Biology Lab Report

Finding the isotonic point of potato cells

Krisha Parekh Year 11 20 September 12

Aim: To find the isotonic point of potato cells.Background Information:

What is an isotonic point? :

An isotonic cellular environment occurs when an equal solute concentration exists inside and outside the cell. Water molecules flow in and out at an equal rate by osmosis, causing the cell size to stay the same.

Osmosis: is the movement of water molecules across a selectively-permeable membrane down a water potential gradient. More specifically, it is the movement of water across a selectively permeable membrane from an area of high water potential (low solute concentration) to an area of low water potential (high solute concentration. Osmosis may occur when there is a partially permeable membrane, such as a cell membrane. Impermeable means that the object doesn't allow things through it, so a membrane has to be permeable so food gets in and waste gets out. When a cell is submerged in water, the water molecules pass through the cell membrane from an area of low solute concentration (outside the cell) to one of high solute concentration (inside the cell); this is called osmosis. The cell membrane is selectively permeable, so only necessary materials are let into the cell and wastes are left out.

When the membrane has a volume of pure water on both sides, water molecules pass in and out in each direction at the exact same rate; there is no net flow of water through the membrane.

In a solution, the concentration of water is diluted (or lowered) by the presence of solute particles. If there is a solution on one side, and pure water on the other, there will be a higher concentration of water molecules on the pure water side of the membrane. Therefore, water molecules pass through the membrane from the pure water side toward the solution side more frequently than from the solution side going to the pure water side. This will result in a net flow of water to the side with the solution. Assuming the membrane does not break, this net flow will slow and finally stop as the pressure on the solution side becomes such that the movement in each direction is equal: dynamic equilibrium. This could either be due to the water potential on both sides of the membrane being the same, or due to osmosis being inhibited by factors such as pressure potential or osmotic pressure.

Hypothesis: When Osmosis occurs, water molecules moves from a higher concentration to a lower concentration through a partially permeable membrane. As the concentration of the salt solution increases from the areas of high concentration from potato cells to the regions of low concentration in the salt solution causing the mass of the potato cells decrease. The isotonic point will be in between 0.2 to 0.3 as its the mid concentration, where as concentrations like 0.1 will be too hypotonic.

Variables:

Dependent:

Mass of the potato cells after the experiment

Independent:

- Concentration of the salt solutions (0,0.1,0.2,0.3,0.4,0.5 mol/dm3)

Mass of the potato cells before the experiment

Control:

-Amount of salt solution (25 cm3)

-Number of potato strips in each test tube (5 per test tube)

Apparatus:

1 raw potato

Knife

Salt Solutions (0.0 M,O.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M)

100 cm3 measuring cylinder

Petridish

Tweezers

Electronic Weighing Balance ( 0.01 gms)

Permanent Marker

6 test tubes

Adhesive paper

Scissor

Diagram:

Procedure:

Using a knife,remove the skin of the potato.

Using a knife, use the potato into 30 strips, approximately the same size.

Divide it into 6 sets with 5 strips in each set.

Label each set in numbers from 1 to 5.

Weigh each strip and note it down, rounded off to 2 decimal places.

Measure 25 cm3 of salt solution with 0.0 M and pour it in a test-tube. Then, do it for the rest of the concentration solutions.

Cover up the test-tube with adhesive paper

Leave the test-tubes for 24 hours

Take out the potato strips the next day,dry them, weigh the potato strips and note down the weight.

Quantitative Data:

Raw Data Table:

Concentration(mol/dm3)Strip NumberInitial Mass ( 0.01 gms)Final Mass ( 0.01 gms)Change in Mass ( 0.01 gms)

o.o m12.953.12+0.17

22.402.44+0.04

32.412.47+0.06

42.502.59+0.09

52.782.80+0.02

0.1 m10.900.88-0.02

20.810.82+0.01

30.980.98+0.01

40.950.93-0.02

50.950.91-0.04

0.2 m11.201.17-0.03

21.251.23-0.02

31.211.18-0.03

41.171.12-0.05

51.301.29-0.01

0.3 m 11.221.16-0.06

21.241.22-0.02

31.191.16-0.03

41.171.12-0.05

51.281.19-0.09

0.4 m11.871.70-0.17

21.721.69-0.03

31.951.84-0.11

42.202.16-0.04

52.021.97-0.05

0.5 m13.193.01-0.18

23.022.89-0.13

33.142.97-0.17

43.263.03-0.23

53.373.33-0.04

Processed Data Table:Concentration(mol/dm3)Strip NumberInitial Mass ( 0.01 gms)Final Mass ( 0.01 gms)Change in Mass ( 0.01 gms)Average ( 0.01 gms)

o.o m12.953.12+0.170.08

22.402.44+0.04

32.412.47+0.06

42.502.59+0.09

52.782.80+0.02

0.1 m10.900.88-0.02-0.01

20.810.82+0.01

30.980.98+0.01

40.950.93-0.02

50.950.91-0.04

0.2 m11.201.17-0.03-0.03

21.251.23-0.02

31.211.18-0.03

41.171.12-0.05

51.301.29-0.01

0.3 m 11.221.16-0.06-0.05

21.241.22-0.02

31.191.16-0.03

41.171.12-0.05

51.281.19-0.09

0.4 m11.871.70-0.17-0.08

21.721.69-0.03

31.951.84-0.11

42.202.16-0.04

52.021.97-0.05

0.5 m13.193.01-0.18-0.15

23.022.89-0.13

33.142.97-0.17

43.263.03-0.23

53.373.33-0.04

Average: Sum of observations/ Number of Observations

example: concentration of 0.5 mol/dm3= = (-0.18)+(-0.13)+(-0.17)+(-0.23)+(-0.04)/5

= -0.15

Qualitative Observations:

There were bubbles present in each test-tube.

-The solution started getting translucent as the concentration of the solution increased.

Graph

Concentration (mol/dm^3)Change in mass ( 0.01 gms)

00.08

0.1-0.01

0.2-0.03

0.3-0.05

0.4-0.08

0.5-0.15

Percentage change in mass:

(initial mass-final mass)/initial mass x 100

example: 0 M : (2.61-2.68)/2.61 x 100

= 2.68%

Concentration(mol/dm^3)% change in mass

0-2.68%

0.1-2.17%

0.22.44%

0.34.09%

0.44.10%

0.54.69%

Conclusion:

The pattern observed from the graph and results are as follows, for the concentration of 0.0 M, the mass increases and so theres a positive change in mass for example, from 2.95 to 3.12 grams and thus the change in mass would be +0.17 grams. The reason for this would be the concentration of water would be higher in the solution rather than the potato tuber, causing endosomosis as water travels from the regions of higher concentration to a lower concentration. On the other hand, as the concentration increases from 0.1 M to 0.5 M, there is a decrease in mass and the higher the concentration, the larger decrease there is. For 0.3 M, the average decrease in mass is, -0.05 grams where as the average change in mass for 0.5 M is -0.15grams. After plotting the graph, the isotonic point is 1.04 as it cuts the x intercept at that point. Therefore, my hypothesis is proved correct as the isotonic point is in between 0 to 0.1 mol/dm^3.

Evaluation:

In my readings of 0.1 M there were a few positive results when there shouldve been a decrease in mass for all the potato strips. The reasons for this couldve been:

while weighing the potato strips, the fluid must not have been cleared thus leading to extra weight.

while noting down the weight, there couldve been some errors thus not providing me with accurate results.

Therefore, to avoid such mistakes, one must cleanse the fluid from the petri-dish while measuring it and the experiment could have been repeated twice for better results.

Safety Measures:

1) While cutting the potato, one should be careful and ensure the knife doesnt slip as it may cause injury.

2) Be careful while handling the scalpel for cutting the potato strips to the required length. EMBED MSGraph.Chart.8 \s

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Krisha Parekh1Krisha Parekh1

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