Author: Biology Junction Team

 

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.

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.

    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.

Introduction: Exercise 6A: Recombinant DNA Technology

     To study the structure and function of a single protein-coding gene, one must prepare the gene in a purified form. Vertebrate cells contain enough DNA to code for more than I 00,000 proteins; therefore it is not very practical to isolate a gene by conventional biochemical procedures. This is why recombinant DNA technology is so important; it can be used to isolate and amplify a specific gene relatively simply.
    Plasmids, small circular DNA molecules, are usually extra chromosomal; they exist apart from the chromosomes in most bacterial species. Plasmids are not necessary for the survival of the host bacteria, but they can contain genes that enable the bacteria to survive in certain environments. If a bacterial cell contains a plasmid carrying a gene that confers resistance to antibiotics, then that cell could survive in the presence of the drug.
Plasmids can be introduced into bacterial cells by the process of transformation. Bacteria placed in a calcium chloride solution can take in plasmid DNA molecules. In this way, large amounts of specific plasmid DNA can be prepared, because one transformed cell gives rise to duplicate cells also containing the plasmid DNA molecule. Plasmids are very important for the molecular biologist because they serve as gene- carrier molecules called cloning vectors. A gene of interest can be joined to vector DNA to form a hybrid or recombinant molecule that can replicate in bacteria. When preparing a recombinant DNA molecule, a procedure is required for cutting cloning vectors and cellular DNA molecules in precise positions.
    Restriction nucleases are important for recombinant DNA technology because they cut DNA at specific sites. These enzymes are usually made by bacterial species in which they degrade invading foreign DNA within the bacterial cell. Most restriction enzymes recognize a specific sequence of nucleotides in DNA and cut a long DNA double helix into restriction fragments, which are measured in the process of agarose gel electrophoresis.

Introduction: Exercise 6B: DNA Fingerprinting

    Electrophoresis is the movement of charged particles in solution under the influence of an electric field. In gel electrophoresis, agarose gel is the stabilizing medium that serves as a matrix for the buffer in which the sample molecules travel. The gel is submerged in buffer within the electrophoretic gel cell. The samples are loaded into the sample wells in the gel, and electric current is passed through the gel.
    Molecules of DNA are negatively charged because of negative charges on the phosphate group. In this exercise, nucleic acids migrate through the pores of the gel from the negative end of towards the positive end. The large DNA molecules move more slowly than smaller molecules, therefore molecules are sorted according to size.

Objective: Exercise 6A

    Investigate basic genetic concepts by transforming bacterial cells by inserting an ampicillin-resistant gene into E. coli cells.

Objective: Exercise 6B

Investigate basic genetic concepts by using restrictive enzymes to digest phage lambda DNA and separate and identify the DNA fragments using gel electrophoresis.

Materials and Methods: Exercise 6A

    The materials used in this exercise included: 2 Luria agar plates, 2 Luria agar plates with ampicillin, 2 microcentrifuge tubes, 1 inoculating loop, 1 Bacti-Spreader, sterile micropipets, calcium chloride, Luria broth, plasmid pUC8, Bunsen burner, hotplate, ice, waterbath.
The two microcentrifuge tubes were marked !I+” and !1-!1, and 250IJI cold calcium chloride was added to each using a pipet. A large colony of bacteria was added to each tube with a sterile inoculating loop. A micropipet was used to transfer 10IJI of the plasmid pUCS solution to the !I+” tube. Both tubes were incubated on ice for 15 minutes, and meanwhile, the two Luria agar plates were labeled “+” and “-” and so were the two Luria plates with ampicillin. The tubes were removed from ice and placed in a 42°C hot waterbath for 90 seconds. The tubes were then removed from the waterbath and placed on ice for two minutes. A micropipet was used to add 250IJI Luria broth to each tube. Another micropipet was used to add 100IJI of the !I+” solution to the two !I+” plates and 1oomicroliters of the  solution to the two “-” plates. The bacteria was flamed to sterilize, and after cooling, was used to spread the cells over the entire surface of the plates. After five minutes, the plates were placed in a 37°C incubator, inverted, overnight.

Materials and Methods: Exercise 6B

    The materials used in this exercise included: 8% agarose gel, 2 electrophoresis chambers, power pack, running buffer- Tris, micropipetter and tips, staining tray, methylene blue dye, gloves, aprons, 4 DNA samples cut with restrictive enzymes, vial tray, microcentrifuge, paper, pencil, distilled water, spatula, plastic container for destaining, masking tape, light box, ruler, semi-Iog graph paper. The gel, on the gel tray, was placed in the center of the chamber, with the well-side of the gel near the black electrode. Approximately 350ml of running buffer were added to the chamber. Of each DNA sample, 10 microliters was loaded into the corresponding gel lane with a micropipet. The power cords were attached to the appropriate connections, and the power supply was turned on, set to 50 volts. The samples were allowed to migrate for three hours. The gel was then removed, stained, and destained overnight. The gel was viewed on a light board, and the band migration distances were measured.

Results: Exercise 6A

  Questions

1. Based on your experimental results, did transformation occur? Why or why not?
Yes, transformation did occur. Colonies of E. coli grew in the presence of ampicillin.

2. What other methods can be used to verify that transformation occurred?
DNA fingerprinting

Results: Exercise 6B

Questions

1. Compare the banding patterns. Do you think the DNA samples were the same?
No, the samples were different sizes

2. Which of the two suspects were the real burglar? Suspect #2

3. Explain the function of each of these steps in DNA fingerprinting:
a. Restriction Enzyme digest – used to cut DNA

b. Gel electrophoresis – used to separate different size pieces of DNA

c. Denaturing into single-stranded DNA – process used to view only one strand of the DNA double helix

d. Southern Blot – DNA bands transferred to a nitrocellulose paper

e. Radioactive DNA probe – used to find & bind to the complementary sequence in one or more RLFP’s

f. Autoradiograph – used to show similarities in DNA samples

Error Analysis

Lab 6A: Not enough agar was poured on the plates.

Lab 6B: Base pair counts for the DNA bands could have been inaccurate as suggested by the best-fit line on the graph.

Conclusion

Lab 6A: This lab showed that genes can be inserted into living bacterial cells thus transforming the cells and giving them new properties such as ampicillin-resistance.

Lab 6B:  Through gel electrophoresis of the DNA samples, it was determined that suspect #2 was guilty. Their DNA banding pattern matched the banding pattern taken of the DNA at the crime scene.

Drosophila Genetics

Introduction

Drosophila Melanogaster, the fruit fly, is a great organism for genetic use because it has simple food requirements, occupies little space, is hardy, completes its life cycle in 12 days, makes a large number of offspring, can be knocked out easily, and it has many types of hereditary variations that can be seen with low power microscopes. Drosophila has a small number of chromosomes, four pairs. They are easily located in the large salivary glands. The Drosophila can be obtained from many places. Research of Drosophilae has led to a lot of knowledge about many of its genes.

Many factors combine to affect the length of the Drosophila life cycle. Temperature affects the life cycle the most. At room temperature the average life cycle of the Drosophila is about 12 days. Eggs of the Drosophila are small, oval shaped, and have two filaments at one end. They are usually laid on the surface of the culture medium, and with practice, can be seen with the naked eye. After one day the eggs hatch into the larva.

The larval stage of the Drosophila eats all the time. Larvae tunnel into the culture medium when they eat. The larva will shed its skin as it increases in size. In the last of the three larval stages, the cells of the salivary glands contain giant chromosomes that can be seen under low power in a microscope.

The pupal stage. Before a larva becomes a pupa it climbs the side of the container. The last larval covering then becomes harder and darker, forming the pupal case. Through this case the later stages of metamorphosis to an adult fly can be seen. In particular, the eyes, the wings, and the legs become visible.

The adult stage. When metamorphosis is over, the adult fly emerges form the pupal case. They are fragile and light in color and their wings are not fully expanded. They get darker in about an hour. They live about a month and then die. A female refrains from mating for about 12 days after she emerges from the pupal case. After she mates her receptacles contain large amounts of sperm and she lays her eggs. Make sure that the first flies you use are virgins.

The experiment will take several weeks. You will be assigned Drosophila with well-defined mutant traits by your teacher. You will keep a close record of what happens as each of these flies mate and pass there traits off to their offspring over a few generations.

There are three types of crosses that are studied in this lab. In monohybrid crosses the mode of inheritance is determined when a single contrasting pair of characteristics is involved. In a dihybrid cross the mode of inheritance is determined when the two pairs of contrasting of characteristics are considered simultaneously. In a sex-linked cross the mode of inheritance is determined when the mutant characteristic is associated with the X chromosome.

Hypothesis

In the sex linked cross of Drosophila Melanogaster, a phenotypic ratio of 1:1 will be obtained.

Materials

The materials used in this lab are as follows: a vile of Drosophilia with c designated trait, vials containing a medium, a refrigerator, ice packs, Petri dishes, a light microscope, a vial of wild type flies, an incubator, a pencil and paper.

Methods

Begin by obtaining a vial of wild type flies. Practice immobilizing and sexing these flies. Make sure to examine the flies and determine the characteristics of their eyes, wings, bristles, and antennae. Next, these are the steps for immobilizing the flies. Hold the vial containing the flies at an angle and place it in a refrigerator for several minutes. When the flies are immobilized, place them into a small plastic Petri dish. Then place the Petri dish on top of the icepack in order to maintain the cool temperature necessary to keep flies immobilized. Use the dissecting microscope to view the flies. Make sure to top the petri dish on when viewing the flies.

You can easily distinguish male flies from females by looking for the following characteristics: males are usually smaller than the females, males have dark blunt abdomens and females have lighter pointed abdomens. The males have sex combs, which are black bristles on the uppermost joint of the forelegs. Next, get a vial containing experimental flies. Make sure to write down the number of the vial that you have. The flies you now have are the P1 generation. The females should have laid eggs. The eggs and larvae are the F1 generation. Then after there are eggs present knock out remove the adult flies from the vial. Sex the adult flies and write down any mutations. Place the flies in the morgue that contains alcohol. Make sure to label the vial with the symbols for the mating.

After about another week has passed knock out and record characteristics of the remaining F1 flies and record the results in table 7.1. Then place the six pairs of these flies in a new vial and place the remaining flies in the morgue. Label the new vial F1, and tell the cross, date and your name.

After another week has passed, remove the F1 flies and put them in the morgue. The F2 generation are the eggs and larvae in the vial. Place the vial back into the incubator. Once again, after another week has passed remove the F2 flies and record their sex and characteristics and place the results in Table 7.2. Recording a greater number of F2 flies will make your results more accurate. Try to collect at least 200 flies. In order to analyze your data you will first have to be able to be able to complete Chi-Square Analysis.

Results

Table 7.1 F1 Generation

1. Describe the observed mutations? In the F1 generation the males had white eyes and the females had red eyes. In the F2 generation the males and females could have had either red or white eyes.

2. Write a hypothesis which describes the mode of inheritance of the trait you studied. This is your null hypothesis ( as described in the Statistical Analysis Section). For a sex linked cross there will always be a one to one ratio of the phenotypes. In the F1 generation there will be a one to one ratio of red eyed females to the number of white eyed males. In the F2 generation there will be a one to one ratio of red eyed females to white eyed females. There will also be a one to one ratio of red eyed males to white eyed males.

3. Refer to a textbook and review Punnett squares. In the space below construct two Punnett squares to predict the expected results of both the parental and F1 crosses from your null hypothesis.

Parental cross

F1 cross

4. Refer to the Punnett squares above. Record the expected ratios for the genotypes and phenotypes of the F1 and F2 in the experiment below.

        5. Do the actual results deviate from what was expected? If so, explain how.
No my results do no deviate much from what was expected. However in the F2 generation there were 67 white females and 52 red females.

6. For the results describe your cross? My cross is a sex linked cross.

7. Are the deviations for the phenotypic ratio of the F2 generation within the limits expected by chance? To answer this question, statistically analyze the data using the Chi-Square-Analysis. Calculate the Chi-Square for the F2 generation in the chart below. Refer to the critical values of the Chi Square distribution table to determine the P value that is associated with your statistic.

(a) Calculate the Chi-Square value for these data.

1. How many degrees of freedom are there? 3 degrees of freedom

2. Chi Square=2.52

3. Referring to the critical values chart what is the probability value for these data? Greater than .05 probabilities that the null hypothesis is right.

(b) According to the probability value, can you accept or reject your null hypothesis? Explain why. I can accept the null hypothesis because my Chi-Square answer is less than the critical value form the table. I have 3 degrees of freedom and my Chi- Square answer was3.0 which is less than 7.82.

        1. Why was it necessary for the females of the parental generation to be virgins? The females store sperm in their receptacles and if they were not virgins we would not be able to tell who the fathers were.

2. Why was it not necessary to isolate virgin females for the F1 cross? The females store sperm in their receptacles and if they were not virgins we would not be able to tell who the fathers were

3. Why the adult flies were removed from the vials at weeks 2 and 4? So, they are not be able to mate with the next generation.

Chi-Square Analysis

Introduction

Statistics can be used to determine if differences among groups are significant, or simply the result of predictable error. The statistical test most frequently used to determine whether data obtained experimentally provide a good fit, or approximation to the expected or theoretical data is the Chi-square test. This test can be used to determine if deviations from the expected values are due to chance alone or to comeother circumstance.

To determine if the observed data fall with in acceptable limits, a Chi-Square analysis is performed to test the validity of a null hypothesis; that there is no statistically significant difference between the observed and expected data. If the Chi-Square analysis indicates that the data vary too much from the expected 3: 1 an alternative hypothesis is accepted.

Methods

The formula for Chi-square is:

X2=E(o-e)2
E

O= observed number of individuals

e= expected number of individuals

E= the sum of the values

The (df) are determined by taking the number of possible phenotypes and subtracting one from it. If the Chi- Square answer is greater than the number from the critical values chart then the null hypothesis is incorrect. The results are said to be significant at .05. This means that only 5 % of the time you would expect to see similar data if the null hypothesis were correct. The probability can also be rejected at .001. This time it means that less than 1 % of the time would you expect to see similar data.

Results

Critical Values Chart

Practice Problem

An investigator observes that when pure-breeding long winged Drosophila are mated with pure breeding short wing flies the F1 have an intermediate wing length. When several intermediate wing length flies are allowed to interbreed the following results are obtained. 230 long wings. 510 intermediate length wings. 260 short wings.

a. What is the genotype of the F intermediate wing length flies? The genotype is Ll.

b. Write a hypothesis describing the mode of inheritance of wing length in Drosophila. There will be 333 long winged flies. 666 intermediate winged flies. There will be 33 short wing flies.

c. Complete the table

Table 7.8

1. How many degrees of freedom are there? There are 2 degrees of freedom.

2. Chi-Square= 84.4

3. Referring to the critical values chart, what is the probability value for these data? Less than .001.

4. According to the probability value can you except or reject the null hypothesis? 

I can reject the null hypothesis because the Chi-square answer is greater than the critical value from the table.

Error Analysis

Results from this lab could have been affected by many things. The constant knocking out of flies could have caused some of the larvae to not hatch therefore affecting our numbers. Also, incorrectly identifying the characteristics of the flies could have also greatly affected the results received. Improper calculation of numbers could have also caused inaccurate results. Finally, some flies could have gotten stuck in the medium and could have been identified.

Conclusion

From the results of the experiment I can conclude that I received results that were close to a 1:1 ratio. The Chi- Square worked from my data was accepted at a possibility greater than .05. The null hypothesis in this case can be accepted.