Cell Communication Lab

 

Purpose:

The purpose of this experiment is to examine the reproduction of yeast cells and how cellular communication plays into it.

 

Intro:

Cellular communication occurs via chemical signals that coordinate functions and respond to stimuli. Chemical signals can come from the environment, other cells in the same organism, or other organisms. Cells can communicate by direct contact, local signaling, or long distance signaling. Yeasts are unicellular fungi that can reproduce sexually or asexually. They alternate between haploid and diploid phases. During asexual reproduction, single haploid cells turn into budding haploid cells, which creates a new daughter cell. During sexual reproduction, in response to a specific chemical signal, haploids can change from an asexually reproducing cell to a gamete. There are two sexes: a-type and alpha-type. When these types mix, they grow towards each other, forming shmoos. The shmoos then fuse to create a zygote, which divides to form new yeast cells. 

 

Methods:

After placing the yeast solution (alpha type, a type and mixed) into seperate test tubes we measured using the light microscope; to find the starting amount of yeast cells. Then took measurements at different increments of time (30 minutes, 24 hours and 48 hours) to see what percent increase the yeast colony experienced periodically.

 

 

                                                      

                                                                   Mixed at 30 minutes

                                                    

 

                                                  

                                                                       Mixed at 48 hours

 

Data: 

 

 

                                                           Data for A and alpha yeast

 

 

                                                                Data for mixed yeast

Graphs:  

 

 

 

 

                                     

 

 

Discussion: 

In this lab we tested the reproduction patterns of alpha type and a type cells.  We also tested a solution of mixed alpha and a type yeast.  What we saw was this:

Alpha type yeast: Over time, the percentage of single haploid cells decreased.  At 0 minutes 77% of the cells were single haploid cells.  At 30 minutes, 76%, and at 24 hours, about 22%.  This can be attributed to the fact that there would be more formation of budding haploid cells due to the progression of time, so single haploid cell percentage would decrease.  But, we saw an increase after 48 hours, where the percentage hit 85%.  There was such a large increase I yeast concentration that the percentage of single haploid cells must have gone up.  After a certain amount of time, there must be a pattern where single haploid cells will produce budding haploid cells, and this is what happened here.  For the budding haploid cells, we saw a rapid increase.  Starting at about 23%, it increased to 24% and 77% during 30 minutes and 24 hours respectively.  But, we also saw a decrease in percentage at the 48 hour mark (14%).  

 

A type yeast: We experienced much of the same trends for the a type yeast.  Over time, single haploid cells decreased at first them increased rapidly.  It started at 89%, at 30 minutes it was at 40%, and at 24 hours it was a 15%.  But, at the 48 hour mark, it increased rapidly (85%).  For budding haploid cells, it started at 10%, increased to 60%, increased again to 84%, then decreased to 14%.  The data we had was very similar for both a and alpha type yeast, leading us to believe that they reproduce in much of the same way.

 

Mixed yeast: This sample was a combination of the a type and aha type yeast cells.  For the single haploid cells, we saw a steady decrease.  It started at a little less than 50% of the total cell count, and ended at about 29%.  This is because single haploid cells will become budding cells and eventually form asci over time, so the cell count will continually decrease.  For budding haploid cells, we observed an overall decrease as well, starting with about 17%, it gradually decreased to 7.4%.  This too, can be attributed to the fact that cells will start to form asci over time, therefore decreasing the number of free floating cells.  For the shmoos, we saw a maximum percentage of 21.8% at 30 minutes.  It then decreased to close its original percentage at 24 hours, but then increased to 17.7% at 48 hours.  For the single zygotes, we saw 7.6% at the 0 minutes. Then we saw 5.9%, 19.5%, and 11%.  For budding zygotes, the data varied by increasing, describing, and increasing once more.  The most important sample from this experiment was the percentage of asci.  It started at a relatively low percentage of 6.4%.  It ended up being at 27.4%.  This is important because as time goes by, yeast are able to communicate as a whole.  When time passes, more yeast communicate with each other,  so they are able to come in close proximity to eachother, and form asci.  

 

Basically, for a and alpha type, when time progresses, we see the overall number of yeast cells increase.  We can see that yeast has the ability to reproduce because of the increase in budding haploid cells, and therefore an increase in overall cell count (yeast cell count must increase because it is reproducing).  For the mixed type, we see an increase in the percentage of asci, which means that there was an an increase in cell communication due to the lapse of time.  Overall, the cells that denote the reproduction of yeast increase over time.  We know that yeast can't physically move because, at zero minutes, it would all conglomerate together.  Rather, it communicates by means of chemical signals.

Conclusion:

From this lab we can conclude that over time the yeast concentration increases. It is able to do so because it communicates bby means of chemical signals called pheromones. When time passes the concentration of yeast increases showing thaat it is able reproduce through chemicall signals. 

 

 

 

 

 

Plant Pigments & Photosynthesis Lab

Plant Pigments Lab

Purpose: 

The purpose of this experiment is to identify the pigmengs in chlorphyll and test how fast the pigments move up the chromatography paper.  This helps establish Rf values. 

Intro: 

Paper chromatography is used to seperate and identify pigments in cell extracts. The solvent moves up the paper because of the attraction of solvent molecules to the paper and the attraction of solvent molecules to each other. The solvent carries the pigment up the paper with it, but the pigments move up at different rates depending on how soluble they are and how much they are attracte to the paper.  Pigments in plants are used to trap light energy in order to perform photosynthesis.  The photons in light excite the electrons created from the splitting of water, and because pigments are able to harness that energy from photons, photosynthesis is able to be carried out. 

Methods:

First, we obtained a 50ml graduated calendar that contained 1cm of solvent in it.  Then, we cut a strip of filter paper and cut a point onto the end.  Next, we drew a line 1.5 cm above the tip of the point and placed a small leaf on it.  We then crushed the leaf onto the filter paper with a coin.  Next, we put the filter paper in the solvent. Once the solvent was within 1 cm of the top of the filter paper, we removed the paper and marked where each line stopped (both the pigment and solvent line).  We then measured the distance between the original and current line.  We repeated these steps for several other pigments as well.  The pigments used were Beta Carotene, Xanthophyll, Chlorophyll A, and Chlorophyll B.

Data/Graphs and Charts: 

Discussion: 

In This lab we used Chromatography paper to test the solublity of the different pigments found in spinach leaves. By placing the chromatography paper into the solvent the pigments where drawn upwards by Capillary action. Through Capillary action the solvent molecules stick to the paper and to each other. Which is quite similar to cohesion and adhesion of water molecules! When then the process was complete the solvent had left behind particles of the different pigments that were in the spinach leaf. We had 5 different lines of pigments that spread across the paper from top to bottom. From the point where we placed the spinach leaf molecules on to the paper was about 20 mm from the tip of the paper, the color of the pigment was a dark yellow color meaning that this is most likely Chorophyll B. Then the next strip of left over pigment molecules was at 40 mm and green in color. This is the band of Chlorophyll A molecules. the chlorophyll molecules are closer to the orginial pigment molecules of spinach because they are bound closer to the paper than the other pigments. The next band of pigment was at 52 mm and was yellow-green in color.This is one of the non-major pigments that are not listed in the lab or this could mean that there was a slight error in our lab. The next was Xanthophyll which is recored at 80 mm from the bottom of the paper and is a pale yellow color. Xanthophyll is higher up the paper because it is more soluble than the chlorophyll pigments, but less soluble than the Beta Caroten, since it contain oxygen and is slowed by forming hydrogen bonds with cellulose. At the top we have Beta Caroten (a yellow-orange color), it is the highest because it is very soluble  and is not slowed by the formation of hydrogen bond with cellulose, moving it up the paper the highest. For the relationship that is shown between the distance the pigment migrated from the front and from the distance solvent front migrated we had: Beta Carotene- 2.142, Xanthophyll- 1.124, Cholorophyll A- 0.571 and Chlorophyll B- 0.286. This data shows that Beta Carotene traveled the furthest from the original band of pigment where the spinach molecules were placed and supports our findings from the experiments. 

Conclusion: We concluded that Beta-Carotene moved the farthest up the chromatography sheet.  The Rf factor for Beta-Carotene was 1.  The pigment that moved the least was Chlorophyll-B.  Its Rf factor was .286.

Photosynthesis Lab 

Purpose: 

This experiment tests the hypothesis that light and chloroplasts are needed in light reactions. DPIP will take the place of NADP, which will be reduced. This will cause the DPIP to turn from blue to colorless. Based on the measurements of light transmittance, we will be able to tell which solutions have the highest rate of photosynthesis.

Intro: 

During photosynthesis, light is absorbed by leaf pigments and the electrons gain energy. Then, NADP is reduced to NADPH. ATP and NADPH are then used in carbon fixation, which turns CO2 into organic molecules.  Photosynthesis needs several things in order to be carried out.  The first thing is the presence of light.  Light excites the electrons that exist from the splitting of water, and those electrons are then able to make their way down the electron transport chain.  Another thing that is necessary for photosynthesis to occur is water.  Water is the source of electrons that help create the difference in proton gradient, which then causes ATP to be created.  Also, a plant must have a pigment that has the ability to absorb light.  The main pigment is chlorophyll-a, but many other photoreceptive pigments exist.  Carotene and chloraphyll-b are examples of other pigments.  The last thing that is needed for photosynthesis is active proteins.  Proteins make up the electron transport chain found in photosystem I and photosystem II, and there are also proteins used in carbon fixation during the Calvin cycle.  

Methods:

Day 1

 First, we prepped five different cuvettes with different solutions.  In the first cuvette, we put only the phosphate buffer, water, and chloroplasts.  In the second cuvette, we put DPIP, water, and unboiled chloroplasts, and phosphate buffer, but we placed a layer of tinfoild around it to mimic a dark environment.  In the third cuvette, we put inDPIP, water, and unboiled chloroplasts, and phosphate buffer.  In the fourth cuvette, we put in DPIP, water, and boiled chloroplasts, and phosphate buffer.  In the fifth cuvette, we put DPIP, water, phosphate buffer, but no chloroplasts.  The chloroplasts weren't put into the solution until they were ready to be put under the light.  Next, we calibrated the colorimeter with the blank cuvette (the one containing only DPIP and water) in order to establish what 0% transmittance was.  Next  we measured the reast of the cuvette in the colorimeter.  We then put the cuvette behind the lamp (with the heat recepticle between the lamp and the cuvettes), and kept time for five minutes.  Once that was complete, we placed each cuvette in the colorimeter to measure the % transmittance.  We repeated these steps for time increments of 5, 10, and 15 minutes as well.  Listed below is the contents of the solutions for each cuvette.

        

        

Day 2
We used the same methods for day 1, but instead we put 2 drops of DPIP in.

Data:

                                                                             Day 1

Day 2

Graphs and Charts:

Discussion:                

Day 1

In this experiment we were testing how chloroplasts and lights are essential for light reactions to take place. We also substituted NADP for DPIP, which is another electron acceptor. When the light hits the chloroplasts the electron levels are raised and will reduce the DPIP. This will cause the chloroplasts to change from blue to colorless, this proves that chloroplasts and light are needed because when the solution changes to colorless that means that the electron acceptors are being used by the electrons brought into the solution by light. Our control would be Sample 1, which was the blank that had no DPIP added but had the unboiled chloroplasts. In this sample we had add amount of transmittance , there was a reaction that was near 100% transmittance. Since this is the cuvette we used to calibrate the spectrometer. The second smaple was the unboiled chloroplasts without light and DPIP, with this there should be no reaction so the solution stayed blue. We placed the foil over the cuvette to block the light which stimulates the reaction. Since we had a low amount of transmittance ( about 98.5% at the end of day 1 and 71% on day 2).  This indicates that human error did occur.  The third sample of unboiled chloroplasts with light had a amount of tranmittance, this is the sample where all the necessary components are available for photosynthesis to function properly (Light, Unboiled chloroplasts & DPIP) then the solution changed to colorless. Sample 4 had boiled chloroplasts with light and DPIP, and at the end of day 1, the percent transmittance was at 94.81, and at the end of day 2, it was 66.49%. Because the chloroplasts weren't able to function due to the denaturing of the proteins involved in photosynthesis, this percentage was lower compared to the rest of the samples. The last sample 5 has no chloroplasts what so ever, meaning that there can't be a reaction because without chloroplasts the percent transmittance was nonexistant on behalf of light having to strike the chloroplasts for this reaction to take place. The research we obatined from this experiment does for sure prove that you need chloroplasts, an electron acceptor and light. 

Day 2

On day two our challenge was to somehow slow the reaction, so that our percent transmittance wasn't so high.  It jumped about 88% to about 98% percent, so tracking the increase in transmittance was more difficult.  We decided in our lab group to increase the DPIP concentration by 1 mL. By doing this we found that the change in transmittance was less drastic in the 15 minutes of testing period. This successfully translates to  slowing the reaction rate. Since we increased the electron acceptor concentration the reaction couldn't go out of control since the electron acceptors are what play a major role in controlling the reaction rate. 

Conclusion: 

Both the dark and light cuvettes (cuvettes 2 and 3) experienced the highest rate of transmittance.  Cuvette 2 had a final transmittance reading of 98.89 on day one, and 71.16 on day two. Cuvette 3 had a final transmittance reading of 98.05 on day one, and 69.70 on day two.  Although cuvette 2 did have a high transmittance rate, it was most likely caused by human error.  The rate of photosynthesis occurs slower in the dark, and eventually doesn't occur at all, so cuvette 2 must have been exposed to light accidentally.  Therefore, chloroplasts that haven't been denatured by boiling and are exposed to light will have the highest rate of photosynthesis.  

Cellular Respiration Lab

Purpose: 

In this lab, we wanted to test the whether germinated seeds and non-germinated seeds would respire, and if so, the rate at which they do so.  Also, we wanted to see how temperature impacted rate of respiration as well.  

Introduction: 

Cellular respiration is the process in which organic molecules are broken down in order to create energy for an organism.  Cell respiration occurs aerobically, which mean that oxygen is necessary for the process to occur.  Cell respiration also occurs in the mitochondria.  Before respiration begins  glucose is broken down into two molecules of pyruvate in a process called glycolosis.  If oxygen is present, then respiration will occur next.  But if oxygen isn't present, fermentation will occur.  When oxygen is present, the two pyruvate molecules are converted to AcetylCoA, which is a coenzyme.  One molecule of CO2 is released in this stage.  Next, the citric acid cycle (or Kreb's cycle) occurs.  In this stage of respiration,  2 molecules of CO2 are released as a byproduct of the breakdown of pyruvate. Two pyruvates are present, so each molecule creates one ATP and 2 molecules of CO2.  Next, the pyruvates undergo oxidative phosphorylation.  In this stage, the electrons carried by NADH (NADH collects extra electrons from molecules throughout respiration) we released onto the electron transport chain, and are subsequently carried down to increasingly more electronegative molecules.  The spillover of extra electrons in this stage fuels the transport of H+ ions across the membrane of the mitochondria, creating a large difference in energy.  When the H+ ions travel across the membrane to once again, Chemiosmosis happens.  The movement of H+ ions across the membrane to a lower concentration create energy for the synthesis of ATP.  32 to 34 ATP are created in this stage.  Overall, 2 ATP are created during glycolosis, 2 ATP are created during the citric acid cycle, and 32 to 34 ATP are created during oxidative phosphorylation.  A way to see if respiration occurs is to monitor the output of CO2 in organisms, because CO2 is emitted in both the conversion of pyruvate to AcetylCoA and the citric acid cycle.

Germination of seeds is the process in which seeds become active and able to grow.  This happens because seeds that are dehydrated intake water, and therefore, enzymes that need water to function are able to do so once again.  A germinated seed can have shoots coming off of it, and a non- germinated seed looks shriveled and water-less.

                                                 

                                                        Non-germinated seeds

                                               

                                                            Germinated seeds

Methods:
First, we collected germinated and non-germinated pea seeds.  In order to germinate the peas, we soaked them in water overnight.  Then weproceeded to test the level of CO2 emitted from several different seed types.  Here's how we did it:

Next, we separated 25 germinated, 25 non-germinated, and 25 glass beads (this was used for a control) into separate containers.

Second, we tested each of those materials for the presence of CO2 over a ten minute time span.

Third, we put the 25 germinated peas into ice-cold water for approximately ten minutes, in order of I test the affect of tempurature of germinated peas on levels of CO2 output.  We then tested the cold peas for presence of CO2 as well.

Lastly, we created graphs showing the CO2 output over ten minutes.

Data:

       

Graphs and Charts:

                                                          CO2 Release vs. Time Graph

    

Discussion: 

In this lab, we tested cellular respiration in different types of seeds. We did this by calculating the amount of CO2 emitted by the seeds, since CO2 is a product of cellular respiration. Our group tested peas that were germinated at room temperature, peas that were germinated in ice water, and dormant seeds. We also used glass balls as a control group. Our results showed that at 0.80 ppm/s, peas germinated at room temperature emitted the most carbon dioxide, showing that they have the highest rate of cellular respiration. The germinated peas in ice water had a lower rate, at 0.71 ppm/s, showing that cellular respirations is still occurring in lower temperatures, just at a lower rate. The dormant peas were drastically lower, with a rate of 0.15 ppm/s. Our control group of glass beads had a rate of 0.10 ppm/s, showing that a small amount of carbon dioxide could have been coming from other places, such as the water trapped in the bottle, it could've been leaking into the bottle, or it could have been trapped in the bottle from the previous test. Overall, our results showed that a germinated cell has a higher rate of cellular respiration than a non-germinated seed, as well as there is a direct relationship between temperature and respiration: as temperate goes up, so does respiration; as temperature goes down, so does respiration. 

Conclusion: From this experiment we can conclude that the non-germinated seeds do not respire, as shown by the very little amount of CO2 that was being given off by the non-germinated seeds. As for the germinated seeds we found that when the seeds were at room temperature that they respired at the highest rate.  The cold seeds had a lower amount of CO2 then the room temperature seeds. In summary when the temperature is higher the respiration of CO2 is higher than those of seeds in colder environments and that non-germinated seeds do not respire.

References: 

http://plantsinmotion.bio.indiana.edu/plantmotion/earlygrowth/germination/germ.html

Campbell Biology, 9th edition

Enzyme Catalysis Lab

Purpose:
The purpose of this experiment is to understand the functions of enzymes and how pH changes the reaction rate. This can be done by observing the enzyme catalase in the break down of hydrogen peroxide (H2O2) into water (H2O) and oxygen gas (O2), and measuring the amount of oxygen generated and the rate of the enzyme-catalyzed reaction.

Introduction:
Enzymes are proteins that act as catalysts in biochemical reactions; catalysts affect the rate of a chemical reaction. In enzyme-catalyzed reactions, the substrate binds to the active site of the enzyme, and the activation energy of the reaction is lowered so that the produced can be formed more easily. Ways in which enzyme action can be affected include salt concentration, temperature, pH, and activations/inhibitors. 

Methods:
The main point in this lab was to understand enzyme activity. We did this by collecting baseline trials of the rate of reaction of the uncatalyzed reactions. Next we would use the same process to collect data of the catalyzed reactions, after we had added the enzyme Catalase. Then we compared the trials of the two types of reactions to one another and observed what happened to the rate of reaction.

Data:

                     

 

Graphs:


                              



Discussion:

 In this lab, we tested the affect of denaturing enzymes and how that relates to the rate of reaction between two substances.  The substances we tested were catalase,which is an enzyme, and hydrogen peroxide, which is a substrate.  In order to denature the enzyme, we added an acid, H2SO4.  The results we got indicated a direct curve, with a few exceptions.  When aci was added after 90 seconds, there was a slight decrease in rate of decomposition, which did not fillow the trend of increasing rate of reaction.  Also, at 180 seconds, there was a slight decrease as well which did not follow the trend.  This could be caused by a few things.  One, we may have put the acid in slightly too early, which would have caused this decrease.  Two, we may have contaminated the beaker with leftover acid from other experiments and the denaturing could have occurred too early, causing less decomposition to occur.  But, on average, we observed that as an enzyme is left with a substrate for a longer time, more decomposition occurs.  Therefore, length of time the substrate is exposed to the enzyme and amount of decomposition are directly related.  

Conclusion:
We can conclude that concentration of enzymes directly related to rate of decomposition of H2O2.

References:

Campell Biology, 2009

 

 

Diffusion and Osmosis Lab

Purpose

1A: The purpose of this experiment is to understand the nature of selectively permeable membranes. This is accomplished by observing how different molecules such as water, glucose, starch, and iodine interact with a selectively permeable membrane.

1B: The purpose of this experiment is to understand the relationship between solute concentration and the movement of water through a selectively permeable membrane. This can be done by testing the osmosis of water using solutions with different molarities of sucrose.

1C: The purpose of this experiment is to understand water potential and what influences change in water potential. As well as understand the components of the water potential equation and how its used.

1D:  The purpose of this experiment is to understand how to calculate water potential.

1E: The purpose of this experiment is to understand plasmolysis and the effect of concentrated solutions on plant cells. This can be done by putting onion cells in different concentrated solutions.

Introduction

1A: Diffusion is the movement of a solute across a membrane.  Diffusion only occurs when a salute travels down its concentration gradient.  A concentration gradient is the difference in concentration of a solute.  To be more specific, dialysis is the diffusion of a solute across a selectively permeable membrane.  A selectively permeable membrane is a membrane in which only some materials can pass through.  An example of this would be the plasma membrane of a eukaryotic cell.  Water molecules and other small substances are able to pass through a selectively permeable membrane freely, but larger molecules are most likely unable to diffuse

Part 1B: Osmosis is the movement of water through a membrane. An isotonic solution includes two solutions that have the same concentration of solutes, therefore there will be no net charge on the water in either solution. When two solutions have different concentrations of solute, the one with less solute is hypertonic and the one with more solute is hypotonic. When you put these two solutions together separated by a selectively permeable membrane, the water will move from the hypotonic solution to the hypertonic solution.

1C: Water potential is the movement of water in or out of cells. The movement of the water is done by the process of Osmosis (movement of water through a membrane). The two components that make up water potential are pressure potential (physicial pressure) and solute potential. Formula is Water potential = pressure potential + solute potential. The potential of pure water is equal to zero. Water always moves from high to low areas of water potential. Water potential can be a positive or negative value, or it can be zero. Pressure influences the movement of water. With an increase of pressure there will be  more positive value, where on the other hand decrease in pressure there will be a more negative value. As for solute, if there is an increase water potential will be more negative. When water leaves the cell it will shrink, If water is added it will grow. In animal cells too much water will cause the cell to swell or burst. But in plant cells the cell will only grow, this is because of the plant cell which holds the cell together. Once the water potential on the inside of the cell and on the outside of the cell are equal, then it will have reached dynamic equilibrium, where there is no water movement.

1D: Water potential equals the ionization constant plus molar concentration plus the pressure constant plus tempurature (in Kelvin).  Water potential shows water's ability to move from one location to another.  

1E: Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out of the cell when in a hypertonic solution. In a hypertonic solution, cells lose water because there is a greater concentration of solute outside the cell. In an isotonic solution, there is no net water movement because the solute concentration is the same inside and outside the cell. In a hypotonic solution, the cell gains water because there is a lower solute concentration outside the cell.

Methods

1A: First, we soaked about 20 cm of dialysis tubing in water in order to make it easier to work with.  Then, we we formed a bag out of the tubing and put a solution made of 15% glucose and 1% starch inside of the bag.  We recorded the color of each solution.  We then tested the glucose/starch solution for the presence of glucose.  Next, we filled 3/4 of a cup with iodine solution (Lugol's solution) and filled the rest with water.  We then tested this solution for the presence of glucose.  Next, we put the bag of glucose/starch solution into the cup of iodine/water solution, and proceeded to wait for approximately 30 minutes.  Once the time was up, we recorded the color of each solution and tested for the presence if glucose in each as well.

                                                                                                                 

1B: First, we filled dyalisis tubing with either water or solutions with different molarities of sucrose. We tied off the ends and left room for water to enter. Then, we recorded the weight of each bag and placed each one in a cup of water. We let them sit for 30 minutes and then took each bag out and weighed them. Finally, we calculated the change in mass for each bag.

                                           

1C: First, we poured the five different molarities of sucrose into their designated beakers as well as the distilled water into its own beaker. Next using a cork borer we made 24 potato cylinders (4 per beaker). Then we weighed the potato cylinders 4 at a time and recorded the intial mass. After weighing them we placed the potato cylinders into the beakers (4 per beaker). Following placing the potato cylinders into their beakers, we let them soak in the solution overnight with a sheet of plastic wrap covering the top of the beakers (this prevents evaporation). The day after we took the potatoes out of the beakers, using a paper towel to dab the excess solution off of the cylinders and we weighed them once more to find the final mass. Then we found the percent change in mass (intial mass/final mass).

                                           
 

1D:

1E:  First, put a piece of an onion on a slide and observe it under a microscope.  Sketch a picture of the image.  Then, add several drops of NaCl solution to the slide and record what occurs.  Next, add water to the slide and observe and record what happens.
                           
 

Data

1A

     1B                                 

 

            

                                       

 

                           

1C

                                       

 

 

1E

                              

                                                       

 

                                                                           Flaccid Cells

 

 

Discussion

1A: The change in color from colorless to black/blue in the glucose/starch solution obviously indicates that iodine passed through the membrane, because iodine changes color when it comes into contact with starch.  Beacause we didn't see any color change from red to black/blue in the iodine solution, it must mean that starch was unable to pass through the membrane.  Also, because the iodine tested negative for glucose before the glucose/starch bag was put in, and positive for glucose after it was put in, it must indicate that glucose was able to pass through the membrane into the iodine solution.  All of this information indicates one thing: the dialysis tubing acted as a selectively permeable membrane.  Because glucose was able to pass through the membrane, we must conlude that it is the smallest molecule (besides water) present in this experiment.  Iodine was able to pass through the membrane as well.  Starch however, has to be the largest of the molecules.  It was unable to pass through the dialysis tubing because its size was bigger than that of the pores in the membrane.  Pore size directly affects the diffusion of molecules, then.  If a substance is larger than the pore size, then it will not be able to diffuse across the membrane.  If the molecule is smaller than the pore size, then is will be able to diffuse across the membrane.

1B: Overall, as the molarity of sucrose goes up, the amount of water that diffused into the bag goes up. The bag put into distilled water had the lowest change in mass, because it was isotonic, so water should stay at equillibrium in the two solutions. Then, as the molarity of sucrose went up, more water diffused into the bag because the solution outside of the bag was hypotonic. The solution with 1.0 M sucrose has the highest change in mass because it had the most water diffuse into it. This occurred because of how hypertonic the outside solution was compared to the solution inside the bag. The only error in our data was with the solution of 0.4 sucrose, because it didn't follow the trend; it had a lower change in mass than it should have. Other than that, the trends of our data and the class data are similar, except that our changes in mass of the higher molarity sucrose solutions are significantly higher. The data trend turned out as we expected, because this simulated diffusion in cells, which when placed in a hypotonic solution, take in water.

1C: In our experiement we found that for the distilled water beaker cylinders gained 1.5g of mass with a +17.2% change in mass. At 0.2M sucrose there was a 0.34g gain in mass. As for 0.4-1.0M sucrose solution there was an overall trend of loss in mass and a negative percent change. At 0.4M we recorded a mass change of -1.53g and -16.7% change in mass. The masses would continue to drop from this point on, until we reached a mass change of -3.08g and a percent change of -33.67% for 1.0M sucrose. From less then 0.2M Sucrose we found a postive mass gain. As for 0.4M sucrose and higher molaritites we found a decrease in mass change. This was on behalf of there being enough solute added to the potato cylinders to cause movement of the water from the potatoes. Since water always moves from high water potential to low water potential. Meaning that there was a high water potential within the potatoes and a low potential of water outside of the potatoes. The mass changed because the water was removed from the potatoes, which decreased the mass. When 0.2M or less of sucrose was added there was not enough solute added to change the water potential.

1D: In this experiment, we found water potential.  We determined that if you multiply the ionization constant, molar concentration, pressure constant, and tempurature, then you get water potential.

1E: In this experiment we found that when we added the drops of 15% NaCl the onion cell shrunk in size. This is on behalf of plasmolysis. The 15% NaCl was a hypotonic solution. We learned in lab 1D that water potential always moves from high to low. So in this case the onion cell had more water potential then the NaCl solution. The added drops caused the water with in the cell to vacate, then shrinking the cell in size. When we added a lot of water to the onion the cell increased in size. The onion increased in size because after adding the NaCl the water potential with in the cell was less then the solution the cell was in, making for a hypertonic solution and a turid onion cell. Since water potential once again moves from high to low concentration, once the water that had a higher water potential was added the water poured into the onion cell until it reached equillibrium. If the onion cell had been placed in an isotonic solution the cell would have remained the same size since the solution that the cell was in would have the same water potential. So in summary, when NaCl was added to the onion the cell shrunk in size and lost water, after the water was added to the cell the onion gained water and increased in size.

Conclusion

1A:  Only molecules that are smaller than membrane pore size are able to diffuse through a selectively permeable membrane.  Molecules larger than the pore size of the membrane are unable to diffuse through a selectively permeable membrane.

1B: The more hypertonic the inside solution is, more water diffuses in. An isotonic solution will have little or no change in the amount of water inside.

1C: At 0.2M or above there isn't enough solute to cause a change in water potential. That's why from 0.4-1.0M they lost mass since there was enough solute to cause the water potential to change, causing water to leave the potato, then lowering the mass of the potatoes.

1D: In order to calculate water potential, you multiply ionization constant, molar concentration, pressure constant, an temperature in Kelvin.

1E: When adding NaCl to the onion cell the cell decreased in size, because it was in a hypotonic solution. When adding the water afterwards the onion cell increased in size because it was now in a hypertonic solution. This is on behalf of plasmolysis where water always moves from high to low concentration.

References

Campbell Biology, 9th Edition

Diffusion and Osmosis Lab, College Board

 

 

 

 

Grace's Restoration Reflection

I wasn't sure what to expect when I arrived at the Glacial Park Restoration Center. As soon as we got out of the bus and stepped into the building I witnessed a stuffed coyote. I thought to myself jokingly, "so is this what we're going to restoring?" I thought it was so funny, I had to take a picture with it. I was also surprised with one thing. The place had wi-fi. I honestly thought that in a this large field, 

there would not be any internet connection. It was a funny thought. 
    When we walked out of the building, we listened to the ecologist speaking about what we were going to do for the rest of the day and what the motto is at the center. He stated that the point of ecological restoration is to speed up the process of restoring what nature would restore in a much longer time span because humans are degrading faster the nature can reproduce. Therefore, we were there to help restore nature quicker. 
    After walking the a beautiful and mesmerizing field of ups and downs, we split up into groups, and I was in the group to remove invasive species in the forest. At first, I couldn't tell the difference between what was hurting the trees and what was helping it. I decided to just cut down everything but the big oak trees. I hope for the most part, that helped out the trees. Although it took a lot of physical labor,
 I had a lot of fun. When our group had finished, I 
truly saw the difference we made. Before the forest
 was very dull and the trees were barely getting 
sunlight. Now the forest was incredibly open and the
 sun was blazing. I felt really good. I felt like I really
 did something to help nature and the ecosystem. 
After taking some pictures, we moved on to the 
next site.

 


At this next stop, I saw a couple of shrubs planted in a field. Our job was to water them. They did not get water from nature in over a week, so it was our job to help speed up the process. Everyone got a gallon of water and found a dry plant to fill up. After this, we planted acorns. That was my favorite part. We learned about how squirrels find these acorns and disperse them around the forest. Without knowing, these two species shared a relationship. But for some reason, the squirrels did not come to this area, so oak trees were scarce. As our job to restore, we planted a thousand acorns in the field, one by one. I had to take my time digging up a hole and planting one specific acorn at a time, but I felt like I contributed more. I was just imagining that it would be my acorn that I planted to grow and become a huge oak tree. I really enjoyed it. 

 
After finishing up at this site, we were on our way to the bus. I could not stop thinking about how much I learned at this field trip. It was a give and take relationship. I was helping to restore nature, while nature was teaching me. I honestly felt great afterwards, and learned a lot from this experience!  

My Restore the Forest Team! =) 

Marley's Restoration Reflection

During this field trip, we got to experience restoration ecology. The point of ecological restoration is to speed up the process of restoring a degraded area. We went out in the field and removed invasive plants that weren't native to the area. The non native plants were taking over, and weren't allowing the native plants to grow. By removing these invasive plants, we made room for the native plants to grow. We then went on to water new, but native, trees that were brought into the area. We also spread seeds in the same area to help the native plants grow.

This field trip exposed me to something I had never heard of before, and I'm glad I got to experience it. I think restoration ecology is amazing because humans are slowly ruining the environment, and this helps to restore some of the damage. Some of the many benefits of restoration ecology include improving biodiversity, improving ecosystem productivity, improved water quality, and increased carbon storage. All of these things are included in restoring natural ecosystems. After experiencing everything during this trip, I would love to get more involved with restoration ecology.