1st Semester 89 - BIOLOGY JUNCTION

AP Sample Lab 2 Catalysis 2

Lab 2 Enzyme Catalysis

Introduction:

Enzymes are proteins produced by living cells. They are biochemical catalysts meaning they lower the activation energy needed for a biochemical reaction to occur. Because of enzyme activity, cells can carry out complex chemical activities at relatively low temperatures. The substrate is the substance acted upon in an enzyme-catalyzed reaction, and it can bind reversibly to the active site of the enzyme. The active site is the portion of the enzyme that interacts with the substrate so that any substrate that blocks or changes the shape of the active sit effects the activity of the enzyme. The result of this temporary union is a reduction in the amount of energy required to activate the reaction of the substrate molecule so that products are formed. The following equation demonstrates this process: E + S ↔ ES ↔ E + P Enzymes follow the law of mass reaction. Therefore, the enzyme is not changed in the reaction and can be recycled to break down additional substrate molecules.

Several factors can affect the action of an enzyme: salt concentration, pH of the environment, temperature, activations and inhibitors. If salt concentration is close to zero, the changed amino acid side chains of the enzyme molecules will attract one another. The enzyme will then denature and form an inactive precipitate. Denaturation occurs when excess heat destroys the tertiary structure of proteins. This usually occurs at 40 to 50º Celsius. If salt concentration is high, the normal interaction of charged groups will be blocked. An intermediate salt concentration is normally the optimum for enzyme activity. The salt concentration of blood and cytoplasm are good examples of intermediate concentrations. The pH scale is a logarithmic scale that measures the acidity or H+ concentration in a solution and runs from 0 to 14, with 0 being highest in acidity and 14 lowest. Amino acid side chains contain groups such as –COOH that readily gain or lose H+ ions. As the pH is lowered an enzyme will tend to gain H+ ions, disrupting the enzyme’s shape. If the pH is raised, the enzyme will lose H+ ions and eventually lose its active shape. Reactions usually perform optimally in neutral environments. Chemical reactions generally speed up as the temperature is raised. More of the reacting molecules have enough kinetic energy to undergo the reaction as the temperature increases. However, if the temperature goes above the temperature optimum, the conformation of the enzyme molecules is disrupted. An activator is a coenzyme that increases the rate of the reaction and can regulate how fast the enzyme acts. It also makes the active site a better fit for the substrate. An inhibitor has the same power of activator regulation but decrease the reaction rate. An inhibitor also reduces the number of S-S bridges and reacts with the side chains near activation sites, blocking them.

The enzyme used in this lab is catalase. It has four polypeptide chains that are each composed of more than 500 amino acids. One catalase function is to prevent the accumulation of toxic levels of hydrogen peroxide formed as a by-product of metabolic processes. Many oxidation reactions that occur in cells involve catalase. The following is the primary reaction catalyzed by catalase, the decomposition of hydrogen peroxide to form water and oxygen:

2 H2O2 → 2 H2O + O2 (gas) Without catalase this reaction occurs spontaneously but very slowly. Catalase speeds up the reaction notably.

The direction of an enzyme-catalyzed reaction is directly dependent on the concentration of enzyme, substrate, and product. For example, lots of substrate with a little product makes more product. Another example is lots of product with a little enzyme forms more substrate. Much can be learned about enzymes by studying the kinetics of enzyme-catalyzed reaction. It is possible to measure the amount of product formed, or the amount of substrate used, from the moment the reactants are brought together until the reaction has stopped.

Hypothesis:

Enzyme catalase, when working under optimum conditions, noticeably increases the rate of hydrogen peroxide decomposition.

Materials:

Exercise 2A

The materials needed for exercise 2A of the lab are: 30 mL of 1.5% (0.44 M) H2O2, a 50- mL glass beaker, 6 mL of freshly made catalase solution, a test tube, boiling water bath, 1 cm³ of liver, a knife for maceration, paper towels, safety goggles, lab apron, pencil, eraser, and paper to record results.

Exercise 2B

The materials needed for exercise 2B are: 10 mL of 1.5% H2O2, two clean glass beakers, 1 mL of H2O, 10 mL of H2SO4, a white sheet of paper, a 5 mL syringe, approximately 5 mL of KMnO4, paper, pencil, eraser, safety goggles, and lab aprons.

Exercise 2C

The materials needed for exercise 2C of the lab are: 20 mL of 1.5% H2O2, two glass beakers, 1 mL of H2O, 10 mL of H2SO4, a white sheet of paper, a 5 mL syringe, approximately 5 mL of KMnO4, paper, pencil, eraser, safety goggles, and lab aprons.

Exercise 2D

For this part of the experiment, the materials needed are 12 cups labeled 10, 30, 60, 120, 180, and 360 on two each, six cups labeled acid, 60 mL of 1.5% H2O2, a clean 50-mL beaker, 6 mL of catalase extract, two 5-mL syringes, KMnO4, a timer, paper, pencil, black marker, eraser, safety goggles, and lab aprons.

Methods:

Exercise 2A

Transfer 10 mL of 1.5% H2O2 into a 50-mL glass beaker and add 1 mL of freshly made catalase solution. Remember to keep the catalase solution on ice at all times. Record the results. Then transfer 5 mL of purified catalase extract to a test tube and place it in a boiling water bath for five minutes. Transfer 10 mL of 1.5% H2O2 into a 50-mL beaker and add 1 mL of the cooled, boiled catalase solution. Again record the results. To demonstrate the presence of catalase in living tissue, cut 1 cm of liver, macerate it, and transfer it into a 50-mL glass beaker containing 10 mL of 1.5% H2O2. Record these results.

Exercise 2B

Put 10 ml of 1.5% H2O2 into a clean glass beaker. Add 1 mL of H2O. Add 10 mL of H2SO4 (1.0 M) using extreme caution. Mix this solution well. Remove a 5 mL sample and place it into another beaker. Assay for the amount of H2O2 as follows. Place the beaker containing the sample over white paper. Use a 5-mL syringe to add KMnO4 a drop at a time to the solution until a persistent pink or brown color is obtained. Remember to gently swirl the solution after adding each drop. Record all results. Check with another group before proceeding to see that results are similar.

Exercise 2C

To determine the rate of spontaneous conversion of H2O2 to H2O and O2 in an uncatalyzed reaction, put about 20 mL of 1.5% H2O2 in a beaker. Store it uncovered at room temperature for approximately 24 hours. Repeat the steps from Exercise 2B, using the uncatalyzed H2O2, to determine the proportional amount H2O2 of remaining after 24 hours. Record the results.

Exercise 2D

If a day or more has passed since Exercise B was performed, it is necessary to reestablish the baseline. Repeat the assay and record the results. Compare with other groups to check that results are similar. To determine the course of an enzymatic reaction, how much substrate is disappearing over time must be measured. First, set up the cups with the times and the word acid up. Add 10 mL of H2SO4 to each of the cups marked acid. Then put 10 mL of 1.5% H2O2 into the cup marked 10 sec. Add 1 mL of catalase extract to this cup. Swirl gently for 10 seconds. (Calculate time using the timer for accuracy.) At 10 seconds, add the contents of one of the acid filled cups. Remove 5 mL and place in the second cup marked 10 sec. Assay the 5-mL sample by adding KMnO4 a drop at a time until the solution obtains a pink or brown color. Repeat the above steps except allow the reactions to proceed for 30, 60, 120, 180, and 360 seconds, respectively. Use the times’ corresponding, marked cups. Record all results and observations.

Results:

Table 1: Test of Catalysis Activity

Experiment	Observations
Hydrogen Peroxide + Fresh Catalase	Bubbling in solution with the release of oxygen.
Hydrogen Peroxide + Boiled Catalase	No reaction occurred.
Hydrogen Peroxide + Liver	Much bubbling in solution with the release of O2.

Table 2: Establishing a Baseline #1

Baseline Calculations (syringe contains KMnO4)	Readings
Final Reading of Syringe	1.2 mL
Initial Reading of Syringe	5.0 mL
Baseline	3.8

Table 3: Uncatalyzed H2O2 Decomposition

(Syringes Contain KMnO4)	Results
Final Reading of Syringe	1.3 mL
Initial Reading of Syringe	5.0 mL
Amount of H2O2 Spontaneously Decomposed	3.7 mL
Percent of H2O2 Spontaneously Decomposed in 24 Hours	94.3%

Table 4: Establishing a Baseline #2

Baseline Calculations (syringe contains KMnO4)	Readings
Final Reading of Syringe	1.5 mL
Initial Reading of Syringe	5.0 mL
Baseline	3.5

Table 5: Time-Course Determination

Potassium Permanganate (mL)	Time in Seconds
	10	30	60	120	180	360
Baseline	3.5	3.5	3.5	3.5	3.5	3.5
Final Reading	1.3	1.6	1.8	2.0	2,4	2.7
Initial Reading	5.0	5.0	5.0	5.0	5.0	5.0
Amount of KMnO4 Consumed	3.7	3.4	3.2	3.0	2.6	2.3
Amount of H2O2 Used	0.2	0.1	0.3	0.5	0.9	1.2

Effect of Time on the Amount of H2O2 Remaining after an Enzyme Catalyzed Reaction

Exercise 2A:

1.a. What is the enzyme in this reaction? The enzyme in this reaction is the catalase solution.

1.b. What is the substrate in this reaction? The substrate is hydrogen peroxide.

1.c. What are the products in this reaction? The products are water and oxygen gas.

1.d. How could you show that the gas evolved is oxygen? Referring to the equation 2H2O2 + Catalase solution→H2O + O2, the only gas released is oxygen.

2. How does the reaction compare to the one using the unboiled catalase? Explain the reason for this difference. With the boiled catalase, there was no sign of bubbling because the catalase was denatured by the heat and caused no reaction.

3.a. What do you observe? I observe quite a bit of gas being released from the solution.

3.b. What do you think would happen if the liver were boiled before being added to the hydrogen peroxide? I think that no signs of a reaction occurring would be shown. The catalase that occurs naturally within the liver would have been denatured.

4. From the formula described earlier recall that rate = G y/G x . Determine the initial rate of the reaction and the rates between each of the time points. Record the rates in the table below.

Time Intervals (seconds)
	Initial 0-10	10-30	30-60	60-120	120-180	180-360
Rates	37/100	-3/200	-1/150	-1/300	-1/150	-1/600

5. When is the rate the highest? Explain why. The rate is the highest in the first ten seconds because the rate decreases as the concentration of the catalase decreases over time.

6. When is the rate the lowest? For what reason is the rate low? The rate is lowest during the last time period of 360 seconds because the most time has passed. The catalase concentration has been reduced and the product amount has increased, blocking the enzymes from reacting with the hydrogen peroxide.

7. Explain the inhibiting effect of sulfuric acid on the function of the catalysis. Relate this to enzyme structure and chemistry. The sulfuric acid’s high concentration of H+ ions gives the acid a low pH. Because enzymes can only function in the pH range of six to eight, the addition of an acidic solution denatures the enzyme, stopping the reaction.

8. Predict the effect of lowering the temperature would have on the rate of the enzyme activity. Explain your prediction. Enzymes generally only work at the between the temperatures of forty and fifty degrees Celsius. Lowering the temperature would slow the reaction until the enzyme is denatured and no longer able to react.

9. Design a controlled experiment to test the effect of varying pH, temperature, or enzyme concentration.

Part One (the effects of a strong acid on enzyme activity): Add 10 mL of 1.5-% hydrogen peroxide to a 50-mL beaker, and add 1 mL of catalase solution. Mix well and then add 1 mL of (0.5 M) HCl to the beaker. Observe the reaction and record the results.

Part Two (the effects of a neutral solution on enzyme activity): Add 10 mL of 1.5-% hydrogen peroxide to a 50-mL beaker, and add 1 mL of catalase solution. Mix well and then add 1 mL of pure water with a pH of 7.0. Observe the reaction and record the results.

Part Three (the effects of a strong base on enzyme activity): Add 10 mL of 1.5-% hydrogen peroxide to a 50-mL beaker, and add 1 mL of catalase solution. Mix well and then add 1 mL of (0.5 M) NaOH to the beaker. Observe the reaction and record the results.

Error Analysis:

Several errors could have occurred throughout the experiment. Miscalculations involving numbers and amounts of solutions would have a severe effect upon the results. Mathematical errors may also have of occurred. When the catalase arrived, it had melted. Because it is to remain on ice at all times, this may have caused errors. The age of the hydrogen peroxide effected results. For example, when calculating the percent of hydrogen peroxide spontaneously decomposed after 24 hours, new hydrogen peroxide yielded a much higher percentage than the aged hydrogen peroxide. Errors occur in every experiment and that is why is it is necessary to repeat an experiment several times for the most accurate results.

Discussion and Conclusion:

Catalase, or enzymes, drastically increases the rate of hydrogen peroxide decomposition. This lab shows how catalase added to hydrogen peroxide leads to the release of oxygen, boiled catalase is denatured, and the presence of catalase in living things can lead to the breaking down of hydrogen peroxide in the body. In the lab it was shown that the natural decomposition hydrogen peroxide is slower than decomposition taking place with the addition of enzymes. If hydrogen peroxide was required to decompose naturally, life could not survive. The addition of catalase increases this decomposition rate allowing life to continue.

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AP Sample Lab 12 Dissolved Oxygen

Dissolved Oxygen and Primary Aquatic Productivity
Laboratory 12

Introduction

Dissolved oxygen levels are an extremely important factor in determining the quality of an aquatic environment. Dissolved oxygen is necessary for the metabolic processes of almost every organism.

Terrestrial environments hold over 95% more oxygen than aquatic environments. Oxygen levels in aquatic environments are very vulnerable to even the slightest change. Oxygen must be constantly be replenished from the atmosphere and from photosynthesis. There are several factors that effect the dissolved oxygen levels in aquatic environments.

Temperature is inversely proportional to the amount of dissolved oxygen in water. As temperature rises, dissolved oxygen levels decrease.

Wind allows oxygen to be mixed into the water at the surface. Windless nights can cause lethal oxygen depletions in aquatic environments.

Turbulence also increases the mixture of oxygen and water at the surface. This turbulence is caused by obstacles, such as rocks, fallen logs, and water falls, and can cause extreme variations in oxygen levels throughout the course of a stream.

The Trophic State is the amount of nutrients in the water. There are two classifications: oligotrophic and eutrophic. Oligotrophic lakes are oxygen rich, but generally nutrient poor. They are clearer and deeper than eutrophic lakes and are younger. Oxygen levels are constant. Eutrophic lakes are more shallow and nutrient rich. The oxygen levels constantly fluctuate from high to low.

Primary production is the energy accumulated by plants since it is the first and basic form of energy storage. The flow of energy through a community begins with photosynthesis. All of the sun’s energy that is used is termed gross primary production. The energy remaining after respiration and stored as organic matter is the net primary production, or growth. The equation for photosynthesis is as follows:

12H2O + 6CO2 → C6H12O6 + 6O2 + 6H2O

There are two ways to measure primary production, the oxygen method and the carbon dioxide method. The oxygen method uses a dark and light bottle to compare the amount of oxygen produced in photosynthesis and used in respiration. Respiration rate is determined by subtracting the dark bottle from the initial bottle. The carbon dioxide method places a transparent plastic bag over one sample and a dark plastic bag over the other. Each bottle is set up so that air is drawn through the enclosure and passes over carbon dioxide-absorbent material. The amount of carbon under the dark bag is respiration, while the amount of carbon under the transparent bag is the amount of photosynthesis minus the amount of respiration.

There are three main gases dissolved in aquatic environments: nitrogen, oxygen, and carbon dioxide. Most gases obey Henry’s law, which says that at a constant temperature, the amount of gas absorbed by a given volume of liquid is proportional to the pressure in the atmosphere that the gas exerts.

c = K ×p

c = Concentration of the gas that is absorbed

K = Solubility factor

p = Partial pressure of the gas

Altitude may affect the p value of the equation. Higher altitudes decrease the solubility of gases in water. Temperature also has an affect, as temperature rises, solubility decreases. Salinity, the occurrence of various minerals in solution, also lowers the solubility of gases in water.

The method used to determine the amount of dissolved oxygen in the water is the Winkler titrametric method. It involves a series of chemical reactions which ends with a quantity of free iodine equal to the amount of oxygen in the sample. The iodine is then titrated with thiosulfate to find this quantity.

Hypothesis

The temperature and amount of light an aquatic environment receives greatly affects the dissolved oxygen levels, along with the amount of primary aquatic productivity.

Materials

Measurement of Dissolved Oxygen

This part of the lab required a sample bottle of water from a natural source, a BOD bottle, thermometer, mangonous sulfate, alkaline iodide, thiosulfate, a 2-mL pipette, sulfuric acid, a 20-mL sample cup, a white piece of paper, starch solution, and a nomograph.

Measurement of Primary Productivity

Part B required a sample bottle of water from a natural source, 7 BOD bottles, aluminum foil, 17 cloth screens, rubber bands, a light, thermometer, concavity slides, light microscope, mangonous sulfate, alkaline iodide, thiosulfate, a 2-mL pipette, sulfuric acid, a 20-mL sample cup, a white piece of paper, starch solution, and a nomograph.

Productivity Simulation

This section required pencil, paper, calculator, and graph paper.

Methods

Measurement of Dissolved Oxygen

The sample bottle was filled completely so that there were no air bubbles in the bottle. The sample bottle was left in the refrigerator until it reached 5° C. A BOD bottle was filled with the sample water until it contained no air bubbles.

Eight drops of mangonous sulfate were added to the bottle. Next, eight drops of alkaline iodide was added and the precipitate manganous hydroxide was formed. The bottle was inverted several times and then allowed to settle until the precipitate was below the shoulders of the bottle. While the solution was settling, a 2mL pipette was filled with thiosulfate. A scoop of sulfuric acid was added, and the bottle was inverted until all of the precipitate dissolved. The sample turned a clear yellow.

20mL of the sample were poured into the sample cup. The cup was placed on a white sheet of paper so that the color changes could be observed. 8 drops of starch solution were added to the sample, making it turn purple. The sample was then titrated with the thiosulfate. One drop of the titrant was added at a time until the color changed to a pale yellow color.

A nomograph was used to determine the percent saturation of dissolved oxygen in the sample.

Measurement of Primary Productivity

A second sample bottle was filled from a natural source making sure there were no air bubbles. Seven BOD bottles were filled completely with the sample with no air bubbles. The first bottle was labeled #1-Initial. The second bottle served as the dark bottle and was labeled #2-Dark. The other five bottles were labeled according to the light intensity: #3-100%, #4-65%, #5-25%, #6-10%, and #7-2%.

Bottle #2 was wrapped completely in aluminum foil so that it received no light. The other five bottles were wrapped in screens to produce the desired light intensity. Bottle #3 had no screens, bottle #4 had 1 screen, bottle #5 had 3 screens, bottle #6 had 5 screens, and bottle #7 had 8 screens. The screens were held in place with rubber bands. Bottles #2-7 were placed under a light source and left overnight.

Bottle #1 was fixed by following the Winkler method. Eight drops of mangonous sulfate were added to the bottle. Next, eight drops of alkaline iodide was added and the precipitate manganous hydroxide was formed. The bottle was inverted several times and then allowed to settle until the precipitate was below the shoulders of the bottle. A scoop of sulfuric acid was added, and the bottle was inverted until all of the precipitate dissolved. The sample turned a clear yellow. It was left at room temperature until the other samples were processed.

A wet mount was observed under a light source, so that the different organisms present could be identified.

The next day, bottles #2-7 were fixed by following the same method used on Bottle #1. The dissolved oxygen levels were determined in each of the seven bottles by titrating. 20mL of the sample were poured into the sample cup. The cup was placed on a white sheet of paper so that the color changes could be observed. 8 drops of starch solution were added to the sample, making it turn purple. The sample was then titrated with the thiosulfate. One drop of the titrant was added at a time until the color changed to a pale yellow color.

Productivity Simulation

The respiration data from Part B was converted to carbon productivity. The data was graphed with comparison to water depths.

Results

A. Measurement of Dissolved Oxygen

Table 1

Dissolved Oxygen Concentration

Temperature

Dissolved Oxygen (mg/l)

% Dissolved Oxygen

5° C

2.0 mg/l

16%

21.5° C

1.28 mg/l

19%

How does temperature affect the solubility of oxygen in water?

As temperature goes up the solubility of oxygen in water goes down. They are inversely proportional.

How does salinity affect the solubility of oxygen in water?

The occurrence of various minerals in solution lowers the solubility of oxygen in water.

Would you expect to find a higher dissolved oxygen content in a body of water in winter or summer?

Oxygen levels would be higher in the winter because the solubility of oxygen in water is higher at lower temperatures.

List and discuss three factors that could influence the dissolved oxygen concentration of a body of water.

Temperature-As temperature goes up solubility goes down.

Pressure- As pressure decreases solubility decreases. Pressure is directly affected by altitude

Salinity-The occurrence of various minerals in solution lowers the solubility of oxygen in water.

Do you think it would be wise to stock a pond with game fish if it had a dissolved oxygen content of 3ppm? Why or why not?

It would not be wise to stock a pond with an oxygen level of 3ppm with game fish because their optimal levels range from 8 to 15ppm. A concentration of dissolved oxygen less than 4ppm is stressful to most forms of aquatic life.

B. Measurement of Primary Productivity

Respiration Rate = 4.6 ml O2/l

Table 3

Gross and Net Productivity/ Respiration Rate

Percent Light	Dissolved Oxygen	Gross Productivity	Net Productivity	Gross Productivity (mg C/m3)
Initial	9.2 ml O2/l	NA	NA	NA
Dark	4.6 ml O2/l	NA	NA	NA
100%	6.4 ml O2/l	1.8 ml O2/l	-2.8 ml O2/hr	0.965 mg C/m3
65%	3.8 ml O2/l	-0.8 ml O2/l	-5.4 ml O2/hr	-0.429 mg C/m3
25%	4.5 ml O2/l	-0.1 ml O2/l	-4.7 ml O2/hr	-0.054 mg C/m3
10%	3.7 ml O2/l	-0.9 ml O2/l	-5.5 ml O2/hr	-0.482 mg C/m3
2%	4.0 ml O2/l	-0.6 ml O2/l	-5.2 ml O2/hr	-0.322 mg C/m3

Were any of the samples light limited? Why?

Each sample was given a certain amount of light by the use of aluminum foil and screen. Bottle #2 received no light, because it was covered with aluminum foil. Bottles #3-7 had varying numbers of screen ranging from 100% to 2% light intensity.

Productivity Simulation

Based on your analysis, which lake is more productive?

Lake 2 would be more productive because there is more oxygen available in the lower layers than in Lake 1.

What is used as the basis for measuring primary productivity?

Primary productivity is measured by the amount of dissolved oxygen available in the water. This shows the amount of oxygen produced by photosynthesis and the amount used by respiration.

Error Analysis

The Part A experiment was affected mainly by human error and inexperience with the Winkler method. The sample may have been over exposed to the air or the temperature may have changed before the fixing procedure was finished.

The original Part B experiment performed was unsuccessful. There were substantially more decomposing bacteria than photosynthetic organisms in the water sample use. The initial dissolved oxygen level was only 0.84 causing the other samples to have little or no oxygen. The amount of oxygen was so low that it was unable to form the free iodine and could not be titrated. This left no quantifiable data to use in graphs and tables.

Discussion and Conclusion

Temperature is inversely proportional to the solubility of gases in water. As temperature rose the dissolved oxygen levels should have decreased. This was qualified in the data obtained from this experiment, as the 5° C water sample measured 2.0 mg/l and the 21.5° C sample measured 1.28 mg/l. The percent saturation showed that even though the 5° C sample contained more oxygen it was still less saturated than the 21.5° C sample.

Part B of the lab was used to measure dissolved oxygen concentration, gross and net productivity, and respiration rate of the water samples. It also demonstrated the effect of light and nutrients on photosynthesis. In aquatic environments oxygen production and oxygen usage must be balanced to prevent anoxia. In the original experiment this balance was interrupted by the limiting of light by screens and aluminum foil. The amount of respiration in all of the bottles exceeded the amount of photosynthesis occurring. This was due to the types of organisms present in the sample, which was mainly decomposing bacteria and protozoan. The experiment was correct in its methods however the data received was not quantifiable. This absence of sufficient oxygen in the water samples is an indicator of poor water quality, which may require further investigation. Excess pollution or dumping of wastes into the water sample is a suspected cause of the poor water quality.

The data used in this report shows that as more light was limited, there was less dissolved oxygen present in the water. This is caused because photosynthesis cannot occur without sufficient light.

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AP Sample 6 Lab 5 – Cellular Respiration

Lab 6 Cellular Respiration

Introduction

Cellular respiration is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. Enzyme mediated reactions are required. The equation for cellular respiration is:

C6H12O6 + 6 O2 à 6 CO2 + 6 H2O + 686 kilocalories of energy/mole of glucose oxidized

Several different measures can be taken from this equation. The consumption of oxygen, which will tell you how many moles of oxygen are consumed during cellular respiration. That is what was measured in this lab. The production of CO2 can also be measured. And of course the release of energy can be measured. Cellular respiration is a catabolic pathway and the mitochondria houses most of the metabolic equipment for cellular respiration. It will break down glucose in what we call an exergonic reaction. Like previously said, the consumption of oxygen molecules will be measured in a gas form. One must know the physical laws of gases when working with them. The laws are summarized by the following equation.

PV=Nrt

Where:

P stands for the pressure of the gas

V is the volume of the gas

n is the number of molecules of gas

R is the gas constant (fixed value)

T is the temperature of the gas ( in K° )

The CO2 produced during cellular respiration will be removed by potassium hydroxide (KOH) and will form a solid potassium carbonate (K2CO3) when the following reaction occurs: CO2 + 2 KOH à K2CO3+ H2O

Since the CO2 is removed, the change in the volume of gas in the respirometer will be directly related to the amount of oxygen consumed. If the water temp and volume stay constant then the water will move toward the region of lower pressure. During respiration, oxygen will be consumed and its volume will be reduced because the CO2 is being converted to a solid. The net result is a decrease in gas volume in the tube and a decrease in pressure of the tube. The vial with beads will detect any atmospheric changes.

Hypothesis

Several different things will affect the rate of O2 consumption. The non germinating peas will have a lower rate than the germinating peas and the coldness of the water will slow the rates.

Materials

The materials used for this lab were: a 100 mL graduated cylinder, 6vials,germinating peas, dry peas, glass beads, 2 water baths, absorbent cotton and non-absorbent cotton, weights, KOH, water, stoppers, pipettes, rubber bands, masking tape, glue, thermometer, ice, a pencil, and paper.

Methods

Set up a 25° C and a 10° C water bath. Ice may be used to obtain 10° C.

Respirometer 1:Obtain a 100 mL graduated cylinder and fill it with 50 mL of H2O.

Drop in 25 germinating peas. Determine the amount of water displaced. Pea volume =11 mL. Take peas out and place on paper towel.

Respirometer 2: refill cylinder with 50 mL of H2O. Drop 25 dry peas into the cylinder. Add glass beads to obtain the same volume that you got in respirometer 1. Remove peas and beads to a paper towel.

Respirometer 3: Add 50 mL of water to the cylinder. Put only beads in to get an equivalent volume to the first 2 respirometers. Put on paper towel when finished. Repeat respirometer 1 steps for respirometer 4. And 2 for 5. And 3 for 6. Listen to your teacher on how and where to set up the respirometers. Now fill your vials with the required items shown in the table and in figure 5.1. Seal the vials after your items have been put in to stop any gas or water leaks. Place a weighted collar onto the bottom of your vials so they will stay submerged in the water baths. During equilibration use masking tape attached to each side of the water baths to hold the respirometers out of water for 7 minutes. Vials 1-3 should be in the 25° C water bath and vials 4-6 should be in the 10° C water bath. Finally submerge totally the respirometers and let them equilibrate for 3 more minutes. Read the water line where the oxygen is and record in intervals of 5 minutes all the way up to 25 minutes. Record in table 5.1.

Results

Table 5.1: Measurement of O2 Consumption by Soaked and Dry Pea Seeds at Room Temperature and 10° C Using Volumetric Methods

Beads Alone

Germinating Peas

Dry Peas and Beads

Reading at time X

Diff.

Reading at time X

Diff.

Corrected Diff.

Reading at time X

Diff.

Corrected Diff.

Initial-0

1.35

1.62

1.32

0-5

1.33

.02

1.20

.42

1.32

.02

5-10

1.33

.02

1.12

.50

.48

1.3

.02

10-15

1.32

.03

1.02

.60

.57

1.29

.03

15-20

1.32

.03

.92

.67

1.3

.02

.01

Initial-0

1.48

1.37

1.46

0-5

1.48

1.15

.22

1.45

.01

.01

5-10

1.45

.03

.98

.39

.36

1.44

.02

.01

10-15

1.43

.05

.84

.53

.48

1.43

.03

.02

15-20

1.41

.07

.70

.67

1.41

.05

.02

In this activity, you are investigating both the effects of germination versus non-germination and warm temperature versus cold temperature on respiration rate. Identify the hypothesis being tested on this activity.
The nongerminating peas will have a slower rate of respiration than the germinating peas and the coldness of the water will slow down the rate as it gets colder.

This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each.
The three controls are the beads in one vial controlling the barometric pressure, the KOH keeps equality in the consumption of CO2, and the time intervals give each vial the same amount of time so the results will not be affected.

Describe and explain the relationship between the amount of oxygen consumed and time.
The relationship was pretty constant, there may have been a gradual rising in O2 consumption.

Condition

Calculations

Rate in mL O2/ minute

Germinating Peas/ 10 oC

(1.62-.92)

.035

Germinating Peas/ 20 oC

(1.37-.7)

.0335

Dry Peas/ 10 oC

(1.32-1.30)

.001

Dry Peas/ 20 oC

(1.46-1.41)

.0025

Why is it necessary to correct the readings from the peas with the readings from the beads?
The beads were just a control, experiencing no gas change.

Explain the effects of germination (versus non-germination) on pea seed respiration.
The germinating seeds had a higher metabolic rate and therefore consumed more oxygen than the nongerminating.

Above is a sample graph of possible data obtained for oxygen consumption by germinating peas up to about 8 oC. Draw in predicted results through 45 oC. Explain your prediction.
Once the temperature gets above about 30 degrees C, the enzymes will denature and that will be the end of respiration.

What is the purpose of KOH in this experiment?
The KOH will take the CO2 and turn it to a precipitant at the bottom of the vial and it will have no affect on the O2 readings.

Why did the vial have to be completely sealed under the stopper?
The vial had to be sealed or gas would leak out and water could leak in and affect the results.

If you used the same experimental design to compare the rates of respiration of a 35g mammal at 10 oC, what results would you expect? Explain your reasoning.
Respiration would be higher in the mammal because they are warm-blooded.

If respiration in a small mammal were studied at both room temperature (21 oC) and 10 oC, what results would you predict? Explain your reasoning.
The rate of respiration would be higher in the 21-degree bath because the mammal would perform better when its body was more comfortable.

Explain why water moved into the respirometer pipettes.
The water moved in the vial because it was fully submerged in water but it came to a stop when it met the oxygen coming out of the vial.

14. Design an experiment to examine the rates of cellular respiration in peas that have been germinating for 0, 24, 48, and 72 hours. What results would you expect? Why?
You could put peas in vials each from a time interval above. You would have a vial with just started germinating peas, one with 24 hour germinating peas, another with 48 hour peas, and the last with 72 hour peas. Place them in a room temp water bath. Take readings at intervals of 5 min up to 20 min. The 72-hour peas should have more O2 consumption because they will use more oxygen because they have been germinating the longest. The just started germinating peas would use the least O2 because they haven’t been germinating vary long. The other two will be in the middle of the “just started peas” and the “72 hour peas”.

Error Analysis

Many errors could have been made in this lab. There could have been miscalculations when trying to equal the pea volumes. The stoppers might not have been sealed and gas could have been lost from the vials affecting the results with vengeance. The water temperatures had to be maintained precisely or the results would not be what they should be. There was also a lot of math in this lab when figuring results and many numbers could have been affected by this poor math.

Disussion and Conclusion

This lab showed many things about thew rates of cellular respiration. This lab showed that germinating peas consume more O2 than nongerminating peas. The colder temperature also slowed the rate of oxygen consumption. The oxygen could be clearly seen because of the following reaction

CO2+2KOH à K2O3 +H2O

This reaction gets rid of the CO2 so that it would not affect the readings of oxygen. It is absorbed by KOH to give you a precipitant K2CO3 + H2O. I conclude that the rate of O2 consumption is directly proportional to the respiration rate in that when the rate increases the gas consumption increases. When the gas consumption is low then the rate is low. Organisms go through cellular respiration more proficiently when the body of the organism is comfortable with its outside temp and environment. This lab showed many things affecting the rate of cellular respiration.

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AP Lecture Guide 07 – Cell Organelles

CELL ORGANELLES

Explain how the following organelles work together to perform the life functions of the cell

1. Nucleus and ribosomes

___________________________________________________________________________

2. Endoplasmic reticulum and Golgi bodies

___________________________________________________________________________

3. Endoplasmic reticulum and ribosomes

___________________________________________________________________________

4. Golgi bodies and lysosomes

___________________________________________________________________________

5. Nucleus and endoplasmic reticulum

___________________________________________________________________________

6. Endoplasmic reticulum and Golgi bodies and vesicles

___________________________________________________________________________

7. Endoplasmic reticulum and cell membrane

___________________________________________________________________________

AP Lab 1 Osmosis Sample 3

Diffusion and Osmosis

Introduction:

Atoms and molecules are constantly in motion. This kinetic energy causes the molecules to bump into each other and move in different directions. This motion is the fuel for diffusion. Diffusion is the random movement of molecules from an area of higher concentration to an area of lower concentration. This will occur until the two areas reach a dynamic equilibrium. When this dynamic equilibrium is reached the concentration of molecules will be approximately equal and there will be no net movement of molecules after this point. The molecules will still be in motion but the concentrations will remain the same.

Osmosis is a special kind of diffusion in which water moves through a selectively permeable membrane. A selectively permeable membrane allows diffusion for only certain solutes (the substance being dissolved) and water, the most common solvent (a dissolving substance). The most common selectively permeable membrane is the cell membrane. Water moves from an area of high water potential to an area of low water potential. Water potential is the measure of free energy of water in a solution and is represented by the symbol ψ (psi). Water potential is affected by two physical factors: the addition of a solute (ψs) and pressure potential (ψp). The addition of solutes to a concentration will lower the water potential of that solute, causing water to move into the area. Water movement is directly proportional to the pressure potential. Water potential can be determined by the equation:

ψ = ψp + ψs

Pure water has a water potential of zero. The addition of solutes will cause the water potential value to be negative, while an increase in pressure potential will cause a more positive water potential value.

There are three relationships that can occur between two solutions. When two solutions have equal solute concentrations, they are isotonic and no net movement of solute occurs. There is also no net movement of water. If the two solutions differ in solute concentrations, they will either be hypertonic or hypotonic. The hypertonic solution has a lower concentration of solute. Water will move out of a hypertonic solution, while solute will move in, (moving up the concentration gradient-similar to water potential). This depends on the selective qualities of the membrane. Pertaining to cells, this will cause the cell to shrivel or become flaccid. The hypotonic solution has a higher concentration of solute, and therefore has less water. This solution will gain water, while losing solute. This movement between the hypotonic and hypertonic solutions will continue until the point of dynamic equilibrium is reached. A hypertonic cell may also undergo plasmolysis. Plasmolysis is the shrinking of the cytoplasm in a plant cell in response to the diffusion of water out of the cell. When a cell is hypotonic it may lyse. In plant cells, it creates turgor pressure against the cell walls keeping the plant from becoming wilted.

Besides osmosis and diffusion, molecules and ions can be moved by active transport. This process includes the use of ATP to drive molecules in or out of a cell. Active transport is generally used to move molecules against a concentration gradient, from an area of low concentration to an area of higher concentration of molecules.

Hypothesis:

In this experiment, diffusion and osmosis will occur until dynamic equilibrium is reached. This experiment is done in a theoretical condition with no other variables affecting the movement of the solute except water potential.

Materials:

Exercise 1A

This exercise requires a 30 cm of 2.5 cm dialysis tubing, 250 ml beaker, distilled water, 2 dialysis tubing clamps, 15 ml of 15% glucose/1% starch solution, 4 pieces of glucose tape, 4 ml of Lugol’s solution (Iodine Potassium-Iodide or IKI), and clock or timer.

Exercise 1B

This experiment requires six strips of 30 cm dialysis tubing, 250 ml beaker, 12 dialysis tubing clamps, distilled water six cups, scale, timer or clock, paper towels, and about 25 ml of each of these solutions: distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose.

Exercise 1C

This experiment requires a large potato, potato corer (about 3 cm long), 250 ml beaker, paper towel, scale, six cups, knife, and about 100 ml of each of these solutions: distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose.

Exercise 1D

This experiment requires a calculator, paper, pencil, and graphing paper.

Exercise 1E

This experiment requires onion skin, dye, microscope, slide, cover slip, salt water (15%), and tap water.

Methods:

Exercise 1A

First, soak the dialysis tubing in distilled water for 24 hours. Remove the tubing and tie off one end using the clamp (twist tubing end about 7 times and fold end on self, slide into the clamp). Next, open the other end of the tubing (rubbing end between fingers) and fill it with the glucose/starch solution. Use the glucose tape and record the color change of the tape and the color of the bag. Tie of the end with the tubing clamp (leave empty space, but no air). Fill the beaker with distilled water and add the 4-ml’s of Lugol’s solution, record the color change. Use glucose tap to test for any glucose in the water (record). Set the dialysis tubing in the beaker and let it sit for about 30 minutes. Remove the bag and record the change in water and bag color. Use the last two pieces of glucose tape to measure the glucose in the water and bag and record results.

Exercise 1B

First, soak the dialysis tubing for about 24 hours. Tie off one end of each tube with the clamps. Next, fill each tube with a different solution (distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose) and tie off the end (leave empty space, but no air). Weigh each tube separately and record the masses. Soak the tubes in separate cups filled with distilled water for about 30 minutes. Remove the tubing, blot dry, reweigh, and record the mass.

Exercise 1C

First, slice the potato into to 3-cm discs. Use the potato corer and core out 24 cores (don’t get any). Weigh 4 cores together and record the mass. Fill each cup with a different solution (distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose). In each cup put 4 potato cores and let it sit over night. Take out the cores and blot them dry. Record the change in mass. Calculate the information and compare.

Exercise 1D

First, determine the solute potential of the glucose solution, the pressure potential, and the water potential. Then, graph the information given about the zucchini cores.

Exercise 1E

First, prepare a wet mount slide of dyed onion skin. Observe under a light microscope and sketch what the cells. Add a few drops of the salt solution, observe, and sketch the change.

Results:

Exercise 1A

Table 1 Change of Color of Dialysis Tubing and Beaker

Solution Color		Presence of Glucose (Glucose Tape)
Initial	Final	Initial	Final
Dialysis Bag	15% Glucose/1% Starch	Clearish White	Indigo	Dark brown	Mahogany
Beaker	Water + IKI	Amber	Amber	Yellow teal	Tan-green

Which substance(s) are entering the bag and which are leaving the bag? What experimental evidence supports your answer?

Glucose is leaving the bag slowly; this is shown buy using the glucose tape in the beaker. Iodine is entering the dialysis bag; this is shown by the change of color in the bag. Water enters the bag; the bag becoming fatter, which shows this.

Explain the results you obtained. Include the concentration differences and membrane pore size in your discussion.

The substances moved and out of the bag, according to the gradient. Some were small substances and moved in and out of the bag quickly and easily. The larger substances, like starch and glucose, were slow or didn’t enter or leave the bag at all.

Quantitative data uses numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffused into the dialysis bag?

The mass of the bag could be recorded before and after soaking.

Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water, IKI, membrane pores, and starch molecules.

Water→ IKI→ Glucose Molecules→ Membrane Pores→ Starch Molecules

What results would you expect if the experiment started with a glucose and IKI solution inside the bag and only starch and water outside? Why?

The IKI would have left the bag and changed the color of the solution in the beaker. Eventually dynamic equilibrium would be reached and there will be no net movement.

Exercise 1B

Table 2 Dialysis Bag Results: Individual Data

Contents of Dialysis Bag	Initial Mass (g)	Final Mass (g)	Mass Difference (g)	Percent Change in Mass*
a) Distilled Water	24.0	24.2	0.2	.83%
b) 0.2 M	26.2	26.7	0.5	1.9%
c) 0.4 M	26.4	27.0	0.6	2.2%
d) 0.6 M	27.9	31.2	3.3	11.8%
e) 0.8 M	28.4	31.9	3.5	12.3%
f) 1.0 M	29.6	34.6	5.0	16.9%

* To Calculate: Percent Change in Mass = Final Mass – Initial Mass

Initial Mass * 100

Table 3

Dialysis Bag results: Class Data

Solution	Group 1	Group 2	Group 3	Average
a) Distilled Water	.77%	1.53%	.83%	1.04%
b) 0.2 M	1.86%	5.30%	1.9%	3.02%
c) 0.4 M	2.4%	2.22%	2.2%	2.27%
d) 0.6 M	12.54%	9.75%	11.8%	11.36%
e) 0.8 M	13.07%	9.64%	12.3%	11.67%
f) 1.0 M	16.55%	18.98%	16.9%	17.48%
Team Members	Tripp & Stephanie	Hudgens & Kris	Elizabeth & Julie	7.81%

Graph 1

Percent Change in Mass of Dialysis Tubing in Glucose Solutions of Different Molarity

Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bags.

They are directly proportional. The percent change of mass goes up as the molarity of the sugar goes up.

Predict what would happen to the mass of each bag in this experiment if all the bags were placed in a 0.4 M sucrose solution instead of distilled water. Explain your response.

There will be no net movement when 0.4 M is in the dialysis bag. When the concentration is above 0.4, the bag will lose water. When the concentration is below 0.4, the bag will gain water.

Why did you calculate the percent change in mass rather than simply using the change in mass?

The volumes of the solutions were not exactly the same.

A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag’s initial mass is 20g, and its final mass is 18g. Calculate the percent change of mass, showing your calculations in the space below.

Percent Change in Mass = Final Mass – Initial Mass X 100 = 18-20 x 100 = 10%

Initial Mass 20

The sucrose solution in the beaker would have been hypertonic to the distilled water in the bag.

Exercise 1C

Table 4 Potato Core: Individual Data

Contents of Beaker	Initial Mass (g)	Final Mass (g)	Mass Difference (g)	% Change in Mass
a) Distilled Water	1.8	21.	.3	16.7
b) 0.2 M	1.5	1.7	.2	13.3
c) 0.4 M	1.5	1.8	.3	20
d) 0.6 M	1.6	1.3	.3	-18.75
e) 0.8 M	1.4	1.1	.3	-21.4
f) 1.0 M	1.6	1.3	.3	-18.75

Table 5 Potato Core Results: Class Data

Contents	Group 1	Group 2	Total	Class Average
Distilled Water	16.7%	28.5%	45.2%	22.6%
0.2 M Sucrose	13.3%	21.4%	34.7%	17.35%
0.4 M Sucrose	20.0%	14.28%	34.28%	17.14%
0.6 M Sucrose	-18.75%	-20.0%	-38.75%	-19.38%
0.8 M Sucrose	-21.4%	-26.66%	-48.06%	-24.03%
1.0 M Sucrose	-18.75%	-21.42%	-40.17%	-20.09%

Graph 2

Percent Change in Mass of Potato Cores at Different Molarities of Glucose

Exercise 1D

If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why?

The water potential of the cells will decrease, the osmotic potential will decrease, and the solute will increase. This occurs because the cells have become dehydrated.

If a plant cell has a lower water potential than its surrounding environment, and if pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the cell gain water or lose water? Explain your response.

The environment is hypotonic, so it will gain water. This is because it has less water than the surrounding environment.

In figure 1.5, the beaker is open to the atmosphere. What is the pressure potential of the system?

The pressure potential is zero.

In figure 1.5, where is the greatest water potential?

The dialysis bag.

Water will diffuse out of the bag. Why?

The water will diffuse out because there is higher water potential inside the bag.

Zucchini cores placed in sucrose solutions at 27° C resulted in the following percent changes after 23 hours:

% Change in Mass Sucrose Molarity

20% Distilled Water

10% 0.2 M

-3% 0.4 M

-17% 0.6 M

-25% 0.8 M

-30% 1.0 M

Graph 3

Percent Change in Mass of Zucchini Cores of Sucrose Solutions of Different Molarity

b) What is the molar concentration of solutes with in the zucchini in cells?

About .36 M

Refer to the procedure for calculating water potential from experimental data.

Calculate solute potential (ψs) of the sucrose solution in which the mass of the zucchini cores doesn’t change. Show work.

ψs =-iCRT
ψs =-(1)(0.35)(0.0831)(295)
ψs =-8.580075

ψ =0+ ψs
ψ =0+(-8.580075)
ψ =-8.580075

Calculate the water potential (ψ) of the solutes within the zucchini cores. Show work here.

ψ = ψs + ψp

-8.580075 = ψs + 0

-8.580075 = ψs

What effect does adding solute have on the solute potential component of that solution? Why?

Adding more solute will increase the solute potential and decrease water potential by making it more negative.

Consider what would happen to a red blood cell placed in distilled water:
a) Which would have the higher concentration of water molecules?

Distilled Water

b) Which would have the higher water potential?

Distilled Water

c) What would happen to the red blood cell? Why?

The red blood cells would pull in water and lyse.

Exercise 1E

Prepare a wet mount of a small piece of epidermis of an onion. Observe under 100x magnification. Sketch and describe the appearance of the onion cells.

Describe the appearance of the onion cells after the NaCl was added.

The plasma membrane shriveled from the cell wall, causing plasmolysis.

Remove the cover slip and flood the onion with fresh water. Observe and describe what happened.

The onion cells absorbed water and increasing in turgor pressure.

What is plasmolysis?

Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out or the cell and into a hypertonic solution surrounding the cell.

Why did the onion cell plasmolyze?

The environment became hypertonic to the cell and the water left the cell running with its concentration gradient because of the salt. With all the water leaving the cell, it shrank, leaving behind its cell wall.

In the winter, grass often dies near roads that have been salted to remove ice. What causes this to happen?

The salt causes the plant cells to plasmolyze.

Error Analysis:

Exercise 1A

Error that could have occurred in this experiment is that some of the glucose/starch solution could leaked into the beaker before the dialysis bag was inserted, but after the glucose test.
Exercise 1B

Error that could have occurred in this experiment is that the sugar could mot have been mixed completely with the water or some sugar could have been lost during the mixing. Another mistake could be that the bags were not bloated dry well enough.
Exercise 1C

Error that could have occurred in this experiment is that the sugar could mot have been mixed completely with the water or some sugar could have been lost during the mixing. Another mistake could be that the potato cores were not bloated dry well enough. Also the measuring of the liquids’ volumes may not have been accurate.
Exercise 1D

Errors that could have been made in this exercise could be that the numbers were put in the calculator wrong.
Exercise 1E

Error that could have been made in this exercise could have been that the onion skin could have dried out before the salt water was added, thus affecting the results.

Discussion and Conclusion:

Exercise 1A

In this experiment, the ability of substances to move across a selectively permeable membrane was viewed. A glucose/starch solution was put in the dialysis bag. The glucose molecules leaked out of the bag (learned from before and after test with glucose tape). Using IKI to test for starch, the change in color for the bag only shows that the starch molecules were too large to escape out of the dialysis bag, but the IKI molecule were small enough to enter the bag.

Exercise 1B

In this experiment, the study of hypotonic and hypertonic solution was tested. When the bags were placed in a hypotonic solution, they gained water. This could be viewed by massing the bags before and after the soaking.

Exercise 1C

In this experiment, potato cores were proven to have some sugar in them. When placed in low or no sugar environments, they gained water. When placed in high sugar environments, the cores lost water.

Exercise 1D

In this experiment, all these conclusion made in previous experiments were reinforced with scientific equations.

Exercise 1E

In this experiment, the turgor pressure of a plant cell’s plasma membrane was observed. The plasma membrane gives the cell shape and form. When salt is added, the cytoplasm loses water and causes the plasma membrane to shrink. This causes the plant to wilt.

Concentration gradient and water potential affected all aspects of this experiment. Water potential is used by many scientists to study the effects of different substance on plants, for good or bad. Pressure potential and solute potential are the two main components. Water moves to different areas based on water potential. Water will move from high water potential to low or low water potential to high. This change is called a gradient. Ψs = -iCRT is the formula used for solute potential. Water potential and solute potential are inversely proportional. When water potential goes up, solute potential is low. All these factors affected the results of these experiments.

Plant and animal cells react differently to different environment. A hypertonic solution will cause an animal cell to shrink in size. A hypotonic solution for an animal cell will cause it to lyse. Isotonic solutions are ideal for animal cells. Plant cell’s plasma membrane shrinks if in a hypertonic solution, causing the cell wall to loose shape or plasmolyze. While an isotonic solution is good, it doesn’t provide quite enough support for the cell wall. Hypotonic solutions are the best for plant cells. The plasma membrane presses against the inside of the cell wall giving it a lot of support. Without this pressure, turgor pressure, the plant will wilt and die. These experiments enable us to better understand living things, including our own bodies, and be able to take care of them.

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