Lab 3 Mitosis and Meiosis

Michelle Seywald

AP Biology

Period 1

January 1, 2010

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Introduction:

Cells divide by two different methods: meiosis and mitosis. Mitosis is generally used for the growth and repair of an organism, while meiosis functions in the production of gametes or spores. Each process involves the division of the cell’s nucleus and of the cytoplasm.

Mitosis is studied in exercise 3A in cells of whitefish blastula and onion root tip, since these cells have a high percentage of cells undergoing mitosis. On roots, the region that contains the highest percentage of cells undergoing mitosis is called the apical meristem. Cells in their non dividing stage are in interphase and remain in this stage the majority of the time. DNA is replicated during this period. At this stage, the nucleus may have one or more dark stained nucleoli and is filled with chromatin. Cell division begins with prophase, during which the chromatin condense into chromosomes, the nuclear envelope disintegrates, and the mitotic spindle begins to form. Next is metaphase, during which the chromosomes line up at the metaphase plat in the middle of the cell. In anaphase, the mitotic spindle pulls the sister chromatids to opposite poles, distributing a diploid amount of chromosomes to each pole. Telophase, the last stage of division, is characterized by formation of the nuclear envelope and nucleoli, and uncoiling of the chromosomes. Following mitosis is cytokinesis, which is division of the cytoplasm into two daughter cells. The overall process of mitosis and cytokinesis results in two daughter cells identical to each other and the parent cell.

Meiosis is a process similar to mitosis in which the nucleus undergoes two divisions and results in 4 haploid gametes. These two rounds of cell division are called meiosis I and meiosis II. Meiosis I depends on the replication of DNA in interphase, just like in mitosis. Beginning with prophase I, homologous chromosomes come together to form a tetrad and crossing over takes place between the nonsister chromatids. This process increases genetic variation in the offspring. At metaphase I, the homologous chromosomes line up at the metaphase plate. During anaphase I, the homologous chromosomes are pulled to opposite poles by the spindle fibers. In telophase I, a haploid set of chromosomes reaches each pole. Cytokinesis usually occurs during this time. Meiosis II is much like mitosis, with chromosomes lining up at the metaphase plate and sister chromatids separating to opposite poles. At the end of telophase II, there are four haploid daughter cells.

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Meiosis and the process of crossing over are explored in exercise 3B with the fungus Sordaria fimicola. These organisms produce a fruiting body called a perithecium containing many asci. These asci each contain 8 ascospores which are either tan or black. The color and orientation of the ascospores within the ascus can signify whether crossing over has taken place. If the ascospores are arranged in a 4:4 pattern, crossing over has not taken place. However, if the ascospores are arranged in either a 2:4:2 or a 2:2:2:2 pattern, crossing over has taken place. By calculating the frequency of crossover asci within a certain region, the map units between the gene for spore color and the centromere can be determined.

Materials:

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3A.1

This exercise requires prepared slides of whitefish blastula and onion root tip, and a light microscope.

3A.2

This exercise requires a prepared slide of onion root tip and a light microscope.

3B.2

This exercise requires 1or more slides/images of hybrid asci of the fungus Sordaria fimicola.

Procedures:

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3A.1

Examine prepared slides of either onion root tips or whitefish blastula. Locate the meristematic region of the onion, or locate the blastula, with the 10x objective and then use the 40x objective to study individual cells. Identify one cell that clearly represents each phase. Sketch and label the cell in the boxes provided.

3A.2

To estimate the relative length of time that a cell spends in the various stages of cell division, examine the meristematic region of a prepared slide of the onion root tip. The length of the cell cycle is approximately 24 hours for the cells in actively dividing onion root tips. It is hard to imagine that one can estimate how much time a cell spends in each phase of cell division from a slide of dead cells, yet this is precisely what will be done in this part of the lab. Since work will be done with a prepared slide, information about how long it takes a cell to divide can not be obtained. Instead, the numbers of cells in each phase will be determined. From this, the percentage of time each cell spends in each phase can be inferred. Observe every cell in one high-power field of view and determine which phase of the cell cycle it is in. This is best done in pairs. The partner observing the slide calls out the phase of each cell while the other partner records. Then switch so the recorder becomes the observer and vice versa. Count at least two full fields of view. If less than 200 cells have been counted, then count a third field of view. Record the data. Calculate the percentage of cells in each phase and record. Consider that it takes, on average, 24 hours (or 1,440 minutes) for onion root tip cells to complete the cell cycle. The amount of time spent in each phase of the cell cycle can be calculated from the percentage of cells in that stage.

3B.2

View the slide and locate a group of hybrid asci (those containing both tan and black ascospores). Count at least 50 hybrid asci and enter the data in Table 3.3. Using the data in Table 3.3, determine the distance between the gene for spore color and the centromere. Calculate the percentage of crossovers by dividing the number of crossover asci (2:2:2:2 or 2:4:2) by the total number of asci x100. To calculate the map distance, divide the percentage of crossover asci by 2. The percentage of crossover asci is divided by 2 because only half of the spores in each ascus are the result of a crossover event. Record the results in Table 3.3.

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Results:

3A.1

Sketches attached.

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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?

First, mitosis replicates the chromosomes in a cell, ensuring that there are two identical copies of DNA. At metaphase, the replicated chromosomes line up at the metaphase plate. At anaphase, the sister chromatids, which are identical to each other, are pulled apart to opposite poles by the spindle fibers. The elongated cell then separates into two diploid cells, each with genetic material identical to each other and to the parent cell. The daughter cells remain diploid because homologous chromosomes do not separate as they do in meiosis, but rather remain together as a diploid set. The genetic material is identical in the parent and the daughter cells because anaphase separates replicated chromosomes and distributes identical copies of chromatids to each daughter cell. In addition, crossing over does not take place during mitosis, preventing sister chromatids from differing from each other.

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

While animal cells use centrioles to direct the spindle fibers, plant cells lack these structures. This, however, does not seem to significantly affect the function of the microtubules of the spindle in plant cells. Another difference is the formation of a cell plate during cytokinesis in plant cells that is not present in animal cells. Rather than pinching off along a cleavage furrow to form two new daughter cells, a new cell wall is simply laid down between the daughter cells in plant cells.

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

The centrosome contains a pair of centrioles and functions as the microtubule organizing center. If this structure were absent from the cell, the microtubules would not assemble, the spindle would

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not form, and the sister chromatids would not separate. In short, mitosis would not be possible without the presence of the centrosome.

3A.2

Table 3.1

Number of Cells Percent of Total

Cells Counted

Time in Each StageField 1 Field 2 Field 3 Total

Interphase

70 64 63 197 65.7%946.08 minutes

Prophase 15 18 17 50 16.7%240.48 minutes

Metaphase

5 8 7 20 6.6%95.04

minutes

Anaphase 4 4 7 15 5%72

minutesTelophas

e6 6 6 18 6%

86.4 minutes

300

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 observations had not been confined to the actively dividing portion of the onion root tip, there would be virtually no cells in prophase, metaphase, anaphase, and telophase. Instead, most cells would be in interphase.

2. Based on the data in Table 3.1, what can you infer about the relative length of time and onion root tip cell spends in each stage of cell division?

Based on the data in Table 3.1, it can be inferred that an onion root tip cell spends the majority of its time in interphase, followed by prophase, then equal amounts in

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Total Cells Counted

metaphase and telophase, and finally the least amount in anaphase.

3. Draw and label a pie chart of the onion root tip cell cycle using the data from Table 3.1

3B.1

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

One major difference between mitosis and meiosis is that crossing over takes place during meiosis but not during mitosis. Also, mitosis produces 2 identical daughter cells, while meiosis produces 4 potentially unique daughter cells. Finally, the nucleus undergoes two divisions in meiosis, but only one division in mitosis.

2. Compare mitosis and meiosis with respect to each of the following in Table 3.2:

Mitosis MeiosisChromosome Number

of Parent CellsDiploid (2n) Diploid (2n)

Number of DNA Replications

1 1

Number of Divisions 1 2Number of Daughter

Cells Produced2 4

Chromosome Number of Daughter Cells

Diploid (2n) Haploid (n)

Purpose/Function Growth and repair Production of

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gametes

3. How are meiosis I and meiosis II different?

Meiosis I undergoes replication of DNA and separates homologous chromosomes into daughter cells, while meiosis II does not replicate DNA before division and separates sister chromatids into daughter cells.

4. How do oogenesis and spermatogenesis differ?Oogenesis results in only one viable egg cell, while spermatogenesis results in 4 functioning sperm cells.

5. Why is meiosis important for sexual reproduction?

Meiosis halves the chromosome number of a cell so that when two gametes fuse together and each contributes their set of chromosomes, the diploid number of chromosomes is preserved. Meiosis also involves crossing over, which increases the genetic variation of the offspring.

3B.2

Table 3.3

Number of 4:4

Number of Asci Showing

CrossoverTotal Asci

% Asci Showing

Crossover Divided by 2

Gene Centromere

distance (map units)

30 51 81 31.4% 31.4

Analysis of Results:

2.

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Meiosis 1 Meiosis 2

Data Analysis:

3A.1:

This experiment focuses on observing the cells of whitefish blastula and onion root tip under a light microscope. Had the microscope light been faulty or weak, the student would not be able to differentiate between the different stages of cell division or the specific structures in each stage. Also, those slides more heavily dyed produced a clearer image of certain structures that groups with lighter dyed slides were unable to see. These possibilities may have contributed to the inaccurate calculation for the duration of telophase. Although it is the shortest stage of mitosis, it was found to the second to last shortest in this lab.

3A.2:

This experiment concentrates on determining the percentage of time spent in each stage of the cell cycle by observing a large number of cells and recording their particular stage. This experiment requires the student to count a large number of cells in the meristematic region of the onion root tip in order to determine the percentage of time spent in each stage of cell division. If the student was unable to differentiate between certain similar stages, such as prophase and interphase, the data may have been miscalculated. In addition, the percentage of cells

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found in each stage may vary from slide to slide, or from region to region observed under the microscope.

3B.2:

This exercise focuses on calculating the map units between a specific gene and the centromere by observing the frequency of crossover asci from the fungus Sordaria fimicola. There was limited room for error, since the images of the hybrid asci were already prepared and the student needed only to tally the number of 4:4 and crossover asci. However, if there was difficulty differentiating between crossovers and non crossovers, or certain asci were counted more than once, the results may have been skewed.

Conclusion:

Mitosis and meiosis were observed throughout this lab under microscope and through simulation which helped clarify their many similarities and differences. In exercise 3A, the different stages of cell division of whitefish blastula and onion root tip were observed and recorded. The frequency of each stage in a region of cells was then calculated and used to determine the amount of time a cell spent in each stage of cell division. From this data, it can be concluded that prophase is the longest stage of mitosis, and telophase is the shortest. In Lab 3B, meiosis was simulated and examined the link between crossing over and map units. It can be seen from observation of mitosis and meiosis that, although similar, the two processes have many differences. For one, the cell nucleus undergoes two divisions in meiosis, but only one division in mitosis. Secondly, crossing over takes place during meiosis but not during mitosis, increasing the genetic variation of offspring. In addition, mitosis produces 2 identical daughter cells, while meiosis produces 4 potentially unique daughter cells. In exercise 3B.2, the frequency of crossover asci (2:4:2 or 2:2:2:2 formation) from the organism Sordaria fimicola is calculated and used to determine the approximate distance from the gene for spore color to the centromere. From the results, it can be concluded that the gene is approximately 31.4 map units from the centromere.

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