Table 2: Mitosis Data

This classical microscope lab has actually been provided in life science classrooms for years. It is likewise a conventional part of the AP Biology curriculum as Investigation #7 in the AP Biology lab hand-operated, and have the right to be a great means to apply a simple knowledge of chi-square tests.

You watching: Table 2: mitosis data




The environment instantly surrounding a cell can have considerable impacts on the procedure of cellular division. Mitosis, one kind of cell division, is the procedure by which a eukaryotic cell separates the chromosomes in its cell nucleus into 2 the same sets, in 2 sepaprice nuclei. This is the basic process that produces the majority of of the cells in multicellular organisms and allows for growth and also repair of an organism.

Fungal pathogens in the soil are known to inhilittle root growth in important farming plants. These fungal pathogens are thmust act via secreting lectin and lectin-favor proteins right into the soil to promote mitosis in root tips past healthy levels. This artificially high level of mitosis is then thneed to damage root tconcern, inevitably slow-moving overall root growth, and also undermine the plant.

In this task, we’ll use lab data to test the hypothesis that lectin promotes mitosis in onion roots. The 2 variables for the activity are Phase and Treatment. Each row in the datacollection is an individual cell observed via a microscope. Phase corresponds to whether the onion cells were oboffered to be in “Interphase” or “Mitosis” and also Treatment corresponds to whether the onion root tips exposed to lectin (“Lectin”) or not (“Control”).

To test for any statistical impacts of lectin on root growth we’re going to usage a chi square test of self-reliance. The null hypothesis of this test is that 2 (or more) variables are independent of one an additional. In other words, the null hypothesis assumes that tbelow is no predictive capacity of one variable on one more. In the instance of this lab, this would suppose that the propercentage of control cells in mitosis are similar to the propercentage of lectin-treated cells in mitosis. If we uncover this to be true, then the phase we observe the onion root pointer cell to be in would be independent of the lectin treatment and also our information would certainly fail to refuse the null hypothesis. However before, if the proportions of cells in mitosis are substantially various between the regulate and also lectin-treated samples, then we would certainly refuse the null hypothesis and conclude that the lectin therapy does affect the propercent of cells observed in mitosis.

In the activity, students will use the “Make-a-graph” attribute to uncover the observed counts of cells in various stperiods of the cell cycle, and also use these oboffered counts to calculate the expected counts for the null hypothesis. They may then either calculate the result by hand also and then use the graph-moved hypothesis test to inspect their outcomes or ssuggest finish the test by hand (the graph-driven chi-square test for independence is a paid feature, but you can begin a 90-day cost-free trial to produce classes and offer your students access).

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Phase: This is a categorical variable via two worths. Each oboffered cell was either in Interphase or Mitosis.

Treatment: This is a categorical variable with 2 worths. The root pointer and its cells were either treated through Lectin or a Control (not treated).

To collect this datacollection, both lectin-infprovided roots and non-exposed lectin root tips (control) were observed making use of microscopes. While looking through the microscopic lense at a root reminder on a stained slide, cells were oboffered and also for each cell it was detailed whether that cell was in interphase or undergoing mitosis. Since the cells were preserved in whatever stage of the cell cycle they were in when the slide was prepared, the slide is choose a picture of what was happening in that root tip at the moment of arsenal.

The null hypothesis is that outcomes will certainly display no distinction between the manage and treated teams in terms of the moment spent in mitosis family member to interphase. The different hypothesis is that the lectin treatment will increase the amount of time that cells in the root pointer spfinish in mitosis family member to the time invested in interphase.


1) Use the “Make a graph” attribute to make a categorical bubble plot (obtainable with the scatter plot icon) to compare the family member proparts within each group. Show Treatment on X and also Phase on Y. What does the graph seem to suggest around the influence of lectin on onion root cell metabolism?

2) Given that a p-worth of .05 or below suggests a strong evidence that lectin does impact cell expansion, give your best estimate for the p-value for a Chi-Square Test of Independence. Without actually doing any type of calculations, execute these results look like a significant deviation from the null expectation for the same proportion of cells in mitosis in both treatment groups?

3) With Use a graph-propelled chi-square test for freedom to have the computer carry out the calculation. After you have made the graph, simply click the Graph Driven Test Button (on the right panel just left of the Appearance button). How close was your estimate of the p-value to the actual calculated value?

By Hand: Determine if tbelow is a statistically substantial distinction in between the observed frequency of cells in interphase and mitosis between the 2 therapy groups. To perform so you will must perform the following:

a) Calculate the expected values assuming independence. For an in-depth explanation of why we calculate independence this means, please check out the Teacher’s Keep in mind listed below. (Hint: e1 have to be around 71.75)

N = full number of cells in the experiment

e1 = supposed worth of cells in the control group in interphase = all cells in interphase * all cells in the regulate group / N.

e2 = supposed value of cells in the control team in mitosis = all cells in mitosis * all cells in the regulate team / N.

e3 = expected worth of cells in the treatment group in interphase = all cells in interphase * all cells in the therapy team / N.

e4 = meant worth of cells in the treatment team in mitosis = all cells in mitosis * all cells in the therapy group / N.

The table listed below is a advantageous means to organize this indevelopment (a modified version of the one offered under Investigation 7 in the AP Biology Lab Manual):