Mitosis Video

http://youtu.be/Zm1UneZuQi8

Photosynthesis Lab

Purpose:

The purpose of this experiment was to test and see if both light and chloroplasts were needed in order for photosynthesis to occur. The Independent variable was how long the the solution had been placed in a light or dark environment with boiled or unboiled chloroplasts. The dependent variable was the amount of transmitance (or light let through) the solution had. 

Introduction:

Before testing with chloroplasts, we used solutions with dye and water to determine absorb acne and transmitance rates using a colorimeter. To make different solutions we started off by putting 2.5 mL of dye solution and 2.5 mL of water in one test tube and then testing it with the colorimeter. Next we would take 2.5 of that original solution and adding 2.5 mL of water, which would make an ever more diluted solution. We did this many more times to insure the solution was very diluted. After looking at the different results of the data using the colorimeter, we found that as the more diluted the solution is the lower the absorbance rate and the higher the transmitance rate. With this in mind, one can see that the easier light can go through the solution, the higher transmitance rate.

Putting the different solutions in the cuvette in the colorimeter to test the absorbance and transmitance of different dilutions.

This is data from the colorimeter say the percent of the orginal solution in the tested solution, the transmitance, and the absorbance.

Procedures, Graphs, and Charts:

     In order to find out how leaf pigments absorbed light, we mixed solutions is boiled (dead) and unboiled (alive) chloroplast along with other solutions (such as water, a phosphate buffer, and DPIP(increases light transmitance). We again used a colorimeter to find out if light and chloroplasts are both needed for photosynthesis. After mixing the different solutions we set them in front of a light and a beaker of water, so that we could see how the chloroplasts absorbed the light. One cuvette(the one wrapped in tin foil) was supposed to be not absorb light since it is the control.

Here is a picture of how we put each cuvette in front of the light.

We first tested each of the cuvettes in the colorimeter, to test transmitance and absorbance, before we put them in front of the light and then again after sitting in the light for 5, 10, and 15 minutes.

Here is the data taken from the colorimeter based upon what type of chloroplast it had, if it was in the light or dark, and how long the cuvette was in the light.

 Results: After looking at the different results of the data using the colorimeter, we found that as the more diluted the solution is the lower the absorbance rate and the higher the transmitance rate. With this in mind, one can see that the easier light can go through the solution, the higher transmitance rate. Because DPIP turns from blue to clear when it is reduced by chloroplasts, the longer and therefore more pale the solution was, the higher the transmitance was. Since the transmitance was higher, more light could get through the solution and the light reactions of photosynthesis could occur from excited electrons from light. This is because the clearer liquid has no pigments for the light to be absorbed in, it can pass through. In order to make sure that the DPIP had a role in this, we made a control group with no DPIP, and photosynthesis did not occur. In general, we had trouble controlling data and storing it in the lab equipment, so some of the data points were lost. For the data points we were able to recover, the the direct relationship between time and percent transmitance is present as we hypthesized. Also, the unboiled chloroplasts were unable to photosynthesize in our exoeriment, which can be explained by the denaturing of crucial proteins needed to in photosynthesis. This experiment could have been more accurate if we had more data to back up our results.

Conclusion: After looking at the results, one can tell that the amount of light transmitted and the presence of chloroplasts both have crucial roles in the photosynthesis. 

Enzyme Lab

Purpose: The purpose of this experiment was to find how much of the hydrogen peroxide decomposes depending how how long the enzyme was able to react with the catalase extract. In shorter terms, we trying to figure out how much product an enzyme could make during a certain time period. 

Introduction: Enzymes, which are made out if proteins, speed up chemical reactions inside a cell; another word to describe an enzyme is a cataylst. When a substrate is added to this enzyme-catalyzed reaction, the energy needed for the reaction to occur is reduced. At any point in a reaction an enzyme-catalyzed reaction could stop if there is a change in pH, temperature, or salt concentration. Sometimes the enzyme-catalyzed reactions can be completed changed due to activators or inhibitors which can change the shape of the enzyme or block certain substrates of going into the active site. 

Methods, Charts, and Graphs: in order to have a baseline of how much unreacted hydrogen peroxide is in 5 mL of  1.5% Hydrogen peroxide, we put 5 ml of unreacted hydrogen peroxide in a cup and titrated it with KMnO4. For the baseline, 3.6 mL of KMnO4 were used. To find how much Hydrogen peroxide is catalyzed over different periods of time, we added yeast, which contains catalase, into the hydrogen peroxide, and let it react for specific times. After he time was up, we added sulfuric acid to denature the enzyme, stopping the reaction. We then took 5 mL of the catalyzed hydrogen peroxide, and titrated it with KMnO4 to see how much hydrogen peroxide was left. We subtracted that number from the baseline to find how much hydrogen peroxide was broken down by the catalase.

Here is the data table that shows how much KMnO2 is consume and H2O2 is used.

Here is a graph showing the amount of H2O2 is used up over time.

Discussion Questions:

     The rate of reaction is highest for the longest time period, 360 seconds, because the catalase has had more time to break apart the H2O2. The lowest rate was the lowest time period, because it hadn't had as much time to react. We stopped the reaction with H2SO4, an acid that denatured the catalase, and as a result, it could no longer break apart the H2O2. Lowering the temperature would also inhibit the reaction, as temperature changes can also denature an enzyme. Enzymes have an optimal pH and temperature, and under these conditions the rate of reaction is highest. An experiment to determine this would be to measure the rate of reaction with the time and temperature held constant, at different pH, and again with time and pH held constant, at different temperatures.

This shows how the pH of an enzyme relates to how efficient proteins can function.