Lab 7 Cell Division
Name ________________________ Date ____________
Prelab:
A. Read the lab for understanding.
B. Complete Prelab questions
C. Complete Part 3
Prelab questions:
1. What is one advantage of sexual reproduction?
2. What are some advantages of asexual reproduction?
3. Why is it important for DNA to be replicated prior to mitosis?
4. How is the cell cycle controlled? Be specific.
5. What would happen if the cell cycle control mechanisms were defective?
Lab:
Part 1: Observing Stages of the Cell Cycle in Root Tips
The purpose of this part of the lab is to explore the timing of the cell cycle.
o
o
o
o
You will be looking at prepared slides of root tips. These show the anatomy of
the growing root necessary for the plant to absorb nutrients.
Examine the slide on low power first.
 Note the zone of division, or meristem. This is the area where mitosis
occurs.
 Note the zone of elongation. The new cells elongate making the root
longer, pushing it into the soil.
 Note the zone of maturation or differentiation. This is where the cells
differentiate into different types of cells. Epithelial cells have
extensions of the cell membrane, called root hair, that are important
for water and nutrient absorption.
Focus and center the zone of division on low power, move to medium power, and focus and center and
area with many mitotic figures (nuclei will look dark). Move to high power and focus with the fine focus
knob only.
Identify the phase of the cell cycle of mitosis for each cell using a search pattern. Do this for at least 50
cells. Record in the data table.
Onion root tip cell phase data
Interphase
Cell #
Prophase
Metaphase
Anaphase
Telophase
Total
% of total
Questions: Compare your results with other groups if possible.
1. Which phase of the cell cycle was the most commonly seen? ______________________________
2. Which phase of mitosis was most commonly seen?
______________________________
3. Prophase is usually the most common phase of mitosis seen. Why?
4. Why would root hairs make water absorption more efficient?
Part 2: Effects of Environment on the Cell Cycle
The purpose of this part of the lab is to determine if a natural plant toxin affects cell division in root tips.
Scientists reported that a fungal pathogen, may negatively affect the growth of soybeans (Glycine max). Soybean
growth decreased during three years of high rainfall, and the soybean roots were poorly developed. Close relatives
of R. anaerobis are plant pathogens and grow in the soil. A lectin-like protein was found in the soil around the
soybean roots. This protein may have been secreted by the fungus. Lectins induce mitosis in some root apical
meristem tissues. In many instances, rapid cell divisions weaken plant tissues.
You have been asked to investigate whether the fungal toxin lectin affects the number of cells undergoing mitosis in
a different plant, using root tips. You will examine onion root meristems (zone of division) from photos. You will be
looking at photographs of root tip meristems. One set of photographs is from roots that were treated with lectin.
The other set is from roots not treated (control group).
o Count 50 cells from each group following a search pattern. In the data table below, record the number of
cells in interphase and the number in mitosis. (You do not need to identify the different phases of mitosis.)
Calculation of chi-square
# observed (o)
(with lectin)
# expected (e)
(without lectin)
(o-e)
(o-e)2
Interphase cells
Mitosis cells
Total cells
(must be the same
for both)
X2 = sum (o-e)2/e =
Calculating Chi-square:
 Use the control group (no lectin) for the expected values.
 Use the lectin -treated group for the observed values.
 Calculate the X2 for your data.
(o-e)2/e
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Calculate the degrees of freedom. The degrees of freedom (df) equals the number of groups minus one. In
this case, there are two groups, interphase and mitosis; therefore, df = 2-1 = 1
Find the chi-square critical (table) value for your degrees of freedom and row 0.05 from the table below.

The p value is 0.05 meaning that we are 95% confident that our data fits in this group.
The “critical value” is 3.84.
If the calculated chi-square value is greater than or equal to this critical value, then the two groups (lectin
and no lectin) are significantly different, and the null hypothesis is rejected.
If the calculated chi-square value is less than this critical value, then the two groups (lectin and no lectin)
are not significantly different, and the null hypothesis is not rejected.
Is the calculated chi-square greater or less than the table value? _____________________

Aret he two groups significantly different or not significantly different? ____________________

1. Write what it means if the null hypothesis is rejected In terms of this part of the investigation.
2. Write a short results section saying if the two groups significantly different or not, and which group had
more mitotic figures. (Be sure to specify calculated chi square value & p value. Based on values, is there
a significant difference between 2 groups? If yes, which one had more mitotic figures? Is null rejected or not rejected.)
3. Write a short conclusions section explaining what conclusions that can be made from this experiment.
Part 3: Loss of Cell Cycle Control in Cancer.
The purpose of this part of the lab is to explore how loss of control of the cell cycle can lead to cancer.
HeLa cells are cervical cancer cells isolated from a
woman named Henrietta Lacks. She was infected
with the human papillomavirus (HPV) and developed
cervical cancer. Shortly before her death, some of the
cancer cells were removed for cell culture. They have
been cultured since 1951 and used in numerous
cancer studies. Her cells are said to be “immortal”
because they have been cultured for so long. Lacks’s
cancer cells contain remnants of human
papillomavirus (HPV), which we know increases the
risk of cervical cancer (there is now a vaccine for this
virus). HPV causes a mutation that inactivates the
tumor suppression proteins p53 and others.
Henrietta’s story is described in the book, The
Immortal Life of Henrietta Lacks by Rebecca Skloot.
Karyotype of Hela 229 cervical adenocarcinoma cell line
A. Examine the picture of the HeLa cells.
1. How do the chromosomes of the HeLa cells differ from those of the noncancerous cells?
2. How could a mutation that affects p53 cause cancer? Be as specific as you can.
B. Examine the pictures of cancer cells with the
Philadelphia chromosome. Philadelphia chromosome or
Philadelphia translocation is a
specific chromosomal abnormality that is associated
with a type of leukemia (cancer of certain blood cells). It is
the result of a reciprocal translocation between
chromosome 9 and 22. A piece of chromosome 9 broke off
and attached to chromosome 22. Likewise, a piece broke
off of chromosome 22 and attached to chromosome 9.
This results in a fusion gene that turns on a protein which
causes cell division.
1. How are normal cells usually prevented from
dividing if there are errors in DNA?
2. How could a loss of this control lead to cancer?
Part 4: Meiosis and Crossing Over in Sordaria.
The fungus Sordaria fimicola exchanges genetic material when two mycelia meet and fuse. The resulting zygote
undergoes meiosis to produce asci; each ascus contains eight haploid spores. A single gene determines the spore
color.
A cross was made between wild type (+; black) and tan (tn) strains. The resulting zygote produces either parental
type asci (which have four black and four tan spores in a row (4:4 pattern)) or recombinant asci (which have a
different pattern due to crossing-over).
1. Examine at least 50 asci and record if they are parental type or represent crossing-over.
Number of Asci
Showing 4:4
Pattern
Number of Asci
Showing
Crossing-over
Total # of Asci
Gene to
Centromere
Distance (Map
Units)
(divide the % by 2)
% Asci Showing
Crossing-over
Group Data
Class Data
2. Do you think that the Philadelphia chromosome examined in Part 3 is due to crossing-over? Justify your
answer using your knowledge of the process of crossing-over.
Grading Rubric: Lab 4
Points
Prelab
questions
Part 1
data table
questions
Part 2
X2 table
null hypothesis
results
conclusions
Part 3
questions
Part 4
data table
question
Total
Possible
5
5
4
4
1
3
2
4
5
2
35