Background opened businesses where they make and package

Background

Agriculture is the backbone of Bhutan’s economy. Until the 1980s, an estimated 95% of the work force was employed in agriculture (Scott, 1983).  Potato is one of the 4 most popular crop produced for its commercial value in Bhutan along with Rice, Maize, and Chilli. Consumed a lot by the natives, the average annual potato consumption is estimated to be 40kg per person compared to India and the whole of Asia which have 17kg and 24 Kg per person respectively (Dorji, Tamang, and Vernooy, 2015). Before globalisation, potatoes within the country was only used to make one of the national dishes of Bhutan called “Kewa Datshi” translated as “potato cheese.” However, recent introduction of many global recipes and cuisines has made the crop more popular within the country. People have also opened businesses where they make and package home-made potato chips and fries. My preferred dish aside from the national dish “Kewa Datshi” had always been French Fries which I always strived to prepare at home. French fries are expected to be firm and crispy and to attain that most recipes suggest to soak the potatoes overnight so that they become crispy and delicious when fried. An article from NY Times mentions how potatoes should be soaked in water for 8 hours before frying as it is the secret to crispy potatoes because “It draws out the starch, making them more rigid and less likely to stick together. (Hesser, 1999). To try make fries salty, I soaked them in Salt solution (NaCl) however, when soaked it became more flexible than rigid. The result of the potatoes becoming flexible was because of osmosis. Therefore, I wanted to explore the concept of the potatoes getting rigid whilst being soaked in water and their absorption capacity when introduced to different concentrations of solutes aside from NaCl. The solute that was decided to explore osmosis is sucrose.

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Osmosis is a passive movement of water molecules, across a partially permeable membrane, from a region of high concentration to a region of low concentration. It is the net movement of free water molecules through these partially permeable membranes until equilibrium is reached.  Osmosis is due to solutes that form bonds with water. Water is considered a universal solvent and will associate and dissolve, polar or charged molecules (solutes). Solutes cannot cross a cell membrane unaided, therefore water moves to equalise the two solutions (Oxford IB Biology, 2014).

Osmosis in plants:

Osmosis in plants is important so that the turgidity of a plant is maintained as they gain water by osmosis through roots. As the cell wall is rigid, internal pressure would increase and push against the cell wall when water enters the cell. This pressure against the cell wall allows the plant to maintain an upright structure. When there is less access to water, the plants start to sag as the cells become flaccid (Anon, 2018).

 

The ability of an extracellular solution to make water move into or out of a cell by osmosis is known as its tonicity. A solution’s tonicity is related to its osmolarity, which is the total concentration of all solutes in the solution. The osmolarity of a solution is the total concentration of osmotically active solutes and cell contains many different osmotically active solutes. 

 A solution with low osmolarity has fewer solute particles per litre of solution, while a solution with high osmolarity has more solute particles per litre of solution. When solutions of different osmolarities are separated by a membrane permeable to water, but not to solute, water will move from the side with lower osmolarity to the side with higher osmolarity (Khan Academy, 2018).

 There are three potential terms used to describe whether the cell or the environment has a higher concentration of solutes. Hypotonic solutions have lower osmolarity and a solute concentration lower than the cell resulting in water entering the cell and the cell grows in size.?? Hypertonic solutions have higher osmolarity and a higher concentration of solute than the cell resulting in water leaving the cell to reach equilibrium and will shrink. ?Isotonic solutions have same osmolarity as the cell and a concentration that is equal to that of the cell, so water enters and leaves the cell at an equal rate and there is no net change (Ib.bioninja.com.au, 2018).

Research Question: What is the water absorption capacity of a potato when introduced to different concentrations of a sucrose?

Aim: To measure the absorption capacity of potato plant in different concentrations of sucrose through osmosis.

Null Hypothesis: The sucrose concentration does not have any effect on the water absorption of potato as the concentration of sucrose increases in the solution.                                                                                                             

Alternative Hypothesis:  Higher the sucrose concentration in solution, lower the mass of the potato.When there is more water concentration in the solution, the potato will absorb more water gaining weight and as the concentration of sucrose increases it will start to decrease as the concentration of water would be higher inside the potato than in the boiling tube making water move out.

Variables:

 

Definition

Variables

Independent Variable

The variable that is changed to test the effects on the dependent variable.

Concentrations of sucrose solutions.
 

Dependent Variable

The variable being tested and measured.

Mass change in potato pieces.

Control Variable

The unchanged variable throughout the course of the experiment.

1.  Size of potato pieces- Surface area to volume ratio affects osmosis, therefore, all potato pieces were cut having 3cm length, 1 cm width and 0.5 cm height.
2. Temperature- It affects the rate of diffusion therefore while conducting the experiment the air conditioning in the room remained the same.
3. Time- If some beakers were provided more time, more transfer of water could occur. To make it fair, the potato pieces were immersed at the same time and kept inside the boiling tubes for 20 minutes.
4. The volume of the solution- To ensure that the only changing variable is the concentration, we maintained the volume of the solution of 20  in the boiling tubes the whole time.
5. Soaking up excess water from potato – The potato pieces were dried by hand towel paper so that they can absorb excess liquid from the washed pieces of potato to fairly calculate their initial mass. They were dried again before calculating final mass as well.

            Materials and Apparatus:

·       2 measuring cylinders 25 (±0.5 ml)           •     Distilled water                        •     6 Boiling tubes

·       1 mole of sucrose                                                  •      Boiling tube rack                   •     15 cm Ruler (± 0.05 cm)

·        Heatproof mat                                                     •      3 Potatoes                              •     Knife

·        Measuring Balance (±0.01g)                              •     Two beakers 250ml               •  Stopwatch (± 0.01 s)    

Risk Assessment:    

The experiment conducted is fairly safe aside from the use of Knives to cut the potato pieces. Caution should be taken while using it since they are very sharp. Since there are no dealings with harmful chemical substances, safety goggles aren’t necessary.

Procedure:

Clear out the table completely and lay down all the materials.
Place all boiling tubes in the test tube holders and mark each of them as 0 mol, 0.2 mol, 0.4 mol, 0.6 mol, 0.8 mol, 1 mol  based on the concentration of the sucrose to be contained inside.
Use one measuring cylinder to measure the volume of distilled water and the other to measure sucrose accordingly to the values presented in Table 2.
From 1 mole of sucrose prepare different concentrations of sucrose mol/mg in a 20 ml volume solution by diluting it using in distilled water while pouring both into boiling tubes accordingly as shown in Table 2. 

Concentration  of sucrose solution/ mol

0

0.2

0.4

0.6

0.8

1

Volume of 1.0 mol  sucrose solution () ± 0.1

0

4

8

12

16

20

Volume of distilled water (cm3) ± 0.1

20

16

12

8

4

0

5.                  Peel the potatoes and wash them in the sink.

6.                  Place the potatoes on the heat mat and cut the potatoes by 3 cm length and 1 cm width and 0.5 cm height by measuring it with a ruler.

7.                  Take 6 cut pieces of the potatoes and dry it on a paper towel to remove any excess water present on the surface.

8.                  Mark a place near the heat proof map and number it 0 mol, 0.2 mol, 0.4 mol, 0.6 mol, 0.8 mol, 1 mol  according to the sucrose concentrations and place the potato pieces in place according to that.

9.                  Take each potato piece and measure the mass, then place it back under the label of the written sucrose concentration.

10.              After finding the initial mass of the cut potato place each piece of potato inside the boiling tubes with matching labels of sucrose concentrations.

11.              Start the timer and time for 20 minutes.

12.              After 20 minutes pour all the liquid out and one by one wipe of the excess liquid with a clean paper towel and measure the final mass of the potato.

13.              Record the initial and final weight in the results table correctly.

14.              Repeat steps 1-13 for another 2 times before calculating the results

Results Table

Concentration of Sucrose/ mol

Trial

Mass of Potato Piece at start/ g

Mass of Potato Piece after 20 mins /g

Change in Mass

%Change in mass /g

0

1

1.80

1.97

0.17

9.44%

2

1.69

1.83

0.14

8.28%

3

1.57

1.69

0.12

7.64%

0.2

1

1.99

2.09

0.10

5.03%

2

1.58

1.69

0.11

6.96%

3

1.78

1.85

0.07

3.93%

0.4

1

1.77

1.78

0.01

0.56%

2

1.81

1.87

0.06

3.31%

3

1.38

1.39

0.01

0.72%

0.6

1

2.02

1.91

-0.11

-5.45%

2

1.87

1.79

-0.08

-4.48%

3

1.82

1.73

-0.09

-4.94%

0.8

1

1.96

1.78

-0.18

-9.18%

2

1.39

1.25

-0.14

-10.07%

3

1.64

1.49

-0.15

-9.15%

1

1

1.80

1.53

-0.27

-15%

2

1.46

1.26

-0.2

-13.70%

3

1.58

1.40

-0.18

-11.39%

Qualitative Analysis

Qualitative Analysis

Aside from the above quantitative data presented in table 3, there were some physical changes in potato pieces that was observed. The sizes of the potato pieces were different after being immersed in different concentrations of sucrose solutions. The pieces in 0 mol of sucrose significantly looked thicker whilst the pieces in 1 mol of sucrose looked like they shrank. There was a change in the rigidity in the potato showing a connection to exploration related to the experiment as the potato pieces in 0 mol of sucrose seemed to be more firm and rigid but from 0.6 mol of sucrose and onwards, the rigidity seemed to be lost and the potato pieces seemed to be flexible indicating that they turned flaccid.

Data Processing.

1. Percentage Change: To first find the absorption of water in the trials conducted for 20 minutes, calculate the change in mass from the initial and final mass of the potato pieces and convert it into a percentage with the following formula:  

% Change in Mass of Potato =  

E.g., Trial 1 % Change in mass of potato in 0 mol of sucrose =  = 9.44%

2.  Mean of Percentage Change: Calculate the mean because the data from the trials have slightly different values every time. To calculate the mean, use the formula:

 =

Where the sum of the trials conducted) is divided by the number of trials conducted (n).

E.g.,   of water in 0 mol of sucrose = 9.44+8.28+7.643 = 8.45 %                                                          

3. Standard Deviation: Use of standard deviation helps to indicate the spread of our values around the mean. It is calculated by    Where, N is the number of data points (3) and ? the mean of all values, whilst  is one of the average data values.

E.g.,   SD of the % of water absorption for 0 mol  concentration of sucrose

 SD (  ) of =   = 0.91

4. Standard Error: It is calculated by the formula , where the   standard deviation and n is the number of data points.

E.g., SE of the % of water absorption at 0 mol  concentration of sucrose.

SE  = 0.53

5. 95% Confidence Limit: If the values of the confidence limit do not overlap, the null hypothesis is 95% rejected.  It is calculated by the formula:  95% Confidence limit = mean ± 2 x SE ()

E.g., 95% Confidence Limit for % of water absorption at 0 mol  concentration of sucrose.

95% Confidence Limit = 8.45 ± (2 x SE) = 8.45 ± (2 x 0.53) = 8.45 ± 1.06

Sucrose concentration/ mol

0

0.2

0.4

0.6

0.8

1

Average % change in mass

8.45%

5.31%

1.53%

-4.89 %

-9.47%

-13.36%

Standard Deviation)

0.91

1.53

1.54

0.49

0.52

1.83

Standard Error

0.53

0.89

0.89

0.28

0.30

1.055

95% Confidence Limit

8.45 ± 1.06

5.31 ± 1.78

1.53 ± 1.78

-4.89 ± 0.56

-9.47 ± 1.06

-13.36 ± 2.11

                                                   

 

Data Presentation

 

Conclusion

Overall, the experimental data in the results table supports the hypothesis as result analysis showed that when there is higher sucrose concentration in solution, the mass of the potato showed a decrease in percentage change of mass of potato piece. It is also depicted by the downward sloping of the above graph which shows a negative correlation.

 

The graph above represents the inverse relation of sucrose concentration with the absorption of water by potato in terms of percentage change in mass of potato. It shows that as the concentration of sucrose increases from 0 mol to 1 mol the mass of the potato pieces decreases accordingly. After being immersed in different concentrations of sucrose, there is a positive value until 0.4 mol   which shows an increase in mass of potato piece. However, by 0.6 mol, a negative value is seen which shows that the mass of potato piece has decreased by around 4.9%.

 

Where the line crosses the X-axis there is no percentage increase or decrease in mass of potato piece so this concentration of sucrose solution is isotonic where the osmolarity is same as the solution and the potato piece.  When the concentration of sucrose (mol) is lower than the sucrose concentration point where the line crosses the X-axis, it is observed to be a hypotonic concentration as there is a percentage increase in mass. This is because hypotonic solutions have a solute concentration lower than the cell resulting in water entering the cell and the cell grows in size.?? Similarly, it is observed that when the concentration of sucrose (mol) is higher than the sucrose concentration point where the line crosses the X-axis, it is observed to be a hypertonic concentration as there is a percentage decrease in mass. This is because hypertonic solutions have a solute concentration higher than the cell resulting in water leaving the cell to reach equilibrium and shrink.??

The hypothesis can be further supported by the qualitative analysis of the experiment as the potato pieces turn from rigid to flexible as the concentration of sucrose concentration increases. Increase in the absorption of water (from hypotonic solution) results in the cell turning turgid which therefore made the potato pieces rigid whilst as the water is lost from the pieces (in hypertonic solution), the cells turn flaccid making the pieces flexible.

The line of best fit on the graph mostly matches the main graph line, however, compared to rest of the concentrations of sucrose (mol), sucrose concentration at 0.4 mol has the biggest gap between but since it meets the decreasing pattern of the negative correlation, it couldn’t be possibly considered as an anomaly due to uncertainties existing in equipment like measuring cylinder and weights. The standard deviation of each data set is very low as it varies from 0.49 to 1.83. This indicates that all data values in the results table for each of the concentrations of sucrose (mol) is close to the mean and isn’t spread out much.

The correlation can be determined by the confidence limit calculated by using standard deviation and standard error. The confidence limit shows whether the null hypothesis is 95% correct or not. However, the confidence limit depicted on the graph with vertical lines on each point does not overlap with each other meaning that the null hypothesis is 95% rejected and therefore proves that the alternative hypothesis is correct. 

Evaluation

Although the experiment carried out was fairly simple, while conducting the trials, it was slightly difficult to maintain similar weights for all potato strips as it was measured with a ruler and cut by hand. Given that there wasn’t much precision in measurement of those strips and the cutting, each of the potato pieces might have ended up with a slightly bigger or smaller surface area to volume ratio. This could have potentially affected the absorption of water by potato pieces during the experiment as the larger the surface area to volume ratio, the more easily solutes can enter the cell through the membrane. 

The trials conducted for the experiment produced slightly different mass changes from one another and it could be due to the fact that the origin, age or type of the potatoes is not known as they were not controlled. However, on the positive note, for each of the trials, only one potato was used and cut into shape for each sucrose concentration. Since this doesn’t create a difference in the type of potato used for a trial, it could be considered as a partial control for trials where each of the pieces of potato for the concentrations of sucrose are from the same potato.

Some of the other weaknesses of the experiment are that the solutions of different sucrose concentrations were prepared by hand without the help of a lab assistant and its accuracy could have been slightly unreliable considering the percentage uncertainty of the measuring cylinders. The air conditioner in the lab was turned on which made the overall temperature of the room similar, however, this can’t be completely relied on because the trials were conducted at different times. Some of the trials were conducted during the break and constant temperature might not have been maintained as the lab door would be opened quite frequently by visitors. To maintain a more constant temperature, the boiling tubes could have been placed in a water bath which is cheap and effective and is available in the school.

 

The time given for the potato pieces to soak in the solutions was just 20 minutes. If kept longer, it would have been possible to get clearer results and observe a much larger change in mass. The qualitative results only observed the change in potato pieces, however, if they were kept longer it is possible to note changes in the solution as well. Therefore, to improve the experiment, we could collect the results after 24 hours and notice if there is any significant change in the difference in change in mass of the potato pieces.  There was also a limited number of 3 trials conducted but despite that the results of the experiment seem fairly reliable as there aren’t many fluctuations in results since the results have small values of standard deviation shown in table 4. However, to get a more precise result and mean, we could improve the experiment by conducting 5 trials in total. Overall, the experiment was a success and the results supported the hypothesis and satisfied the exploration to find out the absorption capacity of potato in different concentrations of sucrose.

 

 

 

References.

1.       Andrew A. and David M. 2014, Oxford IB Diploma Program Biology, Oxford UK, Oxford University Press  

 

2.     Anon, (2018). Online Available at: Dokimi Science, website: http://www.dokimiscience.com/b—osmosis.html Accessed 5 Jan. 2018.

 

 

3.      Dorji, T.Y., Tamang, A.M. and Vernooy, R., 2015. The history of the introduction and adoption of important food crops in Bhutan. Rice, maize, potato and chili.

 

4.      Hesser, A.1999, website: http://www.nytimes.com/1999/05/05/dining/deep-secrets-making-perfect-fry-potato-moment-often-soggy-disappointment-time.html

 

 

5.      Ib.bioninja.com.au. (2018). Osmolarity | BioNinja. Online Available at: http://ib.bioninja.com.au/standard-level/topic-1-cell-biology/14-membrane-transport/osmolarity.html Accessed 4 Jan. 2018.

 

6.      Khan Academy. (2018). Khan Academy. Online Available at: https://www.khanacademy.org/science/biology/membranes-and-transport/diffusion-and-osmosis/a/osmosis Accessed 28 Jan. 2018.

 

7.      Scott, G.J., 1983. Marketing Bhutan’s potatoes: present patterns and future prospects. International Potato Center.