Download MEASURING WATER POTENTIAL IN POTATO TUBERS

Grade Level/Subject
AP Biology
Unit
Big Idea 2
Enduring
Understanding
2.B. Growth, reproduction and dynamic
homeostasis require that cells create and
maintain internal environments that are
different from their external environments.
2.B.1 Cell membranes are selectively permeable
due to their structure
Essential knowledge
Title
Determine the water potential of a potato tuber
Learning Objective
2.10 The student is able to use representations
and models to pose scientific questions about
the properties of cell membranes and selective
permeability based on molecular structure.
2.11 The student is able to construct models
that connect the movement of molecules across
membranes with membrane structure and
function
2 (Question and Methods Given; Solution Open)
Inquiry Level
Materials Required
1 M sucrose solution diluted to 0.1, 0.2, 0.3,
0.4, 0.5, 0.6 and 0.7 M; potato, knife, beakers,
scale in g, timer, paper towels,
MEASURING WATER POTENTIAL IN POTATO TUBERS
OBJECTIVE
The objective of this lab is to measure the water potential of potato tuber tissues. We will
monitor water uptake or loss by tissues in different sucrose solutions to identify an
isotonic solution with water potential equal to that of the tissue.
INTRODUCTION
Water potential () is a measure of the driving force that governs the movement of water
from the soil into plants and finally into the atmosphere. Water potential is the amount of
energy per unit volume (or pressure) contained in a system (like a plant cell, tissue, or
soil) and is expressed in units of megapascals (Mpa). For reference, pure water in a free
standing solution has a water potential of zero.
Water potential of a plant cell is made up of two important components, and the
relationship among these components is expressed mathematically as:
 = s + p
 is the overall water potential of a cell.
s is the solute or osmotic potential and represents the contribution made by dissolved
solutes to . Adding dissolved solutes to a system always decreases water potential, thus
this component is always negative. In a plant cell, important contributors to solute
potential include mineral ions, sucrose, starch, amino acids, proteins, and anything else
that can accumulate to high levels in the cytosol or vacuoles.
p is the pressure potential and represents the contribution made by pressure to . Fully
turgid cells whose plasma membranes are pressing against the cell wall have a positive
p. Cells at incipient plasmolysis (the point at which the membrane is just barely
touching the cell wall) have a p of zero. Cells under tension, like those in the xylem
during active evapotranspiration, have a negative p.
The objective of this experiment is to measure the  of potato tubers. When placed in a
free standing sucrose solution, water will move into or out of a plant tissue depending
upon its water potential relative to the solution. Gain or loss of water can be detected by
weighing the plant tissue before and after immersion in the solution. By incubating
tissues in a series of sucrose solutions of different concentrations, the solution that causes
no change in tissue weight can be identified. The water potential of this isotonic solution
is assumed to equal the water potential of the tissue.
PROCEDURE
1) Pick up one of the sucrose concentrations:0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7 M (mol/L
solution) sucrose. Add 75 mL of each solution to the appropriate labeled beaker using a
clean graduated cylinder for each solution.
2) Obtain 2 pieces of potato without skin, trim to 4 cm. Work quickly to prevent the
tissues from drying out as you cut them. What problems would tissue drying cause in the
experiment?
3) Quickly blot the potato pieces on paper towels to remove any excess moisture and
weigh them together, sets of two, to the nearest 0.01g. Record the weights in Table 1.
After weighing, quickly transfer them to the sucrose solution beaker and ensure that they
are fully immersed.
4) After 45 minutes, remove the potato pieces, blot excess moisture with paper towels,
and reweigh them in sets of two, exactly as in step 3. Record the weights in Table 1.
Calculations
Complete the following in your noteboooks.
1. Subtract the initial tissue weights from the final weights. Second, divide the
difference by the initial weight and multiply by 100 to get the percent weight
change. Make Table 1 in your notebooks.
Molarity [Sucrose]
Initial weight
Final weight
 weight
%  weight
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
2. Create a graph by plotting the percent change in weight (y-axis) vs. sucrose
concentration (x-axis) in your notebook. Calculate the regression line (y=mx+b).
Determine the exact concentration of sucrose that would cause no change in
weight in the potato tubers (isotonic). The water potential of this solution will
equal the water potential of the potato tissue.
3. In an open solution where there is no turgor pressure, therefore p is equal to
zero. Thus, the  of a solution is equal to the s of a solution.
 = s + 
Calculate the s of the solution causing no change in weight of the potato tissues
using the following formula:
(Eq. 2) s = -miRT
m = mol L-1
i = ionization constant = 1 for sucrose
R = gas constant = 8.31 kPa L K -1 mol -1
T = room temperature in K (ºC + 273 = K)
Table 1. Weight change in potato tissues in sucrose solutions of different concentration.
Due next week in lab:
1. Excel graph with regression line.
2. Table 1.
3. Value of s with calculations shown.
4. Answers to the following questions:
 Why did we use sucrose as the solute in our solutions? How might using another
solute influence the results?
 Tissues in which treatments have a water potential equal to that of their solution
after the incubation period? How can you tell?
 What influence would increasing temperature have on our calculation of water
potential?
 The results of the calculations of water potential usually vary among years, among
lab sections, and even among groups within one lab. Why might this be?
 What deficiencies were present in our experimental design? Disregarding human
error, how could even better data be obtained?