Mitosis Onion Root Tip Lab

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 during interphase? During interphase, the cell is undergoing the normal cell processes. For example, if the cell is a muscle cell, the cell is functioning as a muscle cell is intended to function. However, also during interphase, a cell is preparing for future division. This preparation includes duplication cellular structures, such as the genetic material (chromosomes) and the centrosomes. After duplication, each cell contains twice the necessary number of chromosomes, and is therefore able to create two identical daughter cells after cell division. Each daughter cell then has an equal number of chromosomes.
  2. How does mitosis differ in plant and animal cells? How does plant mitosis accommodate a rigid, inflexible cell wall? Mitosis is essentially the same in animal cells and plant cells, with a few important differences. While both plant and animal cells contain a centrosome, plant cells lack the centrioles that are found within the centrosome in animal cells. In addition, plant cells have cell walls, and animal cells do not. This makes plant cells incapable of cleaving the plasma membrane at cytokinesis. Instead, a cell plate forms across the center of the cell where the chromosomes were once aligned, and forms the cell wall between the two daughter cells.
  3. What is the role of the centrosome (the area surrounding the centrioles)? Is it necessary for mitosis? The centrosome is the formation center for the microtubules, or spindle fibers. These spindle fibers attach to the chromosomes, and after the chromosomes align in the middle of the cell, the spindle fibers pull the chromosomes apart during anaphase. Without the centrosome, microtubules would not be formed, and the chromosomes would not be pulled apart. This would mean that the cell could not separate the duplicated DNA, and two daughter cells could not be formed.AP Biology Test Study Help


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  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 observations had been taken at a section of the onion root that was higher up than the tip, the cells would not have been actively dividing. This would likely result in a much higher number and percentage of cells spent in interphase, and very few cells in the other phases of mitosis. If the observations had been taken lower on the onion root, on the root cap, the observed cells would be dead or not undergoing mitosis, and so the cells would primarily, if not entirely, be in interphase.
  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? Interphase is the phase in which cells in the onion root tip spend the majority (91.8%) of the time. Of the phases in  cell division, prophase is the longest, consuming 6.4% of the total cell cycle time. Cells spend the next most amount of time (0.8%) in metaphase, then telophase (0.6%), then anaphase (0.4%). Cells spend a significantly longer period of time in prophase than in the other three cell division phases.
  3. Draw and label a pie chart of the onion root tip cell cycle using the data.

Mitosis Lab Graph



ChromatogramsChromatography is a process that separates the different pigments of a substance across a porous kind of paper called chromatography paper. Molecularly, this process separates molecules by the molecules’ molecular mass. In this process, the chromatography paper allows the solvent and pigments to diffuse, and the solvent and pigments are drawn up the paper. This mimics capillary action in plants, where plants pull water up from the roots to the stems and leaves of the plant. Capillary action is made possible by the forces of cohesion and adhesion of both the pigment and the solvent. When the pigment molecules have a larger molecular mass and have strong cohesive forces and adhesive forces to the chromatography paper, the molecules will stay where the molecules were placed or not move very far. In contrast, when the molecules of a certain pigment have a smaller molecular mass and adhere more to the solvent than the paper, then these molecules will move farther up the chromatography paper with the solvent. There are also many other factors that affect the rate of a pigment travelling up the chromatography paper. In addition to the pigment’s molecular mass and the the pigment’s adhesive and cohesive forces, some factors that affect the rate of the pigment during chromatography include the polarity of the solvent, the pH of the solvent, the temperature of the experiment, and many other factors.

The Rf value stands for the average distance that the pigment travelled, divided by the total distance that the solvent travelled. Dunknown signifies the average distance that the pigment travelled. This means the farthest distance that the pigment travelled, minus the closest distance that the pigment travelled, divided by two. This results in a point in the middle of the pigment range. Dsolvent signifies the total distance that the solvent travelled. This is not an average distance, but rather the specific distance that the solvent travelled up the chromatography paper. Dunknown  / Dsolvent equals the Rf value. This value gives a number without units, that is always less than 1.0. This number is very useful because it can be used by scientists to get an idea of the molecular mass of the particular pigment, as well as the typical distance that the pigment will travel with a certain substance. For example, a pigment with a high Rf value will travel relatively far compared to the other pigments. This means that pigments with small molecular masses have large Rf values.

In the experiment performed in class, both green leaves and non-green leaves were tested. In the green leaf chromatogram (Figure 1), three pigments were present after the solvent separated the pigments (light orange, yellow, and green). However, in the non-green leaf chromatogram (Figure 2), there were four pigments clearly present (maroon, yellow, red, and green). This showed that although the leaves looked all one individual color, there were actually several pigments within each leaf. The non-green leaf actually contained more pigments than the green leaf. Surprisingly, the red-maroon leaf actually seemed to have about an equal concentration of green pigment as the bright green leaf, of two completely different species. This shows that while the leaves may appear to be a certain color, the leaves really contain a large array of pigments that together make up the external appearance.

After learning from my own photosynthesis Prezi presentation and those of my classmates, I learned that plants are green because the chlorophyll absorbs all other wavelengths of light, but reflects green light. However, I still wonder whether there are different wavelengths of green light that are absorbed more than others, or if all wavelengths of green are reflected. ⚘

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Catalase Lab

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

Purpose: To test the effects of pH on the rate of reactions catalyzed by catalase.

Background: Enzymes are proteins within cells that act as catalysts in biochemical reactions. The enzyme catalase speeds up the decomposition reaction of hydrogen peroxide (H2O2) into water (H2O) and oxygen gas (O2) by lowering the reaction’s activation energy. The equation of the reaction is:

2H2O2 → 2H2O + O2

Like all catalysts, catalase is not used up or changed in the reaction, and can be reused. One factor that can affect the rate of an enzymatic reaction is the pH of the reaction. A pH level that varies greatly from the optimal pH of the reaction often causes a slowed reaction rate. The optimal pH level for enzymatic reactions usually matches the standard pH for the organism in which the enzymatic reaction typically occurs. Some of the other factors that affect the rate of an enzymatic reaction include temperature, presence of inhibitors, concentration of enzyme, and concentration of substrate, which is the molecule on which the enzyme acts.

 The substrate of catalase is H2O2, which breaks down into water and oxygen gas significantly faster with catalase present. This enzyme is found in nearly all living cells, because catalase is needed to break down excess H2O2 within the cell. Otherwise, the H2O2 could accumulate, and this would be toxic to the cell and the organism of which the cell is a part. Catalase is particularly concentrated in the livers of animals.

Hypothesis: If pH has an effect on the rate of catalase reactions, then as the pH increases in acidity, the rate of the reaction will decrease.

Materials: See apparatus set-up diagram.

Materials Diagram

(Original Diagram)

Procedure (for standard pH of 4.7):

  1. Fill right-side-up graduated tube with 10mL H2O2 and add 350μL H2O. Leave the lid off.
  2. Fill beaker with water, and place the upside-down graduated tube with a hole in the lid into the beaker of water.
  3. Attach one end of the tube to the top of the upside-down graduated tube.
  4. Attach the other end of the tube to the lid of the right-side-up graduated tube.
  5. Clamp binder clips near each end of the tube.
  6. Add 100μL of the catalase mixture to the H2O2.
  7. Quickly screw on the lid of the graduated tube containing catalase and H2O2. Ensure that the tube is secure and not leaking.
  8. Place the graduated tube containing catalase and H2O2 into an empty beaker.
  9. Holding the upside-down graduated tube steady, remove the binder clips and start the timer. Be sure that both beakers and graduated tube stay level and do not change positions.
  10. Record the volume of water (mL) dropped for every time interval.
  11. Rinse equipment and repeat steps 1 – 10 twice for the second and third trials.


Graph Chart

(Original Chart and Graph)

Analysis: As the pH grew more acidic, oxygen was produced at a slower rate. When the pH of the catalase solution was 4.7, 5.50 mL of oxygen were produced over 160 seconds. When the pH of the catalase solution was 2.5, 0.00 mL of oxygen was produced over the entire 400 seconds.

Conclusion: The data collected supported the hypothesis that pH would have an effect on the rate of catalase reactions. The data indicated that as the pH grew more acidic, the rate of the reaction decreased. The reaction with a pH of 4.7 produced oxygen about 16 times faster than the reaction with a pH of 3.4. Additionally, the reaction with a pH of 2.5 did not produce any oxygen. Oxygen was not produced as quickly during the experiements using solutions of higher acidity because if the optimal conditions for an enzymatic reaction is altered too drastically, the enzyme will be denatured.

Denaturization of enzymes causes a change in structure, which then disables the substrate from attaching to the enzyme’s active site. The enzyme is then denatured and can no longer serve as a functional enzyme. The data collected supports the idea that the catalase was denatured when put into contact with highly acidic pH levels. The standard baseline pH level of this bovine catalase reaction is 4.7.

Some of the potential errors that may have been made in completing this lab include user errors, like misinterpreting the data and slightly inaccurate time readings. This misinterpretation of the water level in the graduated tube and of the time reading may have caused each group’s data to be shifted slightly off from the others by around 1 mL or 1 s.

Another potential error could have resulted from the fact that the data from the 4.7 pH was an average of multiple trials, and the 2.5 pH data and 3.4 pH data were each taken from individual trials. Data from one single trial could include outliers or other user errors, and without using an average of several trials, the data could possibly have been inaccurate or inconsistent with expected results.

In addition, the data from the 3.4 pH and the 4.7 pH did not contain all of the data points that pH 2.5 contained. This left room for estimation, and the estimations that were made could have potentially been inaccurate. If any of the data points that were collected were in any way inaccurate, the estimations that were based off of this data would therefore also be inaccurate. ✎

Works Cited:

Worthington, Krystal. “Catalase.”Worthington Biochemical Corporation. N.p., n.d. Web. 11 Oct. 2013.

Scandalios, John, Lingqiang Guan, and Alexios Polidoros. “Catalases in Plants.”North Carolina State University. N.p., n.d. Web. 10 Nov. 2013.

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