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. 

Cell Respiration Lab

Purpose: The purpose of this experiment was to see the effect of temperature and germination on the rate of cellular respiration of peas. The independent variable was how the cold the peas and also how germinated they were. The dependent variable was the chwnge in the amount of oxygen and carbon dioxide.

Introduction:
In cellular respiration, a oxygen is used and carbon dioxide is created. During cellular respiration, carbohydrates and oxygen are converted into carbon dioxide, water, and energy.

Methods, Charts, and Graphs:

    We used Carbon Dioxide and Oxygen sensors to measure the amount of carbon dioxide and oxygen in the container.

An example of how the sensors looked like.

Then We used a control group of glass beads to measure CO2 and O2 in e container, then measured the levels of the gases for non-germinating seeds. Above of the graphs of the Oxyegn and Carbon Dioxide for non-germinated.

      Then we put germinating seeds in the container, and measured the levels of the gases for 10 minutes. 

These are the graphs for the Levels for germinating beans.

     After that, we soaked the beans in ice water, and measured the rate of respiration for ten minutes for the cold beans. 

    O2 and CO2 levels for soaked beans.

We then repeated the process with peas.

Discussion:

Throughout the experiment, one could see that the amount of Oxygen in the containers with peas was decreasing while the amount of Carbon Dioxide was increase. With this information we can see that cell respiration occurred in the peas, because according to the cell respiration equation plants take in oxygen and take out Carbon Dioxide. When the peas are germinated, the levels of Oxygen and Carbons Dioxide changed throughout the time period, but when the peas were not germinated the levels of Oxygen and Carbin Dioxide remained somewhat constant. When the germinated peas were put into cold water, the decrease of Oxygen and increase of Carbon Dioxide was more dramatic compared to when the germinated peas were tested at room temperature. It is necessary for these germinated peas to go through cell respiration because they need to make enough energy needed for their growth and development. These peas can no longer go through photosynthesis to make energy, because they no longer capture the sunlight needed for photosynthesis.

Conclusion: 

Throughout this experiment we were trying to figure out how the cellular respiration equation was correct in peas. After doing the experiment we figured out the result of cellular respiration by seeing that the plant takes in Oxygen and gives out Carbin Dioxide in germinated peas. 

Diffusion Lab

1A;

The purpose of this experiment is to test the permeability of glucose and starch through a selectively permeable membrane. The inside of the bag has a higher concentration of glucose, and because of diffusion some glucose will go outside of the bag, where there is a lower concentration.

Glucose is a monosaccharide, which means its particles are small enough to pass through a selectively permeable membrane without assistance. Because of this, diffusion of glucose will occur from areas of high concentration to low concentration. The solution's ultimate goal is to reach equilibrium, where the outside and inside of the membrane are at equal concentrations of solute. However, large particles like polysaccharides will be unable to pass through a selectively permeable membrane without facilitation. Starch is a polysaccharide of long chains of glucose.

We filled a bag of dialysis tubing with 15mL of a solution of 15% sucrose and 1% starch. This was put in a cup of water 4 ML of iodine. We let the bag sit the water for about 30 minutes before removing the bag. After letting it sit we tested for the presence of glucose in both the bag and cup.

This is what the diaylsis of the solution looked inside the cup of water and iodine.

This is the data table used to collect date in the experiment.

Before the bag went into the water, we tested the water for starch and glucose. It tested negative for both. Afterwards, the water stayed orange, which showed that no starch left the bag, while glucose was found to be present in the solution. This outcome supported what we know about diffusion because since starch didn't pass through the membrane into the water, it shows that it is a particle too large to pass through. The presence of glucose in the solution from the bag after 30 minutes shows that its structure is small enough to pass through a selectively permeable membrane. It could have been improved if we made sure that the bag was sealed as tightly as possible so that movement of particles was purely from diffusion and not seeping out of the top of the bag.

We concluded that glucose could pass through the membrane. Starch on the other hand is unable to do so and stays on the side of the membrane it started on.

1B:

The purpose of this experiment is to test the osmosis of water in different concentration gradients of sucrose, to see if sucrose can pass through the membrane and the osmosis of water at different concentrations of solute.The independent variable is the amount of sucrose in the bag at the start, and the defendant variable is the percent change of the bag after 30 minutes. 

In this experiment we know that the dialysis bag has a semi-permeable membrane, which means sucrose cannot go through the bag. The only thing that can go through the bag is water so when the weight of the bags change it shows how much water went and out of the bag. When water goes in and out if a membrane it is called osmosis.

We created 6 bags made out of dialysis tubing that were each about 30 cm each. After we cut them we each filled them with a substance of either distilled water, 0.2 M sucrose, 0.4M  sucrose, 0.6 M sucrose, 0.8 M Sucrose, and 1 M Sucrose. After filling the bags we measured the mass of each of them. After closing the bags with rubber bands, we placed all of them in their own cup full of water. We let the bags sit for 30 minutes and after taking the bags out of the water we massed the bags again.

This is a picture of the sucrose solution inside the cup of distelled water.

Here is the data table we used to find both masses for the dialysis bags before and after the experiment plus their mass difference. 

Here is a graph showing the change in mass difference with the dialysis bags.

The reason we calculating percent changes was because they can be easily compared to other tests without having all the initial masses be the same. After looking at the class data we can see that their is a direct relationship between the mass and molarity, because as the molarity of the solution went up the mass difference increased as well. f each of bags were placed in a 0.4 M Sucrose instead of distilled water, the mass differences would change because sucrose would not be able to pass through the semipermeable membrane of the bag. The mass difference for the 0.2M and 0M bags would increase since the solution would be hypotonic. The mass difference for the 0.4M would become zero, because the solution would be isotonic. The mass difference for the 0.6M, 0.8M, and 1M would decrease since the solution was hypertonic.  

1C:


We first filled up 6 different cups with each different solutions (0M, 0.2M, 0.4M, 0.6M, 0.8M,1M). Then by using a cork borer we cut out 4 potato cores for each cup of solution (so 24 potato cylinder in total). Before putting the cores in the solutions we first took the mass of the four cores together. After letting the potato cores soak overnight in the solutions, we dried off the potato cores and then weighed them together.

Here are the potato cores in the 0M, 0.2M, and 0.4 solutions. Notice how they are on the bottom of the cup.

Here are the potato cores in the 0.6M, 0.8M, and 1M solutions. Notice that the potato cores are all floating.

Here is the data from when we massed the potato cores before putting them in the solutions and the mass after they were soaked in the solution overnight. 

Here is the graph representation of the decrease or increase in percent in mass of the potato cores.
1E:


Plasmolysis occurs when water rushes out in a hypertonic solution around a plant cell. As water rushes out of the plant cell the cell shrivels. The cell wall separates from the membrane and the cytoplasm even begins to shrink inside. In the onion cell, this process if plasmolysis occurred because far more water was outside than in the cell. As a plant cell, the rigid cell walls like to be in a hypotonic solution to bear pressure, but on the opposite spectrum of that causes flaccid es and limpness. If water continues to come out of the cell, plasmolysis will occur. In the onion cell, this process if plasmolysis occurred because far more water was outside than in the cell. As a plant cell, the rigid cell walls like to be in a hypotonic solution to bear pressure, but on the opposite spectrum of that causes flaccid es and limpness. If water continues to come out of the cell, plasmolysis will occur.

Here is an example of plasomysis in an onion cell.

http://www.csun.edu/~aef21890/coursework/695/microscopy/microscopy.htm

Milk Lab

     Greetings, Ladies and Gentlemen. To kick off the first lab of All About That Base, No Acid,  we worked with something you might have heard of, something called Milk. We used acetic acid to curdle milk and weighed the resulting precipitate to measure protein content of Skim Milk. The purpose of this was to discover how proteins can be denatured, and changed from their usual state, and also to see how some proteins are not affected by acid.

To start the experiment by we added 15 mL of milk to a 50 mL beaker. After massing both the empty (30.3grams) and filled beaker(45.6 grams), we added 30 drops of concentrated acetic acid to the milk and then waited 5 minutes to let the new mixture settle. Next, we created a funnel using filter paper and we put the settled milk through the filter to catch the developing milk curds. One in awhile we would have to stir/ scrape out things in order to make sure all the liquid went through. After we decided most of the milk and milk curds were filtered out, we set the filter in a box to sit overnight. Before leaving the classroom we poured 1 mL of water into a test tube and 1 mL of the filtered milk into another test tube. Then we added Biuret Reagent to see if there were proteins left in the certain mixture and we had water just as a control. It turned out some of the proteins ended up filtering into the milk instead of staying in the filter paper. 

Above is the milk filtering out the curds in the filter paper.

Above this is how we we stirring/scraping out some of the curds in the filter paper.

Above is us comparing water and filtered mileafter adding the Biuret Reagent. The water is on the right while the filtered milk is on the left. Since the filtered milk turned purple it means there were still proteins in the solution.

After returning to school we weighed the dried milk curds on the filter paper and it weighed 1.3 grams. 

To determine the amount of proteins we collected we subtracted the dried milk curds and the filter paper minus the weight of just the filter paper (1.3-.7) and we got a total of .6 grams of protein. Since we had 15.3 grams of regular milk to begin with we figured out that 4% of the milk was composed of proteins.after doing some other calculations we figured out that the milk was only supposed to have .509 grams of proteins and somehow we collected more than just proteins in our experiment. The expected value was .091 grams lower than the value we got, showing that we filtered out more particles from the milk than just protein. This result could have happened because the filter also filtered out non protein particles, meaning that not just protein was denatured. It could also mean that particles from our environment might have transferred to our paper plate. In the future, this might be prevented if we separated the plates from different groups more as they were laid to dry so that other people's filtered out substance didn't get on ours. These results support the expected outcome because we expected the acetic acid to denature the milk proteins, thereby causes the denatured proteins to be filtered out of the substance with a high success rate.

The purpose of this lab was to see how proteins can be denatured, and this lab illustrated the idea that pH can dramatically effect a substance by causing the protein particles to clump together.