Cube argumentation by Adil and Harold

Claim

Our claim was that Explanation one which is: Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients, is correct,

Evidence  

5 minute trial (figure 1)

side length	penetration depth	surface area to volume ratio	%of cube penetrated
.5cm	0.4cm	12	100%
1cm	0.4cm	6	98.30%
2cm	0.4cm	3	78.40%

2.5 minute trial (figure 2)

side length	penetration depth	surface area to volume ratio	%of cube penetrated
.5cm	.2cm	12	99.20%
1cm	.2cm	6	78.4%
2cm	.2cm	3	48.80%

rate of diffusion .08cm per minute `

Reasoning

From our evidence we concluded that explanation 1 was correct. This is because we found that the cube with the highest surface area to volume ratio was the most efficient at diffusing the vinegar throughout itself.  This may seem questionable considering the results in the last column of figure 1 but it is not. When the vinegar penetrates the cube it does so from all sides of the cube. If one wanted to calculate the volume of the cube the vinegar penetrated they would  get a side length. Since the cubes got penetrated .4 cm one might think that is the side length however since the cube is penetrated on all sides that length needs to be doubled. It then becomes apparent that, the volume the vinegar penetrated is more than the volume of the smallest cube. This means this cube was clear while the rest of the cubes had more time in the vinegar. So to make things more clear we halved the time and put new cubes of the same sizes in vinegar. This resulted in half the penetration of the 5 minute trial this and the fact that the different sized cubes got the same amount of penetration in the same amount of time supports the idea that the rate of diffusion is constant. So since the rate of diffusion is constant then the only thing that could affect the efficiency of the cell is the surface area to volume ratio. We found in every case the cell with the highest surface area to volume ratio was the most efficient. In both trials the cell wit the surface area of 12 (the highest) had nearly 100% penetration. The smallest cell has more places for the vinegar to enter relative to how much space the vinegar has to fill this is why it is more efficient.

Alternative claim

In our Cell size and diffusion experiment, through data analysis and calculations, we said that claim 1, i.e. Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients was correct . However, our alternative claim is also stated below.

Alternative Claim: The rate of diffusion is related to cell size. Nutrients diffuse at a faster rate through small cells than they do through larger cells.

refutation

The rate of diffusion does not rely on the size of the cell. From our experiment, we came to a conclusion that no matter what the size is, vinegar would diffuse into the agar cube with the same rate on all 6 sides.  We found this rate to be .08 cm per minute. However, we tend to see that smaller cubes become clear much faster than others, this is a true statement because these cubes tend to have a higher surface area to volume ratio,  but the rate of diffusion was constant . We did two trials and from both of them, we calculated the rate of diffusion for all 3 cubes and the value was approximately the same. The next time trial we did the same thing but for only 2.5 minutes. The rate of diffusion was still the same. Therefore, we can make a conclusion that the efficiency of the cell depends more on the surface area to volume ratio than to the size of the cell because if it relied on the size of the cell, than we would have got different rates for each cube depending on their size, but instead we got same type of rates. That is why the Alternative claim is not the right choice.

Diffusion Cube Argument - Raquel and Sana

Research Question:

Why are cells so small?

Claim:

Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.

Evidence:

Cubes	Volume (cm3)	Surface Area (cm2)	SA:V ratio	Rate of Diffusion	% Cell Receiving Nutrients
1 (1cm x1cm x1cm)	1	6	6:1	.5cm/20mins	100
2 (1.5cm x1.5cm x1.5cm)	3.4	9	2.6:1	.4cm/20mins	93.6
3 (2cm x2cm x2cm)	8	12	1.5:1	.4cm/20mins	78.4

Justification:

Based on the evidence it proves that the largest surface area to volume ratio was the most efficient at diffusing the vinegar. The smallest cube (cube 1) had the largest surface area to volume ratio which was 6:1. After the 20 minute trial, the cube was completely clear and had 100% diffusion. Cubes 2 and 3 had smaller ratios and were not completely clear after the 20 minute trial. They only diffused 93.6% for cube 2 and 78.4% for cube 3. This means that cube 1 was more efficient at diffusing because it diffused the most amount of vinegar in the same amount of time compared to cubes 2 and 3. This can be related to its higher surface area to volume ratio because cube 2, which had a slightly larger ratio than cube 3, was able to diffuse more than cube 3.

Alternative Claim:

The rate of diffusion is related to cell size.  Nutrients diffuse at a faster rate through small cells than they do through large cells.

Challenge:

Based on the evidence it shows that the rate of diffusion for all three cubes showed no significant difference. Cube 1’s rate was .5cm/20s, Cube 2’s rate was .4/20s, and Cube 3’s rate was .4/20s. The rates were within one centimeter of each other. This disproves the claim because the rates were relatively the same for all three cubes despite their differences in dimension, therefore concluding that diffusion is not related to cell size. Nutrients were diffusing at similar rates to one another no matter what cube size they were.

Diffusion Cubes Argumentation- Connor and Natalia

Question:

Why are cells so small?

Claim:

Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients

Evidence:

Efficiency of Cell Diffusion in 16 minutes and 25 seconds

Cell size	2 x 2 x 2	1.5 x 1.5 x 1.5	1 x 1 x 1
Surface area to volume ratio	3:1	4:1	6:1
% cell receiving nutrients	87.5%	96.3%	100%
Rate of diffusion	.024 cm/min	.028 cm/min	0.3 cm/min

The smallest cube had the largest surface area to volume ratio and was also the most efficient at diffusing.

Reasoning/Justification:

Based on our evidence it was clear that our claim was correct because the cells that had the largest surface area to volume ratios were also the most efficient at diffusing nutrients. The 1 x 1 x 1 cube had the largest surface area to volume ratio and also diffused 100% of the nutrients in the time the other cells diffused only 96.3 and 87.5 percent.  This means that the 1 x 1 x 1 cube was the most efficient at diffusing the nutrient, and this can be accounted for by the large surface area to volume ratio.

Alternative Claim:

The ratio of diffusion is related to cell size. Nutrients diffuse at a faster rate through small cells than they do in larger cells.

Challenge Against the Alternative Claim:

The data that was collected did not support the alternative claim. All of the rates were fairly close to each other. The cells were all timed for the same amount of time (16 minutes and 25 seconds) and we were able to see that the diffusion into each cube were only millimeters away from each other. If it the rate of diffusion was related to cell size, a significant difference in the diffusion rate of all three cells would have been observed.

Cell Diffusion Argument: Russell and T'ea

Why are cells so small ?

Claim:

We have concluded that the reason why cells are so small because if a cell has a large Surface area to volume ratio they are more efficient to diffusing essential nutrients. After conducting an experiment with our cubes we were able to come to this conclusion. In our experiment we had 3 cubes each with different dimensions. The first cube was 1 x 1 x 1, the next cube was 1.5 x 1.5 x1.5, and the last one 2 x 2 x2. For each cube we placed them in their own solution of vinegar for a total of 15 minutes. After that time, we measured how much of the cube was diffused by the vinegar. If there were some parts that were not completely diffused we measured the dimensions of how much of the cube was not clear.

Cell Diffusion Lab Poster

Cell Diffusion Lab Data

Cube Number	SA:V	Starting Volume (cm3)	Ending Volume (cm3)	% Diffused
One (1x1x1 cm)	6:1	1	0	100
Two (1.5x1.5x1.5 cm)	4:1	3.375	.48	85.8
Three (2x2x2 cm)	3:1	8	1.32	83.5

(Note: The volume is related to how much of the cube had a pink tint to it. The ending volume is less because parts of the cube were clear. )

Based on our data we noticed that the only cube that was completely diffused was cube 1 during the 15 minute trial. This cube was also the one with the highest surface area to volume ratio to other cubes. As the surface area became higher it would take longer for diffusion to go on. For cubes 2 and 3 they were only able to diffuse 85.8% and 83.5% respectively. Both of those cubs had a smaller surface area to volume ratio then cell 1. Therefore, cells with a larger surface area to volume ratio are more efficient at diffusing essential nutrients.

Alternate Claim:

Another possible claim as to why cells are so small could be the following: The rate of diffusion is related to cell size, and nutrients diffuse at a faster rate through small cells than they do through larger cells. This claim is not accurate because, over 15 minutes, the rate of diffusion for all three cells was the same. In order to calculate the diffusion rate of the cells, we used the depth of diffusion, which was measured in each of the cells. The depth of diffusion was measured to be 0.4-0.5 cm in all sides of each of the cells. Due to the depth of diffusion being similar in each of the cells, the rate of diffusion would also be similar in each of the cells. Therefore, the claim that diffusion rate depends on the size of the cell is not accurate because we recorded data that proves that the rate of diffusion was the same within cells of different sizes.

Molarity Mixup Lab

Connor Delaney

12/12/16

Part I

The findings of this lab yielded results as follows: The yellow solution is the 1 Molar solution, the red solution is the .8 Molar solution, the purple solution is the .6 Molar solution, the blue solution is the .4 molar solution, the orange solution is the .2 Molar solution, and the green solution is the 0 Molar solution. In this experiment we placed different colored bags of dialysis tubing into different colored solutions. For example the yellow solution may have had a red dialysis tubing bag in it. The bags of dialysis tubing were weighed before they were placed in the solutions and after they had been in the solutions for twenty four hours. Using deductive reasoning from which solutions were more concentrated than others (determined by whether the bags gained or lost weight) the results were determined.

    Water will always flow from an area of high concentration to an area of low concentration. This means that when the two different solutions interacted with each other, the solution with the higher concentration of water would be the one to lose weight, and the solution with the lower concentration of water would be the one to gain weight. The solution with the highest molar concentration would also be the solution with the lowest concentration of water. Weighing these bags before and after allowed for the deductive reasoning of which solution between the bag and the solution it was placed in was more concentrated. This led us to conclude that yellow > red> purple> blue> orange> green as far as concentration.

Part II

    For part two of the molarity mixup lab, we set out to calculate the water potential of a parsnip. To do this we placed pieces of parsnip into the different solutions (1 Molar to 0 Molar in increments of .2 Molar). The parsnip pieces were measured for their weight in grams before and after soaking 24 hours with one in each solution. Their percent change in mass was calculated and graphed, in order to determine the molarity of the the parsnip.

Parsnip Samples of Each Molarity Solution Initial Weight (g) Final Weight (g) percent change in mass.

Using the equation of this graph’s trendline, it can be determined that the molarity of the parsnip is around 1.78 Molars. This is when the percent change would reach 0, meaning that the parsnip and solution would have the same molarity at 1.78 Molars. This number can then be used to calculate the water potential of this parsnip using the equation -iCRT. This comes out to -43.64 bars in water potential for the parsnip.

   

Molarity Mixup and Water Potential Argument

Activity 1: Molarity Mixup

Claim:

In this lab my partner and I were given six sucrose solutions of concentrations spanning from 0.0M to 1.0M in 0.2M increments in different colors (0.0M, 0.2M, 0.4M, 0.6M, 0.8M, 1.0M). We had to determine the molarity of each solution with our understanding of water potential and tonicity. We argue that the green solution was 0.0M, orange was 0.2M, purple was 0.4M, blue was 0.6M, red was 0.8M, and yellow was 1.0M.

Evidence:

	Initial Mass(g)	Final Mass(g)	% Change in Mass
Bag 1- Red	11.94	5.3	-56 %
Bag 2- Orange	13.56	8.9	-34%
Bag 3- Yellow	10.85	4	-63%
Bag 4- Green	14.61	14.5	-11%
Bag 5- Blue	13.1	7.9	-39%
Bag 6- Purple	13.7	8.5	-38%

My partner and I weighed each bag before and after putting them in the sucrose solutions. We did this because we would be able to detect the change in mass of each bag after it was put into the solution. The change in mass was calculated by subtracting the final mass from the initial mass divided by the initial mass and that number multiplied by 10
.

Reasoning:

By examining the change in mass percentages, my partner and I matched up the percentages with the molarities of each solution. The heavier the bag weighed the higher its molar sucrose solution was. Bag 4 (Green) had the lowest percentage of change in mass, -11%, and we matched that to the 0.0M sucrose solution. Bag 2 (Orange) had -34% change in mass and we matched that to the 0.2M sucrose solution. Bag 6 (Purple) had -38% change in mass and we matched that to the 0.4M sucrose solution. Bag 5 (Blue) had -39% change in mass and we matched that to the 0.6M sucrose solution. Bag 1 (Red) had -56% change in mass and we matched that to the 0.8M sucrose solution. Bag 3 (Yellow) had -63% change in mass and we matched that to the 1.0M sucrose solution.

Water moves from areas of high concentration to areas of low concentration. Sucrose solutions that had more sucrose, 1.0M for example,  had fewer amount of water left in it’s dialysis tube after the trial. This is because there was high concentration of water in the bag in the beginning of the lab and moved out into the lower water concentration solution. Sucrose solutions that had less sucrose, 0.2M for example, had more water left in its dialysis tube after the experiment compared to higher sucrose solutions. The water concentration in the dialysis tube is still higher compared to its sucrose solution however less water moves out because there is not a dramatic difference in water concentration.

Activity 2: Water Potential of a Turnip

Claim:

In this activity my partner and I had to determine the water potential of a turnip. The water potential of the turnip was -11.03 bars.

Evidence:

	Initial Mass(g)	Final Mass(g)	% Change in Mass
Slice 1	3.81	2.8	-27%
Slice 2	4.65	4.8	3%
Slice 3	3.42	2.3	-33%
Slice 4	3.45	4	16%
Slice 5	2.84	2.7	-4%
Slice 6	4.99	5.1	2%

My partner and I weighed each slice of turnip before and after putting them into the sucrose solutions to determine their percent change in mass. The change in mass was calculated by subtracting the final mass from the initial mass divided by the initial mass and that number multiplied by 10.

Shown above is a graph displaying the change in mass percentages of all six sucrose solutions. The molar concentration that would be in equilibrium with no percent change in mass would be 0.45 because that was the point at which the line crossed the x-axis.

Reasoning:

Using the -iCRT equation we determined the water potential of the turnip. Our ionization constant  was 1 because we were using sucrose, molar concentration was 0.45, R was 0.0831 because it is our constant, and temperature was 295 (20 degrees plus 273). We multiplied all of these components and received the water potential of the turnip which was -11.03 bars.