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Why does K+ move out of the cell?

Why does K+ move out of the cell?


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I was reading this page: Equilibrium Potentials when I found the following example at the end of the page: "If the K+ equilibrium potential is -90 mV and the membrane potential is -70 mV, in what direction will K+ move through open K+ channels?".

The site gives the answer, and it turns out that K+ moves out of the cell. I'm not entirely sure if I understand why this is the case.

(1) Because the equilibrium potential for K+ is -90mV, this means that the intracellular region must be negatively charged, at -90mV, to have zero net flux of K+ across the membrane. Therefore K+ would leave the cell, making the interior more negative from -70mV to -90mV.

(2) But the resting membrane potential is still -70mV. Will this membrane potential be established by other ions? If not the case, K+ will be pulled inside the cell again, then we wouldn't be able to tell that K+ goes out.

(3) I find the starting conditions somewhat confusing. "K+" goes out means that there needs to be a potential difference. Is the question assuming we have, as starting conditions, the same concentration of K+ at both sides? Otherwise it is not clear why K+ should stop leaving the cell at any moment (even when it is at the state of equilibrium).

(4) Can we conclude, as a general rule, that if MRP is the membrane resting potential and EP is the X+ equilibrium potential, then X+ leaves the cell if EPMRP. Would the roles be reversed for X-?


The other answer is a bit misleading.

"Another cause is the the intracellular K+ concentration"

No, this is exactly the same cause, the differing concentrations is what causes the equilibrium potential. You can think of the equilibrium potential for one ion as "how much voltage does there need to be to prevent this ion from flowing down its concentration gradient." The Nernst equation gives you the equilibrium potential for any one ion. The equilibrium potential is also known as the "reversal potential" because if the voltage exceeds the equilibrium potential, ions will flow the opposite way, against their concentration gradient.

Don't get confused by the resting potential, because it is only a potential that the cell arranges for itself to maintain a stable environment. This cost a lot of energy and attention in form of protein synthesis to keep this potential. This mechanism is achieved by the Na+-K+ ion-pump

The resting potential is not to "maintain a stable environment" or anything like that. The resting potential is the net potential that a cell reaches due to all of the conductances of various ions and the flowing current of each ion governed by its own equilibrium potential.

This mechanism is achieved by the Na+-K+ ion-pump, which exchanges those two ions

This is partly true, but again, misleading. The Na+/K+ pump establishes the relative concentration of ions, which only leads to a certain resting potential because the conductances of different ions vary at rest. Because the conductance of K+ is higher at rest than the conductance of Na+, the resting potential is closer to the equilibrium potential for K+. If the Na+/K+ pump was functioning exactly as it does, pumping Na+ out and K+ in, but the membrane was more permeable to Na+, then the resting potential would be positive rather than negative. All that matters for resting potential is relative ion concentrations and conductivity. The Goldman equation is the way to calculate this resting potential.

What is happening to K+ when a cell has a resting potential of -70mV but a K+ equilibrium of -90mV?

Because you need -90mV to 'hold in' the potassium against it's concentration gradient, at -70mV K+ will flow out of the cell. However, by definition, resting potential is the potential at which the net current will be zero. That means that, yes, other ions have to be involved. This is where the Goldman equation is very useful.

Although K+ may be the ion with the highest membrane conductance, there will also always be some leak of Na+ and Cl- ions (usually others ions are ignored because these three are the major players; other ions can be important in some situations). Therefore, if -70mV is rest, there must be at least as many Na+ ions flowing in or Cl- ions flowing out as there are K+ ions flowing out.

You can calculate these currents if you know the conductance of each ion and the equilibrium potential for each ion using Ohm's Law: I = V/R, where R is 1/conductance and V is the difference between the current voltage and the equilibrium potential. Try it out with the Goldman equation! If you use the voltage the Goldman equation gives you (the resting potential) you will find that the net current is zero!

Okay, so far we've answered your questions (1) and (2). Now (3): you state "Otherwise it is not clear why K+ should stop leaving the cell at any moment". It is important to note that equilibrium potential doesn't mean no ions are moving: it means the voltage isn't changing. If you just left the cell at -70mV and waited some time, eventually the concentrations of the different ions would change. This is where the Na+/K+ pump comes in: this ATPase is constantly pumping some K+ out and Na+ in to counteract the leak that occurs at resting potential. Cl- ions move, too, but mostly passively. It's also important to note that very few ions have to move for the potential to change on the order of millivolts. -70mV might sound like a lot, but electrical forces are very powerful, so very few ions relative to all the ions available have to move.

I don't quite understand what you are asking with (4) but if you can edit the question I will try my best to answer that as well. It's possible I already answered your question by talking about what "equilibrium potential" means for one ion. For a Cl- ion, if the equilibrium potential for Cl- is -65mV, then Cl- will flow into the cell when the membrane voltage is -60mV (tending to make the cell closer to -65mV), and it will flow out of the cell when the membrane voltage is -70mV (still tending to make the cell closer to -65mV).


The Equilibrium potential(-90 mV) of Potassium-ions(K+) will drive the ions to flow out of the cell if K+ channels are opened, because K+ ions try to establish this potential instead of the resting potential(-70 mV). Another cause is the the intracellular K+ concentration that is around 140 mM compared to the extracellular concentration, which is only 5 mM. Don't get confused by the resting potential, because it is only a potential that the cell arranges for itself to maintain a stable environment. This cost a lot of energy and attention in form of protein synthesis to keep this potential. This mechanism is achieved by the Na+-K+ ion-pump, which exchanges those two ions (pumps K+ in and Na+ out to achieve the -70 mV). If K+ ions flow out of the cell the membrane potential will get lower than the resting potential, also referred to as hyperpolarisation. This event occurs at the end of an action potential when the outflow of the K+ ions "undershoots" the resting potential.


Why is the sodium-potassium pump important?

The sodium potassium pump is important for the functioning of most cellular processes.

Explanation:

It is a specialised transport protein found in the cell membranes. It is responsible for movement of potassium ions into the cell while simultaneously moving sodium ions into the cell. This is important for cell physiology.

It has special significance for excitable cells such as nervous cells, which depend upon this pump for responding to stimuli and transmitting impulses. Transmission of nerve cells would've been impossible without the aid of this pump.

The sodium/potassium helps maintain resting potential,effect transport and regulate cellular volume. It also functions as a signal transducer/ integrator to regulate MAPK pathway, as well as intracellular calcium.

In the kidneys the sodium potassium pump helps to maintain the sodium and potassium balance.

It also plays a role in maintaining blood pressure and control cardiac contractions.


Sodium-potassium Pump

Pure water does not conduct electricity. Potassium, sodium and other electrolytes dissolve in water and help carry an electrical charge. The cells in your body need electrolytes to transport and maintain electrical impulses. The sodium-potassium pump describes a mechanism in which sodium and potassium ions move in and out of your cells. Each time this happens, an electrical charge is produced. The sodium-potassium pump also responds to power requests from your nervous system. Proper electrolyte balance is important for your blood, hydration and helps maintain other vital body functions.


A level biology help: Resting potential in the axon

I know that the axon has a constant resting potential of -70mv.

During resting potential 3 Na+ ions constantly leave the axon while 2 K+ ions constantly enter the axon. This gives the axon a net negative potential difference. K+ ions can exit the axon afterwards and Na+ ions cannot come back into the axon. This makes the potential difference even less. But doesn't that mean that the potential difference in the axon can go down to negative infinity. I get why the inside of the axon is more negative than the outside, but I don't get what keeps it at constant -70mv?

Not what you're looking for? Try&hellip

(Original post by muhammad0112)
I know that the axon has a constant resting potential of -70mv.

During resting potential 3 Na+ ions constantly leave the axon while 2 K+ ions constantly enter the axon. This gives the axon a net negative potential difference. K+ ions can exit the axon afterwards and Na+ ions cannot come back into the axon. This makes the potential difference even less. But doesn't that mean that the potential difference in the axon can go down to negative infinity. I get why the inside of the axon is more negative than the outside, but I don't get what keeps it at constant -70mv?

The sodium-potassium pulp actually has very little effect on the resting membrane potential. The resting membrane potential is more to do with the relative permeability of the cell membrane to Na+ and K+ ions. The membrane is around 40x permeable to K+ ions compared to Na+ ions. Since the membrane is relatively impermeable to Na+ ions and the Na+ concentration is much higher outside the cell, the voltage inside the cell is much lower than outside the cell, hence the negative resting membrane potential.

The sodium-potassium pump is only important for maintaining the concentration gradients of Na+ and K+ ions - it is NOT responsible for the resting membrane potential.

(Original post by Jpw1097)
The sodium-potassium pulp actually has very little effect on the resting membrane potential. The resting membrane potential is more to do with the relative permeability of the cell membrane to Na+ and K+ ions. The membrane is around 40x permeable to K+ ions compared to Na+ ions. Since the membrane is relatively impermeable to Na+ ions and the Na+ concentration is much higher outside the cell, the voltage inside the cell is much lower than outside the cell, hence the negative resting membrane potential.

The sodium-potassium pump is only important for maintaining the concentration gradients of Na+ and K+ ions - it is NOT responsible for the resting membrane potential.

I've done some more research and correct me if I'm wrong, but the axon is very permeable to potassium but without the sodium potassium pump, the electrical gradient will equal to the chemical gradient. (the reason why there's an electrical gradient is because the K+ ions would want to come back in the axon as the outside is more positive and the inside is negative). Therefore, the sodium potassium pump increases the chemical gradient of the potassium (as 2 potassium would enter in the axon). And the chemical and electrical gradient would balance out when the membrane is -70mv.

But can you explain why the sodium potassium pump doesn't make a difference to the resting potential?I don't understand it. If the Na/K pump went off once, wouldn't the net pd be -1 in the axon? - If it went off 10 times, that would be 30 sodium ions out and 20 K+ ions in - Thats a net pd of -10mv?. Obviously I understand that 1 Na+ ion isn't equal to 1mv, but wouldn't it be the same idea?

(Original post by muhammad0112)
I've done some more research and correct me if I'm wrong, but the axon is very permeable to potassium but without the sodium potassium pump, the electrical gradient will equal to the chemical gradient. (the reason why there's an electrical gradient is because the K+ ions would want to come back in the axon as the outside is more positive and the inside is negative). Therefore, the sodium potassium pump increases the chemical gradient of the potassium (as 2 potassium would enter in the axon). And the chemical and electrical gradient would balance out when the membrane is -70mv.

But can you explain why the sodium potassium pump doesn't make a difference to the resting potential?I don't understand it. If the Na/K pump went off once, wouldn't the net pd be -1 in the axon? - If it went off 10 times, that would be 30 sodium ions out and 20 K+ ions in - Thats a net pd of -10mv?. Obviously I understand that 1 Na+ ion isn't equal to 1mv, but wouldn't it be the same idea?

As you say, 1 Na+ ions does not equal 1 mV. A huge number of Na+ ions are needed to exit the axon to create a 1 mV change in membrane potential, in the grand scheme of things, the sodium-potassium pumps do not move enough ions to make a difference - though they are important in maintaining high extracellular Na+ concentrations/high intracellular K+ concentrations.

At the resting membrane potential (say -70 mV in an axon), the 3 Na+ ions that get pumped out by the Na/K pump simply move back into the cell along their electrochemical gradient (as they are in equilibrium). Likewise, the K+ ions that get pumped into the axon simply diffuse out of the axon along their electrochemical gradient.

The Na/K pump is important in maintaining the concentration gradients for Na+ and K+, but it has little to no effect on the resting membrane potential, that is all to do with the differential permeability of the membrane to different ions, as reflected in the Nernst equation.

(Original post by Jpw1097)
As you say, 1 Na+ ions does not equal 1 mV. A huge number of Na+ ions are needed to exit the axon to create a 1 mV change in membrane potential, in the grand scheme of things, the sodium-potassium pumps do not move enough ions to make a difference - though they are important in maintaining high extracellular Na+ concentrations/high intracellular K+ concentrations.

At the resting membrane potential (say -70 mV in an axon), the 3 Na+ ions that get pumped out by the Na/K pump simply move back into the cell along their electrochemical gradient (as they are in equilibrium). Likewise, the K+ ions that get pumped into the axon simply diffuse out of the axon along their electrochemical gradient.

The Na/K pump is important in maintaining the concentration gradients for Na+ and K+, but it has little to no effect on the resting membrane potential, that is all to do with the differential permeability of the membrane to different ions, as reflected in the Nernst equation.

Okay, let&rsquos try to figure out why the membrane potential is -70mV.

Firstly you have a high concentration of organic anions (negatively charged protein) in the axon. This is balanced by K+ ions. The K+ concentration [K+] is higher inside the axon compared to outside, therefore the chemical gradient drives K+ ions out of the cell. As K+ ions move out of the cell, this creates a negative membrane potential, which causes K+ ions to move back into the axon (electrical gradient). As more K+ ions leave the axon, the electrical gradient gets larger (as more K+ ions leave the axon) and the chemical gradient remains relatively unchanged (the number of ions moving across the membrane has a negligible impact on the chemical gradient for short periods of time). The number of K+ ions leaving the axon gets progressively less and less as the electrical gradient gets larger until the electrical gradient = chemical gradient. At this point, there is no net movement of K+ ions - this is the equilibrium potential for K+

Now let&rsquos consider Na+. Na+ has a much higher concentration outside the cell, therefore the chemical gradient drives Na+ ions into the cell. As Na+ ions enter the cell, this creates a positive membrane potential (electrical gradient), and this causes the number of Na+ ions entering the cell to slow down as the electrical gradient gets larger. When the electrical and chemical gradients are equal, there is no net movement of Na+ ions - the membrane potential when this occurs is around + 60 mV for Na+ (equilibrium potential).

So if the equilibrium potential for Na+ is

-80mV, why isn&rsquot the resting membrane potential somewhere in the middle. This is because the membrane is far more permeable to K+ ions compared to Na+ ions, therefore the resting membrane potential is far closer to the equilibrium potential for K+ (-80mV) compared to Na+ (+60mV).

At the resting membrane potential, the movement of Na+ into the cell and the movement of K+ out of the cell are equal. Even though there is a large electrical gradient driving Na+ into the cell (there is a large difference between the equilibrium potential for Na+ and the resting membrane potential), the cell is relatively impermeable to Na+ ions. While the electrical gradient is small for K+ ions (K+ equilibrium potential is not far from the resting membrane potential), the membrane is highly permeable to K+ ions.

As you can see, at no point have I discussed the Na/K pump, because it operates on such a small level, it has very little effect on the resting membrane potential. Far more ions are moving across the membrane through leak channels compared to the N/K pump. The Na/K pump is only important in maintaining the chemical gradients, but not for establishing them.


ELI5: Why does K+ leak through or cross the cell membrane wall at all?

If K+ can travel quite easily through a cell membrane wall and has both electrical and chemical driving force one in each direction, then why is it moving at all and why isn’t the concentration of K+ just the same on both sides and K+ not crossing at all?

I can’t seem to find an answer that makes sense to me on google!

Two things. It'll reach an electrical equilibrium, not necessarily a physical one. The cell will loose charge and the outside will gain charge, this will start to pull some of the potassium back. Second, our cells actually have tiny pumps that can force potassium back into the cell.

The resting membrane potential is due to the balance of so many ion channels and pumps. You have the sodium potassium pump, which pumps 2 K+ in for 3 Na+ out. And a bunch more. But of significance is that the membrane is more leaky to K+, because it has channels for it (no gating, just always open). There's not too many of these channels (not enough to completely kill its gradient), but just enough to reduce the overall resistance for it. With sodium being practically completely blocked from exiting without a gated channel (it still leaks but far far less), when you plug everything into the equation (concentrations of all ions in and out, and the membrane resistance of each ion) you get the resting membrane potential at around -90 to -50 mV depending on the cell. So to simply answer your question, the membrane is leaky for K+ sure, but when it comes to ions and cells, you have to think in terms of statistics and populations, the membrane is not completely transparent for K+, it just is a bit leaky, like a bucket with water (K+) inside and a few holes in it, and also some motors pumping the water that left back in. The pumps and the leak channels are not equivalent in terms of bidirectional current so you don't just cancel out, but at some point, the water inside will be low enough for example that less water leaves through the channels, making the efflux of water via holes equal to influx via pumps, so the water level stays constant.


Cellular

Structurally, the Na+ K+ ATPase is composed of a catalytic alpha subunit and an auxiliary beta subunit.[7] Some Na-K ATPases include a subunit that is tissue-specific and belongs to the FXYD protein family.[8] The alpha subunit contains a transmembrane region which is composed of 10 helices, referred to as MA1-M10. Within these ten helices, ion binding sites, specifically three binding sites that bind to Na+ in the E1 state and two binding sites that bind to K+ in the E2 state.[9][10][11][12] The structure of the Na-K ATPase is composed of three sites. Site one and two overlap within both the E1 and E2 states. However, site three is exclusively in the E1 state and is between the M5, M6, and M8 transmembrane helices, which bind to Na+ and catalyze H+ transport as well,[13][14]ꃞpendent on the Na+, K+, and H+ concentrations.[15] According to previous studies, the pump’s E2 state selectivity for K+ may be due to ion binding pocket protonation.[16]


Homeostasis & Transport

I. Hypertonic Solution
1. Solute concentration outside the cell is higher (less water)
2. Water diffuses out of the cell until equilibrium is reached
3. Cells will shrink & die if too much water is lost
4. Plant cells become flaccid (wilt) called plasmolysis

J. Hypotonic Solution
1. Solute concentration greater
inside the cell (less water)
2. Water moves into the cell until equilibrium is reached
3. Animal cells swell & burst (lyse) if they take in too much water
4. Cytolysis is the bursting of cells
5. Plant cells become turgid due to water pressing outward against cell wall
6. Turgor pressure in plant cells helps them keep their shape
7. Plant cells do best in hypotonic solutions

K. Isotonic Solutions
1. Concentration of solutes same inside & outside the cell
2. Water moves into & out of cell at an equal rate so there is no net movement of water
3. Animal cells do best in isotonic solutions

IV. How Cells Deal With Osmosis

A. The cells of animals on land are usually in isotonic environment (equilibrium)

B. Freshwater organisms live in hypotonic environments so water constantly moves into their cells

C. Unicellular freshwater organisms use energy to pump out excess water by contractile vacuoles

D. Plant cell walls prevent plant cells from bursting in hypotonic environments

E. Some marine organisms can pump out excess salt

A. Faster than simple diffusion

B. Considered passive transport because extra energy not used

C. Occurs down a concentration gradient

D. Involves carrier proteins embedded in a cell’s membrane to help move across certain solutes such as glucose

E. Carrier molecules change shape when solute attaches to them

F. Change in carrier protein shape helps move solute across the membrane

G. Channel proteins in the cell membrane form tunnels across the membrane to move materials

H. Channel proteins may always be open or have gates that open & close to control the movement of materials called gated channels

I. Gates open & close in response to concentration inside & outside the cell

A. Requires the use of ATP or energy

B. Moves materials against their concentration gradient from an area of lower to higher concentration

C. May also involve membrane proteins

D. Used to move ions such as Na+, Ca+, and K+ across the cell membrane

E. Sodium-Potassium pump moves 3 Na+ out for every 2 K+ into the cell
1. Causes a difference in charge inside and outside the cell
2. Difference in charge is called membrane potential

F. Ion pumps help muscle & nerve cells work

G. Plants use active transport to help roots absorb nutrients from the soil (plant nutrients are more concentrated inside the root than outside)

A. Moves large, complex molecules such as proteins across the cell membrane

B. Large molecules, food, or fluid droplets are packaged in membrane-bound sacs called vesicles

C. Endocytosis moves large particles into a cell

D. Phagocytosis is one type of endocytosis
1. Cell membrane extends out forming pseudopods (fingerlike projections) that surround the particle
2. Membrane pouch encloses the material & pinches off inside the cell making a vesicle
3. Vesicle can fuse with lysosomes (digestive organelles) or release their contents in the cytoplasm
4. Used by ameba to feed & white blood cells to kill bacteria
5. Known as “cell eating”

E. Pinocytosis is another type of endocytosis
1. Cell membrane surrounds fluid droplets
2. Fluids taken into membrane-bound vesicle
3. Known as “cell drinking”

F. Exocytosis is used to remove large products from the cell such as wastes, mucus, & cell products

G. Proteins made by ribosomes in a cell are packaged into transport vesicles by the Golgi Apparatus


Why does K+ move out of the cell? - Biology

The Na+/K+ pump is found in the membranes of many types of cells. In particular, it plays a very important role in nerve cell membranes. Notice that 3 positive ions (Na+) are pumped out of the cell (towards ECF) for every 2 positive ions (K+) pumped into the cell (towards ICF). This means that there is more positive charges leaving the cell than entering it. As a result, positive charge builds up outside the cell compared to inside the cell. The difference in charge between the outside and inside of the cell allows nerve cells to generate electrical impulses which lead to nerve impulses.

The Na+/K+ pump illustrates "active transport" since it moves Na+ and K+ against their concentration gradients. That is because there is already a high concentration of Na+ outside the cell and a high concentration of K+ inside the cell. In order to move the ions (Na+ and K+) againts their gradients, energy is required. This energy is supplied by ATP (adenosine triphosphate). An ATP molecule floating inside the cell, binds to the pump transferring some energy to it. As the energy is used, the ATP falls off and having lost its energy it is converted into ADP (adenosine diphosphate).

Notice in the diagram that there are 3 binding sites for the 3 Na+ ions on the inner surface of the pump and 2 binding sites for the 2 K+ ions on the outer surface of the pump. The shape of these binding sites ensures that only Na+ and K+ can bind and be transported.

Since the pump requires an ATP every time it works, ATP must be constantly supplied to the cell. ATP is created during the processes called "cellular respiration" which occur inside the cell (in the ICF). Part of cellular respiration happens in the cytoplasm and part happens in the mitochondrion. Since these organelles are part of the nerve cell, as the ATP is made, it floats up to the pump and provides its energy. More ATP is made and the pump continues to do its job. If something interferes with the production of ATP, the pump will stop working and the nerve cell will also stop working. This can cause serious loss of nerve function and even death. Since cellular respiration requires oxygen, if you were to stop breathing, ATP could not be produced and you would die. Of course ATP is needed by many processes in the body so it is not only the Na+/K+ pump that would stop.

There are poisons or toxins that also interfere with the pump. One is called "oubain", an arrow poison. Oubain works by attaching to the pump and blocking its action. A branch of science called "pharmacology" deals with how drugs affect the body. Pharmacologists have designed drugs that, if administered fast enough, can travel to the cells and attach to the oubain removing it from the Na+/K+ pumps allowing them to function properly.

This is just one example of active transport in which ATP is used to pump materials against a concentration gradient. Your body stores glucose (a sugar) in your liver and muscles. In order to stockpile the glucose for when you might need it, the glucose must be pumped into cells building up a high concentration there. Even though it uses up ATP to do this, every glucose molecule can be broken down by cellular respiration to produce 38 ATP's! So it's a worthwhile process.


Transport of Ions through Cell Membrane

The living organisms can be resolved into organs, glands, tissues, cells and organelles. It is very inter­esting in biology to know how solutes and water get into and out of cells and organelles. Most at­tention is to be paid to erythrocytes and mitochon­drion. The cell membrane is a complex lipoprotein structure.

Some channels are continuously open, whereas others are gated i.e. they have gates that open or close. Some are gated by alterations in membrane potential (voltage gated) whereas others are opened or closed when they bind a ligand (ligand gated).

The ligand is often external (neurotransmitter or hormone) or internal (intracellular Ca ++ , cAMP). Other transport proteins are carriers that bind ions and other molecules and then change their configuration, moving the bound molecule from one side of the cell membrane to the other.

Molecules move from areas of high concen­tration to areas of low concentration (down their chemical gradient). Cations move to negatively charged areas whereas anions move to positively charged areas (down their electrical gradient), ligand gated chan­nel.

Some of the carrier proteins are called uniports because they transport only one substance. Others are called symports because transport requires the binding of more than one substance to the trans­port protein and the substances are transported across the membrane together.

In the intestinal mucosa that is responsible for the cotransport by facilitated diffusion of Na + and glu­cose from the intestinal lumen into mucosal cells. Other transporters are called anti-ports because they exchange one substance for another. Example: Na + – K + ATPase.

It catalyses the hydrolysis of ATP to ADP and uses the energy to extrude 3Na + from the cell and take 2K + into the cell for each mole of ATP hydrolysed. The pump is said to have a coupling ratio of 3/2. Its activity is inhibited by ouabain and related to digitalis glycosides used in the treatment of heart failure.

Na + -K + ATPase is a heterodimer made up of α and β subunit.

Na + and K + transport occurs through a subunit.

β subunit is a glycoprotein.

Substances passing through the lipid bilayer of the cell membrane by simple diffusion are:

1. All lipid soluble substances.

2. Lipid soluble gases mainly CO2, O2 and N2.

3. Water—though not lipid soluble—passes because of small molecular size and high kinetic energy.

Substances passing through protein channels of cell membrane by simple diffusion are:

1. Ions mainly Na + , K + and Ca ++ .

A. Passive Diffusion:

1. Some solutes pass through cell membrane by simple diffusion with the concentra­tion gradient.

This can be expressed by the modification of Fick’s law:

where, P = the permeability coefficient.

C0 and Ci = the concentration of solution outside and inside the membrane, respec­tively.

ds/dt = rate of movement of solute.

2. Lipid-soluble solutes pass more readily through cell membranes than lipid-insoluble solutes. Because the cell membrane consists of small water-filled pores of ra­dius about 0.4 nm. through which water- soluble solute of suitable molecular size pass, surrounded by lipid areas through which lipid-soluble solutes penetrate.

3. Water diffuses through the cell pores from a solution of low concentration to a solu­tion of high concentration and this “bulk flow” of liquid across the membrane will speed up molecules diffusing in the direc­tion of the flow and slow down those mov­ing in the opposite direction. This “drag” effect is a second force acting in passive diffusion.

4. The third force which may operate is an electric potential across the membrane. Many cell membranes can maintain po­tential difference between their inside and outside and the potential gradient acts as a driving force for passive transport across the cell. The membrane acts as a passive barrier.

B. Facilitated Transfer:

1. Some compounds, e.g., sugar, amino acids, pass through membranes at a greater rate than expectations. This is because of the effect of a carrier.

2. The carrier in the membrane combines with the substance to be transported and in some way ferried through the membrane and released on the other side.

3. In case of enzymic reactions, there is a “saturation effect”. The rate of transport of the solute increases when the carrier, enzyme, is saturated. This type is some­times termed “catalysed diffusion”.

4. Another mechanism is that the substance to be transferred is converted into another which will penetrate the membrane more easily, e.g., the mitochondrial membrane is impermeable to acyl coenzyme A deriva­tives. The acyl group is transferred to car­nitine to form acyl carnitine derivative which can pass through the membrane. The acyl coenzyme A derivative is then reformed on the other side of the mem­brane.

Fatty acids can also be transferred into and out of mitochondria.

Acetyl-CoA within the mitochondria can be transferred to oxaloacetate to yield citrate to which the mitochondrial mem­brane is permeable. The citrate passes out into the cytoplasm where it is split enzymically to give acetyl-CoA again.

1. The cell membrane forms pockets or invaginations which can draw materials on the outside towards the cell interior.

2. The vesicles extend into the cell where they are pinched off and finally release their contents into the cell by some un­known way.

3. This process occurs in the foetal and new­born animals and helps the absorption of intact protein from the gut.

D. Transport of Ions:

1. The membrane itself contains polar groups and is, therefore, electrically charged.

2. The transport of most ions occur more slowly than the non- electrolytes. But H + , OH − penetrate all cell membranes easily. The red cell is easily penetrated by Cl − and HCO − 3.

3. In the case of ions, especially, Na + and K + , the permeability is very small. The high concentration of K + and low concentra­tion of Na + which are often found in cells are maintained by special mechanism which involve the expenditure of energy.

E. Active Transport:

1. The process by which solutes can often pass through membranes against their con­centration gradient requires energy. This process is termed active transport.

2. Active transport is involved in the absorp­tion from the small intestine of glucose and galactose, amino acids and other sub­stances important to the body.

3. An active transport device which forces Na + out and K + in has been referred to as the “Sodium Pump”.

4. The mechanism requires a carrier which can exist in two forms with different af­finities for Na + and K + . ATPase is involved in it (see active transport of glucose).


Explain the role of K+ in stomatal movement. Describe the potassium ion pump theory.

Opening of stomata in light: Stomata open in light due to following reactions.

  1. In light starch in the guard cells is metabolised into Phospho Enol Pyruvate (PEP). It is later converted into organic acids, particularly malic acid. The reaction takes place in the presence of an enzyme phospho enol pyruvic carboxylase.
  2. Malic acid dissociates into malate and hydrogen ions in the guard cells.
  3. Hydrogen ions from guard cells are transported to epidermal cells and K+ ions from epidermal cells are absorbed into the guard cells.
  4. In the guard cells, K+ ions are balanced by malate anions. Besides, small amount of Cl-ions are also absorbed which neutralize a small percentage of K+ ions.
  5. The process of K+ ions exchange requires ATP and thus, it is an active process.
  6. Increased K+ ions and anion concentration in the guard cells increases their osmotic concentration. Hence, water enters the guard cells by endosmosis.
  7. Turgor pressure of the guard cells increases due to endosmosis and the stoma opens.

Closing of stomata in dark: Closure of stomata in darkness is due to the following reactions:

  1. As carbon dioxide is not utilized in photosynthesis during night, hence its concentration in the sub stomatal cavity increases.
  2. Abscissic acid (ABA) functions in the presence of carbon dioxide. It is an inhibitor hormone. It inhibits K+ uptake by changing the diffusion and permeability of guard cells.
  3. The K+ is now transported back to the epidermal or subsidiary cells from the guard cells. The osmotic concentration of the guard cells decreases. This results in the movement of water out of the guard cells (exosmosis). The guard cells now become flaccid. This causes stomatal closure.

POTASSIUM ION PUMP THEORY

The main features of the theory were put forward by Levitt (1974). Levitt explained the influx of K+ in the guard cells and their role in the stomatal movement. He demonstrated a relationship between stomatal opening and accumulation of K+ ions. According to the hypothesis, pH of the guard cells rises due to active uptake of H+ ions by guard cell. The steps involved are as follows.

  • The light induced proton transport from the cytoplasm into the chloroplast creates a negative potential. This negative potential may lead to influx of positively charged K+ ion from the surrounding cells so that the negative potential is decreased and the proton transport is not ceased. This results in the development of maximum pH.
  • The pH of cytoplasm is raised to 8-9 and that of the chloroplast is lowered to 5. The rise in the pH causes hydrolysis of starch to form organic acid especially phospho enol pyruvate. Phospho enol pyruvate in the presence of enzyme PEP carboxylase combines with CO2 to produce oxalic acid. This acid is then converted to malic acid.
  • Malic acid disassociates into H+ ion and malate ion. H+ ions comes out and in exchange K+ ions enter into the guard cells, thereby increasing the concentration of K+ ion and decreasing the H+ ions in guard cells. This is an active ionic exchange and requires ATP and cytokinin.
  • The pH of the cell sap in the guard cells increases simultaneously. The pH becomes more than 7 and the medium becomes alkaline.
  • There is also an increased uptake of Cl- anions by the guard cells to maintain the electrical and ionic balance inside and outside guard cells.
  • The malate anions formed in the guard cells are neutralized by the K+ ions. K+ ions react with malate to form potassium malate.
  • Potassium malate enters into the cell sap reducing the water potential and increasing the osmotic concentration and osmotic pressure of the cell sap.
  • Endosmosis occurs and the guard cells become turgid and the stoma opens.

STOMATAL CLOSURE:

When the darkness sets in, H+ ion starts diffusing into the cytoplasm. H+ ion reacts with malate ion to form malic acid. Malic acid undergoes decarboxylation and gets converted into pyruvic acid and carbon dioxide. Pyruvic acid is consumed in respiration. Carbon dioxide gets dissolved in water to form carbonic acid. When potassium malate is converted into malic acid, osmotic pressure of the guard cells decreases .Exosmosis starts and the stomata closes.


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