laporan fistum unit i
TRANSCRIPT
Measurement of Osmotic Potential () by Incipisnt Plasmolysis
GROUP III
Yunandar/1114040181Surhayanti Amir/1114040192
Sri Vianita/1114040195Sri Wahyuningsih/1114040199
BIOLOGY ICP
UNIVERSITAS NEGERI MAKASSAR TAHUN AJAR 2012/2013
CHAPTER IINTRODUCTION
A. Background
Water potential (Ψw, psi), which is a measure of the energy state of water
is affected by dissolved solutes, pressure and matrix particles. The contribution to
water potential by dissolved solutes, termed osmotic potential (Ψs ), is always
negative in sign. In other words, solutes decrease the water potential. The
contribution of pressure (Ψp) may be positive, negative or zero, but is generally
positive since most plant cells are turgid (turgor pressure). The contribution due to
the binding of water to colloidal particles (matric) and surfaces, termed matric
potential (Ψm), also lowers the water potential. Although it is often small enough
to be ignored, matrix potential is important when considering soil water relations.
Thus, the water potential of a plant system can be arithmetically represented by
the equation:
Ψw = Ψs + Ψp + Ψm
B. Purpose
Measurement of osmotic potential by incipisnt plasmolysis
CHAPTER IIBASIC THEORY
Now, if we were to take two containers of water, and separate them by a
semi-permeable membrane (one that allows water to go through its pores, but not
solutes like salt or sugar), and add sugar to one side, this would result in a
lowering of the kinetic energy of the water-sugar solution. Thus, from a statistical-
probability point of visw, we would expect the molecules of pure water to
encounter the membrane more often than the lower energy water molecules on the
solution side, and thus, over time, water will move from the pure water to the
solution. Of course some water molecules do go the other way, but the net
exchange favors movement into the solution. This is known as osmosis. It is a
special case of diffusion (Anonymous, 2013).
If a cell is placed in a solution which has a Ψ that is higher than that of the
cell, there will be a net movement of water into the cell. However, if the
surrounding solution has a lower Ψ than in the cell, there will be a net movement
of water out of the cell. If this latter situation continues, the plasma membrane and
cytoplasm will pull away from the cell wall, a condition known as 3 plasmolysis.
By trial and error, a concentration of bathing solution can be found that just
produces plasmolysis, and this is known as “incipisnt plasmolysis”. Thus,
incipisnt plasmolysis is defined as when @50% of the cells are plasmolyzed. At
incipisnt plasmolysis, there is no longer a pressure potential exerted by the wall
(i.e., Ψρ = 0), and therefore, under that condition, Ψ = Ψп. It should also be noted
that for solutions, Ψ = Ψп. A solution which just causes incipisnt plasmolysis thus
has a water potential (and osmotic potential) of the cell cytoplasm. Finally, since
the cell we use are highly vacuolated, it can also be assumed that the osmotic
potential of the cell is basically the vacuolar osmotic potential (Ismail, dkk. 2013).
Thus, the solution will increase in volume, and become more diluted. Over
time, this will slow the flux of water into the solution, but not stop it entirely.
However, eventually, the weight of the water will exert a backpressure on the
solution, which, if given enough time (and large enough container) will increase
the pressure on the membrane and force water molecules to go back into the pure
water. If the pressure is great enough, it can totally balance the number coming in,
and the net flux of water will cease. The amount of pressure needed to totally
balance the flows of water is known as the osmotic pressure and symbolized
aswith units of pressure (e.g., pounds per square inch, atmospheres, bars,
Megapascals) (Anonymous, 2013).
A slightly more complex theory that is often found in general biology
books (including your text, p. 117) is the “bound water” explanation. This says
that any hydrophilic solute (like sucrose or NaCl) will bind up hydrating water
and prevent it from moving freely. Therefore, the side of a semipermeable
membrane with pure water has a higher “free” water concentration than the side
with the solute molecules. According to this explanation, “free” water moves into
hypertonic solutions simply because it is diffusing down its concentration
gradisnt. Although it is popular in introductory texts, this theory is not even
mentioned in several revisws (Baumgarten and Feher, 1998; Weiss, 1996, pp.
216-222).
If the bound water explanation were true, we would expect that a greater
mass of hydrophilic solute would bind more water. Whether a certain mass of
solute is present in a few large molecules or in many small ones shouldn’t matter.
Also, when predicting osmosis, we would have to carefully consider how
hydrophilic the solute is (that is, how many water molecules it binds per
molecule). In fact, the number of molecules present does affect osmosis, and we
can predict osmosis without considering how hydrophilic the solute molecules are
(Anonymous, 2013).
Water is a simple molecule, consisting of one atom of oxygen (0) and two
hydrogen atoms (H), so that the molecular weight of only 18 g / mol. In spite of
the simplicity of the composition of the constituent atoms and small molecular
size, the water molecule has several unique characteristics. These characteristics
caused by a seriss of two H atoms on atom 0 (in center) do not form a straight
line. This circuit makes an angle of 1050. The magnitude of this angle is always
the same if the water is in solid form (ice), but rather variss if water is in liquid
form, although the average angle remains 1050. Water can dissolve more types of
chemicals compared to other liquids (Lakitan, Benjamin. 2011).
If plant cell are placed in pure water, water will initially move into the cell.
After are period of time the cell will become turgid. Turgor pressure is the
pressure exerted against the cell wall by contents of the cell. At first most water
movement is into the cell. As the turgor pressure increases water will begin to
diffuse out of the cell at a greater rate, eventually equilibrium will be reached and
water will enter and leave the cell at the same rate. This stage is used to find the
water potential of a particular cel (Anonymous, 2013).
Intake or water net expenditure by a cell occurs by osmosis, is passive
transport of water through a membrane. The combined effect of these two factors
solute concentration and pressure are called water potential. In the water potential
is important to understand is the water will move through the membrane from a
solution with high water potential to a solution with a lower potential IAR.
Components potential in water potential refers to the potential energy, which is
the capacity to perform work when water moves from areas with higher to areas
with lower (Campbell, 2000).
Potato cell contain polysaccharides starch and glycogen they are good for
storage. The potato cell is surrounded by plasma membrane it is a fluid mosaic
model, which is mosaic of phospholipids and proteins moving around they are not
solid. This is why plant cell can become turgid and flaccid because their walls
(plasma membrane) can stretch. The plasma membrane is a selectively permeabel
barrisr between the cell and the extra cellular environment. Water enters in the
cell through phospholipids (Anonymous, 2013).
Like molecular diffusion and pressure –deriven bulk flow, osmosis occurs
spontaneously in response to a driving force. In simple, diffusion, substances
move down a concentration gradisnt; in pressure-driven bulk flow, substances
move down a pressure gradisnt; in osmosis, both types of gradisnts influence
transport, he is say the direction and rate of water flow across a membrane are
determined not solely by the concentration gradisnt of water or by pressure
gradisnt, but by the sum of these two driving (Finkelstein, 1987).
CHAPTER IIIPRACTICUM METHOD
A. Place and Date
Day / date : Wednesday, March 21st 2013
Time : 10.50 Wita – 12.30 Wita
Place : Biology Laboratory third floor at west FMIPA UNM
B. Tools and Materials
1. Tools
a. 12 reaksion tube
b. Reaksion tube rack
c. Marker pens
d. Graph paper
e. Kitchen knives
f. 10 cm3 Pipettes
g. 100 cm 3 beaker
h. Pasteur pipette
i. Petri dish
2. Material
a. 0.10 M, 0.15 M, 0.20 M, 0.25 M, 0.30 M, 0.40 M Sucrose solution
b. Methylene blue
c. Potato tissue
C. Work Procedure
Chardakov’s Method
1. Prepare 12 reaksion tube and reaksion tube rak.
2. Give the label each of reaksion tube based the contentration sucrose
solution is filled ex 2 tubes labelled 0.10 M, 2 tubes labelled 0.20 M etc....
3. Filled reaction tube with sucrose solution ex 2 tube for 0.10 M, 2 tube for
0.20 M etc...
4. Prepare the potato tissue with the larger is 0.5 CM
5. Inserted potato tissue into each of reaction tube.
6. Add 1 drop of Methylene blue solution to each tube with potato tissue in
and mix it.
7. Leave the tubes for 15 minutes.
8. Take methylene blue with a Pasteur pipette done with caution. enter
pasteur pipette containing methylene blue into a test tube containing a
solution of sucrose. Pasteur pipette try right in the middle of the reaction
tube. Methylene blue spray and observe the movement of methylene blue
in the test tube. do these activities in test tubes containing 0.10 M, 0.20 M
etc .....
9. Observe the movement of methylene blue in the test tube, whether
methylene blue floats, float or sink.
Gravimetric Technique
1. Prepare 6 petri dish.
2. Give the label each of petri dish based the contentration sucrose solution is
filled ex 1 petri dish labelled 0.10 M, 1 petri dish labelled 0.20 M etc....
3. Filled petri dish with sucrose solution ex 1 petri dish for 0.10 M, 1 petri
dish for 0.20 M etc...
4. Prepare the potato tissue with the larger is 2 CM as 6 piece.
5. Weight potato tissue.
6. Inserted potato tissue into each of petri dish.
7. Leave the petri dish for 15 minutes, swirling occasionally.
8. Weigh the potatoes after being soaked in a solution of sucrose.
9. Make a table of initial weight, final weight, and final weight/initial weight
for each concentration.
CHAPTER IVRESULT
A. Result of Practicum
Chardakov’s Methods
CONCENTRATIPON OBSERVATION
0.10 +
0.15 +
0.20 +
0.25 +
0.30 +
0.40 ++
Gravimetric Technique
Concentration
Berat Berat Akhir
Berat awalAwal Akhir
0.10 M 0.30 0.30 0.30/0.30 = 1
0.15 M 0.35 0.30 0.35/0.30 = 1.16
0.20 M 0.30 0.30 0.30/0.30 = 1
0.25 M 0.30 0.30 0.30/0.30 = 1
0.30 M 0.25 0.25 0.25/0.25 = 1
0.40 M 0.40 0.30 0.40/0.30 = 1.33
B. DiscussedChardakov’s MethodGravimetric Technique
Osmosis ialah proses pergerakan molekul pelarut (contoh:air) dari satu larutan yang cair (larutan hipotonik) ke satu larutan yang lebih pekat (larutan hipertonik) melalui membran semipermiabel. Potensial osmosis selalu bernilai negative. Karena titik nol dari potensial osmosis di ambil dari air murni yang bebas ion.
Dari hasil pengamatan yang telah dilakukan, kami mendapatkan data bahwa terjadi perubahan berat. Potongan kentang yang direndam dalam larutan 0.10 M, 0.20 M, 0.25 M, 0.30 M tidak mengalami perubahan berat. Kemungkinan hal tersebut terjadi karena terjadi peristiwa isotonis yaitu jaringan kentang yang tidak mengalami penambahan maupun pengeluaran air atau tidak ada pergerakan molekul air. Sedagkan pada potongan kentang yang direndam dilarutan 0.15 M dan 0.40 M mengalami penurunan berat setelah dilakukan perendaman. Perubahan berat yang terjadi yaitu potongan kentang yang direndam di larutan 0.15 M, berat awal : 0.35 g dan berat akhir 0.30 g sehingga terjadi penurunan berat sebanyak 0.05 g sedangkan pada potongan kentang yang direndam di larutan sukrose 0.40 M, berat awal : 0.40 g dan berat akhir 0.30 g sehingga terjadi penurunan berat sebanyak 0.10 g. Kami mengamati hal ini terjadi, karena kentang bersifat hipotonik dan gula bersifat hipertonik maka air yang berada pada kentang bergerak keluar sehingga kadar air pada kentang berkurang. Dan semakin besar zat terlarut yang diserap oleh umbi kentang, makin besar air yang keluar dari umbi kentang tersebut. Hal ini di tandai dengan semakin besar selisih berat umbi setelah di rendam dalam larutan sukrosa dan di timbang kembali.Osmosis is the movement of solvent molecules (eg, water) from a liquid solution (hypotonic solution) to a solution that is more concentrated (hypertonic solution) through the membrane semipermiabel. potential osmosisalways worth negative. Since the zero point of potential osmosis purified water is taken from the free ion.From the observations that have been made, we obtain data that weight changes. Potato pieces were soaked in a solution of 12:10 M, 12:20 M, 12:25 M, 00:30 M weight unchanged. The possibility of that happening because the event occurs isotonic the potato tissue unchangedand the expenditure of water or no movement of water molecules. While the potato pieces are soaked in a solution of 0:15 M and 0:40 M experience weight after immersion. Weight changes that occur are pieces of potato soaked in a solution of 0:15
M, initial weight: 0.35 g and 0:30 g final weight resulting in weight loss as much as 0.05 g while the potato pieces are soaked in a solution of 0:40 M sucrose, initial weight: 0.40 g and weight final 0:30 g resulting in weight loss as much as 0:10 g. We observed this to happen, because potatoes are hypotonicand sugar is hypertonic, the water that is on the potato moves out so that the water content in potatoes is reduced. And the larger the solute is absorbed by the potato tuber, the greater the water out of the potato tuber. It is on the mark with the greater difference in weight of tubers after a soak in a solution of sucrose and weigh again.
CHAPTER VCONCLUSION AND SUGGESTION
A. Conclusion
A plant if soaked in a solution, the water potential in plant cells is changed depending on
the concentration and viscosity of the solution used. If a high concentration of the
network,
the water out of the network so that the weight is reduced network, the network is
called hypertonic conditions. and vice versa (hypotonic conditions). While isotonic
solution
displacement does not occur so that the weight of the water molecules fixed.
B. Suggestion
1. Laboratory should provide tools that fit the needs of that practice can be
implemented with a conducive and comfortable.
2. Assistant should accompany each group to support the implementation of
practical activitiss in accordance with the desired.
3. My frisnds should understand the working procedures before entering the lab
room
BIBLIOGRAPHY
Anonymous, 2013. http://courseworkbank.info/journal. Accesed 26th march 2013
Anonymous, 2013. Lecture Water. http://employees.csbsju.edu. Accesed 26th
march 2013
Anonymous, 2013. http://biology.clemson.edu. Accesed 26th march 2013
Anonymous, 2013. http://appstate.edu. Accesed 26th march 2013.
Campbell. 2000. Biologi Campbel edisi 3. Jakarta: Erlangga.
Lakitan, Benyamin. 2011. Dasar-dasar Fisiologi Tumbuhan. Jakarta: Rajawali Pers
Finkelestein, A. (1987) Water Movement through Lipid Bilayer, Pores, and Plasma Membranes: Theory and Reality. Wiley, New York.
Taiz, Zeiger. 2002. Plant Physiology edtion 3. Sinauer Associates: England
Questions:
1. What concentration of surcose resulted in incipisnt plasmolysis, and how
did you know when it occurred?
2. Based on the above, what was the osmotic potential of the cells? Show
your calculation.
3. What were possible sources of error in this experiment?
Answer
1. In our observation, of all the sucrose concentration given all the impact
plasmolisis. But the presentation of the different plasmolisis. Experisncing
the highest plasmolisis is sucrosa solution 0:30 and 0:40 then sequentially
is 0:25, 0:10, 0:20 and presentations that have the lowest plasmolisis is
0.15 m sucrose solution.We known plasmolisis happen becouse the water
molecules move from epidermal cells Rhoe discolor leading to the solution
of sucrose, resulting protoplasts epidermal cells lose water, shrink volume
(cells become wrinkled) and finally detached from the cell wall, the events
that occur in epidermal cells Rhoe discolor is commonly called the
Plasmolisis.
2. Observation result of with use abbreviation:
osmotic potential (Ψπ) for 0.15 M in 27°C :-Ψπ = miRT-Ψπ = (0.15)(1)(0.082)(273+27)-Ψπ = -3.69Ψπ = 3.69
3. Errors that may occur in this lab are:
a. Aprentice inaccuracy when determining or calculating the number of
cells undergoing plasmolisis and the number of cells that do not
undergo plasmolisis.
b. Errors in taking the epidermis rhoe discolor, possible incision
epidermis has taken bold measures to normal size in the experiment to
be performed.
c. Sucrose solution used was not valid due to the mixture of sucrose
solution with each other this is caused by the use of a Pasteur pipette
solution simultaneously for all becouse Pasteur pipette is used only
one solution for all.
osmotic potential (Ψπ) for 0.1M in 27°C :-Ψπ = miRT-Ψπ = (0.1)(1)(0.082)(273+27)-Ψπ = -2.46Ψπ = 2.46
osmotic potential (Ψπ) for 0.20M in 27°C :-Ψπ = miRT-Ψπ = (0.20)(1)(0.082)(273+27)-Ψπ = -4.96Ψπ = 4.96
osmotic potential (Ψπ) for 0.25M in 27°C :-Ψπ = miRT-Ψπ = (0.25)(1)(0.082)(273+27)-Ψπ = -6.15Ψπ = 6.15
osmotic potential (Ψπ) for 0.30M in 27°C :-Ψπ = miRT-Ψπ = (0.30)(1)(0.082)(273+27)-Ψπ = -7.38Ψπ = 7.38
osmotic potential (Ψπ) for 0.40M in 27°C :-Ψπ = miRT-Ψπ = (0.40)(1)(0.082)(273+27)-Ψπ = -9.84Ψπ = 9.84