Lab 7 - Cell Division: Mitosis and Meiosis
How do eukaryotic cells divide to produce genetically identical cells or to produce gametes with half the normal DNA?
2. Pre-Lab Questions:
1. Starting as a fertilized egg, or a zygote, the cells divides countless times over an approximately nine month period in order to be born healthy. After 3-4 days, the cell (and subsequent daughter cells) has gone through mitosis several times (4-5) and a morula has formed of 16-32 cells. Within the next few days a blastocyst of more than one hundred cells has formed, with the cells of its inner cell mass being undifferentiated stem cells from which all the body's organs will form. Over time, the cells differentiate and continue to divide according to instructions within the DNA, and eventually, a full human baby is formed and born. This baby will then grow for the next 17-18 years via cell division until the final, adult form is reached. From there, the cells will continue to grow and divide to repair systems and maintain homeostasis, but the growth process is over. In order for one cell (or zygote) to reach 100 trillion cells, the first cell would have to divide about 37 times. log(2)100000000000 = x; x = 36.54 divisions
2. Cell division (often through binary fission) is important to a single celled organism because it is the only way for the organism to reproduce. By dividing, the organism keeps its species alive and ensures that its genes remain within the population. More importantly, according to the tenants of natural selection, the cell is capable of dividing because it has genes that have allowed it to survives in its environment, and creating identical cells will foster the creation of a healthy population. Amongst bacteria, this is also important because it keeps the plasmid genes available for transformation that will allow the genes to be spread throughout the population.
3. To ensure successful cell division, a variety of processes must occur within the cell to ensure that its daughter cells will be normal. First, the cell must grow larger in order to provide enough material for two cells to form, and cell structures must be duplicated, with mitochondria and chloroplasts (in autotrophs) dividing via binary fission. Then DNA must be replicated properly in the cell to ensure that both daughter cells receive identical, functioning copies of DNA. Finally, replication must be checked for errors and these errors must be corrected before the G2 checkpoint allows the cell to proceed to mitosis.
4. Genetic information within each of our body cells is exactly the same, 23 pairs of chromosomes with the genes coding for the exact same proteins (assuming perfect conditions) located on them. What makes the cells different however, is which genes are actually being expressed in a given cell. Transcription factors released by cells cause certain crucial genes to be activated in each type of cell, which causes the differentiation and coils up/deactivates unneeded DNA in specialized cells. Micro-RNAs play a role in "turning off" certain genes, and although different cell types serve different functions involving different activated genes, each cell still had the exact same genetic information as another(except for sex cells, which only have 23 sing chromosomes as opposed to 23 pairs).
5. There are several advantages to asexual reproduction in plants that jell them in various ways. One advantage would be the speed at which the reproduction occurs. Asexual reproduction does not require gamete formation, and thus, the plant can spread its genes more quickly. Additionally, asexual reproduction allows for offspring without the need for pollination, which is especially important in smaller populations. Another, advantage would be the fact that asexual reproduction creates an identical daughter plant with the exact same traits as the parent. As the parent was capable of surviving in the environment, it follows that a genetically identical plant would do just as well, allowing the plant's offspring to be successful by maintaining its advantageous traits without having to mix genes via sexual reproduction. (eduction-portal.com helped to answer this question)
6. It is important for DNA to be replicated prior to cell division because each daughter cell must receive the proper, functional amount of DNA to be identical to its parent and to survive. By going through DNA replication, a parent cell ensures that both daughter cells will have all the genetic material they require to function. If DNA replication did not occur, cell division could result in a cell with only half the required genetic information, or none of it, and this could lead to serious issues and/or probably death. For a cell to be able to code for its necessary proteins, it must have all the required DNA to do just that, and without DNA replication, successful cell division would be jeopardized.
7. During cell division, mitosis spindle fibers emitted by centrioles on either end of the cell, attach to the kinetochores of replicated (4n) DNA, now condensed into chromosomes. Then, the spindle fibers, along with motor proteins on the chromosomes direct the chromosomes to alignment along the center of the cell during prophase and metaphase. Then, during anaphase, the mitotic spindle fibers pull the chromosomes apart towards opposite poles where they will serve as the genetic material for the two new daughter cells after cytokinesis finishes the final segregation of the two cells.
8. Progression in the cell cycle is controlled by complexes composed of specific proteins and enzymes known as cyclin-dependent kinases(CDks). Different CDKs, such as mitosis-promoting factor (MPF), once reacted with cyclin, instruct the cells to move forward with the cell cycle, in this case, the movement from G2 to mitosis. Although these CDKs instruct for progression, cell checkpoints along the way ensure that a cell has properly completed processes necessary for division. For example. The G1 checkpoint ensures that a cell is large enough, in the proper environment, and subject to the correct growth factors for it to enter the DNA synthesis stage. G2 checkpoint ensures that DNA replication has occurred and that it has occurred without damage or mutation, and the metaphase spindle checkpoint makes sure that microtubules are attached to chromosomes properly and that the chromosomes are aligned. If the controls were defective and unready cells were allowed to divide, the cell would either trigger an apoptotic pathway to destroy itself, or if that functionality is damaged, the cell would divide uncontrollably (especially if POGs and TSGs). This could lead to cancer if the abnormal cells have mutations that are malignant.
3. Cell Division Control
Cell Division is controlled by molecules comprised of several specific proteins that contain enzymes that are known as cyclin-dependent kinases (CDKs). These CDKs play an important role in activating or deactivating the processes and reactions that cause cell division when they react with cyclins. When cyclins and CDKs combine, they react with other proteins to allow the cell to proceed along the cell cycle, depending on which complex is triggered (such as the mitosis promoting factor that moves the cell from G2 to begin mitosis. Inextricably connected to the workings of CDKs and cyclins are cell "checkpoints" that serve various purposed in ensuring that the cells able to advance in the stages of the cell cycle. First of these checkpoints is the G1 checkpoint, between G1 and S, that ensures that growth factors are present, that the cell is in a suitable environment, and that the cell is large enough to successfully proceed. Next, between G2 and mitosis is the G2 checkpoint. At the end of G2, the replicated DNA is checked for errors and variable such as cell size are accounted for, if the cell detects errors in replication, mitosis will not begin in a properly functioning cell. Lastly, the metaphase spindle checkpoint ensures that the spindle fibers are properly attached to the kinetochores on the chromosomes and that the chromosofmes are aligned across the center of the cell; if anything is awry, the process will not proceed until corrections are made. Cells that are irreversibly damaged or otherwise defective may undergo programmed cell death, or apoptosis, to ensure that abnormal cells are not created by irregular cell division. This is the final control on cell division, as normal cells die before dividing abnormally.
Part 1: Mitosis
4. Eukaryotic Cell Division
Eukaryotic cell division begins when signals indicate that a cell must grow and divide due to factors in the environment and within the cell. This triggers the beginning of the path to division as the cell moves from regular functioning in the "G0" stage to the first stage of the process, G1. During G1, the cell begins its initial preparations driven by G1 cyclin-dependent kinases(CDKs) as the cell begins to grow in size, duplicate its vital structures(organelles), and copy its chloroplasts(if an autotroph) and mitochondria by a process known as binary fission. Once the cell is ready, and it passes the checks for growth factors and size at the G1 checkpoint, the cell is free to move into the S stage of the cell cycle, triggered by S-phase CDKs. S phase is marked by the synthesis of duplicate DNA chromosomes identical to the ones from which they are replicated, resulting in identical sister chromatids attached to each other by centromeres. Once the replication is complete, the cell moves to the final part of interphase ((G0) G1-S-G2), which is G2. During G2, the cell continues to grow and the replicated DNA is checked for errors and the errors are fixed where appropriate. Finally, the cell reaches the G2 checkpoint, where the cell determines if the DNA replication was done correctly, and if the cell is large enough to divide, at which point the cell can begin the process of mitosis when mitosis-promoting factor (mitosis CDKs and cyclin) bind to trigger the start of final division. Sister chromatids condense in the nucleus and the nucleus subsequently dissolves as the cell goes through prophase (and prometaphase). From here, spindle fibers emitted by centrioles that have migrated to either pole of the cell attach to the kinetochores of the sister chromatids and align them along the cell's equator, during metaphase. One final checkpoint (the mitotic spindle checkpoint) ensures that the chromosomes are aligned correctly and that the fibers are all properly attached to either side of the sister chromatids (at the kinetochores). From here, the spindle fibers pull the chromatids apart during anaphase and the sister chromosomes move to either end of the cell. Then, in the next stage, telophase, two new nuclei form around each pair of chromosomes and the cell begins to separate. In a final stage, not part of mitosis, the cytoplasm (and organelles) of the cell are split equally between the two genetically identical daughter cells, the end of the process. From here, the cell can either enter G0 or go about normal cell functioning, or it can immediately enter G1, repeating the process known as the cell cycle until the cell's death via apoptosis or other means.
5. Page 87 Questions for Part 1
Part 1: Modelling Mitosis
1. If a cell contains a set of duplicated chromosomes, it does in fact contain more genetic information than the cell before duplication, twice as much. However, as the duplicated genetic information is exactly the same, there is no NEW genetic information in the cell. No new genes that code for different proteins are added, and there is no added variation to the cell's genome. While there is twice as much DNA, there is not any different genetic information coded in a cell with duplicated DNA than in one with just the normal chromosome amount.
2. By condensing chromosomes before they are moved, damage to chromosomes while they are in transit becomes far less likely. Instead of loose strands of DNA becoming tangled together, breaking, or being otherwise damaged, the DNA is neatly packed to allow for movement. This ensures that the DNA will be in proper condition by the time it is arranged for final division during anaphase.
3. After duplicated chromosomes are aligned along the center of the cell by the mitotic spindle fibers during metaphase, the sister chromatids prepare to separate. Centrioles on either pole of the cell direct the spindle fibers to pull the chromosomes apart after the metaphase checkpoint has affirmed proper alignment. These fibers, attached to the kinetochores of the chromosomes, then pull toward their pole, separating the sister chromatids at the centromere during anaphase and these sister chromosomes then provide the new cell with its genetic material.
4. Should sister chromatids fail to separate, one spindle fiber will pull both chromatids to its pole where a new cell is forming. When two daughter cells then form during cytokinesis, one cell with have a duplicate chromosome while the other will be left without its genetic information. This could lead to problems in both cells, particularly the one with the missing chromosomes, and this would result in irregular cells that could be harmful.
Part 2: Effects of Environment on Mitosis
6. Experimental Design
Part 2: Effects of Environment on Mitosis
"Various fungi can negatively affect the growth of soybeans by producing a lectin-like protein. Latinos can induce mitosis in the root apical meristem tissue which will often weaken the plant tissue."
Testable Question: Does the fungal pathogen lectin affect the number of cells undergoing mitosis in a plant's root tips?
Experimental Hypothesis: Adding the fungal pathogen lectin to the root tips of a plant will induce rapid cell division in the root tips, increasing the number of cells undergoing mitosis.
Null Hypothesis: Adding the fungal pathogen lectin to the root tips of a plant will have no discernable effect on the rate of cell division or the number of cells undergoing mitosis.
These hypotheses, the experimental and null, are not the same. This experimental hypothesis makes a prediction that the addition of lectin will have an effect to increase the number of cells undergoing mitosis, while the null hypothesis predicts that lectin will have no effect on the number of cells undergoing mitosis. If an experiment were to be conducted, a scientist, in most cases, would hope that the results would prove the tenants of the experimental hypothesis while simultaneously disproving the validity of the null hypothesis.
Experimental Design(and what is being measured and how):
In order to test whether lectins increase the number of onion cells undergoing mitosis, one would first need to gather onion root tips to be used as experimental units. These tips must then be randomly assigned to 1 of 2 treatment groups, lectin free, and lectin exposed. Those in the lectin exposed group must be exposed to lectin for a period of time, then, put under a microscope and the number of cells undergoing mitosis(oberserving the visible effects of division) must be counted. Those in the lectin free group must sit for the same period of time in water as the lectin group, but must remain unexposed, and then they will be put under a microscope and the number of cells undergoing mitosis will be counted. For both treatment groups, the field of view for the microscope should be identical, as should the number of fields of views used to count.
There are several controls that should be placed on the experiment to ensure that any conclusions regarding differences in cells undergoing division can be attributed solely to the effects of lectin. One control that must be taken is that if the lectin exposed group must be exposed to lectinfor a certain amount of time, it is important that the control group is let sit (unexposed), for the same period of time. Additionally, the field of view on the microscope must remain the same for both experimental groups so that the amount of total visible cells remains constant and does not affect the results. Additionally, temeprature, light exposure, and all other possible variables should be controlled, even if they may not seem likely to have an effect on the results.
7. Data Tables
8. Chi-Squared Calculations:
9. Critical Values:
Statistically accepting or rejecting the null hypothesis requires that the experimenter perform a chi-square test. To do this, one must first determine the chi squared value of the results of the experiment. This value is equal (observed value - expected value)squared divided by expected value summed for each treatment group. Upon obtaining this value, the experimenter must also calculate the degrees of freedom in the experiment, which is (# of treatment groups - 1) * (# of phase group - 1). At this point, the researcher must find the appropriate significance probability that he seeks on a chi-square table (seen below), usually .05 and match it with the approiate degree of freedom to find the critical value. From here, the experimenter compares his chi-square value with the critical value, if the chi squared value equals or exceeds the critical value the null hypothesis is rejected, if it is below the value, the null hypothesis is rejected. Bascially, by exceeding the critical value, the researcher can say with the given levele of confidence that the difference between variables cannot be attributed solely to chance, disproving the null hypothesis.
Chi-Squared Critical Values Table
10. Null Hypothesis:
In the case of our experiment, it can be see above that our calculated chi squared value is .9715. Additionally, using the formula above we can calculate the degrees of freedom in this experiment to be (2-1) * (2-1) = 1. Therefore, using the the probability at .05, the critical value necessary to reject the null hypothesis is at least 3.841. As our chi squared value is lower than 3.841 (.9715), we must accept the null hypothesis. In terms of our our experimental hypothesis, that the presence of lectin increases the number of dividing cells in onion root tips, we can say that this is disproven. Our null hypothesis, that lectin has no effect or does not increase the number of dividing cells in onion root tips, is therefore proven.
11. Post Lab Questions:
1. Collecting data from the entire class is important because it provides for enough repetition and total data so that the results of the experiment can be used to draw an overall conclusion. Having more data tends to even out the natural variation in response found among different onion root tips that may be slightly different genetically or may have other variables affecting the results. By ensuring that there is a wide base for the data, definitive overall conclusions can be drawn regarding the relationship between the independent and dependent variable, in this case, the presence of lectin and its effects on the rate of mitosis in onion root tips.
2. No there was not a statistically significant difference between the control H2O group and the lectin exposed group regarding the number of cells undergoing mitosis. Using the chi-square test, the critical value needed to indicate a significant difference would be 3.84(for this experiments particular degrees of freedom), while the data obtained only reached .9715. Thus, it can be concluded that any difference in the groups (lectin/no lectin) can be attributed solely to natural variation among onion root tips and not the independent variable.
3. Statistically, the fungal pathogen lectin did not increase the number of onion root tip cells undergoing mitosis. While lectin exposed cells did have a slightly higher amount of cells undergoing mitosis, the chi-square value of the effect of lectin, .9715, did not meet the critical threshold of 3.84 to be considered to have an effect and the null hypothesis was held to be proven in this case. Thus, it can be concluded that the presence of the fungal pathogen lectin did not increase the number of onion root tip cells undergoing mitosis.
4. There are several other experiments that could be performed to verify that lectin does not have an effect on the rate of cell division in onion root tips. One possible experiment to verify this would be to add a third group of half-concentrated lectin exposed root tips and measure the differences to be checked against the Chi-square table. Another possible experiment that could be done would be to expose the onions bulbs to lectin, and then allow them to grow to maturity, measuring the time it takes and the health of the plant, although this experiment could be difficult to measure. One last experiment that could be done would be to, instead of exposing plant bulbs, expose the cells directly and measure in the lab how many of a set amount enter mitosis and how many do not, and then check for statistical significance.
5. An increased number of cells undergoing mitosis does not necessarily mean that those cells are dividing faster than those in the roots with less dividing cells. While it could certainly indirectly indicate that more growth is needed, the actual rate of mitosis cannot be determined in this manner. Just because more cells are growing doesn't mean that mitosis itself is moving quicker. Thus it is impossible to draw definitive conclusions about the fundamental rate of mitosis based on just the number of cells in mitosis at a given time.
6. An interesting way to determine how fast the rate of mitosis is occurring in root tips would be to record through a microscope how long the cells in each experimental group take to divide. By timing how quickly cells in each group divide, starting with the beginning of prophase and ending with the final throes of telophase, it is possible to determine the length of time it takes one, or multiple recorded cells, to undergo mitosis. Averaging the times within each experimental group will then give a general idea regarding the general time it takes in lectin and non-lectin exposed cells, and then statistical analysis can be done to determine whether or not the null hypothesis has been proven regarding absolute time for mitosis to occur.
Part 3: Loss of Control of the Cell Cycle
12. Case 2 Activity Questions:
1. When mutations occur in a normal cell, the DNA has changed, and the gene that once coded for a specific protein may now code for something slightly different, or something unrecognizable, depending on the severity. If this cell, through its normal regulatory processes, recognizes irreparable damage to the DNA, it may begin an apoptotic pathway. At this point, a series of events governed by genes known as tumor suppressor genes will cause the cell to commit suicide rather than replicate with recognizably faulty DNA. However, some mutations in certain genes may bypass or override suppressor genes, and this could cause the creation of malignant cancer cells that grow and divide without control.
2. If cells with mutated DNA replicated and passed on the mutated genes to daughter cells, a cancerous tumor could begin to form. With the normal regulatory pathways not inhibiting these mutated cells from dividing, these mutated cells could continue to divide uncontrollably into a mass of cells that serve no role other than to deprive other cells of resources. As each generation passes, the number of cells in the tumor with the uncontrolled mutation grows exponentially and eventually, the cancer may metastasize and cause harm throughout the body.
3. Cells monitor DNA integrity through a variety of processes, many of which are tied to the G2 checkpoint in the cell cycle. Enzymes and proteins are equipped to ensure the integrity of the DNA and fix any issues using a variety of regulatory proteins before the cell is allowed to continue throughout the cell cycle. If a cell deems itself to be beyond repair, it may go through programmed cell death, or apoptosis, resulting in the destruction of potentially harmful cells and preventing their duplication. Natural selection has ensured that cells with effective regulatory pathways have survived, and the G2 checkpoint ensures that no damaged cells with DNA that contains errors is allowed to reproduce in most case.
4. Chromosomes in the cancer cells can suffer from a wide range of abnormalities when compared to normal cells. Some cancer cells may have more or less than the normal number of chromosomes, resulting in unhealthy cells without the proper genes to regulate themselves and carry out proper cell functions. Other cancer cell chromosomes may contain mutated DNA, especially those that carry proto-oncogenes and tumor suppressor genes that code for growth factors and regulatory proteins. Additionally, abnormalities may occur in a manner similar to that of the Philadelphia Chromosome, whereby a translocation causes pieces of chromosomes to switch places, which can cause certain forms of leukemia to arise. Overall, there are many different possible ways that cancer cells can differ in comparison to normal cells.
5. Differences between normal cells and cancer cells can lead to cancer due to abnormalities in the biochemical growth pathways that promote or regulate cell division. Mutations in proto-oncogenes or tumor suppressor genes can cause malfunctioning proteins whereby (in the case of TSGs), a regulatory protein may be unable to perform its function and stop cell growth and division. If certain genes are missing entirely, this could obviously be a problem if it contained regulatory genes as well as other important genes. Additionally, a translocation such as that in the Philadelphia effect could cause malfunctions in important genes that cause uncontrolled cell growth and make the cells unable to perform normal functioning. Despite the cell's best efforts, improperly formed proteins can result in the proliferation of cancer cells, and this can occur in several different ways.
Nondisjunction is a phenomenon involving the failure of homologous chromosomes or sister chromatids to split during the anapahse of either meiosis or mitosis. During meiosis, this occurs when, during anapahase I, the homologous chromosomes fail do not split and move toward either pole, resulting in one cell with the 4n number of that chromosome and a another with none. This will go on to cause further complications in the next phases of mitosis, and fertilization will never properly occur with these gametes. Additionally, nondisjunction can also occur during anaphase II of meiosis when sister chromatids fail to separate, leaving one cell as 2n and the other with no chromosomes of that type. Once again, this nondisjunction will probably never see any of these abnormal gametes combine with any other to become a zygote. However, if somehow fertilization does occur, the offspring could have issues with having too many chromosomes, such as Down's Syndrome which is caused by an extra 21st chromosome (trisomy 21). During mitosis, nondisjunction can also occur when sister chromatids fail to separate during anaphase, resulting in a cell without its proper genetic material. This could lead to the cell's death in some cases or lead to abnormalities that may give rise to a cancerous tumor. Boiled down, nondisjunction is the failure of homologous chromosomes or sister chromatids to separate during meiosis or mitosis.
Part 4: Modeling Meiosis
14. Meiosis and Mitosis:
15. Questions on Bottom Half of S93:
1. Much like in mitosis, DNA is replicated prior to meiosis, during the S phase of interphase. This ensures that each parent cell will have the necessary amount of chromosomes (4n) to be able to divide twice into four genetically unique gametes that are each 1n at the end of the entire process of meiosis.
2. Homologous pairs of chromosomes are not exact copies of one another. Each of the homologous chromosomes comes either from one's mother or one's father, and although the genes code for proteins that perform the same function, they may have different alleles coding for different variations of the same gene that affect phenotype and functioning. This fact (along with others) is what makes meiosis capable of producing highly variable gametes from the same DNA.
3. Crossing over is a variation adding process that occurs during prophase I of meiosis. During prophase I, homologous chromosomes pair up to form a structure consisting of four total chromatids. This may lead to the adjacent chromatids of the homologous pair to wind around each and break off portions of the DNA on either chromosomes. These broken off potions can then reattach to the opposite chromosome, creating a unique chromatid that will be found in one gamete.
4. Distance of a specific gene from the center or edge of a chromosome places a restraint on the crossover frequency of eligible chromosomes (adjacent). A gene that is a the the very tip of the chromatid probably breaks off with much greater frequency than a gene or gene fragment that is located near the centromere. This is because crossing over only occurs where the chromatids wrap around each other, which becomes more difficult the more firmly anchored the gene is to the center.
5. Independent assortment is the doctrine that states that the pole orientation of the maternal and paternal chromosomes in a homologous pair is random for each chromosome. This ensures that each gamete is almost guaranteed(statistically) to have DNA from both the maternal and paternal chromosome, which adds a great deal of variation to the process of meiosis as a whole.
6. Assuming that individual genes are sorted independently(which is not always true due to linked genes (I looked this up)), one can calculate the total possible number of gametes by multiplying the number of alleles for each gene times the number of alleles for all other genes. For example, if a certain parent cell contains 10 genes, four of which are homozygous, and of the remaining six are heterozygous, the total number of possible gametes would be 1^4 * 2^6 = 64 possible total gametes.
7. If homologous pairs of chromosomes failed to separate during meiosis (nondisjunction), the final gametes would have either twice as many of the needed chromosomes or none. Resultantly, if these gametes were to form a zygote with a normal gamete(fertilization), the result would be offspring with either three or one total chromosome of a pair. Nondisjunction in certain chromosomes may lead to children with down syndrome (trisomy 21) whereby one of the gametes that was involved in fertilization received twice as many chromosomes due to nondisjunction, resulting in a 3n chromosome. On the other hand, if a gamete with a missing chromosome were to form a zygote, the offspring would have only one chromosome from the unaffected parent gamete, and could result in a disorder such as cri du chat(monosomy 5) syndrome, which results in this case.
8. At the fundamental level mitosis and meiosis are different in their purpose and function. While mitosis aims to create two genetically identical, diploid(2n) daughter cells, meiosis is designed to create four, genetically varied, haploid (1n) gametes. Mitosis is a form of asexual reproduction which is used by somatic cells for growth and repair, while meiosis is involved in creating 1n sex cells for use in sexual reproduction. In one sentence, mitosis creates identical cells that sustain life in a living organism, meiosis creates varied gametes that will contribute to creating life.
Part 5: Meiosis and Crossing Over
16. Lab Bench Activity:
17. Evaluating Results Questions Page S96:
1. It is necessary to divide by two because each crossover produces two spores from the crossover and two spores that are identical to the parent chromosomes. By dividing by 2, we can therefore find the total number as only half the spores in each ascus result from crossing over.
2. Our "class data" from the lab bench showed a distance of 31.8 map units between the spore color gene and the centromere. This is 5.8 map units greater than the published map units at 26 map units.
3. Disparities between the "class data" and the published data can be accounted for by natural variation among ascus. Not all all ascus are identical, and thus, the distance we found may be different than the average distance that was found among all the other ascus. Also, all of the experimenters counted the same eleven ascus, providing a very small sample size to get an accurate estimate of the mean. Had all the experimenters sampled different ascus, the average results would most likely have fallen closer to the average accepted published result of 26 map units.
4. To produce the results that I found in the lab bench activity, the ascus gametes must have crossed over more often than they didn't. Basically, when the homologous chromosomes paired up during prophase I, DNA on the ends of the adjacent chromatids wound around each other. These pieces broke segments off of each other and recombined on the other chromosome, creating the differing results that were found in the spore (gametes) that were produced.
5. A Philadelphia Chromosome is most likely the result of some other chromosomal abnormality rather than the result of some type of crossing over. Crossing over only naturally occurs between homologous chromosomes as they contain the code for the same gene, albeit different alleles. Thus, for the Philadelphia Chromosome to form, some other abnormal process must take place. Most likely, during crossing over, pieces of these two chromosomes break off and float throughout the cell until they translocate to the 22nd and 9th chromosome, forming the Philadelphia Chromosome in a manner not consistent with normal crossing over.
6. Applying the cell cycle to meiosis in the same way as it is applied to mitosis is not advised. Cells that enter into mitosis create replica somatic, 2n daughter cells of themselves that fulfill the same role as the parent and will eventually undergo mitosis themselves after being induced and completing the set processes of the cell cycle. At the end of meiosis, 1n gametes are formed that will be used in sexual reproduction. These gametes can not then undergo meiosis again as one would expect in a cycle, marking the distinction between the two.
18. Explain Everything Meiosis Assignment:
See Google Drive