Friday, December 20, 2013

Cellular Communication Lab

Purpose: The purpose of this lab was to determine the effects of varied amounts of time on each type of yeast cell. The different types of yeast cells were A type, alpha type, and mixed. The independent variable was the amount of time the cells were left, and the dependent variable was the number of cells left at the end.

Introduction: Yeast cells use cellular communication to mate and form shmoos. Cellular communication is a method that cells use to talk to each other and mate. Specifically in yeast cells, the A type cells send out a signaling molecule specific to the receptor of the alpha type, and this forms a diploid zygote, which then turns into a diploid budding zygote and forms an ascus. An ascus is a mixture of alpha and A type cells. These types of cells  are only seen when a and alpha type yeast cells are mixed together. When on their own, we only see single cells and budding cells.
Mixed a and alpha cells are shown here. This we can see because there are many more cells viewable through the microscopes.

This shows the same mixture, yet fewer yeast cells can be seen. 

This is alpha type at 48 hours.

Methods:
To begin the experiment, we got culture tubes and labeled them alpha-type, a-type, and mixed. We had 4 mL of sterile water in our culture tube, where then we put a small amount of yeast into each tube and mixed it around thoroughly to make sure all then yeast got into the water.  We were not allowed to use the same pick to get the yeast out for the alpha and a type, but for the mixed we could use either one. Then we went back to our lab table and used a piper to put 5 drops of yeast onto the designated slides. Once again, we couldn't use the same piper for alpha and a type. We put the coverslip over the drops of yeast that was in the slide next. We had to record the yeast at 0 time, 30 minutes, 24 hours and then 48 hours. Once we took out our microscopes and had them all set up, we put one of the slides and focused the yeast using the 10x lens, then once we saw the yeast cells, we changed the lens to objective 40x. Next, we took pictures through the microscope lens so then its would be easier to count all of the cells. We repeated these steps 3 other times per each type of yeast cell. 

Data:
First, we have the results for the mixture of a and alpha yeast cells. 
Then, we have the data for each type individually.


Graphs:

Below is the graph for alpha type yeast over time. The two lines depict single haploid cells and budding haploid cells, as per the key above. 
Then, we have the graph for a-type yeast. 
Below is the key for the mixed yeast culture. 



Discussion:
 We had two types of yeast cells, a and alpha, that were pre-labelled for our lab. If we did not know which was which, but we had one known sample, we could determine each type. For example, say we had a sample of a type yeast cells. When we combine it with other a type cells, we will only see single and budding cells. But if it is combined with alpha cells, shmoos would be visible under the microscope. This is because yeast cells have g-coupled protein receptors. In yeast, these receptors can receive signaling molecules from the opposite type of cell. Once this happens, kinases are activated that stimulate the growth of the cytoskeleton in the direction of the signal. This is where shmoos is created. For our results, there were some obvious patterns. First of all, the graphs of single and budding haploid cells for the a and alpha type cells were always inversely related. As one type's concentration went up, the other decreased. For a-type, the last reading could be a mistake. The microscope itself maybe have been too dirty or malfunctioning, but no other cells were visible as the screen was very dark. Looking at the mixed culture, we can see that the concentration of single haploid cells decreases sharply, which was expected. The number of budding zygotes and asci cells increased over time, which makes sense because a and alpha cells will be communicating with each other in the mixture. If they are mixed, then their proximity allows them to receive signals and mate. The only surprising result was that of the shmoos, which we thought would increase over time rather than stay relatively flat. 




Conclusion:
The conclusion that we had was that obviously the mixed had a lot more yeast cells than the alpha and a type since it was a mixture of both of them. The other observation that we made was that the more time that elapsed, the less percentage of each cell in the mixed, except for the schmoos. Which could be what was supposed to happen or it could be a calculation error in some way. Also I believe the single haploid was eventually supposed to be less than the budding haploid, because they would all turn into those toward the end. That ended up happening in the alpha type and the a type, but the mixed culture seemed to have some differing results.






Monday, December 9, 2013

Plant Pigments and Photosynthesis Lab

Purpose: The purpose of this lab was to measure the different rates of photosynthesis in isolated chloroplasts using DPIP. We used a dye-reduction technique, which shows that light and chloroplasts are necessary for the light reactions to occur. The independent variable is boiled versus unboiled chloroplasts and the amount of time spent in the light. The dependent variable is  the percent transmittance. The control group is the group with no DPIP and the group with no chloroplasts. We were trying to see if the amount of time that each group of chloroplasts was left in the light would effect the percent transmittance.
For the chromatography portion, we wanted to see the distance that pigments in spinach leaves would travel with a solvent. 

Introduction: DPIP was the compound we used in place of the electron acceptor NADPH in photosynthesis. The dye-reduction technique was what showed us how many reactions took place in each group. Each time the DPIP accepted electrons, or was reduced, DPIP changed from blue to colorless.  This means that the more color was lost, the more reactions were taking place. The percent transmittance shows how much light was absorbed and how much was allowed to pass through. 

Methods: To set up the lab we had a flood light, in front of a heat sink, then eventually our five cuvettes would be lined up behind the heat sink. We first got our Labquest 2 set up and attached our colorimeter to it. We had five cuvettes that were all filled differently with our solutions. The 1st cuvette was the blank control and we were using that to collaborate our colorimeter. That had 1 mL of phosphate buffer, 4mL of distilled water, and 3 drops of unboiled chloroplast. The next was one that had unboiled chloroplasts dark which had 1 mL of phosphate buffer, 3 mL of distilled water, 1 mL of DPIP, and 3 drops of unboiled chloroplasts. Cuvette 2 was then covered in foil so that no light could get through to the solution. The 3rd cuvette was unboiled chloroplasts light, which had 1 mL of phosphate buffer, 3 mL of distilled water, 1 mL of DPIP, and 3 drops of unboiled chloroplasts. Cuvette 4 boiled chloroplasts light which had 1 mL of phosphate buffer, 3 mL of distilled water, 1 mL of DPIP, and 3 drops of boiled chloroplasts. The 5th cuvette was the no chlorplasts light, which was our negative control group. This had 1 mL of phosphate buffer, 3mL + 3 drops of distilled water and 1 mL of DPIP. We set the colorimeter to 1 by using the 1st. We weren't able to put the chlorplasts in until we were about to start the procedures. So we put all the chlorplasts in th colorimeter one at a time and at 0 min we measured the transmittance for all 5 cuvettes. Then we put all five in front of the heat sink where the light was shining through. Then at 5, 10, and 15 minutes, we took the cuvettes away from the light to measure the transmittance again and then put it back after each test. 
Test tubes receiving DPIP


Colorimiter being calibrated before the cuvettes were tested for time 0
























The cuvettes and flood lamp




















































Below are pictures shown from the Chromatography Lab. Here, we used cells from spinach leaves on paper, with the end dipped in a solvent, to assess the pigment distribution.




Data:
For the chromatography lab, we have measurements of each band seen in the picture above. 

 This means that the Rf factor for the first band is 0.14, the second band is 0.27, the third is .35, the fourth is .51, and the final is the solvent front.
Then, we have our measurements from the Photosynthesis lab. In the first chart, we have our measurements from the first round of trials. Then, we have measurements from the second round of trials. The significant differences In results will be explained. 



Graphs and Charts:

Using this key, two graphs are shown. The first graph is from the first round of results, and we know these are inaccurate since transmittance can not be greater than 100%. The second graph depicts results from the second round of trials. 



Discussion:
In our first run through of the photosynthesis lab, we immediately knew something was wrong. In our first measurements at time 0, we had transmittance readings over 100%, which is not possible. For this reason, we increased the concentration of chloroplasts from our first round of trials. This way, the reactions would not be nearly finished taking place before we even took measurements in the colorimeter. After making this change, we saw an improvement in our results. Test tube 2, with unboiled dark chloroplasts, remained fairly constant. This makes sense due to the necessity of light for photosynthesis to take place and DPIP to be used. The transmittance of test tube 3, unboiled light chloroplasts, increased over time. This also makes sense due to the fact that the chloroplasts are functioning, and provided light to fuel their reactions. Test tube 4, boiled light chloroplasts, increased initially and then decreased. This is interesting because boiled chloroplasts become denatured, and therefore do not react. The graph should be relatively flat, and more trials could be taken to determine if a mistake was made. Finally, test tube 5 had no chloroplasts, and therefore no reactions taking place, which resulted in little or no change in the amount of transmittance. Excluding test tube 4, these results turned out the way we had hypothesized. 

In the chromatography lab, we used the below formula to determine the Rf factor for each pigment. The results come out different for each pigment because they are not equally soluble in the solvent. Beta carotene traveled the furthest because it was the most soluble, forming no hydrogen bonds. Xanthophyll, on the other hand, is less soluble and forms hydrogen bonds with cellulose, causing it to be found further from the solvent front. We did not have extensive prior knowledge about pigments such as these, but the information provided supported our results.



Conclusion: Cuvette 1 stayed similar in it's transmittance rates because it didn't have DPIP, and acted as the NADP+, which is the electron carrier in photosynthesis. Cuvette 2 didn't have any light going through to it so therefore there were no electrons being produced,so the color didn't change. Then Cuvette 4, which had unboiled chlorplasts which didn't cause much electron production or color change since some proteins had probably been denatured through the process of boiling them. Then, Cuvette 5 was our negative control that didn't have either type of chlorplast, so therefore it didn't have anything to react with and it basically had very similar transmittance each time like Cuvette 2 had. 











Friday, November 15, 2013

Cellular Respiration Lab

Purpose: The purpose of this lab was to determine how much carbon dioxide gas was produced during cellular respiration for each different type of seed. We used both germinated and non-germinated seeds at room temperature and at cold temperatures. Our independent variable was the germinated versus non-germinated, and cold temperatures versus room temperature. The dependent variable was the amount of carbon dioxide produced. We used glass beads as the control group. 

Introduction: Cellular respiration is the catabolic pathway that breaks down glucose into carbon dioxide, water, and energy. It has three stages-glycolysis, the citric acid cycle, and oxidative phosphorylation. During glycolysis, glucose and other sugar molecules are broken down into pyruvate. Pyruvate then loses a molecule of carbon dioxide and attaches to a coenzyme A, and becomes Acetyl CoA. Acetyl CoA then goes through the citric acid cycle where more carbon dioxide is released and Acetyl CoA attaches to an existing 4-carbon molecule called oxaloacetate to form citrate. The next steps of the cycle decompose citrate back to oxaloacetate so that the process can start over with another Acetyl CoA molecule. While the citrate was being decomposed, it produced NADH and FADH2, which are electron carriers. They then carry electrons to the electron transport chain, which uses the energy given off by transporting the electrons to power the proton-motive force. The protons that are pumped into the intermembrane space then flow through a transport protein called ATP synthase. The flow of H+ ions through ATP synthase powers the synthysis of ATP. Cells use this ATP to carry out different functions, and it is necessary to life. Germination is when a seed has been dormant for a period of time, but when its need of water, oxygen,and sunlight are met, the seed begins to grow and undergo cellular respiration. Non-germinated seeds, however, will not. 

Methods: First off, we collected the germinated corn seeds at room temperature and the non-germinated corn seeds by getting 25 of each type. We connected the CO2 gas sensor to the lab quest so we could collected our data. We then took the room temperature and stated it in a data table. Then, we had to dry the germinated seeds at room temperature and then put them in the respiration chamber and put the CO2 gas sensor in it. We let it sit for about one minute then started collecting data for 10 minutes. A graph of carbon dioxide vs. time is then showed and we had to figure out the slope of each trial. We took out the sensor and put the germinated seeds in ice cold water and let it sit for about 20-25 minutes, while we set up and started the trial of the non-germinated seeds. We did the same process as we did before for the room temperature germinated seeds. After the non-germinated seeds were done, we dried off the germinated seeds that were in the ice water and put them in the respiration chamber, and went through the same process a 3rd time. The next day, we used glass beads and used the CO2 sensor and the respiration chamber to figure out our control group, and lastly we used the lab quest to graph all of the trials so we could compare them. 


Data
Below are the respiration rates of germinated corn at room temperature (23 degrees Celsius) and cooled (15 degrees Celsius). Then, we provided the respiration rate of glass beads as a control group, as they would undergo no respiration. 

Graphs and Charts
Below is the graph of our first trial, where germinated seeds were used at room temperature. The rate of respiration is .680 ppm/s.
Then, we have our second trial using no germinated seeds. As you can see, though the graph looks similar, the scale is much smaller. The rate of respiration is only .071 ppm/s.
For the third trial, the cold germinating seeds were used. The rate of respiration was higher than that of the room temperature, as the slope was .699 ppm/s.
Our last trial was the glass beads. Again, the scale is much smaller than the others, and the slope is much smaller than it appears. Our control group's rate came out to .029 ppm/s.
Finally, we put each graph on the same screen. The cold germinated seeds a re on top, followed by the room temperature germinated seeds, the no germinated seeds, and the glass beads.
Discussion
There is one product of cellular respiration that we measured in this lab: CO2. Since rates of CO2 production changed depending on each trial, we were able to deduce that cellular respiration was occurring. As we can see through our measurements, the germinating seeds had much higher rates of cell respiration. This is because they have begun to use cellular respiration with the available amounts of water and oxygen in order to carry out cellular functions and growth. The non-germinated seeds have not met their requirements of water and oxygen, and will not be able to undergo cellular respiration like the germinated do.  Because of this, their rate is much lower, and represents that of the glass beads. Since glass beads are not living, they will not release any CO2. Temperature also affected our results. We thought that cooling the seeds would decrease their rate of respiration, due to slowing down cell functions. However, this was not the case. The rate for the cooled seeds was higher than the rate for the room temperature seeds. This could be because we did not keep the cool seeds at the same temperature during the six minute testing process. If the seeds started to warm up, their rates of respiration would increase more dramatically than seeds remaining at the same temperature. To amend this in the future, we could put the respiration chamber in a cool environment and then take measurements. 

Conclusion: The different types of seeds all had different amounts of carbon dioxide produced during cellular respiration depending on whether or not they were germinated and whether the germinated seeds were cold or at room temperature. For the room temperature seeds, corns slope was .68 and radish was .61, so those two outcomes were very similar. Peas rate of respirations slope was .32. For the rate of respiration at cold temperature, corn and radish both had .70 for their slope which shows this similarity in their cellular respiration rates and the peas had a slope of .97. Then for the non-germinated seeds the corn and peas were actually more similar, the corn was at .07 and the peas were .05 and the radish was .20. The rate of cellular respiration for the glass beads, which was our control group was different, for all 3 of our lab groups, but close in the number. For group 1, which was our group, our slope for glass beads was .03, for group 2 it was .00 and for group 3 it was -.02. Due to the results, we know that the glass beads should all be around zero. However, it is difficult to say which type of seed is the most efficient, as both the radish and pea seeds have high numbers for different trials.







Monday, November 4, 2013

Enzyme Lab



Purpose
The purpose for experiment 2A was to determine how the enzyme and substrate acted in the reaction. The enzyme in this experiment was the catalase from yeast, and the substrate was the hydrogen peroxide. We added the catalase to the hydrogen peroxide, and oxygen and water were released in the form of bubbles.

In experiment 2C, the purpose was to find out the rate of uncatalyzed hydrogen peroxide decomposition. We let a cup of hydrogen peroxide sit out overnight, and the next day tested it to see how much of the solution was left, and how much of it had spontaneously reacted to form water and oxygen gas.

For experiment 2D, the purpose was to find the effect of an enzyme on the rate of hydrogen peroxide decomposition. To accomplish this, we measured how much subrate was disappearing over time. We did this by titrating the solution with potassium permanganate to determine whether leaving the catalase in the hydrogen peroxide longer would increase, decrease, or not effect the rate of the decomposition of hydrogen peroxide. 

Below is the equation of the spontaneous reaction of hydrogen peroxide into water and oxygen gas. Using catalase speeds up this reaction.



Introduction
An enzyme is a macromolecule that speeds up a reaction without being consumed. A substrate is the reactant that the enzyme acts on. When you bond them together, it becoms an enzyme-substrate complex. Hydrogen peroxide is made of two hydrogen molecules and two oxygen molecule. When an enzyme, catalase, is added to the hydrogen peroxide, it releases water and oxygen. The rate of reaction means how quickly the reaction happens. Using hydrogen peroxide and potassium permanganate, we established a baseline, or the calculation that served as a comparison for all later tests. 
  

Methods
First off, we had to do a baseline for the uncatalzyed rate of hydrogen peroxide decomposition by putting a small amount quantity of it into a beaker, letting it sit overnight uncovered at room temperature. Then we put 1 mL of water into it and 10 mL of sulfuric acid . We mixed it and took out 5 mL to titrate and find the baseline. We then found the base line for an enzyme-catalyzed rate of hydrogen peroxide, doing the same steps as before if it had been a day or more since the practice baseline we did the day before the lab. Next, we tried to determine the course of an enzymatic reaction, by measuring the amount of substrate that disappeared. So we determined the reaction rate after 10, 30, 60, 90, 120,180 and 360 seconds. We put 10 mL of 1.5% hydrogen peroxide in a clean beaker and add 1mL of catalase and swirl for 10 seconds each time. Two of us would time it and stir the catalase/H2O2 solution and then add the 10 mL of sulfuric acid to stop the reaction, then The third would take 5 mL out of each solution then titrate it with potassium permanganate until it was pink or brown. Then we recorded it in a data table to compare the amount of hydrogen peroxide used in each test and the baseline.

Titrating to test hydrogen peroxide concentration

Before yeast

Getting catalase from yeast

Sulfuric Acid Being Inserted to end reactions on time.

Data
Below is the measurement for the baseline. 3.9 mL of potassium permanganate was necessary to react with 5 mL of hydrogen peroxide.
 
Now, we can see the baseline compared to other tests. For example: When catalase reacted with hydrogen peroxide for ten seconds, 2.6 mL of KMnO4 was necessary to react with the remaining H2O2, whereas 3.9 mL was used in the baseline.
 
Graphs and Charts
  Here is the graph of the amount of hydrogen peroxide used in the enzyme- fueled reaction for each amount of time.
 
  
Discussion
  If we look at the data for the timed trials, two patterns are clear. First, as the amount of time catalase and H2O2 react increases, the amount of potassium permanganate needed to filtrate the sample decreases. Second, as the amount of time increases, the amount of H2O2 used increases. (This increase can also be seen in the graph showing hydrogen peroxide decomposition.). These results go along with our predictions. Since catalase is the enzyme we are using, on our H2O2 substrate, as time goes on we expect more reactions to occur. As these reactions continue, hydrogen peroxide is used. The more time allotted for the enzyme and substrate to react, the more product is produced and substrate is used up. It is for this reason that we see the increase in the amount of H2O2 used. As more hydrogen peroxide is broken down by catalase, less is left to react with potassium permanganate  in the titration. Therefore, it makes sense for the amount of KMnO4 needed to decrease as time intervals increase. To stop the reactions, sulfuric acid is used. This works because the ph of sulfuric acid is very far from the optimum ph of catalase. The enzyme begins to denature in the harsh environment, and reactions cease to occur between the H2O2 and catalase. The results went along with our knowledge of the characteristics of enzymes on all fronts.
 

 
Conclusion
One conclusion that we came up with was that if we boiled the catalase solution, the proteins in the enzymes would be denatured and no reaction would have occurred. Then, for the time samples as the amount of time increased, the amount of potassium permanganate decreased because the yeast, which was the catalase solution, was allowed to react with the hydrogen peroxide for a longer period of time. For the same reason,  the amount of hydrogen peroxide used increased because of the amount of reactions that was going on. In short, as time goes on, more reactions between enzymes and their substrates can occur.

 







Monday, October 21, 2013

Diffusion and Osmosis Lab

 
Purpose: 
The purpose of 1A was to find which substances the dialysis bag was permeable to. We were testing the concepts of diffusion and osmosis. We controlled the placement of each solution, which changed the color of the substance in the bag. The dependent variable was the color of the substance inside the dialysis bag, and the independent variable was where each substance was placed. We specifically wanted to determine if the glucose, starch, IKI, and water were permeable to the bag. 
 
For 1C, the purpose was to find the percent change in mass of the potato cores when placed in different solutions. This experiment was testing the concept of the difference between hypertonic, hypotonic, and isotonic solutions. The independent variable was the different substances that the potato cores were placed in, and the dependent variable was the mass of the potato cores. We wanted to see if the potato cores would gain or lose mass in each solution.
 
In 1E, we were also testing hypertonic, hypotonic, and isotonic solutions. We were testing how each solution would effect the diffusion and cellular contents of the onion. The independent variable was the solution that the onion was placed in, and the dependent variable was the effect that the solution had on the onion. 



Introduction: 
In experiment 1A, we were testing the permeability of the dialysis bag. To be selectively permeable means to allow only certain substances through the plasma membrane. The concept of diffusion is that when a solution has a higher concentration of a substance than its solvent, the substance will spread out evenly throughout the solution. Osmosis is when water diffuses across a membrane from a region with lower solute concentration (more water) to a region with higher solute concentration (less water) to evenly balance out the concentration of water on both sides. IKI is iodine. 
 
For experiment 1C, to calculate the percent change in mass, we used the formula: final mass minus initial mass divided by initial mass multiplied by one hundred. The percent change in mass means how much mass was gained or lost through the experiment. A hypertonic solution would be a solution that has less water in it than the cell does, meaning that the water would diffuse out of the cell, causing it to shrivel. A hypotonic solution is one where there is more water outside of the cell, causing the water to diffuse into the cell, and the cell would swell. In an isotonic solution, the concentration of water is the same both inside and outside of the cell. Diffusion still occurs in an isotonic solution, but at the same rate both going in and out. 
 
Experiment 1E was basically testing the same concepts as experiment 1C, but this time instead of looking for percent change in mass, we were looking at the cellular contents. The cellular contents are composed of the organelles inside the cell, the cell wall, and the plasma membrane. 


Methods: 
For 1A, we were testing diffusion so we filled a dialysis bag with 15% glucose and 1% starch solution then tests it for glucose. Next, we took a cup and filled it 2/3 with water and then the rest of the way with IKI. Then we tested the solution in the cup for glucose. Next, we but the dialysis bag into the IKI and water solution and let it sit for about 30 minutes or until the bag went through a significant color change. Lastly, we tested the the IKI and water for glucose and the 15% glucose and 1% starch solution.
 
              
                                

For 1C, we were testing water potential, first off, we had to core the potato. We had to get 4 cores, of the same length, for each of the six cups of sucrose that we were testing them in. We then had to determine the mass of the potato cores. Next, we had to cover the cups with plastic wrap to make sure there is no evaporation. We had to let it stand overnight, then take the mass again of the potato cores and figure out the percent change in mass. 
 
For 1E, we didn't actually do the experiment, but the point of it was to test onion cell plasmolysis. Onion cells were observed in hypotonic, isotonic, and hypertonic solutions.  


Data: 
For 1A, we were able to test for the presence of glucose in and outside of the bag both before and after the experiment was run.  The original solution colors and glucose presence are shown.

For 1C, we have a chart with the initial and final mass of potato cores. By using the formula below, we were able to find the percent change in mass. 


 
For 1E, we researched to find pictures of the status of onion cells in different solutions. The first shows the behavior of a cell in a hypotonic, isotonic, and hypertonic solution. The second picture shows onion cells in isotonic and hypertonic solutions. The top half shows central vacuoles that are in no state of expansion or water loss, while the bottom half shows shriveled vacuoles that have lost too much water to their environment.
 



Graphs and charts:
1A: Below is the graph of the percent change in mass for potato cores at each solution. If the percent change in mass is positive, that means the solution is hypotonic. This is because there is a lower concentration of sugar outside the potato cells and water will rush in.  When the change in mass is negative, the outside solution is hypertonic. This is because the surrounding solution has a higher concentration of sugar, and water will leave the cells. The point where the change in mass is zero, around .22 or .24 M for the potato, is when the solution has the same molars as the cells. This would be an isotonic solution, and there would be no net mass change.



Discussion:
For 1A, we began with a glucose and starch solution inside the dialysis bag, and an IKI solution outside the bag.  As the experiment progressed, the interior of the bag turned blue. This blue coloration is an effect of reactions between IKI and starch, so we knew that IKI had diffused into the bag. Since the outside of the bag did not change color, we also determined that starch had not been able to diffuse through the bag; if it had, the exterior solution would have also turned blue. The second piece of data is the presence of glucose. Initially, the interior of the bag contained a solution with glucose, and the exterior did not. After the experiment was finished, we tested for glucose on the exterior of the bag, and the reading came out positive. Because of this, we were able to determine that glucose was able to diffuse through the bag. Going into this experiment, we did not know what the dialysis bag would be permeable to. However, in the duration of our trial, we were able to observe a significant color change, and knew that our results would show diffusion of at least one substance. Also, the fact that starch could not diffuse through the bag coincided with common sense; looking at all of the molecules present in this experiment, starch was the largest.
 
For 1C, we gathered percent changes in mass for potato cores in different solutions. As the molarity of the solution increased, the percent change in mass decreased. With a 0 M solution, the change in mass was 17.3%. It is positive because of the effects of a hypotonic solution. With a 0.4 M solution and above, the change in mass ranged from -15.7% to -42.6%. This occurs because of the effects of a hypertonic solution. This means that somewhere in between 0.2 and 0.4 M solutions is the molarity of the potato. The graph of percent change in mass versus solution molarity would have to reach zero at some point between these values, meaning that there would be no net mass change and the solution would be isotonic. Though we did not predict this actual point, we assumed that this would occur. Some point between a 0 M and 1 M solution, we thought the concentration of starch in a potato would be passed.
 
For 1E, though we did not actually complete this experiment, the diagrams and pictures above lined up with our knowledge of solutions.  When the onion cell was in an isotonic solution, no change in size was observed, and the flow of water into and out of the cell were equal. When the onion cell was in a hypertonic solution, water exited the cell and the vacuole became plasmolyzed. When the onion cell was in a hypotonic solution, it took in water from the environment and the vacuole swelled.  This went along with our previous knowledge of the effects of a solution on a cell.
 
 
Conclusion:
For 1A, our conclusion is that the IKI can pass through the selectively permeable membrane, but the membrane is not permeable to starch. We can tell this because when starch reacts with IKI that is when it becomes blue. Glucose, on the other hand, was able to get through the membrane; we started with glucose inside the bag, and we ended with glucose outside the bag .
 
For 1C, the conclusion that we came to was that the mass change in cells will depend on the type of solution. For hypotonic solutions, cells will gain mass as water rushes into the cells. For hypertonic solutions, the cells will lose mass as water is lost to the environment. In an isotonic solution, water will move in and out of the cell at equal rates, and there will be no net mass change.
 
For 1E, the conclusion is that if a cell is in a hypertonic solution, then the water exits the cell. This makes the central vacuole collapse, which causes the cell to go through plasmolysis. There is more salt on the outside, meaning more water wants to go out. For hypotonic the water enters the cell, and becomes turgid and when it enters the cell th vacuole will expand. For a hypotonic cell, there is more salt on the inside of the cell, so the water wants to go in. For an isotonic solution, the water would enter and leave the cell at equal rates, which makes it flaccid. For the salt, the concentrations will be equal on both the inside and the outside of the cell. 

 


Monday, September 30, 2013

Cassie's Restoration Ecology Reflection

I do think that restoration ecology is important for our environment, because it helps everything that's been messed up by humans get back to the way it's supposed to be. By us cutting down the invasive species in the forest and clearing out all of the unwanted plants and whatnot, it helps the forest get itself back to the way it's supposed to be and it helps the natural plants to grow back and flourish. 
This is a picture of the forest before we cleared it out

This is what it looked like after we got rid of all the invasive species. As you can see, we made a lot of progress on the forest, but there's still a lot that needs to be done. I think this experience really helped to show us how hard the forest has to work to get itself back to the normal state after humans mess it up and bring in unwanted species. It also shows the importance of maintaining the environment, and why we should take care of our earth. 

We also helped by planting acorns and watering trees. 
This is a type of oak tree, and the reason that we had to water it is because its roots aren't big enough yet to get its own water from the soil. Most tree's roots are big enough because when they start out as a seedling in the ground, they can establish their roots much better than if they were planted there by humans. That's why it's very important to give a lot of water to newly planted trees. 

Overall, I think this was a very good experience, and I learned a lot about the environment and the importance of helping it in any way that we can.