Rameia's Cell Diffusion Argumentation

Cell Diffusion Argumentation

Rameia Ramsey

   In this experiment, my partners and I were challenged with how the size of a cell effects the rate of diffusion within it.  We were given two possible explanations and asked to choose one and use our experiment to confirm why the statement we chose was the correct explanation and also use our experiment to refute the other explanation.  Therefore, cells are small because cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.

    In order to prove this explanation, we cut out rectangular prisms of agar to represent our cells.  We cut out two small rectangular prisms and two large rectangular prisms of agar.  The small rectangular prisms had the dimensions of 2cm for the length, 1.5 cm for the width, and 1.5 cm for the height.  The large rectangular prisms had the dimensions of 4cm for the length, 1.5cm for the width, and 2.5cm for the height.  The volume of the small rectangular prisms was 4.5cm³ and the surface area was 16.5cm^2.  The volume of the larger rectangular prisms was 15cm³ and the surface area was 39.5cm^2.  After measuring and calculating all the dimensions of the prisms we then submerged them into a container of weak acid, in this case our weak acid was vinegar.  In the making of the agar to construct the cells a chemical called Phenolphthalein was added.  This chemicals allows us to see the diffusion of the vinegar into the prism as time went on.  The prisms soaked in the vinegar for a total of 27 minutes.  After taking the cells out of the container we measured how much of the blue colored section of the agar remained.  In the small cells what was left was 1cm in length, 0.5cm in height, and 0.5cm in width.  In the larger prisms the blue that remained was a length of 3cm, width of 1cm, and a height of 0.5cm.  The new volume of the small prism would now be 0.25cm³ and the surface area would be 2.5cm^2.  The volume of the large prisms would now be 1.5cm³ and the surface area is now 10cm^2.  By using these volumes of  the cells we see that in the small cell about 5.6% of the cell did not diffuse while about 94.4% of the cell did diffuse.  In the larger cell we see that 10% of the cell did not diffuse and 90% did diffuse.  As shown in the numbers the larger cell diffused less than the smaller cells which supports our explanation.  The smaller cells were able to diffuse more because they have a larger surface area to volume ration which was a ratio of 3.67 while the larger cells had a smaller surface area to volume ratio of 2.63, in turn making the diffusion in smaller cells faster.

    Our evidence refutes explanation two which states that the rate of cell diffusion is related to cell size.  Nutrients diffuse as a faster rate through small cells than they do through large cells.  In order to calculate the rate of diffusion you divide distance of diffusion by the time it took to diffuse.  The large cells and small cells had the same distance of diffusion which was 0.5cm, and when divided by the total time of submergence which is 27 minutes the rate of diffusion for all the cells was 0.019.  If all of the cells have the same rate of diffusion then the size of a cell can not be attributed to how it diffuses materials.  In conclusion,  cells are their small size because having a larger surface area to volume ratio is more beneficial to diffusing nutrients.  

 The cells made of agar placed into the container with weak acid(vinegar)

 Side view of the cells in the container with the weak acid

The cells after being submerged for 27 minutes.  The yellow parts of the cells show where diffusion has occurred. The blue part shows where diffusion has not reached

Rana's Cell Size and Diffusion Argumentation

Rana Srouji

AP Biology

Mr. Hammer

December 1, 2014

Cell Size and Diffusion Argument

Diffusion is the movement of molecules from a high concentration to a low concentration.  Almost all living organisms have cells that are dependent on diffusion to obtain the essential nutrients needed to survive.  Once the cells take in these nutrients, they break them down to create more components for the cell.  The cell size will then increase, but cells always remain small.  An explanation for this is, cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.

My lab partners and I proved this explanation by creating and testing models of cells using agar.  Originally, the agar was blue in color.  We created four model cells of agar, two of which represented small cells while the other two represented larger cells.  The cells were cut into rectangular prisms for easy calculations.  The small cells were cut with length of 2 cm, width of 1.5 cm, and height of 1.5 cm.  Their volumes were 4.5 cm cubed, and their surface areas were 16.5 cm squared.  The larger cells were cut with length of 4 cm, width of 1.5 cm, and height of 2.5 cm.  The volume of the large cells were 15 cm cubed and their surface area were 39.5 cm squared.  The surface area to volume ratio for the small cells is 3.67 while the ratio for the large cells is 2.63.  Afterwards, all of the cells were then placed into a plastic container filled with vinegar.   All of the cells were placed in the vinegar for 27 minutes.  Within seconds the acid began to diffuse into the agar.  Once the cells were taken out of the vinegar, we took the measurements of the blue colored agar remaining, which is the amount that had not been diffused.  The remaining blue agar in the small cells had a length of 1 cm, width of 0.5 cm, a height of 0.5 cm, and a volume of 0.25 cm cubed.  The amount remaining in the large cells had a length of 3 cm, width of 1 cm, height of 0.5 cm, and volume of 1.5 cm cubed.  In addition, using the volumes, we found that of the small cells only 5.6% of each cell was not diffused within the 27 minutes, and 94.4% of each cell was diffused.  Of the larger cells, 10% of each cell was not diffused, while 90% was diffused.  

Additionally, we used the agar in our experiment because it is a gel-like substance that chemicals can diffuse through, and that can be cut into different shapes.  We created the model cells in two different sizes to show the difference in the amount diffused within the same amount of time, and the difference in the surface area to volume ratios.  It was to show that the diffusion was directly related to the ratio.  The larger cells have a smaller ratio, while the smaller cells have a larger ratio.  Also, we created two cells per size for both the small and the large cells in case any error were to occur.  Phenolphthalein was previously added to the agar before we used it.  The phenolphthalein is a chemical indicator that changes color when coming into contact with acid, so that the amount of acid diffused through the agar would be easily seen.  Once the cells were placed in vinegar, their color would change from blue to yellow because vinegar is an acid which will diffuse through the cell and change the color once it comes in contact with the phenolphthalein.  In conclusion, cells are small because they have a large surface area to volume ratio, which allows them to diffuse most efficiently.  The larger cells had a less amount diffused, only 90% of each large cell diffused, and a smaller surface area to volume ratio, (a ratio of 2.63).  Unlike the small cells which had a larger surface area to volume ratio, of 3.67, and diffused 4.4% more than the large cells, within the same amount of time.  Thus, the cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.  

However, the reasoning for why cells are so small cannot be because the rate of diffusion is related to cell size.  Nutrients do not diffuse at a faster rate through small cells than they do through large cells.  This is because, as my group tested the agar cells, we collected data that showed the distance of diffusion through the cells.  We gathered this data in order to find the rate of diffusion, since the rate of diffusion is the smallest distance of diffusion per cell divided by the amount of time diffused.  Both the small and large cells had the same distance of diffusion, which was 0.5 cm.  The rate of diffusion then is 0.5 divided by the 27 minutes, which then equals 0.019.  Thus, the rate of diffusion cannot be related to the cell size since both the small and large cells have the same rate of diffusion.  

This is an image of the "cells"/agar as soon as they were placed into the container of vinegar.  The acid began to diffuse into the cells as seen by the light yellow coloring on the outer edges of the agar.  

This is an image of the cells/agar after the 27 minutes, once they have been removed from the acid.  The blue inside is the amount that has not been diffused, the yellow coloration is the acid that diffused though the agar and has come into contact with the phenolphthalein.  

Rameia's Mitochondria Organelle Op-Ed

Mitochondria

The Right Choice for President of The Cell

     I, Ms. Mitochondria am the right choice for The President of The Cell.  Let me tell you about myself. I originally came about from a process called endosymbiosis.  This process resulted in a nucleated cell engulfing an aerobic prokaryote.  This cell relied on protection from the environment of the host and the host relied on on the cell to produce energy.  Eventually the descendants of this cell developed into mitochondria, and because of this process I am here today!

     

     I am shaped in a way that allows me to be productive to the fullest extent.  My outer membrane covers me as like it was skin and my inner membrane folds to create layered structures inside of me called cristae, with fluid inside called the matrix.  The folds in my structure allow me to have a greater surface area.  This greater surface area allows me to have more space for reactions such as cellular respiration to occur.  With this extra space, I can give this cell country the opportunity to get more things accomplished.  The more the better.  I am also a better candidate for this position than these other organelles because I have my DNA and ribosomes.  How many of these organelles can say that, not many. 

    As stated in my poster, I am the powerhouse of the cell.  I provide energy for the cell to make everything run efficiently and smoothly.  Most of my energy comes from ATP (adenosine triphosphate).  This energy is produced inside of me using energy that was stored in foods.  I produce the ATP that supplies the energy for the processes of life.  An animal cell can have from anywhere between 1000 and 2000 mitochondria that help to produce high amounts of energy.  In order to make ATP to produce energy for the cell, I go through the process of cellular respiration.  This involves using oxygen to produce energy.  The benefit of this aerobic process is that it improves your body to make ATP faster.  Did I also forget to mention that I create more than 90% of the energy needed by the human body to grow, as well as metabolize amino acids, fatty acids and steroids.  

    As you can see I am the best woman for this position!  My position in the cell is very important and each and every day I am working hard to produce energy.  A vote for me means a vote for energy, a vote for me means a vote for efficiency, and lastly a vote for me means a vote for a better cell country.  Vote me, Ms. Mitochondria for President of The Cell!

Works Cited

• http://www.mitochondrial.org/index_files/page0001.htm
• http://hyperphysics.phy-astr.gsu.edu/hbase/biology/mitochondria.html
• http://www.biology4kids.com/files/cell_mito.html
• http://www.nature.com/scitable/topicpage/mitochondria-14053590
• http://biology.about.com/od/cellanatomy/ss/mitochondria.htm
• http://piktochart.com

Sara's Cell Size and Diffuion Argumentation

Sara Bearden

Cell Size and Diffusion

Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients. For this experiment my group and I used five objects that varied in shape and size. We timed how long the shapes took to fully diffuse the weak acid and what we found was that the small cube was the first one to fully diffuse the acid at 15.28 minutes. It's surface area to volume ratio was 6:1 which, was the largest ratio compared to the others. The small rectangle had a 5:1 surface area to volume ratio and it was fully diffused at 16.14 minutes which is only a minute and twenty-six seconds after the small cube was done diffusing the weak acid. With the higher ratio it allowed for the two shapes to diffuse the weak acid easier and allow the acid to reach the center faster. The higher ratio means that there was more membrane for the weak acid or nutrients to enter and be able to move into the cell to the center. That then allows for that acid or another nutrients to be utilized for functions in the cell a lot quicker.  One of the shapes was a plus sign and it also had one of the larger ratios of 4.4:1 however, the diffusion of the weak acid took 33 minutes and that was the last one too finish. This evidence contradicts the claim that a cell with a larger the surface area to volume ratio a cell is more efficient at diffusion essential nutrients.  During the experiment this plus sign was buried under the other four shapes we had which, caused only for a limited amount of membrane to be exposed to the weak acid directly.  The areas that were under the shapes caused the acid to take longer to reach that area, leading to a longer time for the shape to fully diffuse the acid. This also occurred with the bigger cube where it has a surface area to volume ratio of 3:1 but it took 24 minutes to fully diffuse the weak acid.  There was another shape that had a smaller ratio but, it was smaller than the big cube and it finished before the big cube which, shouldn't be according to my claim.  The shape was the big plus sign that had a surface area to volume ratio of 2.2:1. That shape should have been the last one to finish diffusing the weak acid since it had less membrane exposed to it which, only allowed  limited acid to enter the cell and move to the center to be utilized.  But, since it covered the shapes with bigger ratios than itself it caused them to have a slower diffusing time than predicted.  If the shapes were spread out instead of clumped together the time of full diffusion and the ratios would have matched up.  The shapes or "cells" with the larger surface to volume ratios like the small plus sign, rectangle, and the small c would have times closer to each like the rectangle and small cube did. Their larger ratios allow more membrane to be exposed to the nutrients (weak acid) so more can enter and travel to the center of the cell to be utilized for organelle functions.

The rate of diffusion may relate to the size of the cell and sometimes the nutrients diffuse faster through smaller cells than they do in larger cells but, that's not always the case.  For instance, we had a cube with the dimensions of  2cm by 2cm by 2cm and a plus sign that was 2cm all around in the weak acid along with the other three shapes we had.  The cube makes up one part of the cross since, depending on how you cut the cross up you would get five cubes that are all 2cm by 2cm by 2cm so the plus sign ends up bigger than the cube. The cube had a diffusion rate of 1cm per 24 minutes and the plus sign had a diffusion rate of 2.5 com per 21.56 minutes. The distance used in the diffusion rate was the shortest one for the weak acid to take in order to reach the center. The rate of the plus sign is higher than the cube because the weak acid traveled a longer distance in a shorter amount of time than it did in the cube.  What that means is the weak acid reached the center of the plus sooner than it did in the cube and that would allow the weak acid to be utilized quicker for the functions of the cell.  Therefore, the rate of diffusion does relate to the size of the cell since that determines how far the nutrients have to travel to the middle of the cell but, it doesn't always means that the smaller cells diffuse the nutrients faster since the plus sign diffused the weak acid which, represented the nutrients faster than the cube which, was smaller than the plus sign.

The shapes after the weak acid was fully diffused.

A bird's eye view of the shapes in the acid and you can see the weak acid starting to diffuse into them.

The shapes in the weak acid as the diffuse it.  With them not spread out it caused for less membrane to be exposed to the acid so the diffusion time was slower for those with larger surface area to volume ratio like the plus sign.

This is the weak acid the shapes were put in.  The acid represent the essential nutrients that cells would have to diffuse. The shapes are the cells.

Rameia's Mitochondria Organelle Campaign Poster

Nida Ali's Argumentation of Cell Size and Diffusion

Diffusion Cube Argumentation

My group members and I are supporting the explanation that cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients. We used a variety of shapes and sizes that ranged from a small cube to a large cross. The cube with dimensions of 1cm by 1cm by 1cm was the first to dissolve at 15:28 minutes and had the largest surface area to volume ratio of 6:1. Quickly after, the small rectangle with the dimensions of 2cm by 1cm by 1cm  was the second to dissolve at 16:14 minutes with the second largest surface area to volume ratio of 5:1. The cubes that had the largest surface area to volume ratio were among the smallest cubes that diffused the weak acid quicker than the rest by reaching the center point of the cube from each side of the cube. Even though the small cross with a large surface to volume ratio of 4.4:1, it was the last to completely diffuse the weak acid at 33 minutes. Due to this occurrence, it refutes the claim that cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients. Since my group members and I decided to compile all of the cubes into one container with the weak acid, some cubes were on top of one another. The small cross was at the bottom of the container, which resulted in limited amount of the cube (membrane) to be completely exposed to the weak acid as seen in the image below.  

Once placed in vinegar, the cells began to diffuse. Cells were piled on top of one another which slowed down the process of diffusion for the cells that were at the bottom of the container. The yellow edges of the cell represents that diffusion has taken place. 

The piled shapes took longer to diffuse since less area of the “cells” were exposed to the weak acid. In doing so, the smallest cross took the longest to diffuse since it was piled beneath the larger cube as well as the large cross. The large cube diffused at 24 minutes with a surface area to volume ratio of 3:1. In addition, the large cross with a surface to volume ratio of 2.2:1 finished before the large cube at 21:56 minutes which goes against the previously stated claim. The large cross should have been the last “cell” that diffused due to its small surface area to volume ratio. Since the small cross was piled on top of by the large cube and then by the large cross, the small cross and large cube took longer to diffuse. According to this claim, the “cells” with the smallest surface area to volume ratio should be the last one to fully diffuse. But this occurrence did not occur due to the piling of “cells”. If my group members and I decided to place each “cell” in their own container and allow it to diffuse, we would have seen that the “cells” with the larger surface area to volume ratio would have diffused first in comparison to the “cells” with the smaller ratios.

As seen here, the piling of cells
displays how the entire surface of the cell
not fully exposed to the vinegar in
comparison to the cells that are being
diffused fully due to complete exposure. 

This is what the cells looked like after they were fully diffused.
The remaining blue seen inside of one cell displays
the cell that diffused last due to remaining at the
bottom of the container. 

As seen in the left image above, the “cell” with the blue still remaining is the “cell” (small cross) that diffused last at 33 minutes due to it being piled at the bottom of container as seen in the image on the right.

The explanation that the rate of diffusion may relate to the size of the cell and sometimes the nutrients diffuse at a faster rate through small cells than they do through larger cells is what my group members and I are refuting against. The diffusion rate actually does relate to the size of the cell, but it does not support the idea that smaller cells are able to diffuse the nutrients at a faster pace than larger cells. Likewise, as seen in the large cross with the dimensions of 2cm of each side of the cross, it was typically larger than the large cube with the dimensions of 2cm by 2cm by 2cm. While figuring out the diffusion rate of the cells, we discovered that the cubes with 2cm by 2cm by 2cm made up a cube in the large cross, which thus states that the large cross is larger than the large cube since five large cubes are able to combine to form the large cross. Since the large cross had a larger diffusion rate of 2.5cm per 21:56 minutes than the large cube, which had a diffusion rate of 1cm per 24 minutes, the weak acid was able to travel faster over a longer distance. The large cube had a shorter distance to travel to the center point of the cube and did so at a slower pace than the larger cross. Even though the large cube is smaller than the large cross, it did not diffuse faster than the large cross which thus goes against the claim that nutrients diffuse faster in small cells than they do in large cells. The diffusion rate allows us to comprehend how far and quick the weak acid (vinegar) travels to the center of the cell, but the smaller the size does not mean that it will be able to diffuse to the center first, as seen in our experiment.

Jazmean and Chigozie's Op-Ed

Smooth Endoplasmic Reticulum and Chloroplast for President and Secretary of 

Energy: The Cleaner Vote

Times are rough nowadays. We are still reeling from the effects of a recession, job growth is slow, and the price for goods is increasing rapidly. All of these can make the cells within your body very stressed, and a stressed cell is not a happy cell. During your time of need, you need organelles that are going to have your back and keep everything functioning smoothly and efficiently. This is where we come in: the Smooth Endoplasmic Reticulum and the Chloroplast! We are two organelles that are key in keeping Eukaryotic cells in a state of homeostasis. Our functions include providing the cell with food, energy, and lipids, detoxifying the cell, and transporting energy to other cells to be broken down and metabolized. We need your support to become President and Secretary of Energy of all the cells. Our plan is to create a greener, cleaner, and more efficient cell without sacrificing ATP in the process. Supporting us will guarantee a better cell!

I, the Smooth Endoplasmic Reticulum, am the main producer of lipids within a cell. Lipids are used for a variety of things, most importantly for the formation of the plasmic membrane. Without lipids, there would be no plasmic membrane which means they would be no protective barrier for the other organelles within the cell. Lipids can also be used a storage container for large amounts of energy that are not used by cells right away. When broken down, these lipids can provide cells with 9 kcal/gram of energy compared to the 4 kcal/gram of energy given off by proteins and carbohydrates. I can also make the lipids that are key in transporting fatty acid chains to the Mitochondria to be broken down and metabolized for energy. Unlike other organelles, I am composed of a lot of tubes that are folded in different ways. This allows for an increase of surface area which is great for storing molecules and important enzymes that are can be used as a catalysis for certain processes within a cell. I also play an important role in detoxifying the cell. When other organelles are busy letting bad things get into the cells, I am the one who cleans up behind them and breaks down those harmful substances into products that are safe enough for the cell to use. This function is very important in the liver of humans as the liver is the main site of detoxification for them.
So let's recap. I make one of the most important macromolecules for the cell and I make sure that the cell stays clean and rid of harmful substances. I am the evolutionary by product of the cell membrane so I've been here for awhile. Why not vote for me for President?

I, the Chloroplast, am the producer of food and energy in a plant cell. My most important function is to make food through photosynthesis for the whole cell. I literally ‘go green’ and efficiently use solar energy to prepare food in the form of sugars. The food and ATP I produce powers all the organelles in the cell. I give the other organelles the energy to carry out the instructions from the Nucleus. Many of my critics enjoy comparing me to my long lost cousin, the Mitochondria. Yes, I am like the Mitochondria. We both produce ATP for the United Republic of Cells. However, I would be a better candidate for Secretary of Energy because I am more versatile. Unlike the Mitochondria, I can work in multiple conditions. The Mitochondria can only produce ATP by using oxygen. I can produce ATP with AND without the use of oxygen.

While other organelles depend on me, I do not depend on any other organelle. I survived on my own as a well-functioning organelle in some of the most extreme conditions on Early Earth before I joined the United Republic of Cells. I had to survive only on what I was made of; I could not depend on any other organelles as many other modern organelles do. While I am self-sufficient, I can also work well with others. Over the past years, I have worked with the Nucleus, Cell Membrane, Smooth Endoplasmic Reticulum, and Rough Endoplasmic Reticulum to protect the United Republic of Cells from pathogens.

Without me, plants would not be able to survive. Plants produce oxygen for many living organisms, so many living organisms would also die. If I were to become Secretary of Energy, I would help develop more efficient ways for cells to produce energy for all cells, not just plants cells! Vote me for Secretary of Energy if you do not want the Earth to look like a barren wasteland! Vote today for tomorrow’s energy!

A vote for Smooth Endoplasmic Reticulum and Chloroplast is a vote for

 a better cell and a better future! Vote today!

Work Cited
http://bscb.org/learning-resources/softcell-e-learning/endoplasmic-reticulum-rough-and-smooth/
http://www.sciences360.com/index.php/functions-of-lipids-in-human-cells-12883/
http://education-portal.com/academy/lesson/smooth-er-definition-functions-structure.html#lesson