m}=\frac{ g _{Na} }{ g _{m} }*E _{Na}+\frac{ g _{K} }{ g _{m} }*E _{k}+\frac{ g_{Cl} }{ g _{m} }*E _{Cl}\]
We here see that the resting membrane potential rely on the conductivity of the membrane, the conductivity of the single ions and the electrical voltage of the ions. We have measured a typical firing neuron and measured the conductivity of the ions as the membrane potential increases (see attachment 1, figure 1). Along with that we also have the conductivity as a action potential is being fired (see attachment 1, figure 2).

Answer 11

I agree

Answer 12

We know from experiments that the Na/K-ATPase is really used when the distribution of ions are getting destroyed (close to cell death), else the neuron can handle the stress depending on the conductivity alone.

Answer 13

thank you friends.

Answer 14

More than welcome... sorry if what I wrote is a bit weird. (I only study the voltage gated proteins my self)

Answer 15

hey you given me a great explanation. thanks a lot.

Answer 16

it really helped me.