biobubbleinvestigations

Agricultural Applications of Biotechnology
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Mitosis LAB

Analysis:

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

A single cell goes through a series of stages in the process of Mitosis. The stages are Prophase, Metaphase, Anaphase and Telophase. In this process genetic info is copied and is distributed evenly amongst the two new daughter cells, resulting in two daughter cells that are genetically identical to the original cell. In prophase, the first phase of mitosis the spindle fibers start to form, the nucleolus and nuclear envelope disappear and the chromosomes condense. In metaphase the spindle fibers have completed forming and the chromosomes align in the middle of the cell. In anaphase the chromosomes split and the chromatids move to the poles of the cell. In telophase the nuclei of the daughter cells form. Cytokinesis is the event in which the cell splits forming two genetically identical daughter cells. After Cytokinesis, Interphase occurs. During Interphase there is high metabolic activity within the cell. Interphase is when the cell is not undergoing mitosis. Interphase is divided into three stages the G1, S & G2 phase. In the G1 phase the cells grows. In this phase of Interphase the cell also carries out the functions the cell is supposed to, for example if the cell is a skin cell the cell carries out the functions of a skin cell. The next stage is the s stage or synthesis where the genetic coding is replicated for cell division. In the G2 phase, the cell grows more like organelles and readies for cell division.

How does mitosis differ in plant and animal cells? How does plant mitosis accommodate a rigid, inflexible cell wall?

Mitosis differs between plant and animal cells because plant cells have rigid cell walls which do not allow cytokinesis to occur as such in animal cells where the cell pinches down the middle and splits. In plant cells the division occurs from inside as a result of expanding cell plates. Also in animals cells there are centrioles, microtubules that form the asters in mitosis, while spindle fibers are present and centrioles are not present in plant cells.

What is the role of the centrosome? Is it necessary for mitosis? Defend your answer.

The centrosome is the location where the centrioles are and where the spindle fibers are organized. The appendix and the centrosome are both results of evolution. Unlike the appendix the centrosome does have a function. Also scientists have seen defective centrosomes in cancerous cells associating cancer and centrosomes.  This is not certain, but suggests that a lacking centrosome could cause lethal consequences. Therefore the centrosome is necessary for mitosis.

Data

Number of Cells Percent of Total cells counted Time in each stage
Field 1 Field 2 Field 3 Total
Interphase 105 239 129 473 78.1 112.46
Prophase 45 30 27 102 16.8 241.92
metaphase 4 4 1 9 1.5 21.400
anaphase 1 2 5 8 1.3 18.720
telophase 5 5 4 14 2.3 33.120
  1. Yes the results would have been different because the cells that are dividing and go through the stages of mitosis are only in the root tips of plants. There are no cells going through the dividing stages of mitosis in any other part of the plant.
  1. Based on the data, what can be inferred about the relative length of time an onion root tip cell spends in each stage of cell division is that the cells spend most of the time in the stage Interphase, second prophase then telophase, metaphase and lastly anaphase.

Time in Stages of Mitosis & Interphase

time in stages of mitosis

NCBI. U.S. National Library of Medicine, n.d. Web. 07 Nov. 2013.

Chromatography Lab Investigations

The importance of chromatography is chromatography’s ability to  separate molecules and this is done through the molecule’s solubility, adhesiveness and mass. The chromatography paper allows molecules to separate. The purpose of the solvent in the lab is to move the pigments up the chromatography paper through capillary action. This is done because the pigments are soluble to the solvent therefore the pigments dissolve in the solvent allowing the pigments to move up the chromatography paper along with the solvent. But the pigments cannot move up the chromatography paper as far as the solvent because the pigment’s molecules are too large to and get stuck in the fibers of the chromatography paper, while the solvent move up the paper higher since the solvent’s molecules do not get stuck. This is how the chromatography paper separates molecules. Rf value stands for the distance of the pigment traveled up the chromatography paper divided by the distance traveled by the solvent up the chromatography paper. The Rf value is useful to scientists because the Rf value allows scientists to identify substances through the colors contained. The D-unknown is the distance traveled by the unknown pigment. The D-solvent is the distance traveled by the solvent. There were two colors on the Green leaf chromatogram, including the green there was yellow. On the Non- Green chromatogram the  leaf which was red had two colors again, including the red there was purple. The Green and the non-green leaf chromatogram both had entirely different colors on them, the red had red and purple, the green had green and yellow.

I figured out that pigments that do not seem to be in a leaf can be there, which can be found out through chromatography. For example there is purple in the red leaf.

One question I have is: What  solvents allow chromatography to occur most efficiently?

red chromatogram:

chrom

green chromatogram:

green chrom

Enzyme Lab

Title: The Influence of Acidity on the Rate of Enzyme Catalase Reactions.

Purpose: The purpose of the lab is to determine what affects increased levels of acidity have on the rate of enzyme catalase reactions.

Background: Hydrogen peroxide is the substrate of the enzyme catalase. The role of an enzyme is to decrease the amount of activation energy needed for a chemical reaction. In the case of catalase, catalase lowers the activation energy required to speed up the decomposition of hydrogen peroxide into water and oxygen gas. The reaction is represented in the following chemical equation: 2H2O2—–> 2H2O + O2. This reaction occurs in most living organisms. Catalase reactions occur in a pH close to neutral. If the pH is either too acidic or basic, the enzyme can become denatured or inactive. On the pH scale seven is neutral, below seven is acidic. As the numbers decrease on the pH scale, acidity increases.

Hypothesis: If the hydrogen peroxide solution has increased levels of acidity, then the reaction rate of the decomposition of hydrogen peroxide will decrease or the reaction will not occur at all.

Procedure (control):

1. Add lemon juice to Hydrogen peroxide getting the desired pH of 4.7.

2. Fill the upside down graduated tube with 10mL of the hydrogen peroxide solution, then add Catalase.

3. Fill a beaker with water and place the upside down graduated tube in the beaker of water. There should be enough water inside the beaker to submerge most of the graduated tube.

4. Attach the end of the tube to the bottom of the upside down graduated tube in the water.

5. The graduated tube that is right side up is attached to the other end of the tube and hand-held, in approximately equal level to the upside down graduated tube in the beaker of water.

6. Fill the right side up graduated tube with 10mL of water.

7. Add two drops of the enzyme catalase to the Hydrogen Peroxide solution in the upside down graduated tube.

8. Close the graduated tube as swiftly as possible and start the timer.

9. Record the level of the mL of Oxygen gas every ten seconds over a period of 160 seconds.

Diagram of equipment setup:

enzyme lab setup1Data:

Data Table:

pH 2.5 3.4 4.7 38 degrees
Time(s) mL
10 0 0 0.15 1
20 0 0 0.4 9
30 0 0 0.63 13
40 0 0 0.9  past 15
60 0 0 1.58
80 0 0 2.48
100 0 0.25 3.48
120 0 0.25 4.3
140 0 no data 5
160 0 0.25            5.5
200 0 0.625
300 0 0.875
400 0 1

Graph:

graph

Analysis: The graph shows the data of rise in the level of oxygen over a period of four hundred seconds. The graph depicts the results with a straight line on the x-axis at zero for the pH 2.5, a line that reaches 1 mL after the 400 seconds for a pH of 3.4 and a line that goes up to 5.5 mL after 160 seconds for a pH of 4.7.

Conclusion: In conclusion, the hypothesis which is if the hydrogen peroxide solution has increased levels of acidity, then the reaction rate of the decomposition of hydrogen peroxide will not occur or occur slowly . This is demonstrated in the procedures testing the rate of catalse reactions with a hydrogen peroxide solution with an acidic pH of 2.5 that as a result had no chemical recation occur at 0 mL/s. The procedure with an acidic pH of 3.4 had a reaction occur extremely slow at approximately 0.0025mL/s. This reaction is slow in comparison with the control, ran with a hydrogen peroxide solution of pH 4.7 which ran at a rate of .034 mL/s.  This is what would happen with catalase in its normal environment. The lesser acidic the pH of the Hydrogen Peroxide solution, the higher the rate of the catalase reaction. This is because the pH, catalase reactions naturally occur in, are closer to neutral and the more acidic the pH of where the reaciton occurs the more likely that the catalase enyzme has been denatured or the enzyme’s shape has changed, not allowing the enzyme catalase to bind to the hydrogen peroxide substrate.

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