Cell communication lab

Purpose: The purpose of this experiment is to study yeast cell communication.  We will study the differences between three different party dishes which will contain "a" type, alpha type, and mixed.  The purpose is to study the differences between the cells and the number of cells and how they act.

Introduction:  Yeast cells have the ability to reproduce sexually as well as asexually.  There are two types of yeast cells, alpha and "a" type.  If these two are isolated then they will reproduce asexually. When reproducing asexually they make a growth that is called a schmoo. In order for sexually reproduction to happen these two types must be mixed together.  When these two are mixed a haploid cell, asexual, quickly turns into a gamete cell, sexual.  Once they change they grow toward eachother finding a mate a reproduce.

Methods: Take three agar plates and label them alpha, "a", and mixed.  Then label your cultured tubes the same as it is appropriate. Add 2 ml of sterile water to each cultured tube then add some of each yeast type to the appropriate tubes. Take that solution and put 5 drops of each on their designated plates.  Then observe these under the microscope over fourt time periods.  Record how many cells are in each field as well as how many schmoos.

Purpose: The purpose of this experiment is to study yeast cell communication.  We will study the differences between three different party dishes which will contain "a" type, alpha type, and mixed.  The purpose is to study the differences between the cells and the number of cells and how they act.

Discussion: In our experiment we were trying to understand cell communication through yeast cells communicating.  In our experiment we used a cells and alpha cells separately and a mix.  When cells reproduce asexually they become budding haploids.  This can only happen with cells of the same type.  When there are a mix of a and alpha cells schmoos can be formed.  In our experiment we found that only schmoos were formed only in the mixed group because schmoos can only be formed when one cell sends out a specific chemical messenger to the opposite cell.  We also found in our experimental group that the cells can travel great distances in order to mate.  In our experimental group we place the different types of yeast on opposite sides of the petriadish and by the next day the yeast cells had formed schmoos.  

Conclusion: Cells of the same type form budding haploids. If you mix the different strains the cels will attempt to mate by forming schmoos. This is due to a signal sent out by each cell. If it senses an opposite strain nearby it signals to that strain. The strain receiving the signal will respond by forming a schmoo. Our experiment found that they can do this across large distances also.

Photosynthesis Lab

Purpose: in this lab we were trying to test the rate if absorb acne and transmitance of light in chloroplasts.  Chloroplasts only absorbs certain amounts of light and we tried to test the amount absorbed over time.  Our dependent variable was time and our independent variable

of our experiment was the rate of absorbance for each of our cuvettes

Introduction: Photosynthesis requires light and co2 to work and the biproducts of photosynthesis is o2 and glucose. If there is deprivation of co2 the rate of photosynthesis will drop.  This would only occur in a closed system.

Methods: First we took five cuvettes, one was filled with just water, two with unboiled chloroplasts, one with boiled chloroplasts, and one with no chloroplasts.  All of these had DPIP in it except for the first cuvettes which had just water. One of the cuvettes with unboiled chloroplast had aluminum rapped around it. We used the cuvette filled with water to calibrate the spectrometer, then we took initial readings of all the other cuvettes.  We then exposed the cuvettes to light for five minutes and took the readings, we repeated this two more times. We also used a large water sink bath to prevent heat from being a factor.

Graph and Charts: 

Discussion: The cuvette with just water, dpip, and the phosphate saw no increase in photosynthetic productivity due to the fact that there was no chloroplast to process the light. The cuvettes with chloroplast that was on ice saw a slow decrease in photosynthetic productivity. This was expected because chloroplasts use the light to create O2 through the light reactions of photosynthesis. While it was not natural light, some of the UV rays are still able to be used by the chloroplast for photosynthesis. The cuvette that was wrapped in aluminum foil, still put in front of the light, and still had chloroplasts and dpip saw a sharp decrease in photosynthetic production because there was really no light going in and so all the light was used up. While it was expected that the light from the lamp would provide enough UV for photosynthesis to occur (which it did) it was surprising that the rate of photosynthetic production declined in all of our cuvettes except for the one with no chloroplast, which saw no decrease or increase in photosynthetic productivity. This was because since there are no chloroplasts, nothing could be photo synthesized. The dpip would turn from blue to colorless if it was used by the chloroplast, which it was. It did not turn from blue to colorless in the cuvette with the boiled chloroplast becasue since the chloroplasts were denatured the dpip couldn't be reduced and used. 

Conclusion: How is it possible that there was no increase in absorbance? The chloroplasts need CO2 to restart the cycle. Over time in a closed container there was no CO2 being made. Only a lot of oxygen. The oxygen could not keep the rate of photosynthesis going so there was a decrease in light absorbance.

Sources: http://www.answers.com/Q/What_is_the_effect_of_boiling_chloroplasts_on_the_subsequent_reduction_of_DPIP

Charles Filipek

Cellular respiration

http://thezuluwarrior.blogspot.com/2014/11/cellular-respiration.html

This has been shared from my personal blog. sorry for the inconvenience Mr. filipek

Enzyme reaction rates

Purpose:  the purpose of the experiment was to test an enzymes ability to catalyze.  We saw the relationship between the amount of enzymes in a solution and the ability of the enzyme to catalyze.  The rate of which the enzyme catalyzes was what we were testing and the time that we left the solution catalyzing was the variable.  

Introduction:  certain enzymes increase the reaction rate of certain solutions by lowering the activation energy.  Certain substrates bind to the enzyme at the activation site.  The enzyme lowers the activation energy and this speeds up the reaction.  The reaction will continue to occur until all of the substrate is used up due to the fact that enzymes never get used up.  Enzymes are proteins so they are affected the same ways as proteins.  Proteins can be denatured when the pH or temperature of the solution is affected.  Once a protein is denatured the function will no longer be carried out so the enzyme will never catalyze.  

Methods:  we began with 10 mL of H2O2 in each cup.  We began the reaction by adding in 1mL of yeast that had enzymes that catalyzed with H2O2.  We timed each reaction and once time ended we added in 10 mL of H2SO4 to stop the reaction.  We did this with each time.  Once we did this we took a 5 mL sample and titrated KMnO4 into the solution until in turned purple.  We started with a baseline test that did not undergo a reaction.  Then we moved on to the other cups.  The KMnO4 reacted with the H2O2 until all of it was gone.  This meant that the our baseline number minus the amount of KMnO4 added to the solution was the amount of H2O2 used up in the reaction.  We recorded and then continued to the next cup.  

Data:

Graphs:

Discussion: We wanted to know how fast the enzyme could catalyze the reaction. The titration of the solution measured this and we found that the longer the reaction was allowed to work the less substrate was left. This is due in part to the enzyme working to lower activation energy and speed up the reaction. Also the optimal pH for the enzyme is very close to neutral. By adding an acid we denatured the protein which is what stopped the reaction from proceeding at its catalyze do speed and due to this the enzymes couldn't lower activation energy.  We are unsure why we have an outliers in our data.  For the most part the end result of our experiment (time 360 seconds) was what we were expecting but the in between didn't quite match up with what we were expecting.  Our numbers increased until 60 seconds then it decreased until 120 seconds then it increased again.  Besides these intervals of three data points the data fit what we were trying to understand.  The longer that an enzyme is left in a solution the less substrate there will be.  

Conclusion:  the longer an enzyme is left in a solution with substrate the less substrate there will be.  The enzyme will never be used up so the reaction will continue going.  For the most part we saw this in our experiment.  The amount of substrate continually decreased except for at two points.

Diffusion/Osmosis

1A
Introduction: Diffusion is the constant movement of solutions from high concentration to low concentration. Selectively permeable membranes can manipulate diffusion by allowing certain solutions through and not allow others.

Purpose: The purpose of this experiment is to observe how a selectively permeable membrane effects the rate of diffusion of different sized molecules.

Methods: first, we tested a 15% glucose/1% starch solution for the presence of glucose and then filled a dialysis bag with the solution.

From that we recorded the color. Then we filled up a beaker with distilled water and put 4ml of lugols solution in it, we then recorded that color. Then we took the dialysis bag, submerged it in the water, and waited for 30 minutes.

We then recorded the colors of the bag and the water after the time, we also tested for glucose and recorded the data.

Data:

Discussion: When we tested for glucose at the end, it was present in the dialysis bag, but not present in the water surrounding the bag. The glucose seemed unable to pass through the dialysis bag. However, the color of the glucose solution in the dialysis bag was different from the start. The original color of the dialysis bag's solution was clear, and the original color of the water with the lugols solution was orange. At the end of the experiment, the color of the solution in the dialysis bag was dark purple, and the water surrounding it was still orange. This shows that the lugols solution diffused into the dialysis bag to make a different color. The dialysis bag proved to have a selectively permeable membrane by slowing lugols solution in, while keeping the glucose in.

Conclusion: this experiment showed us an example of selectively permeable membranes. The membrane allowed lugols solution to diffuse in and out, but did not allow glucose to diffuse out.

1B
Purpose: the purpose of this lab was to observe osmosis of water molecules interacting with different molarities of sucrose. We did this by looking at the flow of water in or out of a semipermeable membrane.

Introduction: Osmosis is the diffusion of water across a membrane. The water will travel from a lower concentration of solute to a high concentration. Animal cells are always trying to maintain an isotonic state (equal concentrations of solute and solvent in and out of the cell). Osmosis moves water across the membrane to equal out the concentration of solvent. 

Methods: We started with six different dialysis bags. We added differing molarities of sucrose into each of the bags.

We weighed and recorded the initial mass of each bag. We then submerged the bags into cups that was filled with water. We then let the bags sit in the cups for half an hour. Once time was up we took the bags out of the cups and measured the final weight. We then compared the initial mass to the final and came up with a percent difference. We chose to use a percent difference rather than a mass difference due to the fact that the amount of solution in each bag was not uniform. Since the initial masses were all different percent difference is a more accurate measurement when comparing the data.

Data

( our group data)

	DISTLLED WATER	0.2 M	0.4M	0.6M	0.8M	1.0M
Group 1	0.45	2.38	5	5.86	6.06	7.93
Group 2	5.15	8.47	8.54	10.43	9.76	17.44
Group 3	0	4.33	8.61	14.41	12	10.05
Group 4	2.17	1.73	3.14	6.9	7.2	11.54
Group 5	0.8	6.8	3.6	6.9	5.4	8.3
Group 6	0.89	2.5	3.1	5.3	6	7.4
Group 7	11.4	9.8	10.7	12.3	13.7	13.4
Group 8	2.7	3	4.9	6.2	4.3	6.1
Group 9	0.63	2.14	4.85	8.19	8.6	2.17

Group 10	1.1	4.95	10.57	16.07	19.09	16.87
CLASS AVG:	2.01	4.61	6.3	7.88	9.21	9.69

(Class data)

Graphs and Charts: 

(Class graph)

Discussion: Our results for the most part were what we expected. Osmosis makes water
travel from high concentration from low concentration of solute to high. In this experiment the bags had a lower concentration water than their environment (except for our control). This means that we expected water to rush into the bags in order to reach an isotonic state. What we witnessed was very similar to what we expected. After we weighed each bag we saw that most of the bags increased in weight by a considerable amount. This meant that water was entering the bag, thus adding weight. We saw that the higher the molarities the greater the percentage weight increased. This means that the higher molarity bags allowed more water in. This makes sense since more water is needed to even out the concentrations. We did see a few inaccuracies in our experiment. Our control group was distilled water. For out control group we saw an increase of .625%. In theory this number should be 0% due to the fact that the bags contents are the same as the solution that it is put in, however this can be accounted for by the fact that we used distilled water in the dialysis bags, but sink water for the cup. The difference between the two can make a difference in the data. Our biggest outlier can be seen with our 1M bag. This bag should have the highest percent change due to the fact that it had the highest molarity. Instead we had a negative mass percentage. This means that the bag actually lost water. We do not have an absolute reason as to why this occurred. Our best speculation that such an unusual measurement would take place is that there could have been a hole in the bag. This would allow water to rush out of the bag at a faster rate than it entered in. If this were the case water it would make sense as to why the bag lost weight.

Conclusion: Our experiment meant to test the process of osmosis. We saw in our data that the concentration of solute has a direct correlation to the rate at which water passes through a semipermeable membrane. The higher the concentration the faster water moves.

1C
Purpose: The purpose of this experiment was to test the molarity of sucrose in potatoes. The experiment tested the effects of diffusion using an organic substance. We used the same idea of osmosis as we did in experiment 1B

Introduction: Potatoes are made out of all organic substances. Potatoes are made out of plant cells which means that they are primarily made out of starch and cellulose. Each potatoes piece contains a certain amount of sucrose in it.

Methods: We cut and measured twenty four pieces of potato.

We took an initial measure of four pieces at a time and then we put those four pieces into a cup each filled with water that had differing molarities of sucrose. We covered the cups with saran wrap then we let the cups sit overnight. The next day we uncovered the the cups and we weighed the potato pieces in the respective groups of four. 

Data:   

 class data)

(Our experiment was group 9)

Graphs and charts:

(class avg graph)

Discussion: The osmosis of water volumized the potatoes and added mass into them. The sucrose. Water potential is higher outside the cell than inside the cell then water will move from high to low. We wanted to know how much sucrose is in a potato. The graph shows the net change in weight of the potato in relation to the sucrose solution it was placed in. A potato with no change in weight would be at zero in the chart meaning it was placed in a solution of equilibrium. To fail safe our data we also used the entire class' experiments to avoid anomalies. The molarity of sucrose in a potato is approximately 2.28 M. This means if it was placed in a 2.28 M solution of sucrose then there would be no net change in the weight of the potato.

Conclusion: The potato has a sucrose concentration of approximately 2.28M as demonstrated by the class average graph.

Reference:  http://en.m.wikipedia.org/wiki/Molarity,   http://www.biologycorner.com/worksheets/labreport.html

1D

Purpose: The purpose for this part of the lab of the lab was to find the solute potential of the sucrose solution

Introduction: To find the solute potential, we used the equation Ys=-iCRT. The R is conastant because the solution was held in an open system. For temperature, we assumed a room temperature of 27 degrees celcius. And for molar concentration we found the solution to have a zero at 1.65 moles. Our ionization constant was 1.0, as sucrose doesn't dissolve in water. 

Methods: In order to find the molar concentration of the solution, we graphed out the molarity of the solutions as x and the percent increase in mass as y. When we drew the graph, the function crossed zero at about x=1.65. We already knew what R was, since the system was open. We knew what the ionization constant was, so the only thing left to find was the temperature. Since it had to be in kelvin, we added 27 degrees celcius to 273 degrees kelvin to get 300 degrees kelvin. We then plugged these value into the equation, 

                   Ys=-(1)(1.65)(1.0)(300)

                   Ys= -41.135

Discussion: Our results were normal and expected. We expected the solute potential to be negative, because anything other than pure water is negative. We also expected a high solute potential due to the molarity and the high temperature. 

Conclusion: After we plugged everything in, we found the solute potential to be -41.135

1E

Purpose:  the purpose of this experiment is to see the use of plasma lysis in plant cells.

Method:  you can see plasmolysis happen underneath the microscope.  Add salt water with the cell and you can see the water exit the cell.

Discussion:  plasmolysis is the exit of water from a plant cell.  Plasmolysis occurs when a plant cell is placed in a hypertonic solution.  The solution causes water exit the cell through diffusion.  This causes the cell to shrink and the cell with die due to lack of water.