## Diagrams of Cells

Below are the diagrams of the cells of an onion root tip in each of the stages of mitosis.

## Data Table

Based on our data, we can infer that an onion root tip spends the most amount of time in interphase, and the least amount of time in interphase.

Below is the pie chart for the time spent in each phase.

Below is the pie chart for the time spent in each phase.

## Analysis Questions

1.) Mitosis leads to two daughter cells because the cell replicates DNA during metaphase and anaphase. They contain the same DNA, each gets chromatids that form an identical chromosome.

2.) During interphase, the cell grows and replicates DNA.

3.) In contrasting mitosis in plant cells and in animal cells, everything is basically the same except that animal cells have cleavage forming in telephase, while plant cell form within the cell walls during telephase.

2.) During interphase, the cell grows and replicates DNA.

3.) In contrasting mitosis in plant cells and in animal cells, everything is basically the same except that animal cells have cleavage forming in telephase, while plant cell form within the cell walls during telephase.

## Hypothesis

If we treat an onion with lectin proteins and observe the root under a microscope, then the data would be different from the untreated root by lessening interphase and greatening mitosis.

## Chi Squared Tables and Calculations

The null hypothesis is that there will be no change if the onions are treated with lectin.

The chi squared value was found by adding the Interphase ((O-E)^2)/E with the Mitosis ((O-E)^2)/E.

5.697 + 0.796 = 6.493. The chi squared value is thus 6.493. Since the chi squared value is larger than that necessary to obtain a p value of .05 or more, we reject the null hypothesis. By rejecting the null hypothesis, we state that treating the onions with lectin would change in data.

The chi squared value was found by adding the Interphase ((O-E)^2)/E with the Mitosis ((O-E)^2)/E.

5.697 + 0.796 = 6.493. The chi squared value is thus 6.493. Since the chi squared value is larger than that necessary to obtain a p value of .05 or more, we reject the null hypothesis. By rejecting the null hypothesis, we state that treating the onions with lectin would change in data.