Today you will perform several experiments that will demonstrate the important relationship between water and plants. You will measure the water potential of a plant tissue in three different ways and look for effects of wind and photosynthesis on transpiration. Be sure to review water movement in plants in your textbook before coming to lab (Chapters 3 & 4).
Water potential is the potential energy of water in a system. Water will move from a region where water potential is high, to a region where water potential is low. Water potential has several components, depending on the system. In plants, the concentration of dissolved particles (solutes) is an important component of water potential. Water potential decreases as solute concentration increases. Pressure is another component of water potential that is very important in plants. This is because plant cells have walls that allow them to build up pressure with expanding.
Each lab table should work as a group for this lab. Each group has a balance. You will need to use the balance for several things, so start by getting the water potential experiment going. At each table there will be a plant tissue (potato tuber, carrot, etc.) for which that group will determine the water potential (y). Water potential will be determined indirectly, by taking advantage of the fact that water potential has two primary components–solute potential (ys) and pressure potential (yp). For our experiment, y = y s + yp. You will find the sucrose concentration at which your tissue neither gains nor loses weight. Because the pressure potential of the sucrose solution is 0, the water potential of the sucrose solution is equal to its solute potential. When the water potential of two solutions is equal, there is no net movement of water between the solutions. So, the sucrose solution in which the tissue pieces neither gain nor lose weight has the same water potential as the tissue pieces. It is essential that you understand this logic! To check yourself, explain it to your lab partners. If you are at all confused about this, ask your instructor for help.
Solutions with different concentrations of sucrose will be available. Label seven Petri dishes and fill each one with the appropriate sucrose solution. (Try not to spill sucrose; if you do clean up please.) Use a cork borer to remove cylinders of your plant tissue. Slice the cylinders into thin disks, the thinner, the better to increase surface area exposed to the sucrose solution. Use enough pieces for each treatment to get a measurable weight, at least 2g. Do not use portions of tissue that are suberized-the outer layers, i.e., the epidermis or peel (why?). Weigh the slices for each treatment all together, and place them into the Petri dishes. Work quickly, placing tissue in a damp paper towel if necessary. The tissue must not ever dry out. After about 90 minutes, remove the slices, blot them on Kimwipes until they are as dry on the surface as they were before they were soaked, and reweigh them. (While you are waiting for the 90 minutes, set up the experiment described below). Calculate the % change in weight for each group of slices, and plot the % weight changes against the sucrose concentrations, be sure to keep the sign.
|m [Sucrose]||Initial Weight||Final Weight||Change in weight||% change|
Use the space below to plot %weight change against sucrose concentration (Molal).
From your graph, estimate the [sucrose] of the solution where the slices would have neither gained nor lost weight. (You can do this by drawing a best-fit line by eye, or by using a graphing program.) At this concentration, you would expect that there would be no net flow of water, so the water potential of the slices would be the same as the water potential of the sucrose solution.
Now you can use this value to calculate the water potential of the tissue slices:
y = y s = –CiRT where:
C=concentration of the solution expressed as molality (moles of solute/ kg H20) i=ionization constant (i=1 for sucrose)
R=the gas constant (0.00831kg x MPa mol-1 K-1
T=absolute temperature (K)=degrees C + 273
y = .3 MPa
What is the sucrose concentration in which the drop stays still?
This would be the solution that has the same water potential as your tissue. Are your results in this experiment in agreement with the results of the experiment where you weigh the tissue slices? In the space below, explain how this method of water potential determination works. In your explanation, discuss briefly which method (this one or the first one) is more accurate and why.
The purpose of this exercise is to demonstrate effects of light, wind, and temperature on transpiration. Plants growing in a lightweight medium will be used for this experiment. Each lab table will subject a plant to a different regimen. One plant will be placed under a light, one will have a fan blowing on it, one will be covered with a black plastic bag, etc. Each table will be responsible for recording the weight of its plants every 15 minutes.
These types of experiments are of some historical importance because they were first performed in the 1720’s by Stephen Hales, one of the fathers of quantitative plant biology. You should be able to explain the results in terms of what you know about water uptake, transpiration, and photosynthesis. Before you leave the lab, figure out the mean % weight change for each set of plants, and write a brief explanation of why you think the weight changed (see below).
Record the mean weights for each treatment. Why did some plants lose more weight than other plants? Give details!
In field situations, water potential is usually measured with a device called a pressure “bomb”. A twig is cut and placed into the bomb with the cut end exposed. When the twig is cut the water in the xylem vessels is pulled back from the surface and is under tension (negative yw) because of the transpirational pull. The pressure in the chamber is gradually increased, and the exact pressure where water bubbles to the surface of the twig is considered to be the yw of the plant. That is, the positive pressure needed to push the water out of the stem is equal to negative water potential of the vessels. Make sure you understand how this works!
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