Practical 3 - A-Level Biology coursework PDF

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Polina Lobacheva November 13th, 2018 Y12 AS-Level Biology Ms. Tooley Group Partners: Ksenia Gerdyush Required practical 3: Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissues Aim The aim of this investigation is to determine the water potential of potato tuber cells. Introduction Osmosis is the net movement of water from an area of high water potential concentration to an area of low water potential concentration through a selectively permeable membrane. Water potential is the pressure created by water molecules and is measured in units of pressure (kPa). At regular temperature and pressure (25°C and 100kPa) the water potential of pure water is 0kPa. The term used to describe the membrane is “selective” because it only allows small molecules to pass, such as water1. A graph will be used to find the water potential of the potatoes by setting up different concentrations of sucrose and water potential, then when drawing a line of best fit, the value on the x-axis, when sucrose solution (y-axis) is 0, is the water potential of the potato. Equipment ● Large potato tuber ● Potato chip cutter ● 1 mol/dm3 sucrose solution ● distilled water ● 6 boiling tubes 1 (Toole and Toole, 1995)

● marker pen ● thermometer ● 10 cm3 graduated pipette and pipette filler ● white tile ● scalpel ● paper towels ● digital balance Risk Assessment Hazard Glassware

Risk2 Control Cut yourself with the sharp Safety goggles should be worn.

Scalpels

edges Cuts that could lead to Protect the sharp end when moving around; cut infection

away from you; use a pad/cutting board when cutting.

Method 1. Label 6 boiling tubes 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm3 2. Use the 1.0 mol/dm3 of sucrose solution and water to make up 20 cm3 of sucrose solution (mol/dm3) of each of the following: 0.2, 0.4 0.6, 0.8, 1.0 3. Using the potato chip cutter, cut six chips from the potato tuber. Make sure you remove any peel on the potatoes. Use a ruler, scalpel and tile to cut all of the chips to the same length. Blot the potato chips dry with a paper towel (don’t squeeze the chips). Put each potato onto a clean square of paper towel which you have numbered in the same way as the boiling tubes. 4. Weigh each potato chip. Record these initial masses in a suitable table. 5. After a few minutes, remove the chips from the boiling tubes. Blot the chips dry, as before. Then reweigh them. Record these final masses in your table. 6. Calculate the change in mass and then calculate its percentage change.

2 (CLEAPSS)

7. Plot a graph of your processed data and use this to determine the concentration of sucrose which has the same water potential as the potato tuber cells. Table showing the test tubes set up Concentration of sucrose 0

0.2

0.4

0.6

0.8

1.0

solution mol/dm3 Volume of 1.0 mol/dm3 0

4

8

12

16

20

16

12

8

4

0

sucrose solution cm3 Volume of water cm3

20

Table of results Sucrose concentration

Initial mass (g)

Final mass (g)

Change in mass (g)

Percentage

3.13 2.81 3.20 2.22 3.44 3.15

3.62 2.99 2.91 1.61 2.4 1.97

0.49 0.18 -0.29 -0.61 -1.04 -1.18

change % 15.7 6.4 -9.1 -27.5 -30.2 -37.5

(mol/dm3) 0.0 0.2 0.4 0.6 0.8 1.0 Graph

Conclusion When the line crosses the x-axis where the sucrose solution is at zero, meaning there is no sucrose concentration, the actual water potential of the potato tuber cells is 0.25. My results

agree with my hypothesis. As you can see from my graph, as the concentration of the sucrose solution increases, the percentage change in mass decreases, hence it’s an inversely proportional correlation. These results show that the potato cells increase in mass in solutions with a highwater concentration and decrease in mass in solution with a low-water concentration (such as the sucrose concentrated solution). The graph is evidence for complete plasmolysis. It happens when plant cells have low water potential and lose water, by osmosis, these cells shrink and pull away from the cell wall, making them flaccid. This means the cells are undergoing plasmolysis, which is a process of a vacuole, in a plant cell, shrinking and the membrane pulling away the wall due to water loss by osmosis. In this experiment, as the solution of the concentration around the potato was getting higher, the cells inside the potato were losing water to create an equilibrium between the internal and external environments3. Evaluation Concentration of sucrose

Initial

Final mass

Change in

Percentage

solution (mol/dm3) 0.0

mass (g) 0.83

(g) 0.94

mass (g) 0.11

change % 13.25

0.2

0.85

0.91

0.06

7.06

0.4

0.81

0.71

-0.10

-12.35

0.6

1.00

0.77

-0.23

-23.00

0.8

0.68

0.41

-0.27

-39.71

1.0

0.7

0.36

-0.34

-48.57

These results are from another group who performed the same experiment. These results are not far off what my group obtained, and they both prove the inversely proportional correlation. The difference in these results might have been caused by the age of the potatoes or the shape that they were cut. The group, whose results are shown above, cut their potatoes into relatively small squares, whereas my group cut them in rectangles, and therefore their ratio of

3 (Osmosis - Supplying the cell - OCR Gateway - GCSE Combined Science Revision - OCR Gateway BBC Bitesize)

volume to surface area was bigger than ours, which supports the reason why their percentage mass change interval values are bigger. Sucrose concentration

Initial mass

(mol/dm3)

(g)

Final mass (g)

Change in

Percentage

mass (g)

change %

0.0

6.54

6.81

0.27

4.13

0.2

6.57

6.63

0.06

0.91

0.4

6.46

6.17

-0.35

-5.42

0.6

6.41

5.96

-0.45

-7.02

0.8

6.21

5.57

-0.64

-10.30

1.0

6.33

5.53

-0.80

-12.60

The results above are the expected results that are given in the handbook. These results differ from mine; however, they match the idea of as the sucrose concentration increases the percentage change in mass decreases. The values for the percentage change, for the expected results, decreases slower than the results I obtained, perhaps that is because the ratio of volume to surface area was bigger or the potatoes were left in the solution for a shorter period of time. Also, this experiment is reproducible as those who did it got similar results to mine. Overall, this experiment was very successful and although my group and I have not obtained the exact same values of the expected results, my group’s informative graph showcases a logical correlation between the two variables. On the other hand, I believe there is a lot space for improvement; such as having a more varied range of concentrations (0.0, 0.5, 1.0, 1.5, etc.) which would help me obtain more varied data and a more precise line of best fit. Also, when cutting the potatoes, it was hard to make sure they were all the same, which would affect the results since there would be a different ratio of volume to surface area for each of the pieces therefore affecting the rate of osmosis. By using some sort of professional cutting utensil, we could make perfect rectangles that would be the exact same and wouldn’t contribute to affecting the data. Also, if I were to repeat this experiment, I would use a burette for an accurate amount of

fluid in each test tube. Next time I would also perform the experiment for a longer period of time to allow more osmosis to occur. There was only one minor anomaly, which is the data received from 0.6 mol/dm3 of sucrose concentration solution; this could have been caused by human error. I would fix this by repeating this experiment, at least, 3 times, which would outdo any anomalies in my data when I calculate my average. References BBC Bitesize. (2018). BBC Bitesize - GCSE Combined Science - Supplying the cell - OCR Gateway - Revision 4. [online] Available at: https://www.bbc.com/bitesize/guides/zwkn7p3/revision/4 [Accessed 12 Nov. 2018]. Cleapss.org.uk. n.d. CLEAPSS. [online] Available at: [Accessed 10 April 2020]. Toole, G. and Toole, S. (1995). AQA biology A level....