Cool Mitosis Video

For all you youngin' who be strugglin' graspin' the stages of mitosis fear not! The four bros have made a video quickly going over each stage in a catchy tune

http://youtu.be/U1mIvhXUytw

Cool Yeast lab

The purpose of this lab was to conduct a study on cell communication. We did this by observing yeast mate and produce schmoos. Over several time intervals we observed the yeast both on a liquid base (sterile water) as well as on a dry base (dry Petri dish). These two surfaces were done to observe the effect they have on cell communication and their chemical signals. 

We labeled three test tubes as either alpha, A or mixed and put in the yeast cells respectively in each of the test tubes. Then we put two milliliters of sterile water in each test tubes to simulate wet environments. Then we added an equal amount of broth in each test tube in order to sustain their life overnight.  Then we left them in an incubator for 24 hours. After the 24 hours had elapsed we put samples of each test tube on a microscope slide and then counted the amount of yeast present.  

Below are the pictures we have taken of the yeast after 24 hours have elapsed:

Mixed (82 yeast are present)

Alpha (86)

A type (40 yeast were present)

Then we waited an additional five minutes for each of the test tubes. Below are the pictures of each test tube at the first time interval (5 minutes)

Mixed (39 were present)

Alpha (43 schmoos were present)

A type (95 schmoos were present)

Then we an additional five minutes and let the yeast culture. 

Mixed (231 schmoos were present)

Alpha (40 schmoos were present)

A type (5 schmoos were present)

 

As demonstrated by this lab, under the wet conditions, the A-type yeast alone went from 40 to 95 to 5 yeast cells. The alpha type went from 86 to 43 to 40 cells, and the culture of both A-type and alpha type cells went from 82 to 39 to 231 cells. Now in retrospect this data tells us very little because the yeast cells in each of the three cultures tended to vary up and down drastically. This error can be attributed to the fact that when we counted each yeast cell, we counted them from a field of view in the microscope, but not only were we onto always sampling the culture from the same exact place in the liquid, we also had no way of looking at the culture in the same place under the microscope. Theoretically, however, all three of the cultures should have increased in size. A type and alpha type should have reproduced asexually when separate, and when mixed, they should have reproduced sexually and moreso.

We also performed similar procedures to grow the yeast cultures under dry conditions. Three cultures were taken from the original test tubes (again A-type, alpha, and mixed) and allowed to dry onto an agar plate in a Petri dish. Then, as time progressed, we sampled the cultures and observed them under microscopes to determine how growth had occurred. Our results are pictured below.

At the beginning of the procedure, our dry cultures started out with the following numbers of cells.

Then at the first time interval, our dry cultures had the following numbers of cells.

At the last time interval, our dry cultures had these final numbers of cells.

As with the cultures grown under wet conditions, the ones grown under dry conditions should also have grown in a similar manner. That is, A-type and alpha type grow asexually when separated and sexually when together, and when they grow sexually, growth is greater than when the grow asexually.

Based on the data from both the wet and dry cultures, we can conclude that yeast cells grow best sexually under dry conditions. This can be attributed to the means by which yeast cells sexually reproduce.

For ease of description, Let's consider A-type and alpha type yeast cells to be like opposite genders, almost like male and female. Each gender releases a pheromone that physically attracts the opposite gender. It does this because it behaves as a ligand that stimulates the cell to change its chape and develop a shmoo that grows toward the location from which the opposite pheromone is coming. In order for the shmoo projection to grow out, the yeast cell must completely change its cytoskeletal structure, which reveals the true power of singalling molecules to really get work done. The shmoo is projected in the direction from which the opposite pheromone was produced, and the rest of the cell follows the projection toward the source. This allows the a and alpha type cells to grow toward each other, in order that they might sexually reproduce. Sexual reproduction is much more favorable for the yeast cells than asexual reproduction because it produces twice the amount of each type of cells. 

Bearing all this in mind, when we say that the yeast cells grow best under dry conditions, it's because the dry environment (as opposed to the liquid one) conducts these pheromones more effectively. And this kind of makes sense if we use an analogy. Colognes, perfumes, and body sprays were developed as a means to attract people. They behave in a similar manner as the yeast cells' pheromones, albeit certainly not so drastically, much to the chagrin of companies like Old Spice and Axe. If we were to spray the aerosol into an empty bucket of air and into a full bucket of water, we can probably guess that the dry bucket of air would conduct the scent better than the full, wet one.

Therefore, these yeast cells demonstrate the behavior of singalling moelecules and receptors and just how important and powerful they can be.

Cool Photosynthesis Lab

The main idea with chromotography paper is to be able to see the separation of pigments. In this case we are seeing the variety of pigments found in the chloroplast to see how it helps with photosynthesis. As seen below with the paper strip, we can see the wide variety of pigments being separated and then the pencil mark is there the solvent ends. The very bottom of the paper is a dark green color and as it progresses to the top it begins to get lighter and lighter. The dark green color can be concluded to be chloroplast-a which is the main pigment used to absorb every light except for green. That has a low Rf value compared to Carotene which has an Rf value of 1 because it flows along with the solvent. What factors can play a role in the distance travels can be based on the polarity of the pigment or even the intermolecular forces of the pigment. The more complex and stronger it in hydrogen bonds, then the harder it is to move along the strip. This concept is great to think about in term of thinking about why a plant needs off of these complex pigments in the chloroplast and how Rf and absorbance and relatively be related. It can be concluded that the lower the Rf, the higher the absorbance. The rest is to be seen in the following data and lab procedure. 

The relationship between absorbency and transmittance is important to understand photosynthesis. In order to calibrate the colorimeter we put regular tap water in a cuvette and zeroed the absorbance and transmittance values. This served as a baseline value for the rest of the solutions.  

This is a picture of the colorimeter that we used.

In order to find this relationship we put a 100% blue dye solution in cuvette and put it in a colorimeter to test the amount of light that absorbed and transmitted. 

This is a picture of the 100% blue dye solution for trial one. 

We then followed these same steps but instead of a 100% blue dye solution, we diluted it to 50%, 25%, 12.5%, 6.25%, 3.125% and again measured the absorbable and transmittance values. 

This is a data table of the values we have obtained. 

From this data table we can conclude that the values between absorbency and transmittance have an inverse relationship.

In the next part of the lab, we conducted trials with either boiled or unboiled chloroplasts (dead or living) and we controlled whether or not the solution was exposed to light or not. This was done in order to observe the effects of how changing these factors would effect the rate of photosynthesis. 

First we made five different test tubes with the following contents:

Test tube one with 1mL phosphate buffer, 4mL distilled water, and 3 drops of unboiled (alive) chloroplasts

Test tube two with 1 mL phosphate buffer, 3 mL distilled water, 1 mL DPIP, and 3 drops of unboiled (alive) chloroplasts

Test tube three with 1mL phosphate buffer, 3 mL distilled water, 1 mL DPIP, and 3 drops of unboiled (alive) chloroplasts 

Test tube four with 1 mL phosphate buffer, 3 mL distilled water, 1 mL DPIP, and 3 drops of boiled (dead) chloroplasts 

Test tube five with 1 mL phosphate buffer, 3 mL+3 drops of distilled water, and 1 mL DPIP

Then we poured the solutions into cuvettes for testing in ur colorimeter to test the change in their transmittance over time. 

Going from left to right, these are cuvettes  1 to 5 

The first cuvette was done in order to calibrate the colorimeter and set a baseline for comparison for the rest of the cuvettes. 

We exposed all the cuvettes to light using the 2 liter Erlenmeyer flask filled with water in front of a heat lamp so that heat wouldn't affect the chloroplasts, but only the light. To simulate dark reactions for cuvette two we wrapped the cuvette in aluminum foil. This prevented light from entering the cuvette. 

This is a picture of our heat lamp and Erlenmeyer flask set up. 

This is a data table of the values we have obtained. 

Unfortunately due to an error with the lab equipment, some of the data was not stored correctly and we could not retrieve it. 

However, in a later discussion of what our results should have been we found that all the cuvettes except the third one should have had steady transmittance. Cuvette three was the only one that should have increased transmittance since it had all the requirements for photosynthesis to occur. As photosynthesis progressed, DPIP would change from blue to colorless because the chloroplasts would reduce the DPIP. As DPIP changes from blue to colorless, the colorimeter perceives it as an increase in the solution's transmission of light. 

The function of DPIP in this experiment is to act as a primary electron acceptor. DPIP replaced P680 as an electron acceptor. The source of electrons for DPIP was the distilled water. 

The colorimeter measured the absorbence and transmittance of the solution in each cuvette.

The darkness doesn't allow for the light to excite the electrons so DPIP cannot be reduced and in turn cannot gain electrons. 

Boiling the chloroplasts kills them causing photosynthesis to not occur. Which means that DPIP cannot occur because there is nothing for it to accept. 

Live chloroplasts kept in the light will have photosynthesis occurring which means they give off electrons for DPIP to accept. As DPIP accepts electrons it's reduced and turns from blue to colorless, so it's transmittance increases. If there's no light, photosynthesis can't occur so there's no electrons for DPIP to accept so DPIP can't reduce which results in the color not changing, and so the transmittance doesn't change. 

The purpose of cuvette one was to serve as a calibration of the colorimeter and to tell the colorimeter what 100% transmittance looks like. 

Cuvette two deprived the chloroplasts of light. This shows us that deprivation of light hinders the rate of photosynthesis. 

Cuvette three contained all the necessary requirements for photosynthesis to occur, living chloroplasts and light. 

Cuvette four contained all the necessary requirements for photosynthesis to occur except living chloroplasts. This tells is that death of chloroplasts hinders photosynthesis. 

Cuvette five contained no chloroplasts which tells us that having no chloroplasts hinders photosynthesis.

Cuvette five specifically behaved as a double negative control to show that DPIP does not reduce on it's own. If it did reduce on its own, that that would reduce the reduction that photosynthesis contributed to DPIP. 

This labs shows us that photosynthesis occurs only when chloroplasts are alive and are exposed to light and transmittance and photosynthesis have a direct relationship. 

Cool Cellular Respiration Lab

In this lab, we measured the effect of various conditions (temperature and germination) on the cellular respiration of mung beans and peas. Cellular respiration occurs in living things according to the following formula: C6H12O6 + O2 = CO2 + H2O. Therefore, because O2 is in the reactants and CO2 is in the products, a decrease of the former and an increase of the latter indicates that cellular respiration is occurring. On the other hand, photosynthesis occurs in autotrophs according to the following formula: Light + CO2 + H2O = O2 + C6H12O6, so a decrease in CO2 and an increase in O2 indicates that photosynthesis is occurring.

In order to observe the effects of these various conditions on cellular respiration, we came up with several groups of peas and mung beans and used Vernier lab probes to create graphs of O2 and CO2 concentration. The groups were glass beads (used as a control group because no cellular respiration or photosynthesis should occur), non-germinating peas at room temperature, non-germinating mung beans at room temperature, germinating peas at room temperature, germinating mung beans at room temperature, germinating peas in ice water, and germinating mung beans in ice water.

We started things off with a control group for the experiment. This was done by placing glass beads in the container with the Verniers. The results show that there is no cellular respiration or photosynthesis going on since they are not living objects. The graph shows this by the straight constant line across the graph. There is an increasing slope but that is due to the Verniers adjusting to the conditions of the container. Overall, no increase or decrease in CO2 and O2 levels. 

In the first portion of the experiment, we tested the CO2 and O2 concentrations for the peas and mung beans at room temperature. 

Here are the results for the peas, with CO2 levels indicated by the red line and O2 indicated by the blue line.

And here are the results for the mung beans with O2 indicated by the red line and CO2 levels indicated by the blue line.

For both the peas and mung beans at room temperature, the O2 levels decreased and the CO2 levels increased, which means that for these plants at room temperature, cellular respiration was occurring. At what rate preceisely? For the peas, the CO2 concentration increased at a rate of .583 ppm/sec and the O2 decreased at a rate of .0002167%/sec. And for the mung beans, the CO2 concentration increased at a rate of .183 ppm/sec and the O2 decreased at a rate of 0.0002667%/sec. This tells us that cellulalar respiration occurred at a slightly faster rate in the peas than in the mung beans, and this makes sens, for the peas were larger and therefore have a greater surface area on which their stoma can open and close to facilitate the movement of CO2 and O2.

We then tested to see of the change in temperature of the peas and beans affected the levels of carbon dioxide and oxygen. First we soaked 25 peas in 100 milliliters of 9 degree celcius ice water  for ten minutes. 

Peas in ice water bath

Then we put the peas in the Vernier apparatus.

After ten minutes had elapsed this was how our data looked.

We then conducted the same procedures for mung beans. 

First by soaking 25 of them in the 9 degree celcius ice water for 10 minutes. 

Then placing them in the Vernier apparatus to measure the carbon dioxide and oxygen levels in the tank as 10 minutes elapse. 

Below is our data:

The conclusion that we can draw is that the chilled beans and peas both went through less cellular respiration. We can infer this because from the data we gathered from the probes, compared to the non chilled beans and peas earlier in this lab, the rate of carbon dioxide being produced is less and the amount of oxygen being used up is less. Carbon dioxide being one of the products of cellular respiration and oxygen being one of the compounds used up in cellular respiration. This explains how locations with colder climates experience less vegetation. In the case of no sunlight, cellular respiration kicks in, but as we have discovered in this lab, cellular respiration does not work well when subjected to colder environments. 

In this part of the experiment we tested dried mung beans and peas to test and see what type of reaction would occur. What we did was take 25 peas and 25 beans and put them in a container with O2 and CO2 sensors. We left them in there for ten minutes and then recorded the data and evaluated the graphs. 

In this picture we have the mung beans in a sealed container with the sensors.

Data we collected: 

These are graphs of the data that the sensors recoreded at room temperature

From this data we can tell that for dried mung beans the overall trend of O2 was decreasing and the overall trend of CO2 was increasing. In cellular respiration Oxygen is being absorbed and carbon dioxide is being released. The mung bean graph follows that trend and therefore we can conclude that the mung beans were preformed get cellular respiration.

In the peas graph the overall trend for O2 is increasing and the overall trend for CO2 is decreasing. This means that carbon dioxide is being absorbed and oxygen is being released. This pattern follows photosynthesis because carbon dioxide is absorbed and oxygen is released when water is split in order to feed electrons into the cytochrome complex.

Cool Enzyme Lab

Part 2B:

In this part of the experiment we determined the baseline amount of H2O2 present in a 1.5% solution. We didn't add any enzymes to this reaction because this initial value that we will get is essentially the control group and so we need to know how much H2O2 was initially present in the solution.

First, we took about 10ml of 1.5% H2O2 and poured it into an empty cup. Then we added 1ml of H2O as a replacement for the enzymes. We need the solution to remain uncatalzyed so that our control value can be as accurate as possible, and so that is why we did not add the enzyme into the solution. After that, we took about 10ml of H2SO4 and poured it into the solution and mixed it all together.

In the picture above we are using a pipette to put 10ml of H2SO4 into the solution.

After we mixed the solution, we took a 5ml sample of it and we slowly added KMnO4, with a burette, to the solution until a pink color was obtained and stayed pink. The amount of KMnO4 used is directly proportional to the amount of H2O2 that was in the solution. When the solution turns pink that means the amount KMnO4 is exactly the amount of H2O2 in the solution because there is enough of each to react with each other.

In the photo above we are using the burette to slowly add drops of KMnO4 to the solution until it turns pink. Before we start to add the KMnO4 we take the initial reading of the burette and after the solution turns pink we take a final reading. If we subtract final-initial we get the total amount of KMnO4 used, thus getting the amount of H2O2 in the solution.

The figure above shows our final and initial readings of the burette. After our calculations we concluded that there was 3.5ml of KMnO4 present and therefore found the amount of H2O2 that was in the solution.

In conclusion, we found that 3.5 ml is our baseline value and that is the value that the rest of our trials will be compared to.

Part 2C:

In this procedure, we are determining the rate of spontaneous conversions of H2O2 to H2O and O2 in an uncatalyzed reaction. To find the data needed, we first needed to take 15mL of H2O2 and let it sit uncovered for 24 hours. Then we repeat the steps seems in 2B. We took 10mL of the H2O2 into a new beaker and added 1 mL of H2O (instead of the enzyme solution) and 10 mL of H2SO4. We mixed it well and then added the KMnO4 titration. Surprisingly enough, it only took less than five drops of the KMnO4 to make the solution pink. 

The image above shows the data collected. It only took .1 mL of KMnO4 to make the solution pink and 3.4 mL of the H2O2 was spontaneously decomposed; 97.14% was decomposed in 24 hours. What does this mean? By letting the solution settle overnight for 24 hours, the majority was decomposed naturally and only leaving a small amount left. This is why it took only a few drops of KMnO4. The less amount of H2O2 you have, the less amount of KMnO4 you need since both need to have equal quantities to keep the solution pink. 

Part 2D:

For this section, our goal was to observe the effect of time on the enzyme-catalyzed reaction that converts hydrogen peroxide into water and oxygen. We used the base line assay from Part 2B as a sort of control group. This showed us the original concentration of hydrogen peroxide in the solutions if the reaction were not permitted to occur. For the rest of the reactions we used the following method. We measured out 7 quantities of 10 mL of 1.5% hydrogen peroxide (the substrate for our enzyme), added 1 mL of catalase (an enzyme in yeast) to each quantity of hydrogen peroxide, and allowed the catalyzed reactions to persist for varying amounts of time (10, 30, 60, 90, 120, 180 and 360 seconds), after which we would stop the reactions by adding 10 mL of sulfuric acid. The acid reduced the pH of the solution, which denatured the enzymes by changing their shapes and behavior, which meant that they were no longer capable of catalyzing the reaction. For each solution, we performed a titration with potassium permanganate, wherein we slowly added the titrate to the solution until it remained pink or brown, at which point the volume of titrate equalled the volume of hydrogen peroxide present in the solution. Our results are in the table below...

The result, as we expected, was that the longer we permitted the reactions to be catalyzed, the less hydrogen peroxide would be left, as more of it would have been converted to water and oxygen. Here's the same information again in a graph of amount of hydrogen peroxide (substrate) used up in the reaction versus time. Note that it is not the amount of hydrogen peroxide left over but the amount that was used up...

In this graph, the data looks a little rough, but it gets the point across: the longer a catalyzed reaction is allowed to persist, the less substrate is left.

All in all, this lab gave us the chance to observe enzymes in action and understand their behavior in reactions. Understanding them in a lab environment like this gave us a better idea of how they function in the context of our bodies to regulate the rate of the reactions that sustain us.

Cool Diffusion and Osmosis Lab

After
The overall point of this lab was to study the processes of diffusion osmosis in a model membrane system and investigate the effect of solute concentration on water potential as it relates to plant tissues.

Part 1A
In this part of the experiment we used dialysis tubing to represent a cell membrane and we filled it with 15ml of 15% glucose/1% starch. We then put the dialysis tubing into a 250ml beaker that was filled two thirds of the way with iodine. The purpose was to study the effects of diffusion of the cell membrane.

This is the dialysis tubing that we used to represent the cell membrane 

This is the 250 ml cup that we used to represent the outside of the cell 

We tested the 250 ml and the dialysis tubing for presence of glucose. We found that the dialysis tubing contained glucose and the iodine solution did not.


Then we put the dialysis tubing in the iodine solution for thiry minutes to allow diffusion to take place.

After thirty minutes we observed the changes in color of both the dialysis tubing and the iodine solution.

We then tested each part separately to see if it contained glucose. We found that both of the parts contained glucose.

Analysis
During the process of diffusion Iodine had to be entering the dialysis tube and glucose had to be leaving. This is proven because of the fact that the initial color of the dialysis bad was clear and the beaker was orange and at the end of the experiment the dialysis tubing was black and the iodine was a lighter orange. To quantify this experiment, we could have measured a more accurate way to measure the color change and put a value on the concentration of the iodine to water solution. The reason that we didn't account for the starch movement is because starch molecules are far to large to diffuse across a cell membrane.


Part 1B

Within the dialysis bag portion of the lab, we see that there is a direct relationship between mass percent change and the molarity of solution within the dialysis bag. To justify this, we place dialysis bags of different molarities of sucrose (and one of distilled water) into cups of water. The simple explanation comes to the system trying to get close to the net equilibrium state as possible. The bags illustrates a hypotonic solution where the amount (molarity) of water outside of the system is higher than within the solution itself. This causes water to reach equilibrium by passing through the membrane of the bag and balance out the molarity of water in and out of the system. The increase of water being introduced into the dialysis bag causes an increase of mass. The higher the molarity of sucrose in the dialysis bag, for example, the higher the amount of water needed to balance out the system. The purpose of calculating the mass percent change instead of the basic change in mass is to show the ratio of change. Two solutions with different molarities can have the same amount of mass change-as seen in 0.2M and 1.0M-but the ratios significantly differ. This whole concept is osmosis, the movement of water through a selectively permeable membrane. Now let's say we place the dialysis bags into cups filled with 0.4M of sucrose solution. Then what would happen is solutions lower than 0.4M would try it's best to dilute the 0.4M by letting the remaining water inside pass out of the dialysis bag. The mass would decrease and so will the molarity of the outside solution until it reaches equilibrium. This would make it a hypertonic solution. Solutions with molarities above 0.4M would have the  outside water pass into the membrane to dilute the solutions until it reaches 0.4M. This stays as a hypotonic solution. The concept of osmosis still applies.


Part 1C

       In this part of this experiment we were attempting to understand the concept of water potential and it's two components, pressure potential and solute potential. More specifically, we wanted to determine the water potential of potato cells. Before we head into a discussion of what exactly we did, let's establish some background information.

       Water potential is described in terms of pressure potential and solute potential by its equation Ψ=Ψ(p)+Ψ(s).  As mentioned before, water potential has two components, pressure potential and solute potential. Let's look at the former in greater detail.

       Water under standard atmospheric pressure is said to have a pressure potential of zero. As you increase the pressure of a system, you increase its pressure potential and by extension, its water potential. This is because under more pressure, water has the potential to move more and do more work. For example, let's say I have a syringe filled with pure water and a bucket of pure water. In this scenario, we have two reservoirs of water: one in the syringe and one in the bucket. Now, I stick the needle of the syringe in the bucket of water and push the plunger. In which direction does water travel? Out of the syringe and into the bucket. The reason for this is that by pushing the plunger, I'm increasing the pressure of the water inside the syringe. By increasing the pressure, I'm increasing the pressure potential and also the water potential to a number greater than 0. Meanwhile, the bucket of water exposed to atmospheric pressure still has a pressure potential of 0 and a water potential of 0. Water travels from areas of high water potential to areas of low water potential, so water will travel out of the syringe and into the bucket. Now what if I pulled the plunger back instead of pushing it forward? We'd have the same situation in reverse. By pulling the plunger, I'm decreasing the pressure in the syringe. Doing so decreases the pressure potential and the water potential to a number less than 0. Again, water will flow from an area of high water potential to an area of low water potential. Since the water in the bucket still has a water potential of 0, water will be sucked into the syringe. This example is performed under the assumption that both the water in the gun and the water in the bucket were pure, that is, there were no dissolved solutes and the solute potential was 0. When we start dealing with solutes in water, we have to account for them in addition to the pressure of the water.

       In most high school freshman biology courses, we learn that solutes and water naturally diffuse from areas of high concentration to areas of low concentration. This concept is still the same for water potential, we're just putting a mathematical spin on things. Just like we said that water at atmospheric pressure has a pressure potential of 0, water without any solutes in it has a solute potential of 0. As you add solutes to water, you decrease the solute potential and by extension, the water potential. Since you can't remove solutes from water that has no solutes, you'll never see a solute potential greater than 0. This makes sense. If you have a bucket of pure water and a bucket of water and sand, the pure water will move more. It has a greater potential to do work. Now we won't be dealing with things as big as sand. We'll typically be dealing with solutes on a much smaller scale, but the concept still holds. Let's take a look at an example. Let's say I have a classic U-tube. Everyone's favorite. The tube is at atmospheric pressure so that the pressure potential in the entire system is always 0; that way we can just isolate the difference that solute potential makes. In the left arm of the U tube is water with a few solutes in it; the solute potential is just slightly less than zero. In the other is water with a lot of solutes in it; the solute potential is extremely less than zero. It doesn't matter what solutes they are, just that they're there. Separating the two arms is a membrane permeable only to water; that is to say that the solutes can't move, only the water. As our years of biology experience will tell us, the system naturally wants to reach equilibrium. It wants to make the solute concentrations of both sides equal. If it can't move the solutes in order to achieve this, then it will move water. We should already know that, based on osmosis, water moves from areas of high concentrations to low concentrations, so water will move from the left to the right. From a water potential standpoint, we can arrive at the same conclusion. The left side has less solute than the right side. Therefore, its solute potential is greater (less negative). Therefore, the left side's water potential is greater than that of the right side, so again we can say that water will move from the left to the right.

Now that we've set straight some background information, we can move onto our process for determining the water potential of the potato cells. In order to do so, we placed identical slices of potato into solutions of varying concentrations of sucrose and observed the change in mass of each slice due to the movement of the sucrose solutions and water in to or out of the potato cells. Here are our findings in a graph of percent change in mass dependent on the sucrose concentrations of the solutions.

Note the second point, where the potato slice was placed in solution in which the solute (sucrose) concentration was 0.2 M. This potato slice had no change in mass. You know what that means? There was no net movement of water in to or out of the potato cells, so the water potential of the potato slice must have been the same as the water potential of the solution. Let's look at pressure potential. Since this was an open system (after all, none of this experiment was performed in an air tight chamber; we just had plastic cups), both the potato slices and the sucrose solutions were exposed to atmospheric pressure. Therefore, the pressure potentials of both were zero. If the water potential of the potato and the solution were the same and the pressure potentials of both were 0, then it follows that their solute potentials, and by extension, their solute (sucrose) concentrations were equal as well--0.2 M. In the next section, part 1D, we used this knowledge to determine what the potato's water potential was, but for right now, it will have to suffice knowing that the the potato's sucrose concentration was 0.2 M.


Part 1E
Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out of the cell and into a hypertonic solution surrounding the cell. The cell membrane pulls away from the cell wall and thus leads to the shrinking of the cytoplasm. The onion cells plasmolyze because the cell was placed in a hypertonic solution of NaCl. So as a result, the water wants to leave the cell and thus causes it to shrink. This experiment can be related to real life because in winter grass often dies near roads that have been salted to remove ice. The salt concentration out of the cell is greater out of the cell-creating a hypertonic solution. This causes water to move out of the grass cell causing plasmolysis and the cell shrivel out and die due to lack of water

Cool Milk Analysis Lab

The purpose of this lab was to determine whether or not the reported amount of protein on the container is accurate. As seen on the picture of the nutrition facts, the amount of protein in the carton is 8 grams.

In order to this, we used concentrated acetic acid to denature the proteins. Denaturing of proteins allows for the curdling of milk into a substance that can be filtered out. This substance is the protein in the milk. After filtering the solution, we would then mass it and that is the experimental value of protein in the skim milk. 

In the picture above, our male model, Than, is measuring out 15 milliliters of skim milk. 

We then added 30 mL of acetic acid to the milk, stirred it in, and let it sit for about 5 minutes to denature the proteins within.

Once the acetic acid denatured the proteins, they were separated from the rest of the milk and we could filter out the whey portion of the milk, leaving behind 0.72 grams of solid protein curds based on the following data we retrieved.

Data:

Mass of Total Milk (g): 15.12

Mass of Filter and Protein (g): 2.44

Mass of Filter (g): 1.72

Mass of Protein (g): .72

Expected Mass of Protein (g): .51

Percent Yield (%): 4.76%

Percent Error (%): 41%

The data above shows that we ended up getting results that seem to say that we filtered more protein than is advertised, but this is unlikely, as the company would probably prefer to say that have as much protein as possible. The most likely cause of our positive percent error was that we were not able to filter out all the filtrate successfully, so some of it would have remained in our filtered protein, thereby increasing its calculated mass. This makes our results rather invalid. What may have caused this was that we ran out of time to fully filter out the whey so it remained with the protein. We expected the calculated amount of protein to be less than what the milk carton reported, but our results said otherwise.

Therefore, we cannot draw a conclusion as to whether the company accurately reports the amount of protein.