AP Lecture Guide 17 – From Gene to Protein

AP Biology: CHAPTER 17

FROM GENE TO PROTEIN

1. How did diseases involving metabolic pathways lead to hypotheses about the nature of

genes?

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2. Identify some genetic diseases that occur along metabolic pathways.

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3. What was Beadle and Tatum’s hypothesis regarding enzymes?

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4. How has that hypothesis been modified?

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5. What occurs during transcription?

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6. What occurs during translation?

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7. How does the protein process differ in prokaryotes and eukaryotes?

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8. Briefly explain how Marshall Nirenberg and Heinrich Matthaei “cracked the genetic code?”

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9. What is the genetic code and why is said to be universal?

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10. List several features about the genetic code.

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11. Give an example of what happens if reading frames are altered?

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12. List the highlights of the three stages of transcription.

a. Initiation _______________________________________________________________

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b. Elongation ______________________________________________________________

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c. Termination _____________________________________________________________

13. What happens to the transcript RNA before it leaves the nucleus?

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14. What is the advantage of the 5’ cap and poly A tail?

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15. Distinguish between exons and introns.

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16. Describe the mechanism for splicing RNA.

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17. What does alternative RNA processing do for cells?

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18. Identify the roles of the players of the translation process.

a. Transfer RNA ___________________________________________________________

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b. Aminoacyl-tRNA synthetase _______________________________________________

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c. Ribosomes _____________________________________________________________

19. Identify and briefly describe the steps of translation. Initiation Elongation Termination

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20. What is the advantage of polyribosomes?

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21. Give an example of how a polypeptide gets into the ER for additional processing.

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22. How does protein synthesis differ between prokaryotes and eukaryotes?

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23. Define point mutations. ______________________________________________________

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24. Define mutations that are:

a. Missense ______________________________________________________________

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b. Nonsense ______________________________________________________________

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c. Insertion or deletion ______________________________________________________

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25. Use the diagram to trace the flow of chemical information from the gene to the protein product.

 

AP Lecture Guide 19 – Control of Eukaryotic Genome

AP Biology: CHAPTER 19

CONTROL OF EUKARYOTIC GENOME

1. Outline the levels of DNA packing within the eukaryote nucleus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2. What is the difference between heterochromatin and euchromatin? Which is transcribed?

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3. Which regions of the chromosome will typically be in the form of hererochromatin?

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4. How do the coding regions and genome sizes of prokaryotes and eukaryotes compare?

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5. Much of mammalian non-coding DNA is in the form of ______________________________

6. What is the cause of Fragile X?

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7. What is the cause of Huntington’s disease?

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8. Discuss an example of interspersed repetitive DNA?

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9. What is a multigene family?

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10. Multigene families are hypothesized to have evolved from…

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11. How is the globulin multigene family an adaptive to mammals?

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12. Explain how gene amplification can regulate gene expression.

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13. How can transposons alter gene expression?

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14. How do immunoglobulin genes code for a seemingly infinite variety of antibodies?

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15. Review the opportunities for gene regulation in eukaryotes in the diagram.

16. Where is the most important step in gene regulation?

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17. Describe the effect of each of the following control mechanisms.

a. DNA methylation ________________________________________________________

b. Histone acetylation _______________________________________________________

c. Transcription factors ______________________________________________________

d. Control elements ________________________________________________________

e. Enhancers _____________________________________________________________

f. Activators ______________________________________________________________

g. DNA-binding domain _____________________________________________________

18. How does alternative RNA splicing affect gene expression?

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19. How does RNA degradation affect gene expression?

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20. How does protein processing and degradation affect gene expression?

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21. Identify the opportunities to regulate gene expression in eukaryotes.

22. Typically, what happens to cell function when cells become cancerous?

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23. What is a proto-oncogene? What happens to them when cancer occurs?

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24. List the three events that can turn proto-oncogenes into oncogenes.

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25. Identify and describe mutations in specific proteins that can lead to cancer.

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26. What is p53?

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27. Why is it said that cancer formation is a multi-step process?

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AP Sample 5 Lab 5 Cellular Respiration

 

 

Lab 5     Cellular Respiration

 

 

Introduction:

 

Cellular respiration is an ATP-producing catabolic process in which the ultimate electron acceptor is an inorganic molecule, such as oxygen. It is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. Carbohydrates, proteins, and fats can all be metabolized as fuel, but cellular respiration is most often described as the oxidation of glucose, as follows:

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

Cellular respiration involves glycolysis, the Krebs cycle, and the electron transport chain. Glycolysis is a catabolic pathway that occurs in the cytosol and partially oxidizes glucose into two pyruvate (3-C). The Krebs cycle is also a catabolic pathway that occurs in the mitochondrial matrix and completes glucose oxidation by breaking down a pyruvate derivative (Acetyl-CoA) into carbon dioxide. These two cycles both produce a small amount of ATP by substrate-level phosphorylation and NADH by transferring electrons from substrate to NAD+ (Krebs cycle also produces FADH2 by transferring electrons to FAD). The electron transport chain is located at the inner membrane of the mitochondrion, accepts energized electrons from reduced coenzymes that are harvested during glycolysis and Krebs cycle, and couples this exergonic slide of electrons to ATP synthesis or oxidative phosphorylation. This process produces 90% of the ATP.

Cells respond to changing metabolic needs by controlling reaction rates. Anabolic pathways are switched off when their products are in ample supply. The most common mechanism of control is feedback inhibition. Catabolic pathways, such as glycolysis and the Krebs cycle, are controlled by regulating enzyme activity at strategic points. A key control point of catabolism is the third step of glycolysis, which is catalyzed by an allosteric enzyme, phosphofructokinase. The ratio of ATP to ADP and AMP reflects the energy status of the cell, and phosphofructokinase is sensitive to changes in this ratio. Citrate and ATP are allosteric inhibitors of phosphofructokinase, so when their concentration rise, the enzyme slows glycolysis. As the rate of glycolysis slows, the Krebs cycle also slows since the supply of Acetyl-CoA is reduced. This synchronizes the rates of glycolysis and the Krebs cycle. ADP and AMP are allosteric activators for phosphofructokinase, so when their concentrations relative to ATP rise, the enzyme speeds up glycolysis, which speeds of the Krebs cycle.

Cellular respiration is measure in three manners: the consumption of O2 (how many moles of O2 are consumed in cellular respiration?), production of CO2 (how many moles of CO2 are produced in cellular respiration?), and the release of energy during cellular respiration.

PV = nRT is the formula for the inert gas law, where P is 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, and T is the temperature of the gas in degrees K. This law implies several important things about gases. If temperature and pressure are kept constant then the volume of the gas is directly proportional to the number of molecules of the gas. If the temperature and volume remain constant, then the pressure of the gas changes in direct proportion to the number of molecules of gas. If the number of gas molecules and the temperature remain constant, then the pressure is inversely proportional to the volume. If the temperature changes and the number of gas molecules is kept constant, then either pressure or volume or both will change in direct proportion to the temperature.

Hypothesis:

 

The respirometer with only germinating peas will consume the largest amount of oxygen and will convert the largest amount of CO2 into K2CO3 than the respirometers with beads and dry peas and with beads alone. The temperature of the water baths directly effects the rate of oxygen consumption by the contents in the respirometers (the higher the temperature, the higher the rate of consumption).

 

Materials:

The following materials are necessary for the lab: 2 thermometers, 2 shallow baths, tap water, ice, paper towels, masking tape, germinating peas, non-germinating (dry) peas, glass beads, 100 mL graduated cylinder, 6 vials, 6 rubber stoppers, absorbent and non- absorbent cotton, KOH, a 5-mL pipette, silicon glue, paper, pencil, a timer, and 6 washers.

 

Methods:
Prepare a room temperature and a 10oC water bath. Time to adjust the temperature of each bath will be necessary. Add ice cubes to one bath until the desired temperature of 10oC is obtained.

Fill a 100 mL graduated cylinder with 50 mL of water. Add 25 germinating peas and determine the amount of water that is displaced. Record this volume of the 25 germinating peas, then remove the peas and place them on a paper towel. They will be used for respirometer 1. Next, refill the graduated cylinder with 50 mL of water and add 25 non-germinating peas to it. Add glass beads to the graduated cylinder until the volume is equivalent to that of the expanded germinating peas. Remove the beads and peas and place on a paper towel. They will be used in respirometer 2. Now, refill the graduated cylinder with 50 mL of water. Determine how many glass beads would be required to attain a volume that is equivalent to that of the germinating peas. Remove the beads. They will be used in respirometer 3. Then repeat the procedures used above to prepare a second set of germinating peas, dry peas and beads, and beads to be used in respirometers 4,5,and 6.

Assemble the six respirometers by obtaining 6 vials, each with an attached stopper and pipette. Then place a small wad of absorbent cotton in the bottom of each vial and, using the pipette or syringe, saturate the cotton with 15 % KOH. Be sure not to get the KOH on the sides of the respirometer. Then place a small wad of non-absorbent cotton on top of the KOH-soaked absorbent cotton. Repeat these steps to make the other five respirometers. It is important to use about the same amount of cotton and KOH in each vial.

Next, place the first set of germinating peas, dry peas and beads and beads alone in vials 1,2, and 3. Place the second set of germinating peas, dry peas and beads, and glass beads in vials 4,5, and 6. Insert the stoppers in each vial with the proper pipette. Place a washer on each of the pipettes to be used as a weight.

 

RespirometerTemperatureContents
1RoomGerminating Peas
2RoomDry Seeds + Beads
3RoomBeads
410oCGerminating Peas
510oCDry Seeds + Beads
610oCBeads

 

Make a sling using masking tape and attach it to each side of the water baths to hold the pipettes out of the water during the equilibration period of 10 minutes. Vials 1,2, and 3 should be in the bath containing water at room temperature. Vials 4, 5, and 6 should be in the bath containing water that is 10oC. After the equilibration period, immerse all six respirometers into the water completely. Water will enter the pipette for a short distance and stop. If the water does not stop, there is a leak. Make sure the pipettes are facing a direction from where you can read them. The vials should not be shifted during the experiment and your hands should not be placed in the water during the experiment.

Allow the respirometers to equilibrate for three more minutes and then record the initial water reading in each pipette at time 0. Check the temperature in both baths and record the data. Every five minutes for 20 minutes take readings of the water’s position in each pipette, and record.

 

Results:

Table 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

1.381.351.47
 

0-5

1.3801.16.19.191.46.01.01
 

5-10

1.3801.04.31.311.44.03.03
 

10-15

1.3800.93.42.421.43.04.04
 

15-20

1.3800.57.78.781.42.05.05
 

Initial-0

1.401.321.40
 

0-5

1.39.011.20.12.111.400.01
 

5-10

1.38.021.11.21.191.400.02
 

10-15

1.38.021.00.32.301.39.01.01
 

15-20

1.38.020.95.37.931.38.020

 

 

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 rate of cellular respiration is higher in the germinating peas in cold than in the beads or non-germinating peas; the cooler temperature in the cold water baths slows the process of cellular respiration in the both germinating and non-germinating peas.

This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each. The constant temperature in the water baths yielding stable readings, the unvarying volume of KOH from vial to vial leading to equal amounts of carbon dioxide consumption, identical equilibration periods for all the respirometers, precise time intervals between measurements, and glass beads acting as a control for barometric pressure all served as controls.

 

Describe and explain the relationship between the amount of oxygen consumed and time. There was a constant, gradual incline in the amount of oxygen consumed over precise passage of time.

 

 

Condition

 

Calculations

 

Rate in mL O2/ minute

 

Germinating Peas/ 10 oC

 

(1.40-1.38)

20 min.

.001
 

Germinating Peas/ 20 oC

 

(1.35-.57)

20 min.

.040
 

Dry Peas/ 10 oC

 

(1.40-1.38)

20 min.

.001
 

Dry Peas/ 20 oC

(1.47-1.42)

20 min.

.003

 

 

Why is it necessary to correct the readings from the peas with the readings from the beads? The beads served as a control variable, therefore, the beads experienced no change in gas volume.

 

Explain the effects of germination (versus non-germination) on pea seed respiration. The germinating seeds have a higher metabolic rate and needed more oxygen for growth and survival. The non-germinating peas, though alive, needed to consume far less oxygen for continued subsistence.

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 reached a certain point, the enzymes necessary for cellular respiration denatured and germination (and large amounts of oxygen consumption) was inhibited.

 

What is the purpose of KOH in this experiment? The KOH drops absorbed the carbon dioxide and caused it to precipitate at the bottom of the vial and no longer able to effect the readings.

 

Why did the vial have to be completely sealed under the stopper? The stopper at the top of the vial had to be completely sealed so that no gas could leak out of the vial and no water would be allowed into the vial.

 

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 since they are warm-blooded and endothermic.

 

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. Respiration would be higher at 21 degrees because it would be necessary for the animal to maintain a higher body temperature. The results would proliferate at 10 degrees because the mammal would be required to retain its body temperature at an even lower temperature in comparison to room temperature.

 

Explain why water moved into the respirometer pipettes. While the peas underwent cellular respiration, they consumed oxygen and released carbon dioxide, which reacted with the KOH in the vial, resulting in a decrease of gas in the pipette. The water moved into the pipette because the vial and pipette were completely submerged into the bath.

 

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? Respirometers could be set up with respirometer 1 containing non-germinating peas, respirometer 2 holding peas that have been germinating 24 hours, 3 would contain the peas that germinated 48 hours, and 4 would hold the peas that germinated 72 hours. All the respirometers should have the KOH added to the bottom in the same manner as in lab described earlier. The respirometers should be placed in baths with the same temperature for all the respirometers. The seeds that have not begun germination would consume very little oxygen. The peas that have been germinating for 72 hours will have the greatest amount of oxygen consumption, while the other two samples will consume a medium (in comparison to respirometers 1 and 4 results) amount of oxygen.

 

Error Analysis:

 

Numerous errors could have occurred throughout the lab. The temperature of the baths may have been allowed to fluctuate, the amounts of peas, beads, KOH, and cotton may have varied from vial to vial damaging the results, and these problems would have occurred only during set up. Air may have been allowed to creep into the vial via a leaky stopper or poorly sealed pipette. Timing for the equilibration of the respirometers and the five-minute time intervals may have been erroneous. It was somewhat difficult to read the markings on the pipettes and so errors are always likely. Mathematical inaccuracies may have taken place when filling out the table and finding the corrected difference by using the formula provided.

 

Discussion and Conclusion:

 

The lab and the results gained from this lab demonstrated many important things relating to cellular respiration. It showed that the rates of cellular respiration are greater in germinating peas than in non-germinating peas. It also showed that temperature and respiration rates are directly proportional; as temperature increases, respiration rates increase as well. Because of this fact, the peas contained by the respirometers placed in the water at 10 oC carried on cellular respiration at a lower rate than the peas in respirometers placed in the room temperature water. The non-germinating peas consumed far less oxygen than the germinating peas. This is because, though germinating and non-germinating peas are both alive, germinating peas require a larger amount of oxygen to be consumed so that the seed will continue to grow and survive.

In the lab, CO2 made during cellular respiration was removed by the potassium hydroxide (KOH) and created potassium carbonate (K2CO3). It was necessary that the carbon dioxide be removed so that the change in the volume of gas in the respirometer was directly proportional to the amount of oxygen that was consumed. In the experiment water will moved toward the region of lower pressure. During respiration, oxygen will be consumed and its volume will be reduced to a solid. The result was a decrease in gas volume within the tube, and a related decrease in pressure in the tube. The respirometer with just the glass beads served as a control, allowing changes in volume due to changes in atmospheric pressure and/or temperature.

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% StarchMilky WhiteMidnight BlueAlgae GreenMahogany
 

Beaker

Water + IKIAmberRusty AmberPear GreenOlive 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

26.026.20.2.77%
 

b) 0.2 M

27.027.50.51.85%
 

c) 0.4 M

25.025.60.62.4%
 

d) 0.6 M

27.931.43.512.54%
 

e) 0.8 M

28.332.03.713.07%
 

f) 1.0 M

28.434.64.716.55%

 

 

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 & StephanieHudgens & KrisElizabeth & JulieNA

 

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 Water1.82.116.7%
0.2 M Sucrose1.51.713.3%
0.4 M Sucrose1.51.820.0%
0.6 M Sucrose1.61.3-18.75%
0.8 M Sucrose1.41.1-21.4%
1.0 M Sucrose1.61.3-18.75%

 

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, & EliHudgens

Kris

NANA

 

 

 

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.

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% StarchClearish WhiteIndigoDark brownMahogany
 

Beaker

Water + IKIAmberAmberYellow tealTan-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

24.024.20.2.83%
 

b) 0.2 M

26.226.70.51.9%
 

c) 0.4 M

26.427.00.62.2%
 

d) 0.6 M

27.931.23.311.8%
 

e) 0.8 M

28.431.93.512.3%
 

f) 1.0 M

29.634.65.016.9%

 

 

* 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 & StephanieHudgens & KrisElizabeth & 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

1.821..316.7
 

b) 0.2 M

1.51.7.213.3
 

c) 0.4 M

1.51.8.320
 

d) 0.6 M

1.61.3.3-18.75
 

e) 0.8 M

1.41.1.3-21.4
 

f) 1.0 M

1.61.3.3-18.75

 

 

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.