SEF031 Form and Function in Biology
Workshop: Water Potential in Action
Dr. Angelika Stollewerk
Family Name
First Names
Student ID (9-digits)
1. Water potential and the movement of water between cells
(20 marks)
The water potential (Ψ) of a cell is a measure of the tendency for water molecules to
enter or leave the cell. Pure water has the water potential (zero), and the effect of
dissolved substances is to lower the water potential. Solutions (at atmospheric
pressure) have negative values of water potential. Water diffuses from regions of high
water potential (less negative Ψ) to regions of lower water potential (more negative
Ψ). In figure 1, the solute potentials (Ψs) and pressure potentials (Ψp) are shown in
kilopascals.
Cell A
Ψs = -130
Ψp = 50
Cell B
Ψs = -110
Cell C
Ψp = 50
Ψs = -100
Ψp = 30
Figure 1. Three adjacent plant stem cells.
a. By means of an equation, state the relationship between water potential (Ψ) and
the solute potential (Ψs) and pressure potentials (Ψp).
(1 mark)
water potential = solute potential + pressure potentials
OR
Ψ = Ψs + Ψp
b. What are the water potentials (Ψ) of cells A, B and C?
(3
marks)
A: R = -130 + 50 = -80 kilopascals
B: R = -110 + 50 = -60 kilopascals
C: R = -100 + 30 = -70 kilopascals
one mark for each correct answer but subtract one mark if there are no units given
c. What will be the direction of net water movements between these three cells (until
equilibrium is reached) and why?
(5 marks)
From B to A and C (2)
From C to A (1)
because there is a net flow of water from higher water potential (less negative) to lower water
potential (more negative). (2)
Two adjacent leaf cells of the same plant have initial solution potentials and pressure
potentials (values in kilopascals) as indicated in Figure 2.
Cell D
Cell E
Ψs = -
Ψs = -
1400
2100
Ψp = 600
Ψp = 900
Figure 2. Two adjacent plant leaf cells.
d. At equilibrium, what will the water potential of the two cells be?
(4
marks)
D: Ψ = -1400 + 600 = -800kilopascals (1)
B: Ψ = -2100 + 900 = -1200kilopascals (1)
At equilibrium the water potential will be (-800 + -1200)/2 = -1000 kilopascals (2)
Subtract one mark if no units are given
e. Why would you expect leaf cells in the light to have lower (more negative) values
solute (Ψs) than those of the stem cells?
(4 marks)
Water potential of the cells of the plant decreases (2) as you go from the roots to the leaf
because of increasing amounts of dissolved substances (sugars, other soluble metabolites
including ions). (2)
f. Animal cells do not have a cell wall to protect them if the pressure potential rises.
How is raised pressure potential avoided in:
i. Amoeba;
(1
mark)
The contractile vacuole.
ii. a mammalian red blood cell;
(2
marks)
the fluid around the red blood cells is kept isotonic (internal and external water pressures are
the same) with the fluid within.
Total:
/20 marks
2. The evidence for active uptake of ions by cells
(19 marks)
The process of uptake of ions by tissues is often studied in plants using 'tissue discs'.
These are thin slices of tissue, about 1 mm thick and typically between 0.5 and 1cm in
diameter, cut from cylinders of storage tissue such as carrot root, potato tuber or
beetroot. After cutting, the discs are washed in running tap water for about three
days, prior to use in experiments. During this time, the metabolic rate of the tissue
increases markedly, and the tissue may develop a respiration rate comparable to that
of immature, fast growing tissues. Samples of about 500 discs may constitute a
typical sample, although smaller quantities are used when the reaction vessel is small,
such as a Warburg flask.
a. What are the advantages to an experimental physiologist of the choice of thin discs
of tissue, rather than larger blocks or long cylinders of tissue?
(2 marks)
Most cells are in close contact with the bathing solution or greater surface area to volume
ratio (1)
conditions are uniform for the experimental cells. (1)
b. What are the possible causes of the increase in metabolic rate of the cells of tissue
discs over that of comparable cells in the intact storage organ?
(2 marks)
Cutting of tissue discs causes damage (1) this is to repair the damage caused by cutting (1).
c. Why are large samples consisting of hundreds of discs often preferred in
experiments?
(2 marks)
To rule out the effects of unrepresentative discs (e.g. discs containing much vascular tissue
and fewer metabolising cells.)
In one investigation, samples of storage tissue discs were incubated in a medium
containing phosphate ions at 0.005 mol dm-3, and aerated with air or nitrogen. After
24 hours it was found that the uptake of ions per gram of fresh mass was as follows:
Uptake of phosphate/µmol l
Tissue treated with air
Tissue treated with nitrogen
33
4
d. What use are phosphate ions to actively metabolising cells?
(3
marks)
To make ATP (1) or nucleic acids (1) or phospholipids (1) if growth is occurring too.
e. What is the likely effect on the metabolism of discs of bubbling nitrogen gas
through the medium, rather than air?
(2 marks)
Nitrogen will replace oxygen in the solution (1) and will prevent aerobic respiration (1).
f. How might you explain the effect of the absence of oxygen on phosphate uptake by
discs?
(2 marks)
Aerobic respiration\production of ATP (1) is required for the active uptake of ions like
phosphate. (1)
In another investigation, samples of storage tissue discs were incubated in a medium
containing sodium chloride solution at 0.2mol dm-3, with some samples of discs held
at 25oC and others at 4oC. After four days it was found that the uptake of ions was as
follows:
Discs at 4oC
Discs at 25oC
Uptake of sodium ions/ µmol
per sample of five discs
Uptake of chloride ions/µmol per
sample of five discs
82
46
180
144
g. Examine the results carefully. What do they suggest about absorption of NaCl by
the tissue discs? What do they suggest about the role of metabolism in ion uptake?
(4 marks)
Ion uptake is a selective process (e.g. more Na+ than Cl-) (1)
It is dependent upon metabolic energy (ATP) (1)
It occurs faster at 25°C than 4° (1) because the reactions of metabolism are faster at 25°C than
4° (1)
h. When experiments of this type are repeated today, the physiologist often includes
antibiotic in the medium. Why is this thought necessary? What effects might
microorganisms have in these experiments?
(2 marks)
Microorganisms themselves take up ions from the medium for use in metabolism (1)
Microorganisms may attack the cells of the tissue discs, altering their metabolism (1)
Total:
/19 marks
3. The water potential inflorescence stalk
(11
marks)
In this investigation narrow strips of tissue, about 2mm wide, were cut from a length
of hollow stalk of a dandelion (Taraxacum officinale) inflorescence, about 2.5cm long.
Some of these short, slightly curved strips were then immersed in sucrose solutions of
different concentrations, and one was immersed in distilled water. The changes in
shape that occurred were observed, and the shapes of the strips after 5mins of
immersion were recorded. The experimental procedure and some of the outcomes are
recorded in Figure 3. Examine Figure 3 and then answer the questions below.
Figure 3. Investigating water potential in inflorescence stalk cells.
a. Care was taken to cut all the strips of stalk tissue from the same region of the
flower stalk. What was the purpose of this precaution?
(2 marks)
So that the physiology of all tissue strips was comparable (e.g. all the cells had the same initial
Ψ). (2)
b. What type of cells form the outer margin of the stalk, and how do they differ in
structure from the cells of the outer cortex (the remaining cells of the tissue strips)?
(2 marks)
The outer margin of the stalk is epidermis. (1) Epidermis is a continuous compact layer of
cells on the surface of the plant body. (1)
c. What would be the shape of the tissue strip that was placed in distilled water, after
about five minutes of immersion?
(2 marks)
Curled almost into a ring (1) exposing more of the inside (1).
d. What changes have occurred in the cells of the tissue strip immersed in 0.5mol dm-3
sucrose solution to cause the observed change in shape?
(1 marks)
Loss of water from the thin-walled inside cells.
e. What is the approximate water potential (Ψ) of the stalk cells? Explain your
reasoning.
(2 marks)
Equivalent to the water potential of the 0.3 mol dm-3 solution (1) since least change of shape
has occurred here (1).
f. If the experiment were repeated with a stalk from a dandelion plant growing high
up on an exposed wall after a period of low rainfall, what difference in the result would
you anticipate?
(2 marks)
More solutes would be concentrated in a decreased volume of cell sap (1), so the water
potential will be higher (1).
Total:
Total Marks:
Percentage =
/50
/11 marks