Catalase Lab

In this lab, I studied the effects of the concentration of catalase on enzymatic activity. I found that a greater concentration of enzyme leads to a greater rate of reaction. The lab measured the rate of the formation of oxygen via the breakdown of hydrogen peroxide. The catalase reaction is very important in cells because it stops the accumulation of hydrogen peroxide, which is lethal to the cell. I hypothesized that as the concentration of the enzyme increased, so would the quantity of the products.
Enzymes such as catalase are catalysts that execute thousands of chemical reactions that occur in living cells. In an enzyme catalyzed reaction, the substrate (the substance to be acted upon), binds to the active site of the enzyme. The enzyme then changes the substrate into the reaction products. The product is then released into solution and the unchanged enzyme is ready to make another enzyme-substrate compound. Both the enzyme concentration and the substrate concentration can affect the reaction rate. The presence of more molecules increases the probability of collisions and enzyme substrate coupling. In this lab, the enzyme catalase accelerated the breakdown of hydrogen peroxide, which is summarized in this reaction.
Figure 1

This reaction is critical in living cells, because it stops the accumulation of hydrogen peroxide, which is lethal to the cell. Hydrogen peroxide is a harmful byproduct of many normal metabolic reactions. In order to prevent damage to the cell, it must quickly be changed into less harmful substances. Therefore, the presence of catalase is important because it breaks down a potential toxin.
Catalase is also used in the production of cheese, as hydrogen peroxide is a byproduct of the cheese-making process, and it must be removed so that the cheese is not spoiled. In addition, some studies have shown that the graying of human hair results from a buildup in hydrogen peroxide due to decreasing amounts of catalase in the aging body.

Materials used in the experiment:
-50 mL beaker
-10 mL and 50 mL graduated cylinders
-Fresh 3% Hydrogen Peroxide Solution
-Forceps
-Water pan
-Catalase solution
-Filter Paper, cut into round dots with whole puncher
-Reaction chambers (Drosophila vials with 1- hole stoppers)
-Stopwatch
-Ice

Setting up the experiment:
1. I put a small amount of the catalase solution into the 50-ml beaker
2. I placed four catalase-soaked filter paper disks high on one interior sidewall of the reaction chamber. (They will stick to the sidewall). To prepare a disk for use in the reaction chamber, I held it by its edge with a pair of forceps and dipped it into the catalase solution for a few seconds. I thin drained the disk against the sidewall of the beaker transferring it to the reaction chamber.

Preparing the reaction chamber:
1. I placed the reaction chamber upright and added 10 mL of 3% hydrogen peroxide solution, without allowing it to touch the filter paper disks.
2. I put a stopper in the chamber so that it fit tightly.
3. I filled a pan (almost full) with water.

Conducting the experiment:
1. I laid the 50-mL graduated cylinder on its side in the pan so that it filled completely with water. As an air bubble was present, I carefully worked it out by gently shaking it.
2. I turned the cylinder upside down into an upright position while keeping it filled with water.
3. I carefully placed the reaction chamber and contents on its side in the pan of water, making certain that the side with the disks faced upward.
4. I moved the graduated cylinder into a position so that its mouth came to lie directly over the tip of the stopper. I held it in this position for the duration of the experiment.
5. I rotated the reaction chamber 180 degrees on its side so that the hydrogen peroxide solution came into contact with the catalase-soaked disks.

Recording data:
1. I measured gas levels in the graduated cylinder at 30-second intervals during a 5 minute period.

Figure 2

Figure 2 illustrates the production of gas bubbles in the test tube, which rose to form a pocket of gas in the tube.

Figure 3

Figure 3 illustrates my table of results for the experiment. From this data, I concluded that when a higher concentration of enzyme is present, the rate of reaction is faster. A notable exception is seen in the discrepancy in the 3 dot category. The one dot solution increased by an average of 0.31 mL per 30 second interval, the two dot solution increased by 0.9 mL every 30 seconds, the three dot solution increased by 0.35 mL every 30 seconds, and the four dot solution increased by 1.7 mL every 30 seconds.

Figure 4

Figure 4 depicts a line graph of the trends of the solutions during the five minute experiment. In general, as noted above, when a higher concentration of enzyme is present, more of the product (oxygen) is produced.

Discussion Questions:
-(Graph Above) Although the amount of oxygen produced eventually slowed down, the action of the catalase remained the same. As catalase is both a reactant and product in the equation (as noted above), the amount of catalase available is constant, and it continued to react with the substrate (hydrogen peroxide) until the substrate was depleted.
-(Table above) Based on the data, I concluded that as the concentration of enzyme increases, so does the quantity of products.
-For optimum enzymatic activity, the surrounding conditions must be set at an appropriate temperature, pH, substrate concentration, and enzyme concentration. Enzymes function well only a specific environment, and within a narrow range of conditions that may vary with the structure of the enzyme. For example, catalase, the enzyme used in this experiment, worked well at room temperature (70 degrees), in water (pH 7), and in a solution in which the available substrate exceeded the active sites on the enzyme.
In this experiment, I studied the effects of the quantity of catalase on enzymatic activity. I proved that my hypothesis was correct, as my data reveals that when a greater concentration of an enzyme is present, a larger amount of products are produced. This is because the greater quantity of enzyme provided more active sites with which the substrate could react, which resulted in a higher probability of a successful reaction. This can be visualized in the following figures:
Figure 5

Figure 6

As illustrated in Figure 4, the four dot solution produced the most oxygen (17 mL after five minutes), followed by the two dot solution (9 mL after 5 minutes), the three dot solution (3.5 mL after 5 minutes), and the one dot solution (3.1 mL after 5 minutes).
Of note, there was an unanticipated result in the three dot solution, as less oxygen was produced than was expected. The amount of oxygen visible during the reaction produced in that reaction remained constant; however, some of the gas was inadvertently not captured in the test tube due to human error. In addition, one of the dots in that solution was not completely soaked in catalase, which likely affected the results by reducing the amount of enzyme available to facilitate the breakdown of hydrogen peroxide.
If I were to repeat the experiment, I would position the test tube so that all of the oxygen is collected. I would also vary the ambient temperature, pH, and salinity in order to determine the effect of changing those variables on the activity of catalase. I would also determine the effect of a competitive inhibitor of catalase on the rate of decomposition of hydrogen peroxide. An interesting potential application of catalase is to restore natural hair pigments, as white hair may result from an increase in hydrogen peroxide in aging hair follicles, and exogenous catalase might reverse pigment degradation. Catalase may also have a role in cancer treatment, as tumor cells could be targeted with a chemical inhibitor of catalase, which might allow naturally produced hydrogen peroxide to preferentially destroy those cells.

http://www.sciencedaily.com/releases/2009/02/090223131123.htm
http://www.princeton.edu/~achaney/tvme/wiki100k/docs/Catalase.html

Interstellar

Last week, my biology honors class and I went to an early showing of IMAX Insterstellar, a film about Cooper, an ex-member of the decommissioned NASA, working with some of his old colleagues to find a new home for people as the Earth becomes uninhabitable.
In the film, the crew of the spacecraft use a wormhole near Saturn to reach a distant galaxy in search for members of an earlier expedition. The wormhole is thought to have two points set at different places and time in the galaxy. Although we have no visual evidence for their existence, the theories of relatively as discussed in the film, suggest they do exist. Even though time travel is considered to be science fiction fantasy popularized by Captain Kirk and Picard, humanity may achieve the ability to travel into the dimension of time.

Time travel revolves around relatively. If wormholes do exist this would point to the possibility of time travel. By accelerating one end of a wormhole to a velocity higher than the other, it would be possible to travel through time, as long as one brought it back to normal velocity during the journey. However, this theory may not be possible due to unknown quantum effects. Furthermore, for humans to travel through time, we would need to develop a way to travel at faster speeds in space. At current speeds, it takes about 300 days to reach Mars, and 80000 to reach Proxima Centauri, our nearest star system. Finding these wormholes could require traveling far beyond Promixa Centauri.
As discussed after the film, Earth is eventually headed into the direction possible destruction. Although this will not occur for thousands of years, the emission of greenhouse gases, the abundant use of Earth’s natural resources, and overpopulation could contribute to such a world. If humanity can change our actions, and become more Eco-friendly, a change in the outcome of our future could arise.
In summary, the film Interstellar raises the possibility of the existence of wormholes and the capability and possible need for time travel.

http://www.nasa.gov/centers/glenn/technologiy/warp-scales_prt.htm

An Egg-Citing Egg-Speriment By William H. Vince F. and Nolan D.

In this lab, we studied the osmosis of water through the semi-permeable membrane of an egg. Osmosis occurs when a solvent moves from an area of high concentration to one of low concentration via a semi-permeable membrane. To illustrate the process of osmosis, we dissolved the eggshell via an acid-base reaction by combining calcium carbonate (a base) and acetic acid, and we then placed the denuded egg in water. The amount of water that passed through the semipermeable membrane was determined by weighing the egg at five minute intervals. We concluded that the egg membrane is semi-permeable, and we determined that the weight of the egg increased by approximately 1.6% of its original mass every five minutes.
In this experiment, we observed the process of osmosis through the semi-permeable membranes of a chicken egg. The semi-permeable membrane is located under the hard calcium carbonate shell, as illustrated in Figure 1.
Figure 1

The outer shell of a chicken egg is composed of 94-97 percent CaCO3. We first dissolved the outer shell by means of an acid-base reaction with vinegar, which contains acetic acid. When calcium carbonate mixes with acetic acid (CH3CO2H) the following chemical reaction occurs: Initially, the carbonate (CO3–) part of calcium carbonate is protonated by acetic acid to make carbonic acid (H2CO3),and the calcium and acetate form calcium acetate.
2 CH3COOH + CaCO3 = H2CO3 + Ca(CH3COO)2.
Next, the carbonic acid breaks down to form carbon dioxide and water.
H2CO3 = H2O + CO2.
The overall reaction can be written as the sum of two reactions:
2 CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2.

After the egg shells are dissolved, the egg’s outermost layer is composed of two semi-permeable membranes. These membranes play a vital role in the development of a chick by allowing air and moisture to pass through to the developing embryo. The membranes are partially made of keratin, and they provide the embryo with a defense against bacterial invasion.

The de-shelled eggs were placed in water in order to simulate the behavior of a cell, which is also surrounded by a semi-permeable membrane, when placed into a hypotonic solution. We predicted that water would flow by osmosis through the semi-permeable membranes down its concentration gradient. Since water is a relatively hypotonic solution, which means that it has a lower concentration of solute when compared to the interior of the egg, water flows from an area of low solute concentration to an area of high solute concentration: into the egg. The absorption of water by the egg can be measured by weighing the egg and observing an increase in circumference.

In this lab, we used the following materials:
-3 eggs of relatively the same size
-white vinegar (apple cider or rice wine vinegar will work too)
-large beaker
-three large plastic cups
-triple beam balance scale
-plastic tray
-300ml of distilled water
-timer

Preparing the eggs:
1.I placed the eggs in the beaker gently without cracking the shell. I covered the eggs in the vinegar and let them sit overnight.
2. After the eggs were soaked, I removed them from the vinegar solution, and I gently rubbed off the white calcium acetate from the egg.

Starting the experiment:
1. I filled a cup with 100 ml of distilled water.
2. I marked each egg with a number (1-3), and I weighed egg 1with the triple beam balance, after placing the egg in the plastic tray.
3. I added egg 1 to the cup.
4. I let the egg sit in the water for five minutes, and I then weighed it again.
5. I repeated steps 1-4 using eggs 2 and 3.

Measuring the rate of osmosis:
1. I weighed egg 1 in the plastic tray after five minutes in the water bath, and I returned it to the bath as soon as possible.
2. After a second five minute interval, I re-weighed the egg 1 and returned it to the water for a third five minute period.
3. I removed egg 1 and weighed it a final time.
4. I repeated steps 1-3 for eggs 2 and 3.
5. I recorded my data in the table.
Note: When recording data, I accounted for the weight of the plastic tray.

Figure 2

Figure 2 illustrates the egg’s CaCO3 shell being dissolved by the surrounding vinegar. Carbon dioxide bubbles surround the egg, because CO2 is a product of this reaction as described previously.
Figure 3

Figure 3 depicts the egg after the shell has disintegrated. A layer of calcium acetate, another product of the reaction, is visible on the egg. After the shell dissolved, the egg became rubbery to the touch.
Figure 4

Figure 5

Figures 4 and 5 depict the eggs being weighed on the triple beam balance. I found that the weight of egg 1 increased from 71.5 g to 74 g after five minutes, it remained at 74 g after ten minutes, and it increased to 75 g after 15 minutes. The weight of egg 2 increased from 76.5 g to 78 g, 79g, and 80 g after five, ten and fifteen minute intervals respectively. Finally, the weight of egg 3 increased from 86.8 g to 88 g, 89.3 g, and 91g after sequential five minute immersions.
Figure 6

The graph in Figure 6 illustrates the increase in weight of each egg (expressed as percent weight gain) during the time they were immersed in distilled water.
Figure 7

The graph in Figure 7 depicts the increase in actual weight of the eggs during the time they were immersed in distilled water. It is apparent that the weight of the eggs increased with a longer duration of immersion in water.
Figure 8

Figure 8 depicts the actual weight gain and percent increase in the weight of each egg during the experiment. As discussed previously, the weight of each egg progressively increases during the experiment, as the eggs absorb water by osmosis through the semi-permeable membranes.
Figure 9

Figure 9 depicts an egg that ruptured after the experiment was complete. At the top of the photo, the gray semipermeable membrane lies next to the egg yolk. We noted that the thickness of the egg membrane is quite small in comparison to the size of an intact egg.

Discussion Questions:
*When the eggs were placed in the vinegar, they swelled because the reaction of vinegar and the eggshell (2 CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2.), results in water, which moved into the egg by osmosis.
*In the first part of the experiment, water moved into the egg. We know that water has moved into the egg, because they gained weight during the immersion in distilled water.
*Chicken egg membranes are semi-permeable because water was able to move into the egg by osmosis but no egg albumin (which is a large molecule) was able to diffuse through the membrane and out of the egg.
*It is possible to calculate an egg’s volume by measuring the amount of water displaced by the egg. If a beaker is filled with water and an egg is submerged in the water, the amount of displaced water can be measured and is equal to the volume of the egg. A change in weight indicates that more water has moved into the egg, and this would likely be accompanied by a change in volume. This is because the egg expands as water passes into it.
*Another possible experiment is to place the de-shelled egg in solutions of different tonicity (isotonic, hypertonic, and hypotonic) in order to determine the effect of those solutions on the water content of the egg. I anticipate that the egg would shrink when placed into a hypertonic solution, and an isotonic solution would have no effect of the egg. It would also be interesting to study active transport of molecules across an egg membrane.
*Saline solution is used to expand blood volume, as it is isotonic. Hypotonic distilled water enters red blood cells via osmosis and causes them to burst open (see Figure 10).
Figure 10

In this experiment, we observed an acid base reaction, osmosis through a semipermeable membrane and the effect of a hypotonic solution on the weight of a chicken egg.
The first part of the experiment involved a chemical reaction between a base (CaCO3), which is found in egg shells, and an acid (acetic acid), which is found in vinegar.
The chemical reaction is:

2 CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2.

We observed carbon dioxide bubbles, the dissolution of the hard calcium carbonate shell, and the formation of calcium acetate. We also noted that the egg became rubbery after immersion in vinegar. This may have occurred because the acid started to denature the albumin found in egg whites. Denaturation of proteins can occur through exposure to acids, bases or high temperature.
The next part of our experiment was to observe the osmosis of water through the semipermeable membranes of the egg, which lie under the hard shell. These membranes are important in the development of the embryo because they are permeable to water and air, and they provide a barrier to dust and bacteria. After we soaked the egg in water (a hypotonic solution), the egg began to enlarge. We weighed the egg at five minute intervals, and we noted an average increase in weight of 2.2% over the first five minutes, then an average increase of 0.8% over the next five minutes, then an average increase of 1.2% during the final five minutes (see Figure 7). The increase may have slowed as the experiment progressed because the concentration gradient decreased over time.
Next, we observed the effect of the tonicity of the solution on the egg as a model for the reaction of a cell to its environment (or a bathing solution). The egg swelled because the hypotonic solution resulted in an osmotic gradient that caused water to flow from a low concentration of solute to a high concentration. In this case, the high solute concentration was provided by egg proteins.
If a cell is in an isotonic environment (the external osmolarity is equal to the internal osmolarity), the cell volume remains unchanged. If a cell is in a hypertonic environment (the external osmolarity is higher than the internal osmolarity), the cell will shrink. Finally, if a cell is in a hypotonic environment (the external osmolarity is less than the internal osmolarity), the cell will swell.
The relative osmolarity of a solution is determined by the number of solute particles, and not by type of particles. An increase in temperature will speed up the motion of molecules and will also increase the rate of diffusion or osmosis. Although diffusion occurs in both directions, the overall net movement will be down the concentration gradient.
If we were to repeat this experiment, I would use solutions with different osmolarity, and I would vary the temperature to see if the rate of osmosis would be affected. Finally, I would study the egg membrane’s ability to let air and water flow osmotically while preventing the movement of bacteria and large molecules. It is possible that eggs could be analyzed for an intact shell and cell membranes to decrease the possibility of eating eggs that are contaminated with salmonella and other disease causing organisms. If one could develop an inexpensive semi-permeable membrane, it might be useful in the desalination of water for use in developing countries.

https://www.exploratorium.edu/cooking/eggs/eggcomposition.html
World’s Poultry Science Journal/Volume 61/Issue 1/March 2005, pp 71-86
Van.physics.illinois.edu/qa/listing.php?id=461
imaginationstationtoledo.org/content/2011/04/how-to-make-a-naked-egg

Diffusion via Dialysis

In this lab, we studied the diffusion of substances through a semi-permeable membrane. We found that smaller molecules like iodine and glucose were able to pass through the dialysis tubing, while larger starch molecules did not diffuse through the membrane. The pore size determined which molecules passed through the membrane. A similar phenomenon is seen in cell membranes, as the phospholipid bilayer also acts as a semi-permeable membrane that limits the passage of certain molecules into and out of the cell.

This experiment allowed me to visualize the diffusion of iodine, glucose, and starch through dialysis tubing. Diffusion is the process by which material moves from regions of higher concentration to areas of lower concentration. Diffusion is important to living cells, as it is one mechanism for moving substances into or out of the cell through the phospholipid bilayer. Diffusion is also used by physicians to treat kidney failure with dialysis.

We used Lugol’s iodine in the experiment for three reasons: it is colored, so it is visible to the naked eye; its molecules are small in size, which allows them to pass through the membrane; and molecular iodine is not soluble in water, but it forms a triiodide molecule (and becomes water soluble) when potassium iodide is added.

Formation of a triiodide:

We used glucose because it is a small molecule and it is easy to measure it in solution with a glucose strip. In addition, we used starch because it is a large molecule and reacts with Lugol’s iodine to cause a color change.

We used the following materials to demonstrate diffusion through a dialysis membrane:
-2 250ml beakers
-2 15cm pieces of dialysis tubing
-Glucose test strips
-Solution of Lugol’s iodine in a dropper bottle
-15ml glucose solution
-15 ml soluble starch solution
-distilled water

Preparation of the starch beaker:
1. I poured enough distilled water (about 125ml) to fill half of the beaker.
2. I tied off one end of a piece of dialysis tubing and added the starch solution.
3. I tied off the other end of the tubing and rinsed the outside with water.
4. I placed the tubing into the beaker of distilled water.
5. I added enough Lugol’s iodine to change the color of the water (20 drops).
6. I waited 20 minutes before recording my results.

Preparation of the glucose beaker:
1. I poured enough distilled water (about 125ml) to fill half of the beaker.
2. I tied off one end of a piece of dialysis tubing and added the glucose solution.
3. I tied off the other end and rinsed the tubing as discussed above.
4. I placed the tubing into the beaker of distilled water.
5. I recorded the results 20 minutes later.

Testing the solutions:
1. In the presence of starch, the Lugol’s iodine will change the solution into a deep blue or black color.
2. In the presence of glucose, a glucose test strip will change color when it is placed into the solution.

Diffusion of Lugol’s iodine and starch:
Figure 1

Figure 1 illustrates the assessment of starch and iodine diffusion through a semi-permeable membrane.

Figure 2

Figure 2 illustrates the addition of iodine to the beaker, which made the distilled water turn a light shade of yellow. The starch solution is inside the dialysis tubing.

Figure 3

Figure 3 illustrates the iodine/starch experiment during monitoring to detect a color change.

Figure 4

Figure 4 illustrates the iodine-starch beaker after twenty minutes. The iodine has diffused into the dialysis tubing and reacted with the starch, which has changed the solution color to blue. It is apparent that the starch did not diffuse through the semi-permeable dialysis tubing, as the color of the solution outside the tubing remains yellow.

Diffusion of glucose:
Figure 5

Figure 5 illustrates the assessment of glucose diffusion through a semi-permeable membrane.

Figure 6

Figure 6 illustrates the glucose diffusion experiment during monitoring for twenty minutes.

Figure 7

Figure 7 illustrates Mr. Wong testing the glucose solution with test strips.

Figure 8

Figure 8 depicts the test strip for our beaker on the right with a distilled water control on the left. The test strip turned yellow, consistent with the presence of glucose, which showed that it diffused through the semi-permeable membrane of the tubing.

In this experiment, I explored the effect of molecule size on diffusion using a selectively permeable membrane. This process is comparable to diffusion through cell membranes, which regulates the substances that enter and exit the cell. Diffusion is caused by the random movement of molecules, and results in substances moving from regions of higher concentration to those of lower concentration.

Lugol’s iodine, which is molecular iodine combined with potassium iodide, was used in the experiment. Molecular iodine is insoluble in water, but mixing it with potassium iodide forms I3-, which is soluble in water. The chemical equation for this reaction is: I2 + I- ——> I3-

*Initially, enough Lugol’s iodine was added to the water in beaker A to give the water a yellow hue. The triiodide diffused through the semipermeable membrane and turned the soluble starch solution blue. The color change occurred because the triiodide slipped into the beta amylose coils of the starch (see Figure 9). Perhaps if the experiment were continued for a longer duration, this binding would remove the iodine ions from solution and make the solution outside the beaker lighter.
Figure 9

*The chemical test for glucose demonstrated that the glucose molecules in the dialysis tubing of beaker B diffused through the semipermeable membrane. This was confirmed by testing distilled water as a negative control, which did not change the test strip. In contrast, glucose solution was used as a positive control and changed the test strip color.

*Although we did not let the beaker sit overnight, and the only substance that we showed to diffuse is glucose, I hypothesize that water would pass through the membrane by the process of osmosis. Osmosis is the movement of water through a semipermeable membrane to form an equal water molecule concentration on both sides of the membrane.

* The starch did not pass through the membrane. This was shown when the solution inside the tubing turned blue after iodine was added due to the reaction between triiodide and starch. The solution outside the tubing remained the yellow color of the iodine, which indicated that the starch did not diffuse outside the tube. This is because the large size of the starch molecules did not permit diffusion through the dialysis membrane.

* I hypothesize that the pore diameter of the dialysis membrane is between the size of a glucose molecule and a starch molecule. This is because the glucose, iodine, and water all passed through the membrane, but the starch remained inside the tubing.

* I assume that the pore diameter of the tubing is between the size of the starch and glucose molecules. Furthermore, the pore diameter did not change, and the molecules diffused based only on size, and not because of polarity or charge.

If I were to repeat this experiment, I would make sure to clearly mark the beakers. My group accidentally placed iodine in both solutions, but we luckily caught our mistake and corrected it promptly. In addition, I would try different dialysis tubing with varied pore diameter to determine the relative size of glucose, starch, and iodine molecules. Furthermore, I would like to test other substances, including lipids and proteins, in place of starch. Finally, I would like to create an experimental membrane with an active transport mechanism in order to visualize the process of active transport, in contrast to diffusion and osmosis.

Chemwiki.ucdavis.edu