Jazmean's Cell Size and Diffusion Argumentation

Cell Size and Diffusion Argumentation

Diffusion is the random motion of atoms or molecules from an area of high concentration to an area of low concentration. Diffusion is a mode of passive transport for cells because no energy is required for it to happen. Cells use diffusion in order to move atoms or molecules in and out of the cell. Through this process, a cell can expel waste or gain vital nutrients and at the same time, grow or shrink in size. Despite this, cells are never bigger than a grain a salt (which is 0.5 mm), but can be as small as 6.5 nano meters (which is the size of hemoglobin). So this begs the question, "Why are cells so small?". This is what my group members and I investigated in class. We were presented with two explanations as to why cells are small.  Explanation 1 stated that cells that have a larger surface area to volume ratio are more efficient at diffusion essential nutrients. Explanation 2 stated that the rate of diffusion is related to cell size and that nutrients diffuse at a faster rate through small cells than they do through large cells. Due to the data that was collected, there is no clear explanation that is the most valid or acceptable. 

Our group had the follow materials available for us to use: Agar (a green/blue gelatinous substance that is obtained from algae), weak acid (vinegar), beakers, a stopwatch, rulers, and plastic knives. The agar was in a container. We took the agar from the container and cut the agar into cube-like shapes (in reality they are rectangular prisms).  

Displayed are the measurements of the cubes: 

Cubes (Before Diffusion) (Outer Cube)	Length	Width	Height
1 (Small)	2 cm	2 cm	1.5 cm
2 (Big)	3.5 cm	3.5 cm	1.7 cm
3 (Big)	3.5 cm	3.5 cm	1.6 cm
4 (Small)	1.9 cm	1.9 cm	1.5 cm

We then placed the cubes into a container filled with vinegar at the same time. When the cubes hit the vinegar, the stopwatch started to keep time. At 7.3 seconds, the color started to change. Due to time restrictions, we were not able to have our cubes be fully diffused. At 32 minutes and 9 seconds, we stopped the experiment and pulled the cubes out.

Because of the time restrictions and our cubes not being fully diffused, we had to take measurements of the cube not diffused yet. 

Not Diffused Cube Measurements (inner cube)	Length	Width	Height
1 (Small)	1 cm	0.8 cm	0.7 cm
2 (Big)	2.5 cm	2.5 cm	0.5 cm
3 (Big)	2 cm	2.5 cm	0.9 cm
4 (Small)	0.9 cm	0.5 cm	0.5 cm

We then had to calculate the diffusion rates. We had to calculate the distance of the length, width, and height out each outer and inner cube by subtraction the big length from the small length and dividing that number by
two. 

Explanation given to my group by Mr. Hammer

How to Calculate Cube Diffusion Rates (Repeat for all Cubes)

side one= big length - small length/ 2

side two= big width - small width/ 2

side three= big height - small height/ 2  

Displayed below is the results.

Cubes	Length	Width	Height
1	0.5	0.6	0.4
2	0.5	0.5	0.6
3	0.75	0.5	0.35
4	0.5	0.7	0.5

We then averaged the length, width, and height in order to make the process easier. And to find the rate, we took the averages and divided it by the time the cubes stayed in the vinegar. The time was changed from minutes into seconds. So instead of it being 32.9 minutes, the time was changed to 1,929 seconds. 

Averaged Diffusion Rates

	Rate of Diffusion
Cube 1	.000259 cm/sec.
Cube 2	.000276 cm/sec.
Cube 3	.000276 cm/sec.
Cube 4	.000294 cm/sec.

In addition to calculating the rates of diffusion of each cube, we also calculated the surface area and volume of each cube. If you compile all the data into one chart, this is what it looks like:

Cubes	Volume	Surface Area	Ratio (SA to V)	Rate of Diffusion
1 (small)	6	20	10 : 3 (3.33)	.000259 cm/sec
2 (big)	20.825	48.3	48.3 : 20.825  (2.32)	.000276 cm/sec
3 (big)	19.6	46.9	46.9 : 19.6   (2.39)	.000276 cm/sec
4 (small)	5.415	18.6	18.6 : 5.415   (3.43)	.000294 cm/sec

As you can see by our final data table, we have conflicting data. For Explanation 1 to be the most valid, the cubes with a larger surface area to volume ratio should have a faster rate of diffusion. This would be true if Cube 1 was not apart of the data table. Cube 1 has a surface area to volume ratio of 10 : 3 and has a rate of diffusion of .000259 cm/sec. However, Cube 2 and 3 have a ratio of 48.3 to 20.825 and 46.9 to 19.6 (respectively) and a rate of diffusion of .000276 cm/sec. Cube 2 and 3's ratios are smaller than Cube 1's but their rate of diffusion is faster than that of Cube 1's. This means that Cube 2 and 3 would be more efficient in diffusion nutrients to the center of the cell than Cube 1. Despite this, Cube 2 and 3 have a smaller surface area to volume ratio and a slower rate of diffusion compared to Cube 4. Cube 2, 3, and 4 support Explanation 1. 

For Explanation 2 to be the most valid, the smaller cubes would have a faster diffusion rate compared to the larger cubes. Cube 4 is the one of the small cubes and it does have a faster diffusion rate compared to the larger cubes. However, Cube 1 is also one of the small cubes but it does not have a faster diffusion rate compared to the larger cubes. 

If you were to ignore Cube 1, Explanation 2 would be the most valid explanation. Cube 3 has a larger surface area to volume ratio, but has the same rate of diffusion has Cube 2. This would disprove Explanation 1. However, because the difference between the ratio is not drastic, one could infer that the small difference would not effect the rate of diffusion drastically either. Because of the conflicting data, I would not feel right as a scientist to call for one side. Further trials need to be conducted in order to solidify the data in order to pinpoint which Explanation is the most acceptable. 

Further Thoughts

These explanations can go hand in hand if you think about it. If you increase the surface area to volume ratio of a cube for example, the size of the cube gets decreases. This is the same for a cell. If you increase the SA to V ratio of a cell, its size decreases. A large SA to V ratio limits how big a cell can get. If a large SA to V ratio makes diffusion more efficient for a cell, it might make sense as to why people think smaller cells diffuse faster than larger cells; it's because their surface area to volume ratio is large. 

Sara's Water Potential and Molarity Mix Up Argumentation

Sara Bearden

Water Potential & Molarity Mix up Argumentation

   The color that has a molarity of 0.0M is dark green, light green is 0.2M, blue is 0.4M, orange is 0.6M, red is 0.8M, and yellow is 1.0M.  Figuring out the molarity of each colored was based off both of the data collected from the first activity and the second activity due to the fact that the light green was done twice in the first part which, led to different results than expected. Also the vegetables give a more realistic reading about the solutions since the vegetables have real cell membrane where the dialysis tubing is just a representation of a cell membrane. From the second activity the dark green showed to have the greatest percent change in mass with a  28.7% change in the turnip and a 25% change in the artichoke. That means that there was a low concentration of water in both vegetables and a higher concentration in the dark green solution so by osmosis the water flowed into the vegetable which, increases the weight. This also made the piece softer since water was absorbed into the vegetable.  Since, it was the softest that means the most water flowed into it and also with a greater change in the weights the dark green solution is 0.0M. The light green solution also had a bigger change in the masses of both vegetables as well and also one of the smallest changes in mass for the dialysis tubing.  There was a 5.2g difference in the mass for the dialysis tubing, a 1.1g difference in the turnip, and a 0.2g difference in the artichoke mass at the end of the experiment. All mass differences were positive which, means that the weight increased which, means more water flowed in into the cell. Since, more water flowed in there was a low concentration of water in the vegetables but, for the dialysis tubing water would have flowed out first then in due to the fact it had more solution than water in it than the pure water. The random motion of the water by osmosis may have resulted in the increase of the mass to be high but, the results were lower than the other colors for the dialysis tubing. Therefore, light green has a molarity of 0.2M. The blue had very similar percent changes in mass in both vegetables with a 11.5% for turnip and 11.1% for artichoke but, for the dialysis tubing there was a 27.2% change. This is due to the water flowing into the cells by osmosis since there is a low concentration there since it's hypertonic.  The dialysis tubing was the data that helped with figuring this one out because the vegetables had similar results so seeing that bigger change in the mass and feeling the tubing and noticing it was a little firm helped conclude that blue was 0.4M.  While the plants were in the orange solution their masses didn’t change as much but, they were still positive with a 0.1g increase in the turnip and in the artichoke. This would be a result due to the plants having a slightly smaller concentration of water than the orange solution so the water flowed in them but, not as much during their random movement.  But, for the dialysis tubing the orange cell had a 11.1g increase and felt firm which, indicates it was a hypertonic solution so, water flowed in due to the low concentration and filling the cell. Therefore, making the orange solution be 0.6M. Red was one of the more hypertonic solutions because it was the first one to show a negative percent change in mass in the turnip which was -9.8% but, the artichoke still had a positive percent change of 11.1%. That could be from the vegetables having different water concentrations where the turnip has more.  With a negative change that means the solution the turnip was in had less water than the turnip and that the water flowed to the solution due to the low concentration there and showing how the red is 0.8M.  Lastly for the yellow the dialysis tubing in the end was very firm and it had the greatest increase in mass with a 14.4g increase.  Also there was a decrease in the mass in the vegetable in the end with a 1.0g decrease for the turnip and a 0.2g decrease for the artichoke. What that means is that there was more solutes in the dialysis tubing than in the water so the water flowed into the bag and for the vegetables there was more water in them than the solution itself so the water flowed from the vegetable into the solution but osmosis.  This is due to the low concentration of water in the solution and a high concentration in the vegetables. The tubing had the lowest concentration of water which, is why more water ended up flowing into it through its random movement and this causes for orange to have the highest molarity which was 1.0M.

    The water potential of the turnip is -21.2 and the water potential for the artichoke is -15.0.  The solute potential had to be calculated first and in order to get the concentration the point at which, the line of the graph crossed the x-axis line was used. The graph was made based the percent change of the pieces and the molarity of the colors they were in. The graph shows that the point would be about 0.86M for the turnip and 0.61 for the artichoke. By picking this point the vegetable would be at equilibrium with the solution so that means there is an equal amount of water in both from the random movement of the water over time. The temperature of the solutions was measured to be 23 degrees Celsius and when converted to Kelvin it was 296. After calculations the solute potential for the turnip was -21.2 and for the artichoke it was -15.0.  The pressure potential is 0 because when using the point at which the line crosses the x-axis in the graph is where the cell would be at equilibrium. When the cell is at equilibrium there is no pressure being exerted on the cell wall or any shrinkage of the membrane away from the cell wall.  So, when adding 0 to both numbers you get the same result for the solute potential as for the water potential.

The green is the artichoke and the black is the turnip.