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Lab 5 Ap Sample 2 Cell Resp

AP Lab 5 Cell Respiration

Introduction:
Cellular respiration is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria in each cell. Cellular respiration involves a number of enzyme mediated reactions. The equation for the oxidation glucose is C6H12O6 + O2 à CO2 + H2O + 686 kilocalories per mole of glucose oxidized. There are three ways cellular respiration could be measured. 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. In this lab, the volume of O2 consumed by germinating and non-germinating peas at two different temperatures will be measured.

PV=nRT is the inert gas law. 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. T is the temperature of the gas in degrees K. This law tells us 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.

In this lab, CO2 , made during cellular respiration will be removed by potassium hydroxide (KOH) and will make potassium carbonate (K2CO3). Carbon dioxide is removed so the change in the volume of gas in the respirometer will be directly proportional to the amount of oxygen that is consumed. In the experiment water will move toward the region of lower pressure. During respiration, oxygen will be consumed and its volume will be reduced to a solid. The result is a decrease in gas volume within the tube, and a related decrease in pressure in the tube. The respirometer with just the glass beads will allow changes in volume due to changes in atmospheric pressure or temperature changes.

Hypothesis:
The respirometer with only germinating peas will have a larger consumption of oxygen and will have a larger amount of CO2 that is converted into K2CO3 than the respirometer with beads and dry peas and the respirometer with beads alone.

Materials:
The materials used in the lab are as follows: a thermometer, 2 water baths, tap water, masking tape, germinating peas, non-germinating (dry) peas, 100 mL graduated n cylinder, 6 vials, 6 rubber stoppers, absorbent and non absorbent cotton, KOH, 5 mL syringe, 6 pipettes, ice, and 6 washers.

Methods:
First, set up both a room temperature 25oC and a 10oC water bath. Make sure you allow time to adjust the temperature in each bath. To obtain a temperature of 10oC add ice to of the baths until the temperature in the bath is 10oC. Next, obtain a 100 mL graduated cylinder and fill it with 50 mL of water. Drop in 25 germinating peas and determine the amount of water that is displaced. Record the volume of the 25 germinating peas. Then remove these peas and place them on a paper towel. They will be used in respirometer 1. Next, refill the graduated cylinder with 50 mL of water and drop 25 non-germinating peas into it. Then drop glass beads into the respirometer until the volume is equivalent to that of the expanded germinating peas. Remove the beads and peas. They will be used in respirometer 2. Next, 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 + 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 put a small wad of absorbent cotton in the bottom of each vial and, using the syringe, saturate the cotton with 15 % KOH making sure not to get the KOH on the sides of the respirometer. Then place a small wad of dry cotton on top of the KOH-soaked absorbent cotton. Repeat these steps to make the other five respirometers. Make sure to use about the same amount of cotton in each vial.

Next, place the first set of germinating peas, dry peas + beads and beads in vials 1,2, and 3. Place the second set of germinating peas, dry peas + beads, and 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.

Make a sling using the 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 of 25o C. Vials 4, 5, and 6 should be in the bath containing water that is 10oC. After the equilibration period completely immerse all six respirometers in the water completely. Water will enter the pipette for a short distance and stop. If it does not stop, there is a leak. Make sure the pipettes are facing so 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 in each pipette time 0. Check the temperature in both baths and record in table 5.1. Every five minutes for 20 minutes, take readings of the water’s position in each pipette, and record the data in table 5.1.

Results:

Table 5.1: the Measurement of Oxygen Consumption by Soaked and Dry Pea Seeds at Room Temperature 25o C and 10oC Using Volumetric methods.

Temp o C	Time (min)	Reading at time X	Diff.	Reading at time X	Diff.	Corrected Diff.	Reading at time X	Diff.	Corrected Diff.
25	Initial- 0	14.4		13.9			14.2
25	0 to 5	14.1	.3	13	.9	.6	14.1	.1	-.2
25	5 to 10	14.0	.4	11.1	2.8	2.4	13.9	.3	-.1
25	10 to 15	13.9	.5	10.3	3.6	3.1	13.7	.5	0
24	15 to 20	13.9	.5	8.8	5.1	4.6	13.5	.7	.2
10	Initial – 0	14.2		14.2			14.7
10	0 to 5	14.8	-.6	14.0	.2	.8	15.2	-.5	.1
10	5 to 10	14.6	-.4	13.5	.7	1.1	15	-.7	-.3
10	10 to 15	14.8	-.6	13.2	.9	1.5	15	-.7	-.1
10	15 to 20	14.9	-.7	12.6	1.6	2.3	15	-.7	0

Graph: Consumption of Oxygen for Germinating Peas and Dry Peas at 10oC and 25o C.

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.
The hypothesis being tested in this activity is that the germinating peas in a water bath of 25 o C will have a higher respiration rate than the other vials.

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

One control is each vial had the same volume. This showed that the volume of the vial did not effect respiration rate. Another control was the vial with beads alone. The beads carried out no respiration. The final control was the 10 minute equilibration period. This allowed the contents of the vials to carry out respiration for a short period of time before they were completely immersed in the water.

3.Graph the results from the corrected difference column for the germinating and dry peas at both room temperature and at 10oC.

4. Explain the relationship between the amount of oxygen consumed and time.

As time increased oxygen consumption increased.

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 at 10o C.

Condition	Show Calculations	Rate in mL O / minute
Germinating peas at 10oC	2.3-1.5=.8/5	.16mL O2 /minute
Germinating peas at room temperature	4.6-3.1/5	.3mL O2 /minute
Dry peas at 10oC	(.1)/5=	.02 mL O2 /minute
Dry peas at room temperature	(.2-0 )/5=	.04 mL O 2 /minute

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

The beads carried out no cellular respiration. The peas did. Changes in atmospheric pressure could have caused changes in respiration rate and correcting the readings provided the most accurate results under the given conditions.

7. Explain the effect of germination versus non-germination on pea seed respiration.

Germination causes a higher rate of respiration than the non-germinating peas.

8. Graph the predicted results through 45o C. Explain your prediction.

As the temperature increased cellular respiration increased, but after a certain temperature the respiration rate will start to go down. The peak is the optimal temperature.

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

KOH removes carbon dioxide formed during cellular respiration.

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

The stopper was completely sealed to prevent water from entering the respirometer.

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

The mammal would carry out a higher rate of cellular respiration. This is because the mammal maintains a constant temperature that is higher than the temperature of the cold blooded reptiles that will have a temperature of 10 C.

12. If respiration in a small mammal were studied at both room temperature 21 o C and 10oC what results would you predict? Explain your reasoning.

The rate of cellular respiration would be higher at 21 degrees C because the 10 degrees C temperature could cause the overall body of the mammal temperature to drop the most.

13. Explain why water moved into the respirometers’ pipettes.

Water moved into the pipettes because oxygen was being consumed and allowed water to move only partially into the pipette.

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?

I would use the same format using respirometers to measure the cellular respiration rate of the peas. The peas that had been germinating for 72 hours would have a higher respiration rate because they have a higher energy demand.

Error Analysis:
Several factors could have caused inaccurate results in this experiment. First, not maintaining a constant temperature in the water bath could have caused inaccurate results. Also moving the vials in the water after the experiment began could have caused inaccurate results. Putting your hands in the water bath while the vials were in the water could have caused inaccurate results. Allowing the peas to come into contact with the KOH could have also caused inaccurate results. Finally not having the same amount of cotton in each vial could have caused an error in the results.

Conclusion:
In this experiment the vial with just germinating peas had the greatest consumption of oxygen. This is because germinating peas carried out a more rapid process of cellular respiration than the non-germinating peas. The beads carried out no cellular respiration. The non-germinating peas require less energy than the germinating peas so the dry peas carry out a slower process of cellular respiration. This in turn caused less oxygen to be consumed in the vials with non-germinating peas than the vials with germinating peas. The higher temperature caused cellular respiration to occur at a higher rate which in turn caused a greater consumption of oxygen.

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Lab 5 Ap Sample 4

Cellular Respiration
AP Biology Lab 5

Introduction:
Cellular respiration is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within a cell. There are a number of physical laws that relate to gases and are important in the understanding of how the equipment in this lab works. These are summarized as general gas laws that state: PV=nRT where: P stands for pressure of the gas, V stands for the volume of the gas, n stands for the number of molecules of gas there are, R stands for the gas constant, and T stands for the temperature of the gas. A respirometer is the system used to measure cellular respiration. Pressure changes in the respirometer are directly relative to a change in the amount of gas there is in the respirometer as long as the volume and the temperature of the respirometer do not change. To judge the consumption of oxygen in two different respirometers you must reach equilibrium in both respirometers.

Cellular respiration is the procedure of changing the chemical energy of organic molecules into a type that can be used by organisms. Glucose may be oxidized completely if an adequate amount of oxygen is present. The equation for cellular respiration is C6H12O6 + 6O2 à 6CO2 + 6H2O + energy. Carbon dioxide is formed as oxygen is used. The pressure due to CO2 might cancel out any changes due to the consumption of oxygen. To get rid of this problem, a chemical will be added that will selectively take out the carbon dioxide put off. Potassium hydroxide will chemically react with the carbon dioxide by this equation: CO2 + K2CO3 + H2O.

Hypothesis:
The rate of cellular respiration will be higher in germinating peas in cold and room temperature water baths than in that of the beads or non-germinating peas. The cooler temperature in the cold water baths should slow the process of cellular respiration in the peas.

Materials:
The materials used in this lab were the following: a water bath, a graduated cylinder, a thermometer, tape, metal washers, beads, germinating peas, non-germinating peas, beads, beakers, another graduated cylinder, ice, paper, and a pencil are needed for this lab.

Methods:
Obtain a room temperature water bath and a 10-degree Celsius water bath. Add ice to room temperature water and watch the thermometer until the temperature has reached 10-degrees Celsius. For respirometer one, obtain a graduated cylinder and fill it with 50mL of water. Drop in 25 germinating peas and determine the water displacement. Record the volume, remove the peas and place them on a paper towel. For respirometer two, obtain the same graduated cylinder, filled again with 50mL of water. Drop twenty-five of the non-germinating peas in the water and continue adding beads to the water until the same water displacement for the non-germinating peas equals the first result. Remove the contents, and drain the water leaving the peas and beads to dry on a paper towel. For respirometer three, fill the 100mL graduated cylinder with 50mL of water and obtain the first water displacement value by adding just beads to the water in the cylinder. Take out the beads, allow the water to drain, and repeat this same procedure for respirometers 4, 5, and 6, which will be placed in the cooler water. For assembly of the respirometers, obtain 6 vials, each with a stopper and a pipette. Place a small wad of absorbent cotton in the bottom of each vial and using a dropper, saturate the cotton with 15% KOH solution. Make sure the vials are dry on the inside. Do not get KOH on the sides of the respirometer. Place a small wad of non-absorbent cotton on top of the KOH saturated cotton, making sure the same amount is used for each respirometer. Place the first set of peas in their respective vials. Do the same for the second set of peas. Insert the stopper with the calibrated pipette. Place a weighted collar on the end of each vial. Make a sling of masking tape attached to each side of the water baths to hold the pipettes out of the water during the equilibration period of seven minutes. Vials 1, 2, and 3, should rest in the room temperature water while 4, 5, and 6, should rest in the 10-degree Celsius water bath. After seven minutes of equilibration, immerse all 6 respirometers entirely in their designated water baths. Water enters the pipette for a short distance and stops. If the water continues to move into a pipette, check for leaks. Working quickly, arrange the pipettes so the can be read through the water at the beginning of the experiment. These should not be shifted during the experiment. Keep hands out of the water bath after the experiment has started. Make sure a constant temperature is maintained. Allow respirometers to equilibrate three more minutes, record the initial position of the water in each pipette to the nearest .01mL. Check the temperature in both water baths and record in table 5.1. Check and record every five minutes for twenty minutes by repeating the procedure for that task.

Data:

Table 5.1

Beads Alone		Germinating Peas		Dry peas and Beads
Read-ing @ time X	Diff.	Reading @ time X	Diff.	Corrected Diff.	Reading @ time X	Diff.	Corrected Diff.
Initial-0	14.0	——-	13.5	——-	—–	14.1	—-	—-
0-5	14.1	-0.1	13.4	0.1	0.2	14.4	-.3	-.2
5-10	14.0	0	13.2	0.3	0.3	14.5	-.4	-.4
10-15	14.1	-0.1	12.8	0.7	0.8	14.6	-.5	-.4
15-20	14.4	-0.4	12.2	1.3	1.7	14.9	-.8	-.4
Initial-0	14.8	——-	14.0	——-	—–	15	—-	—-
0-5	14.8	0	13.0	1.0	1.0	14.8	.2	.2
5-10	14.7	0.1	12.2	1.8	1.7	14.6	.4	.3
10-15	14.4	0.4	10.3	3.7	3.3	14.4	.6	.2
15-20	14.3	0.5	9.8	4.2	3.7	14.3	.7	.2

Graph 5.1

Table 5.2

Condition	Show Calculations Here	Rate in mL water/minutes
Germinating peas/ 10degrees Celsius	Sloped downward steadily, bigger drop off at the end	Rise=1.3 Rate=0.052
Germinating peas/ room temperature	Steady drop downward.	Rise=4.2 Rate=0.168
Non-germinating peas/ 10degrees Celsius	Steadily gained height.	Fall=1.5 Rate=0.06
Non-germinating peas/ room temperature	Steady fall in rate.	Fall=0.7 Rate=0.028

Graph 5.2

Questions:
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 hypothesis of this experiment was: The rate of cellular respiration will be higher in germinating peas in cold and room temperature water baths than in that of the beads or non-germinating peas. The cooler temperature in the cold water baths should slow the process of cellular respiration in the peas.

This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each. First of all, the water baths held a constant temperature. Secondly, the volume of KOH was constant from vial to vial. Lastly, the equilibration period was identical for all the respirometers.

Describe and explain the relationship between the amount of oxygen consumed and time. The amount of oxygen that was consumed was the greatest in the warmer water. The oxygen consumption increased over time in the germinating peas.

Why is it necessary to correct the readings from the peas with the readings from the beads? This is necessary to show the actual rate at which cellular respiration occurs in peas. The beads served as a control variable.

Explain the effects of germination versus non-germination on peas seed respiration. The germinating seeds are alive and growing, therefore respirate to grow.

Explain the results shown in the sample graph in your lab manual. As the temperature increased, the enzymes denatured so germination was inhibited.

What is the purpose of KOH in this experiment? The KOH drops absorbed the carbon dioxide so that it would not cause the put off of that gas to make the readings equilibrate.

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 100 degrees Celsius, what results would you expect? Explain your reasoning. Respiration would be higher in the mammal since they are warm-blooded.

If respiration in a small mammal were studied at both room temperature, 21-degrees Celsius and 10-degrees Celsius, what results would you predict? Explain your reasoning. Respiration would be higher at 21 degrees because the animal would have to keep its body temperature up. The results would multiply at 10-degrees because the mammal would have to keep its body that much warmer in comparison to the 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? A person could set up four respirometers which have one of the following: Seeds that have not begun to germinate, seeds germinating for one day, seeds germinating for two days and seeds germinating for three days. The results would probably be that there would be no oxygen used by the seeds that have not germinated yet. The seeds that have been germinating for three days will have the greatest amount of oxygen consumption.

Error Analysis:
Error in this lab could have occurred if the seals on the vials weren’t tight and there was a leak of water into the vials. Another source of error could have been if the KOH had touched the sides of the vial. Also, the absorbent cotton balls that were used for the KOH could have been too saturated. Another source of error could be at the temperature of the water baths. If a close eye wasn’t kept on the temperature, the ten degree Celsius would have fallen in degrees.

Conclusion:
Oxygen consumption in respirometers with germinating peas is greater than in the respirometers with non-germinating peas. Respiration was affected by the temperature of the water bath as well. Respiration occurs faster in the warmer water baths.

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Lab 4 Plant Pigments

Lab 4 Plant Pigments & Photosynthesis

Introduction:
The purpose of this lab experiment was to separate plant pigments using paper chromatography, and to measure the rate of photosynthesis in isolated chloroplasts. Because of capillary action the solvent moves up the paper causing the pigments to become visible at certain distances.

The substances visible on the paper are called pigments. Chlorophyll a is the main pigment that makes up about 75% of the pigmentation in plants. Chlorophyll b makes up about 25% of the pigmentation. And carotenes and xanthophylls are accessory pigments that make up the rest of the pigmentation. Carotene is the most soluble of the pigments and as a result will be carried the farthest by the solvent. The paper will display a spectrum of the pigments found in the spinach leaves. Using the formula Rf one can determine the relationship between the distance the solvent traveled to the distance the pigment traveled.

Rf=distance l2igment migrated (mm) distance solvent front migrated

Light is a part of a continuum of radiation or energy waves. The energy from visible light is used in the photosynthetic process. Light is absorbed in the leaf pigments, electrons within each photosystem are boosted to a higher energy level to produce ATP and to reduce NADP and NADPH. The ATP is then used in carbon fixation. This is the incorporation of CO2 into organic molecules.
To measure light transmittance in chloroplasts a spectrophotometer will be used. The reason behind measuring the light transmittance is to calculate the rate of photosynthesis in the chloroplasts. A solution called DPIP will be used in place of NADP to judge the color change of the chloroplast solutions. This technique is known as dye- reduction and it tests the hypothesis that light and chloroplasts are required for light reactions to occur.

Hypothesis:
In this experiment it is hypothesized that the cuvette with boiled chloroplasts and the cuvette kept in the dark containing unboiled chloroplasts will have very slight changes in light transmittance, whereas the cuvette containing unboiled chloroplasts that have been exposed to light will have an increasingly higher % transmittance over the course of time.

Materials and Methods:
Lab 4A:
The materials used in this section of the lab were: filter paper, glass vial, small amount of solvent, a quarter, and spinach leaves. The first step was to cut a point on one end of the filter paper and draw a pencil line 1.5cm from the tip of this point. Next a spinach leave was placed on the strip of paper and rolled over with a quarter on top of the pencil line. This gives a green line across the paper, which contains the pigments of the leave. Then the strip of paper was placed into the vial with the point down in the bottom. When the pigment reached the point 1 cm from the top of the vial then it was removed. The solvent front was then quickly marked with a pencil and then each pigment front was marked as well. From the distance the pigment traveled and the distance the solvent traveled the Rf value was calculated.

Lab 4B:
The materials used in this lab were: a spectrophotometer, 4 cuvettes, phosphate buffer, distilled water, boiled chloroplasts, unboiled chloroplasts, and DPIP .First the cuvettes were labeled 1-4 and cleaned with lens paper because even the oil from your hands can affect the transmittance of light through the cuvette. Cuvette 2 was then wrapped with foil to keep the contents in the dark. Next 1 ml of phosphate buffer was added to all four cuvettes, 4ml of distilled water was added to cuvette 1, 3rn1 of distilled water was added to cuvettes 2,3, and 4, and Iml of DPIP was added to cuvettes 2,3, and 4. Then 3 drops of unboiled chloroplasts were added to cuvette 1, it was covered with parafilm, placed into the spectrophotometer, and set to 100% transmittance. This cuvette was used to recalibrate between readings as well. Three drops of unboiled chloroplasts were placed in cuvette 2 and 3, and three drops of boiled chloroplasts were placed in cuvette 4. The cuvettes were then covered with parafilm. Each was placed in the spectrophotometer and the % transmittance of each, every five minutes for 15 minutes, was recorded.

Data:
Table 4.1

Questions:

1. Which pigment migrated the furthest and why? Carotene, it was the most soluble and didn’t form bonds with the filter paper.
2. Which of the 2 types of chlorophyll is more soluble? Chlorophyll a
3. Why do leaves change color in autumn? The chlorophyll production in the leaves slows down.
4. What is the function of the chlorophyll in photosynthesis? They absorb red and blue light rays.
5 .What are the accessory pigments and what are their functions? Carotene and xanthophylls both absorb different wavelengths of light than chlorophyll does.
6. What are some other ways chromatography is used to separate plant pigments? There are three types: Column, Paper, and Thin Layer chromatography.
7 .What does the & value represent? The distance the pigment traveled and the distance the solvent traveled expressed as one value.
8. What factors involved in the separation of the pigments? In this test it was solubility.
9 .Would you expect the Rf value of a pigment to be same if a different solvent were used? Explain. No, for a different solute there would be a different solvent rate.
10. What kind of chlorophyll does the reaction center contain? What are the roles of the other pigments? Chlorophyll a” the other pigments catch different wavelengths of light.

Error Analysis:

Fingerprints on either the filter paper or the cuvettes may have affected the experiment because the oil from your hands can get on these things and affect the results. The spectrophotometer may have not been calibrated correctly because this was the first time this particular one had been used. Other than these there were few places for error in this lab.

Conclusion:

From lab 4a we discovered that the many pigments found in chloroplasts are all involved in gathering energy from sunlight. The spectrum of color displayed on the filter paper showed the pigments and the solubility of each. In lab 4b the spectrophotometer measured the light transmittance through the various cuvettes and the chloroplast solutions in each. The actual purpose of this was to observe the DPIP go from a blue color to a clear color. This indicated that photosynthesis was occurring and at what rate it was occurring. The cuvette with the unboiled chloroplasts that had been exposed to light showed the biggest change in % transmittance, which indicates that the amount of light available has a very big effect on the rate at which the light reactions of photosynthesis occur .

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Lab & AP Sample 2

Plant Pigments and Photosynthesis

Introduction:
Photosynthesis has two main parts, which are the light dependent and the light –independent. In the light-dependent reactions pigments trap energy from light, and this energy is used to split water molecules (photolysis). The light-independent reactions or dark phase of photosynthesis involve the fixing of carbon dioxide. It makes glucose and fructose chains and also releases oxygen , which passes through the stomata of the plant.

Organisms that carry out photosynthesis making their own organic molecules are called autotrophic. Some autotrophic organisms include plants, algae, and blue-green bacteria. Plants have many varieties of pigments, all of which absorb different colors of light. Chlorophyll a is the primary plant pigment and makes up about three-fourths of all the plant pigments. It absorbs red and blue light and is not found in photosynthetic bacteria.

Chlorophyll b is another plant pigment. It absorbs blue-green and orange-red light. Carotenoids are a type of accessory pigment that absorb blue and blue-green light. These pigments are fat soluble and usually masked by chlorophyll a. Anthocyanin is another accessory pigment that absorbs bright red colors. There is also chlorophyll c and d that sometimes take the place of chlorophyll b.

Chromatography is a process used to separate mixtures that can separate plant pigments. This lab uses paper chromatography where a piece of paper is used to wick solvent up to the pigments and separate them according to solubilities. The rate of migration on a chromatogram is the Rf value.

Hypothesis

Plants contain several different pigments, and the rate of photosynthesis in plant cells is directly related to light and temperature.

Materials

Exercise 4A: Plant Pigment Chromatography

This exercise required 1 50-mL graduated cylinder, a small amount of a solvent, a stopper, filter paper, scissors, a pencil, spinach leaves, and a quarter.

Exercise 4B: Photosynthesis/The Light Reaction

The materials needed for this part of the lab were a spectrophotometer, a light, a water flask, a test tube rack, ice, 5 labeled cuvettes, lens tissue, foil, and parafilm. The substances put in the cuvettes were 5 mL of phosphate buffer, approximately 16 mL of distilled water, 9 drops of unboiled chloroplasts, and 3 drops of boiled chloroplasts.

Methods

Exercise 4A: Plant Pigment Chromatography

A 50-mL graduated cylinder was filled with about 1 cm of solvent and then tightly stoppered. The filter paper was then cut to a point on one end, and a line was drawn 1.5 cm above the point. Using the ribbed edge of a quarter, spinach cells were extracted onto the pencil line. This procedure was repeated 8-10 times using a new portion of the leaf each time. The filter paper was then placed in the cylinder with the tip barely touching the solvent and none of the edges touching the sides. When the solvent reached 1 cm below the top of the paper, it was removed from the cylinder. The solvent location was immediately marked, and then the bottom of each pigment band was also marked.

Exercise 4B: Photosynthesis/The Light Reaction

The spectrophotometer was set to 605 nm and allowed to warm up. The chloroplast suspensions were prepared the previous day, part of which were boiled, and stored on ice until they were ready for use. An incubation area was prepared with a flood light, water flask, and test tube rack, by using the flask as a heat sink between the light and the rack. Five cuvettes were numbered respectively and then wiped with lens tissue. The walls and bottom of cuvette 2 were covered with foil and a foil cap was made for the top.

To each cuvette 1 mL of phosphate buffer was added. Then, to cuvette 1 4 mL of distilled water was added, but to cuvettes 2, 3, and 4 3 mL of distilled water was added. Next, 1 mL of DPIP was added to cuvettes 2, 3,and 4. To cuvette 5, 3 mL plus 3 drops of distilled water were added and 1 mL of DPIP. To cuvette 1, 3 drops of unboiled chloroplasts were added.

The spectrophotometer was brought back to zero and the contents of cuvette 1 were mixed by inverting and placed in the sample holder. Cuvette 1 was used periodically through this experiment to recalibrate the spectrophotometer. Three drops of unboiled chloroplasts were added to cuvette 2. After removing the foil sleeve, it was placed in the sample holder and the transmittance was recorded. Additional readings were also taken at 5, 10, and 15 minutes. Next, three drops of unboiled chloroplasts were transferred to cuvette 3. The percent transmittance was recorded at 0, 5, 10, and 15 minutes. Three drops of boiled chloroplasts were added to cuvette 4, and the transmittances were recorded at the same times. Finally, cuvette 5 was mixed and placed in the sample holder. The transmittance readings were recorded.

Results

Table 4.1 Distance Moved by Pigment Band

Band Number	Distance (mm)	Band Color
1.	0 mm	Yellow-brown
2.	5 mm	Light green
3.	30 mm	Green
4.	48 mm	Yellow

Distance Solvent Front Moved 60 mm.

Table 4.2 Rf Values

0.8 = Rf for xanthophyll (yellow)

0.5 = Rf for chlorophyll a (bright green to blue green)

0 = Rf for chlorophyll b (yellow green to olive green)

Table 4.4 Transmittance (%)

Cuvette	0	5	10	15
2 Unboiled/Dark	41%	43%	44%	43%
3 Unboiled/Light	35%	38%	39%	37%
4 Boiled/Light	51%	52%	53%	54%
5 No Chloroplasts	57%	57%	56%	55%

Lab 4B Color Chart

Cuvette	Initial Color	Final Color
1	Clear	Clear
2	Light clear blue	Blue/green
3	Light clear blue	Dark clear blue
4	Light clear blue	Light clear blue
5	Light clear blue	Dark clear blue

Questions:
Exercise 4A: Plant Pigment Chromatography

What factors are involved in the separation of the pigments?

The solubility, size of particles, and their attractiveness to the paper are all involved in the separation.

Would you expect the Rf value of the pigment to be the same if a different solvent were used? Explain.

No, the different solubilities of the pigments would change the Rf values. For example chlorophyll b is only soluble to fat solutions.

What type of chlorophyll does the reaction center contain? What are the roles of the other pigments?

The reaction center contains chlorophyll a. The other pigments collect different light waves and transfer the energy to chlorophyll a.

Exercise 4B: Photosynthesis/The Light Reaction

What is the purpose of DPIP in this experiment?

DPIP is the electron acceptor in this experiment.

What molecule found in chloroplasts does DPIP “replace” in this experiment?

DPIP substitutes for the NADP molecules.

What is the source of the electrons that will reduce DPIP?

The electrons come from the photolysis of water.

What was measured with the spectrophotometer in this experiment?

The spectrophotometer measures the percentage of light transmittance through the cuvette due to DPIP reduction.

What is the effect of darkness on the reduction of DPIP? Explain.

The effect of darkness is that no reaction will occur.

What is the effect of boiling the chloroplasts on the subsequent reduction of DPIP? Explain.

Boiling denatures the protein molecules and stops the reduction.

What reasons can you give for the difference in the percent transmittance between the live chloroplasts that were incubated in the light and those that were kept in the dark?

In the dark cuvette, there was no light energy available, so there was no flow of electrons and no photolysis of water, while in the lighted cuvette these processes were allowed to continue.

Error Analysis
In Lab 4A, several mistakes could have been made with the chromatography paper. Too much handling, bending, or allowing the paper to touch the sides of the cylinder could all have affected the outcome of this experiment. The different pigment bands were also difficult to distinguish.

In Lab 4B, the procedure was very complicated and required a lot of pre-lab planning and reading. Incorrect usage of the spectrophotometer or neglecting to recalibrate often enough could have caused errors in this portion of the lab.

Discussion and Conclusion

Lab 4A demonstrated the different plant pigments by chromatography and showed how to calculate Rf values and explained their importance. There are 4-5 main pigments present in plants ranging from green to yellow in color. Lab 4B proves that light and chloroplasts are required for the light reactions of photosynthesis to occur. It showed the effects of boiling, denaturing, which did not allow photosynthesis to occur.

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Lab & Ap Sample 2 Mitosis & Meiosis

Mitosis & Meiosis -AP lab 3

Introduction
Cells come from preexisting cells. New cells are formed during cell division which involves both replication of the cell’s nucleus, karyokinesis, and division of the cytoplasm, cytokinesis. The two kinds of cellular division are mitosis and meiosis. Mitosis usually makes body cells, somatic cells. Making an adult organism from an egg, asexual reproduction, regeneration, and the maintenance and repair of body parts are performed during mitotic cell division. This process called meiosis makes gametes, in animals, and spores, in plants. Gamete or spore cells have half the chromosomes that the parent cell has.

In plants mitosis takes place in the meristems which are normally found at the tips of stems or roots. However, in animal cells cell division takes place every where as new cells are formed and old ones are replaced. Studying mitosis can be accomplished by looking at tissues where there are many cells in a process of meiosis. Two examples are an onion root tip, or developing embryos, in animals such as whitefish blastula. A blastula is formed after an egg is fertilized and the egg begins to divide. There are several phases of the mitotic cell cycle. A precursor to mitosis is interphase. The actual steps of the mitotic cell cycle are prophase, metaphase, anaphase, and telophase. Interphase is a stage in the cell cycle in which the cell is not dividing. The nucleus contains a nucleolus and also contains chromatin. During interphase DNA replication occurs. The first phase of mitotic cell division is prophase. During prophase the chromatin begins to thicken until noticeable chromosomes are formed. Each chromosome has two chromatids that are joined at the centromere. During the later part of prophase, the nuclear envelope and nucleolus disappear. Mitotic spindle fibers, composed of microtubules, also become apparent. Following prophase is metaphase. By the time the cell has reached metaphase the chromosomes have moved to the center of the mitotic spindle. The centromere of the chromosome attaches to the spindle. The centromeres of each chromosome line up on an area called the metaphase plate. Metaphase is followed by anaphase. In the beginning of anaphase, the centromeres of each pair of chromatids separate and moved by the spindle fibers to the opposite ends of the cell. When the daughter chromosomes reach the ends of the cell the form a clump at each spindle pole. The final phase of mitosis is telophase. Telophase is identified by a recognizable condensation of the chromosomes, which is followed by the formation of a new nuclear envelope. The chromosomes slowly uncoil into chromatin once again and the nucleoli and nuclear envelope reform. It is then possible for cytokinesis, the division of the cytoplasm into two cells, to occur. In an animal cells a cleavage furrow forms and the cell pinches off into two new daughter cells.

The process of meiosis involves two nuclear divisions that result in the formation of four haploid cells. Meiosis I, a reduction division, is the first division to reduce the chromosome number from diploid to haploid and separates the homologous pairs. Meiosis II separates the sister chromatids resulting in four haploid gametes. Unlike mitosis meiosis increases genetic variation. In meiosis I each pair of homologous chromosomes come together which is known as a synapse. Chromatids of homologous chromosomes may exchange parts which is called crossing over. The distance between two genes on a chromosome may be estimated by calculating the percentage of crossing over that takes place between them. Meiosis I is preceded by interphase. During interphase DNA synthesis occurs and each chromosome is made of two chromatids joined at the centromeres. The first step of meiosis I is prophase I. During prophase I homologous chromosomes come together and synapse. A tetrad consisting of four chromatids is also formed. Prophase I is followed by metaphase I. In metaphase I the crossed over tetrads line up in the center of the cell. In anaphase I the homologous chromosomes separate and are moved to opposite ends of the cell. The final phase of meiosis I is telophase I. During telophase I centriole duplication is completed. Most of the time cytokinesis and formation of the nuclear envelope occur in order two make to cells. Meiosis II a second mitotic cell division then takes place in order to separate the chromatids in the two daughter cells made in meiosis I. This reduces the amount of DNA to one strand per chromosome. This is the only difference between meiosis I and II. Before meiosis II there is period called interphase or interkenesis. DNA replication does not take place in interphase II. Interphase II is followed by prophase II, No DNA replication occurs in prophase II and replicated centrioles separate and move to opposite sides of the chromosome groups. During metaphase II the chromosomes are centered in the middle of each daughter cell. During anaphase II the centromere regions of the chromatids are separate. The last stage of meiosis II is telophase II. In telophase II the chromosomes are at opposite ends of the cell and a nuclear envelope forms, and sometimes the cytoplasm divides.

Sordaria fimicola is fungus that may be used to show the results of crossing over during meiosis. Sordaria throughout most of its life is haploid, but becomes diploid after the fusion of two different types of nuclei, which forms a diploid nucleus. In Sordaria meiosis results in the making of eight haploid ascospores found in a sac called an ascus. Most asci are found in a perithecium. The life cycle of Sordaria fimicola is as follows: a spore is discharged through an ascus. The ascospore then undergoes mitosis, which forms a filament. The filament then undergoes mitosis, which forms a mycelium. Mycelial fusion and fertilization then takes place. This forms a diploid zygote. The zygote undergoes mitosis to form four haploid nuclei. The nuclei also undergo mitosis and form eight haploid nuclei, which then form eight ascospores. When mycelia of a mutant strain of Sordaria and a wild type of Sordaria undergo meiosis four black and four tan ascospores form. The arrangement of the ascospores reflects whether crossing over has occurred or not. Gametes, egg and sperm, are made during meiosis. Each egg and sperm cell contains half the total chromosomes a normal cell of that species would have. When the egg and sperm unite during fertilization the total chromosome number is restored.

Exercise 3A.1 – Hypothesis
While looking at prepared slides of onion root tip cells and whitefish blastula cells under a microscope I will be able to identify and draw the stages of mitosis in these cells.

Materials
The materials used in this experiment were a light microscope and prepared slides of onion root tip cells and whitefish blastula cells.

Methods
Using the microscope examine the slides of onion root tip cells and whitefish blastula cells. Begin by locating the merismatic region of the onion or the blastula using the 10 X objective. Then use the 40 X objective to study individual cells. Identify one cell that clearly represents each phase. Sketch and label the cell on a separate piece of paper.

Exercise 3A.2 – Hypothesis
When undergoing mitosis most of the cells in an onion root tip will be in interphase. More cells will be in the stage of prophase than metaphase. More cells will be in metaphase than anaphase and more cells will be in anaphase than telophase.

Materials
The materials used in this experiment were a prepared slide of an onion root tip and a light microscope.

Methods
Obtain a prepared slide of an onion root tip and observe every cell in one high power field of view and determine which phase of the cell cycle it is in. Make sure to do this in pairs so one person can observe the cells and the other person can record which phase the cell is in. Make sure to count three full fields of view and at least 200 cells. Then, record your data in table 3.1. Next, calculate the percentage of cells in each phase by using the equation; percentage of cells in stage X 1,440 minutes =_________ minutes of cell cycle spent in stage.

Exercise 3B.1 – Hypothesis
Using beads it will be possible to show the stages of meiosis I and meiosis II.

Materials
The materials used in this exercise were chromosome simulation kits containing four strands of beads. Two strands will be one color and the other two strands should be another color.

Methods
To show the process of interphase place one strand of each color near the center of your work area. Next, simulate DNA replication by bringing the magnetic centromere region of one strand in contact with the centromere region of the other of the same color. Do the same with its homolog. Next, to show crossing over in prophase I pop the beads apart on one chromatid at the fifth bead. Do the same with the other chromatid. Then reconnect the beads to those of the other color. Proceed through prophase I of meiosis and note how crossing over results in recombination of genetic information. Then to show metaphase I place the chromosomes near the middle of the cell. During anaphase I, the homologous chromosomes separate and are “pulled” to opposite ends of the cell. Next, to show telophase I place each chromosome at opposite sides of the cell.

In prophase II of meiosis II replicated centrioles separate and move to opposite sides of the chromosome groups. Next, to show metaphase II arrange the chromosomes so they are centered in the middle of each daughter cell. Then, separate the chromatids of the chromosomes and pull the daughter chromosomes toward the opposite sides of the daughter cell in order to show anaphase II. Finally in order to show telophase II, place the chromosomes at opposite sides of the dividing cell.

Exercise 3B.2 – Hypothesis
There will be more asci that maintain a 4:4 relationship of not crossing over than asci that do cross over.

Materials
The materials used in this exercise were a prepared slide of Sordaria fimicola and a light microscope.

Methods
Begin by obtaining a prepared slide that contain asci of Sordaria fimicola. Then, using the 10 X objective, view the slide and locate a group of hybrid asci. Make sure to count at least 50 hybrid asci and enter your data in table 3.3.

Results
Exercise 3A.1

Exercise 3A.2

Table 3.1

Number of Cells
Field 1	Field 2	Field 3	Total
Interphase	42	36	47	125	61.27	14 hours 42 min
Prophase	10	14	18	32	20.10	4 hours 49 min
Metaphase	6	5	4	15	7.35	1 hour 45 min
Anaphase	2	3	2	7	3.43	49 min
Telophase	7	5	4	16	7.84	1 hour 52 min

Exercise 3B.2

Table 3.3

Number of 4:4	Number of Asci showing crossover	Total Asci	% Asci showing crossover divided by 2	Gene to centromere distance (map units)
60	45	105	21.4 %	21.4 map units

Questions:
Exercise 3A.1

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

In mitosis two new nuclei are being formed. Also cytokinesis is actually a part of mitosis.

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

A blastula is a hollow ball of cells that forms when an egg divides quickly; a large amount of mitosis is taking place here. The onion root tip is the place where growth occurs in the onion so a large amount of mitosis is taking place here.

Exercise 3A.1

1. 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?

The tips are located in the meristem. Cells that are not in the meristem do not divide as quickly but they are elongating and differentiating. None of the phases would have been visible.

2. 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 and telophase is the shortest.

Exercise 3B.1

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

Mitosis has one nuclear division and meiosis has two nuclear divisions. Mitosis makes two identical daughter cells. Meiosis makes four daughter cells that half the number of chromosomes that their parent cells had. Crossing over and the exchange of genes occurs in meiosis but not in mitosis.

2.Compare mitosis and meiosis with respect to the following:

	Mitosis	Meiosis
Chromosome number of parent cells	2n	2n
Number of DNA replications	1	1
Number of divisions	1	2
Number of daughter cells produced	2	4
Chromosome number of daughter cells	2n	n
Purpose	Growth and repair	Gamete and spore production

3. How are meiosis I and meiosis II different?

Meiosis I starts with a tetrad and separates the homologs. In meiosis the strands separate into 4.

4. How do oogenesis and spermatogenesis differ?

In oogenesis an egg is formed and three cells called polar bodies die. In spermatogenesis sperm are formed.

5. Why is meiosis important for sexual reproduction?

In meiosis the chromosome number is reduced by half. When fertlization occurs chromosome number is restored. Gene exchange causes variation.

6.Using your data in table 3.3 determine the distance between the gene for spore color and the centromere. Record your results in table 3.3.

Table 3.3

Number of 4:4	Number of Asci showing crossover	Total Asci	% Asci showing crossover divided by 2	Gene to centromere distance (map units)
60	45	105	21.4 %	21.4 map units

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

Error Analysis
In exercise 3A.2 inaccurate results were received. There should have been fewer cells in telophase than any of the other phases and cells should have spent less time in telophase than in any of the other phases. The results received are inaccurate because the number of cells that were in telophase were improperly counted. We received results that more cells were in telophase and spent more time in telophase than anaphase. In exercise 3B.2 improperly identifying some of the asci as crossover or non-crossover might have caused the results that were received to be inaccurate.

Conclusions:

From these experiments one can conclude that it is possible to look at mitotic stages of onion root tip cells and whitefish blastula through a microscope and draw them. Also, from these experiments one can conclude that most of the cell cycle in an onion root tip is spent in interphase. Prophase is after interphase in time spent in each cycle. Metaphase is after prophase. Anaphase is after metaphase. The least amount of time is spent in telophase. Also, a person can simulate the chromosomes in meiosis I and meiosis II using a chromosome simulation kit. Finally, one can conclude form the results of the experiments that more asci do not cross over in Sordaria fimicola than the number of asci that do cross over.

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