Cellular Respiration

Purpose: In this experiment was to measure the rate of cellular respiration in germinated and non germinated peas.

Introduction: In the experiment we were testing the rate of cellular respiration in germinated and ungerminated peas. Cellular respiration is when chemical energy is released and changed into ATP, which creates energy.

Procedure: To begin the experiment we had to measure the room temperature using a thermometer and then record the temperature in table 1.  Then we gathered 23 germinated peas and placed them into the respiration chamber. Once the CO2 shaft was in place we let the peas sit for a minute before we began to collect the data, which lasted for 10 minutes. Once a lll the data was collected a graph of CO2 gas vs. time was displayed. After that we soaked the germinated peas in ice water and then repeated the same process and graphed the data.

Discussion/Conclusion: In our experiment, our graph showed an inverse relationship between the amount of carbon dioxide and oxygen in our container.  Oxygen went down as time increased while the amount of carbon dioxide increased as time increased.  This showcases cell respiration.  The germinating seeds used the oxygen to oxidize the sugars inside and create more ATP.  They did this in order to continue germinating and allow themselves to grow.  The only way to do this is with energy.  As a result, as the seeds continued to make more and more energy, the amount of oxygen in the container decreased.  Therefore, the amount of carbon dioxide increased considering the carbons of the sugar molecules were released as carbon dioxide.  

When we submerged our seeds in water and later put them in our container, we got the same inverse relationship.  However, the amount of oxygen depleted and carbon dioxide gained did not occur at as rapid of a speed.  This is because in cellular respiration, the oxygen used is reduced to water.  Therefore, the seeds were already covered in water, making it harder for them to receive the oxygen they needed to create more energy.  This left more oxygen in the container and did not allow carbon dioxide to be created as quickly as when they were dry. 

The experiment's results could have been tampered by us not closing the container fully.  Thus, it was an open system and changes in temperature could have affected it as well as more oxygen rushing in, making our graph constant, with the amount of oxygen in the container staying the same.  When a temperature is warmer, cellular respiration occurs at a faster rate.   

Graph showing amount of oxygen and carbon dioxide in the container with dry germanating seeds.

Graph showing amount of oxygen and carbon dioxide in container filled with wet germanating seeds. 

Germanating seeds soaking in a beaker of water.

Enzyme Catalysis

Purpose:

In the experiment the purpose was to determine the rate at which a 1.5% H2O2 solution decomposes when catalyzed by the purified catalase extract.

Intro:

In the experiment enzymes play a huge role because they are also known as catalysts. A catalyst affects the rate of a chemical reaction. An enzyme-catalyst reaction occurs when the substrate attaches to the active site of the enzyme, which then causes the reaction. The more time that the enzyme has to catalyze the substrate the less of the 1.5% H2O2 solution will be used.

2B : To begin the experiment we added 10 mL of 1.5% H2O2 into a cup. Then we added 1 mL of H2O, 10 mL of H2SO4 and proceeded to mix it. Then we removed a  5 mL sample and put it into a different cup. After that we used a burette to add one drop at a time of KMnO4 to the solution. After each drop we swirled the solution until the pink or brown color remained.  We ended up getting a baseline of four after putting potassium permanganate into our solution.  Originally, we had four mL of hydrogen peroxide but it took four mL of potassium permanganate to keep the substance at a pink shade making us have a total of 8mL now.  We subtracted the amount we have now to what we started with to get our baseline of four. We put the sulfur in our solution in order to stop the enzymes from catalyzing anymore.  The pH of sulfur is too low for an enzyme to function in and ends up resulting in a denatured enzyme.  Our baseline could have been less considering we were not careful with the amount of potassium permanganate we put into our solution. This could have resulted in an excess amount of potassium permanganate.   

Solution after sulfur was added

2C:

After leaving our solution overnight, we found that 3.5 mL of hydrogen peroxide had been catalyzed.  This means that a total of 12.5% of the solution was catalyzed overnight.  This could have been altered by our setting of the room however.  The temperature of the room could have made it harder for the hydrogen peroxide to catalyze as it may not have been in its optimal temperature.   

2D:

After performing this experiment, our results ended up all over the place.  The amount of hydrogen peroxide was catalyzed using varied  time intervals.  At times, we received a positive result and at others a negative.  However, there was no pattern to our data.  This could be because of the way we handled the potassium permanganate in an uncontrolled manner.  Instead of carefully adding drop after drop, we let a stream flow into our beaker perhaps resulting in an excess of potassium permanganate.  We should have noticed that as the time interval increased, the amount of hydrogen peroxide used increased as well.  This is because the enzyme should have catalyzed the hydrogen peroxide into water and oxygen gas.  As the time increased, the enzyme had more time to do this.   

Hydrogen peroxide before potassium permanganate was added.

Diffusion/Osmosis

1A: Diffusion

This experiment tested the diffusion of small molecules through dialysis tubing, a selectively permeable membrane.  We tested the presence of glucose in a 15% glucose/1% starch solution.

Diffusion is the movement of molecules from an area of high concentration to an area of low concentration.  Diffusion through a selectively permeable membrane occurs until the solutions have reached dynamic equilibrium: the solute concentrations on both side of the membrane are the same.

First we tested a 15% glucose/1% starch solution for the presence of glucose using Testape and recorded the color of the Testape.  We put 15 mL of this solution into dialysis tubing.  Then, we filled up a cup 2/3 of the way full with distilled water, added about 4 mL of an Iodine solution to the distilled water and then tested this solution for the presence of glucose and recorded the color.  We then put the tied dialysis bag into the solution and let it sit for about half an hour.  After half an hour, we  took the dialysis bag out and recorded the color of the solution in the bag and in the cup.  Finally, we tested each solution again for the presence of glucose and recorded the results in the table.

 Testape after testing the concentration of glucose

 Dialysis bag in the water/Iodine solution starting to diffuse

Dialysis bag after half an hour of being submerged in the water/Iodine solution.

Table 1.1

	Initial Contents		Solution Color		Presence of Glucose
			Initial	Final	Initial	Final
Bag	15%glucose/1%starch		clear	purple	brown	green
Beaker	H20 & IKI		yellow/orange	same	green	green

Based on the colors observed in the table above, we can conclude that glucose is leaving the bag and the Iodine solution is entering the bag.  This means that the concentration of glucose was higher in the bag than it was in the cup.  The concentration of the Iodine solution moved from outside the bag to inside the bag to make the concentrations equal.  This movement caused the change in color of the solution inside the bag.  If we had weighed the bag and cup with the different solutions we could have determined the percent change in mass (if there was one) and had numbers to support the observation that molecules are moving from high to low concentration.  Based on our observations, we believe that the Iodine solution molecules were smallest because they were able to get through the membrane of the dialysis tubing and change the color of the solution drastically in just half an hour.  The glucose molecules follow the Iodine solution in size, because based on our Testape results, we see that glucose moved out of the dialysis tubing: high to low concentration.  The membrane pores and starch molecules are the two largest.  The membrane pores were big enough to let solutions in and out.  Starch cannot pass through this semipermeable membrane, making them the biggest molecules.  If we would have started with glucose and Iodine solution inside the bag, they would have moved out of the bag into the starch solution in an attempt to reach dynamic equilibrium.  The large starch molecules would not be able to penetrate the membrane so nothing would enter the bag.

In this experiment, the Iodine solution diffused from a high to low concentration through a selectively permeable membrane, as expected.  In our specific experiment, there was too much water in the cup with the Iodine solution so not as much of our Iodine was able to diffuse through the membrane.  Also due to this, the color of the water/Iodine solution in the cup was more of a yellow/orange color than a red color, what it should have been.

1B: Osmosis

This experiment tested the relationship between solute concentration and the movement of water through a selectively permeable membrane by the process of osmosis. We were trying to find the net movement of water through a selectively permeable membrane.

Osmosis is the movement of water from a higher to lower water concentration through a selectively permeable membrane. Osmosis moves down the concentration gradient, which is when there is a high concentration of water in one area and it moves to a lower area of water concentration.

First we formed six bags out of dialysis tubing and filled them with 15-25 ML of distilled water, .2M sucrose, .4M sucrose, .6M sucrose, .8M sucrose, and 1.0M sucrose. Then we weighed each bag separately and recorded the weight. We submerged each bag in a separate cup of distilled water and let them sit for 30 minutes. After 30 minutes we took the bags out and re-massed them.

Potato and tools used to core the potato

Potato cores submerged in the different solution concentrations

Group Data:

Contents	Initial Mass (g)	Final Mass (g)	Mass Difference	Percent Change
0.0 M Water	29.8	30.04	0.24	0.8
0.2 M Sucrose	17.18	18.35	1.17	6.8
0.4 M Sucrose	29.6	30.66	1.06	3.6
0.6 M Sucrose	12.6	13.48	0.88	6.9
0.8 M Sucrose	24.1	25.4	1.3	5.4
1.0 M Sucrose	25.4	27.5	2.1	8.3

Class Data, Percent Change in Mass:

	Distilled Water	0.2 M	0.4M	0.6M	0.8M	1.0M
Group 1	-0.45	2.38	5	5.86	6.06	7.93
Group 2	5.15	8.47	8.54	10.43	9.76	17.44
Group 3	0	4.33	8.61	14.41	12	10.05
Group 4	-2.17	1.73	3.14	-6.9	7.2	11.54
Group 5	0.8	6.8	3.6	6.9	5.4	8.3
Group 6	0.89	2.5	3.1	5.3	6	7.4
Group 7	11.4	9.8	10.7	12.3	13.7	13.4
Group 8	2.7	3	4.9	6.2	4.3	6.1
Group 9
Group 10	0.63	2.14	4.85	8.19	8.6	-2.17
Group 11
Group 12	1.1	4.95	10.57	16.07	19.09	16.87
Group 13
Group 14
CLASS AVG:	2.01	4.61	6.30	7.88	9.21	9.69

Our data suggests that sucrose molarity determines whether water will move in our out of the cell.  The molecules want to move so that concentrations are equal inside and outside of the dialysis bag.  We can see that as the molarity of the solution increased, the mass also increased because the water was moving out and the sucrose was moving in.  Osmosis is present here.  In the group of bags that were in the cup of distilled water, the mass barely changed at all because there was water on either side of the membrane so they did not want to move across their concentration gradient.

Our data shows that water does move through a selectively permeable membrane through the process of osmosis.  Though the results are fairly consistent across the board, each bag did not weigh the exact same amount when we started so the percent change can also be somewhat attributed to that.

1C: Water Potential

In this experiment we tested water potential in potato cores placed in different molar concentrations of sucrose. We were trying to calculate how much water moves in and out of a potato cell

Water potential is the likeliness for water to be diffused from one place to another. Solute potential decreases water potential and pressure potential increases water potential that a cell can have. A cell that is lacking water has a higher water potential than a cell that has a abundance of water.

We used a potato core borer to cut 24 potato cylinders and removed and excess skin. We grouped them into fours and weighed them. After that we filled six cups with 2/3 full of distilled water, .2M sucrose, .4M sucrose, .6M sucrose, .8M sucrose, 1.0M sucrose and submerged the potato cores. We covered them with plastic wrap and let them sit overnight. Then the following day, we took the potato cores out and massed them. We then found the percent change in mass of the potato cores.

Group Data:

Contents	Initial Mass (g)	Final Mass (g)	Mass Difference (g)	Percent Change in Mass
0.0 M Water	13.7	15.2	1.5	10.9
0.2 M Sucrose	12.1	12.3	0.2	1.7
0.4 M Sucrose	10.7	9	-1.7	-15.9
0.6 M Sucrose	13.1	9.4	-3.7	-28.2
0.8 M Sucrose	7.3	5	-2.3	-31.5
1.0 M Sucrose	11.2	7	-4.2	-37.5

Class Data, Percent Change in Mass:

	DISTLLED WATER	0.2 M	0.4M	0.6M	0.8M	1.0M
Group 1	10	4.8	-12	-32.2	-36.4	-36
Group 2	12.99	1.65	-12.5	-29.69	-38.46	-34.65
Group 3	3.06	2.2	-14.47	-39.39	-75	-70.37
Group 4
Group 5	10.9	1.7	-15.9	-28.2	-31.5	-37.5
Group 6	11.59	0.22	-8.6	-6.2	-46.52	-55.31
Group 7	10.1	2.9	-12.5	-27.4	-40	-36.5
Group 8	9	0.9	-9.4	-27.1	-29.1	-34.3
Group 9	13.11	3.51	-12.28	-31.57	-36.84	-33.33
Group 10	17.57	-2.47	-8.93	-27.34	-32	-37.52
Group 11	2.94	-1.39	-13.2	-30.41	-34.85	-31.34
Group 12	9.24	-0.28	0.46	-.27.27	-35.26	-33.7
Group 13
Group 14
Class Avg.	10.05	1.25	-10.85	-27.95	-39.63	-40.05

Based on our data we can conclude that potato cores had a high water potential before we put them into the solutions, meaning that there was more water inside the potato than outside the potato.  If potatoes were left to dehydrate, they would have a lower water potential because water would be rushing into the cell since it moves from high to low concentration.  A high water potential means that water will flow out of the cell instead of in.  If the environment a cell is in has a high water potential, the water will enter the cell and as a result the cell will be hypotonic.  We can also conclude that potatoes contain sucrose molecules because when the cores were placed in distilled water, change in mass was positive meaning they took in water.

Our data shows that potatoes do have a high water potential before being placed in a sucrose solution.  Our results were consistent with everyone else's results, however, our potato cores weren't a uniform size, so the overall mass was affected.  Some cups had very short cores and other cups had larger cores which could account for a very large percent change.

1E: Onion Cell Plasmolysis

Plasmolysis is when cytoplasm of a plant cell separates from the cell wall which is caused by water loss. In other words it is when a plant gets dehydrated to a point when the insides begin to separate resulting in wilting of the plant and then eventually death. Plasmolysis typically will not occur in nature unless under harsh conditions. Plasmolysis works most efficiently when an a plant cell is emerged in a strongly concentration of saline or sugary solution resulting in water loss by osmosis.

Plasmolysis occurs in an onion cell because the large vacuole in the center of the cell contains a solution with lower osmotic pressure than the solution outside of the membrane. Due to this the vacuole loses water and reduces in size. The cell membrane and cell wall start to get further apart which causes the plasma membrane and protoplasm to move to the center of the cell.

Plasmolysis in a red onion cell


Sources:

Chapter 6&7 Powerpoint
http://www.biology-online.org/dictionary/Plasmolysis
Youtube

Milk Lab

The purpose of the experiment was to calculate how much protein is in a 15 mL sample non-fat (skim) milk. We were testing the amount of protein through denaturation by altering pH. Our independent variable was the acid being added to the milk and our dependent variable was the amount of protein in the milk after we added the acid. We were looking to see if the amount of protein we found in the sample of denatured milk was proportional to the amount of milk in the full carton.

Proteins have four different structural levels. The primary structure is the sequence of amino acids, the secondary structure are coils that result from hydrogen bonds between the backbone of the protein. The tertiary structure is a result of interactions between R groups and the quaternary structure is when two or more polypeptide chains form a macromolecule. Denaturing a protein affects the secondary, tertiary and quaternary structures by breaking the bonds needed to form them. Denaturation is when the native structure of a protein is lost due to a change in pH, salt concentration, temperature or other various environmental factors.

We weighed the milk before we began to denature it. Then we added acid and waited for the milk to denature and coagulate. Once we saw chunks in the milk we weighed a piece of filter paper and folded it into a funnel. From there, we put the milk into the funnel and separated the milk from the denatured protein. We calculated the percentage of protein in the milk after we massed the amount of protein we caught on our filter paper.

Mass of filter paper alone (g)		.9 g
Mass of paper and dry protein (g)		1.1 g
Mass of dry protein (g)		.2 g
Mass of empty beaker (g)		26.7 g
Mass of 15 mL of milk (g)		10.1 g

Based on the amount of protein in a 236 mL carton of milk (8 grams), in our 15 mL sample should have yielded .5 grams, which amounts to 5%, after we denatured the milk with the acetic acid. In our milk, we found .2 grams of protein, amounting to 2%. We calculated a -60% error after we denatured the milk. Our high error was most likely due to the fact that we got impatient with our milk and very aggressively stabbed a hole in the filter paper, releasing all of the milk and most of the protein that was still left. After this, we had a lot of milk in our beaker which more than likely had most of that other 3% of protein we should have gotten in the filter paper. In the future, the investigation could be improved by a different means of separating the protein from the milk. Many groups ran into problems when the chunks of protein in the milk got stuck in the hole at the bottom of the filter paper. Instead, those groups used a coffee filter to squeeze the milk out and from what we saw, that yielded more protein. We were  expecting to find less than 5% protein because we thought that the acid would eat away at the protein in the milk. We found less than 5% but for more obvious reasons than the acid eating the proteins.


In a 15 mL sample of milk there should have been .5 grams of protein. Our data does not support this claim, due to the fact that we lost some protein after we poked a large hole in the filter paper.

Fitzpatrick, Kathleen, and Erin Barley. "The Structure and Function of Large Biological Molecules." 23 Sept. 2014. Web.