Question

A cell is immersed in water at 0 ^o C for a considerable period of time. It is then taken out and the contents of its vacuole are found to be approximately 0.2 M sucrose [assume this is the overall molarity for the cell].

            (a) At equilibrium, what was the cell's Y_p?

            (b) What would its Y_p have been had it contained 0.2 M NaCl?

Answer 1

Here water where is placed is a hypotonic to the cell concentration and these two solutions are separated by semi-permeable membrane of cell . The solution that is hypertonic to the other must have more solute and therefore less water. At standard atmospheric pressure, the water potential of the hypertonic solution is less than the water potential of the hypotonic solution, so the net movement of water will be from the hypotonic solution into the hypertonic solution. The solute potential (Ys) 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:

Yp = - 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.

at 0^oC R *T = 2.271 MPa Mol^-1

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

Water potential in cell simplifies to,

Y = Ys + Yp

Since cell is about 0.2M at 0^oC

Solute potential at 0^oC Ys = 0.2 * -2.271 MPa Mol^-1 = -0.4542MPa

Now we will assume the cell is full but has no turgor pressure initially. Here is how that cell would look: Ys = -0.4542MPa

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 -0.4542MPa,

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:

Y = Ys + Yp

Y = -0.4542MPa + 0.4542MPa

Ys= 0 MPa

Yp= 0 MPa

Y= 0 MPa

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

Answer 2) now when water contains 0.2M of salt

Ys = - 0.2* 2.271 at 0^oC = - 0.4542 MPa

Since here cell contains 0.2M of sucrose and outside solution contains 0.3M NaCl , both the solutes can not cross the membrane water will now leave the cell and moves from area of greater to lesser water potential ,

Ys = - 0.4542 MPa,

From the above equation cell potential is Yp= -0.4542MPa at equilibrium,

Therefore here also, Y = Ys+ Yp = 0