Rana's Surviving Winter in the Dust Bowl Argument

Rana Srouji

Mr. Hammer

AP Biology

4 May 2015

Dust Bowl Argumentation

Over the past years, Oklahoma has been in a devastating state for farming.  We received less than ten inches of rain and the crops did not grow well at all.  Out of what my family has planted we were only able to conserve five hundred bushels of wheat for the winter and five hundred gallons of potable water.  The next time that we will be able to harvest is in June.  Thus, in order to survive this eight month winter in the dust bowl, my family and I must eat our bull, keep the cow alive, feed it, drink the milk, and then eat the rest of the wheat.  The bull would be the first to be eaten for obvious reasons.  According to Table 1, it would consume 46,000 calories a day, and take about 45 gallons of water per day.  Keeping the bull alive would only reduce the chances of survival for my family and I because the bull would consume all of the food and water that we have within a short period of time.  Also, eating the bull would provide us with plenty of protein and fat to keep us healthy and would allow us to reserve the limited amount of wheat that we have for when we run out of meat.  However, the bull meat does not provide any carbohydrates, but since we are keeping the cow alive, we would milk the cow and use the milk to receive the carbs necessary to live.  This will be fine considering the cow produces about 6 gallons of milk per day which is more than necessary for the daily serving size.  However, since only 41% of the bull’s total weight is actually consumable, this means that of the 1200 pounds that it may weigh, we could only eat about 708 pounds and each pound is considered a serving size.  Also, considering that we have about 240 days in this winter, this is not sufficient enough to provide the correct dietary needs to each family member daily.  Thus, once we run out of bull, we could resort to consuming the wheat.  The five hundred bushels of wheat is meant to last eight months to feed the family and the cow, which means it must be conserved and neither the cow nor the family will be able to get their usual servings from it.  This also goes for the water intake, there is only five hundred gallons of water for both the cow and the family to drink from, thus the cow would have to get less than its usual servings from the water as well but the family would get their usual servings of water because humans cannot survive more than three to five days without the potable water.  The cow consumes about fifty gallons of water a day, but it would only get a portion of that amount because if it received fifty gallons, then the cow alone would receive water for only ten days and there would be no water for the family.  Overall, my family would only be able to survive this eight month winter by eating the bull, keeping the cow alive, feeding the cow, drinking the milk, and eating the wheat, until June for the next harvest.  

Surviving Winter in the Dust Bowl Argument

Joshua Everett
Mr. Hammer
AP Biology
May, 4 2015

Surviving Winter in the Dust Bowl 

In the 1930’s, there were many states that suffered through the era of the “Dust Bowl”. Many often suffered from the inevitable droughts, high temperatures, and high winds. In October of 1934, I was forced to survive the winter with the limited resources the Dust Bowl put on my farm. In order to survive the time period from October until June, when the next harvesting season is, I have decided to eat my bull and keep the female cow alive only to starve it out later. I would drink the cow’s milk but once milk production ceases I would proceed to eat the wheat from the previous harvest.

I would kill the bull in order to conserve on the resources that I have. The bull consumes the most amount of calories a day when compared to myself and the female cow. The bull would need to consume 45 gallons of water a day in order to be healthy which is not ideal when I need to drink water in order to survive as well. If the bull was to consume 45 gallons of water a day, my 500 hundred gallons of water that I have in total, would be gone in approximately 11 days. That would mean during those 11 days I would not be able to have any water which means I would die within the first 3-5 days. The bull would be most useful as food since it weighs the most and would have the most consumable meat. Only 41% of the cow’s weight is consumable and since the bull weighs 1,200 pounds, there would be approximately 1,260 grams of meat to eat a day. This still does not reach the amount of calories a male human should consume a day. The bull meat is a great source in protein and fat but lacks carbohydrates.

Good sources of carbohydrates include wheat and milk. The purpose of keeping the cow alive would be to continue the process of milk production. A lactating cow produces 6 gallons of milk a day which is equivalent to 21,965 grams a day. By continuing milk production, I would expect to obtain the minimal amount of milk needed to replenish the nutrients I am not receiving from the bull meat. In order to keep this female cow alive, I would have to feed it water and food. The recommended amount of water intake for a lactating female cow is 50 gallons a day. Again, this is too much water for the cow and I to survive. If I wanted to keep the cow and I alive I would have to cut the cows amount of water significantly and cause the cow to be unhealthy and unable to produce the milk needed in order for me to survive until June. I would instead collect as much milk as I can from the cow before it dies, and ration the milk over the rest of the time until June. It makes more sense to kill the cow because it would provide additional cow meat that could be consumed as well.

With the cow meat and the milk, I would have good sources of proteins and fat but limited sources of carbohydrates. In order to make sure I receive the correct amount of carbohydrates, I would eat the wheat. Wheat is the only resource that contains the most carbs. I have 500 bushels of wheat from the previous harvest which is equivalent to 53,364 grams of wheat a day. This huge number is more than enough to survive the Dust Bowl and to replenish the missing carbohydrates.

Without the cows having to worry about giving them water, I would be able to drink 2 gallons of water a day. The minimum humans need to drink is 0.4 gallons of water. I would only drink this amount because I would want to conserve as much water as I can because even though the harvest of food will be coming in June, I do not know when the next rainfall would come. I want to be prepared as possible just in case rainfall does not come until a later date. With the extra wheat and meat, I would also save those just in case the harvest is not as productive. I would always want to be prepared for the worst so I have better survival chances in the future. If I were to follow this plan of action, I should be able to survive the winter with little to no problems at all.

Nida, Daphne and Sara's Effect of Environmental Stimuli on Cell Division Arguement

Rana, Jube, and Rameia's Effects of Environmental Cell Division Powerpoint Argumentation

Joshua, Chigozie, and Jazmean's Effect of Enviornmental Stimuli on Cell Division Argument

Daphne's Virus Argumentation

When attempting to define whether or not viruses are living creatures, it is important we consider what it truly means to be alive. In order to qualify as a living, a being must possess all of the following characteristics:

1. Living things are composed of cells
2. They possess different levels of molecular/cellular organization
3. They take in energy to use for maintenance and/or growth
4. They respond to stimuli presented by their environments
5. They reproduce, sexually or asexually
6. They experience growth
7. They adapt to their environment over time (as a species) through evolution

Because of these standards, viruses cannot be considered to be alive. When scientists crystallized the tobacco mosaic virus, they determined that the virus lacked the genes for metabolism. This indicates that they do not take in energy for maintenance and/or growth. Since it takes just one of these qualities being absent to disqualify something from being alive, viruses cannot be considered living. Furthermore, viruses cannot reproduce on their own.  When not in contact with a host cell, viruses are no more active than free organic matter. In this state, they are inactive and no internal biological activities occur within the virus. Viruses consist of no more than membrane-bound DNA or RNA. Only when they infect a host cell are they able to multiply and divide through the lytic cycle. In order to be alive, organisms must be able to reproduce on their own. Also, viruses do not respond to stimuli in their environments, which is another factor required for living creatures.
People (especially scientists) like to create organized categories and imagine that all of the world fits into them, checking off all the boxes. Such is not necessarily the case with viruses. They almost straddle the definition of life, seeing as they do not check off some of these qualifications. However, they seem to possess lifelike qualities, which is what makes classifying them so difficult. As far as classifying them by the preset standards established by scientific laws, viruses are most definitely not alive.

Jazmean's Virus Argumentation

If we follow the standards set by scientists that constitute organisms as living, we cannot classify viruses as living things. Scientists use seven standards to determine if an entity is living: are they made up of cells, do they have levels of organization, can they obtain and use energy, do they grow and develop, can they reproduce, do they respond to their environment, and can they adapt to their environment. If an organism does not meet all of the seven qualifications, they are considered to be nonliving. Viruses only meet two of the seven qualifications: they can reproduce and they can adapt to their environment. And even though they meet these two qualifications, they cannot reproduce nor adapt to their environment without a host cell. This means that viruses are not independently efficient unlike other organisms such as Amoebas and Algae because a majority of their survival depends on their host cell.  Most scientists consider viruses outside of their host cell to be dormant and not functional.

Viruses do not conduct chemical or metabolic processes within them. They are not able to produce energy like plants and they are not able to use molecules like glucose to undergo cellular respiration like animal cells. They have no organelles that can help them do that. They only possess genetic information that is used to produce more viruses. They are not able to use their genetic information by themselves. Viruses have to employ the use of a host cell in order for the genetic information to be put to use. They also do not have the ability to respond to external stimuli when outside of their host cell.

 Because viruses are not able to meet all of the qualifications of what makes a living thing living, they are not able to be classified as living things.

Rameia's Virus Argumentation

Should Viruses be Considered Living Things?

Rameia Ramsey

          In order to be classified as a living thing there are seven characteristics that must be possessed by an object. A living thing must be composed of cells, there must be different levels of organization, it must use energy, it must respond to its environment(stimuli), it must grow and develop, it must be able to reproduce, and it must be able to adapt to its environment.  These seven characteristics make up a living organism.  Based off of these characteristics and the information provided in the data table I come to the decision that a virus should not be classified as a living thing.  A virus should not be classified as a living thing because they do not contain all of the properties that are associated with living things and would not be able to function properly without the aid of a host cell. 

          All living things must be composed of cells, a virus automatically violates this characteristic. A virus is a tiny bundle of genetic material that is carried in a protein shell called a capsid. Some viruses have an additional layer around the coat called an envelope that is made of a lipid. A virus in no shape or form is composed of cells. It is simply a bundle of DNA or RNA. This shows how a virus does not have all the characteristics needed to be a living thing

          A virus can also not be considered a living thing because it has no energy source.  On the data table the Influenza Virus and the Adenovirus are listed with having an energy source as none. Living things need an energy source to be able to take in energy and use it for maintenance and  growth. As a comparison on the chart listed is a plant called Elodea.  This plant uses the sunlight as its energy source. With the aid of sunlight this plant is able to carry out the photosynthesis in order to make itself food and help it grow and develop.  Viruses do not have an energy source which is why they rely on host cells to help them function. Host cells allow viruses to carry out their destructive behaviors. Since there is no actual source of energy aiding the virus in  developing and growing it is not a living thing.

          A virus will not respond to external stimuli which  makes it unfit to be a living thing. If we compare the Influenza Virus and the Adenovirus with another organism on the chart such as the Coriander Seeds we can see that a virus is not able to respond to stimuli. If the Coriander seed comes into contact with water or sun it will respond to that stimuli by growing. Living things will make changes in response to their environment but a virus does not change. Its form will always stay the same, without a host cell a virus won't do anything. Therefore it can not be classified as a living thing.

          Reproduction is not important for a living thing to survive by itself, but it must reproduce in order for a species to survive. Living things either reproduce sexually(joint of sex cells) or asexually(without the use of gametes) . Viruses can not reproduce by themselves, there needs to be a host cell for a virus to reproduce.  The Influenza Virus and Adenovirus use its host cell to survive. When those viruses enter its host cell the genetic material that it carries allows the virus to force the now infected cell to make copies of the virus. Without this host cell a virus can not reproduce to survive.  Other organisms on the chart such as dogs are able to reproduce sexually and create more offspring, if that one individual dog doesn't reproduce it will not harm it but if that one virus does not have a host cell it will not reproduce at all and become stagnant.

          Another reason why a virus should not be classified as a living thing is because it does not grow.  Living things grow through the process of cell division and cell enlargement.  As an organism gets larger the number of cells it contains increases. As I already established, viruses are not composed of cells, this means that they do not have the ability to grow.  A virus needs a host cell to grow, a virus by itself can not grow at all. As a comparison, the dog that is depicted in the chart is able to grow.  As the dog carries out its life cycle it grows and developed from a puppy into an adult dog. Viruses do not have the capability to grow and develop by itself, thus making a virus not a living thing because it lacks the ability to reproduce on its own.

          Viruses should not be considered a living thing because they lack many of the properties that make up a living organism.  As stated on the chart the Influenza Virus and he Adenovirus do not have an energy source, not respond to stimuli, can not reproduce without a host cell, and can not grow.  Without a host cell a virus can not do anything and is no harm or threat to anyone or anything. Its the host cell that allows the virus to function properly. Not possessing anyone of the characterizes makes something unfit to be living and the fact that a virus lacks several of these properties is proof that a virus should not be classified as a living thing.

   

Nida Ali's Should A Virus Be Classified As A Living Thing?

Viruses should not be classified as a living thing. Viruses should be able to use their energy, grow/develop, reproduce, respond to their environment, have waste products and adapt to their environment. Unfortunately, not all viruses meet all of these characteristics. Based off of the data found in Information about Viruses and other objects found on Earth, several viruses such as the Influenza Virus and Adenovirus have no energy source, do not have any waste production, does not have a carbon source, does not respond to external stimuli and do not grow. In order for a virus to be considered “alive” it must reproduce, obtain/use energy, grow/develop and respond to the environment. Since viruses are not cells there is no need for food since no activity occurs. Viruses do not use their own energy nor do they produce energy. Also, viruses do not grow and conduct no activity inside of its protein coat. Since the Influenza Virus and the Adenovirus are not able to respond to stimuli, under certain conditions like being presented to sunlight, they would not be able to grow in comparison to Amoeba, which is a single celled animal, is able to grow since it is able to respond to external stimuli. The Influenza Virus and the Adenovirus do not have an energy source, which thus results in no growth. However, the Coriander seeds has an energy source and carbon source which thus results in growth. Due to these lacking characteristics, viruses reply on host cells to carry out all of their functions.

The Influenza Virus and Adenovirus both require a host cell in order for reproduction to occur. The Plasmodium Falciparum parasite has a form of reproduction of sexual and asexual but reproduction can only occur inside of a host cell. Viruses cannot use their genetic material by themselves, in order to be effective, the virus needs to be present in a living cell. These viruses are not considered living since they are unable to reproduce. Viruses are unable to divide by binary fission. Viruses can not divide or reproduce by themselves without a host cell, as seen in the Influenza Virus, Adenovirus and the Plasmodium Falciparum parasite. However, while comparing viruses to other organisms such as dogs, coriander seeds, Amoeba, Sponges, E.Coli Elodea and Tube worms, these organisms are considered alive since they are able to reproduce either by sexually, asexually or even both. Since viruses are unable to reproduce without a host cell, they are not considered alive.  Viruses need a living cell in order to function and reproduce. Viruses are not living when they remain outside of the cell. The virus contains genetic material and when it is outside of a living cell, it does not carry out its functions. Once the virus is inside the cytoplasm of a living cell, the virus can activate. The virus' DNA takes control of the cell once it's within the cytoplasm. To add, viruses do not move themselves even though they do react to some changes in their environment.

Moreover, since viruses lack characteristics such as being unable to grow/develop , lack energy, unable to reproduce, inability to produce waste products, and are unable to respond to external stimuli, then viruses should not be considered a living thing. Viruses rely on host cells to be productive and efficient. Viruses are unable to perform/ function without being present in a living cell, as seen with the Influenza Virus and Adenovirus.

Chigozie's Water Potential and Molarity Mixup

Chigozie Amonu

Mr. Hammer

AP Biology

21 November 2014

Water Potential & Molarity Mixup

            Argument One

For this lab, my lab group and I had to determine the unknown molarities of the six unlabeled solutions. The molarities of each different colored solution could be 0.0M, 0.2M, 0.4M, 0.6M, 0.8M, or 1.0M. After the conduction of experiments and compilation and analysis of data from Activity One and Two, we determined that the order of molarity of the solutions, from least to greatest (0.0M to 1.0M), was orange, dark green, light green, blue, yellow, and red. To get these results, two separate experiments were done. One was done with dialysis tubing bags, and the other was done using two vegetables, squash and parsnip. My lab group and I analyzed all of the data that we got from both experiments. We eventually decided to use the data from the experiments done with squash and parsnip because these two results proved this claim while the data from the dialysis tubing did not. Simply, the majority of our evidence supported this claim.

For the vegetable experiment, my lab group and I cut six pieces from each vegetable and weighed each piece. We then measured out 75 mL of solution to put the vegetables in. The six pieces of each vegetable were put into the six different solutions, and the cup and beakers that they were in were covered with foil and left to sit overnight. The next day in class, my lab group and I weighed the vegetables and recorded the weight in data tables. The percent change for each piece of vegetable was calculated (percent change= [final mass minus initial mass] divided by [initial mass] times 100). For the parsnip, the percent changes were -25.32% for red, 47.62% for orange, -20.96% for yellow, 10.64% for light green, 32.09% for dark green, and -13.24% for blue. For the squash, the percent changes were -45.68% for red, -6.94% for orange, -43.04% for yellow, -4.55% for light green, 20.37% for dark green, and -31.43% for blue.

Parsnip Data

Color of Solution in Which Parsnip was Placed	Initial Weight of Parsnip	Weight of Parsnip After Overnight Stay	Percent Change
Red	15.8 g	11.8 g	-25.32%
Orange	12.6 g	13.2 g	47.62%
Yellow	16.7 g	13.2 g	-20.96%
Light Green	14.1 g	15.6 g	10.64%
Dark Green	13.4 g	17.7 g	32.09%
Blue	13.6 g	11.8 g	-13.24%

Squash Data

Color of Solution in Which Squash was Placed	Initial Weight of Squash	Weight of Squash After Overnight Stay	Percent Change
Red	8.1 g	4.4 g	-45.68%
Orange	7.2 g	6.7 g	-6.94%
Yellow	7.9 g	4.5 g	-43.04%
Light Green	8.8 g	8.4 g	-4.55%
Dark Green	5.4 g	6.5 g	20.37%
Blue	7.0 g	4.8 g	-31.43%

We determined that a negative percent change indicated the shrinking of the cell. This would mean that the cell was in a hypertonic solution that had a high concentration of water in the cell and a low concentration of water outside of the cell. The water would move across the concentration gradient from high concentration to low concentration through osmosis, a type of diffusion. This information led us to believe that the most negative percent change of the vegetable equaled the highest molarity of the solution in which it is placed since there would be less water and more solute, in this case sucrose, outside of the cell (in the solution). On the other hand, a positive percent change would indicate the swelling of a cell. This would mean that the cell was in a hypotonic solution that had a low concentration of water in the cell and a high concentration of water outside of the cell. The water would move into the cell through osmosis, causing it to swell, and proving that there was a less solute concentration and higher water concentration in the solution. The most positive percent change in the vegetable equaled the lowest molarity of the solution in which the vegetable is placed.

My lab group and I then arranged these percent changes from greatest to least. For the parsnip, the order was orange, dark green, light green, blue, yellow, and red. For the squash, the order was dark green, light green, orange, blue, yellow, and red. We compared these orders and found that the blue, red, and yellow were 0.6M, 0.8M, and 1.0M, respectively, for each vegetable. The dark green, light green, and orange were in different orders on each greatest to least list. For the squash, the order was dark green, light green, orange. For the parsnip, the order was orange, dark green, light green. We decided to use the parsnip order because it seemed more logical to move one color (orange) to the top since the dark and light green were ordered consecutively on each list and would move up and down the final list as one entity. After analyzing all of our data, we concluded* that the order of molarities is orange (0.0 M), dark green (0.2M), light green (0.4M), blue (0.6M), yellow (0.8M), and red (1.0M).

As I mentioned before, we did not use the data from the dialysis tubing to support our claim. There was more room for error in this experiment. The bags of solution may not have been properly tied and could have leaked, causing a discrepancy in the dialysis tubing bag’s final mass. Also, we did not use the same scale to weigh the bags on the second day of the lab. We also concluded to not use the data from this experiment because the evidence did not support our claim. The percent changes were -6.25% for red, 19.94% for orange, 9.94% for yellow, 0.97% for light green, 0.68% for dark green, and 30.37% for blue. For this experiment, a more negative percent change would represent the shrinking of the cell, and therefore a hypertonic solution and a lower molarity. A more positive percent change would represent the swelling of the cell and therefore a hypotonic solution and a higher molarity. Based on this information, the order of molarity of the solutions would be red (0.0M), dark green (0.2M), light green (0.4M), yellow (0.6M), orange (0.8M), and blue (1.0M). All in all, this data did not support our original claim, and we determined the squash and parsnip experiment to be more accurate.

Dialysis Tubing Data

Color of Solution in Bag	Initial Weight of a Bag	Weight After One Day	Percent Change
Red	32.0 g	30.0 g	-6.25%
Orange	32.1 g	38.5 g	19.94%
Yellow	34.2 g	37.6 g	9.94%
Light Green	30.9 g	31.2 g	0.97%
Dark Green	29.5 g	29.7 g	0.68%
Blue	32.6 g	42.5 g	30.37%

Argument Two

            The water potential for parsnip is -12.29 bars, and the water potential for squash is -3.69 bars. To find this, my lab group and I had to first determine the solute and pressure potential. To calculate the solute potential, we used the formula Ѱs= -iCRT (solute potential equals the number of particles the molecule will make in water times molar concentration times pressure constant times temperature in degrees Kelvin). We had to find C, the molar concentration. To do this, we had to graph the percent changes of each vegetable from Activity Two (squash and parsnip) compared to the molarity of each solution. The x-axis is molarity, and the y-axis is percent change. The molar concentration ended up being where the graph of the percent changes hit the x-axis. These molar concentrations were 0.5M for the parsnip and 0.15M for the squash. We used room temperature (23ºC) as the temperature for the solute potential and converted it to Kelvin (273+23ºC). We then plugged these values into Ѱs= -iCRT. The equation for parsnip was –(1)(0.5)(0.0831)(296) and it equaled -12.29 bars, the solute potential. The equation for squash was –(1)(0.15)(0.0831)(296) and it equaled -3.69 bars, the solute potential. We then determined that the pressure potential was zero, and used the equation Ѱ= Ѱs+Ѱp (water potential= solute potential plus pressure potential) to calculate the water potential. The equation for parsnip would be Ѱ= -12.29+0, and would equal -12.29 bars, the water potential. The equation for squash would be Ѱ= -3.69+0, and would equal -3.69 bars, the water potential. The squash has a higher water potential than the parsnip. Because of this, water can move more freely in and out of the squash cells during osmosis.

Title: Percent Changes of Parsnip in Comparison to Molarities of Solution; x-axis represents the molarity, and the y-axis represents the percent change

Title: Percent Changes of Squash in Comparison to Molarities of Solution; x-axis represents the molarity, and the y-axis represents the percent change

Joshua's Water Potential and Molarity Mix Up Arguments

Joshua Everett

AP Biology

Mr. Hammer

November 21, 2014

Water Potential and Molarity Mix Up

Argument 1:

In activity one of the experiment my lab partner and I were able to conclude, through the results we received, which molarity concentration each solution was that was used in the experiment. There were six different solutions that were represented by food coloring, each a different color. Going up in increments of 0.2M, the possible molarities of these solutions ranged from 0.0M to 1.0M . After interpreting the results and drawing general conclusions, the order of molarity of the solutions were orange, dark green, light green, blue, yellow and red in respect for the range of possible molarities from least molarity to greatest molarity (0.0M to 1.0M increasing by 0.2M). In the experiment there were two processes that were done, one with a dialysis tubing cell and the other with squash and parsnip. After fully analyzing the results and making calculations, we decided to use the vegetable experiment to support our order of molarity because there were two vegetables that displayed the same order while the one dialysis tubing cell had different results.

In the vegetable experiment, we cut up each vegetable into six pieces that were  all relatively the same size.  We weighed each of the pieces for both vegetables and recorded them in a data table. We also measured out 75 mL of each solution twice into separate cups. We placed one of the squash pieces into each of the different solutions. The same process was done for the parsnip pieces as well. Once they were submerged into the solutions, we covered each of the cups with a piece of foil and we let them sit overnight. The next day we took out each piece of vegetable from the solutions and weighed them again. We recorded the weights into a data table and we started developing patterns between the two vegetables.  With the two sets of weights for each piece of vegetable we were able to calculate the percent change between the original weight and the final weight. To calculate the percent change, you subtract the final mass minus the initial mass all divided by the initial mass. Then you multiply the answer by 100 to get a percent.  For the squash, the percent change for solution red was -45.68%, orange being -6.94%, yellow being -43.04%, dark green being -4.55, light green being 20.37%, and blue being -31.43%.  For parsnip the percent change for red was -25.32%, orange was 47.62%, yellow was -20.96, dark green was 10.64%, light green was 32.09% and blue was -13.24%. We arranged the percent changes for each of the vegetables into separate lists  in order from positive percent change (+) to negative percent change (+) to see a better comparison between them.  The negative percent changes represent that the cell shrunk and the positive percent changes represents the cell swelling. When a cell shrinks it means that the cell is in a hypertonic solution in which there was a high concentration of water inside the cell and a low concentration of water outside of the cell. Through the process of diffusion, water would move across the concentration gradient and move from high to low concentration. If there is not a lot of water outside of the cell, where the solution is, that means that most of the solution contained sucrose solutes. This would mean that the solution with the vegetable that had the most negative percent change would have the greatest molarity. However, when a cell swells, it is in a hypotonic solution in which  there is a higher concentration of water on the outside of the cell where the solution is located and a lower water concentration on the inside of the cell. The water will move from outside the cell to inside the cell, making the cell swell but also making a positive percent change. If there is more water on the outside of the cell then that suggests that there is not a lot of solute of sucrose present in the water but rather more of the sucrose will be present inside the cell where less water is. This would mean the pieces of vegetables with a solution that had the most positive percent change was have the smallest molarity. The order of most positive percent change to most negative percent change for the squash was the dark green, light green, orange, blue, yellow, and red. For parsnip it was orange, dark green, light green, blue, yellow, and red. Next, we made a combined list where the placement of the solutions was determined by their similar placement from their separate lists. The blue, yellow, and red solutions were the same in placement for both vegetables. The orange and both greens moved in the lists so we decided that since orange was at the top once it would stay there even though dark green was once at the top too. But we thought that since dark green and light green are always other, it would make sense that orange starts off and the dark and light green will just follow. The list starts off with the most positive percent change and decreases into the most negative percent change as the list goes down. The order of the molarity would be orange being 0.0M, dark green being 0.2M, light green being 0.4M, blue being 0.6M, yellow being 0.8M, and red being 1.0M.

For the dialysis tubing part, we made six different cells by tying off the ends of the tubing and placing 30 mL of a  different solution in each cell. We weighed each cell and recorded it in a data table. We submerged the each cell in a different plastic cup that that had 250 mL of water. We covered each cup with foil and let them sit overnight. The next day, we removed the cells from each cup, weighed them, and recorded the weight in a data table. We then calculated the percent change of each cell and there was only one negative number and the rest were positive. The percent change for red was -6.3%, orange was 19.94%, yellow was 9.94%, dark green was 29.7%, light green was 0.97%, and blue was 30.37%. The negative percent change represents that the cell was possibly in a hypertonic solution where there would be a high concentration of water inside the cell than outside the cell. The water will move from inside the cell to outside the cell because of the process of osmosis which accounts for the decrease in weight and the negative percent change for the red solution. Because the solution is now in the cell and not in the environment, like in the vegetable experiment, the cell will only shrink if there was more water in the cell and a small amount of sucrose is present. The red solution will only have the smallest concentration of sucrose because of the water movement and negative percentage change. The positive percent changes depicted that the cell was in a hypotonic solution and that there is a  greater amount of solutes in the cell. With the greater amount of solutes, there is less water in the cell which will cause water to flow inside. As the positive percent change increases, that will determine the molarity of each solution will increase from red continuing on. For dialysis tubing the molarities would be red being 0.0M, dark green being 0.2M, light green being 0.4M, yellow being 0.6M, orange being 0.8M, and blue being 1.0M. Because red is considered 0.0M in this lab, the solution in the cell would have contained no solutes in the cell. Technically, the cell losing water from the cell would have came from the cell being isotonic to its environment. A human error that can account for this hypertonic situation could be the use of tap water and not distilled water. Tap water could contain chemicals that could have caused water to move from low concentration to high concentration which would explain why osmosis occurred.  Overall, this is not my claim because this order is not confirmed by any other experiment. I used the claim for the vegetable experiment because it was supported by both squash and parsnip and seemed more accurate.  

Argument 2:

The water potential for squash is -3.69 bars and the water potential for parsnip is -12.29 bars. My lab partner and I calculated the percent change in mass using the recorded weights of the vegetable pieces before they were each put into 75 mL of a different sucrose solution and after. Afterwards, we proceeded to make a graph comparing the percent change in both squash and parsnip to the concentration of each solution. The percent change in mass was on the y axis and the molarity was on the x axis. The purpose for this graph was to determine where a cell would be isotonic where there is zero percent change in mass. This will only occur then there is an equal amount of water entering and leaving the cell at the same time. In this isotonic stage this is no pressure being exerted on the cell, so we said there's no pressure on the cell. Next we located the point on the graph where the line crosses the x axis. This point will tell us the molarity of the solution that the cell will have at an isotonic state. For squash the molarity is 0.15M and for parsnip it is 0.50M.  We used these numbers, 0 for the pressure and 0.15M and 0.50M for the molarities, to plug into an equation to determine the water potential of the cell. The water potential equation is Ψ=Ψp+Ψs which means that the water potential equals the pressure potential plus the solute potential. The pressure potential is zero but to determine the solute potential we had to use this solute potential equation, Ψs=-iCRT. This equation means the solute potential equals to the product of negative number of particles the molecule will make in water, molar concentration, the pressure constant, and the temperature in Kelvin. The number of particles sucrose makes in water is 1, so -i equals to -1.The pressure constant is 0.0831 liter bars/mole K and the temperature is 23 degrees celsius plus 273 in order to convert it to Kelvin. After multiplying all the numbers, not forgetting the 0.15 molar concentration, I got -3.69 bars for squash. To figure out the solute potential for parsnip, I followed the same process but instead I substituted the molar concentration from 0.15M to 0.50M. The solute potential for parsnip is -12.29 bars. In order to find the water potential I had to plug the solute potential for the parsnip and squash back into the Ψ=Ψp+Ψs. The pressure is 0 for each vegetable so I just plugged in the solute potential into the equations and I got the water potential. The water potential for the squash came out to be -3.69 bars and the water potential for the parsnip was -12.29 bars. 

The title is The Water Potential of the Squash. The x axis is the molarity of the solutions and the y axis is the percent change in mass.

The title is The Water Potential of the Parsnip. The x axis is the molarity of the solutions and the y axis is the percent change in mass.