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Aim: The aim of
the experiment is to understand osmosis and diffusion using potato cell
membrane to show the movement of water across the cell membrane in different
concentrations of sucrose solution.

Introduction:

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A cell is the basic unit of all life and cells require very
specific conditions to be able to function properly. Temperature, the amount of
water and nutrients must all be perfect in order for a cell to be healthy, and
these optimal conditions vary depending on the organism. The amount of fluid inside
and outside a cell is extremely important, and this fluid amount is often
determined by the amount of solutes outside of the cell. Solutes are the
particles that are dissolved in a solvent (e.g. water), and together they form
a solution. In the body, solutes are ions like sodium and potassium.

The three types of solutions that can occur in the body based
on solute concentration are: isotonic, hypotonic, and hypertonic. An isotonic
solution is one where the concentration of solutes is the same both inside and
outside of the cell. A hypotonic solution is one where the concentration of
solutes is greater inside the cell than outside of it, and a hypertonic
solution is one where the concentration of solutes is greater outside the cell
than inside it.

Diffusion is a movement of atom, molecules and ions from a place
of high concentration to a place of low concentration. Diffusion
happens when the particles are free to move; in gases and particles that are dissolved
in solutions. Homeostasis controls the internal conditions of the body so diffusion
can take place. In order to have a high diffusion rate there needs to be a short
distance, large surface area and a big concentration difference (Fick’s Law). Fick’s
laws state that the rate of diffusion is directly proportional to the membrane
surface area and the concentration gradient and is inversely proportional to
the membrane thickness. High temperatures increase diffusion and large
molecules slow diffusion. Examples of diffusion include:

1.      Gas
exchange in alveoli

In the
lungs oxygen diffuses from the alveoli and moves into the blood stream. Carbon
dioxide moves from the blood stream around the lungs into the alveoli through
diffusion.

 

 

 

 

Fig 1: Diagram showing the gas exchange
that takes place in the lungs.

2.     
Diffusion during respiration in cells

Oxygen, carbon dioxide and glucose enter and
leave cells by diffusion. During respiration in cells the concentration of
oxygen and glucose inside the cell is lower than in the blood stream so these
substances move into the cell. As carbon dioxide is produced the concentration
inside the cell increases to a level higher than the surrounding blood outside the
cell, so carbon dioxide diffuses out of the cell.

 

 

 

 

Fig 2: Diagram showing diffusion
during respiration in cells

3.      Gas
exchange in plants; Photosynthesis

The higher concentration of carbon dioxide in
the air diffuses into the leaf. Oxygen produced during photosynthesis builds up
to higher concentrations and diffuses out of the leaf into the air which has a
lower concentration of oxygen.

 

 

 

 

 

 

Fig 3: Diagram showing gas exchange in plants

Osmosis is the movement of water from low concentration to
high concentration through a partially permeable membrane. Examples of osmosis
include:

1.      Osmosis in
plant roots

They gain water by osmosis through their roots. Water moves
into plant cells by osmosis, making them stiff so that they are able to hold
the plant upright.

 

 

 

 

Fig 4: Diagram showing osmosis in plants

2.      Osmosis in red
blood cells

When a red
blood cell is placed in water, water enters the cell by osmosis. The cell membrane
is weak so it causes the cell to burst as the volume and pressure in the cell
increases. Red blood cells shrink when placed in concentrated solutions of sugar
or salt as water moves out of them by osmosis. This makes the cells appear
wrinkled when viewed through a microscope.

 

 

 

 

Fig 5: Diagram showing osmosis in red blood cells

3.     
Osmosis in plant cells

Plant cells have a strong
rigid cell wall. This stops
the cell bursting when it absorbs water by osmosis. The increase in pressure
makes the cell rigid. If plant cells lose too much water by osmosis they become
less rigid and eventually the cell membrane shrinks away from the cell wall.

 

 

 

 

 

Fig 6: Diagram showing osmosis in plant cells

 

 

Hypothesis:

The movement of water molecules into and out of the potato
cuboids will depend on the different concentrations of the sucrose solution
used in the experiment. This will result in the percentage change in the mass of
the potato varying for each sucrose solution concentration. The following
results are expected:

1.       In the hypertonic solution the solute
concentration is greater outside the cell, water flows out of the cell which
causes the cell to shrink. The sucrose solution is more concentration causing
the water to flow out of the potato cuboid which makes it shrink and decrease
in mass.

2.      In the
hypotonic solution the solute concentration is greater inside the cell. Water
flows into the cell, causing the cell to swell. The sucrose solution
concentration is higher in the potato cuboid so water enters the cell and makes
it swell which increases the mass.

3.      In the
isotonic solution the solute concentration is equal inside and outside the
cell. Water flows in and out of the cell at the same rate as a result the shape
of the cell remains the same. This means the mass of the potato cuboid will
stay the same.

Methodology:

1.      The potato
was peeled and cut into six cuboids with a knife in a 2cm length, 1cm width and
a 1cm height.

2.      Each potato
cuboid was paired with one of various concentrations of sucrose solution
ranging from 0.0M to 1.0M (0.0M, 0.2M, 0.4M, 0.6M, 0.8M, and 1.0M). Filter
paper was also labelled with each concentration of the sucrose solution and a
‘B’ for the before mass and an ‘A’ for the after mass.

3.      The potato
cuboids were weighed in grams (g) using a balance ensuring not to move the
balance whilst it was switched on.

4.      The filter
papers were placed onto the balance pan and ‘zero’ was pressed to clear the
mass by turning it to 0.

5.      The potato cuboids
were transferred into each solution into a beaker using forceps. The stop clock
was then started by pressing the green button.

6.      After
approximately 40 minutes (recording each time) each potato cuboid was removed
one at a time. The excess solution was drained with tissue paper, without
squeezing. Then one at a time, each filter paper was placed on the balance pan
and the balance was zeroed. The corresponding potato cuboid that was placed in
the matching concentration was weighed and the mass was recorded in table 1.
This was repeated for each concentration.

  

 

Results:

(X-Axis)
Solution Concentration
(M)

Mass (g)
Before Soaking
(B)

Mass (g)
After Soaking
(A)

(Y-Axis)
% Change of mass (g)
(A-B/B)*100

Description of the solution:

0.0

2.83

2.98

(2.98-2.83)/2.83*100=5.3003

Hypotonic

0.2

2.73

2.88

(2.88-2.73)/2.73*100=5.494505

Hypotonic

0.4

3.66

3.67

(3.67-3.66)/3.66*100=0.27322

Isotonic

0.6

2.72

2.67

(2.67-2.72)/2.72*100=-1.838235

Hypertonic

0.8

2.69

2.60

(2.60-2.69)/2.69*100=-3.345724

Hypertonic

1.0

2.94

2.85

(2.85-2.94)/2.94*100=-3.06122

Hypertonic

Table 1: Table of results showing different concentrations of
Sucrose Solution, change in mass and description of the solution

(X-Axis)
Sucrose Solution (M)

(Y-Axis) % Change of mass (g)
1 DP

0.0

5.3

0.2

5.5

0.4

0.3

0.6

-1.8

0.8

-3.3

1.0

-3.1

Table 2: Table of results showing sucrose solution in
different concentration and % change in mass in 2 DP

 

 

 

 

 

 

 

 

Graph 1: Graph showing % Change in mass against Sucrose
Solution concentration (M)

Discussion:

The results show that the potato cuboids in the solution 0.0M
gained the greatest increased mass (5.3%) as a result of water moving from the
solution to the potato cells through osmosis. This confirms that the solution
is hypotonic and the concentration of the sucrose solution was higher outside than
inside the potato cuboid. The same results are seen in the solution 0.2M as the
% increase is even higher (5.5%) meaning it is also a hypotonic solution. The %
change of mass was highest in the solutions that had the lowest concentrations
suggesting that the higher the concentration, the lower the % change of mass.
They are inversely proportioned.

The results for the solution 0.4M had a small percentage
change as it only increased by 0.3%. This suggests that the solution was
isotonic as the mass stayed the same because it only increased by 0.01g. The
solute concentration was equal inside and outside the potato cell so water
flowed in and out of the cell at the same rate and as a result the mass of the
potato cuboid stayed the same. The three hypertonic solutions were for the concentrations
0.6M, 0.8M and 1.0M. These were hypertonic because the % change of mass was in
the minus as the mass became lower after soaking them in the sucrose solution. The
solute concentration was greater outside the potato cell so water molecules
flowed out of the cell which caused it to shrink and lose mass. The sucrose
solutions were more concentrated in the hypertonic solution which caused the
potato to shrink. These solutions had the highest concentrations and therefore
lost the most mass.

The concentration of the potato cuboid cell at the isotonic
concentration can be calculated using the equation of the line when y=0:

Y=mx+c

Y=-10.061x+5.499

0=-10.061x+5.499

-5.499=-10.061x

X=-5.499/-10.061

=0.55M

The rate of change is the speed at which the potato cuboids
change in mass over time (the 40 minutes they were soaked in the solution); it
is represented by the slope of the line on the graph. Using the equation
y=-slope x+ number this is -10.061. The rate of change is important because it
shows how fast the dependant variable (% change in mass) is changing and by how
much at each interval of time. For example at 0.2M the rate of change is 8.25%
per minute. This is calculated by:

40 minutes= 40/60 =2/3= 0.66…

5.5%/ 0.66…= 8.25% per minute

This means at the concentration of 0.2M the % change of mass
increased by 8.25% every minute for the total of 40 minutes (8.25% * 0.66…=
5.5%).

Evaluation:

The experiment went well and no
crucial problems were encountered during the experiment because human errors
were minimised by measuring the potatoes and cutting them with a sharp knife,
ensuring the balance pan was on zero before weighing the potatoes and making
sure the potato cuboids were patted dry and properly drained without squeezing.
Also the concentrations were written on individual pieces of filter paper to
minimise the risk of confusion. To further improve the experiment I suggest
repeating the experiment three times for each concentration of solution to
ensure there are no anomalies or mistakes which will improve the accuracy and
reliability of the results. Another suggestion improve the experiment is too
use the same batch of potatoes and same brand throughout the entire experiment.
In the experiment I could’ve also tried using distilled water as a variable or
left the potatoes in the solution for longer.

Conclusion:

In conclusion the experiment
confirmed the hypothesis and proved it correct; the movement of water molecules
into and out of the potato cuboids did depend on the different concentrations
of the sucrose solution used in the experiment. The results show that the
relationship between % change in mass and concentration of the solution is
inversely proportioned; the higher the concentration, the lower the % change of
mass. The concentration of the solution also impacted the rate of osmosis; the
lower the concentration the higher the rate.

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