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.

AP Lab 5 Sample 7

 

Cellular Respiration

Blake Lockwood

Introduction:

 

The human body has to have energy in order to perform the functions that allow life. This energy comes from the process of cellular respiration. Cellular respiration releases energy that the body can use in the form of ATP from carbohydrates by using oxygen. Cellular respiration is not just one singular reaction, it is a metabolic pathway made up of several reactions that are enzyme mediated. This process begins with glycolysis in the cytosol of the cell. In glycolysis, glucose is split into two three-carbon compounds called pyruvate, producing a small amount of ATP The final two steps of cellular respiration occur in the mitochondria. These final two steps are the electron transport system and the Krebs Cycle. The overall equation for cellular respiration is

C6H12O6 + 6O2 -> 6CO2 + 6H2O + 686 kilocalories of energy per mole of glucose oxidized.

There are three ways to measure the rate of cellular respiration. These three ways are by measuring the consumption of oxygen gas, by measuring the production of carbon dioxide, or by measuring the release of energy during cellular respiration. In order to measure the gases, the general gas law must be understood. The general gas law state: PV=nRT 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 K). The gas law also shows concepts 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 present. 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 of volume will change in direct proportion to the temperature.

In this experiment, the rate of cellular respiration will be measured by measuring the oxygen gas consumption by using a respirometer in water. This experiment measures the consumption of oxygen by germinating and non-germinating at room temperature and at ice water temperature. The carbon dioxide produced in cellular respiration will be removed by potassium hydroxide (KOH). As a result of the carbon dioxide being removed, the change in the volume of gas in the respirometer will be directly related to the amount of oxygen consumed. The respirometer with glass beads alone will show any changes in volume due to atmospheric pressure changes or temperature changes.

 

Hypothesis:

 

The germinating peas will have a higher rate of respiration, than the beads and non-germinating peas.

 

Materials:

 

This lab requires two thermometers, two water baths, beads, germinating and non-germinating peas, beads, six vials, twelve pipettes, 100 mL graduated cylinder, scotch tape, tap water, ice, KOH, absorbent and non-absorbent cotton, six washers, six rubber stoppers, scotch tape, and a one mL dropper.

 

Methods:

 

Start the experiment by setting up two water baths, one at room temperature and the other at 10 degrees Celsius. Then, find the volume of twenty-five germinating peas. Next, put 50 mL of water in a graduated cylinder and put twenty-five non-germinating peas in it. Then, add beads until the volume is the same as twenty-five germinating peas. Next, pour our the peas and beads, refill the graduated cylinder with 50 mL of water, and add only beads until the volume is the same as the twenty-five germinating peas. Repeat these steps for another set of peas and beads. Also, put together the six respirometers by gluing a pipette to a stopper and taping another pipette to the pipette for all six respirometers. Then, put two absorbent cotton balls, several drops of KOH, and half of a piece of non-absorbent cotton into all six vials. Next, add the peas and beads to the appropriate respirometers. Place one set of respirometers into the room temperature water bath and the other set in the ice water bath. Elevate the respirometers by setting the pipettes onto masking tape and allow them to equilibrate for five minutes. Next, lower the respirometers into the water baths and take reading at 0, 5, 10, 15, and 20 minutes. Record the results in the table.

 

Results:

 

Table:

 

Beads AloneGerminating Peas

Dry Peas and Beads

Reading at time XDiff.Reading at time XDiff.Corrected Diff.

 

Reading at time XDiff.Corrected Diff.
Initial13.212.712.9
0 to 511.02.210.52.20.011.11.8-0.4
5 to 10103.29.03.70.510.02.8-0.3
10 to 159.24.08.04.70.79.43.5-0.5
15 to 209.14.07.55.11.29.33.6-0.4
Initial14.013.514.0
0 to 513.30.712.11.40.713.60.4-0.3
5 to 1012.91.111.02.51.413.20.8-0.3
10 to 1512.61.410.03.52.112.91.1-0.3
15 to 2012.201.89.04.52.712.51.5-0.3

 

 

Questions

1. In this activity, you are investigating both the effect of germination versus non-germination and warm temperature versus cold temperature on respiration rate. Identify the hypothesis being tested in this activity. Increasing the temperature could increase the oxygen consumption. Germinating peas have a higher respiration rate than non-germinating.

 

2. This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each control. One control was the respirometer with only beads in it because it didn’t use respiration. Another control was that the water temperatures were constant. The final control was that there was the same amount of KOH in each vial.

 

3. Graph the results from the corrected difference column for the germinating peas and dry peas at both room temperature and at 10 degrees Celsius. On graph paper.

 

4. Describe and explain the relationship between the amount of oxygen gas consumed and time. The oxygen gas consumed increased fairly constantly in respect to time.

 

5. From the slope of the four lines on the graph, determine the rate of oxygen gas consumption of germinating and dry peas during the experiments at room temperature and at 10 degrees Celsius. Recall that rate = _y over _x.

 

ConditionShow Calculations HereRate in mL oxygen gas/minute
Germinating Peas/10°C(1.2-0.7)/50.1
Germinating Peas/ Room Temperature(2.7-2.1)/50.12
Dry Peas/10°C(-.4-0)/5-0.08
Dry Peas/Room Temperature(-.3-(-.3))/50

 

6. Why is it necessary to correct the readings from the peas with the readings from the beads? The gas changes in the beads were only due to pressure and temperature, and not gas consumption, so the beads act as a control.

 

7. Explain the effect of germination (versus non-germination) on pea seed respiration. Germinating peas consumed more oxygen than non-germinating.

 

8. Below is a sample graph of possible data obtained for oxygen consumption by germinating peas up to about 8 degrees Celsius. Draw in predicted results through 45 degree Celsius. Explain your predictions.

 

 

 

 

 

 

 

 

 

 

 

 

The amount of oxygen consumed will steadily increase until the temperature reaches a point at which the enzymes become denatured.

 

9. What is the purpose of KOH in this experiment? The purpose of the KOH was to remove the effect of carbon dioxide from the readings.

 

10. Why did the vial have to be completely sealed around the stopper? The vial had to be completely sealed so that gases couldn’t escape and water couldn’t leak into the respirometer.

 

11. If you used the same experimental design to compare the rates of respiration of a 25 g. reptile and a 25 g. mammal, at 10 degrees Celsius, what results would you expect? Explain your reasoning. The reptile would use less oxygen because it is cold-blooded and wouldn’t be as active at a colder temperature as the mammal would.

 

12. If respiration in a small mammal were studied at both room temperature (21°C) and 10°C, what results would you predict? Explain your reasoning. The respiration of the small mammal would be higher at 10 degrees Celsius because it would need more energy to keep its normal body temperature.

 

13. Explain why water moved into the respirometers’ pipettes. Water moved into the respirometer’s pipettes because pressure decreased when the amount of oxygen was decreased.

 

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?

 

Set up five respirometers containing beads, non-germinating peas, peas that have germinated for one day, peas that have germinated for two days, and peas that have germinated for three days. Measure the water readings in intervals of five minutes for twenty minutes. The peas that have been germinating for three days will have the highest rate of respiration and the beads will have the lowest rate of respiration.

 

15. According to your graph, what happens to the rate of oxygen consumed by germinating peas over time? What does this indicate to you? The rate of oxygen consumption is fairly constant.

 

16. How did the KOH affect the water movement in the respirometer? It allows more water into the pipette.

 

17. Which of the two pea types, germinating or non-germinating, consumes the most oxygen? Why? Germinating peas consume more oxygen because they are growing and are more active than non-germinating peas.

 

18. What was the effect of temperature on pea respiration? Warmer temperatures allow for the peas to respire at a faster rate.

 

19. During aerobic respiration, glucose is broken down to form several end products. Which end products contain the carbon atoms from glucose? The hydrogen atoms from glucose? The oxygen atoms from glucose? The energy stored in the glucose molecules? Carbon dioxide contains the carbon, water contains the hydrogen, both carbon dioxide and water contain the oxygen, and ATP contains the energy.

 

20. What is fermentation? What are the two types of fermentation? What organisms use fermentation? Fermentation is a catabolic process that makes a limited amount of ATP from glucose without an electron transport chain and that produces an end-product such as ethyl alcohol or lactic acid. The two types of fermentation are alcoholic and lactic acid fermentation. Plants use alcoholic while animals use lactic acid.

 

21. Draw a Venn diagram showing how respiration and fermentation are similar and how they differ.

 

 

 

 

 

 

 

 

 

 

22. What are the three pathways involved in the complete breakdown of glucose to carbon dioxide and water? What reaction is needed to join two of these pathways? What are the substrates and products of this reaction and where does it take place? The three pathways are glycolysis, the electron transport chain, and the Krebs Cycle. The reaction of the pyruvate joining with CoA enzyme and NAD to produce acetyl CoA, NADH, and carbon dioxide. The acetyl CoA goes to the Krebs Cycle and NADH to the electron transport chain in the mitochondria.

23. Write the letter of the pathway that best fits each of the following processes.

Pathway

a. Glycolysis

b. Krebs Cycle

c. Electron Transport System

Process

1. Carbon dioxide is given off b.

2. Water is formed c.

3. PGAL a.

4. NADH becomes NAD+ c.

5. Oxidative phosphorylation c.

6. Cytochrome carriers c.

7. Pyruvate a.

8. FAD becomes FADH2 b.

24. Calculate the energy yield of glycolysis and cellular respiration per glucose molecule. Distinguish between substrate-level phosphorylation and oxidative phosphorylation. Where does the energy for oxidative phosphorylation come from? 36 ATPs are formed per glucose moelcule. Four of the ATPs are formed from substrate level and 32 from oxidative.

 

  

Substrate

OxidativeTotal
Glycolysis246
Transition066
Krebs22224
Total43236

 

 

Error Analysis:

Some of the errors that could have occurred during the experiment were water leaking into the respirometers, gas escaping the respirometers, water temperature in the bath changing, and mathematical mistakes.

Discussion and Conclusion:

This lab showed that germinating peas consumed more oxygen at a faster rate than the non-germinating peas and the beads did. The non-germinating peas and the beads didn’t consume hardly any oxygen at all. It also showed that the respiration rate of germinating peas was faster than the respiration rate of non-germinating peas. Finally, this experiment showed that respiration rates increase as the temperature increases. This shows that temperature and respiration rates are directly proportional to each other.

AP Lab 3 Sample 4 Mitosis

 

 

Lab 3     Mitosis & Meiosis

 

 

Introduction

 

All cells come from other cells. New cells are formed during cell division which involves both the replication of the cell’s nucleus and division of the cytoplasm. The two kinds of cellular division are mitosis and meiosis. Mitosis usually makes body cells, somatic cells. Mitosis is used in adult cells for asexual reproduction, regeneration, and the maintenance and repair of body parts. The process called meiosis makes gametes, sperm and eggs, and spores in plants. Gamete or spore cells have half the chromosomes that the parent cell has.

Mitosis is the first of the cell divisions studied in this lab. It is easily observed in cells that are growing at a fast paces such as whitefish blastula or onion root tips, which are used in this lab. The onion root tips have the highest percentage of cells going through mitosis. The whitefish blastula is formed directly after the egg is fertilized. This is a period of a fast paced growth and numerous cellular divisions where mitosis can be observed. Just before mitosis the cell is in interphase, a part of the cell cycle where the cell has a distinct nucleus and nucleoli. Next is prophase, where the chromatin thickens into distinct chromosomes and the nuclear envelope breaks open releasing them into the cytoplasm. The first signs of the spindle apparatus begin to appear. Next the cell begins metaphase, where the spindle attaches to the centromere of each chromosome pair and moves them to the middle of the cell. This level position is called the metaphase plate. Then anaphase begins when the chromatids are separated and pulled to the opposite poles. The final stage is telophase where the nuclear envelope is reformed and the chromosomes gradually uncoil. Cytokinesis then may occur forming a cleavage furrow and then the two daughter cells will separate.

Meiosis is more complex and involves two nuclear divisions. The two divisions are called Meiosis I and Meiosis II. These two divisions result in the production of four haploid gametes. This process allows increased genetic variation due to crossing over where genes can be exchanged. The process, like mitosis, depends on interphase to replicate the DNA. Meiosis begins with prophase I. In this stage, homologous chromosomes move together to form a tetrad. This is where crossing over occurs resulting in the recombination of genes. Metaphase I moves the tetrads to the metaphase plate in the middle of the cell, and anaphase I reduces the tetrads to their original two stranded form and moves them to opposite poles. Telophase I then prepares the cell for its second division. Meiosis II is just like mitosis except that the daughter cells are haploid instead of diploid. DNA replication does not occur in interphase II, and prophase II, metaphase II, anaphase II, and telophase II occur as usual. The only change is the number of chromosomes.

 

Hypothesis

 

Mitosis is easily observed in the whitefish blastula and the onion root tip. Meiosis and crossing over occurs in the production of gametes, in animals, and spores, in plants.

 

Materials

 

Lab 3A. 1

The materials used in this lab are as follows: light microscopes, prepared slides of whitefish blastula and onion root tips, pencil, and paper.

Lab 3A. 2

The materials used in this lab are as follows: light microscopes, prepared slides of onion root tips, paper, and pencil.

Lab 3B. 1

The materials used in this section of the lab are as follows: a chromosome simulation kit, pencil and paper.

Lab 3B. 2

The materials used in this section of the lab are as follows: light microscopes, prepared slides of Sordaria fimicola, pencil, and paper.

 

Methods

 

Lab 3A. 1

Observe prepared slides of whitefish blastula and onion root tips under the 10X and 40X objectives. Sketch and identify each section of cell division.

Lab 3A. 2

Observe every cell and determine what stage the cell is in. Count at least 200 cells total, separating them into groups of the same phase. Consider it takes 24 hours for the onion root-tip cells to complete the cell cycle.

Lab 3B. 1

Use the lab book to show how to make the chromosomes. The simulation kit has plenty of beads to use. There are red and yellow beads to be used to show the different chromatids. There is also a piece that resembles half of a centromere which has a magnet to connect to another one.

Lab 3B. 2

Use a light microscope to observe the prepared slide and record all data.

Results

Lab 3A. 1

The sketches below show the phases of mitosis for the onion root-tip.

 

 

The sketches below show the phases of mitosis for the whitefish blastula.

 

Why is it more accurate to call mitosis “nuclear replication” rather than “cellular division”?

 

In mitosis, two new nuclei are formed and the cytokinesis is just part mitosis.

 

Explain why the whitefish blastula and onion root tip are selected for a study of mitosis.

 

The blastula is generally a ball of cells that are rapidly going under mitosis. The onion root-tip is an area where mitosis also occurs very rapidly.

Lab 3A. 2

The table below shows the data collected for this lab.

 

Number of Cells

Field 1Field 2Field 3Total
Interphase86385317778.671132.8
Prophase957219.33134.4
Metaphase625135.7883.2
Anaphase31262.6738.48
Telophase31373.1144.8
225

 

 

If your observations had not been restricted to the area of the root tip that is actively dividing, how would your results have been different?

 

There would have been less cell division since most of the plant’s growth occurs at the root-tip.

 

Based on the data in the table above, what can your infer about the relative length of time an onion root-tip cell spends in each stage of cell division?

 

The most time of an onion root-tip cell’s life is in interphase. Prophase is the next most common phase that the cells spend in.

Lab 3B. 1

 

List three major differences between the events of mitosis and meiosis.

 

Mitosis has only one nuclear division while meiosis has two. Mitosis also makes two diploid cells and meiosis makes four haploid cells. Crossing over is also only found in meiosis.

 

Compare mitosis and meiosis with respect to each of the following:

 

 

MitosisMeiosis
Chromosome number of parent cells2n2n
Number of DNA replications11
Number of divisions12
Number of daughter cells produced24
Chromosome number of daughter cells2nn
PurposeRepair and growthGamete and spore production

 

 

How are Meiosis I and Meiosis II different?

 

Meiosis I ends in two chromosomes with two chromatids and Meiosis II ends in four chromosomes with only one chromatid.

 

How do oogenesis and spermatogenesis differ?

 

Oogeneis forms the eggs and spermatogenesis forms the sperm.

 

Why is meiosis important for sexual reproduction?

 

Meiosis makes the chromosome number come out in half so that fertilization can come back and restore the diploid number.

Lab 3B. 2

 

Number of 4:4Number of Asci Showing CrossoverTotal Asci% Asci Showing Crossover Divided by 2Gene to Centromere Distance (Map Units)
3919589.59.5 map units

 

 

Draw a pair of chromosomes in MI and MII, and show how you would get a 2:4:2 arrangement of ascospores by crossing over.

 

 

Error Analysis

 

No errors were reported in this lab but there could have been. The cells in Lab 3A. 1 could have been miscounted or not counted.

 

Conclusion

 

From the data collected in this experiment, it can be concluded that the mitotic stages of the whitefish blastula and the onion root-tip can be observed with a light microscope. The time spent in each phase of mitosis can be recorded and it is concluded that the most time spent in a stage is in interphase. It can also be concluded that the least time spent in a stage is in telophase. It is also understood that someone can simulate meiosis using a chromosome simulation kit. On the last part of the lab, Lab 3B. 2, one could conclude that more asci do not cross over than do the number of asci that do cross over.

AP Lab 3 Sample 3 Mitosis

 

 

Lab 3    Mitosis and Meiosis

 

 

Introduction:

 

All new cells come from previously existing cells. New cells are formed by karyokinesis- the process in cell division which involves replication of the cell’s nucleus and cytokinesis-the process in cell division which involves division of the cytoplasm. Two types of nuclear division include mitosis and meiosis. Mitosis typically results in new somatic, or body, cells. Mitotic cell division is involved in the formation of an adult organism from a fertilized egg, asexual reproduction, regeneration, and maintenance or repair of body parts. Meiosis results in the formation of either gametes in animals or spores in plants. The cells formed have half the chromosome number of the parent cell.

Mitosis is best observed in cells that are growing at a rapid pace, such as in the whitefish blastula or onion root cell tips. The root tips contain a special growth region called the apical meristem where the highest percentage of cells are undergoing mitosis. The whitefish blastula is formed immediately after the egg is fertilized, a period of rapid growth and numerous cell divisions where mitosis can be observed.

There are several stages included in before, during, and following mitosis. Interphase occurs right before a cell enters mitosis. During interphase, the cell will have a distinct nucleus with one or more nucleoli, which is filled with a fine network of threads of chromatin. During interphase, DNA replication occurs. After duplication the cell is ready to begin mitosis. Prophase is when the chromatin thickens until condensed into distinct chromosomes. The nuclear envelope dissolves and chromosomes are in the cytoplasm. The first signs of the microtubule-containing spindle also begin to appear. Next the cell begins metaphase. During this phase, the centromere of each chromosome attaches to the spindle and are moved to the center of the cell. This level position is called the metaphase plate. The chromatids separate and pull to opposite poles during the start of anaphase. Once the two chromatids are separate, each is called a chromosome. The last stage of mitosis is telophase. At this time, a new nuclear envelope is formed and the chromosomes gradually uncoil, forming the fine chromatin network seen in interphase. Cytokinesis may occur forming a cleavage furrow that will form two daughter cells when separated.

Meiosis is more complex than mitotic stages and involves two nuclear divisions called Meiosis I and Meiosis II. They result in the production of four haploid gametes and allow genetic variation because of crossing over of genetic material. Prior the process, interphase replicates the DNA. During prophase I, the first meiotic stage, homologous chromosomes move together to form a tetrad and synapsis also begins. This is where crossing over occurs, resulting in the recombination of genes. In Metaphase I, the tetrads move to the metaphase plate in the middle of the cell as on mitotic metaphase. Anaphase I brings the tetrads back to their original two stranded form and moves them to opposite poles. During Telophase I, the centriole is finished and the cell prepares for a second division. In Meiosis II, in Prophase II, centrioles move to opposite ends of the chromosome group. In Metaphase II, the chromosomes are centered within the center of each daughter cell. Anaphase II involves the centromere of the chromatids separating. Telophase II occurs when the divided chromosomes separate into different cells, known as haploid cells.

Sordaria fimicola, an ascomycete fungus, can be used to demonstrate the results of crossing over during meiosis. It spends most of its life haploid and only becomes diploid when the fusion of the mycelia of two different strains results in the fusion of two different types of haploid nuclei to form a diploid nucleus. Meiosis, followed by mitosis, in Sordaria results in the formation of eight haploid ascospores contained within a sac called an ascus. They are contained in a perithecium, a fruiting body, until mature enough to be released. The arrangement of spores directly reflects whether or not crossing over occurred. If an ascus has four tan ascospores in a row and four black ascospores in a row -4:4 arrangement, then no crossing over has taken place. If the asci has black and tan ascospores in sets of two -2:2:2:2 arrangement, or two pairs of black ascospores and four tan ascospores in the middle -2:4:2 arrangement, then crossing over has taken place.

 

Hypothesis:

 

The stages of mitosis can be examined in whitefish blastula and onion root cell tips by using a microscope. The process of crossing over and the stages of meiosis only occur during the creation of gametes and spores.

 

Materials:

 

Exercise 3A

The materials necessary for this exercise are a light microscope, prepared slides of whitefish blastula, onion root cell tips, pencil, and paper.

Exercise 3B

For this portion of the lab, materials needed are a bag of color-coded connecting beads and magnetized “centromeres,” several trays, and labels marked interphase, prophase, metaphase, anaphase, and telophase.

 

Methods:

 

Exercise 3A.1: Observing Mitosis

During this experiment, prepared slides of whitefish blastula and onion root tips should be observed under the 10X and 40X objectives of a light microscope. A cell in each stage of mitosis should be identified and sketched.

Exercise 3A.2: Time for Cell Replication

In this section of the lab, use the highest power objective on the microscope to observe and count every cell in the field of view. The cells should be counted according to the stage of mitosis they are in. At least 200 cells and 2 fields of view should be examined and counted. The percentage of cells in each stage is then recorded and the amount of time spent in each phase is calculated.

Exercise 3B.1: Simulation of Meiosis

For this portion of the experiment, a chromosome simulation kit will be used to demonstrate meiosis. Two sets of two strands with each set a different color, are connected to simulate DNA replication in both of the homologous pairs, the stage called interphase. Next, the chromosomes were entwined to represent synapsis in the stage known as prophase. Sections of beads were entwined between the pairs as in crossing over and aligned at the equator. Beads of each pair exchange places, representing metaphase. Next, anaphase was simulated by the homologous pairs being separated to opposite sides of the tray, or in terms of the “chromosomes,” the cell. Pushing the chromosomes into two separate cells, or trays, mimicked telophase.

Meiosis II was simulated as well. Prophase II is shown by the separation of the two beads, but no true change. The chromosomes again move to the equator during metaphase II, and in anaphase II, the two chromatids are separated and moved to opposite poles. Telophase II separates the chromosomes into four different cells.

Exercise 3B.2: Crossing Over during Meiosis in Sordaria

Prepared slides of Sordaria fimicola were observed under a light microscope. The asci were identified as either 4:4 or asci showing crossover. These readings were recorded. The percentage of each and map units were calculated.

 

Results:

 

Exercise 3A

 

Whitefish Blastula

Onion Root Cell Tips

 

 

Why is it more accurate to call mitosis “nuclear replication” rather than “cellular division”? It is more accurate to describe mitosis as “nuclear replication” because the cell does not divide in any of the mitotic steps. The entire process of mitosis is a series of steps that divides the nucleus into two separate nuclei at opposite poles. When a cell is truly split, the process is known as cytokinesis.

 

Explain why the whitefish blastula and onion root tips are selected for a study of mitosis. The blastula is what is formed directly following fertilization and, therefore, the cell is growing and many of the phases can be seen at this time. Onion root tip cells are also specimens that include a large amount of cell growth and a high percentage of cells experiencing mitotic activities.

 

Table 1: Number of Cells in Each Stage of Mitosis and Amount of Time Spent in Each Stage

 

 

 

Number of Cells

 

Field 1

 

Field 2

Total
 

Interphase

7110117273.2%1054.0
 

Prophase

13152812.0%171.6
 

Metaphase

12132510.6%153.2
 

Anaphase

1231.3%18.4
 

Telophase

3473.0%42.9
 

Total Cells Counted

235

 

 

If your observations had not been restricted to the area of the root tip that is actively dividing, how would your results differ? The majority of the cells would be in the stage of interphase and the results would be more difficult to gain and inaccurate.

 

Based on the data in Table 3.1, what can you infer about the relative length of time an onion root-tip cell spends in each stage of cell division? Prophase is the longest stage of mitosis (though Interphase, which occurs prior mitosis, takes up the most time of the cell’s life). Then, based on the data gained, the time spent in each stage decreases as you go further along.

 

Exercise 3B

 

List three major differences between the events of mitosis and meiosis. In mitosis, the nucleus divides once, and in meiosis, the nucleus is divided twice. Mitosis produces two identical daughter cells and meiosis produces up to four different cells. Synapsis and crossing over do not take place in mitosis, but do in meiosis.

 

Compare mitosis and meiosis with respect to each of the following.

Table 2: Comparing Mitosis and Meiosis

 

 

 

Topic Being Compared

 

Mitosis

 

Meiosis

 

Chromosome number of Parent Cells

Diploid (2n)Diploid (2n)
 

Number of DNA Replications

OnceOnce
 

Number of Divisions

OneTwo
 

Number of Daughter Cells

TwoFour
 

Chromosome Number of Daughter Cells

Diploid (2n)Haploid (n)
 

Purpose

Growth and repairProduction of gametes or spores

 

 

 

How are Meiosis I and Meiosis II different? Meiosis I begins with a tetrad and separates the homologous pairs. Meiosis II separates the two sister chromatids into haploids.

 

How do oogenesis and spermatogenesis differ? Oogenesis produces egg cells and spermatogenesis produces sperm cells.

 

Why is meiosis important for sexual reproduction? In meiosis the chromosome number is reduced to n so that it can be fertilized and void of any related (fertilized 2n) defects. Crossing- over occurs during meiosis, allowing for variations in the organisms created.

 

Table 3: The Number of Crossovers and Non-Crossovers

 

 

Number of 4:4

 

Number of Asci Showing Crossover

 

Total Asci

 

% Asci Showing Crossover Divided by 2

 

Gene to Centromere Distance (Map Units)

596812726.8%(1)

 

2. Draw a pair of chromosomes in MI and MII, and show how you would get a 2:4:2 arrangement of ascospores by crossing over.

Error Analysis:

 

Because the results gathered in the lab were based mostly on observations and sketching, chances of error are slim. However, when counting the number of cells in specific stages in Exercise 3A, mistakes could have occurred. When identifying these stages in Exercise 3A, mistakes were also possible.

 

Discussion and Conclusion:

 

The stages of mitosis were observed and timed in Exercise 3A. These stages are prophase, metaphase, anaphase, and telophase. Prophase is the most time-consuming phase, while anaphase is the least time-consuming. Mitosis is just one portion of a cell’s life. The longest time of a cell’s life (73% to be exact) is spent in interphase, a phase just prior to prophase. During this phase, DNA replication takes place. Prophase involves the first signs of cell division with a thickening of the chromatin threads until the chromatin is condensed to chromosomes. In metaphase the chromosomes move to the center of the spindle and the centromere attaches to the spindle. During anaphase the chromatids are separated and moved to opposite ends of the poles. The final stage, telophase, involves the condensation of the chromosomes and the formation of a new nuclear envelope. Following telophase, cytokinesis may occur and the cytoplasm will be divided into two cells.

During the first section of Exercise 3B, the stages of meiosis were simulated using magnetic beads and centromeres with trays serving as the “cell.” Crossing over in Sordaria was observed using a microscope in the second portion of Exercise 3B. Using the information, the map units were then determined. The distance of the gene relative to the centromere in the Sordaria was 26.8 map units.

AP Lecture Guide 06 – Metabolism & Enzymes

 

AP Biology: Chapter 6 

 

METABOLISM & ENZYMES

 

1. Define the following terms:

a. Catabolic pathway ________________________________________________________

b. Anabolic pathway _________________________________________________________

c. Kinetic energy ____________________________________________________________

d. Potential energy __________________________________________________________

2. The First Law of Thermodynamics is the principle of… _______________________________

___________________________________________________________________________

___________________________________________________________________________

3. The Second Law of Thermodynamics involves changes in… __________________________

___________________________________________________________________________

___________________________________________________________________________

4. What is meant by a change in free energy? ________________________________________

___________________________________________________________________________

5. Compare reactions that are…

a. Exergonic _______________________________________________________________

________________________________________________________________________

b. Endergonic ______________________________________________________________

________________________________________________________________________

6. Sketch the ATP cycle:

 

 

7. How does ATP “couple reactions”? ______________________________________________

___________________________________________________________________________

___________________________________________________________________________

8. Sketch the profile of an exergonic reaction.

 

 

9. How do enzymes affect the energy profile? ________________________________________

___________________________________________________________________________

10. Define activation energy. ______________________________________________________

___________________________________________________________________________

11. Why are enzymes said to be specific? ____________________________________________

___________________________________________________________________________

___________________________________________________________________________

12. List factors that influence the rate of enzyme reactions. ______________________________

___________________________________________________________________________

___________________________________________________________________________

13. Label the diagram of the catalytic enzyme cycle.

14. How do competitive and noncompetitive inhibitors differ in their enzyme interactions?

___________________________________________________________________________

___________________________________________________________________________

15. What happens during allosteric regulation? ________________________________________

___________________________________________________________________________

___________________________________________________________________________

16. Describe feedback inhibition. ___________________________________________________

___________________________________________________________________________

___________________________________________________________________________

17. Define enzyme cooperativity. ___________________________________________________

___________________________________________________________________________

___________________________________________________________________________

Cell Respiration

 

Cell Respiration

Overview:
In this experiment, you will work with seeds that are living but dormant. A seed contains an embryo plant and a food supply surrounded by a seed coat. When the necessary conditions are met, germination occurs, and the rate of cellular respiration greatly increases. In this experiment you will measure oxygen consumption during germination. You will measure the change in gas volume in respirometers containing either germinating or non-germinating pea seeds. In addition, you will measure the rate of respiration of these peas at two different temperatures.

Objectives:
Before doing this laboratory you should understand:

  • how a respirometer works in terms of the gas laws; and
  • the general processes of metabolism in living organisms.

After doing this laboratory you should be able to:

  • calculate the rate of cell respiration from experimental data.
  • relate gas production to respiration rate; and
  • test the effect of temperature on the rate of cell respiration in ungerminated versus germinated seeds in a controlled experiment.

Introduction:
Cellular respiration is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. Cellular respiration involves a series of enzyme-mediated reactions. The equation below shows the complete oxidation of glucose. Oxygen is required for this energy-releasing process to occur.

C6H12O6 + 6O2 —–> 6 CO2 + 6 H2O + 686 kilocalories of energy / mole of glucose oxidized

By studying the equation above, you will notice there are three ways cellular respiration could be measured. One could measure the:

1. Consumption of O2 ( How many moles of oxygen are consumed in cellular respiration?)

2. Production of CO2 ( How many moles of carbon dioxide are produced by cellular respiration?)

3. Release of energy during cellular respiration.

In this experiment, the relative volume of O2 consumed by germinating and non-germinating (dry) peas at two different temperatures will be measured.

Background Information:
A number of physical laws relating to gases are important to the understanding of how the apparatus that you will use in this exercise works. The laws are summarized in the general gas law that states:

PV = nRT

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 ( its value is fixed), and

T is the temperature of the gas (in K0).

This law implies the following important concepts about gases:

1. If temperature and pressure are kept constant, then the volume of the gas is directly proportional to the number of molecules of gas.

2. If the temperature and volume remain constant, then the pressure of the gas changes in direct proportion to the number of molecules of gas present.

3. If the number of gas molecules and the temperature remain constant, then the pressure is inversely proportional to the volume.

4. 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.

It is also important to remember that gases and fluids flow from regions of high pressure to regions of low pressure.

In this experiment, the CO2 produced during cellular respiration will be removed by potassium hydroxide (KOH) and will form solid potassium carbonate (K2CO3) according to the following reaction.

CO2 + 2 KOH —-> K2CO3 + H2O

Since the carbon dioxide is being removed, the change in the volume of gas in the respirometer will be directly related to the amount of oxygen consumed. In the experimental apparatus if water temperature and volume remain constant, the water will move toward the region of lower pressure. During respiration, oxygen will be consumed. Its volume will be reduced, because the carbon dioxide produced is being converted to a solid. The net result is a decrease in gas volume within the tube, and a related decrease in pressure in the tube. The vial with glass beads alone will permit detection of any changes in volume due to atmospheric pressure changes or temperature changes. The amount of oxygen consumed will be measured over a period of time. Six respirometers should be set up as follows:

RespirometerTemperatureContents
1RoomGerminating seeds
2RoomDry Seeds and Beads
3RoomBeads
4100CGerminating Seeds
5100CDry Seeds and Beans
6100CBeads

Procedure:
 1.Prepare a room-temperature bath (approx. 25 degrees Celsius) and a cold-water bath (approx. 10 degrees Celsius).

2.Find the volume of 25 germinating peas by filling a 100mL graduated cylinder 50mL and measuring the displaced water.

3.Fill the graduated cylinder with 50mL water again and drop 25 non-germinating peas and add enough glass beads to attain an equal volume to the germinating peas.

4.Using the same procedure as in the previous two steps, find out how many glass beads are required to attain the same volume as the 25 germinating peas.

5.Repeat steps 2-4. These will go into the 10-degree bath.

6.To assemble 6 respirometers, obtain 6 vials, each with an attached stopper and pipette. Number the vials. Place a small wad of absorbent cotton in the bottom of each vial and, using a dropper, saturate the cotton with 15% KOH (potassium hydroxide). It is important that the same amount of KOH be used for each respirometer.

7.Place a small wad of dry, nonabsorbent cotton on top of the saturated cotton.

8.Place the first set of germinating peas, dry peas & beads, and glass beads in the first three vials, respectively. Place the next set of germinating peas, dry peas & beads, and glass beads in vials 4, 4, and 6, respectively. Insert the stopper with the calibrated pipette. Seal the set-up with silicone or petroleum jelly. Place a weighted collar on each end of the vial. Several washers around the pipette make good weights.

9.Make a sling of masking tape attached to each side of the water baths. This will hold the ends of the pipettes out of the water during an equilibration period of 7 minutes. Vials 1, 2, and 3 should be in the room temperature bath, and the other three should be in the 10 degree bath.

10.After 7 min., put all six set-ups entirely into the water. A little water should enter the pipettes and then stop. If the water continues to enter the pipette, check for leaks in the respirometer.

11.Allow the respirometers to equilibrate for 3 more minutes and then record the initial position of the water in each pipette to the nearest 0.01mL (time 0). Check the temperature in both baths and record. Record the water level in the six pipettes every 5 minutes for 20 minutes.

Table 5.1: Measurement of O2 Consumption by Soaked and Dry Pea Seeds at Room Temperature (250C) and 100C Using Volumetric Methods.

Temp
(oC)
Time
(min)
Beads AloneGerminating Peas

Dry Peas and Beans

Reading at time XDiff*Reading at time XDiff*Corrected Diff. ^Reading at time XDiff*Corrected diff ^
Initial – 0
0-5
5- 10
10 -15
15-20
Initial – 0
0-5
5- 10
10 -15
15-20

* difference = ( initial reading at time 0) – ( reading at time X )

^ corrected difference = ( initial pea seed reading at time 0 – pea seed reading at time X) – ( initial bead reading at time X).

Analysis of Results:
1. In this investigation, you are investigating both the effect of germination versus non-germination and warm temperature versus cold temperature on respiration rate. Identify the hypothesis being tested in this activity.

_______________________________________________________________________

_______________________________________________________________________

2. This activity uses a number of controls. Identify at least three of the control, and describe the purpose of each control.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

3. Graph the results from the corrected difference column for the germinating peas and dry peas at both room temperature and 100C.

a. What is the independent variable? ____________________________________________________

b. What is the dependent variable? ______________________________________________________

Graph Title: _____________________________________________________________________

Graph 5.1

 

4. Describe and explain the relationship between the amount of oxygen consumed and time.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

5. From the slope of the four lines on the graph, determine the rate of oxygen consumption of germinating and dry peas during the experiments at room temperature and 100C. Recall that rate = delta Y/delta X.

Table 5.2

ConditionShow Calculations HereRate in ml.O2 / min
Germinating Peas/100C 

 

 

 

Germinating peas /Room Temperature 

 

 

 

 

Dry peas/100C 

 

 

 

Dry Peas /Room Temperature 

 

 

 

 

6. Why is it necessary to correct the readings from the peas with the readings from the beads?

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

7. Explain the effect of germination ( versus non-germination) on peas seed respiration.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

8. What is the purpose of KOH in this experiment?

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

9. Why did the vial have to be completely sealed around the stopper?

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

10. If you used the same experimental design to compare the rates of respiration of a 25 g. reptile and a 25 g. mammal, at 100C, what results would you expect/ Explain your reasoning.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

11. If respiration in a small mammal were studied at both room temperature (210C) and 100C, what results would you predict? Explain your reasoning.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

12. Explain why water moved into the respirometer pipettes.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

________________________________________________________________________

AP LAB PAGE

 

DNA & Protein Synthesis Chapter 10 Worksheet

 

    DNA & Protein Synthesis

 

Section 10-1 DNA

1. What does DNA stand for?

2. What is DNA’s primary function?

3. What is the function of proteins?

4. What are the repeating subunits called that make up DNA?

5. Name the 3 parts of a DNA nucleotide.

6. Sketch and label a DNA nucleotide.

7. Name the 4 nitrogen bases on DNA.

8. What is the difference between a purine & a pyrimidine?

9. Name 2 purines.

10. Name 2 pyrimidines.

11.Who is responsible for determining the structure of the DNA molecule & in what year was this done?

12. The model of DNA is known as a ____________________________ because it is composed of two ___________________ chains wrapped around each other.

13. What makes up the sides of a DNA molecule?

14. What makes up the “steps” of a DNA molecule?

15. How did Rosalind Franklin contribute to determining the structure of DNA?

16. What type of bonds holds the DNA bases together? Are they strong or weak bonds?

17. What makes up the “backbone” of the DNA molecule?

18. On DNA, a ____________________ base will always pair with a __________________ base.

19. What is the most common form of DNA found in organisms?

20. How many base pairs are in a full turn or twist of a DNA molecule?

21. Name the complementary base pairs on DNA.

22. How many hydrogen bonds link cytosine & guanine? adenine & thymine?

23. How does the nucleotide sequence in one chain of DNA compare with the other chain of DNA?

24. Why must DNA be able to make copies of itself?

25. Define DNA replication.

26. What is the first step that must occur in DNA replication?

27. What acts as the template in DNA replication?

28. What is a replication fork?

29. What enzymes help separate the 2 strands of nucleotides on DNA? What bonds do they break?

30. What is the function of DNA polymerases?

31. ____________________ are joined to replicating strands of DNA by ________________ bonds.

32. If the sequence of nucleotides on the original DNA strand was A – G – G – C – T – A, what would be the nucleotide sequence on the complementary strand of DNA?

33. Does replication of DNA begin at one end and proceed to the other? Explain.

34. Why does DNA replication take place at many places on the molecule simultaneously?

35. When replication is complete, how do the 2 new DNA molecules compare to each other & the original DNA molecule?

36. Is DNA replicated (copied) before or after cell division?

37. Sketch & label DNA replication. (Figure 10-5, page 188)

38. What is the error rate in DNA replication? What helps lower this error rate to 1 in 1 billion nucleotides?

39. What is a mutation?

40. Name several things that can cause DNA mutations.

 

Section 10-2 RNA

 

41. What sugar is found on DNA?

42. What base is missing on RNA, & what other base replaces it?

43. Uracil will pair with what other on DNA?

44. Is RNA double or single stranded?

45. Name the 3 types of RNA and tell the shape of each.

46. Which type of RNA copies DNA’s instructions in the nucleus?

47. Which type of RNA is most abundant?

48. What does tRNA transport?

49. What 2 things make up ribosomes?

50. Define transcription.

51. In what part of a cell are proteins made?

52. What is RNA polymerase & tell its function.

53. What are promoters?

54. Where does RNA polymerase bind to the DNA it is transcribing?

55.What makes the beginning of a new gene on DNA in eukaryotes?

56. What do promoters mark the beginning of on prokaryotic DNA?

57. When a promoter binds to DNA, What happens to the double helix?

58. Are both strands of DNA copied during transcription?

59. As RNA polymerase moves along the DNA template strand, what is being added?

60. What bases pair with each other during transcription?

61. What is the termination signal?

62. What happens when RNA polymerase reaches the termination signal?

63. What are the products of transcription called?

64. Transcripts are actually ____________________________ molecules.

65. In transcription, ________________________’s instructions for making a protein

are copied by _______________________.

66. Which RNA molecules are involved in the synthesis (making) of a protein?

67. What happens to the newly made mRNA molecule following transcription in the nucleus?

 

Section 10-3 Protein Synthesis

 

68. What makes up proteins, what are the subunits called, & what bonds them together?

69. How many different kinds of amino acids make up proteins?

70. What determines how protein polypeptides fold into 3-dimensional structures?

71. Why does a protein need a 3-dimensional structure?

72. What is the genetic code & why is it important?

73. What is a codon & what does each codon code for?

74. How many codons exist?

75. Name the amino acid coded for by each of these codons:

a. UUA

b. AUU

c. UGU

d. AAA

e. GAG

f. UAA

76. What codon starts protein synthesis?

77. What codons stop protein synthesis?

78. Proteins are synthesized (made) at what organelle in the cytosol?

79. Sketch and label a tRNA molecule & tell its function.

80. Define translation & tell how it starts.

81. Where are amino acids found in a cell & how are they transported?

82. What is an anticodon & where is it found on tRNA?

83. What codon on mRNA would bind with these anticodons: (use table 10-1, page 194)

a. AAA

b. GGA

c. UAC

d. CGU

84. What are ribosomes made of and in what 2 places can they be found in a cell?

85. What is the difference between proteins made by free ribosomes & those made by attached, membrane proteins on the ER?

86. How many binding sites are found on the ribosomes and what does each site hold?

87. To start making a protein or _________________________________, a ribosome attaches to the ______________________________ codon on the __________________ transcript.

88. The start codon, AUG, pairs with what anticodon on a tRNA molecule?

89. What amino acid does the start codon always carry?

90. What type of bonds are the ones that attach amino acids to each other in a growing polypeptide?

91. __________________________ are linked to make proteins as a ______________________ moves along the mRNA transcript.

92. What ends translation?

93. Can more than one ribosome at a time translate an mRNA transcript? Explain.

94. What determines the primary structure of a protein?

95. What would the translation of these mRNA transcripts produce?

a. UAA CAA GGA GCA UCC

b. UGA CCC GAU UUC AGC