BIOL 221 – Concepts of Botany Fall 2010
A. Water Relations
Water plays a critical role in plants. Water is the universal solvent that allows biochemical reactions to occur in all organisms, but that is not the only importance in plants. Water moves from the roots to the shoots and to every cell in between. Water diffusion across membranes
( osmosis ) results in the development of turgor pressure (positive pressure) in plant cells. The turgor pressure is what keeps the primary plant body upright on land and when it decreases, the plant wilts (cells become flaccid ). In addition, the turgor pressures in young developing plant cells are required for the expansion of these cells. Therefore, growth of the plant depends on the movement of water to growing apical regions of the shoot and roots.
In addition, water movements facilitate nutrient transfer (minerals) from the roots via xylem to the living shoot tissues that need them. Water in the phloem keeps important photosynthate compounds in solution for transport from sources (photosynthetic leaves) to sinks (nonphotosynthetic organs, rapidly growing regions). So obviously water is extremely important to the health of a plant. In today’s lab you will have a chance to investigate the importance of water uptake and movement at the cellular and whole plant levels. Enjoy!
Water moves through the plant continuously due to severe differences in water potential between the soil, the plant and the atmosphere. The xylem vascular tissue takes advantage of the physical properties of water ( surface tension, cohesion, adhesion ) in order to move water from the roots to the shoot. No energy is expended in this process but an important sacrifice is made: that is, water must be lost through transpiration (transpiration is the loss of water vapor through stomata). Water availability is a major limiting factor for growth of a plant, and if the rate of transpiration exceeds the uptake of water from the soil, then wilting can occur. In the wilted state, the plant cells do not have positive pressure in them and thus cannot grow. So the need for regulation of transpiration is present.
Guard cells regulate the flow of water out of primarily the leaves (where most stomata are), but due to the nature of photosynthesis, guard cells need to be open for CO
2
to be available.
This interaction results in a complex regulation of stomatal aperture in order to maximize photosynthesis while minimizing water loss.
Like it or not, water is going to be lost (and a large amount of it at that) but the rate will depend on other environmental factors in addition to the anatomy of the plant.
Fig. 1 Water movement from the xylem to the atmosphere.
Concepts of Botany, (page 1 of 9)

C. Water Potential gradients drive water movement in plants.
Water moves in the direction of high to low water potential and we can speak of the water potential (

W
) at any point along that continuum as consisting of two components as follows:

W =

S +

P
The simple rule is that water flows from an area of high

W to low

W
(that is, down the
 gradient). As a reference point, pure water at atmospheric pressure has a

W of zero.
W

S refers to solute potential (also known as osmotic potential). Pure water has a solute potential of zero. As solute is added, the value for solute potential becomes negative and then more negative. This causes water potential to decrease also. All else being equal, as solute is added, the water potential of a solution drops, and water will tend to move into that solution from its surroundings, down the water potential gradient created.

P
refers to pressure potential which corresponds to any mechanical pressure placed on a volume of solution. In a plant cell, pressure exerted by the rigid cell wall against the expansive pressure of the plasma membrane ( turgor pressur e) increases the

W
thereby limiting further water uptake.
Let’s describe how this relates to water transport locally across cellular membranes
(osmosis):
Scenario 1 : If you have two adjacent cells in some tissue, both of which have the same solute concentration and the same mechanical pressure applied to each cell, there will be no

W gradient between the two cells and therefore no net flow of water between them.
Scenario 2: When solutes like sodium ions (Na+) or chloride ions (Cl-) enter a cell through some mechanism (the mechanism here is irrelevant), the total solute concentration increases in the cell, thereby making the

S of that cell more negative and, all else being equal, decreasing the

W
of that cell. If the

W
inside that cell becomes lower (i.e., more negative) than outside that cell, water will flow into the cell, down the

W
potential gradient.
Scenario 3 : Two adjacent cells, same solute concentrations but you squeeze one (exerting a positive pressure to one cell’s volume), you will increase the squeezed cell’s
 other and there will be a net flow of water down that

W relative to the
W gradient until equilibrium is reached.
Scenario 4: One cell surrounded by some atmosphere: The solute potential of the atmosphere is lacking (no solution to speak of) yet the atmospheric pressure is relatively extremely negative relative to the intracellular pressure (turgor pressure). Thus, water will tend to flow out of the cell. Plants, however, have a cuticle to protect them from loss of water across membranes in this way. Instead, water loss from cells is restricted primarily to the atmosphere of intercellular spaces and then further water loss from these spaces to the external atmosphere is restricted by the guard cells surrounding stomata.
Let’s describe how this relates to water transport globally (bulk flow, transpiration stream) along the soil-plant-atmosphere continuum:
From leaf to atmosphere: As long as stomata are open, there will be water lost from intercellular spaces of the spongy mesophyll primarily b/c steep water potential gradient between the humid leaf interior spaces and the external atmosphere with its extremely negative

W .
Concepts of Botany, (page 2 of 9)

From cell to intercellular spaces in the leaf: Subsequently, this transpiration reduces the
 of the intracellular spaces and this pulls water from mesophyll cells down the

W gradient
W created.
From xylem to mesophyll cells in the leaf: This steepens the high-to-low

W gradient from leaf xylem across the mesophyll cells to the intracellular spaces. As water is lost from mesophyll cells to adjacent intercellular spaces, the
 decreased turgor. In the xylem,

S
S becomes more negative and the

is highest (closest to zero)

P
P decreases due to
is higher than in the cells further down along the continuum that have lost water (turgor) to the intercellular space. Thus, water is effectively pulled from the xylem down a

W gradient.
Bulk-flow up xylem vessels or tracheid-paths: An negative pressure is thus created in the xylem and water is effectively pulled up the xylem vessels from the roots.
Water –flow root from soil:
Diffusion is a major driver of water flow to the roots. Osmosis occurs as water moves across the cell membrane into the cytosol. Root cells are packed with solutes relative to interparticulate spaces in the soil. Moreover, when transpiration is occurring, there is a negative pressure in the xylem that pulls water from the surrounding cortical cells of the root.
D. The Transpiration Stream:
1.
Observe the plant material that will be used in lab today. Notice how long the petiole of the leaf is. Add a volume of 0.1% toluene blue into a glass test tube that is sufficient to submerge the cut end of the leaf petiole (as described in steps 2 and 3 below).
2.
Freshly cut a leaf at the base of the petiole from the provided plant source.
3.
Immediately insert the cut shoot into the toluene blue and place the tube in the rack near the high light environment.
4.
Allow the treatment to continue for a minimum of 30 - 40 minutes.
5.
Observe the plant. Look for the presence of stain. Answer the following questions: a) Measure the furthest distance (into leaves) that the dye is visible (measure in cm). b) Can you discern a pattern regarding the presence of the dye in the leaves or stem?
Describe it. c) Cut a section of the shoot/petiole about an inch from the original cut and look at it with
a dissecting scope.
Concepts of Botany, (page 3 of 9)

1) Do you see the vascular tissues? How can you tell them apart from other tissues?
2) Describe the dye pattern present in the stem/petiole.
3) Draw what you see:
4) Explain how the dye pattern relates to your knowledge of vascular organization. . e) Cut thin sections of various leaves – both stained and apparently unstained and
1) What staining pattern can you detect?
2) Draw what you see:
2) Is there any variation between leaves?
3) Is dye present in all of the leaves? Explain why.
Concepts of Botany, (page 4 of 9)

E. “Transpirometer” Measurements under Environmental Conditions
1. Remove the plunger and needle (if present) from a 1ml syringe.
2. Seal the plastic “needle end” with parafilm.
3. Add 0.8ml of water to the sealed syringe using a second syringe with needle.
4. Place the syringe in a glass test tube

this is your transpirometer!
5. Cut a leaf from the supplied plant material and IMMEDIATELY place it into the prepared syringe. Do this very quickly. If you take too long, air will enter the xylem = problems with water uptake.
6. Label the tube so you can recognize it (don’t cover the number scale on the syringe).
7. Prepare a total of four transpirometers as described above.
8. Place one plant/transpirometer set-up in each of the conditions below for 60-75 minutes: a. High Light (back bench light apparatus) b. Medium Light (on your lab bench) b. Darkness (place in cabinets under lab benches) c. Medium Light + Wind (fan station)
9. Record the amount of water lost (beginning volume minus final volume).
10. Weigh the leaf tissue for each experiment (do this last) and calculate area as cm
2
.
Hint: Determine total leaf area by first calculating the weight/cm
2
of the bean leaf by cutting a square leaf section 3 cm X 3 cm and weighing this leaf section. Divide the leaf section weight by 9 cm
2
to find the weight of 1cm
2 leaves by the mass of 1cm
2
section of leaf. By dividing the total mass of the
, you will determine the surface area in cm
2
of the leaves on your bean plant.
11. Calculate the rate of transpiration as ml/min/cm
2
.
Transpiration Rate Data:
High
Light
Low
Med Light
Dark
Wind
Starting H
2
O
Volume
Final H
2
O
Volume
H
2
O Lost
Time
Total Leaf Weight
Weight/cm
2
Total Leaf Area
Transpiration
Rate (ml/min/cm
2
)
Concepts of Botany, (page 5 of 9)

16. NOW, plot the transpiration rates for each treatment on the graph below. Determine what graph format would be the best for comparison. Label the graph appropriately.
Which environment caused the most water loss? Why?
How does light environment affect transpiration?
How does a windy environment affect transpiration?
Why does total surface area need to be considered for all treatments?
When would you expect the most transpiration to occur naturally?
What plant adaptations can you think of that might decrease leaf water loss?
Concepts of Botany, (page 6 of 9)

F. Turgor Pressure at the Cellular Level
1.
Prepare 8 beakers with 20ml of the following sucrose solutions:
0 M; 0.2M; 0.4M; 0.6M; 0.8M; 1.0M; 1.2M; 1.4 M
2.
Using the vegetable or fruit available on your table, prepare five 1cm
3
squares for each solution (for a total of 55 squares). Use the French fry cutter to make sticks and cut these sticks in 1cm lengths to obtain 1cm
3
squares. Do not use squares that have skin or dead spots on them.
3.
Weigh the squares prior to adding to a solution beaker (this is the Wt
I
). Be sure to correlate weight to the specific solution.
4.
Imbibe the tissues in the solutions for a minimum of 60 minutes.
5.
After 60 minutes, remove the squares, blot dry and weigh them again (this is the Wt
F
).
Record this weight in the provided table.
Wt
I
Wt
F
% Change Wt
I
Wt
F
% Change Wt
I
Wt
F
% Change
0 M
0.2 M
0.4 M
0.6 M
0.8 M
1.0 M
1.2 M
1.4 M
Wt
I
Wt
F
% Change Wt
I
Wt
F
% Change Wt
I
Wt
F
% Change
0 M
0.2 M
0.4 M
0.6 M
0.8 M
1.0 M
1.2 M
1.4 M
6.
Calculate the % weight change for each solution: (Wt
F
– Wt
I
)/Wt
I
Concepts of Botany, (page 7 of 9)

7.
Graph your results below. Report your findings to the rest of the class.
8.
Record and graph the results from the other groups.
What are the hypotonic and hypertonic solutions for your specific tissue?
What are the isotonic points for each different tissue?
Which tissue had the highest amount of solutes? Explain your rationale.
Which tissue had the lowest amount of solutes? Explain your rationale.
G. Relationship between Solute Content, Solute Potential and Water Potential:
As previously described, the concentration of solutes has a strong impact on solute potential and therefore water potential. The equation below shows the relationship of water potential to solute potential:
Concepts of Botany, (page 8 of 9)


W
=

S +

P
We can simplify this equation for today’s lab due to the use of containers/solutions open to the environment so that it looks like this:

W
=

S
Where water potential equals solute potential (this is a unique situation to open containers).
Solute potential can be calculated with the following equation:

S
= miRT
Where: R = gas constant = 0.00831 L*MPa/°K*mol
T = temperature in °K (= 273 + °C) (assume room temperature = 24°C) m = molarity of solution i = dissociation constant (equals 1 in today’s lab using sucrose)
1. Calculate the solute potential for each solution used in the previous experiment:

S
0.0 =
0.2 =
0.4 =
0.6 =
0.8 =
1.0 =
1.2 =
1.4M
2. Which solution come was closest to the isotonic point?
The isotonic point is the point where equilibrium exists between the solution and the tissue, and based on our experimental setup. This means that the solute concentration in the tissue is equivalent to the solute concentration in the isotonic solution. Therefore, the solute potentials of the solution and the tissue are equivalent.
3. What is the solute potential of your tissue?
4. What is the water potential of your tissue (remember that at the isotonic point, no pressure gradient exists).
Concepts of Botany, (page 9 of 9)