Curriculum Map 134 - 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 – The Cell

AP Biology: CHAPTER 7

THE CELL

1. The tool that lead to the understanding that cells are the basic unit of life was the…

___________________________________________________________________________

2. The smallest structures visible with the light microscope are the ….

___________________________________________________________________________

3. What is the advantage of the electron microscope?

___________________________________________________________________________

4. How do biologists isolate cell components?

___________________________________________________________________________

5. What are four things all cells have in common?

a. _________________________________________________________________________

b. _________________________________________________________________________

c. _________________________________________________________________________

d. _________________________________________________________________________

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 4

Diffusion and Osmosis

Introduction:

Atoms and molecules are the building blocks of cells. Both have kinetic energy and are constantly in motion. They continually bump into one another and bounce off into new directions. This action results in two important processes, diffusion and osmosis.

Diffusion is the random movement of molecules from an area of higher concentration of those molecules to an area of lower concentration. Cells have selectively permeable membranes that only allow the movement of certain solutes. Diffusion is vital for many of life’s functions in a cell. It allows oxygen and carbon dioxide exchange in the lungs and between the bodies of intracellular fluid and cells. Diffusion also aids in the transport of nutrients and water in the xylem and phloem of plants. In those plants, it permits for the absorption of water into roots. An example of this process is the diffusion of a smell in a room. Eventually dynamic equilibrium will be reached. This means that the concentration of the molecules carrying the smell will be approximately equal through out the surrounding enclosed area and no net movement of the molecules will occur from one area to another.

Osmosis is special kind of diffusion. It is the diffusion or movement of water through semi-permeable membranes from a region of higher water potential (hypotonic solute) to a region of lower water potential (hypertonic solute). Water potential is the measure of free energy of water in a solution. There are three types of solutions. Isotonic solutions have an equal concentration of solute on both sides of the membrane, and dynamic equilibrium has been reached in the solution. Hypertonic solutions have a higher concentration of solute on one side of the membrane than the other. Hypotonic solutions are the opposite of hypertonic solutions. A solute is what is being dissolved by the solvent (water is the most common solvent) in a solution.

Water will always move from an area of higher water potential to an area of lower water potential. An important factor effecting of diffusion and osmosis is water potential. Water potential measures the tendency of water to leave one place in favor of another place. Water potential is affected by two physical factors. One factor is the addition of solute, which lowers the water potential. The other factor is pressure potential. An increase in pressure raised the water potential. The water potential of pure water at atmospheric pressure is defined as being zero. The Greek letter psi is used to represent water potential. The following formula can be used for calculations:

ψ (Water potential) = ψp (Pressure potential) + ψs (Solute potential)

Movement of water into and out of a cell is influenced by the solute potential on one side of the cell membrane relative to the other side. Plasmolysis is a phenomenon in walled plant cells in which the cytoplasm shrivels and the plasma membrane pulls away from the cell wall when the cell loses water to a hypertonic environment. This leads to a loss of turgor pressure (the force directed against a cell wall after the influx of water and the swelling of a walled cell due to osmosis) and eventual death of the plant. If water moves into the cell, the cell may lyse, or burst (in animal cells, plant cells are equipped to handle large intakes of water). Water movement is directly proportional to the pressure on a system. Pressure potential is usually positive in living cells and negative in dead ones.

Diffusion and osmosis are not the only processes responsible for the movement of ions or molecules in an out of cells. Active transport is process that uses energy from ATP to move substances through the cell membrane. Normally, active transport moves a substance against its concentration gradient, that is to say from a region of low concentration to an area of higher concentration.

Hypothesis:

Osmosis and diffusion will continue until dynamic equilibrium is reached and net movement will no longer occur. Diffusion is effected by the solute size and concen-tration gradient across a selectively permeable membrane. Water potential greatly determines the results in sections of the experiment.

Materials:

Exercise 1A

For this exercise, the following materials are required: a 30 cm of 2.5 cm dialysis tubing, 250 ml beaker, distilled water, funnel, 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), a timer, paper and pencil.

Exercise 1B

This exercise of the experiment requires six strips of 30 cm dialysis tubing, 250 ml beaker, 12 dialysis tubing clamps, funnel, six cups, distilled water, an electronic balance, timer, paper towels, and about 25 ml of each of these solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 M glucose, and 1.0 M glucose. For recording results, paper and pencil are necessary.

Exercise 1C

This part of the experiment requires a large potato, potato corer (about 3 cm long), 250 ml beaker, paper towels, scale, six cups, knife, paper, pencil and about 100 ml of each of these solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 glucose, and 1.0 M glucose.

Exercise 1D

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

Exercise 1E

This section of the experiment requires paper, pencil, paper towels, onionskin, 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. Before handling the tubing, wash dirty hands thoroughly to prevent getting oils on the dialysis tubing and changing the results. Remove the tubing and tie off one end using the clamp. To use the clamp, twist the end of the bag several times and then fold it onto itself. Next, open the other end of the tubing by rubbing the end between two fingers. Fill it with the glucose/starch solution using a funnel. Use the glucose tape by dipping it into the solution. Record the color change of the tape and the color of the bag. Tie of the end with the tubing clamp. It is necessary to leave space for expansion but no air. Fill the beaker with distilled water and add the 4-ml of Lugol’s solution. Record the color change. Use glucose tap to test for any glucose in the water. Record these results. 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. Record results.

Exercise 1B

First, soak the dialysis tubing for about 24 hours. Again be sure to cleanse hands. Tie off one end of each tube with the clamps. Next, fill each tube with a different solution (distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 glucose, and 1.0 M glucose) with the funnel and tie off the end again leaving empty space, but no air. Weigh each bag separately on the electronic balance and record the masses. Soak the bags in separate cups filled with distilled water for about 30 minutes. Remove the bags and gently blot dry with paper towel. Reweigh, and record the mass.

Exercise 1C

First, slice the potato into to 3-cm discs. Use the potato corer to core out 24 cores. Weigh 4 cores together and record their mass. Fill each cup with one of the following solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 glucose, and 1.0 M glucose. In each cup put 4 potato cores, and allow them to sit over night. Take out the cores and blot them dry. Again weigh them on the electronic scale. Record the change in mass. Calculate the information for the table. Compare the results with another group.

Exercise 1D

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

Exercise 1E

Prepare a wet mount slide of dyed onion skin. Observe under a light microscope and sketch how the cells appear. Add a few drops of the salt solution using a paper towel to wick the solution under the slip. Observe how the cells are effected and make another sketch.

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	Milky White	Midnight Blue	Algae Green	Mahogany
Beaker	Water + IKI	Amber	Rusty Amber	Pear Green	Olive Green

Initial Glucose Tests Final Glucose Tests

Which substance(s) are entering the bag and which are leaving the bag? What experimental evidence supports your answer? Iodine Potassium Iodide and water enter the bag. This is proven by the color change (starch test) and the increase in the size of the bag. Glucose left the bag and this is proven by a positive test on the surrounding water.

Explain the results you obtained. Include the concentration differences and membrane pore size in your discussion. The results show that the water, glucose, and IKI molecules were small enough to pass through the selectively permeable membrane. The starch didn’t leave the beaker because its molecules were too large to pass through the selectively permeable membrane’s pores.

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 bags could be massed before and following their immersion in the solution. The volume of the solution in the beaker could be found before and after the immersion of the bag by using a graduated cylinder.

Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water, IKI, membrane pores, and starch molecules. The smallest substance was water, then the IKI molecules, glucose, the membrane pores, and the largest substance was the 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? Based on the size of the molecules, the glucose and IKI would move out of the bag and the water would go in. The large starch molecules would be left in the beaker.

Exercise 1B

Table 2: Dialysis Tubing Mass Change Results: Individual Data

Contents of Dialysis Tubing	Initial Mass (g)	Final Mass (g)	Mass Difference (g)	Percent Change in Mass
a) Distilled Water	26.0	26.2	0.2	.77%
b) 0.2 M	27.0	27.5	0.5	1.85%
c) 0.4 M	25.0	25.6	0.6	2.4%
d) 0.6 M	27.9	31.4	3.5	12.54%
e) 0.8 M	28.3	32.0	3.7	13.07%
f) 1.0 M	28.4	34.6	4.7	16.55%

Table 3: Dialysis Tubing Mass Change Results: Group 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	NA

Graph 1: Percent Change in Mass of Dialysis Tubing in Sucrose Solutions of Different Molarities

Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bags. The molarity is directly proportional to the percent change in mass. As the mass percentage increases, so does the molarity.

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. They are inversely proportional because whenever the sucrose molarity inside the bag is more concentrated, it will become more dilute and vice versa. The solutions will reach equilibrium somewhere between the two concentrations.

Why did you calculate the percent change in mass rather than simply using the change in mass? Each group began with different amounts of solution for their initial mass. Therefore, results cannot be based on those numbers. The differences in mass don’t deal with the proportional aspect of the solutions, making the real results less accurate. The percent was calculated to give the exact difference, along with considering the quantities of solution.

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. 18g (final mass) – 20g (initial mass) / 20g (initial mass) = 2/20g x 100 = 10% change of mass

Exercise 1C

Table 4: Potato Core: Individual Results

Contents in Beaker	Initial Mass (g)	Final Mass (g)	% Change in Mass
Distilled Water	1.8	2.1	16.7%
0.2 M Sucrose	1.5	1.7	13.3%
0.4 M Sucrose	1.5	1.8	20.0%
0.6 M Sucrose	1.6	1.3	-18.75%
0.8 M Sucrose	1.4	1.1	-21.4%
1.0 M Sucrose	1.6	1.3	-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%
Team Members	Stephanie, Tripp, & Eli	Hudgens Kris	NA	NA

Graph 2: Percent Change in Mass of Potato Cores at Different Molarities of Sucrose

Exercise 1D

Graph 3: Percent Change in Mass of Zucchini Cores at Different Molarities of Sucrose

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 potato would decrease because water moves from a high water potential region to a low potential region, and a dehydrated potato cell is hypertonic in comparison with the environment. The concentration of solute would increase and osmotic potential would decrease.

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. If the plant cell has lower water potential, that means the water will come into the cell; the cell is hypertonic to its environment. This cell will gain water because water follows its concentration gradient.

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 greatest water potential is within the dialysis bag.

Water will diffuse_________the bag. Why? Water will diffuse out of the bag because the inside of the bag has the highest water potential.

Calculate solute potential of the sucrose solution in which the mass of the zucchini cores does not change. Show work. ψ s = -iCRT ψ s = (-1)(0.36 mole/liter)(0.0831 liter bar/mole K)(300 K) ψ s = -8.975 bars

Calculate the water potential of the solutes within the zucchini cores. Show work. ψ = ψ s+ ψ p ψ =0 + -8,975 , ψ = -8.975 bars

What effect does adding solute have on the solute potential component of that solution? Why? Adding solute to a solution would increase the solute potential and decrease the water potential.

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

distilled water would have the higher concentration of water molecules.

Which would have the higher water potential? The distilled water would also have the higher water potential.

What would happen to the red blood cell? Why? The red blood cell would take in a lot of water and might lyse due to pressure inside. This is because animal cells lack tolerance under hypotonic situations.

Exercise 1E

Describe the appearance of the onion cells. The onion cells appear to have great turgor pressure, spread out, thick and bright in the inside. The cell walls were very defined and it was clear where one cell ended and another began.

Describe the appearance of the onion cells after the NaCl was added. The plasma membrane shriveled from the cell wall, or in other words, plasmolysis occurred.

Remove the cover slip and flood the onion with fresh water. Observe and describe what happened. The onion cells were again hypertonic to their environment and were restored to their original state of appearance.

What is plasmolysis? Plasmolysis is the separation of the plasma membrane from the cell wall in a plant cell.

Why did the onion cell plasmolyze? The environment around the cell was hypertonic to the cell so water left the cell to reach dynamic equilibrium with the NaCl solution. With all the water leaving the cell, the cell membrane separated from 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 grass’s environment to become hypertonic, and the water leaves the plant cells, causes withering and eventually death of the plant. The high concentration of salt in the soil also speeds the death of the plant.

Sketch of Onion Cells Onion Cells + NaCl

Error Analysis:

Several could have possibly been made throughout the lab.

Exercise 1A

The data collected in this lab experiment did not appear to contain any errors, however, an error in the results may have unknowingly occurred. If there was a leak where the tubing was twisted shut or a tear in the dialysis tubing, all of the data would be inaccurate.

Exercise 1B

This section of the lab had to be repeated because of incorrect data (that is to say it did not “harmonize” with the other groups’ data). If the person handling the dialysis tubing did not wash their hands thoroughly and accidentally touched the portion of the tubing to serve as the permeable membrane, the oils from their hands could have blocked pores on the tubing, effecting the data.

Exercise 1C
Some mistakes that could have taken place are mathematical miscalculations while finding the initial and final masses. A piece of potato skin could have been left in the beakers along with the potato. This causes problems in the data tables. Another possible source of error could be that the students did not pat dry the potato sample well enough and increased the masses of the cores. Numerous may have occurred while using the electronic balance.

Exercise 1D

In this part of the lab, only calculations were made. Simple mathematical errors are bound to occur in this section of the lab.

Exercise 1E

In part 1E, after adding the NaCl solution to the onion cells, the cells should have reduced in size, but no reaction appeared to take place. This may have occurred in part because the onion itself was already dried out and dehydrated, or while the onion was being looked at through the microscope, the heat from it may have caused the cells to loose water. Another possibility is that the reaction took place so quickly that those witnessing could not see it.

Discussion and Conclusion:

Exercise 1A

The data shows what molecules can and cannot diffuse across a selectively permeable membrane. The color change showed that the Iodine Potassium Iodide was small enough to pass through the pores of the membrane. It is shown that the water and glucose solution moved out of the dialysis bag because water is small enough to pass through the membrane and the Testape tested positive for glucose inside the beaker. The glucose started out inside the bag and tested negative with the Testape inside the beaker before the immersion.

Exercise 1B

It can be concluded from the results gathered during the experiment that sucrose cannot pass through the selectively permeable membrane, but instead water molecules must move across the membrane to the area of lower water potential to reach dynamic equilibrium.

Exercise 1C

The results provided information that leads us to conclude that potatoes do contain sucrose molecules. This is known because the cores took in water while they were emerged in the distilled water. This means they had a lower water potential and higher solute potential than the distilled water.

Exercise 1D

The calculations made it evident that all of the results could be determined and proven correct with the simple equations and formulas. Performing these mathematical computations helped give a better understanding of water and solute potential.

Exercise 1E

This particular part of the lab illustrated the shrinking of the plasma membrane from the cell wall in a plant cell, or, in other words, plasmolysis. It shows how plant cells react in a hypertonic environment, the NaCl solution. The turgor pressure decreases as water leaves the cell. This shows how the onion cells had high water potential so water moved to the area outside the cell with lower water potential. Then, after adding water back to the cells, water moved back into the cells, restoring turgor pressure.

Overall

Water potential and concentration gradients are the two phenomenons that effected the results of the experiments. There are many important facts pertaining to water potential. Water potential is used by botanists to determine the movement in and out of a cell. It is effected by two components, pressure and solute potential. Water moves from areas of higher water potential (higher free energy and more water molecules) to areas of lower water potential (lower free energy and less water molecules). Water diffuses down a water potential gradient. Pure water has an atmospheric pressure of zero which is important when using the formula ψs = -iCRT. Water potential is inversely proportional to solute potential. These facts led to or effected the results gained in each section of the lab.

In plant and animal cells, loss or gain of water can have different effects. In an animal cell, it is ideal to have an isotonic solution. If the solution is hypertonic, the cell will shrivel from lack of water intake. Inversely, if the solution is hypotonic the cell could take in too much water and the cell will lyse and break open. For a plant cell, the ideal solution is a hypotonic solution because the cell takes in water increasing turgor pressure. Turgor pressure is important for plant support and maintaining shape. If the solution is hypertonic, the cell will plasmolyze and died from lack of water. In an isotonic solution, the plant cell does not have enough turgor pressure to prevent wilting and possible death. The information gained through this lab is important in understanding the effects of different solutions on organisms in our environments, including ourselves.

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