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OpenStudy (anonymous):

what would happen with the Vrest if Sodium–Potassium Exchange Pump were broken?

OpenStudy (abhisar):

u mean resting potential ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

@Abhisar

OpenStudy (abhisar):

Ohkay..u knw what resting potential is ?

OpenStudy (abhisar):

I want you to reply fast man !!!

OpenStudy (abhisar):

你明白静息电位?

OpenStudy (abhisar):

快速回复

OpenStudy (abhisar):

钾泵钠的功能是维持静息膜电位。因为(3娜)泵出,每2 k移动有钠高了很多梯度比英寸以外因此,如果此操作可关闭娜将在移动,比K快将迁出,细胞会去极化。

OpenStudy (anonymous):

so if the Sodium–Potassium Exchange Pump were inaction, the cell' rest potential will increase? @Abhisar btw please just type the answer in English, because yr Chinese translation isn't correct.

OpenStudy (abhisar):

:D i knw .. u were not responding at english so i thought it would be better to reply in chinese...also it looks like u used google translator to post ur question which in addition forced me to believe u dont understand english

OpenStudy (anonymous):

so am i right?

OpenStudy (abhisar):

@alphadxg

OpenStudy (anonymous):

thx :D

OpenStudy (abhisar):

Anytime !

OpenStudy (abhisar):

But do cross check !

OpenStudy (abhisar):

@Somy

OpenStudy (anonymous):

O oh, the answer in my textbook is The activity of the sodium-potassium pump is essential to maintain the resting potential. With the pump inactivated, the sodium and potassium concentration gradients would gradually disappear, resulting in a greatly reduced resting potential. but i think if the sodium and potassium concentration gradients disappear, the rest potential must trend to zero, the rest potential increase! @Abhisar

OpenStudy (abhisar):

ohkay now wait....i'll hv to take out my physiology book..i studied them long time back

OpenStudy (somy):

are you guys talking about cell surface membrane? @Abhisar

OpenStudy (abhisar):

see the negative sign of the resting potential (-70mV) shows that inside the cell is negative whle outside the cell is positive

OpenStudy (abhisar):

@Somy yes, like axolemma

OpenStudy (abhisar):

resting potential means the potential difference across the membrane

OpenStudy (somy):

im not really sure whats the answer but logically thinking if we look at membrane and at sodium potassium pump the basic function of Na and K is to maintain the charges in the cell thus the gradient. If this pump is inactive that means the gradient of charges will either increase or decrease. So this maintained gradient would be destroyed. Since Na is actually needed to pump out and K is needed to constantly come in or the opposite way round then that means some functions will basically stop, example Glucose that used to come in along with Na cannot come like that anymore now then no exchange happens, cell is alive so its continuously active, thus obviously K and Na that are in the cell will be used up in some functions right? so thats why the gradient will drop in the cell

OpenStudy (somy):

im not sure if its right, because im just analyzing this just by looking at the picture of the pump and knowing sever roles of this pump so i came up with this

OpenStudy (somy):

the gradient will drop since number of K and Na in the cell will be less then outside the cell

OpenStudy (somy):

@Abhisar i hope i didn't misunderstand the question?

OpenStudy (somy):

if these pumps are active i don't think the gradient would increase rather it would just be maintained unless there is abnormalities in the pump itself

OpenStudy (abhisar):

Yes ur ryt @Somy ! Different types of ion channels are present on the neural membrane. These ion channels are selectively permeable to different ions. When a neuron is not conducting any impulse, i.e., resting, the axonal membrane is comparatively more permeable to potassium ions (K+) and nearly impermeable to sodium ions (Na+). Similarly, the membrane is impermeable to negatively charged proteins present in the axoplasm. Consequently, the axoplasm inside the axon contains high concentration of K+ and negatively charged proteins and low concentration of Na+. In contrast, the fluid outside the axon contains a low concentration of K+, a high concentration of Na+ and thus form a concentration gradient. These ionic gradients across the resting membrane are maintained by the active transport of ions by the sodium-potassium pump which transports 3 Na+ outwards for 2 K+ into the cell. As a result, the outer surface of the axonal membrane possesses a positive charge while its inner surface becomes negatively charged and therefore is polarised. The electrical potential difference across the resting plasma membrane is called as the resting potential.

OpenStudy (somy):

@13xyhuang2 i hope you get it?

OpenStudy (abhisar):

So basically there are two types of transport i.e active and passive going on simultaneously. Former tries to increase the gradient and later tries to decrease it. So it kinda maintains an equilibrium

OpenStudy (somy):

hehe actually i haven't learnt about nervous system in this much details i've just learnt the working of membrane itself so thats why i came up with this @Abhisar :D

OpenStudy (abhisar):

Anyways...!! Thanx a lot @Somy for taking the pain and helping us out :)

OpenStudy (somy):

nah no problem i got to learn some stuff also :D @Abhisar

OpenStudy (anonymous):

@Abhisar @Somy i know this, but what would happen with the rest potential ?

OpenStudy (somy):

it will decrease

OpenStudy (abhisar):

Yes it start fading away gradually i.e potential difference will decrease as said by @Somy

OpenStudy (abhisar):

Don worry bou the negative sign. its just to show that cytoplasm is more negative than outside

OpenStudy (anonymous):

@Somy you said, the gradient will drop since number of K and Na in the cell will be less than outside the cell so the K+ Na+ will come in, but the potential will increase

OpenStudy (somy):

dear this is the problem, if the pump is broken how can any of them come in?

OpenStudy (abhisar):

@shrutipande9

OpenStudy (somy):

@13xyhuang2 K and Na are charged ions and they cannot just diffuse through phospholipid bilayer the only way for them to come in is protein channel specifically for them since its facilitated diffusion not just diffusion that happens

OpenStudy (somy):

so if these K and Na channels are broken, there is no other way Na or K can come in

OpenStudy (anonymous):

but there are still some K and Na channals active

OpenStudy (somy):

did the question say that other channels are okay? if so then NO it again will not increase because now other channels will take up the work of the broken channel and thus again maintain the gradient (tho it'll be a bit harder)

OpenStudy (anonymous):

only Sodium–Potassium Exchange Pump inactivated

OpenStudy (somy):

yes dear, that mean ALL sodium-potassium pumps are broken

OpenStudy (abhisar):

@13xyhuang2 i think u hav not gone through ur textbooks properly !

OpenStudy (abhisar):

It would be better if u read it carefully once again !

OpenStudy (somy):

look in the membrane when it comes to protein channels then they are VERY SPECIFIC they let ONLY sever type of ions like in sodium potassium pump it will let ONLY Na and K to be transported Cl for example CANNOT pass through this K and Na channel

OpenStudy (somy):

do you know how enzymes work?

OpenStudy (abhisar):

I would like to correct one thing @Somy , Na-K pump is an active transport and not just facilitated diffusion. It requires ATP

OpenStudy (somy):

yeah thats right, i just didn't mention it :D @Abhisar

OpenStudy (abhisar):

u said this "K and Na are charged ions and they cannot just diffuse through phospholipid bilayer the only way for them to come in is protein channel specifically for them since its facilitated diffusion not just diffusion that happens"

OpenStudy (anonymous):

i do know. XD but in the membrane, it have Sodium–Potassium Exchange Pump, Na+ channal, K+channal and other channals. when the Na+ k+ exchange pump inactivated, you said the concentration of K+ and Na+ inside the cell will less than outside, so because of this, K+ and Na+ will diffuse inside the cell through K+,Na+channals, the positive charges inside the cell the rest potential increase. @Somy

OpenStudy (somy):

yeah coz i was talking about diffusion :P @Abhisar

OpenStudy (abhisar):

LOL....probably i got it wrong :D

OpenStudy (somy):

thats would be right if the channel of Na and K were not broken but still the potential would not increase in any case since the channels are BROKEN then neither facilitated diffusion nor active transport can happen so none can come in that is why the potential will drop

OpenStudy (abhisar):

@13xyhuang2 due see, in a Na-K pump by the use of ATP three Na+ are exported while 2 K+ are imported now this creates a potential difference as well as a conc gradient i.e Na+ conc is more outside and K+ conc is more inside.

OpenStudy (abhisar):

Now if the pump stops the only operating process will be the diffusion (or facilitated diffusion as @Somy said :P). Now this will gradually lead to depolarisation or decrease in potential difference

OpenStudy (anonymous):

but soon the conc gradient will disappear, i think the Na+, k+ conc will become equivalence both outside and inside the cell

OpenStudy (abhisar):

so ??

OpenStudy (somy):

BUT CELL IS LIVING so IT NEEDS Na and K so of course the number of K and Na will decrease

OpenStudy (anonymous):

this is what i can't figure out. π-π btw, depolarization, the potential increase.....@Abhisar

OpenStudy (anonymous):

i mean from negative increase to positive

OpenStudy (abhisar):

that sign is just to indicate that inner side is more negative than outer

OpenStudy (somy):

by any chance do you not understand terms 'more negative' "less negative" ? @13xyhuang2

OpenStudy (abhisar):

do u think na-K pump ever stops in a normal cell?

OpenStudy (anonymous):

i can understand this.. @Somy

OpenStudy (anonymous):

I think what Somy is trying to say is that: Imagine everything that has to do with K is destroyed (like the channels) so that means outside the K+ will keep increasing while inside the cell is less so (more) Na will be taken in to compensate for the lack of positive charge in the Cell However if this explanation is confusing I will delete my answer immediately ^^'

OpenStudy (somy):

then i'll say it like this the gradient inside the cell will be more negative then outside the cell

OpenStudy (abhisar):

@Somy do u think na-K pump ever stops in a normal cell?

OpenStudy (anonymous):

only if it disposes with some drugs.... @Abhisar

OpenStudy (somy):

oh that was to me? emm i do not think so @Abhisar unless it is effected by a disease or genetic reasons

OpenStudy (abhisar):

u get ur reasons then @Somy ! If it stops-----number of K and Na will decrease-----cell dies

OpenStudy (anonymous):

@Abhisar umm reasons for what??

OpenStudy (somy):

well yeah it does eventually die @Abhisar

OpenStudy (anonymous):

the point is what happen with the rest potential...π-π

OpenStudy (anonymous):

I feel so sorry for everyone here, right now TT_TT

OpenStudy (abhisar):

@shreehari499

OpenStudy (shrutipande9):

is d prob solved or not yet?

OpenStudy (anonymous):

Nope

OpenStudy (somy):

no since the question guy is not getting it @shrutipande9

OpenStudy (shrutipande9):

ohh well then lets start from d beginning? @13xyhuang2

OpenStudy (anonymous):

ok

OpenStudy (shrutipande9):

generally cells have a higher conc of potassium and low conc of sodium inside....this is to maintain the membrane potential and to have charge separation accross the membrane..ok?

OpenStudy (anonymous):

yep

OpenStudy (shrutipande9):

generally what happens through a few pores there is generally influx of sodium inside and potassium outside by diffusion along the conc gradient..

OpenStudy (anonymous):

then?

OpenStudy (shrutipande9):

so what d pump does is pushes the sodium outside and gets potassium in...so if thats not working then the charge separation accross the membrane is hampered and this in turn affects many processes....for eg receiving a signal from a neuron...

OpenStudy (anonymous):

so?

OpenStudy (shrutipande9):

so thats it..:P isnt this enuf? or u need more?

OpenStudy (anonymous):

i want know what will the rest potential change since the Na+K+pumps are distroied.

OpenStudy (shrutipande9):

u want to know d value?

OpenStudy (anonymous):

i want to know whether the potential will be more negative or less negative

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