Osmosis

Osmosis is the passive movement of water across a membrane

The actual movement of water through a cell membrane is the result of two processes: diffusion and bulk flow. As you recall a membrane is the thickness of a phospholipid bilayer. The size of a water molecule permits it to pass through the bilayer. This would be largely a diffusion movement subject to Fick's Law. The membrane also posesses integral proteins; the one involved with water transport is called an aquaporin. The aquaporin protein serves as a water-filled pipe across the membrane. The flow through this channel is accomplished by submicroscopic bulk flow subject to Poiseuille's Equation. These two pathways are shown below.

Generally the movement of water across the membrane is not treated by plant physiologists as either diffusion or bulk flow. Rather plant physiologists focus upon the driving force for osmosis...energy.

Osmosis is driven by water potential

Osmosis is a spontaneous process, so it must be the result of a downhill energy system. We call this energy system water potential. Water spontaneously moves from an area of higher water potential (energy) to an area of lower water potential (energy). For modern plant physiology the water potential is generally expressed in MPa, but other units have been used historically (bars, atm, etc.).

The water potential of a solution is generated by a combination of four contributing factors:

Ψ = Ψs + Ψp + Ψg + Ψm

We will examine each of these factors in turn.

Ψs is the solute potential

The solute potential (ψs) is the effect of dissolved substances on the potential energy of a solution. It is defined as 0 MPa for distilled water. For solutions the solute potential is determined by the Van't Hoff Equation:

Ψs = - CiRT

where C is the molar concentration of the solute, i is the ionization constant for the solute, R is a constant and T is the absolute temperature (°K). The negative sign indicates that solutes decrease the potential energy of a solution.

For simple problem solving uses, here are handy values for RT:

RT = 2.271 @ 0 C
2.436 @ 20 C
2.478 @ 25 C
L MPa mol-1

An example solute potential would be:
1 M Sucrose @ 20 C = -2.436 MPa

Obviously the solute potential relates most directly to the diffusion process. Solute potential influences much about solutions...it decreases the freezing point and elevates the boiling point. For osmosis, solute decreasing the water potential tends to cause water to enter the area of high solute concentration.

ψp is the pressure potential

The pressure potential (ψp) is the effect of hydrostatic pressure on the potential energy of a solution. It is defined as 0 MPa for STP (absolute pressure of 1 atm = 0.1 MPa). For a case of a partial vacuum or tension as in transpiration, the pressure potential is <0. For a case of turgor pressure the pressure potential would be >0.

Increasing the pressure of an area will increase the water potential and water will tend to leave that area. This component of water potential relates most directly to bulk flow.

ψg is the gravitational potential

The gravitational potential (ψg) is the effect of height of a system above sea level. It is defined as 0 MPa at sea level. Basically raising a system 10 meters will increase its water potential energy by 0.1 MPa, water will then tend to move down from there. As most laboratory biology is done all at one level, this component is often considered negligible.

ψm is the matric potential

The matric potential (ψm) is the effect of colloids (adhesion) in soil or as a result of polymers in the cell wall.

Because matric potential is limited in cells, and because the height of the cell in the lab is negligible, the water potential expression simplifies to:

ψ = ψs + ψp

Now we will take an example through a calculation or two. A cell is about 0.3 M, what is its solute potential at 20C ? = -CiRT

- 0.3 * 2.436 = -0.731 MPa

Now we will assume the cell is full but has no turgor pressure initially. Here is how that cell would look:


ψs = -0.731 MPa
ψp = 0 MPa
ψ = -0.731 MPa
 

Now we put this cell into a large volume of pure water at STP (ψ=0 MPa). We will assume that the cell walls are rigid (a reasonable but not perfect idea), so we expect no changes in volume or concentration inside cell. Water moves by osmosis from an area of higher water potential potential (0 MPa) to area of lower potential (-0.731 MPa); in this case water moves into the cell. At equilibrium (when there is no more net movement of water) the water potentials of the cell and the solution are equal:

ψs = 0 MPa
ψp = 0 MPa
ψ = 0 MPa

ψs = -0.731 MPa
ψp = 0.731 MPa
ψ = 0 MPa
 

This cell was placed in a hypotonic solution, and it gained water but mostly the turgor pressure of the cell was increased (it became crispy).

Now what if we put that cell into a large volume of 0.1 M sucrose (ψs = - 0.1 * 2.436 = -0.244 MPa)? We will assume for the moment that sucrose cannot go into or out of the cell. The water will now leave the cell as it moves from the area of greater to lesser water potential. At equilibrium water potentials will be equal:

ψs = -0.244 MPa
ψp = 0 MPa
ψ = -0.244 MPa

ψs = -0.731 MPa
ψp = 0.487 MPa
ψ; = -0.244 MPa
 

As you can see, the cell lost water, but the loss mostly reduced the pressure inside the cell. The cell is not as crispy as it was in the distilled water.

Now we will move the cell into 0.3 M Sucrose (ψs = - 0.3 * 2.435 = -0.731 MPa). Again we will assume that water moves from an area of greater water to lesser water potential; it will move into the cell. At equilibrium we will observe:

ψs = -0.731 MPa
ψp = 0 MPa
ψ = -0.731 MPa

ψs = -0.731 MPa
ψp = 0 MPa
ψ = -0.731 MPa
 

We are back to the original cell conditions. The solution is isotonic with respect to the original cell.

What would happen if we put the cell into a really strong solution, say 0.6 M Sucrose (ψs = - 0.6 * 2.435 = -1.461 MPa)? This would be a hypertonic solution. Since the cell already has no turgor and there is no internal structure to withstand tension (vacuum = negative potential), the water will continue to move out of the cell until the solute concentration is increased sufficiently to reach equilibrium:

ψs = -1.461 MPa
ψp = 0 MPa
ψ = -1.461 MPa
 
 
ψs = -1.461 MPa
ψp = 0 MPa
ψ = -1.461 MPa
 

How much cell volume is lost as the water moves out? At equilibrium the solute potential will have to match that of the solution, so it will have become twice as concentrated as it was originally (from 0.3 to 0.6 M). So the equilibrium volume will be about 50% of the original. This cell is plasmolyzed in this hypertonic solution, meaning that the cell membrane will have pulled away from the relatively rigid cell wall. The protoplast will occupy about half of the wall volume, the other half of the volume will be occupied by the bathing solution.

Now that we have seen how water potential can be used to understand the movement of water into and out of cells, we might be wise to consider what assumptions we made in this analysis. Obviously we only considered the movement of water across the membrane. We must have assumed that the solute of the solution outside the cell is impermeant to the cell membrane. Likewise, we must have assumed that the cell's range of solutes are all also impermeant to the membrane. We also must have assumed there was no vesicular transport (exocytosis or endocytosis) going on. We further assumed that the volume of the cell is negligible compared to some vast volume for the solution...the gain or loss of water was significant to the cell, but was insignificant to the solution in the outer container. We assumed there was no evaporation from the surface of the solution that would increase the solute concentration during our project. Finally we assumed our project was done at STP (sea level) and out of soil so that we could ignore gravitational potential and matric potential.

How fast does the water move in osmosis?

As osmosis in cases like the cells above is mostly a diffusion problem, we can approximate the rates by using Fick's law:

Js = -Ds • ΔC • Δx-1

where:

Js = rate
Ds = Diffusion coefficient
ΔC = concentration difference
Δx = the distance needed to cross

One way to think of the diffusion coefficient in the case of crossing a membrane is the membrane's "permeability." Some important questions we might ask in that thinking include:

In our worked example we ASSUMED that solutes cannot move into and out of cells. Is that realistic? In our example we used sucrose. Is sucrose permeant?

A phospholipid bilayer can be traversed only by very small compounds such as O2, CO2 and water. It could never be crossed by sucrose. How DO cells acquire molecules of larger size, such as sucrose? Hopefully you are thinking now about:

Another pondering: How might we alter the permeability of a membrane? Perhaps these ideas lead you in your interpretation of our diffusion lab project.