Mitosis Lab

In my AP Biology class, I observed dividing onion root tip cells through a light microscope to see the phases of mitosis that the cells were in and calculate the length of time the cells spend in each phase of mitosis. Below is the lab format.

Analysis Questions

1. Explain how mitosis leads to two daughter cells, each of which is diploid and genetically identical to the original cell. What activities are going on in the cell in interphase?

Mitosis consists of five stages. The first stage is known as interphase. In this phase, a cell prepares to divide. The cell duplicates its DNA, centrosomes, and, in the case of animal cells, centrioles. The replicated DNA is enclosed in the nucleus in the form of chromatin and the nucleus also has one or more nuclei. Also, each centrosome contains two centrioles. In the second phase known as prophase, the chromatin in the nucleus condenses into chromosomes, the nucleoli disappear, and the mitotic spindle develops. Each chromosome is made up of sister chromatids that carry the same alleles. The nuclear envelope begins to break down as well. In the third phase known as metaphase, the nuclear envelope has completely broken down and the chromosomes line up in the center of the cell with the centrosomes at opposite poles. The mitotic spindle attaches to the centromeres, or the connecting point of the sister chromatids, of the chromosomes. In the fourth phase known as anaphase, sister chromatids are pulled apart by the mitotic spindle. The last phase is known as telophase, in which the cell begins to split into two daughter cells that have the same number of chromosomes as the original cell and are each complete with a nucleus. Cytokinesis occurs the moment that the daughter cells each become an individual cell.

2. How does mitosis differ in plant and animal cells? How does plant mitosis accommodate a rigid, inflexible cell wall? Screenshot 2013-11-07 20.41.50

Plant cells do not have centrioles like animal cells, just centrosomes. Centrioles are structures made of microtubules that help organize the mitotic spindle, and centrosomes are areas where the mitotic spindle meets. In plant cells, when cytokinesis, or the moment two daughter cells form from a single cell, occurs, a cell wall forms between the two cells. This wall separates the cells. This occurs because the cell wall is rigid and inflexible. In comparison, animal cells do not have cell walls so the daughter cells can completely separate due to the flexibility of the plasma membrane.

3. What is the role of the centrosome (the area surrounding the centrioles)? Is it necessary for mitosis? Defend your answer.

The centrosome is the area where the mitotic spindle meets. The centrosome acts as an organizing center as the mitotic spindle grows out from this area. The mitotic spindle is important in that the spindle separates the chromosomes of a cell during anaphase in mitosis. The mitotic spindle pulls sister chromatids apart at the centromere of chromosomes and the daughter chromosomes move towards opposite poles of the cell. Without centrosomes, mitotic spindles are not organized and chromosomes cannot be separated correctly to make sure that each daughter cell has the necessary chromosomes it needs. For example, if the mitotic spindle was not organized then one daughter cell could have multiple copies of the same allele carried by sister chromatids and not have the other alleles carried by other sister chromatids. This could lead to mutation, which could cause the organism as a whole to not function properly.

Lab Results

Screenshot 2013-11-08 06.10.28

Pie chart mitosis 2

Questions

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?

If the area of observation in the experiment had not been restricted to the area of the root tip that is actively dividing, then the results of the experiment would have been different because the cells would supposedly be spending their entire existence in interphase. Since the cells would not be dividing, the cells would all seem to be in interphase. No other phases of mitosis would be observed so the time cells spend in the  phases of prophase, metaphase, anaphase, and telophase would be 0 minutes, while the time spent in interphase would equal the total number of cells present multiplied by 1440 minutes, which would equal the maximum range of time the experiment was preformed for (1 day).

2. Based on the data, what can you infer about the relative length of time an onion root tip cell spends in each stage of cell division?

The percentage of total onion root tip cells in each phase observed in this experiment was multiplied by the number of minutes in one day to determine the time spent in each phase. 76.9 % of the cells observed were in interphase. Due to the fact that most of the cells were in interphase, the time calculated for the time cells spend in the phase, which was 1107.4 minutes, was greater than all the other phases. It can therefore be inferred that cells spend the most time in interphase during mitosis. The amount of time spent in each phase decreases in the order that the phases occur. 16.6% of the cells were in prophase, 2.8% in metaphase, 2.2 % in anaphase, and 1.5% in telophase. The amount of cells in each phase decreased, so the time spent in each phase decreased proportionally. The time calculated for prophase was 239.0 minutes, for metaphase was 40.3 minutes, for anaphase was 31.7 minutes, and for telophase was 21.6 minutes. Therefore, cells must spend more time in prophase than metaphase, more time in metaphase than anaphase, and more time in anaphase than telophase.

Resources:

The University of Arizona. “Online Onion Root Tips.” The Biology Project. The University of Arizona, 20 Aug. 1998. Web. 07 Nov. 2013. <http://www.biology.arizona.edu/cell_bio/activities/cell_cycle/activity_description.html&gt;.

Chromatography

chromset
Chromatography Setup

Chromatography is the separation of the molecules of a particular substance that is represented by different colored bands on chromatography paper. The solubility of the molecules in the solvent, the adhesion of the molecules to the paper, and the mass of the molecules allow chromatography to happen. The solvent moves against the force of gravity by capillary action, which occurs due to the adhesion of the solvent to the paper and the cohesion of the solvent molecules. The chromatography paper participates in the separation of the molecules of a substance by drawing up the solvent and allowing the movement of the molecules of the substance. The separation of these molecules on the paper is determined by variations in the mass and solubility of the molecules. The paper also provides a platform by which the distance from the initial site of application and color of the pigments can be observed.

The Rf value (reference front) is represented by the equation: rfeq. The Rf value must be useful to scientists as the values allow scientists to identify the molecules that are present in a particular substance because each molecule has a specific Rf value, and the values can be used to compare the masses of molecules based on the distance travelled. D unknown signifies the unknown distance travelled by a pigment from the initial site of application. D solvent is the distance travelled by a solvent from the point of contact with the paper.

Two pigments were identified on both the green leaf chromatogram and the non-green leaf chromatogram. Yellow and green were the two pigments identified on the green leaf chromatogram. The yellow pigment was very close to the initial site of application, whereas the green pigment was not. Purple and red were the two pigments identified on the non-green leaf chromatogram. The purple pigment was very close to the initial site of application, whereasthe red pigment was not. From this experience, I learned that there are different types of pigment in leaves that capture the light energy from the sun that is used in photosynthesis. These pigments absorb and reflect certain wavelengths of light. Chlorophyll a is the most common and has a green color. To learn more about which wavelengths certain pigments absorb and reflect, click here to view my Prezi on the topic: How Do Different Colored Lights Affect Photosynthesis?

Screen Shot 2013-10-10 at 6.41.00 AM
Red Chromatogram
Screen Shot 2013-10-10 at 6.43.38 AM
Green Chromatogram

References:

Chromatography setup. Digital image. Pearson Education Inc., Web. 9 Oct. 2013. http://www.phschool.com/science/biology_place/labbench/lab4/design1.html.

Clark, Jim. “Paper Chromatography.” Chem Guide. 2007. Web. 9 Oct. 2013. http://www.chemguide.co.uk/analysis/chromatography/paper.html.

Reece, Jane B., Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, and Robert B. Jackson. Campbell Biology. 9th ed. San Francisco: Pearson Benjamin Cummings, 2011. Print.

Catalase Lab

Title: The Effect of Low pH on the Rate of Catalase Catalyzed Reactions

Purpose: The purpose of this experiment is to test how low pH or acidity affects the rate of reactions catalyzed by catalase.

Background: Enzymes are catalysts that speed up the rate of chemical reactions by lowering the activation energy, or the energy needed to start a reaction. The baseline of a reaction is the standard for the reaction, which includes rate, color, and other variables. Enzymes are not consumed by reactions, but are instead recycled. A particular enzyme only has an active site that will suit a particular molecule. Catalase is an enzyme that specifically participates in the breakdown of hydrogen peroxide, which is produced naturally by cellular metabolic processes, into oxygen gas and water. This reaction is represented as: 2 H2O2 → 2 H2O + O2. Hydrogen peroxide is harmful to cells, which is why this reaction occurs. The optimal pH for catalase in humans is 7, or neutral, and the optimal temperature is 37 °C. Catalase is found the peroxisomes of plant and animal cells; the livers of humans and animals have a great supply of catalase.

Hypothesis: If catalase is added to a solution of lemon juice and hydrogen peroxide with a pH of 3.4, then the reaction will occur at a slower rate than when catalase is added to hydrogen peroxide with a pH of 4.7 and at a faster rate than when catalase is added to lemon juice with a pH of 2.5.

Materials (for the test when catalase is added to hydrogen peroxide with a pH of 4.7):
• Two graduated tubes connected to a piece of plastic tubing [one tube has a hole in the cap (graduated tube 1) and the other has a hole in the cap and an extension at the bottom of the tube that connects the tube to the plastic tubing (graduated tube 2)]
• Two binder clips
• Two 400mL beakers
• 10 mL of hydrogen peroxide with a pH of 4.7
• 400 mL of distilled water
• One P1000 micropipette with tips
• 100 μL of animal liver tissue containing catalase (referred to as just catalase in this lab for simplicity)
• Timer

Procedure (for the test when catalase is added to hydrogen peroxide with a pH of 4.7):
1) Clip the plastic tubing with the binder clips close to where the tubes connect so no oxygen gas flows through the tube.
2) Fill one of the beakers with 400 mL of distilled water and place graduated tube 2 in the beaker with the hole in the cap pointed downwards. Record the initial presence of oxygen gas in the tube by reading the water level.
3) Take the cap off of graduated tube 1 and add 10 mL of hydrogen peroxide to the tube.
4) Using a micropipette, add 100 μL of catalase to graduated tube 1. Quickly screw the cap back on and place the tube into the empty beaker.
5) Remove the binder clips and start the timer. Record the water level at each time interval to determine the amount of oxygen gas produced.
6) Clean all the materials and preform additional trials (three recommended) and choose the most accurate to include as data.

Lab Setup x

Data:

Table

The Rate of Catalase Catalyzed Reactions at Different pHs

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Graph

The Rate of Catalase Catalyzed Reactions at Different pHs

Lab graph final

Lab Setup y

Analysis: As the pH of the catalase catalyzed reactions decreased, the amount of oxygen produced by the reactions decreased. The reaction that occurred at a pH of 4.7 produced the most oxygen. The reaction that occurred at a pH of 3.4 produced less oxygen than the reaction that occurred with a pH of 4.7, but more than the reaction that occurred at a pH of 2.5. The reaction that occurred at a pH of 2.5 produced no oxygen at all. The evidence that the reaction that occurred at a pH of 2.5 produced no oxygen is that there was no difference in water level in graduated tube 2 over time, therefore, the water level was constant and was not decreased by the pressure of oxygen on the water.

Conclusion:
The data reveals that low pH or acidity affects the rate of catalase catalyzed reactions. The data shows that the rate of the reaction that occurred at a pH of 3.4 was greater than the rate of the reaction that occurred at a pH of 2.5, but less than than the rate of the reaction that occurred at a pH of 4.7. The hypothesis is, therefore, supported by the data. As the pH became more acidic, the catalase catalyzed reactions occurred at a slower rate. The reaction that occurred at a pH of 2.5 produced 0 mL of oxygen gas, evident by the constant water level in graduated tube 2. If no oxygen was being produced, then the rate of the reaction was either extremely slow or the reaction was not occurring. The reaction that occurred at a pH of 3.4 produced 0 mL of oxygen up until 80 seconds. Then, the amount of oxygen produced was constantly 0.25 mL up until 160 seconds. Then, the amount of oxygen produced continued to increase, as the amount of oxygen produced at 200 seconds was 0.625 mL, at 300 seconds was 0.875, and at 400 seconds was 1 mL. The rate of the reaction continuously increased over time. The reaction that occurred at a pH of 4.7  also increased continuously over time.  The reaction began increasing a slow rate and later on, between 60 seconds and 100 seconds, the amount of oxygen produced began increasing at a greater rate. The amount of oxygen produced at 60 seconds was 1.58, at 80 seconds was 2.48, and at 100 seconds was 3.48; the water level difference is greater to or equal to 1 mL. After 100 seconds occurred, the rate of the reaction began to slow down, evident by the lower water level difference after 100 seconds. At 120 seconds, the amount of oxygen produced was 4.3 mL, at 140 seconds was 5 mL, and at 160 seconds was 5.5 mL; the water level differences for these times are less than 1 mL. 

The baseline control of this experiment was the reaction that occurred when catalase was added to the hydrogen peroxide with a pH of 4.7. This set the standard rate of the breakdown of hydrogen peroxide into oxygen gas and water for the experiment. When catalase was added to the solution of lemon juice and the hydrogen peroxide, the high pH must have denatured some individual enzymes, but not all, so the reaction occurred at a slower rate than the baseline reaction. Denaturation changes the shape of an enzyme, including the enzyme’s active site, so a particular cannot bind to the enzyme anymore. This supports the idea that enzymes are specific to the molecules enzymes bind to. Since the baseline reaction occurred under a more acidic environment than the optimal pH for catalase, the rate of the breakdown of hydrogen peroxide into oxygen gas and water will be faster under a neutral pH. A possible error that could have occurred in this experiment could include misreading the water level. If the water levels were incorrectly read, then the amount of oxygen produced by each reaction recorded in the data chart are inaccurate. As a result, the rate of the reactions could have been faster or slower. This is why  multiple trials were preformed and the most accurate was chosen to graph.

References:

“Catalase.” GMO Compass. 7 July 2010. Web. 3 Oct. 2013. <http://www.gmo-compass.org/eng/database/enzymes/89.catalase.html&gt;.

Enzymes. Perf. Paul Anderson. YouTube. YouTube, 26 Nov. 2011. Web. 3 Oct. 2013. <http://www.youtube.com/watch?v=ok9esggzN18&gt;.

“Peroxisomes.” BSCB: The British Society for Cell Biology :: www. bscb.org. BSCB, 2013. Web. 3 Oct. 2013. <http://www.bscb.org/?url=softcell/peroxi&gt;.

Reece, Jane B., Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, and Robert B. Jackson. Campbell Biology. 9th ed. San Francisco: Pearson Benjamin Cummings, 2011. Print.