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Investigation: The Effect of Salt on a Potato - Biology

Investigation: The Effect of Salt on a Potato - Biology


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What do you think will happen if you soak a potato in salt water overnight?

Pre Lab Discussion:

  1. Consider what you know about osmosis and transport across the membrane and PREDICT what will happen to potato slices soaked in salt water. Describe your predictions.
  2. If you are going to place potato slices in salt water, what do you need to set up as your CONTROL?

Procedure and Data Collection:

Use a scale to weigh your potato slices in grams. Mass of Potato Slice 1 ___________ Mass of Potato Slice 2 _________

Place potato slice 1 in a cup of salt water, place potato slice 2 in a cup of distilled water.

Use a sharpie to label the cups as SALT or DI water and write your initials. Let sit overnight.

Weigh each of the slices the next day. Mass of Potato Slice 1 ___________ Mass of Potato Slice 2 _________

Analysis

  1. Describe what happened to the mass of each slice (did it go up, down, or stay the same). Write in complete sentences.
  2. Refer to your notes or your understanding about osmosis. Explain WHY you got these results. Write in complete and thoughtful sentences, use the back of this page, and include the following words in your explanation: osmosis, cells, solution

Investigating the factor affecting osmosis in potatos

Disclaimer: This work has been submitted by a university student.

Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of AUEssays.com.

Osmosis is diffusion of water molecules moving from a particularly permeable membrane from a high concentration to a lower concentration. The water molecules can pass both ways through the membrane in two-way traffic, but there is a steady net flow of water molecules into the lower concentration of water. In figure 1 is a diagram showing how osmosis takes place.

What affects the rate of osmosis?

There are many factors that can affect the rate of osmosis these are concentrations of solutions in osmosis, the surface area and temperature. In this specific experiment, there are specific factors that can affect the rate of osmosis.

The concentration of solutions (the water potential) can affect rate of osmosis as more differences between the concentrations of the solutions means a steeper concentration gradient, meaning a faster rate of osmosis. In this particular experiment, if there is more

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concentration (potential) of the solution outside the potato than inside, then when osmosis takes place water molecules will go through the permeable membrane into the potato making the potato heavier and more saturated, while if there is a higher concentration inside the

potato than outside, the opposite will occur, lastly, if the concentration inside and outside are similar or equal, then the rate of reaction will be quite slow or no reaction will occur as then reach equilibrium. Concentration can be altered by diluting the solution but keeping it at the same volume, (i.e. from 10cm³ of sucrose solution, to 2cm³ of distilled water with 8cm³ of sucrose solution, equal to 10cm³ of a mixture.)

In osmosis with this particular experiment, this is the permeability of the potato, whether the water molecules can easily get through the permeable membrane or not during osmosis, the easier for the water molecules to pass through means faster rate of osmosis. The permeability of the potato can be determined by the age of the potato, as older the potato is the less permeable it is, meaning a slower rate of osmosis.

The higher the temperature is the faster the rate of osmosis, this is because high temperature gives the particles from the solutions more kinetic energy, therefore there they can collide faster and have more successful collusions to give a faster rate of osmosis.

The concentration of the surrounding solution affects the length and mass of the potato. This is due to the different concentration of water molecules between the potato and its surrounding solution. When there is a lower concentration inside the potato than its surrounding, the water molecules from the outside will move through the permeable

membrane and into the potato, causing the potato to become saturated, thus adding the weight of the potato at the end of the reaction. However, if the concentration outside is lower than the concentration inside the potato, then water molecules from the potato will move through the permeable membrane and into the outer solutions, thus the potato is unsaturated and its weight is lost at the end of reaction. However, when there is equal amount of water molecules in and outside the potato, then no osmosis will take place as it reaches equilibrium and follows the theory of osmosis. Here is a predict graph and analysis of what will happen as the concentration of sugar solution increases to the rate of osmosis.

1. Potato is saturated and water molecules move at maximum rate of osmosis, therefore it increase the mass of potato.

2. In case of slightly fewer water molecules and lower rate of reaction, therefore only little increase in mass of potato.

3. If the concentration of water molecules in and outside of the potato is at equilibrium, then no net movement and no change in mass.

4. If there are more water molecules inside the potato and moves out of the potato, then the mass will decrease and become unsaturated.

5. More water molecules inside the potato than outside, therefore the potato becomes dehydrated and unsaturated with light weight.

Knife, Potato borer and Ruler

1.Prepare 6 different sugar solutions of different concentration. This is because different concentration of the distilled water and will affect the rate of osmosis. The following table shows the amount of distilled water and sucrose solution to be added to prepare various concentrations of sucrose solutions.

Concentration of Sugar Solution (%)

Volume of sugar solution (cm3)

Volume of distilled water (cm3)

Thus, 6 different solutions are prepared and experimented to yield more accurate and reliable results. The length of potato will be the same that is each potato strip will be of 2 cm long and the size of it will always be at whatever potato borer used, so that it doesn’t disturb the result and only the concentration is affecting the result.

Cut out 6 pieces of potato with the help of potato borer.

Put test tubes onto a test tube rack.

Pour out all 6 sugar solutions prepared in each test tube with the help of a pipette.

Weigh the mass of potatoes on an electronic balance before putting them into the test tubes.

Put them in different solutions at the same time and start the stop watch time it up to 10 minutes.

Pour out the mixtures out of the test tubes immediately after 10 minutes and measure their individual mass after osmosis. Note down you results onto the table.

Repeat the steps 1-6 for two more times to yield more data which could give us accurate and precise results.

Data Collection and Processing:

Concentration of sugar solution (%)

Weight of potato BEFORE (g) +0.01

Change in mass of the potato before and after the reaction is calculated by subtracting the mass of the potato before by mass of the potato after, so in the first trial of 0% sugar solution change in mass is as follows:

Change in mass= Mass of potato before – mass of potato after

% change in weight is calculated by multiplying the ratio of difference to the mass of potato before into 100. This would give us the %change in weight. For example,

% change in weight= (Difference in mass/ Initial mass)* 100

The mean % change is calculated by adding the sum of all the trials and dividing the answer by 3. For instance in 0% sugar solution the mean% change is calculated as given below:

Mean % change= Trial 1 + Trial 2 + Trial 3

Graph 1 – Graphing Analysis

Conclusion and Evaluation:

Graph 1 above shows on the whole that as the concentration of sugar solution increases, the percentage change in mass decreases (i.e. from 5.51 at 2% and concentration to 3.89 at 5% concentration). This is because of the fact that as the concentration of sugar solution increases the amount of water present in the solution or the concentration of water would be less. Hence, the concentration gradient of the water inside and outside of the potato is less steep. The graph shows a gentle decreasing slope from 0 to 5% sugar solution. This means that the rate of osmosis decrease gently as the concentration of water molecules in and outside are nearer to equilibrium, which is at approximately 4% in this case. After 5% there is a sudden decrease in the slope, this could be due to the reason that there was more movement of water molecules in and out of the potato. However, at 15%, there is an increase in mass (from -0.56 to 0.32) and this proves against the hypothesis. Thus we can say that the data obtained is anomalous. An error found out in this is the differences of average of mass between each concentration there is less difference between the results of 0, 2 and 5%. The highest significant difference between the mass of two trials of the same concentration is 0.05 in 5% concentration where as the highest difference among the 10, 15 and 20% concentration is 0.09in 10% solution. This can be seen in graph 3 shown below.

Graph 3- Comparison of significant difference in results

This states that the slope between concentrations 0 to 2% is less steep than concentration 5 to 10% and this concludes that saturating the potato gives a faster rate of osmosis as the concentration or the amount of water molecules in 0, 2 and 5% sugar solution is higher. This suggests that there is an uneven net flow in osmosis. But according to me, this investigation is not reliable on whether the relationship between rate of osmosis and % of sugar concentration are proportional.

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Graph 2 shows the % mean change in mass of potato strips at different concentrations. It shows that the % change in mass of potato in 0% sugar solution is highest as the sugar solution is absent in the mixture. This is because of the reason that water can easily pass through the semi permeable membrane and also because the water molecules inside the potato were less in number. The graph also shows that the % mean change in mass of potato is negative in case of 10% sugar solution. This could be because of the reason that there are more water molecules inside the potato and would come out during the reaction making the potato lighter and unsaturated.

From the results and graph one could easily observe a pattern between the concentration of sugar solution and change in mass. As the concentration of sugar solution increases, change in mass of the potato decreases. That is the rate of osmosis decreases with the decrease in concentration of water molecules. However, there is an anomalies data at concentration 15% as the change in mass is not negative but at 0.32 this concludes that the hypothesis was wrong and so it is not in favour of osmosis theory. In order to improve on this I will have to investigate the change in mass results in all three experiments for concentration at 15%, it should come out to be in negative as the mass of potato should decrease according to osmosis theory. In trial 2 and 3 for 15% sugar solution the change in mass is 1.57 and 0.83 instead of been negative and also there is no calculation mistake. This proves that the there has been error in preparing the 15% sugar solutions.

The experiment should be performed more times to collect a good yield of data which could lead us giving an accurate and reliable result. This would also decrease the anomalous data when drawing a graph. The volumes of the solution should be accurate and equal before pouring it into a mixture, or else the concentration of a specific concentration would be altered. The start and end time for osmosis should be taken care off so that osmosis does not take place for different duration for every trial. Weight the potatoes all together at the same time on individual electronic balance, so that osmosis does not take place to one potato while another potato is being weighed.

Osmosis is diffusion of water molecules moving from a particularly permeable membrane from a high concentration to a lower concentration. The water molecules can pass both ways through the membrane in two-way traffic, but there is a steady net flow of water molecules into the lower concentration of water. In figure 1 is a diagram showing how osmosis takes place.

What affects the rate of osmosis?

There are many factors that can affect the rate of osmosis these are concentrations of solutions in osmosis, the surface area and temperature. In this specific experiment, there are specific factors that can affect the rate of osmosis.

The concentration of solutions (the water potential) can affect rate of osmosis as more differences between the concentrations of the solutions means a steeper concentration gradient, meaning a faster rate of osmosis. In this particular experiment, if there is more

concentration (potential) of the solution outside the potato than inside, then when osmosis takes place water molecules will go through the permeable membrane into the potato making the potato heavier and more saturated, while if there is a higher concentration inside the

potato than outside, the opposite will occur, lastly, if the concentration inside and outside are similar or equal, then the rate of reaction will be quite slow or no reaction will occur as then reach equilibrium. Concentration can be altered by diluting the solution but keeping it at the same volume, (i.e. from 10cm³ of sucrose solution, to 2cm³ of distilled water with 8cm³ of sucrose solution, equal to 10cm³ of a mixture.)

In osmosis with this particular experiment, this is the permeability of the potato, whether the water molecules can easily get through the permeable membrane or not during osmosis, the easier for the water molecules to pass through means faster rate of osmosis. The permeability of the potato can be determined by the age of the potato, as older the potato is the less permeable it is, meaning a slower rate of osmosis.

The higher the temperature is the faster the rate of osmosis, this is because high temperature gives the particles from the solutions more kinetic energy, therefore there they can collide faster and have more successful collusions to give a faster rate of osmosis.

The concentration of the surrounding solution affects the length and mass of the potato. This is due to the different concentration of water molecules between the potato and its surrounding solution. When there is a lower concentration inside the potato than its surrounding, the water molecules from the outside will move through the permeable

membrane and into the potato, causing the potato to become saturated, thus adding the weight of the potato at the end of the reaction. However, if the concentration outside is lower than the concentration inside the potato, then water molecules from the potato will move through the permeable membrane and into the outer solutions, thus the potato is unsaturated and its weight is lost at the end of reaction. However, when there is equal amount of water molecules in and outside the potato, then no osmosis will take place as it reaches equilibrium and follows the theory of osmosis. Here is a predict graph and analysis of what will happen as the concentration of sugar solution increases to the rate of osmosis.

1. Potato is saturated and water molecules move at maximum rate of osmosis, therefore it increase the mass of potato.

2. In case of slightly fewer water molecules and lower rate of reaction, therefore only little increase in mass of potato.

3. If the concentration of water molecules in and outside of the potato is at equilibrium, then no net movement and no change in mass.

4. If there are more water molecules inside the potato and moves out of the potato, then the mass will decrease and become unsaturated.

5. More water molecules inside the potato than outside, therefore the potato becomes dehydrated and unsaturated with light weight.

Knife, Potato borer and Ruler

1.Prepare 6 different sugar solutions of different concentration. This is because different concentration of the distilled water and will affect the rate of osmosis. The following table shows the amount of distilled water and sucrose solution to be added to prepare various concentrations of sucrose solutions.

Concentration of Sugar Solution (%)

Volume of sugar solution (cm3)

Volume of distilled water (cm3)

Thus, 6 different solutions are prepared and experimented to yield more accurate and reliable results. The length of potato will be the same that is each potato strip will be of 2 cm long and the size of it will always be at whatever potato borer used, so that it doesn’t disturb the result and only the concentration is affecting the result.

Cut out 6 pieces of potato with the help of potato borer.

Put test tubes onto a test tube rack.

Pour out all 6 sugar solutions prepared in each test tube with the help of a pipette.

Weigh the mass of potatoes on an electronic balance before putting them into the test tubes.

Put them in different solutions at the same time and start the stop watch time it up to 10 minutes.

Pour out the mixtures out of the test tubes immediately after 10 minutes and measure their individual mass after osmosis. Note down you results onto the table.

Repeat the steps 1-6 for two more times to yield more data which could give us accurate and precise results.

Data Collection and Processing:

Concentration of sugar solution (%)

Weight of potato BEFORE (g) +0.01

Change in mass of the potato before and after the reaction is calculated by subtracting the mass of the potato before by mass of the potato after, so in the first trial of 0% sugar solution change in mass is as follows:

Change in mass= Mass of potato before – mass of potato after

% change in weight is calculated by multiplying the ratio of difference to the mass of potato before into 100. This would give us the %change in weight. For example,

% change in weight= (Difference in mass/ Initial mass)* 100

The mean % change is calculated by adding the sum of all the trials and dividing the answer by 3. For instance in 0% sugar solution the mean% change is calculated as given below:

Mean % change= Trial 1 + Trial 2 + Trial 3

Graph 1 – Graphing Analysis

Conclusion and Evaluation:

Graph 1 above shows on the whole that as the concentration of sugar solution increases, the percentage change in mass decreases (i.e. from 5.51 at 2% and concentration to 3.89 at 5% concentration). This is because of the fact that as the concentration of sugar solution increases the amount of water present in the solution or the concentration of water would be less. Hence, the concentration gradient of the water inside and outside of the potato is less steep. The graph shows a gentle decreasing slope from 0 to 5% sugar solution. This means that the rate of osmosis decrease gently as the concentration of water molecules in and outside are nearer to equilibrium, which is at approximately 4% in this case. After 5% there is a sudden decrease in the slope, this could be due to the reason that there was more movement of water molecules in and out of the potato. However, at 15%, there is an increase in mass (from -0.56 to 0.32) and this proves against the hypothesis. Thus we can say that the data obtained is anomalous. An error found out in this is the differences of average of mass between each concentration there is less difference between the results of 0, 2 and 5%. The highest significant difference between the mass of two trials of the same concentration is 0.05 in 5% concentration where as the highest difference among the 10, 15 and 20% concentration is 0.09in 10% solution. This can be seen in graph 3 shown below.

Graph 3- Comparison of significant difference in results

This states that the slope between concentrations 0 to 2% is less steep than concentration 5 to 10% and this concludes that saturating the potato gives a faster rate of osmosis as the concentration or the amount of water molecules in 0, 2 and 5% sugar solution is higher. This suggests that there is an uneven net flow in osmosis. But according to me, this investigation is not reliable on whether the relationship between rate of osmosis and % of sugar concentration are proportional.

Graph 2 shows the % mean change in mass of potato strips at different concentrations. It shows that the % change in mass of potato in 0% sugar solution is highest as the sugar solution is absent in the mixture. This is because of the reason that water can easily pass through the semi permeable membrane and also because the water molecules inside the potato were less in number. The graph also shows that the % mean change in mass of potato is negative in case of 10% sugar solution. This could be because of the reason that there are more water molecules inside the potato and would come out during the reaction making the potato lighter and unsaturated.

From the results and graph one could easily observe a pattern between the concentration of sugar solution and change in mass. As the concentration of sugar solution increases, change in mass of the potato decreases. That is the rate of osmosis decreases with the decrease in concentration of water molecules. However, there is an anomalies data at concentration 15% as the change in mass is not negative but at 0.32 this concludes that the hypothesis was wrong and so it is not in favour of osmosis theory. In order to improve on this I will have to investigate the change in mass results in all three experiments for concentration at 15%, it should come out to be in negative as the mass of potato should decrease according to osmosis theory. In trial 2 and 3 for 15% sugar solution the change in mass is 1.57 and 0.83 instead of been negative and also there is no calculation mistake. This proves that the there has been error in preparing the 15% sugar solutions.

The experiment should be performed more times to collect a good yield of data which could lead us giving an accurate and reliable result. This would also decrease the anomalous data when drawing a graph. The volumes of the solution should be accurate and equal before pouring it into a mixture, or else the concentration of a specific concentration would be altered. The start and end time for osmosis should be taken care off so that osmosis does not take place for different duration for every trial. Weight the potatoes all together at the same time on individual electronic balance, so that osmosis does not take place to one potato while another potato is being weighed.

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What is Osmosis?

Say you have a container, with a separator down the middle. The separator is a semipermeable membrane, or a biological membrane that allows for the passage of certain molecules. The container is filled with water on both sides, with the side on the left containing a low concentration of salt, and the container on the right containing a high concentration of salt.

This particular solution is saltwater, where the water is the solvent and the salt is the solute. All this means is that the saltwater is more water than salt, and while that’s true for both sides, one side has more salt than the other.

Over time, the water on the left, containing less salt, will diffuse through the semipermeable membrane into the side of the container with a higher concentration of salt. Water molecules can pass through this membrane, but the salt can’t. The concentration of salt will remain the same on both sides, but now the concentration of water on the side with more salt will be greater.

This is due to a process called osmosis. You can learn more about a related process, diffusion, in this guide. Knowing basic cell structure is important for fully understanding osmosis. Check out this course on GCSE Biology for more in depth instruction.


"The Effect Of Salt Concentration On Osmosis In Potato Cells" Essays and Research Papers

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OSMOSIS EXPERIMENT: POTATO, WATER, SALT

An experiment that involves potatoes, water, and salt are conducted to prove that there’s an osmosis reaction in a plant cell, which is in this case it’s a potato. Osmosis is a process where there’s a movement of molecules in dissolve component such as water from a high concentration to a lower concentration with a membrane selective permeable. It operates like a gate where it allows and separates certain things to go inside specifically dissolve substance such as water. The experiment started by putting both same sized potato to each beaker that filled with water and salt water solution. Due to osmosis water, it will flow to where there is a higher concentration of solvent and the solvent will flow to where there is less concentration. When putting the salt water, the water will flow out of the cell because it will dilute the extra solvent (salt) outside the cell. And when putting the fresh water, It will rush in because there is more solvent in the cell.

It is predicted before the experiment was conducted, the potato that was places inside the beaker with salt water will shrink due to molecules of salt solution are joined more tightly together than in fresh water. A large number of salt entered the cells of the potato, more water exits the cell ( like in the hypertonic solution) causing the cell to snivel or die causing the potato to shrink.

Proving the osmosis theory, only uses a couple of things which includes independent variables Beaker, measuring tube, Measuring weight, Stopwatch and dependent variables Potato, Salt and Water. The first and most important step is that to cut the potato into the same size, because at the end, it’ll show the difference of size between both potatoes. Making sure that the potato is the same size, it’s better to measure it with measuring weight to make sure that they’re in the same weight to begin with. Then, measure the amount of salt that’s going to be inserted in one of the beaker. In this case, it’s 5 tea spoon of salt which is the same as 30g. Then prepare the two beakers and fill it with 200ml of water each. Making sure that it’s the same amount of water in each beaker, using the measuring tube to be more precise. Stir the beaker until the salt is dissolve and begin to put the potatoes to each beaker at the same time while starting the timer to 60 minutes. Since it’s quite a long time, from this process it is necessary to observe both of the potatoes.

After an hour, immediately put the water inside the measuring tube to figure out the difference between both water after it was surrounded by the salt solution and also after the potato was inside the water (with no salt). Do it one by one and slowly is more effective because it’s less likely to spill. If so, it’ll ruin the entire experiment. After both water has already been measured moving on to wight the potato to find out what happened to the potato when it’s surrounded by both water and salt water. Here’s the result after the experiment was conducted:


In osmosis, water which is uncharged molecules passes through the cell membrane. A cell membrane is a semi-permeable membrane, that is, it only allows small particles to go through it. Osmosis depends on its concentration gradient, from a region of low concentration as it moves to a region of high concentration. It is dependent on temperature, size of the molecule, thinness of the membrane and the concentration gradient. In this activity, osmosis in potato cell was studied. Salty water with Cl- (aq) is concentrated as compared to the cell sap, that is, hypertonic solution. The cell sap, therefore, loses water by osmosis and shrink. From the results, as the concentration of salts increases, the mass of potato reduces, therefore solution with 0.0 molarity of NaCl (aq) recorded an increase in mass while the ones with 1.0 molarity of NaCl (aq) recorded a great decrease in mass. The process where a plane cell loses water, shrink and become flaccid is called deplasmolysis. A plasmolyzed cell can be made turgid by immersing it in distilled water for some time (deplasmolysis) (Odom et al, 2017).

Therefore for osmosis to occur, the two solutions must be separated by a semi-permeable, membrane, for this case, the potato cell membrane is only permeable to water molecules from potato cell but is not permeable to salt molecules from Cl-(aq) ions that are present in water solution. At 30 min, all of them showed a decrease in mass because water is drawn from potato tissue to the salt solution. As the time increases, the cell is continually losing water to offset the osmotic imbalance. The overall change in mass is a decrease, in the end, the solutions become isotonic, that is, all solution s have attained the same concentration, and there is no net movement of water molecules from one solution to another.

The main aim of osmosis is to offset osmotic imbalances in bodies of living organisms to provide optimum conditions for body functions. Osmosis continues to take place until when the two solutions are almost equal in concentration.

As osmosis is taking place, a force will be created in the hypertonic solution that aims to prevent osmosis from taking place such a force is called osmotic pressure. If osmotic pressure of a solution is high, it will draw a lot of water from the adjacent solution, and if it is low, it will draw little waiter. Osmotic pressure is, therefore, a measure of the total amount of dissolved salts in water (Xu et al, 2017).


Methods

Plant materials and vector construction

This study was performed at the Gansu Provincial Key Lab for Aridland Crop Sciences, Gansu Agricultural University in 2012 and 2013. The microtubers of the potato cultivar ‘Shepody’ were used in all the tests, which were from Gansu Key Laboratory of Crop Genetic and Germplasm Enhancement. The culture methods were the same as those described in a previous publication [49].

The 1521 bp cDNA of the AtHKT1 gene (AY685182) isolated from Arabidopsis thaliana was cloned into the BamH I-Sac I site of the plasmid pBI121 containing the CaMV 35S promoter, thus replacing the β-glucuronidase (GUS) gene with the AtHKT1 gene. The recombinant plasmid pAtHKT1 was introduced into Agrobacterium tumefaciens strain LBA4404 using the freeze-thaw method [50]. The presence of the plasmid was verified using restriction enzyme digestion and PCR amplification.

Transformation, PCR, PCR-southern, southern blot and reverse transcription RT-PCR analysis

The detailed procedures of the potato transformation, plant genomic DNA extraction, total RNA extraction, reverse transcription, PCR-Southern, Southern Blot and RT-PCR were conducted as described previously [49]. In brief, the slices of the ‘Shepody’ microtubers were used as the receptor for Agrobacterium-mediated transformation. The AtHKT1 gene was amplified using the forward primer 5′- CGGGATCCATGGACAGAGTGGTGGCAAAAATAG -3′(1 - 31 bp) and the reverse primer 5′- CGGAGCTCTTAGGAAGACGAGGGGTAAAGTATCC - 3′(1490 - 1521 bp), which generated an expected PCR product fragment of 1521 bp. The amplification was performed in a thermal cycle (UNO II, Biometra) programmed for one cycle of 1 min at 94 °C, followed by 35 cycles of 50 s at 94 °C, 60 s at 55 °C, and 90 s at 72 °C. A final extension step was performed for 10 min at 72 °C.

Gene expression assays using quantitative real-time PCR

Total RNA was extracted from the plantlets of fifteen transgenic potato lines and a non-transgenic control. The RNA quality, reverse transcription and qRT-PCR amplification were performed as described by Wang [49]. efla was the internal control gene with the forward primers 5′-CAA GGATGA CCC AGC CAA G − 3′ and the reverse primers 5′-TTCCTT ACC TGAACGCCT GT-3′. The forward and reverse primers of the AtHKT1gene were 5′ - ATC TGG CTC CTA ATC CCT CAA - 3’and 5′ - CCG TCA CTC CAA GAA GAA CAC - 3′, respectively. Each Quantitative Real-Time PCR was independently repeated three times. A blank control was included in which no cDNA was added to the reaction mixtures. After each reaction, melt curve analysis was used to verify the specificity of amplification, and the relative expression levels were calculated by 2 −ΔΔCt .

Physiochemical evaluations, yield and biomass of transgenic potato transformed with the AtHKT1 gene

In controlled-environment greenhouses, mini-tubers of the AtHKT1 transgenic potato lines were evaluated for their levels of stress resistance. The mini-tubers were outputted from test-tube seedlings as previously described [51]. In brief, test-tube seedlings were transferred in a plate (30 × 60 cm and 10 cm in height) filled with vermiculite [(NH4)2HPO4 26.7 g kg − 1 , KH2PO4 13.3 g kg − 1 ] under room temperatures. The vermiculite moisture was maintained at 70–80% field capacity. A nutrient solution at 100 ml MS (pH 6.0) was added to each plate every seven days. Mini-tubers were harvested at maturity and stored at 4 °C for 3 months. After recovery from dormancy, each mini-tuber was planted in a pot (32 cm in diameter and 25 cm in height) filled with 50% loessial soil and 50% vermiculite (v/v) under room temperatures. All pots were watered daily to maintain the vermiculite moisture at 70–80% field capacity with 200 ml nutrient solution (KNO3 9.89 mmol, (NH4)2SO4 1.29 mmol, MgSO4 2.08 mmol, KH2PO4 2.57 mmol and FeSO4 0.20 mmol per liter, pH 5.8) added to each pot every seven days. Healthy plants with a similar physical size at the budding stage were selected for the NaCl treatments, and those plants that survived the NaCl challenge were measured further.

Two transgenic potato lines (T3 and T13) had the highest expression of the AtHKT1 gene, three had a moderate expression (T1, T2 and T5), and the remaining two had a low expression (T11 and T15). These transgenic lines, along with the non-transgenic control, were tested under four NaCl treatments: 0, 50, 100, and 150 mmol L − 1 NaCl. These salt levels were maintained by adding 1500 ml NaCl (pH 6.0) and 500 ml nutrient solution (pH 5.8) to each pot every seven days and a daily watering at 500 ml to maintain the vermiculite moisture at 70–80%. NaCl irrigation was started from potato initial flowering stage and continued for 30 days. The entire experiment with three replications was repeated three times, using three planting dates from April through August in 2013.

At each of the three runs, the treatments with three replicates were completely randomized. In each replicate, each treatment had 10 pots, and one plant per pot, resulting in a total of 960 pots (4 stress-treatments × 3 replicates × 8 lines × 10 pots each). The relative humidity in the greenhouse was at 70 to 75%, and the day/night temperatures were 25/15 °C. The photon flux density was 47.25 μmol m − 2 s − 1 and the day/night photoperiod was 14/10 h. At day 30 of NaCl treatments, the Na + content in vermiculite was tested (Additional file 1: Table S1) and the following measurements were taken for each plant: plant height, stem thickness expansion rate, leaf area expansion rate, PN, Tr, Gs, Ci, total chlorophyll, proline, soluble sugar contents, RWC, MDA, electrolyte leakage, SOD, CAT and POD. These measurements were taken following the protocols previously described [49]. The leaves was used to test the K + and Na + contents. The roots were sampled to determine the root activity. The K + and Na + contents were measured as described by Ghars [52].

The root activity was determined using the triphenyl tetrazolium chloride (TTC) method [49]. The survival rate was counted as the ratio of the number of plants that survived the NaCl treatments to the total number of plants before the NaCl treatments at day 65 after the NaCl treatments and. At 90 days after seedling emergence (full maturity), the tubers, shoots, and roots were hand-harvested and weighed for each plant. The plant sample dry weights were determined after being oven-dried at 70 °C to a constant weight.

Genetic stability analysis and quantification of salt tolerance

Plant genomic DNA was extracted from 0.1 g leaves of the control and all the transgenic plants. The AtHKT1 gene was amplified using PCR techniques. All the methods were the same as those described in the above.

To quantify the degree of tolerance to NaCl, we calculated evaluation indices for all plant lines using an integrated cluster analysis in which standard deviation coefficients were integrated with allocation weighted values. The detailed methodology was described previously [49]. In brief, an evaluation index was developed using the following three steps:

A subordinator function was formed and used to standardize the measured raw data of each parameter

where j is the variable of six plant-growth traits, including daily gain of plant height, leaf area expansion rate, stem thickness expansion rate, plant tissue water content, tuber yield, and plant survival rate in salt treatment i is the genotype Xij is the value of the jth variable of the ith genotype Xmin and Xmax are the minimal or maximal value among the Xij values for the jth variable, respectively. Equation (1) was to calculate positive correlation between a particular growth variable and salt tolerance, while eq. (2) was to calculate negative correlation.

A weighted coefficient was expressed by the standard deviation coefficient for each of the six plant-growth traits, as follow:

where n represents the number of genotypes, and m represents the plant growth traits.

An integrated evaluation index was generated using the equation below:

where D represents the integrated evaluation index that quantifies the response of the potato genotypic lines to the different levels of salt stress [49].

Statistical analysis

The data from the three runs of experiments were pooled together in the analysis, as the initial analysis showed that there were no significant treatment by run interactions for most of the variables measured, and the three runs of the experiment resembled the trend of treatment effects. All the data were analyzed using Tukey’s HSD test with the SPSS package (SPSS Software, 19.0, SPSS Institute Inc., USA). Cluster analysis and the quantitative assessment indices were used to categorize the seven genetic lines into the three levels of salt tolerance: low, medium, and high. The significances among treatments in the ANOVA or between the three levels of salt tolerance were determined at P < 0.05. The salt tolerance was assessed using parameters, including plant morphological and physiological traits, enzyme activities, and yield-related variables. The contributions of AtHKT1 gene expression, salt tolerance evaluation index and plant physiochemical parameters to the overall variance were further analyzed using principal component analysis (PCA) and a two-dimensional PCA-biplot was generated to graphically demonstrate PC1 and PC2 scores. The angles between two vectors and the origin represent their relationship, with acute angles indicating positive correlations, right angles indicating no correlation, and obtuse angles indicating a negative correlation.


Catalase activity in potato

The aim of this investigation is to understand and monitor the effects pH levels have on catalase activity within controlled potato samples. Catalase is a type of enzyme that is found widespread among organisms that grow in the presence of oxygen (almost all living things) its specific use is to speed up the decomposition of Hydrogen peroxide (H2O2) within cells.

Hydrogen peroxide is a bi-product of chemical processes such as respiration and is harmful if not removed or broken down into less harmful molecules.

The enzyme (Catalase) acts as a catalyst, vastly accelerating this chemical reaction and allowing cells to function at a level where they can support life. Catalase Hydrogen-peroxide Water + Oxygen 2H2O2 2H2O + O2

Enzymes are a protein made up of Amino Acids, they are essential to all living organisms as well as providing many commercial applications across multiple industries (including detergents, food and beverages, pharmaceuticals, biofuels, biodegradable plastics, leather and adhesive removal), within bio-detergents for example, the enzyme Protease is used to allow a deeper cleaning of textile materials at lower temperatures, while Pectinase (another form of enzyme found in plant cell walls) is used to partially digest fruit and vegetables in baby food (I).

Each type of enzyme is unique in that it will only react with the substrate that complements the enzymes ‘Active Site’. The way the enzyme and substrate combine to react can be described as a ‘lock and key’ model, as the shape of each substrate (e. g. Hydrogen-peroxide), will only fit with the complementing enzyme (e. g. Catalase). (II) Having very specific and varying chemical reactions to make, each enzyme also exists in very different environments, for example Gastric lipase is an enzyme in the stomach of mammals that speeds up the digestion of foods, being in a very acidic environment their optimum pH range is 3-6 (III).

When enzymes are exposed to pH levels above or below their optimum pH, the Active Site of the enzyme can become distorted or warped to the point where the complementary substrate can no longer fit in the active site, when this occurs the enzyme is known to have become ‘denatured’ (IV), this is very important to enzymes as they are reusable and become useless once denatured.

Temperature also affects the activity of catalase as well as cause enzymes to become denatured, however this Investigation will monitor the activity of catalase across the pH spectrum and identify where activity is at its optimum, and at what pH the catalase enzyme becomes denatured (temperature will therefore be closely controlled). Factors that might affect catalase activity: pH Levels Temperature Concentration of potato (catalase) Potato type (source of catalase) Concentration of Hydrogen-peroxide (Reactant) Time (measure reaction rates)

Factor to monitor (independent variable) – pH: pH is one of the most important factors affecting catalase and all enzyme activity, by analysing reaction rates over a range of pH levels (pH3 – pH11), this investigation will show at what pH catalase activity will be at its optimum. As well as show at which levels the catalase enzyme becomes denatured (where environmental factors such as temperature or pH will irreparably damage the enzyme, essentially stopping the protein from functioning). A buffer solution is created for each of the pH levels included in the investigation.

Research and preliminary data obtained from a smaller experiment (figure 2 & table 2) utilising yeast as a source of catalase, suggest that enzymes will become denatured below pH3 and above pH11, for this reason and due to time restraints, the pH levels tested will be pH3, pH5, pH7, pH9 and pH11. This cross section of pH levels will provide tangible data as each step is equally separated, and within the environmental settings catalase can support reactions. Strategy: There are a number of ways that could be used to measure the rate of catalase activity: Measuring the appearance of a product (O2) (V)

Measuring the rate of a disappearance of substrate (H2O2) Measuring the pressure of the product as it appears (O2) (VI) For this investigation I will be measuring the rate of the appearance of a product (in this case O2), the reason I chosen this method is that preliminary research suggest this is a wildly used and reliable study with the limitation of time and equipment available (V). O2 could also be measured by simply counting oxygen bubbles, however by measuring O2 in cm3 I can obtain more accurate and scientific results.

The production of water (H2O) could also be monitored, however this would be difficult to measure as the amount of product is relatively small and translucent as with Hydrogen-peroxide. Measuring the disappearance of substrate (H2O2) would be extremely difficult to measure by sight or with the equipment available to us, which also applies to measuring pressure. Potato samples (catalase source) will be placed in a pH buffer solution before being exposed to the Hydrogen-peroxide (substrate), O2 production will then be measured in cm3 across a controlled period of time.

The experiment will be conducted at a temperature of 25o as temperature has such a big effect on enzyme activity (as shown below). (VII) Hypothesis: Preliminary research (figure 2 & table 2) and my understanding of how enzymes function have led me to predict that when pH levels are altered, the rate of catalase activity will also change. My understanding of the subject tells me that each type of enzyme has a specific duty and will only react with its complementary substrate, research also identifies that the active site of enzymes can become denatured when exposed to extreme changes to pH or temperature outside to their natural environment.

This shows me that catalase (as a type of enzyme) will have an optimum pH level for activity and that a pH level much higher or lower than this will cause the enzymes to become denatured, causing activity levels to fall and eventually stop as the substrate no longer fits in the active site. Based on this knowledge it’s therefore essential to maintain a steady even temperature throughout the experiment as well as control other variables such as sample size/surface area, and the time these samples are exposed to both the buffering solution and substrate (Hydrogen-peroxide, H2O2).

Preliminary research (figure 2 & table 2) has been conducted on yeast samples where 3 reading were taken at pH levels of 3, 7 and 11. Data readings suggest that catalase activity did not begin until after pH3 and peaked between pH7 & pH10. This evidence along with research from other sources supports my hypothesis that optimum activity will occur between pH8 and pH10. Catalase enzymes will become denatured and unreactive at and below pH3 and also denatured when reaching high pH alkali levels above pH10. HHApparatus: Apparatus Image

Justification Boiling Tube – (with delivery arm) Contains the reactants (potato samples and Hydrogen-peroxide), and where the reaction is made. Arm is required for the Delivery tube allowing the passage of gas product (Oxygen) Rubber Bung Seals boiling tube from outside gasses, ensures all products from the reaction are unable to escape and follow their path to the measuring cylinder via the delivery tube 2 x Measuring Cylinder Where product (Oxygen) is captured and used to measure the volume of oxygen (cm3), and thus the rate of the reaction.

A further is used to measure accurate levels of substrate (Hydrogen-peroxide) Delivery Tube Connects boiling tube to the measuring cylinder, allowing the passage of gas product (Oxygen) for measurement Beaker (filled with water) Ensures the Measuring Cylinder is airtight allowing accurate readings of product (Oxygen) captured Boiling Tube Rack To safely hold Boiling Tube in position throughout the experiment Digital Timer Accurately measure how long potato sample is submerged in both buffer solution and Hydrogen-peroxide, ensuring a fair investigation Thermometer

Ensure that temperature conditions remain stable, temperature could affect catalase activity and therefore should be controlled to ensure fair results Scalpel Precisely cut potato samples to required length, sample sizes should remain the same to ensure concentration of catalase within the experiment remain the same Size 5 borer Precisely measure potato sample sizes, ensuring catalase concentration remains constant throughout the experiment White tile White tiles provide a safer environment to cut samples of potato to required size and dimensions Control Variables:

Control Variable Why it needs controlling How it’s controlled Catalase concentration (potato sample) Catalase concentration will affect the rate of reaction and ultimately the quantity of product (Oxygen) captured and measured. All potato samples will be cut from using a size 5 borer and measured to 3cm in length. Each test will use 3 of these samples (total of 9cm), both to maximise the concentration of catalase and increase surface area of exposed potato flesh. All samples taken from the same potato species (Maris Piper) Substrate concentration (Hydrogen-peroxide)

Hydrogen-peroxide (reactant) concentration will affect reaction rates being the substrate that enables catalase to drive reactions Each test will contain 25cm3 of 6% Hydrogen-peroxide solution (H2O2) Time measured To ensure the investigation is fair, each sample needs the same amount of time exposed to both the buffering solution and the Hydrogen-peroxide (substrate), a greater exposure to either would alter the results of that test For each test the 3 potato examples are submerged in buffer solution for precisely 1 minute.

Once the first oxygen bubble is delivered to the measuring cylinder, volume is recorded after precisely 1 minute. Should no oxygen be measured after 3 minutes, record as 0. Temperature Temperature is a factor that greatly affects the activity levels of all enzymes, to ensure this doesn’t affect readings, temperature needs to remain constant Temperature is controlled by conducting the investigation at a stable temperature of 25o Risk Assessment: Risk What is the risk? How to control Risk level Hydrogen-peroxide (H2O2) Corrosive – Irritant to skin and eyes.

Damage to clothing Both goggles and lab coat to be worn at all times. The collection of the solution in small quantities Medium Glass Breakages – Injuries – lacerations Careful handling and the correct use of apparatus, keep glass away from table edge Low Scalpel Injuries – lacerations Careful handling, Only use with white tile Low Water & Solutions Spill/Slip Hazard – Electric shock Careful handling, ensuring all liquids are away from table edges and electrical sockets Low Equipment diagram: Method: 1. Collect all equipment and clear all work surfaces. 2.

Prepare boiling tube, measuring cylinder, delivery tube and all apparatus as shown in diagram above. 3. Using a cork borer cut 15 pieces of potato to 3cm long 4. Place 3 pieces of the potato samples into a pH 3 buffer for exactly 1 minute. 5. Place buffered potato samples into the boiling tube. 6. Measure out 25ml of 6% Hydrogen-peroxide (H2O2) 7. Add Hydrogen-peroxide and immediately place rubber bung to the top of the boiling tube, as soon as the first bubble of oxygen leaves the delivery tube, time for 1 minute. If no oxygen has been released after 3 minutes record the volume of 0cm3 8.

After 1 minute from the time the first oxygen bubble has left the delivery tube, record the volume of oxygen captured within the measuring cylinder and record 9. Repeat steps 4-8 with the remaining buffer solutions (pH 5, 7, 9 & 11) (To ensure accuracy and gather a greater field of data, repeat the experiment a further two times. ) Results: Primary data: This table shows how much product (O2) was measured after one minute of measured production: Table 1: Test 1 Test 2 Test 3 pH Level Cm3 of product (O2) collected Cm3 of product (O2) collected Cm3 of product (O2) collected

Mean (cm3) of product (O2) 3 0 0 0 0 5 2. 0 2. 0 2. 0 2. 0 7 2. 5 2. 75 3. 0 2. 75 9 4. 5 4. 0 3. 5 4. 0 11 3. 0 3. 25 4. 0 3. 42 (Figure 1) shows these results plotted onto a graph for analysis: Preliminary data: This table provides the results from preliminary research where catalase activity was measured in yeast: Table 2: Test 1 Test 2 Test 3 pH Level Cm3 of product (O2) collected Cm3 of product (O2) collected Cm3 of product (O2) collected Mean (cm3) of product (O2) 3 0 0 0 0 7 0. 5 2 0. 85 11 1 3 1. 55 (Figure 2) shows these results plotted onto a graph: Secondary data (1):

The table below shows results obtained from a peer (Sean McDonnell) conducting the same experiment under the same conditions Table 3: (Figure 3) shows Sean McDonnells’ data plotted onto a graph: Secondary data (2): The following table and supporting graph monitor O2 produced in a similar experiment (the only exception being that bubbles were counted rather than cm3 of O2 measured). This data has been collected from peers across the college. Table 4: Test 1 Test 2 Test 3 pH Level Bubbles of product (O2) counted Bubbles of product (O2) counted Bubbles of product (O2) counted

Mean number of bubbles product (O2) counted 3 0. 0 0. 0 0. 0 0. 0 5 4. 0 7. 0 5. 5 5. 5 7 10. 0 15. 0 11. 0 12. 0 9 8. 0 10 8. 0 8. 6 11 0. 0 1. 0 0. 2 0. 4 (Figure 4) – Shows this data plotted onto a graph Secondary data (Temperature): The following article is added as a supporting article to demonstrate the effect temperature has on catalase activity. http://practicalbio. blogspot. co. uk/2012/03/easy-enzyme-experiment-potato-catalase. html Results analysis: Primary results collected along with supporting evidence confirm that pH levels are extremely important to the rates catalase/enzymes can react.

For example, every data piece taken has suggested no activity at pH3, this means the enzymes have become denatured and can no longer drive reaction rates as the substrate (in this case H2O2 – Hydrogen-peroxide) no longer fits into the active site of the enzyme (catalase) rendering it useless. There are however some discrepancies in the data. Both my preliminary data and primary experiment have provided more volume of product (O2 – oxygen) at pH11 than expected, which is confirmed by the secondary data showing a much lower production of the gas at this high alkali level.

The full set of results also provides questions on where the enzymes are reacting at their optimum level. Although every graph shows the same trends in terms of ‘limited to no’ activity at pH3, and correlate in shape to where reaction levels begin to drop, actually determining at what pH the catalase enzyme is at its optimum is not so clear. Table 4 suggests pH7, Table 3 suggests pH9 whilst Table 2 suggest a pH of 11 which is a very broad spectrum almost taking up half of the full pH spectrum.

The full spread of data combined does give me confidence in my prediction, even though the investigation could have provided more consistent results. Evaluation: Contamination in the primary experiment – conducting every test of the pH at the same level rather than doing the series and repeating Temperature – temperature rates where different as over 2 day – due to previous remark pH 11 was taken in different conditions than the previous pH levels Eye level – reading of product level – inaccurate in the grand scheme of things – more


Make a Potato Shrink--with Saltwater

Introduction
Have you ever wondered how plants "drink" water from the soil? Water uptake in plants is quite complicated. A process called osmosis helps the water move from the soil into the plant roots&mdashand then into the plant's cells. In this activity you will see for yourself how you can make water move with osmosis!

Background
Most water in the ground is not pure water. It usually contains dissolved mineral salts. Animals and plants need these salts (which include calcium, magnesium, potassium and the sodium you might be familiar with as table salt) to grow, develop and stay healthy. Different water sources carry different amounts of these salts. Nature wants to balance a system that is not balanced. So if you mix water with two different salt concentrations, the salts don't stay separated but spread out evenly through the solution until the salt concentration is the same throughout.

You'll find a similar reaction if you separate two salt solutions with a semipermeable membrane. A semipermeable membrane is a type of barrier that only lets certain particles pass through while blocking others. This type of membrane usually lets water pass through but not the salts that are dissolved in the water. In this situation, because only water can move through this membrane, the water will start moving from the area of lower salt concentration (which has more water and less salt) to the area of higher salt concentration (which has less water and more salt). This water movement will only stop once the salt and water concentration on both sides of the membrane is the same.

The process of moving water across a semipermeable membrane is called osmosis. Plants use this process to their advantage for water uptake. They create an environment of high salt concentration in their root cells that are in contact with the soil. The cell walls act as a semipermeable membrane that only let water through. Because the water outside the root cells has a lower salt concentration, water starts moving into the root cells due to osmosis. The water entering the plant fills up the cells and can travel to the rest of the plant. Osmosis, however, works in both directions. If you put a plant into water with a salt concentration that is higher than the concentration inside its cells, water will move out of the plant to balance out the concentration difference. As a result the plant shrinks and eventually dies. You will see this effect with your own eyes in this activity using potatoes and different saltwater solutions.

  • Distilled water
  • Measuring cup with milliliters (mL)
  • Table salt
  • Weight scale with gram measurements
  • Three plastic cups or glasses
  • Spoon
  • At least three potatoes
  • Apple corer. (Alternatively, you can have an adult help you use a cutting board and knife.)
  • Knife (and an adult helper to help you use it)
  • Ruler
  • Paper
  • Pen or pencil
  • Timer
  • Paper towels
  • Graphing paper (optional)
  • Other vegetable(s) or fruit (optional)


Preparation

  • Prepare three different saltwater solutions. Create labels for the three cups: "0 grams," "2 grams" and "4 grams."
  • To each of the cups add 100 mL of distilled water.
  • Weigh out 2 grams of table salt, and add it to the cup that says "2 grams." Then weigh out 4 grams of table salt, and add it to the cup labeled "4 grams." Use a spoon to mix the solutions until all the salt is dissolved.
  • Draw a table in which you can enter the starting measurements (length and diameter or width) and end measurements of each potato strip for every salt concentration (0, 2 and 4 grams).
  • Prepare at least three potato cores. Carefully push the corer all the way through the potato, and remove the core carefully so the potato piece stays intact. (Alternatively, you can have an adult help cut the potato into strips that all have the same dimensions.) The potato pieces should be at least one-half inch thick and two inches long. (Ideally you will be able to prepare nine matching cores or strips so you can test three pieces in each solution to compare the results thoroughly.)
  • Use a knife to carefully remove any potato skin from your cores, and rinse the cores quickly with water.
  • Use a ruler to ensure each potato piece is the same size (ideally to the millimeter). Carefully use a knife to trim any pieces as needed.
  • Measure the dimensions (length and diameter or width) of each potato strip in millimeters, and write the information in the table.
  • Optionally, you can also weigh each potato piece and record their weights.
  • Put one potato strip (or three if you made nine pieces) into each of the cups. While you do that feel the potato strips with your fingers and try to flex them a little bit. How do they feel? Are they easy to bend?
  • Start your timer for 30 minutes. Let the potato strips sit in the different solutions for the whole time. What do you think will happen to the strips in each of the cups?
  • After 30 minutes inspect the potato strips inside the solutions. Do you see any changes?
  • Take the potato strip(s) out of the "0 grams" cup and place on a paper towel. While doing that feel the potato pieces again and try to bend them slightly. How do they feel? Are they easier or more difficult to bend than before?
  • Use the ruler to measure the exact length and diameter or width (in millimeters) of each of the potato strips, and write the results in your table. What do you notice about the potato strip measurements? Optionally you can weigh these pieces and record their weights.
  • Next take the potato strips from the "2 grams" cup, and place them on a paper towel as you do this feel them. Measure their lengths and diameters or widths. Write your results in the table. Optionally you can weigh these pieces and record their weights. What changed about these potato strips?
  • Repeat the same steps with the potato strips in the "4 grams" cup. Write your results in the table. Are your results for these similar or different compared with the other ones?
  • How did the feeling of the strips compare based on what solution they were in? Why do you think this is?
  • Compare the results in your table. How did the length and diameter or width of the potato strips change in each cup? What about the weights if you took them? Can you explain your results?
  • Extra: If you weighed each of your strips before and after soaking them, compare the weights. How does the mass of the potato strips change in each solution?
  • Extra: Leave the potato strips in the solutions for a longer time period. How do they look if you let them soak in the saltwater for one hour or overnight?
  • Extra: If you have graphing paper, make a graph of your results with the salt concentration on the horizontal axis and the potato strip length or diameter after soaking on the vertical axis. Draw two lines to make your graph. For the first, connect each of the data points you found. For the second, draw a horizontal line starting at the point on the vertical axis that shows the original length of your potato strip. Based on your graph can you find a salt concentration at which the potato strip length should not change at all?
  • Extra:How does the activity work with other vegetables or fruit? Try it to find out!


Observations and Results
Did your potato strips shrink and expand? At the beginning all the potato strips should have had the same length and should have all felt the same. When you put them into the different solutions, however, this starts to change. Whereas the potato strips in the "0 gram" cup probably got larger in size, the other potato strips probably got shorter after leaving them in the saltwater for 30 minutes. (If you didn't see any significant changes after 30 minutes, leave the potato strips in the saltwater solutions longer.)

The shrinking and expanding of the potato strips is due to osmosis. Potatoes are made of cells, and their cell walls act as semipermeable membranes. The 0 grams solution contains less salts and more water than the potato cells (which have more salts and less water). To balance out these concentration differences, the water from the cup moves into the potato cells. The incoming water in the potato cells pushes on the cell walls and makes the cells bigger. As a result the whole potato strip gets bigger. The opposite is the case in the higher concentration salt solutions. If the salt concentration in the cup is higher than inside the potato cells, water moves out of the potato into the cup. This leads to shrinkage of the potato cells, which explains why the potato strips get smaller in length and diameter. Due to the shrinking of the potato cells the potato strip also becomes less rigid. If you bent the potato strips, you should have noticed that those that had been in the solution with the highest amount of salt were much easier to bend than the potato strips in the water without salt.

If you made the graph you probably noticed that there is a salt concentration at which the potato strip neither expands nor shrinks. This should be where your data curve and your start length line intersect. At this point the salt concentration inside the potato cells and inside the cup are the same. Because the concentrations are already balanced no water moves.

Cleanup
Discard the saltwater solutions in the sink. Throw the potato strips into the compost, and clean up your workspace. You can cook with the other pieces of unused potato.

More to Explore
Osmosis, from Biology Dictionary
Do Fish Drink? from McGill University's Office for Science and Society
Cucumber Chemistry: Moisture Capture with Desiccants, from Scientific American
Suck It Up! How Water Moves Through Plants, from Science Buddies
STEM Activities for Kids, from Science Buddies


Results:

Onion:
In a solution of 4% NaCl, the onion absorbed only some of the solution, creating more dark purple spots under the microscope observation. This solution was a hypertonic solution.
In a solution of 8% NaCl, the onion absorbed more of the solution, creating even more dark purple spots under the microscope observation. This solution was a hypotonic solution.
In a solution of 0% NaCl, the onion absorbed more solvant than solute, creating less purple spots under the microscope observation. This solution was an isotonic solution.

Egg:
In a solution of 4% NaCl, the egg's final mass was 96 grams (whereas its original mass was 82 grams), and became very squishy. It absorbed less solvant than it did the solute. This solution was an isotonic solution.
In a solution of 8% NaCl, the egg's final mass was 91.5 grams (whereas its original mass was 76 grams), and it became less squishy than the other eggs. This egg absorbed the solute more than it did the solvant. This solution was a hypertonic solution.
In a solution of 0% NaCl, the egg's final mass was 92 grams (whereas its original mass was 75 grams), and it became much more squishy than the other two eggs. It absorbed more solvant than it did solute. This solution was a hypotonic solution.

Potato:
The results of the potato were lost.

Conclusion:

For something that already has salt levels, whenever the solution it was soaked in tried to equal the levels on the inside and outside of the membrane, the substance became more inflated because the water was passing into it. This occurred when there was less salt in the solution than in the substance. For a substance that has salt, but is soaked in a solution with no NaCl percentage, the water will attempt to carry the salt from the substance, thus making it smaller. These examples were shown in the egg's mass--how much it increased per solution (in grams). In the onion samples, the increase and decrease in salt levels was shown by the darker and lighter spots of links under the microscope. When the links were fewer and farther between, with lighter spots of color, it had absorbed more solvant than solute. When the links were closer together and darker in color, the onion had absorbed more solute than solvant.
A few uncertainties that could have varied the results were that the onion samples were all from the same onion, whereas the eggs were all different in mass (and probably molecular makeup). Also, we could not get multiple trials of the same substance, given the time that we had, and we also misplaced the results from the potatos, which made it impossible made it impossible to create any theories.
Further experimentation would be good to have, as well as more trials. Specific information would make it much easier to prescribe any sort of results. prescribe any sort of results.


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