Part a
Diagrams:
Part 1a
Data:
Part 1a
Analysis Questions:
1. Based on your data, what can you infer about the ralatice length of time an onion root tip spends in each stage of cell division?
Based on the data, it can be infered that the more cells there are in each phase, the longer the phase will take. The cell spends the most time in interphase. After finishing interpahse, the cell spends the second most amount of time in prophase, followed by telophase. In both anaphase and metapahse the cell spends the least time.
The reason the cell is likely to spend the least amount of time in both metaphase and anaphase, is because in those phases the cell only lines up chromosomes and pulls aways from the metaphase plate. However, in interphase the cell grows and replicates DNA before Mitosis, so it is logical that the cell would spend the most time in interphase.
2. Draw and label a pie chart of the onion root tip cell cycle using the data from your table.
The amount of hours spent in each phase of the Cell Cycle-
Based on the data, it can be infered that the more cells there are in each phase, the longer the phase will take. The cell spends the most time in interphase. After finishing interpahse, the cell spends the second most amount of time in prophase, followed by telophase. In both anaphase and metapahse the cell spends the least time.
The reason the cell is likely to spend the least amount of time in both metaphase and anaphase, is because in those phases the cell only lines up chromosomes and pulls aways from the metaphase plate. However, in interphase the cell grows and replicates DNA before Mitosis, so it is logical that the cell would spend the most time in interphase.
2. Draw and label a pie chart of the onion root tip cell cycle using the data from your table.
The amount of hours spent in each phase of the Cell Cycle-
Part 1b
Hypothesis:
If we treat an onion with lectin proteins and observe the root under a microscope, the data would be different from that of the untreated onion in part 1a because it would spend less amount of time in telophase and more in mitosis.
Null Hypothesis: The cells in the cell group, if treated with lectin, would exhibit no change.
Chi Square Tables/Calculations
X²= (O-E)²/E
X²= 6.49
This value rejects the null hypothesis