Circuit Training: Water Potential for AP Biology
Name: __________KEY______________
Directions: Beginning with the first cell marked #1, determine what biomolecule or subunit is shown. To advance in the
circuit, hunt for your answer and mark that cell #2. Continue working in this manner until you complete the circuit.
Unless otherwise stated, assume the systems are open and at room temperature (21 degrees celsius).
**Disclaimer, the numbers in the problems below have been selected to highlight the problem solving process involved
in determining water potential and may not be actual value ranges expected in nature.
Ans: -75.155
# __1___
An animal cell is at equilibrium with a beaker of
saline solution with a  = -1.22. If an animal cell
bursts at when its p= 6.5 bars, calculate the
molarity of saline solution that the cell must be
placed in to make it lyse.
Ans: -23.04
# __12___
Because of a cell wall, plant cells have a finite
volume that they can achieve. A plant cell at max
volume with a s =-7.49 is placed in a solution with
s =-0.5. Calculate the resulting p.
Ans: -3.0
# __10___
If a plant is droopy and its cells have a p= 0 bars
with a s = -5.5 and is placed in a solution of 2M
sucrose, what would its resulting  be if allowed to
reach equilibrium?
p= 6.99
= -21.65
Ans: 0.5
# __3___
If the cell from the previous square is placed in a
beaker in which  = -4.0 bars, in which direction
will the net flow of water be? Answer in terms of


 = -4.0
Ans: -23.48
# ___7__
A cell has a s = -1.2 MPa and p= 0 MPa. If 1 MPa
= 10 bars, what is in bars?
Ans: -5.83
# _16____
If a cell’s p= 3 bars and  = -4.0 bars, calculate it’s
s


s = -7.0
Ans: -4.0
# __4___
A cell has a s = -3 MPa and p= 1.5 MPa. If 1 MPa
= 10 bars, what is in bars?
Ans: -21.65
# __11___
If the plant from the previous square was removed
from the beaker it was in and placed in a beaker
with a 1M sucrose solution, calculate the resulting
.
 = -7.72
 = -15
 = -12.0
 = -23.04
Ans: -7.0
# _17____
What is the s of a 2.5M solution of sodium
chloride (salt) at 27 degrees Celsius in an open
container?
s =124.65
Ans: -4.886
# _14____
A mesophyll cell contains 0.2M glucose. If the cell is
at 25 degrees Celsius. If its p= 1 bar, calculate 


Ans: 
# __15___
A cell has a s = -0.72 MPa and p= 0.137 MPa,
determine its in bars.
Ans: 124.65
# __18___
A dialysis tubing of 0.5M NaCl solution is place in a
solution of 2.3M NaCl solution at 30 degrees
Celsius. Calculate the dialysis tubing’s s.
Ans: -12.0
# __8___
The cell from the previous square is placed in a
beaker of water with s=-5.0 bars and allowed to
reach equilibrium. What will be the resulting ?
Ans: 6.99
# __13___
A selectively permeable dialysis tubing containing a
0.7M sucrose solution does not change volume
when put in an open beaker of 0.5M glucose
solution. What is the dialysis tubing’s p?
s= -75.155

Ans: -15
# __5___
If a cell with a solute potential of - 0.25 MPa is
placed in a chamber filled with distilled water. A
plunger is depressed to create a pressure of 0.4
MPa, which direction will the net flow of water be?
Answer in terms of 

 = -0.25
Ans: 
# __9___
A cell with a = -3.0 bars is placed in a 1.2M NaCl
solution at room temperature (21 degrees Celsius),
what direction will will the net flow of water be?
Answer in terms of 

= -3.0
Ans: -0.25
# __6___
A cell at 17 degrees Celsius with 0.5M glucose is
placed in an open beaker with 0.75M sodium
chloride at 12 degrees Celsius. The cell and beaker
are allowed to reach equilibrium. What is the cell’s
resulting ?
= -5.83
 = -23.48
 = -4.886
Ans: -7.72
# ___2__
If a plant cell’s p= 3 bars and its s = -2.5, what is
the resulting 

 = 0.5