1. When placed in a solution with a lower water potential, the appearance of a plant cell through a microscope changes considerably. What are these changes and why do they occur?

SEE LAB REVIEW PAGE FOR ANSWER

2. A plant cell that has no turgor pressure and  an osmotic potential of  -0.3 Mpa is placed in a beaker containing a solution with an osmotic potential of  -0.7 Mpa.

What is the water potential of the cell and the solution prior to immersing the cell into the solution? Show your work/explain you answer.

Water potential = osmotic potential + pressure potential

For the cell,
  pressure potential = 0 MPa (cell has no turgor pressure)
  water potential = osmotic potential + pressure potential
therefore,
  water potential =  -0.3 MPa  + 0 MPa
  water potential =  -0.3 MPa

For the solution,
  pressure potential = 0 MPa (beaker open to atmosphere)
  water potential = osmotic potential + pressure potential
therefore,
  water potential =  -0.7 MPa  + 0 MPa
  water potential =  -0.7 MPa
 

 Wednesday and  Thursday

1. In a simple osmometer like the one you constructed in the lab, water will eventually stop flowing into the dialysis tubing bag containing sucrose. Although the solution in the bag remains hypertonic to the distilled water in which it is immersed, the water potentials are now equal. How did this happen?
 

SEE LAB REVIEW PAGE FOR ANSWER

2. A turgid plant cell that has an osmotic potential of  -0.3 Mpa is placed in a beaker containing a solution with an osmotic potential of  -0.7 Mpa.

What is the water potential of the cell and the solution prior to immersing the cell into the solution? Show your work/explain you answer.

Water potential = osmotic potential + pressure potential

For the cell,
  pressure potential = + MPa (cell has turgor pressure)
  water potential = osmotic potential + pressure potential
therefore,
  water potential =  -0.3 MPa  + (+) MPa
  water potential >  -0.3 MPa

For the solution,
  pressure potential = 0 MPa (beaker open to atmosphere)
  water potential = osmotic potential + pressure potential
therefore,
  water potential =  -0.7 MPa  + 0 MPa
  water potential =  -0.7 MPa